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FCSYRKqNo2ber6iq | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let the population of rabbits surviving at time $$t$$ be governed by the differential equation $${{dp\left( t \right)} \over {dt}} = {1 \over 2}p\left( t \right) - 200.$$ If $$p(0)=100,$$ then $$p(t)$$ equals: | [{"identifier": "A", "content": "$$600 - 500\\,{e^{t/2}}$$ "}, {"identifier": "B", "content": "$$400 - 300\\,{e^{-t/2}}$$"}, {"identifier": "C", "content": "$$400 - 300\\,{e^{t/2}}$$"}, {"identifier": "D", "content": "$$300 - 200\\,{e^{-t/2}}$$"}] | ["C"] | null | Given differential equation is
<br><br>$${{dp\left( t \right)} \over {dt}} = {1 \over 2}p\left( t \right) - 200$$
<br><br>By separating the variable, we get
<br><br>$$dp\left( t \right) = \left[ {{1 \over 2}p\left( t \right) - 200} \right]dt$$
<br><br>$$ \Rightarrow {{dp\left( t \right)} \over {{1 \over 2}p\left( t \right) - 200}} = dt$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ Integrating on both the sides,
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int {{{d\left( {p\left( t \right)} \right)} \over {{1 \over 2}p\left( t \right) - 200}}} = \int {dt} $$
<br><br>Let $${1 \over 2}p\left( t \right) - 200 = s \Rightarrow {{dp\left( t \right)} \over 2} = ds$$
<br><br>So, $$\int {{{d\,p\left( t \right)} \over {\left( {{1 \over 2}p\left( t \right) - 200} \right)}}} = \int {dt} $$
<br><br>$$ \Rightarrow \int {{{2ds} \over s} = \int {dt} } \Rightarrow 2\log s = t + c$$
<br><br>$$ \Rightarrow 2\log \left( {{{p\left( t \right)} \over 2} - 200} \right) = t + c$$
<br><br>$$ \Rightarrow {{p\left( t \right)} \over 2} - 200 = {e^{{1 \over 2}}}k$$
<br><br>Using given condition $$p\left( t \right) = 400 - 300{e^{t/2}}$$ | mcq | jee-main-2014-offline |
Pbkc9ctwhFJQbquk | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | If a curve $$y=f(x)$$ passes through the point $$(1,-1)$$ and satisfies the differential equation, $$y(1+xy) dx=x$$ $$dy$$, then $$f\left( { - {1 \over 2}} \right)$$ is equal to : | [{"identifier": "A", "content": "$${2 \\over 5}$$ "}, {"identifier": "B", "content": "$${4 \\over 5}$$"}, {"identifier": "C", "content": "$$-{2 \\over 5}$$"}, {"identifier": "D", "content": "$$-{4 \\over 5}$$"}] | ["B"] | null | $$y\left( {1 + xy} \right)dx = xdy$$
<br><br>$${{xdy - ydx} \over {{y^2}}} = xdx \Rightarrow \int { - d\left( {{x \over y}} \right) = \int {xdx} } $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ - {x \over y} = {{{x^2}} \over 2} + C\,\,$$
<br><br>as $$\,\,\,y\left( 1 \right) = - 1 \Rightarrow C = {1 \over 2}$$
<br><br>Hence, $$y = {{ - 2x} \over {{x^2} + 1}} \Rightarrow f\left( {{{ - 1} \over 2}} \right) = {4 \over 5}$$ | mcq | jee-main-2016-offline |
Od8wAhtRzk2KVNfWDLjdG | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | If f(x) is a differentiable function in the interval (0, $$\infty $$) such that f (1) = 1 and
<br/><br/>$$\mathop {\lim }\limits_{t \to x} $$ $${{{t^2}f\left( x \right) - {x^2}f\left( t \right)} \over {t - x}} = 1,$$ for each x > 0, then $$f\left( {{\raise0.5ex\hbox{$\scriptstyle 3$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}} \right)$$ equal to : | [{"identifier": "A", "content": "$${{13} \\over 6}$$"}, {"identifier": "B", "content": "$${{23} \\over 18}$$"}, {"identifier": "C", "content": "$${{25} \\over 9}$$"}, {"identifier": "D", "content": "$${{31} \\over 18}$$"}] | ["D"] | null | $$\mathop {\lim }\limits_{t \to x} {{{t^2}f\left( x \right) - {x^2}f\left( t \right)} \over {t - x}} = 1$$
<br><br>It is in $${0 \over 0}$$ form
<br><br>So, applying L' Hospital rule,
<br><br>$$\mathop {\lim }\limits_{t \to x} {{2tf\left( x \right) - {x^2}f'\left( x \right)} \over 1} = 1$$
<br><br>$$ \Rightarrow $$ 2xf(x) $$-$$ x<sup>2</sup>f '(x) = 1
<br><br>$$ \Rightarrow $$ f '(x) $$-$$ $${2 \over x}$$f(x) $$=$$ $${1 \over {{x^2}}}$$
<br><br>$$ \therefore $$ I.F = $${e^{\int {{{ - 2} \over x}dx} }} = {e^{ - 2\log x}} = {1 \over {{x^2}}}$$
<br><br>$$ \therefore $$ Solution of equation,
<br><br>f(x)$${1 \over {{x^2}}}$$ = $$\int {{1 \over x}\left( { - {1 \over {{x^2}}}} \right)} \,dx$$
<br><br>$$ \Rightarrow $$ $${{f(x)} \over {{x^2}}} = {1 \over {3{x^3}}} + C$$
<br><br>Given that,
<br><br>f(1) = 1
<br><br>$$ \therefore $$ $${1 \over 1}$$ = $${1 \over 3}$$ + C
<br><br>$$ \Rightarrow $$ C $$=$$ $${2 \over 3}$$
<br><br>$$ \therefore $$ f(x) $$=$$ $${2 \over 3}$$ x<sup>2</sup> + $${1 \over {3x}}$$
<br><br>$$ \therefore $$ f$$\left( {{3 \over 2}} \right) = {2 \over 3} \times {\left( {{3 \over 2}} \right)^2} + {1 \over 3} \times {2 \over 3} = {{31} \over {18}}$$ | mcq | jee-main-2016-online-9th-april-morning-slot |
4ZhNCSi2n4lmKWbz65tcb | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let y = y(x) be the solution of the differential equation $${{dy} \over {dx}} + 2y = f\left( x \right),$$ <br/><br/>where $$f\left( x \right) = \left\{ {\matrix{
{1,} & {x \in \left[ {0,1} \right]} \cr
{0,} & {otherwise} \cr
} } \right.$$
<br/><br/>If y(0) = 0, then $$y\left( {{3 \over 2}} \right)$$ is : | [{"identifier": "A", "content": "$${{{e^2} + 1} \\over {2{e^4}}}$$ "}, {"identifier": "B", "content": "$${1 \\over {2e}}$$ "}, {"identifier": "C", "content": "$${{{e^2} - 1} \\over {{e^3}}}$$ "}, {"identifier": "D", "content": "$${{{e^2} - 1} \\over {2{e^3}}}$$ "}] | ["D"] | null | When x $$ \in $$ [0, 1], then $${{dy} \over {dx}}$$ + 2y = 1
<br><br>$$ \Rightarrow $$ y = $${1 \over 2}$$ + C<sub>1</sub>e<sup>$$-$$2x</sup>
<br><br>$$ \because $$ y(0) = 0 $$ \Rightarrow $$ y(x) = $${1 \over 2}$$ $$-$$ $${1 \over 2}$$e<sup>$$-$$2x</sup>
<br><br>Here, y(1) = $${1 \over 2}$$ $$-$$ $${1 \over 2}$$ e<sup>$$-$$2</sup> = $${{{e^2} - 1} \over {2{e^2}}}$$
<br><br>When $$x \notin \left[ {0,1} \right]$$, then $${{dy} \over {dx}}$$ + 2y = 0 $$ \Rightarrow $$ y = c<sub>2</sub> e<sup>$$-$$2x</sup>
<br><br>$$ \because $$ y(1) = $${{{e^2} - 1} \over 2}$$ $$ \Rightarrow $$ $${{{e^2} - 1} \over 2}$$ = c<sup>2</sup>e<sup>$$-$$2</sup> $$ \Rightarrow $$ C<sub>2</sub> = $${{{e^2} - 1} \over 2}$$
<br><br>$$ \therefore $$ y(x) $$\left( {{{{e^2} - 1} \over 2}} \right){e^{ - 2x}} \Rightarrow y\left( {{3 \over 2}} \right)$$ = $${{{e^2} - 1} \over {2{e^3}}}$$ | mcq | jee-main-2018-online-15th-april-morning-slot |
VuYSuUBxWNrRV8TBiQUqF | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | The curve satifying the differeial equation, (x<sup>2</sup> $$-$$ y<sup>2</sup>) dx + 2xydy = 0 and passing through the point (1, 1) is : | [{"identifier": "A", "content": "a circle of radius one. "}, {"identifier": "B", "content": "a hyperbola."}, {"identifier": "C", "content": "an ellipse."}, {"identifier": "D", "content": "a circle of radius two. "}] | ["A"] | null | (x<sup>2</sup> $$-$$ y<sup>2</sup>) dx + 2xydy = 0
<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}}$$ = $${{{y^2} - {x^2}} \over {2xy}}$$
<br><br>Let y = vx
<br><br>$${{dy} \over {dx}}$$ = v + x $${{dv} \over {dx}}$$
<br><br>$$ \Rightarrow $$ v + x$${{dv} \over {dx}}$$ = $${{{v^2}{x^2} - {x^2}} \over {2v{x^2}}}$$ $$ \Rightarrow $$ v + x$${{dv} \over {dx}}$$ = $${{{v^2} - 1} \over {2v}}$$
<br><br>$$ \Rightarrow $$ x$${{dv} \over {dx}}$$ = $${{ - {v^2} - 1} \over {2v}}$$
<br><br>$$ \Rightarrow $$ $${{2vdv} \over {{v^2} + 1}}$$ = $$-$$ $${{dx} \over x}$$
<br><br>After intergrating, we get
<br><br>$$\ln \left| {{v^2} + 1} \right|$$ = $$-$$ ln$$\left| x \right|$$ + lnc
<br><br>$${{y{}^2} \over {{x^2}}}$$ + 1 = $${c \over x}$$
<br><br>As curve passes through the point (1, 1), so 1 + 1 = c
<br><br>$$ \Rightarrow $$ c = 2
<br><br>x<sup>2</sup> + y<sup>2</sup> $$-$$ 2x = 0, which is a circle of radius one. | mcq | jee-main-2018-online-15th-april-evening-slot |
nQ2qPM0JLVO25SzPZYJlB | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let f : [0,1] $$ \to $$ <b>R</b> be such that f(xy) = f(x).f(y), for all x, y $$ \in $$ [0, 1], and f(0) $$ \ne $$ 0. If y = y(x) satiesfies the differential equation, $${{dy} \over {dx}}$$ = f(x) with y(0) = 1, then y$$\left( {{1 \over 4}} \right)$$ + y$$\left( {{3 \over 4}} \right)$$ is equal to : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "5"}] | ["A"] | null | If f(xy) = f(x) f(y) $$\forall $$ x, y $$ \in $$ R and f(0) $$ \ne $$ 0
<br><br>put x = y = 0
<br><br>$$ \Rightarrow $$ f(0) = [f(0)]<sup>2</sup>
<br><br>$$ \Rightarrow $$ f(0) = 1
<br><br>put y = 0 $$ \Rightarrow $$ f(0) = f(x) f(0)
<br><br>$$ \Rightarrow $$ f(x) = 1
<br><br>given that $${{dy} \over {dx}}$$ = f(x)
<br><br>$$ \therefore $$ $${{dy} \over {dx}}$$ = 1 $$ \Rightarrow $$ y = x + k
<br><br>given that y(0) = 1
<br><br>$$ \therefore $$ k = 1
<br><br>hence y = x + 1
<br><br>y$$\left( {{1 \over 4}} \right)$$ + y$$\left( {{3 \over 4}} \right)$$ = $$\left( {{1 \over 4} + 1} \right)$$ + $$\left( {{3 \over 4} + 1} \right)$$ = 3 | mcq | jee-main-2019-online-9th-january-evening-slot |
20Dy2KcdpKNLohKcHbCgg | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | If $${{dy} \over {dx}} + {3 \over {{{\cos }^2}x}}y = {1 \over {{{\cos }^2}x}},\,\,x \in \left( {{{ - \pi } \over 3},{\pi \over 3}} \right)$$ and $$y\left( {{\pi \over 4}} \right) = {4 \over 3},$$ then $$y\left( { - {\pi \over 4}} \right)$$ equals - | [{"identifier": "A", "content": "$${1 \\over 3} + {e^6}$$"}, {"identifier": "B", "content": "$${1 \\over 3}$$"}, {"identifier": "C", "content": "$${1 \\over 3}$$ + e<sup>3</sup>"}, {"identifier": "D", "content": "$$-$$ $${4 \\over 3}$$"}] | ["A"] | null | $${{dy} \over {dx}} + 3{\sec ^2}x.y = {\sec ^2}x$$
<br><br>I.F. = $${e^{3\int {{{\sec }^2}xdx} }} = {e^{3\tan x}}$$
<br><br>or $$y.e{}^{3\tan x} = \int {{{\sec }^2}x.{e^{3\tan x}}} $$
<br><br>or $$y.{e^{3\tan x}} = {1 \over 3}{e^{3\tan x}} + C$$
<br><br>Given
<br><br>$$y\left( {{\pi \over 4}} \right) = {4 \over 3}$$
<br><br>$$ \therefore $$ $${4 \over 3}.{e^3} = {1 \over 3}{e^3} + C$$
<br><br>$$ \therefore $$ C = e<sup>3</sup> | mcq | jee-main-2019-online-10th-january-morning-slot |
6DatEBrFx1erLJNC8l48B | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | The curve amongst the family of curves represented by the differential equation, (x<sup>2</sup> – y<sup>2</sup>)dx + 2xy dy = 0 which passes through (1, 1) is : | [{"identifier": "A", "content": "a circle with centre on the y-axis"}, {"identifier": "B", "content": "an ellipse with major axis along the y-axis"}, {"identifier": "C", "content": "a circle with centre on the x-axis"}, {"identifier": "D", "content": "a hyperbola with transverse axis along the x-axis"}] | ["C"] | null | (x<sup>2</sup> $$-$$ y<sup>2</sup>) dx + 2xy dy = 0
<br><br>$${{dy} \over {dx}} = {{{y^2} - {x^2}} \over {2xy}}$$
<br><br>Put $$y = vx \Rightarrow {{dy} \over {dx}} = v + x{{dv} \over {dx}}$$
<br><br>Solving we get,
<br><br>$$\int {{{2v} \over {{v^2} + 1}}dv = \int { - {{dx} \over x}} } $$
<br><br>ln(v<sup>2</sup> + 1) = $$-$$ ln x + C
<br><br>(y<sup>2</sup> + x<sup>2</sup>) = Cx
<br><br>1 + 1 = C $$ \Rightarrow $$ C = 2
<br><br>y<sup>2</sup> + x<sup>2</sup> = 2x | mcq | jee-main-2019-online-10th-january-evening-slot |
rJ5WtqxjQSBlMYGc2qARC | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | If y(x) is the solution of the differential equation $${{dy} \over {dx}} + \left( {{{2x + 1} \over x}} \right)y = {e^{ - 2x}},\,\,x > 0,\,$$ where $$y\left( 1 \right) = {1 \over 2}{e^{ - 2}},$$ then | [{"identifier": "A", "content": "y(log<sub>e</sub>2) = log<sub>e</sub>4"}, {"identifier": "B", "content": "y(x) is decreasing in (0, 1)"}, {"identifier": "C", "content": "y(log<sub>e</sub>2) = $${{{{\\log }_e}2} \\over 4}$$"}, {"identifier": "D", "content": "y(x) is decreasing in $$\\left( {{1 \\over 2},1} \\right)$$"}] | ["D"] | null | $${{dy} \over {dx}} + \left( {{{2x + 1} \over x}} \right)y = {e^{ - 2x}}$$
<br><br>I.F. $$ = {e^{\int {\left( {{{2x + 1} \over x}} \right)dx} }} = {e^{\int {\left( {2 + {1 \over x}} \right)dx} }} = {e^{2x + \ell nx}} = {e^{2x}}.x$$
<br><br>So, $$y\left( {x{e^{2x}}} \right) = \int {{e^{ - 2x}}.x{e^{2x}} + C} $$
<br><br>$$ \Rightarrow xy{e^{2x}} = \int {xdx + C} $$
<br><br>$$ \Rightarrow 2xy{e^{2x}} = {x^2} + 2C$$
<br><br>It passes through $$\left( {1,{1 \over 2}{e^{ - 2}}} \right)$$ we get C $$=$$ 0
<br><br>$$y = {{x{e^{ - 2x}}} \over 2}$$
<br><br>$$ \Rightarrow {{dy} \over {dx}} = {1 \over 2}{e^{ - 2x}}\left( { - 2x + 1} \right)$$
<br><br>$$ \Rightarrow f(x)$$ is decreasing in $$\left( {{1 \over 2},1} \right)$$
<br><br>$$y\left( {{{\log }_e}2} \right) = {{\left( {{{\log }_e}2} \right){e^{ - 2({{\log }_e}2)}}} \over 2}$$
<br><br>$$ = {1 \over 8}{\log _e}2$$
<br><br> | mcq | jee-main-2019-online-11th-january-morning-slot |
MKMQSvnXL3Er22G6tn5Wg | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let y = y(x) be the solution of the differential equation,<br/><br/> $${({x^2} + 1)^2}{{dy} \over {dx}} + 2x({x^2} + 1)y = 1$$ <br/><br/>such
that y(0) = 0. If $$\sqrt ay(1)$$ = $$\pi \over 32$$ , then the value of
'a' is :
| [{"identifier": "A", "content": "$${1 \\over 2}$$"}, {"identifier": "B", "content": "$${1 \\over 16}$$"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "$${1 \\over 4}$$"}] | ["B"] | null | $${({x^2} + 1)^2}{{dy} \over {dx}} + 2x({x^2} + 1)y = 1$$
<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}} + \left( {{{2x} \over {1 + {x^2}}}} \right)y = {1 \over {{{\left( {1 + {x^2}} \right)}^2}}}$$
<br><br>I.F = $${e^{\int {{{2x} \over {1 + {x^2}}}dx} }} = {e^{\ln \left( {1 + {x^2}} \right)}} = 1 + {x^2}$$
<br><br>$$ \therefore $$ $${d \over {dx}}\left( {\left( {1 + {x^2}} \right)y} \right) = {1 \over {1 + {x^2}}}$$
<br><br>Integrating both sides we get,
<br><br>$${\left( {1 + {x^2}} \right)y}$$ = $${\tan ^{ - 1}}x$$ + C
<br><br>When x = 0 then y = 0
<br><br>So, 0 = 0 + C
<br><br>$$ \Rightarrow $$ C = 0
<br><br>$$ \Rightarrow $$ $${\left( {1 + {x^2}} \right)y}$$ = $${\tan ^{ - 1}}x$$
<br><br>Put x = 1,
<br><br>y.(2) = $${\pi \over 4}$$
<br><br>$$ \Rightarrow $$ y = $${\pi \over 8}$$ = y(1)
<br><br>Given,
<br><br>$$\sqrt ay(1)$$ = $$\pi \over 32$$
<br><br>$$ \Rightarrow $$ $$\sqrt a. {\pi \over 8}$$ = $$\pi \over 32$$
<br><br>$$ \Rightarrow $$ $$\sqrt a = {1 \over 4}$$
<br><br>$$ \Rightarrow $$ $$a = {1 \over {16}}$$ | mcq | jee-main-2019-online-8th-april-morning-slot |
w6OVOrATIUVFYzelDa3rsa0w2w9jxad1o9s | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | The general solution of the differential equation (y<sup>2</sup>
– x<sup>3</sup>)dx – xydy = 0 (x $$ \ne $$ 0) is :
(where c is a constant of integration) | [{"identifier": "A", "content": "y<sup>2</sup>\n + 2x<sup>3</sup>\n + cx<sup>2</sup>\n = 0"}, {"identifier": "B", "content": "y<sup>2</sup>\n + 2x<sup>2</sup>\n + cx<sup>3</sup>\n = 0"}, {"identifier": "C", "content": "y<sup>2</sup>\n\u2013 2x + cx<sup>3</sup>\n = 0"}, {"identifier": "D", "content": "y<sup>2</sup>\n\u2013 2x<sup>3</sup>\n + cx<sup>2</sup>\n = 0"}] | ["A"] | null | $$({y^2} - {x^3})dx - xydy = 0$$<br><br>
$$ \Rightarrow y(ydx - xdy) = {x^3}dx$$<br><br>
$$ \Rightarrow {y \over x}\left( {{{ydx - xdy} \over {{x^2}}}} \right) = dx$$<br><br>
$$ \Rightarrow - {y \over x}d\left( {{y \over x}} \right) = dx$$<br><br>
$$ \Rightarrow - {1 \over 2}{\left( {{y \over x}} \right)^2} = x + k$$<br><br>
$$ \Rightarrow - {y^2} = 2{x^3} + 2{x^2}k$$<br><br>
$$ \Rightarrow {y^2} + 2{x^3} + c{x^2} = 0$$
| mcq | jee-main-2019-online-12th-april-evening-slot |
FcgPlMWlchDinxtwe8jgy2xukfuup2te | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | The general solution of the differential equation
<br/><br>$$\sqrt {1 + {x^2} + {y^2} + {x^2}{y^2}} $$ + xy$${{dy} \over {dx}}$$ = 0 is :
<br/><br/>(where C is a constant of integration)
</br> | [{"identifier": "A", "content": "$$\\sqrt {1 + {y^2}} + \\sqrt {1 + {x^2}} = {1 \\over 2}{\\log _e}\\left( {{{\\sqrt {1 + {x^2}} - 1} \\over {\\sqrt {1 + {x^2}} + 1}}} \\right) + C$$"}, {"identifier": "B", "content": "$$\\sqrt {1 + {y^2}} - \\sqrt {1 + {x^2}} = {1 \\over 2}{\\log _e}\\left( {{{\\sqrt {1 + {x^2}} - 1} \\over {\\sqrt {1 + {x^2}} + 1}}} \\right) + C$$"}, {"identifier": "C", "content": "$$\\sqrt {1 + {y^2}} + \\sqrt {1 + {x^2}} = {1 \\over 2}{\\log _e}\\left( {{{\\sqrt {1 + {x^2}} + 1} \\over {\\sqrt {1 + {x^2}} - 1}}} \\right) + C$$"}, {"identifier": "D", "content": "$$\\sqrt {1 + {y^2}} - \\sqrt {1 + {x^2}} = {1 \\over 2}{\\log _e}\\left( {{{\\sqrt {1 + {x^2}} + 1} \\over {\\sqrt {1 + {x^2}} - 1}}} \\right) + C$$"}] | ["C"] | null | $$\sqrt {1 + {x^2} + {y^2} + {x^2}{y^2}} $$ + xy$${{dy} \over {dx}}$$ = 0
<br><br>$$ \Rightarrow $$ $$\sqrt {\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)} $$ + xy$${{dy} \over {dx}}$$ = 0
<br><br>$$ \Rightarrow $$ $$\sqrt {\left( {1 + {x^2}} \right)} \sqrt {\left( {1 + {y^2}} \right)} $$ = -xy$${{dy} \over {dx}}$$
<br><br>$$ \Rightarrow $$ $$\int {{{ydy} \over {\sqrt {1 + {y^2}} }}} = - \int {{{\sqrt {1 + {x^2}} } \over x}} dx$$ .....(1)
<br><br>Now put 1 + x<sup>2</sup>
= u<sup>2</sup>
and 1 + y<sup>2</sup>
= v<sup>2</sup>
<br><br>2xdx = 2udu and 2ydy = 2vdv
<br><br>$$ \Rightarrow $$ xdx = udu and ydy = vdv
<br><br>Substitute these values in equation (1)
<br><br>$$\int {{{vdv} \over v}} = - \int {{{{u^2}du} \over {{u^2} - 1}}} $$
<br><br>$$ \Rightarrow $$ $$\int {dv} = - \int {{{{u^2} - 1 + 1} \over {{u^2} - 1}}} du$$
<br><br>$$ \Rightarrow $$ v = $$ - \int {\left( {1 + {1 \over {{u^2} - 1}}} \right)} du$$
<br><br>$$ \Rightarrow $$ v = -u - $${1 \over 2}{\log _e}\left| {{{u - 1} \over {u + 1}}} \right|$$ + C
<br><br>$$ \Rightarrow $$ $$\sqrt {1 + {y^2}} $$ = $$ - \sqrt {1 + {x^2}} $$ + $${1 \over 2}{\log _e}\left| {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right|$$ + C
<br><br>$$ \Rightarrow $$ $$\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C$$ | mcq | jee-main-2020-online-6th-september-morning-slot |
Alm5jY7wF8yGlWtkEhjgy2xukfg6y95k | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | If y = y(x) is the solution of the differential <br/><br>equation $${{5 + {e^x}} \over {2 + y}}.{{dy} \over {dx}} + {e^x} = 0$$ satisfying
y(0) = 1, then a value of y(log<sub>e</sub>13) is :</br> | [{"identifier": "A", "content": "-1"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "2"}] | ["A"] | null | $${{5 + {e^x}} \over {2 + y}}.{{dy} \over {dx}} + {e^x} = 0$$
<br><br>$$ \Rightarrow $$ $${{5 + {e^x}} \over {2 + y}}.{{dy} \over {dx}} = -{e^x} $$
<br><br>Integrating both sides,
<br><br>$$ \Rightarrow $$ $$\int {{{dy} \over {2 + y}}} = \int {{{ - {e^x}} \over {{e^x} + 5}}} dx$$
<br><br>$$ \Rightarrow $$ ln (y + 2) = – ln(e<sup>x</sup>
+ 5) + k
<br><br>$$ \Rightarrow $$ (y + 2) (e<sup>x</sup>
+ 5) = C
<br><br>$$ \because $$ y(0) = 1
<br><br>$$ \Rightarrow $$ C = 18
<br><br>$$ \therefore $$ y + 2 = $${{{18} \over {{e^x} + 5}}}$$
<br><br>At x = log<sub>e</sub>13
<br><br>y + 2 = $${{{18} \over {13 + 5}}}$$ = 1
<br><br>$$ \Rightarrow $$ y = -1 | mcq | jee-main-2020-online-5th-september-morning-slot |
CQkbH0sdSO5KhngAxZjgy2xukfak59l5 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | The solution of the differential equation <br/><br/>
$${{dy} \over {dx}} - {{y + 3x} \over {{{\log }_e}\left( {y + 3x} \right)}} + 3 = 0$$ is:<br/><br/>
(where c is a constant of integration)
| [{"identifier": "A", "content": "$$x - {1 \\over 2}{\\left( {{{\\log }_e}\\left( {y + 3x} \\right)} \\right)^2} = C$$"}, {"identifier": "B", "content": "$$y + 3x - {1 \\over 2}{\\left( {{{\\log }_e}x} \\right)^2} = C$$"}, {"identifier": "C", "content": "x \u2013 log<sub>e</sub>(y+3x) = C"}, {"identifier": "D", "content": "x \u2013 2log<sub>e</sub>(y+3x) = C"}] | ["A"] | null | $${{dy} \over {dx}} - {{y + 3x} \over {\ln (y + 3x)}} + 3 = 0$$<br><br>Let $$\ln (y + 3x) = t$$<br><br>$${1 \over {y + 3x}}.\left( {{{dy} \over {dx}} + 3} \right) = {{dt} \over {dx}}$$<br><br>$$ \Rightarrow \left( {{{dy} \over {dx}} + 3} \right) = {{y + 3x} \over {\ln (y + 3x)}}$$<br><br>$$ \therefore $$ $$\left( {y + 3x} \right){{dt} \over {dx}} = {{y + 3x} \over t}$$<br><br>$$ \Rightarrow tdt = dx$$<br><br>$${{{t^2}} \over 2} = x + c$$<br><br>$${1 \over 2}{\left( {\ln (y + 3x)} \right)^2} = x + c$$ | mcq | jee-main-2020-online-4th-september-evening-slot |
ZIxbIoXkm0TQKzuPeRjgy2xukf7fd26d | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let y = y(x) be the solution of the differential equation, <br/>xy'- y = x<sup>2</sup>(xcosx + sinx), x > 0. if y ($$\pi $$) = $$\pi $$ then <br/>$$y''\left( {{\pi \over 2}} \right) + y\left( {{\pi \over 2}} \right)$$ is equal to : | [{"identifier": "A", "content": "$$1 + {\\pi \\over 2}$$"}, {"identifier": "B", "content": "$$2 + {\\pi \\over 2} + {{{\\pi ^2}} \\over 4}$$"}, {"identifier": "C", "content": "$$2 + {\\pi \\over 2}$$"}, {"identifier": "D", "content": "$$1 + {\\pi \\over 2} + {{{\\pi ^2}} \\over 4}$$"}] | ["C"] | null | $$xy' - y = {x^2}(x{\mathop{\rm cosx}\nolimits} + sinx),\,x > 0,\,y(\pi ) = \pi $$<br><br>$$y' - {1 \over x}y = x\{ x\cos x\, + \,\sin x\} $$<br><br>$$I.F. = {e^{ - \int {{1 \over 2}dx} }} = {e^{ - \ln x}} = {1 \over x}$$<br><br>$$ \therefore $$ $$y.{1 \over x} = \int {{1 \over x}.x(x\,cos\,x + sin\,x)<br>dx} $$<br><br>$${{y \over x}}$$ = $$\int ( x\,\cos \,x + \sin \,x)dx$$<br><br>$$ = x\,\sin \,x + C$$<br><br>$${{y \over x}}$$ = $$x\,\sin \,x + C$$<br><br>$$ \Rightarrow y = {x^2} = \sin \,x + cx$$<br><br>x = $$\pi $$, y = $$\pi $$<br><br>$$\pi $$ = $$\pi $$c $$ \Rightarrow $$ C = 1<br><br>$$y = {x^2}\sin x + x \Rightarrow y\left( {{\pi \over 2}} \right) = {{{\pi ^2}} \over 4} + {\pi \over 2}$$<br><br>$$y' = 2x\sin x + {x^2}\cos x + 1$$<br><br>y" = $$2\sin x + 2x\cos x + 2x\cos x - {x^2}\sin x$$<br><br>y" $$\left( {{\pi \over 2}} \right) = 2 - {{{\pi ^2}} \over 4} $$
<br><br>$$\Rightarrow y\left( {{\pi \over 2}} \right)$$ + y" $$\left( {{\pi \over 2}} \right) = 2 + {\pi \over 2}$$ | mcq | jee-main-2020-online-4th-september-morning-slot |
MGtqO8KBavSCQxZMCxjgy2xukf49d1w5 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | If x<sup>3</sup>dy + xy dx = x<sup>2</sup>dy + 2y dx; y(2) = e and
<br/>x > 1, then y(4) is equal to : | [{"identifier": "A", "content": "$${{\\sqrt e } \\over 2}$$"}, {"identifier": "B", "content": "$${1 \\over 2} + \\sqrt e $$"}, {"identifier": "C", "content": "$${3 \\over 2} + \\sqrt e $$"}, {"identifier": "D", "content": "$${3 \\over 2}\\sqrt e $$"}] | ["D"] | null | $${x^3}dy + xy\,dx = {x^2}dy + 2y\,dx$$<br><br>$$ \Rightarrow dy({x^3} - {x^2}) = dx(2y - xy)$$<br><br>$$ \Rightarrow $$ $$ - \int {{1 \over y}dx} $$ = $$\int {{{x - 2} \over {{x^2}(x - 1)}}dx} $$<br><br>$$ \Rightarrow $$ $$ - \ln y = \int {\left( {{A \over x} + {B \over {{x^2}}} + {C \over {(x - 1)}}} \right)dx} $$<br><br>Where A = 1, B = +2, C = $$ - $$1<br><br>$$ \Rightarrow - \ln y = \ln x - {2 \over x} - \ln (x - 1) + \lambda $$<br><br>As $$ y(2) = e$$<br><br>$$ \Rightarrow - 1 = \ln 2 - 1 - 0 + \lambda $$<br><br>$$ \therefore $$ $$\lambda = - \ln 2$$<br><br>$$ \Rightarrow \ln y = - \ln x + {2 \over x} + \ln (x - 1) + \ln 2$$<br><br>Now put x = 4 in equation<br><br>$$ \Rightarrow \ln y = - \ln 4 + {1 \over 2} + \ln 3 + \ln 2$$<br><br>$$ \Rightarrow {\mathop{\rm lny}\nolimits} = ln\left( {{3 \over 2}} \right) + {1 \over 2}\ln e$$<br><br>$$ \Rightarrow y = {3 \over 2}\sqrt e $$ | mcq | jee-main-2020-online-3rd-september-evening-slot |
tmiZPxBRjz9RFOLOorjgy2xukf0wlgsw | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | The solution curve of the differential equation,
<br/><br/>(1 + e<sup>-x</sup>)(1 + y<sup>2</sup>)$${{dy} \over {dx}}$$ = y<sup>2</sup>,
<br/><br/>which passes through
the point (0, 1), is : | [{"identifier": "A", "content": "y<sup>2</sup> + 1 = y$$\\left( {{{\\log }_e}\\left( {{{1 + {e^{ - x}}} \\over 2}} \\right) + 2} \\right)$$"}, {"identifier": "B", "content": "y<sup>2</sup> + 1 = y$$\\left( {{{\\log }_e}\\left( {{{1 + {e^{ x}}} \\over 2}} \\right) + 2} \\right)$$"}, {"identifier": "C", "content": "y<sup>2</sup> = 1 + $${y{{\\log }_e}\\left( {{{1 + {e^{ - x}}} \\over 2}} \\right)}$$"}, {"identifier": "D", "content": "y<sup>2</sup> = 1 + $${y{{\\log }_e}\\left( {{{1 + {e^{ x}}} \\over 2}} \\right)}$$"}] | ["D"] | null | Given (1 + e<sup>-x</sup>)(1 + y<sup>2</sup>)$${{dy} \over {dx}}$$ = y<sup>2</sup>
<br><br>$$ \Rightarrow $$ $$\left( {{{{y^2} + 1} \over {{y^2}}}} \right)dy = {{{e^x}dx} \over {{e^x} + 1}}$$
<br><br>Integrating both sides,
<br><br>$$\int {\left( {{{{y^2} + 1} \over {{y^2}}}} \right)dy = } \int {{{{e^x}dx} \over {{e^x} + 1}}} $$<br><br>$$ \Rightarrow $$ $$y - {1 \over y} = ln\left| {{e^x} + 1} \right| + c$$<br><br>It passes through (0, 1)<br><br>$$ \Rightarrow $$ c = $$ - ln2$$<br><br>$$ \Rightarrow $$ $${y^2} - 1 = yln\left( {{{{e^x} + 1} \over 2}} \right)$$<br><br>$$ \Rightarrow $$ $${y^2} = 1 + yln\left( {{{{e^x} + 1} \over 2}} \right)$$ | mcq | jee-main-2020-online-3rd-september-morning-slot |
7QEBB1pQGhFomq6Gq07k9k2k5khh7r2 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | If $${{dy} \over {dx}} = {{xy} \over {{x^2} + {y^2}}}$$; y(1) = 1; then a value of x
satisfying y(x) = e is : | [{"identifier": "A", "content": "$$\\sqrt 2 e$$"}, {"identifier": "B", "content": "$${1 \\over 2}\\sqrt 3 e$$"}, {"identifier": "C", "content": "$${e \\over {\\sqrt 2 }}$$"}, {"identifier": "D", "content": "$$\\sqrt 3 e$$"}] | ["D"] | null | $${{dy} \over {dx}} = {{xy} \over {{x^2} + {y^2}}}$$
<br><br>Let y = vx
<br><br>$$ \therefore $$ $${{dy} \over {dx}}$$ = v + x.$${{dv} \over {dx}}$$
<br><br>$$ \Rightarrow $$ v + x.$${{dv} \over {dx}}$$ = $${{x\left( {vx} \right)} \over {{x^2} + {v^2}{x^2}}}$$ = $${v \over {1 + {v^2}}}$$
<br><br>$$ \Rightarrow $$ x.$${{dv} \over {dx}}$$ = $${v \over {1 + {v^2}}}$$ - v = $${{v - v - {v^3}} \over {1 + {v^2}}}$$ = $$ - {{{v^3}} \over {1 + {v^2}}}$$
<br><br>$$ \Rightarrow $$ $$\int {{{1 + {v^2}} \over {{v^3}}}} dv = - \int {{{dx} \over x}} $$
<br><br>$$ \Rightarrow $$ $$ - {1 \over {2{v^2}}} + \log v$$ = $$ - \log x + C$$
<br><br>$$ \Rightarrow $$ $$ - {1 \over 2}{{{x^2}} \over {{y^2}}} + \log \left( {{y \over x}} \right)$$ = $$ - \log x + C$$ ......(1)
<br><br>putting x = 1, y = 1 we get
<br><br>$$ \Rightarrow $$ C = $$ - {1 \over 2}$$
<br><br>From eq. (1)
<br><br>$$ - {1 \over 2}{{{x^2}} \over {{y^2}}} + \log \left( {{y \over x}} \right)$$ = $$ - \log x - {1 \over 2}$$
<br><br>Put y = e
<br><br>$$ - {1 \over 2}{{{x^2}} \over {{e^2}}} + \log \left( {{e \over x}} \right)$$ = $$ - \log x - {1 \over 2}$$
<br><br>$$ \Rightarrow $$ x<sup>2</sup> = 3e<sup>2</sup>
<br><br>$$ \Rightarrow $$ x = $$ \pm $$3$$\sqrt e $$ | mcq | jee-main-2020-online-9th-january-evening-slot |
epwEIvt88KyxFkSHJc7k9k2k5iu0u4n | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | If for x $$ \ge $$ 0, y = y(x) is the solution of the
differential equation<br/>
(x + 1)dy = ((x + 1)<sup>2</sup> + y – 3)dx, y(2) = 0,
then y(3) is equal to _______. | [] | null | 3 | (x + 1)dy = ((x + 1)<sup>2</sup> + y – 3)dx
<br><br>$$ \Rightarrow $$ (1 + x)$${{dy} \over {dx}}$$ - y = (1 + x)<sup>2</sup> - 3
<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}} - {y \over {1 + x}} = \left( {1 + x} \right) - {3 \over {1 + x}}$$
<br><br>I.F = $${e^{ - \int {{{dx} \over {1 + x}}} }}$$ = $${1 \over {1 + x}}$$
<br><br>Solution of the differential equation,
<br><br>$$y\left( {{1 \over {1 + x}}} \right)$$ = $$\int {\left( {\left( {1 + x} \right) - {3 \over {1 + x}}} \right)\left( {{1 \over {1 + x}}} \right)dx} $$
<br><br>$$ \Rightarrow $$ $${y \over {1 + x}}$$ = $$\int {{{{x^2} + 2x + 1 - 3} \over {{{\left( {x + 1} \right)}^2}}}dx} $$
<br><br>$$ \Rightarrow $$ $${y \over {1 + x}}$$ = x + $${3 \over {1 + x}}$$ + C
<br><br>As y(2) = 0 $$ \Rightarrow $$ x = 2, y = 0
<br><br>$$ \therefore $$ 0 = 2 + $${3 \over {1 + 2}}$$ + C
<br><br>$$ \Rightarrow $$ C = -3
<br><br>So solution is $${y \over {1 + x}}$$ = x + $${3 \over {1 + x}}$$ - 3
<br><br>y(3) means x = 3 and find value of y.
<br><br>$${y \over {1 + 3}} = 3 + {3 \over {1 + 3}} - 3$$
<br><br>$$ \Rightarrow $$ y = 3
| integer | jee-main-2020-online-9th-january-morning-slot |
lM7C61I45F1lddLYt07k9k2k5grpj4s | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let y = y(x) be a solution of the differential
equation,
<br/><br>$$\sqrt {1 - {x^2}} {{dy} \over {dx}} + \sqrt {1 - {y^2}} = 0$$, |x| < 1.
<br/><br>If $$y\left( {{1 \over 2}} \right) = {{\sqrt 3 } \over 2}$$, then $$y\left( { - {1 \over {\sqrt 2 }}} \right)$$ is equal to :</br></br> | [{"identifier": "A", "content": "$$ - {{\\sqrt 3 } \\over 2}$$"}, {"identifier": "B", "content": "None of those"}, {"identifier": "C", "content": "$${{1 \\over {\\sqrt 2 }}}$$"}, {"identifier": "D", "content": "$$-{{1 \\over {\\sqrt 2 }}}$$"}] | ["C"] | null | Given $$\sqrt {1 - {x^2}} {{dy} \over {dx}} + \sqrt {1 - {y^2}} = 0$$
<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}} = - {{\sqrt {1 - {y^2}} } \over {\sqrt {1 - {x^2}} }}$$
<br><br>$$ \Rightarrow $$ $${{dy} \over {\sqrt {1 - {y^2}} }} + {{dx} \over {\sqrt {1 - {x^2}} }} = 0$$
<br><br>$$ \Rightarrow $$ sin<sup>-1</sup> y + sin<sup>-1</sup> x = c
<br><br>Given that $$y\left( {{1 \over 2}} \right) = {{\sqrt 3 } \over 2}$$
<br><br>So when x = $${1 \over 2}$$ then y = $${{\sqrt 3 } \over 2}$$
<br><br>$$ \therefore $$ sin<sup>-1</sup> $${{\sqrt 3 } \over 2}$$ + sin<sup>-1</sup> $${1 \over 2}$$ = c
<br><br>$$ \Rightarrow $$ c = $${\pi \over 3}$$ + $${\pi \over 6}$$ = $${\pi \over 2}$$
<br><br>So sin<sup>-1</sup> y + sin<sup>-1</sup> x = $${\pi \over 2}$$
<br><br>$$ \Rightarrow $$ sin<sup>-1</sup> y = $${\pi \over 2}$$ - sin<sup>-1</sup> x
<br><br>$$ \Rightarrow $$ sin<sup>-1</sup> y = cos<sup>-1</sup> x
<br><br>Now $$y\left( { - {1 \over {\sqrt 2 }}} \right)$$ means x = $$ - {1 \over {\sqrt 2 }}$$ and find y.
<br><br>Putting x = $$ - {1 \over {\sqrt 2 }}$$
<br><br>sin<sup>-1</sup> y = cos<sup>-1</sup> $$\left( { - {1 \over {\sqrt 2 }}} \right)$$ = $${{3\pi } \over 4}$$
<br/><br/>y (at x = $$ - {1 \over {\sqrt 2 }}$$) = sin($${{3\pi } \over 4}$$) = $${{1 \over {\sqrt 2 }}}$$ | mcq | jee-main-2020-online-8th-january-morning-slot |
nXpEQlCFiYbA3EetYk7k9k2k5fmqk1a | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let y = y(x) be the solution curve of the differential equation,
<br/><br>$$\left( {{y^2} - x} \right){{dy} \over {dx}} = 1$$, satisfying y(0) =
1. This curve intersects the x-axis at a point whose abscissa is :
</br> | [{"identifier": "A", "content": "2 + e"}, {"identifier": "B", "content": "-e"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "2 - e"}] | ["D"] | null | $$\left( {{y^2} - x} \right){{dy} \over {dx}} = 1$$
<br><br>$$ \Rightarrow $$ $${{dx} \over {dy}}$$ + x = y<sup>2</sup>
<br><br>I.F = $${e^{\int {dy} }}$$ = e<sup>y</sup>
<br><br>Solution is given by
<br><br>xe<sup>y</sup> = $${\int {{y^2}{e^y}dy} }$$
<br><br>$$ \Rightarrow $$ xe<sup>y</sup> = (y<sup>2</sup> – 2y + 2)e<sup>y</sup> + C
<br><br>y(0) = 1 means x = 0, y = 1
<br><br>$$ \therefore $$ C = -e
<br><br>$$ \therefore $$ xe<sup>y</sup> = (y<sup>2</sup> – 2y + 2)e<sup>y</sup> - e
<br><br>put y = 0
<br><br>$$ \therefore $$ x = 0 – 0 + 2 – e
<br><br>$$ \Rightarrow $$ x = 2 - e | mcq | jee-main-2020-online-7th-january-evening-slot |
f8xBCCaONb1L3fooym7k9k2k5e4i3rk | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | If y = y(x) is the solution of the differential equation, $${e^y}\left( {{{dy} \over {dx}} - 1} \right) = {e^x}$$ such that y(0) = 0, then
y(1) is equal to:
| [{"identifier": "A", "content": "2 + log<sub>e</sub>2"}, {"identifier": "B", "content": "log<sub>e</sub>2"}, {"identifier": "C", "content": "1 + log<sub>e</sub>2"}, {"identifier": "D", "content": "2e"}] | ["C"] | null | Given,
$${e^y}\left( {{{dy} \over {dx}} - 1} \right) = {e^x}$$
<br><br>$$ \Rightarrow $$ $${e^y}{{dy} \over {dx}} - {e^y} = {e^x}$$ ....(1)
<br><br>Let $${e^y} = t$$ $$ \Rightarrow $$ $${e^y}{{dy} \over {dx}} = {{dt} \over {dx}}$$
<br><br>$$ \therefore $$ $${{dt} \over {dx}} - t = {e^x}$$
<br><br>So here p = -1 and q = e<sup>x</sup>
<br><br>We know, IF = $${e^{\int {pdx} }}$$
<br><br>= $${e^{\int { - 1dx} }}$$ = $${e^{ - x}}$$
<br><br>$$ \therefore $$ t.$${e^{ - x}}$$ = $$\int {{e^x}.{e^{ - x}}dx} $$
<br><br>$$ \Rightarrow $$ t.$${e^{ - x}}$$ = x + c
<br><br>Putting value of t, we get
<br><br>$${e^y}{e^{ - x}}$$ = x + c
<br><br>$$ \Rightarrow $$ $${e^{y - x}} = x + c$$ .....(2)
<br><br>Given y(0) = 0 means y = 0 when x = 0.
<br><br>Putting in equation (2), we get
<br><br>e<sup>0</sup> = 0 + c
<br><br>$$ \Rightarrow $$ c = 1
<br><br>$$ \therefore $$ $${e^{y - x}} = x + 1$$ ....(3)
<br><br>Now we have to find y(1), which means when x = 1 find the value of y in the above equation.
<br><br>Putting x = 1 in equation (3)
<br><br>$${e^{y - 1}} = 1 + 1$$
<br><br>$$ \Rightarrow $$ y - 1 = log<sub>e</sub>2
<br><br>$$ \Rightarrow $$ y = 1 + log<sub>e</sub>2 | mcq | jee-main-2020-online-7th-january-morning-slot |
CJeQnQDQYP7JggZKtwjgy2xukezevy28 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | If a curve y = f(x), passing through the point
(1, 2), is the solution of the differential equation,
<br/>2x<sup>2</sup>dy= (2xy + y<sup>2</sup>)dx, then $$f\left( {{1 \over 2}} \right)$$ is equal to : | [{"identifier": "A", "content": "$${1 \\over {1 - {{\\log }_e}2}}$$"}, {"identifier": "B", "content": "$${1 \\over {1 + {{\\log }_e}2}}$$"}, {"identifier": "C", "content": "$${{ - 1} \\over {1 + {{\\log }_e}2}}$$"}, {"identifier": "D", "content": "$${1 + {{\\log }_e}2}$$"}] | ["B"] | null | $$2{x^2}dy = \left( {2xy + {y^2}} \right)dx$$<br><br>
$$ \Rightarrow 2{x^2}{{dy} \over {dx}} = 2xy + {y^2}$$<br><br>
$$ \Rightarrow {{2{x^2}} \over {2{x^2}{y^2}}}{{dy} \over {dx}} = {{2xy} \over {2{x^2}{y^2}}} + {{{y^2}} \over {2{x^2}{y^2}}}$$<br><br>
$$ \Rightarrow {1 \over {{y^2}}}{{dy} \over {dx}} - {1 \over x}{1 \over y} = {1 \over {2{x^2}}}$$<br><br>
Let $$ - {1 \over y} = t$$<br><br>
$$ \Rightarrow {1 \over {{y^2}}}{{dy} \over {dx}} = {{dt} \over {dx}}$$<br><br>
$$ \Rightarrow {{dt} \over {dx}} + t{1 \over x} = {1 \over {2{x^2}}}$$<br><br>
This is linear differentiatial equation.<br><br>
$$ \therefore I.F = {e^{\int {{1 \over x}dx} }} = {e^{\ln x}}$$<br><br>
$$ \therefore t.{e^{\ln x}} = \int {{1 \over {2{x^2}}}.{e^{\ln x}}dx} $$<br><br>
$$ \Rightarrow - {1 \over y}.x = \int {{1 \over {2{x^2}}}.xdx} $$<br><br>
$$ \Rightarrow - {x \over y} = {1 \over 2}\int {{{dx} \over x}} $$<br><br>
$$ \Rightarrow - {x \over y} = {1 \over 2}\ln x + c$$<br><br>
This curve passes through the point (1, 2)<br><br>
$$ \therefore - {1 \over 2} = 0 + c$$<br><br>
$$ \Rightarrow c = - {1 \over 2}$$<br><br>
$$ \therefore - {x \over y} = {1 \over 2}\ln x - {1 \over 2}$$<br><br>
$$ \Rightarrow - {{2x} \over y} = \ln x - 1$$<br><br>
$$ \Rightarrow y = {{2x} \over {1 - \ln x}}$$<br><br>
$$ \Rightarrow f\left( x \right) = {{2x} \over {1 - \ln x}}$$<br><br>
So, $$f\left( {{1 \over 2}} \right) = {{2 \times {1 \over 2}} \over {1 - \ln \left( {{1 \over 2}} \right)}} = {1 \over {1 + {{\log }_e}2}}$$ | mcq | jee-main-2020-online-2nd-september-evening-slot |
vjTquFf7G6Ehes6d2o1klretipy | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | The population P = P(t) at time 't' of a certain species follows the differential equation
<br/><br>$${{dP} \over {dt}}$$
= 0.5P – 450. If P(0) = 850, then the time at which population becomes zero is :</br> | [{"identifier": "A", "content": "$${\\log _e}18$$"}, {"identifier": "B", "content": "$${1 \\over 2}{\\log _e}18$$"}, {"identifier": "C", "content": "2$${\\log _e}18$$"}, {"identifier": "D", "content": "$${\\log _e}9$$"}] | ["C"] | null | $${{dp} \over {dt}} = {{p - 900} \over 2}$$<br><br>$$\int\limits_{850}^0 {{{dp} \over {p - 900}} = \int\limits_0^t {{{dt} \over 2}} } $$<br><br>$$ \Rightarrow $$ $$\ln |p - 900|_{850}^0 = {t \over 2}$$<br><br>$$ \Rightarrow $$ $$\ln 900 - \ln 50 = {t \over 2}$$<br><br>$$ \Rightarrow $$ $${t \over 2} = \ln 18$$<br><br>$$ \Rightarrow t = 2\ln 18$$ | mcq | jee-main-2021-online-24th-february-morning-slot |
XwxnZARFYQL78w2sdv1kluh41v8 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | The rate of growth of bacteria in a culture is proportional to the number of bacteria present and the bacteria count is 1000 at initial time t = 0. The number of bacteria is increased by 20% in 2 hours. If the population of bacteria is 2000 after $${k \over {{{\log }_e}\left( {{6 \over 5}} \right)}}$$ hours, then $${\left( {{k \over {{{\log }_e}2}}} \right)^2}$$ is equal to : | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "4"}] | ["D"] | null | $${{dx} \over {dt}} \propto x$$<br><br>$${{dx} \over {dt}} = \lambda x$$<br><br>$$\int\limits_{1000}^x {{{dx} \over x} = \int\limits_0^t {\lambda dt} } $$<br><br>$$\ln x - \ln 1000 = \lambda t$$<br><br>$$\ln \left( {{x \over {1000}}} \right) = \lambda t$$<br><br>Put t = 2, x = 1200<br><br>$$\ln \left( {{{12} \over {10}}} \right) = 2\lambda \Rightarrow \lambda = {1 \over 2}\ln {6 \over 5}$$<br><br>Now, $$\ln \left( {{x \over {1000}}} \right) = {t \over 2}\ln \left( {{6 \over 5}} \right)$$<br><br>$$x = 1000{e^{{t \over 2}\ln \left( {{6 \over 5}} \right)}}$$<br><br>$$x = 2000$$ at $$t = {k \over {\ln \left( {{6 \over 5}} \right)}}$$<br><br>$$ \Rightarrow 2000 = 1000{e^{{k \over {2\ln (6/5)}} \times \ln (6/5)}}$$<br><br>$$ \Rightarrow 2 = {e^{k/2}}$$<br><br>$$ \Rightarrow \ln 2 = {k \over 2}$$<br><br>$$ \Rightarrow {k \over {\ln 2}} = 2$$<br><br>$$ \Rightarrow {\left( {{k \over {\ln 2}}} \right)^2} = 4$$ | mcq | jee-main-2021-online-26th-february-morning-slot |
HhbaQBEeiFLbPqxKu21kmhzaf0v | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let the curve y = y(x) be the solution of the differential equation, $${{dy} \over {dx}}$$ = 2(x + 1). If the numerical value of area bounded by the curve y = y(x) and x-axis is $${{4\sqrt 8 } \over 3}$$, then the value of y(1) is equal to _________. | [] | null | 2 | Given, $${{dy} \over {dx}}$$ = 2(x + 1)
<br><br>Integrating both sides, we get
<br><br>$$y = {x^2} + 2x + c$$
<br><br>Let the two roots of the quadratic equation $$\alpha $$ and $$\beta $$
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267191/exam_images/idoyhyhdrqzpe0cyaljz.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 16th March Morning Shift Mathematics - Differential Equations Question 135 English Explanation"><br><br>As parabola intercept the x axis so D > 0
<br><br>From figure, AB = |$$\alpha $$ - $$\beta $$| = $${{\sqrt D } \over {\left| a \right|}}$$ = $$\sqrt D $$
<br><br>and BC = $$ - {D \over {4a}}$$ = $$ - {D \over 4}$$
<br><br>$$ \therefore $$ Area of rectangle (ABCD) = AB $$ \times $$ BC = $$\sqrt D \times {D \over 4}$$
<br><br>From property we know,
<br><br>Area of parabola with the x axis = $${2 \over 3}$$(Area of rectangle)
<br><br>$$ \Rightarrow $$ $${{4\sqrt 8 } \over 3}$$ = $${2 \over 3} \times \sqrt D \times {D \over 4}$$
<br><br>$$ \Rightarrow $$ $$D\sqrt D $$ = $$8\sqrt 8 $$
<br><br>$$ \Rightarrow $$ D = 8
<br><br>$$ \therefore $$ b<sup>2</sup> - 4ac = 8
<br><br>$$ \Rightarrow $$ 4 - 4c = 8
<br><br>$$ \Rightarrow $$ 1 $$-$$ c = 2 $$ \Rightarrow $$ c = $$-$$ 1<br><br>Equation of f(x) = x<sup>2</sup> + 2x $$-$$ 1<br><br>$$ \therefore $$ f(1) = 1 + 2 $$-$$ 1 = 2 | integer | jee-main-2021-online-16th-march-morning-shift |
VXhDzKmEPb3kiJilUw1kmiwnrph | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let C<sub>1</sub> be the curve obtained by the solution of differential equation <br/><br>$$2xy{{dy} \over {dx}} = {y^2} - {x^2},x > 0$$. Let the curve C<sub>2</sub> be the <br/><br>solution of $${{2xy} \over {{x^2} - {y^2}}} = {{dy} \over {dx}}$$. If both the curves pass through (1, 1), then the area enclosed by the curves C<sub>1</sub> and C<sub>2</sub> is equal to :</br></br> | [{"identifier": "A", "content": "$${\\pi \\over 4}$$ + 1"}, {"identifier": "B", "content": "$$\\pi$$ + 1"}, {"identifier": "C", "content": "$$\\pi$$ $$-$$ 1"}, {"identifier": "D", "content": "$${\\pi \\over 2}$$ $$-$$ 1"}] | ["D"] | null | $${{dy} \over {dx}} = {{{y^2} - {x^2}} \over {2xy}}$$<br><br>Put $$y = vx$$<br><br>$$v + x{{dv} \over {dx}} = {{{v^2}{x^2} - {x^2}} \over {2v{x^2}}} = {{{v^2} - 1} \over {2v}}$$<br><br>$$x{{dv} \over {dx}} = {{{v^2} - 1 - 2{v^2}} \over {2v}} = - {{({v^2} + 1)} \over {2v}}$$<br><br>$$ \Rightarrow {{2v} \over {{v^2} + 1}}dv = - {{dx} \over x}$$<br><br>$$\ln ({v^2} + 1) = - \ln x + \ln c \Rightarrow {v^2} + 1 = {c \over x}$$<br><br>$$ \Rightarrow {{{y^2}} \over {{x^2}}} + 1 = {c \over x} \Rightarrow {x^2} + {y^2} = cx$$<br><br>It pass through (1, 1)<br><br>$$ \therefore $$ $${x^2} + {y^2} - 2x = 0$$<br><br>Similarly for second differential equation $${{dy} \over {dx}} = $$$${{2xy} \over {{x^2} - {y^2}}}$$<br><br>Equation of curve is x<sup>2</sup> + y<sup>2</sup> $$-$$ 2y = 0<br><br>Now required area is <br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267708/exam_images/i68tklovz24o6pflbuin.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 16th March Evening Shift Mathematics - Differential Equations Question 133 English Explanation"><br><br>$$ = 2\int\limits_0^1 {\left( {\sqrt {2x - {x^2}} - x} \right)} dx $$
<br><br>$$= ({\pi \over 2} - 1)$$ sq. units | mcq | jee-main-2021-online-16th-march-evening-shift |
LuPXuk1HHbq1nBRiX91kmjaiiev | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Which of the following is true for y(x) that satisfies the differential equation <br/><br/>$${{dy} \over {dx}}$$ = xy $$-$$ 1 + x $$-$$ y; y(0) = 0 : | [{"identifier": "A", "content": "y(1) = 1"}, {"identifier": "B", "content": "y(1) = e<sup>$$-$$$${1 \\over 2}$$</sup> $$-$$ 1"}, {"identifier": "C", "content": "y(1) = e<sup>$${1 \\over 2}$$</sup> $$-$$ e<sup>$$-$$$${1 \\over 2}$$</sup>"}, {"identifier": "D", "content": "y(1) = e<sup>$${1 \\over 2}$$</sup> $$-$$ 1"}] | ["B"] | null | $${{dy} \over {dx}} = (x - 1)y + (x - 1)$$<br><br>$${{dy} \over {dx}} = (x - 1)(y + 1)$$<br><br>$${{dy} \over {y + 1}} = (x - 1)dx$$<br><br>Integrating both sides, we get<br><br>$$\ln (y + 1) = {{{x^2}} \over 2} - x + c$$<br><br>$$x = 0,y = 0$$<br><br>$$ \Rightarrow c = 0$$<br><br>$$ \therefore $$ $$\ln (y + 1) = {{{x^2}} \over 2} - x$$<br><br>putting $$x = 1,\ln (y + 1) = {1 \over 2} - 1 = - {1 \over 2}$$<br><br>$$y + 1 = {e^{ - {1 \over 2}}}$$<br><br>$$y = {e^{ - {1 \over 2}}} - 1$$<br><br>$$ \therefore $$ $$y(1) = {e^{ - {1 \over 2}}} - 1$$ | mcq | jee-main-2021-online-17th-march-morning-shift |
4O6TNg78va9YhAAeY01kmm45bsy | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let y = y(x) be the solution of the differential equation <br/><br/>xdy $$-$$ ydx = $$\sqrt {({x^2} - {y^2})} dx$$, x $$ \ge $$ 1, with y(1) = 0. If the area bounded by the line x = 1, x = e<sup>$$\pi$$</sup>, y = 0 and y = y(x) is $$\alpha$$e<sup>2$$\pi$$</sup> + $$\beta$$, then the value of 10($$\alpha$$ + $$\beta$$) is equal to __________. | [] | null | 4 | $$xdy - ydx = \sqrt {{x^2} - {y^2}} dx$$<br><br>dividing both sides by x<sup>2</sup>, we get<br><br>$${{xdy - ydx} \over {{x^2}}} = {{\sqrt {{x^2} - {y^2}} } \over {{x^2}}}dx$$<br><br>$$ \Rightarrow d\left( {{y \over x}} \right) = {1 \over x}\sqrt {1 - {{\left( {{y \over x}} \right)}^2}} dx$$<br><br>$$ \Rightarrow {{d\left( {{y \over x}} \right)} \over {\sqrt {1 - {{\left( {{y \over x}} \right)}^2}} }} = {{dx} \over x}$$<br><br>Integrating both side, we get<br><br>$$ \Rightarrow \int {{{d\left( {{y \over x}} \right)} \over {\sqrt {1 - {{\left( {{y \over x}} \right)}^2}} }} = \int {{{dx} \over x}} } $$<br><br>$${\sin ^{ - 1}}\left( {{y \over x}} \right) = \ln (x) + C$$<br><br>Given, y(1) = 0 $$ \Rightarrow $$ at x = 1, y = 0<br><br>$$ \therefore $$ $$ \Rightarrow {\sin ^{ - 1}}(0) = \ln (1) + C$$<br><br>$$ \Rightarrow $$ C = 0<br><br>$$ \therefore $$ $${\sin ^{ - 1}}\left( {{y \over x}} \right) = \ln (x)$$<br><br>$$ \Rightarrow $$ y = x sin(ln(x))<br><br>$$ \therefore $$ Area $$ = \int_1^{{e^{\pi {} }}} {x\sin (\ln (x))} dx$$<br><br>Let, lnx = t<br><br>$$ \Rightarrow $$ x = e<sup>t</sup><br><br>$$ \Rightarrow $$ dx = e<sup>t</sup> dt<br><br>New lower limit, t = ln(1) = 0<br><br>and upper limit t = ln$$({e^{\pi {} }})$$ = $${\pi {} }$$<br><br>$$ \therefore $$ Area = $$\int_0^{^{\pi {} }} {{e^t}\sin (t).{e^t}} dt$$<br><br>$$ = \int_0^{^{\pi {} }} {{e^{2t}}\sin t\,} dt$$<br><br>$$ = \left[ {{{{e^{2t}}} \over {({1^2} + {2^2})}}(2\sin t - 1\cos t)} \right]_0^{\pi {} }$$<br><br>$$ = {\left[ {{{{e^{2\pi {} }}} \over 5}(0 - ( - 1)) - {1 \over 5}( - 1)} \right]}$$<br><br>$$ = {{{e^{2\pi {} }}} \over 5} + {1 \over 5}$$<br><br>$$ = \alpha {e^{2\pi {} }} + \beta $$<br><br>$$ \therefore $$ $$\alpha = {1 \over 5},\beta = {1 \over 5}$$<br><br>So, $$10(\alpha + \beta ) = 4$$ | integer | jee-main-2021-online-18th-march-evening-shift |
1krpvi530 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let y = y(x) be the solution of the differential equation $$x\tan \left( {{y \over x}} \right)dy = \left( {y\tan \left( {{y \over x}} \right) - x} \right)dx$$, $$ - 1 \le x \le 1$$, $$y\left( {{1 \over 2}} \right) = {\pi \over 6}$$. Then the area of the region bounded by the curves x = 0, $$x = {1 \over {\sqrt 2 }}$$ and y = y(x) in the upper half plane is : | [{"identifier": "A", "content": "$${1 \\over 8}(\\pi - 1)$$"}, {"identifier": "B", "content": "$${1 \\over {12}}(\\pi - 3)$$"}, {"identifier": "C", "content": "$${1 \\over 4}(\\pi - 2)$$"}, {"identifier": "D", "content": "$${1 \\over 6}(\\pi - 1)$$"}] | ["A"] | null | We have,<br><br>$${{dy} \over {dx}} = {{x\left( {{y \over x}.\tan {y \over x} - 1} \right)} \over {x\tan {y \over x}}}$$<br><br>$$\therefore$$ $${{dy} \over {dx}} = {y \over x} - \cot \left( {{y \over x}} \right)$$<br><br>Put $${y \over x} = v$$<br><br>$$ \Rightarrow y = vn$$<br><br>$$\therefore$$ $${{dy} \over {dx}} = v + {{ndv} \over {dx}}$$<br><br>Now, we get<br><br>$$v + n{{dv} \over {dx}} = v - \cot (v)$$<br><br>$$ \Rightarrow \int {(\tan )dv} = - \int {{{dx} \over x}} $$<br><br>$$\therefore$$ $$\ln \left| {\sec \left( {{y \over x}} \right)} \right| = - \ln \left| x \right| + c$$<br><br>As, $$\left( {{1 \over 2}} \right) = \left( {{y \over x}} \right) \Rightarrow C = 0$$<br><br>$$\therefore$$ $$\sec \left( {{y \over x}} \right) = {1 \over x}$$<br><br>$$ \Rightarrow \cos \left( {{y \over x}} \right) = x$$<br><br>$$\therefore$$ $$y = x{\cos ^{ - 1}}(x)$$<br><br>So, required bounded area<br><br>$$ = \int\limits_0^{{1 \over {\sqrt 2 }}} {\mathop x\limits_{(II)} (\mathop {{{\cos }^{ - 1}}x}\limits_{(I)} )dx = \left( {{{\pi - 1} \over 8}} \right)} $$ (I. B. P.)<br><br>$$\therefore$$ Option (1) is correct. | mcq | jee-main-2021-online-20th-july-morning-shift |
1krpwxwmo | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let y = y(x) be the solution of the differential equation $${e^x}\sqrt {1 - {y^2}} dx + \left( {{y \over x}} \right)dy = 0$$, y(1) = $$-$$1. Then the value of (y(3))<sup>2</sup> is equal to : | [{"identifier": "A", "content": "1 $$-$$ 4e<sup>3</sup>"}, {"identifier": "B", "content": "1 $$-$$ 4e<sup>6</sup>"}, {"identifier": "C", "content": "1 + 4e<sup>3</sup>"}, {"identifier": "D", "content": "1 + 4e<sup>6</sup>"}] | ["B"] | null | $${e^x}\sqrt {1 - {y^2}} dx + {y \over x}dy = 0$$<br><br>$$ \Rightarrow {e^x}\sqrt {1 - {y^2}} dx + {{ - y} \over x}dy$$<br><br>$$ \Rightarrow \int {{{ - y} \over {\sqrt {1 - {y^2}} }}} dy = \int_{II}^{{e^x}} {_1^xdx} $$<br><br>$$ \Rightarrow \sqrt {1 - {y^2}} = {e^x}(x - 1) + c$$<br><br>Given : At x = 1, y = $$-$$1<br><br>$$\Rightarrow$$ 0 = 0 + c $$\Rightarrow$$ c = 0<br><br>$$\therefore$$ $$\sqrt {1 - {y^2}} = {e^x}(x - 1)$$<br><br>At x = 3 <br><br>$$1 - {y^2} = {({e^3}2)^2} \Rightarrow {y^2} = 1 - 4{e^6}$$ | mcq | jee-main-2021-online-20th-july-morning-shift |
1krrw09ja | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let a curve y = y(x) be given by the solution of the differential equation $$\cos \left( {{1 \over 2}{{\cos }^{ - 1}}({e^{ - x}})} \right)dx = \sqrt {{e^{2x}} - 1} dy$$. If it intersects y-axis at y = $$-$$1, and the intersection point of the curve with x-axis is ($$\alpha$$, 0), then e<sup>$$\alpha$$</sup> is equal to __________________. | [] | null | 2 | $$\cos \left( {{1 \over 2}{{\cos }^{ - 1}}({e^{ - x}})} \right)dx = \sqrt {{e^{2x}} - 1} dy$$<br><br>Put cos<sup>$$-$$1</sup>(e<sup>$$-$$x</sup>) $$\theta$$, $$\theta$$ $$\in$$ [0, $$\pi$$]<br><br>$$\cos \theta = {e^{ - x}} \Rightarrow 2{\cos ^2}{\theta \over 2} - 1 = {e^{ - x}}$$<br><br>$$\cos {\theta \over 2} = \sqrt {{{{e^{ - x}} + 1} \over 2}} = \sqrt {{{{e^x} + 1} \over {2{c^x}}}} $$<br><br>$$\sqrt {{{{e^x} + 1} \over {2{c^x}}}} dx = \sqrt {{e^{2x}} - 1} dy$$<br><br>$${1 \over {\sqrt 2 }}\int {{{dx} \over {\sqrt {{e^x}} \sqrt {{e^x} - 1} }} = \int {dy} } $$<br><br>Put $${e^x} = t,{{dt} \over {dx}} = {e^x}$$<br><br>$${1 \over {\sqrt 2 }}\int {{{dx} \over {{e^x}\sqrt {{e^x}} \sqrt {{e^x} - 1} }} = \int {dy} } $$<br><br>$$\int {{{dt} \over {t\sqrt {{t^2} - t} }} = \sqrt 2 y} $$<br><br>Put $$t = {1 \over z},{{dt} \over {dz}} = - {1 \over {{z^2}}}$$<br><br>$$\int {{{ - {{dz} \over {{z^2}}}} \over {{1 \over z}\sqrt {{1 \over {{z^2}}} - {1 \over z}} }} = \sqrt {2y} } $$<br><br>$$ - \int {{{dz} \over {\sqrt {1 - z} }} = \sqrt 2 y} $$<br><br>$${{ - 2{{(1 - z)}^{1/2}}} \over { - 1}} = \sqrt 2 y + c$$<br><br>$$2{\left( {1 - {1 \over t}} \right)^{1/2}} = \sqrt 2 y + c$$<br><br>$$2{(1 - {e^{ - x}})^{1/2}} = \sqrt 2 y + c\buildrel {(0, - 1)} \over
\longrightarrow \Rightarrow c = \sqrt 2 $$<br><br>$$2{(1 - {e^{ - x}})^{1/2}} = \sqrt 2 (y + 1)$$, passes through ($$\alpha$$, 0)<br><br>$$2{(1 - {e^{ - \alpha }})^{1/2}} = \sqrt 2 $$<br><br>$$\sqrt {1 - {e^{ - \alpha }}} = {1 \over {\sqrt 2 }} \Rightarrow 1 - {e^{ - \alpha }} = {1 \over 2}$$<br><br>$${e^{ - \alpha }} = {1 \over 2} \Rightarrow {e^\alpha } = 2$$ | integer | jee-main-2021-online-20th-july-evening-shift |
1krubsd7e | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let y = y(x) be the solution of the differential equation $$\left( {(x + 2){e^{\left( {{{y + 1} \over {x + 2}}} \right)}} + (y + 1)} \right)dx = (x + 2)dy$$, y(1) = 1. If the domain of y = y(x) is an open interval ($$\alpha$$, $$\beta$$), then | $$\alpha$$ + $$\beta$$| is equal to ______________. | [] | null | 4 | Let y + 1 = Y and x + 2 = X<br><br>dy = dY<br><br>dx = dX<br><br>$$\left( {X{e^{{X \over X}}} + Y} \right)dX = XdY$$<br><br>$$ \Rightarrow {{XdY - YdX} \over {{X^2}}} = {{{e^{{Y \over X}}}} \over X}dX$$<br><br>$$ \Rightarrow {e^{ - {Y \over X}}}d\left( {{Y \over X}} \right) = {{dX} \over X}$$<br><br>$$ \Rightarrow - {e^{ - {Y \over X}}} = \ln |X| + c$$<br><br>$$ \Rightarrow - {e^{ - \left( {{{y + 1} \over {x + 2}}} \right)}} = \ln |x + 2| + c$$<br><br>$$\because$$ (1, 1) satisfy this equation<br><br>So, $$c = - {e^{ - {2 \over 3}}} - \ln 3$$<br><br>Now, $$y = - 1 - (x + 2)\ln \left( {\ln \left( {\left| {{3 \over {x + 2}}} \right|} \right) + {e^{ - {2 \over 3}}}} \right)$$<br><br>Domain :<br><br>$$\ln \left| {{3 \over {x + 2}}} \right| > {e^{ - {e^{ - {2 \over 3}}}}}$$<br><br>$$ \Rightarrow {3 \over {\left| {x + 2} \right|}} > {e^{ - {e^{ - {2 \over 3}}}}}$$<br><br>$$ \Rightarrow \left| {x + 2} \right| < 3{e^{{e^{ - {2 \over 3}}}}}$$<br><br>$$ \Rightarrow - 3{e^{{e^{ - {2 \over 3}}}}} - 2 < x < 3{e^{{e^{ - {2 \over 3}}}}} - 2$$<br><br>So, $$\alpha + \beta = - 4$$<br><br>$$ \Rightarrow \left| {\alpha + \beta } \right| = 4$$ | integer | jee-main-2021-online-22th-july-evening-shift |
1krvymhco | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let y = y(x) be the solution of the differential equation $${{dy} \over {dx}} = 1 + x{e^{y - x}}, - \sqrt 2 < x < \sqrt 2 ,y(0) = 0$$<br/><br/>then, the minimum value of $$y(x),x \in \left( { - \sqrt 2 ,\sqrt 2 } \right)$$ is equal to : | [{"identifier": "A", "content": "$$\\left( {2 - \\sqrt 3 } \\right) - {\\log _e}2$$"}, {"identifier": "B", "content": "$$\\left( {2 + \\sqrt 3 } \\right) + {\\log _e}2$$"}, {"identifier": "C", "content": "$$\\left( {1 + \\sqrt 3 } \\right) - {\\log _e}\\left( {\\sqrt 3 - 1} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {1 - \\sqrt 3 } \\right) - {\\log _e}\\left( {\\sqrt 3 - 1} \\right)$$"}] | ["D"] | null | $${{dy - dx} \over {{e^{y - x}}}} = xdx$$<br><br>$$ \Rightarrow {{dy - dx} \over {{e^{y - x}}}} = xdx$$<br><br>$$ \Rightarrow - {e^{x - y}} = {{{x^2}} \over 2} + c$$<br><br>At x = 0, y = 0 $$\Rightarrow$$ c = $$-$$1<br><br>$$ \Rightarrow {e^{x - y}} = {{2 - {x^2}} \over 2}$$<br><br>$$ \Rightarrow y = x - \ln \left( {{{2 - {x^2}} \over 2}} \right)$$<br><br>$$ \Rightarrow {{dy} \over {dx}} = 1 + {{2x} \over {2 - {x^2}}} = {{2 + 2x - {x^2}} \over {2 - {x^2}}}$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265056/exam_images/fendqfkdavmkbahhpwrf.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th July Morning Shift Mathematics - Differential Equations Question 120 English Explanation"><br>So minimum value occurs at $$x = 1 - \sqrt 3 $$<br><br>$$y\left( {1 - \sqrt 3 } \right) = \left( {1 - \sqrt 3 } \right) - \ln \left( {{{2 - \left( {4 - 2\sqrt 3 } \right)} \over 2}} \right)$$<br><br>$$ = \left( {1 - \sqrt 3 } \right) - \ln \left( {\sqrt 3 - 1} \right)$$ | mcq | jee-main-2021-online-25th-july-morning-shift |
1krxjweg5 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let y = y(x) be the solution of the differential <br/><br/>equation (x $$-$$ x<sup>3</sup>)dy = (y + yx<sup>2</sup> $$-$$ 3x<sup>4</sup>)dx, x > 2. If y(3) = 3, then y(4) is equal to : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "16"}] | ["B"] | null | $$(x - {x^3})dy = (y + y{x^2} - 3{x^4})dx$$<br><br>$$ \Rightarrow xdy - ydx = (y{x^2} - 3{x^4})dx + {x^3}dy$$<br><br>$$ \Rightarrow {{xdy - ydx} \over {{x^2}}} = (ydx + xdy) - 3{x^2}dx$$<br><br>$$ \Rightarrow d\left( {{y \over x}} \right) = d(xy) - d({x^3})$$<br><br>Integrate<br><br>$$ \Rightarrow {y \over x} = xy - {x^3} + c$$<br><br>given f(3) = 3<br><br>$$ \Rightarrow {3 \over 3} = 3 \times 3 - {3^3} + c$$<br><br>$$ \Rightarrow c = 19$$<br><br>$$\therefore$$ $${y \over x} = xy - {x^3} + 19$$<br><br>at $$x = 4,{y \over 4} = 4y - 64 + 19$$<br><br>$$15y = 4 \times 45$$<br><br>$$ \Rightarrow y = 12$$ | mcq | jee-main-2021-online-27th-july-evening-shift |
1krygiv8j | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let y = y(x) be the solution of the differential equation dy = e<sup>$$\alpha$$x + y</sup> dx; $$\alpha$$ $$\in$$ N. If y(log<sub>e</sub>2) = log<sub>e</sub>2 and y(0) = log<sub>e</sub>$$\left( {{1 \over 2}} \right)$$, then the value of $$\alpha$$ is equal to _____________. | [] | null | 2 | $$\int {{e^{ - y}}} dy = \int {{e^{\alpha x}}} dx$$<br><br>$$ \Rightarrow {e^{ - y}} = {{{e^{\alpha x}}} \over \alpha } + c$$ ..... (i)<br><br>Put (x, y) = (ln2, ln2)<br><br>$${{ - 1} \over 2} = {{{2^\alpha }} \over \alpha } + C$$ ..... (ii)<br><br>Put (x, y) $$ \equiv $$ (0, $$-$$ln2) in (i)<br><br>$$ - 2 = {1 \over \alpha } + C$$ ..... (iii)<br><br>(ii) $$-$$ (iii)<br><br>$${{{2^\alpha } - 1} \over \alpha } = {3 \over 2}$$<br><br>$$\Rightarrow$$ $$\alpha$$ = 2 (as $$\alpha$$ $$\in$$ N) | integer | jee-main-2021-online-27th-july-evening-shift |
1krzqf052 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let y = y(x) be the solution of the differential <br/><br>equation xdy = (y + x<sup>3</sup> cosx)dx with y($$\pi$$) = 0, then $$y\left( {{\pi \over 2}} \right)$$ is equal to :</br> | [{"identifier": "A", "content": "$${{{\\pi ^2}} \\over 4} + {\\pi \\over 2}$$"}, {"identifier": "B", "content": "$${{{\\pi ^2}} \\over 2} + {\\pi \\over 4}$$"}, {"identifier": "C", "content": "$${{{\\pi ^2}} \\over 2} - {\\pi \\over 4}$$"}, {"identifier": "D", "content": "$${{{\\pi ^4}} \\over 4} - {\\pi \\over 2}$$"}] | ["A"] | null | $$xdy = (y + {x^3}\cos x)dx$$<br><br>$$ \Rightarrow $$ $$xdy = ydx + {x^3}\cos xdx$$<br><br>$$ \Rightarrow $$ $${{xdy - ydx} \over {{x^2}}} = {{{x^3}coxdx} \over {{x^2}}}$$<br><br>$$ \Rightarrow $$ $${d \over {dx}}\left( {{y \over x}} \right) = \int {x\cos xdx} $$<br><br>$$ \Rightarrow {y \over x} = x\sin x - \int {1.\sin xdx} $$<br><br>$$ \Rightarrow $$ $${y \over x} = x\sin x + \cos x + C$$<br><br>$$ \Rightarrow 0 = - 1 + C \Rightarrow C = 1,x = \pi ,y = 0$$<br><br>so, $${y \over x} = x\sin x + \cos x + 1$$<br><br>$$y = {x^2}\sin x + x\cos x + x$$<br><br>$$x = {\pi \over 2}$$<br><br>$$y\left( {{\pi \over 2}} \right) = {{{\pi ^2}} \over 4} + {\pi \over 2}$$ | mcq | jee-main-2021-online-25th-july-evening-shift |
1ks08wi98 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let y = y(x) be solution of the differential equation <br/><br/>$${\log _{}}\left( {{{dy} \over {dx}}} \right) = 3x + 4y$$, with y(0) = 0.<br/><br/>If $$y\left( { - {2 \over 3}{{\log }_e}2} \right) = \alpha {\log _e}2$$, then the value of $$\alpha$$ is equal to : | [{"identifier": "A", "content": "$$ - {1 \\over 4}$$"}, {"identifier": "B", "content": "$${1 \\over 4}$$"}, {"identifier": "C", "content": "$$2$$"}, {"identifier": "D", "content": "$$ - {1 \\over 2}$$"}] | ["A"] | null | $${{dy} \over {dx}} = {e^{3x}}.{e^{4y}} \Rightarrow \int {{e^{ - 4y}}dy = \int {{e^{3x}}dx} } $$<br><br>$${{{e^{ - 4y}}} \over { - 4}} = {{{e^{3x}}} \over 3} + C \Rightarrow - {1 \over 4} - {1 \over 3} = C \Rightarrow C = - {7 \over {12}}$$<br><br>$${{{e^{ - 4y}}} \over { - 4}} = {{{e^{3x}}} \over 3} - {7 \over {12}} \Rightarrow {e^{ - 4y}} = {{4{e^{3x}} - 7} \over { - 3}}$$<br><br>$${e^{4y}} = {3 \over {7 - 4{e^{3x}}}} \Rightarrow 4y = \ln \left( {{3 \over {7 - 4{e^{3x}}}}} \right)$$<br><br>$$4y = \ln \left( {{3 \over 6}} \right)$$ when $$x = - {2 \over 3}\ln 2$$<br><br>$$y = {1 \over 4}\ln \left( {{1 \over 2}} \right) = - {1 \over 4}\ln 2$$ | mcq | jee-main-2021-online-27th-july-morning-shift |
1ks0d8x6g | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | If $$y = y(x),y \in \left[ {0,{\pi \over 2}} \right)$$ is the solution of the differential equation $$\sec y{{dy} \over {dx}} - \sin (x + y) - \sin (x - y) = 0$$, with y(0) = 0, then $$5y'\left( {{\pi \over 2}} \right)$$ is equal to ______________. | [] | null | 2 | $$\sec y{{dy} \over {dx}} = 2\sin x\cos y$$<br><br>$${\sec ^2}ydy = 2\sin xdx$$<br><br>$$\tan y = - 2\cos x + c$$<br><br>$$c = 2$$<br><br>$$\tan y = - 2\cos x + 2 \Rightarrow $$ at $$x = {\pi \over 2}$$<br><br>$$\tan y = 2$$<br><br>$${\sec ^2}y{{dy} \over {dx}} = 2\sin x$$<br><br>$$ \therefore $$ $$5{{dy} \over {dx}} = 2$$ | integer | jee-main-2021-online-27th-july-morning-shift |
1ktcz25fq | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let y(x) be the solution of the differential equation <br/><br/>2x<sup>2</sup> dy + (e<sup>y</sup> $$-$$ 2x)dx = 0, x > 0. If y(e) = 1, then y(1) is equal to : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "log<sub>e</sub> 2"}, {"identifier": "D", "content": "log<sub>e</sub> (2e)"}] | ["C"] | null | $$2{x^2}dy + ({e^y} - 2x)dx = 0$$<br><br>$${{dy} \over {dx}} + {{{e^y} - 2x} \over {2{x^2}}} = 0 \Rightarrow {{dy} \over {dx}} + {{{e^y}} \over {2{x^2}}} - {1 \over x} = 0$$<br><br>$${e^{ - y}}{{dy} \over {dx}} - {{{e^{ - y}}} \over x} = - {1 \over {2{x^2}}} \Rightarrow $$ Put $${e^{ - y}} = z$$<br><br>$${{ - dz} \over {dx}} - {z \over x} = - {1 \over {2{x^2}}} \Rightarrow xdz + zdx = {{dx} \over {2x}}$$<br><br>$$d(xz) = {{dx} \over {2x}} \Rightarrow xz = {1 \over 2}{\log _e}x + c$$<br><br>$$x{e^{ - y}} = {1 \over 2}{\log _e}x + c$$, passes through (e, 1)<br><br>$$ \Rightarrow C = {1 \over 2}$$<br><br>$$x{e^{ - y}} = {{{{\log }_e}ex} \over 2}$$<br><br>$${e^{ - y}} = {1 \over 2} \Rightarrow y = {\log _e}2$$ | mcq | jee-main-2021-online-26th-august-evening-shift |
1kteisxmb | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let y = y(x) be the solution of the differential equation<br/><br/> $${{dy} \over {dx}} = 2(y + 2\sin x - 5)x - 2\cos x$$ such that y(0) = 7. Then y($$\pi$$) is equal to : | [{"identifier": "A", "content": "$$2{e^{{\\pi ^2}}} + 5$$"}, {"identifier": "B", "content": "$${e^{{\\pi ^2}}} + 5$$"}, {"identifier": "C", "content": "$$3{e^{{\\pi ^2}}} + 5$$"}, {"identifier": "D", "content": "$$7{e^{{\\pi ^2}}} + 5$$"}] | ["A"] | null | $${{dy} \over {dx}} - 2xy = 2(2\sin x - 5)x - 2\cos x$$<br><br>IF = $${e^{ - {x^2}}}$$<br><br>So, $$y.{e^{ - {x^2}}} = \int {{e^{ - {x^2}}}(2x(2\sin x - 5) - 2\cos x)dx} $$<br><br>$$ \Rightarrow y.{e^{ - {x^2}}} = {e^{ - {x^2}}}(5 - 2\sin x) + c$$<br><br>$$ \Rightarrow y = 5 - 2\sin x + c.{e^{{x^2}}}$$<br><br>Given at x = 0, y = 7<br><br>$$\Rightarrow$$ 7 = 5 + c $$\Rightarrow$$ c = 2<br><br>So, $$y = 5 - 2\sin x + 2{e^{{x^2}}}$$<br><br>Now, at x = $$\pi$$,<br><br>y = 5 + 2$${e^{{x^2}}}$$ | mcq | jee-main-2021-online-27th-august-morning-shift |
1ktiqwm6u | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | If $${{dy} \over {dx}} = {{{2^{x + y}} - {2^x}} \over {{2^y}}}$$, y(0) = 1, then y(1) is equal to : | [{"identifier": "A", "content": "log<sub>2</sub>(2 + e)"}, {"identifier": "B", "content": "log<sub>2</sub>(1 + e)"}, {"identifier": "C", "content": "log<sub>2</sub>(2e)"}, {"identifier": "D", "content": "log<sub>2</sub>(1 + e<sup>2</sup>)"}] | ["B"] | null | $${{dy} \over {dx}} = {{{2^{x + y}} - {2^x}} \over {{2^y}}}$$<br><br>$${2^y}{{dy} \over {dx}} = {2^x}({2^y} - 1)$$<br><br>$$\int {{{{2^y}} \over {{2^y} - 1}}dy = \int {{2^x}\,dx} } $$<br><br>$${{\ln ({2^y} - 1)} \over {\ln 2}} = {{{2^x}} \over {\ln 2}} + C$$<br><br>$$ \Rightarrow {\log _2}({2^y} - 1) = {2^x}{\log _2}e + C$$<br><br>$$\because$$ $$y(0) = 1 \Rightarrow 0 = {\log _2}e + C$$<br><br>$$C = - {\log _2}e$$<br><br>$$ \Rightarrow {\log _2}({2^y} - 1) = ({2^x} - 1){\log _2}e$$<br><br>put x = 1, $${\log _2}({2^y} - 1) = {\log _2}e$$<br><br>2<sup>y</sup> = e + 1<br><br>y = log<sub>2</sub>(e + 1) Ans. | mcq | jee-main-2021-online-31st-august-morning-shift |
1ktk8vcgu | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | If $$y{{dy} \over {dx}} = x\left[ {{{{y^2}} \over {{x^2}}} + {{\phi \left( {{{{y^2}} \over {{x^2}}}} \right)} \over {\phi '\left( {{{{y^2}} \over {{x^2}}}} \right)}}} \right]$$, x > 0, $$\phi$$ > 0, and y(1) = $$-$$1, then $$\phi \left( {{{{y^2}} \over 4}} \right)$$ is equal to : | [{"identifier": "A", "content": "4 $$\\phi$$ (2)"}, {"identifier": "B", "content": "4$$\\phi$$ (1)"}, {"identifier": "C", "content": "2 $$\\phi$$ (1)"}, {"identifier": "D", "content": "$$\\phi$$ (1)"}] | ["B"] | null | Let, $$y = tx$$<br><br>$${{dy} \over {dx}} = t + x{{dt} \over {dx}}$$<br><br>$$\therefore$$ $$tx\left( {t + x{{dt} \over {dx}}} \right) = x\left( {{t^2} + {{\varphi ({t^2})} \over {\varphi '({t^2})}}} \right)$$<br><br>$${t^2} + xt{{dt} \over {dx}} = {t^2} + {{\varphi ({t^2})} \over {\varphi '({t^2})}}$$<br><br>$$\int {{{t\varphi '({t^2})} \over {\varphi ({t^2})}}} dt = \int {{{dx} \over x}} $$<br><br>Let $$\varphi ({t^2}) = p$$<br><br>$$\therefore$$ $$\varphi '({t^2})2tdt = dp$$<br><br>$$ \Rightarrow \int {{{dy} \over {2p}}} = \int {{{dx} \over x}} $$<br><br>$${1 \over 2}\ln \varphi ({t^2}) = \ln x + \ln c$$<br><br>$$\varphi ({t^2}) = {x^2}k$$<br><br>$$\varphi \left( {{{{y^2}} \over {{x^2}}}} \right) = k{x^2},\varphi (1) = k$$<br><br>$$\varphi \left( {{{{y^2}} \over 4}} \right) = 4\varphi (1)$$ | mcq | jee-main-2021-online-31st-august-evening-shift |
1l544jesq | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let the solution curve of the differential equation</p>
<p>$$x{{dy} \over {dx}} - y = \sqrt {{y^2} + 16{x^2}} $$, $$y(1) = 3$$ be $$y = y(x)$$. Then y(2) is equal to:</p> | [{"identifier": "A", "content": "15"}, {"identifier": "B", "content": "11"}, {"identifier": "C", "content": "13"}, {"identifier": "D", "content": "17"}] | ["A"] | null | <p>Given,</p>
<p>$$x{{dy} \over {dx}} - y = \sqrt {{y^2} + 16x} $$</p>
<p>$$ \Rightarrow x{{dy} \over {dx}} = y + \sqrt {{y^2} + 16x} $$</p>
<p>$$ \Rightarrow {{dy} \over {dx}} = {y \over x} + \sqrt {{{\left( {{y \over x}} \right)}^2} + 16} $$</p>
<p>This is a homogenous different equation.</p>
<p>Let $${y \over x} = v$$</p>
<p>$$ \Rightarrow y = vx$$</p>
<p>$$ \Rightarrow {{dy} \over {dx}} = v + x{{dv} \over {dx}}$$</p>
<p>$$\therefore$$ $$v + x{{dv} \over {dx}} =v+ \sqrt {{v^2} + 16} $$</p>
<p>$$ \Rightarrow $$ $$ x{{dv} \over {dx}} = \sqrt {{v^2} + 16} $$</p>
<p>$$ \Rightarrow {{dv} \over {\sqrt {{v^2} + 16} }} = {{dx} \over x}$$</p>
<p>Integrating both sides, we get</p>
<p>$$\int {{{dv} \over {\sqrt {{v^2} + 16} }} = \int {{{dx} \over x}} } $$</p>
<p>$$ \Rightarrow \ln \left| {v + \sqrt {{v^2} + 16} } \right| = \ln x + \ln c$$</p>
<p>$$ \Rightarrow v + \sqrt {{v^2} + 16} = cx$$</p>
<p>Now putting, $$v = {y \over x}$$, we get</p>
<p>$${y \over x} + \sqrt {{{{y^2}} \over {{x^2}}} + 16} = cx$$</p>
<p>$$ \Rightarrow {y \over x} + \sqrt {{{{y^2} + 16{x^2}} \over {{x^2}}}} = cx$$</p>
<p>$$ \Rightarrow y + \sqrt {{y^2} + 16{x^2}} = c{x^2}$$ ...... (1)</p>
<p>Given, $$y(1) = 3$$</p>
<p>$$\therefore$$ When x = 1 then y = 3.</p>
<p>Putting in equation (1) we get,</p>
<p>$$3 + \sqrt {9 + 16} = c.\,1$$</p>
<p>$$ \Rightarrow c = 8$$</p>
<p>$$\therefore$$ Solution of equation,</p>
<p>$$y + \sqrt {{y^2} + 16{x^2}} = 8{x^2}$$</p>
<p>Now, y(2) means when x = 2 then y = ?</p>
<p>$$\therefore$$ $$y + \sqrt {{y^2} + 16 \times 4} = 8 \times 4$$</p>
<p>$$ \Rightarrow y = 15$$</p> | mcq | jee-main-2022-online-29th-june-morning-shift |
1l54b68mh | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>If y = y(x) is the solution of the differential equation $$\left( {1 + {e^{2x}}} \right){{dy} \over {dx}} + 2\left( {1 + {y^2}} \right){e^x} = 0$$ and y (0) = 0, then $$6\left( {y'(0) + {{\left( {y\left( {{{\log }_e}\sqrt 3 } \right)} \right)}^2}} \right)$$ is equal to</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$$-$$2"}, {"identifier": "C", "content": "$$-$$4"}, {"identifier": "D", "content": "$$-$$1"}] | ["C"] | null | <p>Given,</p>
<p>$$(1 + {e^{2x}}){{dy} \over {dx}} + 2(1 + {y^2}){e^x} = 0$$</p>
<p>$$ \Rightarrow {{dy} \over {dx}} = {{ - 2(1 + {y^2}){e^x}} \over {1 + {e^{2x}}}}$$</p>
<p>$$ \Rightarrow \int {{{dy} \over {1 + {y^2}}} = \int {{{ - 2{e^x}dx} \over {(1 + {e^{2x}})}}} } $$</p>
<p>$$ \Rightarrow {\tan ^{ - 1}}(y) = \int {{{ - 2{e^x}dx} \over {(1 + {e^{2x}})}}} $$</p>
<p>Now,</p>
<p>Let $${e^x} = t$$</p>
<p>$$ \Rightarrow {e^x}dx = dt$$</p>
<p>$$ \Rightarrow {\tan ^{ - 1}}(y) = \int {{{ - 2dt} \over {1 + {t^2}}}} $$</p>
<p>$$ \Rightarrow {\tan ^{ - 1}}(y) = - 2{\tan ^{ - 1}}(t) + C$$</p>
<p>$$ \Rightarrow {\tan ^{ - 1}}(y) = - 2{\tan ^{ - 1}}({e^x}) + C$$ [Putting value of t]</p>
<p>Given, y(0) = 0 means when x = 0 the y = 0</p>
<p>$$\therefore$$ $${\tan ^{ - 1}}(0) = - 2{\tan ^{ - 1}}({e^o}) + C$$</p>
<p>$$ \Rightarrow 0 = - 2 \times {\pi \over 4} + C$$</p>
<p>$$ \Rightarrow C = {\pi \over 2}$$</p>
<p>$$\therefore$$ $${\tan ^{ - 1}}(y) = 2{\tan ^{ - 1}}({e^x}) + {\pi \over 2}$$</p>
<p>$$ \Rightarrow y = \tan \left( { - 2{{\tan }^{ - 1}}({e^x}) + {\pi \over 2}} \right)$$</p>
<p>Differentiating both sides, we get</p>
<p>$$y' = {\sec ^2}\left( { - 2{{\tan }^{ - 1}}({e^x}) + {\pi \over 2}} \right) - 2\,.\,{{{e^x}} \over {1 + {e^{2x}}}}$$</p>
<p>$$ = {\sec ^2}\left( { - 2{{\tan }^{ - 1}}({e^x}) + {\pi \over 2}} \right) \times {{ - 2{e^x}} \over {1 + {e^{2x}}}}$$</p>
<p>$$\therefore$$ $$y'(0) = {\sec ^2}\left( { - 2{{\tan }^{ - 1}}({e^o}) + {\pi \over 2}} \right) \times {{ - 2{e^o}} \over {1 + {e^o}}}$$</p>
<p>$$ = {\sec ^2}\left( { - 2 \times {\pi \over 4} + {\pi \over 2}} \right) \times {{ - 2} \over 2}$$</p>
<p>$$ = {\sec ^2}(0)x - 1$$</p>
<p>$$ = 1 \times 1$$</p>
<p>$$ = - 1$$</p>
<p>And</p>
<p>$$y\left( {\log _e^{\sqrt 3 }} \right) = \tan \left( { - 2{{\tan }^{ - 1}}\left( {{e^{\log _e^{\sqrt 3 }}}} \right) + {\pi \over 2}} \right)$$</p>
<p>$$ = \tan \left( { - 2{{\tan }^{ - 1}}(\sqrt 3 ) + {\pi \over 2}} \right)$$</p>
<p>$$ = \tan \left( { - 2 \times {\pi \over 3} + {\pi \over 2}} \right)$$</p>
<p>$$ = \tan \left( { - {\pi \over 6}} \right)$$</p>
<p>$$ = - \tan \left( {{\pi \over 6}} \right)$$</p>
<p>$$ = - {1 \over {\sqrt 3 }}$$</p>
<p>$$\therefore$$ $$6\left( {y'(0) + {{\left( {y(\log _e^{\sqrt 3 })} \right)}^2}} \right)$$</p>
<p>$$ = 6\left( { - 1 + {{\left( {{{ - 1} \over {\sqrt 3 }}} \right)}^2}} \right)$$</p>
<p>$$ = 6\left( { - 1 + {1 \over 3}} \right) = 6 \times {{ - 2} \over 3} = - 4$$</p> | mcq | jee-main-2022-online-29th-june-evening-shift |
1l55hl138 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let x = x(y) be the solution of the differential equation <br/><br/>$$2y\,{e^{x/{y^2}}}dx + \left( {{y^2} - 4x{e^{x/{y^2}}}} \right)dy = 0$$ such that x(1) = 0. Then, x(e) is equal to :</p> | [{"identifier": "A", "content": "$$e{\\log _e}(2)$$"}, {"identifier": "B", "content": "$$ - e{\\log _e}(2)$$"}, {"identifier": "C", "content": "$${e^2}{\\log _e}(2)$$"}, {"identifier": "D", "content": "$$ - {e^2}{\\log _e}(2)$$"}] | ["D"] | null | <p>Given differential equation</p>
<p>$$2y{e^{{x \over {{y^2}}}}}dx + \left( {{y^2} - 4x{e^{{x \over {{y^2}}}}}} \right)dy = 0,\,x(1) = 0$$</p>
<p>$$ \Rightarrow {e^{{x \over {{y^2}}}}}[2ydx - 4xdy] = - {y^2}dy$$</p>
<p>$$ \Rightarrow {e^{{x \over {{y^2}}}}}\left[ {{{2{y^2}dx - 4xydy} \over {{y^4}}}} \right] = {{ - 1} \over y}dy$$</p>
<p>$$ \Rightarrow 2{e^{{x \over {{y^2}}}}}d\left( {{x \over {{y^2}}}} \right) = - {1 \over y}dy$$</p>
<p>$$ \Rightarrow 2{e^{{x \over {{y^2}}}}} = - \ln y + c$$ ...... (i)</p>
<p>Now, using x(1) = 0, c = 2</p>
<p>So, for x(e), Put y = e in (i)</p>
<p>$$2{e^{{x \over {{e^2}}}}} = - 1 + 2$$</p>
<p>$$ \Rightarrow {x \over {{e^2}}} = \ln \left( {{1 \over 2}} \right) \Rightarrow x(e) = - {e^2}\ln 2$$</p> | mcq | jee-main-2022-online-28th-june-evening-shift |
1l566lm5l | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let the solution curve $$y = y(x)$$ of the differential equation</p>
<p>$$\left[ {{x \over {\sqrt {{x^2} - {y^2}} }} + {e^{{y \over x}}}} \right]x{{dy} \over {dx}} = x + \left[ {{x \over {\sqrt {{x^2} - {y^2}} }} + {e^{{y \over x}}}} \right]y$$</p>
<p>pass through the points (1, 0) and (2$$\alpha$$, $$\alpha$$), $$\alpha$$ > 0. Then $$\alpha$$ is equal to</p> | [{"identifier": "A", "content": "$${1 \\over 2}\\exp \\left( {{\\pi \\over 6} + \\sqrt e - 1} \\right)$$"}, {"identifier": "B", "content": "$${1 \\over 2}\\exp \\left( {{\\pi \\over 6} + e - 1} \\right)$$"}, {"identifier": "C", "content": "$$\\exp \\left( {{\\pi \\over 6} + \\sqrt e + 1} \\right)$$"}, {"identifier": "D", "content": "$$2\\exp \\left( {{\\pi \\over 3} + \\sqrt e - 1} \\right)$$"}] | ["A"] | null | <p>$$\left( {{1 \over {\sqrt {1 - {{{y^2}} \over {{x^2}}}} }} + {e^{{y \over x}}}} \right){{dy} \over {dx}} = 1 + \left( {{1 \over {\sqrt {1 - {{{y^2}} \over {{x^2}}}} }} + {e^{{y \over x}}}} \right){y \over x}$$</p>
<p>Putting y = tx</p>
<p>$$\left( {{1 \over {\sqrt {1 - {t^2}} }} + {e^t}} \right)\left( {t + x{{dt} \over {dx}}} \right) = 1 + \left( {{1 \over {\sqrt {1 - {t^2}} }} + {e^t}} \right)t$$</p>
<p>$$ \Rightarrow x\left( {{1 \over {\sqrt {1 - {t^2}} }} + {e^t}} \right){{dt} \over {dx}} = 1$$</p>
<p>$$ \Rightarrow {\sin ^{ - 1}}t + {e^t} = \ln x + C$$</p>
<p>$$ \Rightarrow {\sin ^{ - 1}}\left( {{y \over x}} \right) + {e^{y/x}} = \ln x + C$$</p>
<p>at x = 1, y = 0</p>
<p>So, $$0 + {e^0} = 0 + C \Rightarrow C = 1$$</p>
<p>at $$(2\alpha ,\alpha )$$</p>
<p>$${\sin ^{ - 1}}\left( {{y \over x}} \right) + {e^{y/x}} = \ln x + 1$$</p>
<p>$$ \Rightarrow {\pi \over 6} + {e^{{1 \over 2}}} - 1 = \ln (2\alpha )$$</p>
<p>$$ \Rightarrow \alpha = {1 \over 2}{e^{\left( {{\pi \over 6} + {e^{{1 \over 2}}} - 1} \right)}}$$</p> | mcq | jee-main-2022-online-28th-june-morning-shift |
1l57od20u | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>If $${{dy} \over {dx}} + {{{2^{x - y}}({2^y} - 1)} \over {{2^x} - 1}} = 0$$, x, y > 0, y(1) = 1, then y(2) is equal to :</p> | [{"identifier": "A", "content": "$$2 + {\\log _2}3$$"}, {"identifier": "B", "content": "$$2 + {\\log _3}2$$"}, {"identifier": "C", "content": "$$2 - {\\log _3}2$$"}, {"identifier": "D", "content": "$$2 - {\\log _2}3$$"}] | ["D"] | null | <p>$${{dy} \over {dx}} + {{{2^{x-y}}({2^y} - 1)} \over {{2^x} - 1}}=0$$, x, y > 0, y(1) = 1</p>
<p>$${{dy} \over {dx}} = - {{{2^x}({2^y} - 1)} \over {{2^y}({2^x} - 1)}}$$</p>
<p>$$\int {{{{2^y}} \over {{2^y} - 1}}dy = - \int {{{{2^x}} \over {{2^x} - 1}}dx} } $$</p>
<p>$$ = {{{{\log }_e}({2^y} - 1)} \over {{{\log }_e}2}} = - {{{{\log }_e}({2^x} - 1)} \over {{{\log }_e}2}} + {{{{\log }_e}c} \over {{{\log }_e}2}}$$</p>
<p>$$ = |({2^y} - 1)({2^x} - 1)| = c$$</p>
<p>$$\because$$ $$y(1) = 1$$</p>
<p>$$\therefore$$ $$c = 1$$</p>
<p>$$ = |({2^y} - 1)({2^x} - 1)| = 1$$</p>
<p>For $$x = 2$$</p>
<p>$$|({2^y} - 1)3| = 1$$</p>
<p>$${2^y} - 1 = {1 \over 3} \Rightarrow 2y = {4 \over 3}$$</p>
<p>Taking log to base 2.</p>
<p>$$\therefore$$ $$y = 2 - {\log _2}3$$</p> | mcq | jee-main-2022-online-27th-june-morning-shift |
1l58aj791 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let the solution curve y = y(x) of the differential equation <br/><br/>$$(4 + {x^2})dy - 2x({x^2} + 3y + 4)dx = 0$$ pass through the origin. Then y(2) is equal to _____________.</p> | [] | null | 12 | <p>$$(4 + {x^2})dy - 2x({x^2} + 3y + 4)dx = 0$$</p>
<p>$$ \Rightarrow {{dy} \over {dx}} = \left( {{{6x} \over {{x^2} + 4}}} \right)y + 2x$$</p>
<p>$$ \Rightarrow {{dy} \over {dx}} - \left( {{{6x} \over {{x^2} + 4}}} \right)y = 2x$$</p>
<p>$$I.F. = {e^{ - 3\ln ({x^2} + 4)}} = {1 \over {{{({x^2} + 4)}^3}}}$$</p>
<p>So $${y \over {{{({x^2} + 4)}^3}}} = \int {{{2x} \over {{{({x^2} + 4)}^3}}}dx + c} $$</p>
<p>$$ \Rightarrow y = - {1 \over 2}({x^2} + 4) + c{({x^2} + 4)^3}$$</p>
<p>When x = 0, y = 0 gives $$c = {1 \over {32}}$$,</p>
<p>So, for x = 2, y = 12</p> | integer | jee-main-2022-online-26th-june-morning-shift |
1l59kklkn | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>If $$y = y(x)$$ is the solution of the differential equation <br/><br/>$$2{x^2}{{dy} \over {dx}} - 2xy + 3{y^2} = 0$$ such that $$y(e) = {e \over 3}$$, then y(1) is equal to :</p> | [{"identifier": "A", "content": "$${1 \\over 3}$$"}, {"identifier": "B", "content": "$${2 \\over 3}$$"}, {"identifier": "C", "content": "$${3 \\over 2}$$"}, {"identifier": "D", "content": "3"}] | ["B"] | null | <p>$$2{x^2}{{dy} \over {dx}} - 2xy + 3{y^2} = 0$$</p>
<p>$$ \Rightarrow 2x(xdy - ydx) + 3{y^3}dx = 0$$</p>
<p>$$ \Rightarrow 2\left( {{{xdy - ydx} \over {{y^2}}}} \right) + 3{{dx} \over x} = 0$$</p>
<p>$$ \Rightarrow - {{2x} \over y} + 3\ln x = C$$</p>
<p>$$\because$$ $$y(e) = {e \over 3} \Rightarrow - 6 + 3 = C \Rightarrow C = - 3$$</p>
<p>Now, at $$x = 1$$, $$ - {2 \over y} + 0 = - 3$$</p>
<p>$$y = {2 \over 3}$$</p> | mcq | jee-main-2022-online-25th-june-evening-shift |
1l5aihggo | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$g:(0,\infty ) \to R$$ be a differentiable function such that <br/><br/>$$\int {\left( {{{x(\cos x - \sin x)} \over {{e^x} + 1}} + {{g(x)\left( {{e^x} + 1 - x{e^x}} \right)} \over {{{({e^x} + 1)}^2}}}} \right)dx = {{x\,g(x)} \over {{e^x} + 1}} + c} $$, for all x > 0, where c is an arbitrary constant. Then :</p> | [{"identifier": "A", "content": "g is decreasing in $$\\left( {0,{\\pi \\over 4}} \\right)$$"}, {"identifier": "B", "content": "g' is increasing in $$\\left( {0,{\\pi \\over 4}} \\right)$$"}, {"identifier": "C", "content": "g + g' is increasing in $$\\left( {0,{\\pi \\over 2}} \\right)$$"}, {"identifier": "D", "content": "g $$-$$ g' is increasing in $$\\left( {0,{\\pi \\over 2}} \\right)$$"}] | ["D"] | null | $$
\int\left(\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g(x)\left(e^x+1-x e^x\right)}{\left(e^x+1\right)^2}\right) d x=\frac{x g(x)}{e^x+1}+c
$$<br/><br/>
On differentiating both sides w.r.t. $\mathrm{x}$, we get<br/><br/>
$$
\begin{aligned}
&\left(\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g(x)\left(e^x+1-x e^x\right.}{\left(e^x+1\right)^2}\right) \\\\
&=\frac{\left(e^x+1\right)\left(g(x)+x g^{\prime}(x)\right)-e^x \cdot x \cdot g(x)}{\left(e^x+1\right)^2} \\\\
&\left(e^x+1\right) x(\cos x-\sin x)+g(x)\left(e^x+1-x e^x\right) \\\\
&=\left(e^x+1\right)\left(g(x)+x g^{\prime}(x)\right)-e^x \cdot x \cdot g(x) \\\\
&\Rightarrow g g^{\prime}(x)=\cos x-\sin x \\\\
&\Rightarrow g(x)=\sin x+\cos x+C
\end{aligned}
$$<br/><br/>
$\mathrm{g}(\mathrm{x})$ is increasing in $(0, \pi / 4)$<br/><br/>
$g "(x)=-\sin x-\cos x<0$<br/><br/>
$\Rightarrow \mathrm{g}^{\prime}(\mathrm{x})$ is decreasing function<br/><br/>
let $h(x)=g(x)+g^{\prime}(x)=2 \cos x+C$<br/><br/>
$$
\Rightarrow \mathrm{h}^{\prime}(\mathrm{x})=\mathrm{g}^{\prime}(\mathrm{x})+\mathrm{g}^{\prime \prime}(\mathrm{x})=-2 \sin \mathrm{x}<0
$$<br/><br/>
$\Rightarrow \mathrm{h}$ is decreasing<br/><br/>
let $\phi(\mathrm{x})=\mathrm{g}(\mathrm{x})-\mathrm{g}^{\prime}(\mathrm{x})=2 \sin \mathrm{x}+\mathrm{C}$<br/><br/>
$$
\Rightarrow \phi^{\prime}(\mathrm{x})=\mathrm{g}^{\prime}(\mathrm{x})-\mathrm{g}{ }^{\prime \prime}(\mathrm{x})=2 \cos \mathrm{x}>0
$$<br/><br/>
$\Rightarrow \phi$ is increasing<br/><br/>
Hence option D is correct. | mcq | jee-main-2022-online-25th-june-morning-shift |
1l5aj3huu | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>If the solution curve $$y = y(x)$$ of the differential equation $${y^2}dx + ({x^2} - xy + {y^2})dy = 0$$, which passes through the point (1, 1) and intersects the line $$y = \sqrt 3 x$$ at the point $$(\alpha ,\sqrt 3 \alpha )$$, then value of $${\log _e}(\sqrt 3 \alpha )$$ is equal to :</p> | [{"identifier": "A", "content": "$${\\pi \\over 3}$$"}, {"identifier": "B", "content": "$${\\pi \\over 2}$$"}, {"identifier": "C", "content": "$${\\pi \\over 12}$$"}, {"identifier": "D", "content": "$${\\pi \\over 6}$$"}] | ["C"] | null | <p>$${{dy} \over {dx}} = {{{y^2}} \over {xy - {x^2} - {y^2}}}$$</p>
<p>Put $$y = vx$$ we get</p>
<p>$$v + x{{dv} \over {dx}} = {{{v^2}} \over {v - 1 - {v^2}}}$$</p>
<p>$$ \Rightarrow x{{dv} \over {dx}} = {{{v^2} - {v^2} + v + {v^3}} \over {v - 1 - {v^2}}}$$</p>
<p>$$ \Rightarrow \int {{{v - 1 - {v^2}} \over {v(1 + {v^2})}}dv = \int {{{dx} \over x}} } $$</p>
<p>$${\tan ^{ - 1}}\left( {{y \over x}} \right) - \ln \left( {{y \over x}} \right) = \ln x + c$$</p>
<p>As it passes through (1, 1)</p>
<p>$$c = {\pi \over 4}$$</p>
<p>$$ \Rightarrow {\tan ^{ - 1}}\left( {{y \over x}} \right)\ln \left( {{y \over x}} \right) = \ln x + {\pi \over 4}$$</p>
<p>Put $$y = \sqrt 3 x$$ we get</p>
<p>$$ \Rightarrow {\pi \over 3} - \ln \sqrt 3 = \ln x + {\pi \over 4}$$</p>
<p>$$ \Rightarrow \ln x = {\pi \over {12}} - \ln \sqrt 3 = \ln \alpha $$</p>
<p>$$\therefore$$ $$\ln \left( {\sqrt 3 \alpha } \right) = \ln \sqrt 3 + \ln \alpha $$</p>
<p>$$ = \ln \sqrt 3 + {\pi \over {12}} - \ln \sqrt 3 = {\pi \over {12}}$$</p> | mcq | jee-main-2022-online-25th-june-morning-shift |
1l6dvn1zt | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>The general solution of the differential equation $$\left(x-y^{2}\right) \mathrm{d} x+y\left(5 x+y^{2}\right) \mathrm{d} y=0$$ is :</p> | [{"identifier": "A", "content": "$$\\left(y^{2}+x\\right)^{4}=\\mathrm{C}\\left|\\left(y^{2}+2 x\\right)^{3}\\right|$$"}, {"identifier": "B", "content": "$$\\left(y^{2}+2 x\\right)^{4}=C\\left|\\left(y^{2}+x\\right)^{3}\\right|$$"}, {"identifier": "C", "content": "$$\\left|\\left(y^{2}+x\\right)^{3}\\right|=\\mathrm{C}\\left(2 y^{2}+x\\right)^{4}$$"}, {"identifier": "D", "content": "$$\\left|\\left(y^{2}+2 x\\right)^{3}\\right|=C\\left(2 y^{2}+x\\right)^{4}$$"}] | ["A"] | null | $\left(x-y^{2}\right) d x+y\left(5 x+y^{2}\right) d y=0$
<br/><br/>
$$
y \frac{d y}{d x}=\frac{y^{2}-x}{5 x+y^{2}}
$$<br/><br/>
Let $y^{2}=t$
$$
\frac{1}{2} \cdot \frac{d t}{d x}=\frac{t-x}{5 x+t}
$$
<br/><br/>
Now substitute, $t=v x$
<br/><br/>
$$
\begin{aligned}
& \frac{d t}{d x}=v+x \frac{d v}{d x} \\\\
& \frac{1}{2}\left\{v+x \frac{d v}{d x}\right\}=\frac{v-1}{5+v} \\\\
& x \frac{d v}{d x}=\frac{2 v-2}{5+v}-v=\frac{-3 v-v^{2}-2}{5+v} \\\\
& \int \frac{5+v}{v^{2}+3 v+2} d v=\int-\frac{d x}{x} \\\\
& \int \frac{4}{v+1} d v-\int \frac{3}{v+2} d v=-\int \frac{d x}{x} \\\\
& 4 \ln |v+1|-3 \ln |v+2|=-\ln x+\ln C \\\\
& \left|\frac{(v+1)^{4}}{(v+2)^{3}}\right|=\frac{c}{x} \\\\
& \left| \frac{\left(\frac{y^{2}}{x}+1\right)^{4}}{\left(\frac{y^{2}}{x}+2\right)^{3}}\right|=\frac{c}{x} \\\\
& \left|\left(y^{2}+x\right)^{4}\right|=C\left|\left(y^{2}+2 x\right)^{3}\right|
\end{aligned}
$$ | mcq | jee-main-2022-online-25th-july-morning-shift |
1l6f3egjv | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=y(x)$$ be the solution of the differential equation</p>
<p>$$\frac{d y}{d x}=\frac{4 y^{3}+2 y x^{2}}{3 x y^{2}+x^{3}}, y(1)=1$$.</p>
<p>If for some $$n \in \mathbb{N}, y(2) \in[n-1, n)$$, then $$n$$ is equal to _____________.</p> | [] | null | 3 | <p>$${{dy} \over {dx}} = {y \over x}{{(4{y^2} + 2{x^2})} \over {(3{y^2} + {x^2})}}$$</p>
<p>Put $$y = vx$$</p>
<p>$$ \Rightarrow {{dy} \over {dx}} = v + x{{dv} \over {dx}}$$</p>
<p>$$ \Rightarrow v + x{{dv} \over {dx}} = {{v(4{v^2} + 2)} \over {(3{v^2} + 1)}}$$</p>
<p>$$ \Rightarrow x{{dv} \over {dx}} = v\left( {{{(4{v^2} + 2 - 3{v^2} - 1)} \over {3{v^2} + 1}}} \right)$$</p>
<p>$$ \Rightarrow \int {(3{v^2} + 1){{dv} \over {{v^3} + v}} = \int {{{dx} \over x}} } $$</p>
<p>$$ \Rightarrow \ln |{v^3} + v| = \ln x + c$$</p>
<p>$$ \Rightarrow \ln \left| {{{\left( {{y \over x}} \right)}^3} + \left( {{y \over x}} \right)} \right| = \ln x + C$$</p>
<p>$$ \downarrow \,y(1) = 1$$</p>
<p>$$ \Rightarrow C = \ln 2$$</p>
<p>$$\therefore$$ for $$y(2)$$</p>
<p>$$\ln \left( {{{{y^3}} \over 8} + {y \over 2}} \right) = 2\ln 2 \Rightarrow {{{y^3}} \over 8} + {y \over 2} = 4$$</p>
<p>$$ \Rightarrow [y(2)] = 2$$</p>
<p>$$ \Rightarrow n = 3$$</p> | integer | jee-main-2022-online-25th-july-evening-shift |
1l6m54xbg | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let the solution curve of the differential equation $$x \mathrm{~d} y=\left(\sqrt{x^{2}+y^{2}}+y\right) \mathrm{d} x, x>0$$, intersect the line $$x=1$$ at $$y=0$$ and the line $$x=2$$ at $$y=\alpha$$. Then the value of $$\alpha$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{1}{2}$$"}, {"identifier": "B", "content": "$$\\frac{3}{2}$$"}, {"identifier": "C", "content": "$$-$$$$\\frac{3}{2}$$"}, {"identifier": "D", "content": "$$\\frac{5}{2}$$"}] | ["B"] | null | <p>$${{xdy - ydx} \over {\sqrt {{x^2} + {y^2}} }} = dx$$</p>
<p>$$ \Rightarrow {{dy} \over {dx}} = {{\sqrt {{x^2} + {y^2}} } \over x} + {y \over x}$$</p>
<p>$$ \Rightarrow {{dy} \over {dx}} = \sqrt {1 + {{{y^2}} \over {{x^2}}}} + {y \over x}$$</p>
<p>Let $${y \over x} = v$$</p>
<p>$$ \Rightarrow v + x{{dv} \over {dx}} = \sqrt {1 + {v^2}} + v$$</p>
<p>$$ \Rightarrow {{dv} \over {\sqrt {1 + {v^2}} }} = {{dx} \over x}$$</p>
<p>OR $$\ln \left( {v + \sqrt {1 + {v^2}} } \right) = \ln x + C$$</p>
<p>at $$x = 1,\,y = 0$$</p>
<p>$$ \Rightarrow C = 0$$</p>
<p>$${y \over x} + \sqrt {1 + {{{y^2}} \over {{x^2}}}} = x$$</p>
<p>At $$x = 2$$,</p>
<p>$${y \over 2} + \sqrt {1 + {{{y^2}} \over 4}} = 2$$</p>
<p>$$ \Rightarrow 1 + {{{y^2}} \over 4} = 4 + {{{y^2}} \over 4} - 2y$$</p>
<p>OR $$y = {3 \over 2}$$</p> | mcq | jee-main-2022-online-28th-july-morning-shift |
1l6m5qs9v | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>If $$y=y(x), x \in(0, \pi / 2)$$ be the solution curve of the differential equation <br/><br/>$$\left(\sin ^{2} 2 x\right) \frac{d y}{d x}+\left(8 \sin ^{2} 2 x+2 \sin 4 x\right) y=2 \mathrm{e}^{-4 x}(2 \sin 2 x+\cos 2 x)$$, <br/><br/>with $$y(\pi / 4)=\mathrm{e}^{-\pi}$$, then $$y(\pi / 6)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{2}{\\sqrt{3}} e^{-2 \\pi / 3}$$"}, {"identifier": "B", "content": "$$\\frac{2}{\\sqrt{3}} \\mathrm{e}^{2 \\pi / 3}$$"}, {"identifier": "C", "content": "$$\\frac{1}{\\sqrt{3}} e^{-2 \\pi / 3}$$"}, {"identifier": "D", "content": "$$\\frac{1}{\\sqrt{3}} e^{2 \\pi / 3}$$"}] | ["A"] | null | <p>$$({\sin ^2}2x){{dy} \over {dx}} + (8{\sin ^2}2x + 2\sin 4x)y$$</p>
<p>$$ = 2{e^{ - 4x}}(2\sin 2x + \cos 2x)$$</p>
<p>$${{dy} \over {dx}} + (8 + 4\cot 2x)y = 2{e^{ - 4x}}\left( {{{2\sin 2x + \cos 2x} \over {{{\sin }^2}2x}}} \right)$$</p>
<p>Integrating factor</p>
<p>$$(I.F.) = {e^{\int {(8 + 4\cot 2x)dx} }}$$</p>
<p>$$ = {e^{8x + 2\ln \sin 2x}}$$</p>
<p>Solution of differential equation</p>
<p>$$y.\,{e^{8x + 2\ln \sin 2x}}$$</p>
<p>$$ = \int {2{e^{(4x + 2\ln \sin 2x)}}{{(2\sin 2x + \cos 2x)} \over {{{\sin }^2}2x}}dx} $$</p>
<p>$$ = 2\int {{e^{4x}}(2\sin 2x + \cos 2x)dx} $$</p>
<p>$$y.\,{e^{8x + 2\ln \sin 2x}} = {e^{4x}}\sin 2x + c$$</p>
<p>$$y\left( {{\pi \over 4}} \right) = {e^{ - \pi }}$$</p>
<p>$${e^{ - \pi }}\,.\,{e^{2\pi }} = {e^\pi } + c \Rightarrow c = 0$$</p>
<p>$$y\left( {{\pi \over 6}} \right) = {{{e^{{{2\pi } \over 3}}}{{\sqrt 3 } \over 2}} \over {{e^{\left( {{{4\pi } \over 3} + 2\ln {{\sqrt 3 } \over 2}} \right)}}}}$$</p>
<p>$$ = {e^{{{ - 2\pi } \over 3}}}\,.\,{2 \over {\sqrt 3 }}$$</p> | mcq | jee-main-2022-online-28th-july-morning-shift |
1l6reae56 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>If the solution curve of the differential equation $$\frac{d y}{d x}=\frac{x+y-2}{x-y}$$ passes through the points $$(2,1)$$ and $$(\mathrm{k}+1,2), \mathrm{k}>0$$, then</p> | [{"identifier": "A", "content": "$$2 \\tan ^{-1}\\left(\\frac{1}{k}\\right)=\\log _{e}\\left(k^{2}+1\\right)$$"}, {"identifier": "B", "content": "$$\\tan ^{-1}\\left(\\frac{1}{k}\\right)=\\log _{e}\\left(k^{2}+1\\right)$$"}, {"identifier": "C", "content": "$$2 \\tan ^{-1}\\left(\\frac{1}{k+1}\\right)=\\log _{e}\\left(k^{2}+2 k+2\\right)$$"}, {"identifier": "D", "content": "$$2 \\tan ^{-1}\\left(\\frac{1}{k}\\right)=\\log _{e}\\left(\\frac{k^{2}+1}{k^{2}}\\right)$$"}] | ["A"] | null | $\frac{d y}{d x}=\frac{x+y-2}{x-y}=\frac{(x-1)+(y-1)}{(x-1)-(y-1)}$
<br/><br/>Let $x-1=X, y-1=Y$
<br/><br/>$$
\frac{d Y}{d X}=\frac{X+Y}{X-Y}
$$
<br/><br/>Let $Y=t X \Rightarrow \frac{d Y}{d X}=t+X \frac{d t}{d X}$
<br/><br/>$t+X \frac{d t}{d X}=\frac{1+t}{1-t}$
<br/><br/>$X \frac{d t}{d X}=\frac{1+t}{1-t}-t=\frac{1+t^{2}}{1-t}$
<br/><br/>$\int \frac{1-t}{1+t^{2}} d t=\int \frac{d X}{X}$
<br/><br/>$$
\begin{aligned}
&\tan ^{-1} t-\frac{1}{2} \ln \left(1+t^{2}\right)=\ln |X|+c \\\\
&\tan ^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{2} \ln \left(1+\left(\frac{y-1}{x-1}\right)^{2}\right)=\ln |x-1|+c
\end{aligned}
$$
<br/><br/>Curve passes through $(2,1)$
<br/><br/>$0-0=0+c \Rightarrow c=0$
<br/><br/>If $(k+1,2)$ also satisfies the curve
<br/><br/>$$
\begin{aligned}
&\tan ^{-1}\left(\frac{1}{k}\right)-\frac{1}{2} \ln \left(\frac{1+k^{2}}{k^{2}}\right)=\ln k \\\\
&2 \tan ^{-1}\left(\frac{1}{k}\right)=\ln \left(1+k^{2}\right)
\end{aligned}
$$ | mcq | jee-main-2022-online-29th-july-evening-shift |
1l6reeqh6 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=y(x)$$ be the solution curve of the differential equation $$ \frac{d y}{d x}+\left(\frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6}\right) y=\frac{(x+3)}{x+1}, x>-1$$, which passes through the point $$(0,1)$$. Then $$y(1)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{1}{2}$$"}, {"identifier": "B", "content": "$$\\frac{3}{2}$$"}, {"identifier": "C", "content": "$$\\frac{5}{2}$$"}, {"identifier": "D", "content": "$$\\frac{7}{2}$$"}] | ["B"] | null | $\frac{d y}{d x}+\left(\frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6}\right) y=\frac{(x+3)}{x+1}, x>-1$,
<br/><br/>Integrating factor I.F. $=e^{\int \frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6} d x}$
<br/><br/>$$
\begin{aligned}
& \text { Let } \frac{2 x^{2}+11 x+13}{(x+1)(x+2)(x+3)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3} \\\\
& A=2, B=1, C=-1 \\\\
& \text { I.F. }=e^{(2 \ln |x+1|+\ln |x+2|-\ln |x+3|)} \\\\
& =\frac{(x+1)^{2}(x+2)}{x+3}
\end{aligned}
$$
<br/><br/>Solution of differential equation
<br/><br/>$$
\begin{aligned}
&y \cdot \frac{(x+1)^{2}(x+2)}{x+3}=\int(x+1)(x+2) d x \\\\
&y \frac{(x+1)^{2}(x+2)}{x+3}=\frac{x^{3}}{3}+\frac{3 x^{2}}{2}+2 x+c
\end{aligned}
$$
<br/><br/>Curve passes through $(0,1)$
<br/><br/>$$
\begin{gathered}
1 \times \frac{1 \times 2}{3}=0+c \Rightarrow c=\frac{2}{3} \\\\
\text { So, } y(1)=\frac{\frac{1}{3}+\frac{3}{2}+2+\frac{2}{3}}{\frac{\left(2^{2} \times 3\right)}{4}}=\frac{3}{2}
\end{gathered}
$$ | mcq | jee-main-2022-online-29th-july-evening-shift |
ldoa5dpt | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let $y=y(x)$ be the solution of the differential equation
<br/><br/>$\left(3 y^{2}-5 x^{2}\right) y \mathrm{~d} x+2 x\left(x^{2}-y^{2}\right) \mathrm{d} y=0$
<br/><br/>such that $y(1)=1$. Then $\left|(y(2))^{3}-12 y(2)\right|$ is equal to : | [{"identifier": "A", "content": "64"}, {"identifier": "B", "content": "$16 \\sqrt{2}$"}, {"identifier": "C", "content": "32"}, {"identifier": "D", "content": "$32 \\sqrt{2}$"}] | ["D"] | null | $\left(3 y^{2}-5 x^{2}\right) y \cdot d x+2 x\left(x^{2}-y^{2}\right) d y=0$
<br/><br/>$$
\Rightarrow \frac{d y}{d x}=\frac{y\left(5 x^{2}-3 y^{2}\right)}{2 x\left(x^{2}-y^{2}\right)}
$$
<br/><br/>Put $\mathrm{y}=\mathrm{mx}$
<br/><br/>$$
\Rightarrow m+x \cdot \frac{d m}{d x}=\frac{m\left(5-3 m^{2}\right)}{2\left(1-m^{2}\right)}
$$
<br/><br/>$$
\begin{aligned}
& x \cdot \frac{d m}{d x}=\frac{\left(5-3 m^{2}\right) m-2 m\left(1-m^{2}\right)}{2\left(1-m^{2}\right)} \\\\
& \Rightarrow \frac{\mathrm{dx}}{\mathrm{x}}=\frac{2\left(\mathrm{~m}^{2}-1\right)}{\mathrm{m}\left(\mathrm{m}^{2}-3\right)} \mathrm{dm} \\\\
& \Rightarrow \frac{d x}{x}=\left(\frac{2}{m}-\frac{\frac{4}{3}}{m}+\frac{\frac{4 m}{3}}{\mathrm{~m}^{2}-3}\right) d m
\end{aligned}
$$
<br/><br/>$\Rightarrow \int \frac{d x}{x}=\int \frac{\left(\frac{2}{3}\right)}{m}+\int \frac{2}{3}\left(\frac{2 m}{m^{2}-3}\right) d m$
<br/><br/>$\Rightarrow \ln |\mathrm{x}|=\frac{2}{3} \ln |\mathrm{m}|+\frac{2}{3} \ln \left|\mathrm{m}^{2}-3\right|+\mathrm{C}$
<br/><br/>Or, $\ln |\mathrm{x}|=\frac{2}{3} \ln \left|\frac{\mathrm{y}}{\mathrm{x}}\right|+\frac{2}{3} \ln \left|\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}-3\right|+\mathrm{C}$
<br/><br/>Put $(\mathrm{x}=1, \mathrm{y}=1)$ : we get $\mathrm{c}=-\frac{2}{3} \ln (2)$
<br/><br/>$\Rightarrow \ln |\mathrm{x}|=\frac{2}{3} \ln \left|\frac{\mathrm{y}}{\mathrm{x}}\right|+\frac{2}{3} \ln \left|\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}-3\right|-\frac{2}{3} \ln (2)$
<br/><br/>$\Rightarrow\left(\frac{\mathrm{y}}{\mathrm{x}}\right)\left[\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}-3\right]=2 .\left(\mathrm{x}^{3 / 2}\right)$ ........(1)
<br/><br/>Put $\mathrm{x}=2$ in equation (1), we get
<br/><br/>$\Rightarrow \mathrm{y}\left(\mathrm{y}^{2}-12\right)=4 \times 2 \times 2 \times 2 \sqrt{2}$
<br/><br/>$\Rightarrow \mathrm{y}^{3}-12 \mathrm{y}=32 \sqrt{2}$
<br/><br/>$\Rightarrow\left|\mathrm{y}^{3}(2)-12 \mathrm{y}(2)\right|=32 \sqrt{2}$ | mcq | jee-main-2023-online-31st-january-evening-shift |
1ldonc42m | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>The area enclosed by the closed curve $$\mathrm{C}$$ given by the differential equation <br/><br/>$$\frac{d y}{d x}+\frac{x+a}{y-2}=0, y(1)=0$$ is $$4 \pi$$.</p>
<p>Let $$P$$ and $$Q$$ be the points of intersection of the curve $$\mathrm{C}$$ and the $$y$$-axis. If normals at $$P$$ and $$Q$$ on the curve $$\mathrm{C}$$ intersect $$x$$-axis at points $$R$$ and $$S$$ respectively, then the length of the line segment $$R S$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{4 \\sqrt{3}}{3}$$"}, {"identifier": "B", "content": "$$2 \\sqrt{3}$$"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$$\\frac{2 \\sqrt{3}}{3}$$"}] | ["A"] | null | $$
\begin{aligned}
& \frac{d y}{d x}+\frac{x+a}{y-2}=0 \\\\
& \frac{d y}{d x}=\frac{x+a}{2-y} \\\\
& (2-y) d y=(x+a) d x \\\\
& 2 y \frac{-y}{2}=\frac{x^2}{2}+\mathrm{ax}+\mathrm{c} \\\\
& \mathrm{a}+\mathrm{c}=-\frac{1}{2} \text { as } \mathrm{y}(1)=0 \\\\
& \mathrm{X}^2+\mathrm{y}^2+2 \mathrm{ax}-4 \mathrm{y}-1-2 \mathrm{a}=0 \\\\
& \pi \mathrm{r}^2=4 \pi \\\\
& \mathrm{r}^2=4 \\\\
& 4=\sqrt{a^2+4+1+2 a} \\\\
& (\mathrm{a}+1)^2=0
\end{aligned}
$$
<br/><br/>$$
P, Q=(0,2 \pm \sqrt{3})
$$
<br/><br/>Equation of normal at $\mathrm{P}, \mathrm{Q}$ are $\mathrm{y}-2=\sqrt{3}(\mathrm{x}-1)$
<br/><br/>$$
\begin{aligned}
& \mathrm{y}-2=-\sqrt{3}(\mathrm{x}-1) \\\\
& \mathrm{R}=\left(1-\frac{2}{\sqrt{3}}, 0\right) \\\\
& \mathrm{S}=\left(1+\frac{2}{\sqrt{3}}, 0\right) \\\\
& \mathrm{RS}=\frac{4}{\sqrt{3}}=4 \frac{\sqrt{3}}{3}
\end{aligned}
$$ | mcq | jee-main-2023-online-1st-february-morning-shift |
ldqy3vn1 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | The solution of the differential equation
<br/><br/>$\frac{d y}{d x}=-\left(\frac{x^2+3 y^2}{3 x^2+y^2}\right), y(1)=0$ is : | [{"identifier": "A", "content": "$\\log _e|x+y|+\\frac{x y}{(x+y)^2}=0$"}, {"identifier": "B", "content": "$\\log _e|x+y|-\\frac{x y}{(x+y)^2}=0$"}, {"identifier": "C", "content": "$\\log _e|x+y|+\\frac{2 x y}{(x+y)^2}=0$"}, {"identifier": "D", "content": "$\\log _e|x+y|-\\frac{2 x y}{(x+y)^2}=0$"}] | ["C"] | null | <p>$$y = vx$$</p>
<p>$$v + x{{dv} \over {dx}} = - \left( {{{1 + 3{v^2}} \over {3 + {v^2}}}} \right)$$</p>
<p>$$x{{dv} \over {dx}} = - \left( {{{1 + 3{v^2}} \over {3 + {v^2}}} + v} \right)$$</p>
<p>$${{dv} \over {dx}} = - \left( {{{{{(1 + v)}^3}} \over {3 + {v^2}}}} \right)$$</p>
<p>$$ \Rightarrow {{3 + {v^2}} \over {{{(1 + v)}^3}}} = {{ - dx} \over x}$$</p>
<p>$$ \Rightarrow \ln |v + 1| + {{2v} \over {{{(v + 1)}^2}}} = C - \ln |x|$$</p>
<p>$$x = 1,v = 0 \Rightarrow C = 0$$</p>
<p>$$ \Rightarrow \ln |x + y| - \ln |x| + {{2xy} \over {{{(x + y)}^2}}} = - \ln |x|$$</p> | mcq | jee-main-2023-online-30th-january-evening-shift |
1ldsuyode | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=f(x)$$ be the solution of the differential equation $$y(x+1)dx-x^2dy=0,y(1)=e$$. Then $$\mathop {\lim }\limits_{x \to {0^ + }} f(x)$$ is equal to</p> | [{"identifier": "A", "content": "$${e^2}$$"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "$${1 \\over {{e^2}}}$$"}, {"identifier": "D", "content": "$${1 \\over e}$$"}] | ["B"] | null | <p>Given,</p>
<p>$$y(x + 1)dx - {x^2}dy = 0$$</p>
<p>$$ \Rightarrow \left( {{{x + 1} \over {{x^2}}}} \right)dx = {{dy} \over y}$$</p>
<p>$$ \Rightarrow {1 \over x}dx + {{dx} \over {{x^2}}} = {{dy} \over y}$$</p>
<p>Integrating both sides, we get</p>
<p>$$\int {{{dx} \over x} + \int {{{dx} \over {{x^2}}} = \int {{{dy} \over y}} } } $$</p>
<p>$$ \Rightarrow \ln |x| - {1 \over x} = \ln |y| + C$$ ..... (1)</p>
<p>Given $$y(1) = e$$</p>
<p>$$\therefore$$ $$x = 1$$ and $$y = e$$</p>
<p>Putting value of x and y in equation (1), we get</p>
<p>$$\ln |1| - {1 \over 1} = \ln |e| + C$$</p>
<p>$$ \Rightarrow 0 - 1 = 1 + C$$</p>
<p>$$ \Rightarrow C = - 2$$</p>
<p>$$\therefore$$ Equation (1) becomes,</p>
<p>$$\ln |x| = - {1 \over x} = \ln |y| - 2$$</p>
<p>$$ \Rightarrow \ln |y| = \ln |x| - {1 \over x} + 2$$</p>
<p>$$ \Rightarrow y = {e^{\ln |x|}}\,.\,{e^{2 - {1 \over x}}}$$</p>
<p>$$ \Rightarrow y = x\,.\,{e^{2 - {1 \over x}}}$$</p>
<p>Now,</p>
<p>$$\mathop {\lim }\limits_{x \to {0^ + }} y$$</p>
<p>$$ = \mathop {\lim }\limits_{x \to {0^ + }} \left( {x\,.\,{e^{2 - {1 \over x}}}} \right)$$</p>
<p>$$ = 0\,.\,{e^{ - \alpha }}$$</p>
<p>$$ = 0$$</p> | mcq | jee-main-2023-online-29th-january-morning-shift |
1ldwwrqe8 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=y(x)$$ be the solution of the differential equation $$(x^2-3y^2)dx+3xy~dy=0,y(1)=1$$. Then $$6y^2(e)$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{3}{2}\\mathrm{e}^2$$\n"}, {"identifier": "B", "content": "$$3\\mathrm{e}^2$$"}, {"identifier": "C", "content": "$$\\mathrm{e}^2$$"}, {"identifier": "D", "content": "$$2\\mathrm{e}^2$$"}] | ["D"] | null | <p>Given,</p>
<p>$$\left( {{x^2} - 3{y^2}} \right)dx + 3xydy = 0$$</p>
<p>$$ \Rightarrow {x^2}dx - 3{y^2}dx + 3xydy = 0$$</p>
<p>$$ \Rightarrow {{{x^2}dx} \over {3xdx}} - {{3{y^2}dx} \over {3xdx}} + {{3ydy} \over {3xdx}} = 0$$</p>
<p>$$ \Rightarrow {x \over 3} - {{{y^2}} \over x} + y{{dy} \over {dx}} = 0$$</p>
<p>Let $${y^2} = t$$</p>
<p>$$2y{{dy} \over {dx}} = {{dt} \over {dx}}$$</p>
<p>$$\therefore$$ $${1 \over 2}{{dt} \over {dx}} - {t \over x} + {x \over 3} = 0$$</p>
<p>$$ \Rightarrow {{dt} \over {dx}} - {{2t} \over x} = - {{2x} \over 3}$$</p>
<p>This is linear Differential Equation.</p>
<p>$$
\text { I.F. }=e^{\int-\frac{2}{x} d x}=e^{-2 \ln |x|}=e^{\ln \frac{1}{x^2}}=\frac{1}{x^2}
$$</p>
<p>So, $t \cdot \frac{1}{x^2}=\int \frac{1}{x^2}\left(\frac{-2 x}{3}\right) d x
$
<br/><br/>$\Rightarrow \frac{y^2}{x^2}=\frac{-2}{3} \ln |x|+C$
<br/><br/>When, $x=1, y=1$
<br/><br/>$$
1=\frac{-2}{3} \ln (1)+C \Rightarrow C=1 $$
<br/><br/>$$ \therefore \frac{y^2}{x^2}=\frac{-2}{3} \ln |x|+1
$$
<br/><br/>At $x=e, \frac{y^2(e)}{e^2}=\frac{-2}{3}+1 $
<br/><br/>$\Rightarrow y^2(e)=\frac{e^2}{3} $
<br/><br/>$\Rightarrow 6 y^2(e)=2 e^2$</p> | mcq | jee-main-2023-online-24th-january-evening-shift |
1lgpxupwa | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=y_{1}(x)$$ and $$y=y_{2}(x)$$ be the solution curves of the differential equation $$\frac{d y}{d x}=y+7$$ with initial conditions $$y_{1}(0)=0$$ and $$y_{2}(0)=1$$ respectively. Then the curves $$y=y_{1}(x)$$ and $$y=y_{2}(x)$$ intersect at</p> | [{"identifier": "A", "content": "no point"}, {"identifier": "B", "content": "two points"}, {"identifier": "C", "content": "infinite number of points"}, {"identifier": "D", "content": "one point"}] | ["A"] | null | <p>The given differential equation is</p>
<p>$$\frac{d y}{d x} = y + 7$$</p>
<p>This is a first order linear differential equation and can be solved using an integrating factor. </p>
<p>Rearrange the equation to the standard form of a linear differential equation :</p>
<p>$$\frac{d y}{d x} - y = 7$$</p>
<p>The integrating factor is $e^{-\int dx} = e^{-x}$.</p>
<p>Multiplying each side of the equation by the integrating factor gives :</p>
<p>$$e^{-x} \frac{d y}{d x} - e^{-x} y = 7e^{-x}$$</p>
<p>The left-hand side of the equation is the derivative of $(e^{-x}y)$ with respect to $x$. So we can write the equation as :</p>
<p>$$\frac{d}{d x}(e^{-x}y) = 7e^{-x}$$</p>
<p>Integrate both sides with respect to $x$ :</p>
<p>$$e^{-x}y = -7e^{-x} + C$$</p>
<p>Multiply both sides by $e^{x}$ to isolate $y$ :</p>
<p>$$y = -7 + Ce^{x}$$</p>
<p>So, the general solution to the differential equation is $y = -7 + Ce^{x}$.</p>
<p>Now, let's apply the initial conditions to find the particular solutions :</p>
<p>For $y_{1}(0)=0$, we substitute into the general solution and solve for $C$ :</p>
<p>$$0 = -7 + C$$</p>
<p>So, $C = 7$, and the solution for $y_{1}$ is $y_{1}(x) = 7e^{x} - 7$.</p>
<p>For $y_{2}(0)=1$, again substitute into the general solution:</p>
<p>$$1 = -7 + C$$</p>
<p>So, $C = 8$, and the solution for $y_{2}$ is $y_{2}(x) = 8e^{x} - 7$.</p>
<p>The two curves intersect when $y_{1}(x) = y_{2}(x)$. Setting these equal and solving for $x$ gives :</p>
<p>$$7e^{x} - 7 = 8e^{x} - 7$$</p>
<p>$$e^{x} = 0$$</p>
<p>But $e^{x} = 0$ has no solution, because the exponential function never equals zero.</p>
<p>So, the curves $y=y_{1}(x)$ and $y=y_{2}(x)$ do not intersect at any point.</p>
| mcq | jee-main-2023-online-13th-april-morning-shift |
1lgsvopsl | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=y(x)$$ be the solution of the differential equation $$\frac{d y}{d x}+\frac{5}{x\left(x^{5}+1\right)} y=\frac{\left(x^{5}+1\right)^{2}}{x^{7}}, x > 0$$. If $$y(1)=2$$, then $$y(2)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{693}{128}$$"}, {"identifier": "B", "content": "$$\\frac{697}{128}$$"}, {"identifier": "C", "content": "$$\\frac{637}{128}$$"}, {"identifier": "D", "content": "$$\\frac{679}{128}$$"}] | ["A"] | null | I.F $=\mathrm{e}^{\int \frac{5 \mathrm{dx}}{\mathrm{x}\left(\mathrm{x}^5+1\right)}}=\mathrm{e}^{\int \frac{5 \mathrm{x}^{-6} \mathrm{dx}}{\left(\mathrm{x}^{-5}+1\right)}}$
<br/><br/>Put, $1+\mathrm{x}^{-5}=\mathrm{t} \Rightarrow-5 \mathrm{x}^{-6} \mathrm{dx}=\mathrm{dt}$
<br/><br/>$$ \therefore $$ $$
e^{\int-\frac{d t}{t}}=e^{-\ln t}=\frac{1}{t}=\frac{x^5}{1+x^5}
$$
<br/><br/>$$
\begin{aligned}
y \cdot \frac{x^5}{1+x^5} & =\int \frac{x^5}{\left(1+x^5\right)} \times \frac{\left(1+x^5\right)^2}{x^7} d x \\\\
& =\int x^3 d x+\int x^{-2} d x
\end{aligned}
$$
<br/><br/>$$
y \cdot \frac{x^5}{1+x^5}=\frac{x^4}{4}-\frac{1}{x}+c
$$
<br/><br/>$$
\text { Given that: } x=1 \Rightarrow y=2
$$
<br/><br/>$$
\begin{aligned}
& 2 \cdot \frac{1}{2}=\frac{1}{4}-1+\mathrm{c} \\\\
& \mathrm{c}=\frac{7}{4} \\\\
& \mathrm{y} \cdot \frac{\mathrm{x}^5}{1+\mathrm{x}^5}=\frac{\mathrm{x}^4}{4}-\frac{1}{\mathrm{x}}+\frac{7}{4} \\\\
& \text { Now put, } \mathrm{x}=2 \\\\
& \mathrm{y} \cdot\left(\frac{32}{33}\right)=\frac{21}{4} \\\\
& \mathrm{y}=\frac{693}{128}
\end{aligned}
$$ | mcq | jee-main-2023-online-11th-april-evening-shift |
1lguu2d04 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=y(x)$$ be a solution curve of the differential equation.</p>
<p>$$\left(1-x^{2} y^{2}\right) d x=y d x+x d y$$.</p>
<p>If the line $$x=1$$ intersects the curve $$y=y(x)$$ at $$y=2$$ and the line $$x=2$$ intersects the curve $$y=y(x)$$ at $$y=\alpha$$, then a value of $$\alpha$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{1+3 e^{2}}{2\\left(3 e^{2}-1\\right)}$$"}, {"identifier": "B", "content": "$$\\frac{3 e^{2}}{2\\left(3 e^{2}-1\\right)}$$"}, {"identifier": "C", "content": "$$\\frac{1-3 e^{2}}{2\\left(3 e^{2}+1\\right)}$$"}, {"identifier": "D", "content": "$$\\frac{3 e^{2}}{2\\left(3 e^{2}+1\\right)}$$"}] | ["A"] | null | We have,
<br/><br/>$$
\begin{aligned}
& \left(1-x^2 y^2\right) d x=y d x+x d y, y(1)=2 \\\\
& d x=\frac{y d x+x d y}{1-(x y)^2}
\end{aligned}
$$
<br/><br/>On integrating both sides, we get
<br/><br/>$$
\begin{aligned}
\int d x & =\int \frac{d(x y)}{1-(x y)^2} \\\\
x & =\frac{1}{2} \log \left|\frac{1+x y}{1-x y}\right|+C
\end{aligned}
$$
<br/><br/>As, $y(1)=2$
<br/><br/>$$
\begin{aligned}
1 & =\frac{1}{2} \log \left|\frac{1+2}{1-2}\right|+C \\\\
\Rightarrow C & =1-\frac{1}{2} \log 3
\end{aligned}
$$
<br/><br/>Now, substitute $x=2$ as $y(2)=\alpha$
<br/><br/>$$
2=\frac{1}{2} \log \left|\frac{1+2 \alpha}{1-2 \alpha}\right|+1-\frac{1}{2} \log 3
$$
<br/><br/>$$
\begin{aligned}
1+\frac{1}{2} \log 3 & =\frac{1}{2} \log \left|\frac{1+2 \alpha}{1-2 \alpha}\right| \\\\
2+\log 3 & =\log \left|\frac{1+2 \alpha}{1-2 \alpha}\right|
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
&\Rightarrow \frac{1+2 \alpha}{1-2 \alpha} \mid =3 e^2 \\\\
&\Rightarrow \frac{1+2 \alpha}{1-2 \alpha}= \pm \frac{3 e^2}{1} \\\\
& \Rightarrow\frac{1}{2 \alpha} =\frac{ \pm 3 e^2+1}{ \pm 3 e^2-1} \\\\
&\Rightarrow 2 \alpha=\frac{ \pm 3 e^2-1}{ \pm 3 e^2+1}
\end{aligned}
$$
<br/><br/>$$
\begin{array}{ll}
\Rightarrow \alpha=\frac{1}{2}\left(\frac{ \pm 3 e^2-1}{1 \pm 3 e^2}\right) \\\\
\therefore \alpha=\frac{3 e^2-1}{2\left(1+3 e^2\right)} \text { or } \frac{3 e^2+1}{2\left(3 e^2-1\right)}
\end{array}
$$ | mcq | jee-main-2023-online-11th-april-morning-shift |
1lgvqhawp | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let the tangent at any point P on a curve passing through the points (1, 1) and $$\left(\frac{1}{10}, 100\right)$$, intersect positive $$x$$-axis and $$y$$-axis at the points A and B respectively. If $$\mathrm{PA}: \mathrm{PB}=1: k$$ and $$y=y(x)$$ is the solution of the differential equation $$e^{\frac{d y}{d x}}=k x+\frac{k}{2}, y(0)=k$$, then $$4 y(1)-6 \log _{\mathrm{e}} 3$$ is equal to ____________.</p> | [] | null | 4 | Let the equation of tangent to the curve at $(x, y)$.
<br><br>Whose slope is $\frac{d y}{d x}$ is
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnkjbalj/b11eb6ae-312f-49ae-887d-c92b77fa1799/cc33a060-6789-11ee-b7ba-e3cd7ea3cf86/file-6y3zli1lnkjbalk.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lnkjbalj/b11eb6ae-312f-49ae-887d-c92b77fa1799/cc33a060-6789-11ee-b7ba-e3cd7ea3cf86/file-6y3zli1lnkjbalk.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 10th April Evening Shift Mathematics - Differential Equations Question 43 English Explanation">
<br><br>$$
\begin{aligned}
&\mathrm{Y}-y=\frac{d y}{d x}(\mathrm{X}-x)\\\\
&\begin{aligned}
& \text { Putting } Y=0 \Rightarrow X=x-\frac{y}{\left(\frac{d y}{d x}\right)} \mathrm{t} \\\\
& \Rightarrow \alpha=x-\frac{y}{\left(\frac{d y}{d x}\right)}
\end{aligned}
\end{aligned}
$$
<br><br>and putting $\mathrm{X}=0 \Rightarrow \mathrm{Y}=y-x \frac{d y}{d x}$
<br><br>$$
\Rightarrow \beta=y-x \frac{d y}{d x}
$$
<br><br>$\because \mathrm{P}$ divides $\mathrm{AB}$ in $1: k$
<br><br>$$
x=\frac{k \alpha+0}{k+1} \text { and } y=\frac{k \times 0+\beta}{k+1}
$$
<br><br>$$
\begin{aligned}
& \Rightarrow x(k+1)=k\left(x-\frac{y}{\frac{d y}{d x}}\right) \\\\
& \Rightarrow x k+x=x k-\frac{y k}{\frac{d y}{d x}} \\\\
&\Rightarrow x \frac{d y}{d x}=-y k
\end{aligned}
$$
<br><br>$$
\begin{aligned}
& \text { or } \int \frac{d y}{y}=-k \times \int \frac{1}{x} d x \\\\
& \Rightarrow \log y=-k \log x+\log \mathrm{C} \\\\
& \text { or } \log y \times x^k=\log \mathrm{C} \\\\
& \Rightarrow y x^k=\mathrm{C} \\\\
& \text { putting } x=1, y=1 \Rightarrow c=1 \\\\
& \text { so } y x^k=1
\end{aligned}
$$
<br><br>Putting $x=\frac{1}{10}, y=100 \Rightarrow 100 \times\left(\frac{1}{100}\right)^k=1 \Rightarrow k=2$
<br><br>so $y x^2=1$ or $y=\frac{1}{x^2}$
<br><br>Now $e^{\frac{d y}{d x}}=k x+\frac{k}{2}$
<br><br>$$
\Rightarrow \frac{d y}{d x}=\log _e\left(k x+\frac{k}{2}\right)=\log _e(2 x+1)
$$
<br><br>On integrating
<br><br>$$
\begin{aligned}
& y=\int 1 \cdot \log _e(2 x+1) d x \\\\
& =x \log _e(2 x+1)-\int \frac{1 \times 2}{2 x+1} \times x d x \\\\
& =x \log _e(2 x+1)-\int 1-\frac{1}{2 x+1} d x \\\\
& y=x \log _e(2 x+1)-x+\frac{1}{2} \log _e(2 x+1)+c
\end{aligned}
$$
<br><br>Put $x=0, y=k=2 \Rightarrow c=2$ putting $x=1$
<br><br>$$
\begin{aligned}
& y(1)=\log _e 3-1+\frac{1}{2} \log _e 3+2=\frac{3}{2} \log _e 3-1+2 \\\\
& \Rightarrow 4 y(1)=6 \log _e 3+4 \\\\
& \Rightarrow 4 y(1)-6 \log _e 3= 4
\end{aligned}
$$ | integer | jee-main-2023-online-10th-april-evening-shift |
1lh2y4an3 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>If the solution curve $$f(x, y)=0$$ of the differential equation
<br/><br/>$$\left(1+\log _{e} x\right) \frac{d x}{d y}-x \log _{e} x=e^{y}, x > 0$$,
<br/><br/>passes through the points $$(1,0)$$ and $$(\alpha, 2)$$, then $$\alpha^{\alpha}$$ is equal to :</p> | [{"identifier": "A", "content": "$$e^{\\sqrt{2} e^{2}}$$"}, {"identifier": "B", "content": "$$e^{2 e^{\\sqrt{2}}}$$"}, {"identifier": "C", "content": "$$e^{e^{2}}$$"}, {"identifier": "D", "content": "$$e^{2 e^{2}}$$"}] | ["D"] | null | We have, $\left(1+\log _e x\right) \frac{d x}{d y}-x \log _e x=e^{y}, x>0$
<br/><br/>Put $x \log _e x=t$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \left(x \cdot \frac{1}{x}+\log _e x\right) \frac{d x}{d y} =\frac{d t}{d y} \\\\
& \Rightarrow \left(1+\log _e x\right) \frac{d x}{d y} =\frac{d t}{d y} \\\\
& \therefore \frac{d t}{d y}-t =e^{y}
\end{aligned}
$$
<br/><br/>Now, IF $=e^{\int-d y}=e^{-y}$
<br/><br/>$\begin{aligned} & \therefore \text { General solution, } t\left(e^{-y}\right)=\int\left(e^y \cdot e^{-y}\right) d y+c \\\\ & \Rightarrow t e^{-y}=\int d y+c \\\\ & \Rightarrow t e^{-y}=y+c \\\\ & \Rightarrow \left(x \log _e x\right) e^{-y}=y+c ..........(i)\end{aligned}$
<br/><br/>Equation (i), passes through the point $(1,0)$
<br/><br/>$$
\begin{aligned}
& \therefore 0=0+c \\\\
& \Rightarrow c=0 \\\\
& \therefore \left(x \log _e x\right) e^{-y}=y \\\\
& \Rightarrow x \log _e x=y e^y
\end{aligned}
$$
<br/><br/>Which passes through $(\alpha, 2)$
<br/><br/>$$
\begin{aligned}
& \therefore \alpha \log _e \alpha=2 e^2 \\\\
& \Rightarrow \log _e \alpha^\alpha=2 e^2 \\\\
& \Rightarrow \alpha^\alpha=e^{2 e^2}
\end{aligned}
$$ | mcq | jee-main-2023-online-6th-april-evening-shift |
lsan508v | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | If $\frac{\mathrm{d} x}{\mathrm{~d} y}=\frac{1+x-y^2}{y}, x(1)=1$, then $5 x(2)$ is equal to __________. | [] | null | 5 | $\frac{d x}{d y}-\frac{x}{y}=\frac{1-y^2}{y}$
<br/><br/>Integrating factor $=\mathrm{e}^{\int-\frac{1}{y} d y}=\frac{1}{y}$
<br/><br/>$\begin{aligned} & x \cdot \frac{1}{y}=\int \frac{1-y^2}{y^2} d y \\\\ & \frac{x}{y}=\frac{-1}{y}-y+c \\\\ & x=-1-y^2+c y\end{aligned}$
<br/><br/>$\begin{aligned} & x(1)=1 \\\\ & 1=-1-1+c \Rightarrow c=3 \\\\ & x=-1-y^2+3 y \\\\ & 5 x(2)=5(-1-4+6) \\\\ & =5\end{aligned}$ | integer | jee-main-2024-online-1st-february-evening-shift |
lsaoufup | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let $y=y(x)$ be the solution of the differential equation
<br/><br/>$\frac{\mathrm{d} y}{\mathrm{~d} x}=2 x(x+y)^3-x(x+y)-1, y(0)=1$.
<br/><br/>Then, $\left(\frac{1}{\sqrt{2}}+y\left(\frac{1}{\sqrt{2}}\right)\right)^2$ equals : | [{"identifier": "A", "content": "$\\frac{4}{4+\\sqrt{\\mathrm{e}}}$"}, {"identifier": "B", "content": "$\\frac{3}{3-\\sqrt{\\mathrm{e}}}$"}, {"identifier": "C", "content": "$\\frac{2}{1+\\sqrt{\\mathrm{e}}}$"}, {"identifier": "D", "content": "$\\frac{1}{2-\\sqrt{\\mathrm{e}}}$"}] | ["D"] | null | $\begin{aligned} & \frac{d y}{d x}=2 x(x+y)^3-x(x+y)-1 \\\\ & \text { Put } x+y=t \\\\ & \Rightarrow \frac{d y}{d x}=\frac{d t}{d x}-1 \\\\ & \frac{d t}{d x}-1=2 x(t)^3-x t\end{aligned}$
<br/><br/>$\begin{aligned} \Rightarrow & \frac{d t}{2 t^3-t}=x d x \\\\ & \int \frac{1}{2 t^3-t} d t=\int x d x \\\\ \Rightarrow & \int \frac{t}{2 t^4-t^2} d t=\int x d x\end{aligned}$
<br/><br/>$\begin{aligned} & t^2=z \\\\ & 2 t d t=d z \\\\ & \frac{1}{2} \int \frac{d z}{2 z^2-z}=\int x d x \\\\ & \ln \left|\frac{z-\frac{1}{2}}{z}\right|=x^2+c\end{aligned}$
<br/><br/>$\ln \left|\frac{(x+y)^2-\frac{1}{2}}{(x+y)^2}\right|=x^2+c$
<br/><br/>$y(0)=1 \Rightarrow c=\ln \left(\frac{1}{2}\right)$
<br/><br/>$\Rightarrow \frac{(x+y)^2-\frac{1}{2}}{(x+y)^2}=e^{x^2} \times \frac{1}{2}$
<br/><br/>$\frac{(x+y)^2-\frac{1}{2}}{(x+y)^2}=\sqrt{e} \times \frac{1}{2}$
<br/><br/>$\Rightarrow(x+y)^2=\frac{1}{2-\sqrt{e}}$ | mcq | jee-main-2024-online-1st-february-morning-shift |
lsbld6tf | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | If the solution of the differential equation <br/><br/>$(2 x+3 y-2) \mathrm{d} x+(4 x+6 y-7) \mathrm{d} y=0, y(0)=3$, is
<br/><br/>$\alpha x+\beta y+3 \log _e|2 x+3 y-\gamma|=6$, then $\alpha+2 \beta+3 \gamma$ is equal to ____________. | [] | null | 29 | <p>$$\begin{array}{ll}
2 x+3 y-2=t & 4 x+6 y-4=2 t \\
2+3 \frac{d y}{d x}=\frac{d t}{d x} & 4 x+6 y-7=2 t-3
\end{array}$$</p>
<p>$$\begin{aligned}
& \frac{d y}{d x}=\frac{-(2 x+3 y-2)}{4 x+6 y-7} \\
& \frac{d t}{d x}=\frac{-3 t+4 t-6}{2 t-3}=\frac{t-6}{2 t-3} \\
& \int \frac{2 t-3}{t-6} d t=\int d x \\
& \int\left(\frac{2 t-12}{t-6}+\frac{9}{t-6}\right) \cdot d t=x \\
& 2 t+9 \ln (t-6)=x+c \\
& 2(2 x+3 y-2)+9 \ln (2 x+3 y-8)=x+c \\
& x=0, y=3 \\
& c=14 \\
& 4 x+6 y-4+9 \ln (2 x+3 y-8)=x+14 \\
& x+2 y+3 \ln (2 x+3 y-8)=6 \\
& \alpha=1, \beta=2, \gamma=8 \\
& \alpha+2 \beta+3 \gamma=1+4+24=29
\end{aligned}$$</p> | integer | jee-main-2024-online-27th-january-morning-shift |
jaoe38c1lscnd6h3 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>If $$y=y(x)$$ is the solution curve of the differential equation $$\left(x^2-4\right) \mathrm{d} y-\left(y^2-3 y\right) \mathrm{d} x=0, x>2, y(4)=\frac{3}{2}$$ and the slope of the curve is never zero, then the value of $$y(10)$$ equals :</p> | [{"identifier": "A", "content": "$$\\frac{3}{1+(8)^{1 / 4}}$$\n"}, {"identifier": "B", "content": "$$\\frac{3}{1-(8)^{1 / 4}}$$\n"}, {"identifier": "C", "content": "$$\\frac{3}{1-2 \\sqrt{2}}$$\n"}, {"identifier": "D", "content": "$$\\frac{3}{1+2 \\sqrt{2}}$$"}] | ["A"] | null | <p>$$\begin{aligned}
& \left(x^2-4\right) d y-\left(y^2-3 y\right) d x=0 \\
& \Rightarrow \int \frac{d y}{y^2-3 y}=\int \frac{d x}{x^2-4} \\
& \Rightarrow \frac{1}{3} \int \frac{y-(y-3)}{y(y-3)} d y=\int \frac{d x}{x^2-4} \\
& \Rightarrow \frac{1}{3}(\ln |y-3|-\ln |y|)=\frac{1}{4} \ln \left|\frac{x-2}{x+2}\right|+C \\
& \Rightarrow \frac{1}{3} \ln \left|\frac{y-3}{y}\right|=\frac{1}{4} \ln \left|\frac{x-2}{x+2}\right|+C
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \text { At } \mathrm{x}=4, \mathrm{y}=\frac{3}{2} \\
& \therefore \mathrm{C}=\frac{1}{4} \ln 3 \\
& \therefore \frac{1}{3} \ln \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\frac{1}{4} \ln \left|\frac{\mathrm{x}-2}{\mathrm{x}+2}\right|+\frac{1}{4} \ln (3) \\
& \text { At } \mathrm{x}=10 \\
& \frac{1}{3} \ln \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\frac{1}{4} \ln \left|\frac{2}{3}\right|+\frac{1}{4} \ln (3) \\
& \ln \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\ln 2^{3 / 4}, \forall \mathrm{x}>2, \frac{\mathrm{dy}}{\mathrm{dx}}<0 \\
& \text { as } \mathrm{y}(4)=\frac{3}{2} \Rightarrow \mathrm{y} \in(0,3) \\
& -\mathrm{y}+3=8^{1 / 4} \cdot \mathrm{y} \\
& \mathrm{y}=\frac{3}{1+8^{1 / 4}}
\end{aligned}$$</p> | mcq | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lscot4sr | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>If the solution curve, of the differential equation $$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x+y-2}{x-y}$$ passing through the point $$(2,1)$$ is $$\tan ^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{\beta} \log _{\mathrm{e}}\left(\alpha+\left(\frac{y-1}{x-1}\right)^2\right)=\log _{\mathrm{e}}|x-1|$$, then $$5 \beta+\alpha$$ is equal to __________.</p> | [] | null | 11 | <p>$$\begin{aligned}
& \frac{d y}{d x}=\frac{x+y-2}{x-y} \\
& \mathrm{x}=\mathrm{X}+\mathrm{h}, \mathrm{y}=\mathrm{Y}+\mathrm{k} \\
& \frac{d Y}{d X}=\frac{X+Y}{X-Y} \\
& \left.\begin{array}{l}
\mathrm{h}+\mathrm{k}-2=0 \\
\mathrm{~h}-\mathrm{k}=0
\end{array}\right\} \mathrm{h}=\mathrm{k}=1 \\
& \mathrm{Y}=\mathrm{vX} \\
& v+\frac{d v}{d X}=\frac{1+v}{1-v} \Rightarrow X-\frac{d v}{d X}=\frac{1+v^2}{1-v} \\
&
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \frac{1-v}{1+v^2} d v=\frac{d X}{X} \\
& \tan ^{-1} v-\frac{1}{2} \ln \left(1+v^2\right)=\ln |X|+C
\end{aligned}$$</p>
<p>As curve is passing through $$(2,1)$$</p>
<p>$$\begin{aligned}
& \tan ^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{2} \ln \left(1+\left(\frac{y-1}{x-1}\right)^2\right)=\ln |x-1| \\
& \therefore \alpha=1 \text { and } \beta=2 \\
& \Rightarrow 5 \beta+\alpha=11
\end{aligned}$$</p> | integer | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lsd4y5kd | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=y(x)$$ be the solution of the differential equation
<br/><br/>$$\sec ^2 x d x+\left(e^{2 y} \tan ^2 x+\tan x\right) d y=0,0< x<\frac{\pi}{2}, y(\pi / 4)=0$$.
<br/><br/>If $$y(\pi / 6)=\alpha$$, then $$e^{8 \alpha}$$ is equal to ____________.</p> | [] | null | 9 | <p>$$\begin{aligned}
& \sec ^2 x \frac{d x}{d y}+e^{2 y} \tan ^2 x+\tan x=0 \\
& \left(\text { Put } \tan x=t \Rightarrow \sec ^2 x \frac{d x}{d y}=\frac{d t}{d y}\right) \\
& \frac{d t}{d y}+e^{2 y} \times t^2+t=0 \\
& \frac{d t}{d y}+t=-t^2 \cdot e^{2 y} \\
& \frac{1}{t^2} \frac{d t}{d y}+\frac{1}{t}=-e^{2 y} \\
& \left(\text { Put } \frac{1}{t}=u \frac{-1}{t^2} \frac{d t}{d y}=\frac{d u}{d y}\right) \\
& \frac{-d u}{d y}+u=-e^{2 y} \\
& \frac{d u}{d y}-u=e^{2 y} \\
& \text { I.F. }=e^{-\int d y}=e^{-y} \\
& u e^{-y}=\int e^{-y} \times e^{2 y} d y \\
& \frac{1}{\tan x} \times e^{-y}=e^y+c \\
& x=\frac{\pi}{4}, y=0, c=0 \\
& \begin{aligned}
& \mathrm{x}=\frac{\pi}{6}, \quad \mathrm{y}=\alpha \\
& \sqrt{3} \mathrm{e}^{-\alpha}=\mathrm{e}^\alpha+0 \\
& \mathrm{e}^{2 \alpha}=\sqrt{3} \\
& \mathrm{e}^{8 \alpha}=9
\end{aligned}
\end{aligned}$$</p> | integer | jee-main-2024-online-31st-january-evening-shift |
jaoe38c1lse53x0f | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=y(x)$$ be the solution of the differential equation $$\frac{d y}{d x}=\frac{(\tan x)+y}{\sin x(\sec x-\sin x \tan x)}, x \in\left(0, \frac{\pi}{2}\right)$$ satisfying the condition $$y\left(\frac{\pi}{4}\right)=2$$. Then, $$y\left(\frac{\pi}{3}\right)$$ is</p> | [{"identifier": "A", "content": "$$\\sqrt{3}\\left(2+\\log _e 3\\right)$$\n"}, {"identifier": "B", "content": "$$\\sqrt{3}\\left(1+2 \\log _e 3\\right)$$\n"}, {"identifier": "C", "content": "$$\\sqrt{3}\\left(2+\\log _e \\sqrt{3}\\right)$$\n"}, {"identifier": "D", "content": "$$\\frac{\\sqrt{3}}{2}\\left(2+\\log _e 3\\right)$$"}] | ["C"] | null | <p>$$\begin{aligned}
& \frac{d y}{d x}=\frac{\sin x+y \cos x}{\sin x \cdot \cos x\left(\frac{1}{\cos x}-\sin x \cdot \frac{\sin x}{\cos x}\right)} \\
& =\frac{\sin x+y \cos x}{\sin x\left(1-\sin ^2 x\right)} \\
& \frac{d y}{d x}=\sec ^2 x+y \cdot 2(\operatorname{cosec} 2 x) \\
& \frac{d y}{d x}-2 \operatorname{cosec}(2 x) \cdot y=\sec ^2 x \\
& \frac{d y}{d x}+p \cdot y=Q
\end{aligned}$$</p>
<p>$$\text { I.F. }=\mathrm{e}^{\int p \mathrm{dx}}=\mathrm{e}^{\int-2 \operatorname{cosec}(2 \mathrm{x}) \mathrm{dx}}$$</p>
<p>Let $$2 \mathrm{x}=\mathrm{t}$$</p>
<p>$$2 \frac{\mathrm{dx}}{\mathrm{dt}}=1$$</p>
<p>$$\mathrm{dx}=\frac{\mathrm{dt}}{2}$$</p>
<p>$$=\mathrm{e}^{-\int \operatorname{cosec}(\mathrm{t}) \mathrm{dt}}$$</p>
<p>$$=\mathrm{e}^{-\ln \left|\tan \frac{t}{2}\right|}$$</p>
<p>$$=\mathrm{e}^{-\ln |\tan x|}=\frac{1}{|\tan x|}$$</p>
<p>$$\begin{aligned}
& y(\text { IF })=\int Q(I F) d x+c \\
& \Rightarrow y \frac{1}{|\tan x|}=\int \sec ^2 x \cdot \frac{1}{|\tan x|}+c \\
& y \cdot \frac{1}{|\tan x|}=\int \frac{d t}{|t|}+c \quad \text { for } \tan x=t \\
& y \cdot \frac{1}{|\tan x|}=\ln |t|+c \\
& y=|\tan x|(\ln |\tan x|+c) \\
& \text { Put } x=\frac{\pi}{4}, y=2 \\
& 2=\ln 1+c \Rightarrow c=2 \\
& y=|\tan x|(\ln |\tan x|+2) \\
& y\left(\frac{\pi}{3}\right)=\sqrt{3}(\ln \sqrt{3}+2)
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-morning-shift |
jaoe38c1lse5genw | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>The solution curve of the differential equation
$$y \frac{d x}{d y}=x\left(\log _e x-\log _e y+1\right), x>0, y>0$$ passing through the point $$(e, 1)$$ is</p> | [{"identifier": "A", "content": "$$\\left|\\log _e \\frac{y}{x}\\right|=y^2$$\n"}, {"identifier": "B", "content": "$$\\left|\\log _e \\frac{y}{x}\\right|=x$$\n"}, {"identifier": "C", "content": "$$\\left|\\log _e \\frac{x}{y}\\right|=y$$\n"}, {"identifier": "D", "content": "$$2\\left|\\log _e \\frac{x}{y}\\right|=y+1$$"}] | ["C"] | null | <p>$$\frac{\mathrm{dx}}{\mathrm{dy}}=\frac{\mathrm{x}}{\mathrm{y}}\left(\ln \left(\frac{\mathrm{x}}{\mathrm{y}}\right)+1\right)$$</p>
<p>Let $$\frac{x}{y}=t \Rightarrow x=t y$$</p>
<p>$$\begin{aligned}
& \frac{d x}{d y}=t+y \frac{d t}{d y} \\
& t+y \frac{d t}{d y}=t(\ln (t)+1)
\end{aligned}$$</p>
<p>$$\mathrm{y} \frac{\mathrm{dt}}{\mathrm{dy}}=\mathrm{t} \ln (\mathrm{t}) \Rightarrow \frac{\mathrm{dt}}{\mathrm{t} \ln (\mathrm{t})}=\frac{\mathrm{dy}}{\mathrm{y}}$$</p>
<p>$$\Rightarrow \int \frac{\mathrm{dt}}{\mathrm{t} \cdot \ln (\mathrm{t})}=\int \frac{\mathrm{dy}}{\mathrm{y}}$$</p>
<p>$$\Rightarrow \int \frac{d p}{p}=\int \frac{d y}{y} \quad$$ let $$\ln t=p$$</p>
<p>$$\frac{1}{\mathrm{t}} \mathrm{dt}=\mathrm{dp}$$</p>
<p>$$\begin{aligned}
& \Rightarrow \ln p=\ln y+c \\
& \ln (\ln t)=\ln y+c \\
& \ln \left(\ln \left(\frac{x}{y}\right)\right)=\ln y+c \\
& \text { at } x=e, y=1 \\
& \ln \left(\ln \left(\frac{e}{1}\right)\right)=\ln (1)+c \Rightarrow c=0
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \ln \left|\ln \left(\frac{x}{y}\right)\right|=\ln y \\
& \left|\ln \left(\frac{x}{y}\right)\right|=e^{\ln y} \\
& \left|\ln \left(\frac{x}{y}\right)\right|=y
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-morning-shift |
jaoe38c1lsf0v366 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>If the solution curve $$y=y(x)$$ of the differential equation $$\left(1+y^2\right)\left(1+\log _{\mathrm{e}} x\right) d x+x d y=0, x > 0$$ passes through the point $$(1,1)$$ and $$y(e)=\frac{\alpha-\tan \left(\frac{3}{2}\right)}{\beta+\tan \left(\frac{3}{2}\right)}$$, then $$\alpha+2 \beta$$ is _________.</p> | [] | null | 3 | <p>$$\begin{aligned}
& \int\left(\frac{1}{x}+\frac{\ln x}{x}\right) d x+\int \frac{d y}{1+y^2}=0 \\
& \ln x+\frac{(\ln x)^2}{2}+\tan ^{-1} y=C
\end{aligned}$$</p>
<p>Put $$x=y=1$$</p>
<p>$$\begin{aligned}
& \therefore C=\frac{\pi}{4} \\
& \Rightarrow \ln x+\frac{(\ln x)^2}{2}+\tan ^{-1} y=\frac{\pi}{4}
\end{aligned}$$</p>
<p>Put $$x=e$$</p>
<p>$$\begin{aligned}
& \Rightarrow \mathrm{y}=\tan \left(\frac{\pi}{4}-\frac{3}{2}\right)=\frac{1-\tan \frac{3}{2}}{1+\tan \frac{3}{2}} \\
& \therefore \alpha=1, \beta=1 \\
& \Rightarrow \alpha+2 \beta=3
\end{aligned}$$</p> | integer | jee-main-2024-online-29th-january-morning-shift |
jaoe38c1lsfknrcy | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>If $$\sin \left(\frac{y}{x}\right)=\log _e|x|+\frac{\alpha}{2}$$ is the solution of the differential equation $$x \cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x$$ and $$y(1)=\frac{\pi}{3}$$, then $$\alpha^2$$ is equal to</p> | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "3"}] | ["D"] | null | <p>Differential equation :-</p>
<p>$$\begin{aligned}
& x \cos \frac{y}{x} \frac{d y}{d x}=y \cos \frac{y}{x}+x \\
& \cos \frac{y}{x}\left[x \frac{d y}{d x}-y\right]=x
\end{aligned}$$</p>
<p>Divide both sides by $$\mathrm{x}^2$$</p>
<p>$$\cos \frac{y}{x}\left(\frac{x \frac{d y}{d x}-y}{x^2}\right)=\frac{1}{x}$$</p>
<p>Let $$\frac{y}{x}=t$$</p>
<p>$$\begin{aligned}
& \cos \mathrm{t}\left(\frac{\mathrm{dt}}{\mathrm{dx}}\right)=\frac{1}{\mathrm{x}} \\
& \cos \mathrm{t~dt}=\frac{1}{\mathrm{x}} \mathrm{dx}
\end{aligned}$$</p>
<p>Integrating both sides</p>
<p>$$\begin{aligned}
& \sin \mathrm{t}=\ln |\mathrm{x}|+\mathrm{c} \\
& \sin \frac{\mathrm{y}}{\mathrm{x}}=\ln |\mathrm{x}|+\mathrm{c}
\end{aligned}$$</p>
<p>Using $$\mathrm{y}(1)=\frac{\pi}{3}$$, we get $$\mathrm{c}=\frac{\sqrt{3}}{2}$$</p>
<p>So, $$\alpha=\sqrt{3} \Rightarrow \alpha^2=3$$</p> | mcq | jee-main-2024-online-29th-january-evening-shift |
1lsg51qnf | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$Y=Y(X)$$ be a curve lying in the first quadrant such that the area enclosed by the line $$Y-y=Y^{\prime}(x)(X-x)$$ and the co-ordinate axes, where $$(x, y)$$ is any point on the curve, is always $$\frac{-y^2}{2 Y^{\prime}(x)}+1, Y^{\prime}(x) \neq 0$$. If $$Y(1)=1$$, then $$12 Y(2)$$ equals __________.</p> | [] | null | 20 | <p>$$\mathrm{A}=\frac{1}{2}\left(\frac{-\mathrm{y}}{\mathrm{Y}^{\prime}(\mathrm{x})}+\mathrm{x}\right)(\mathrm{y}-\mathrm{xY} / \mathrm{x})=\frac{-\mathrm{y}^2}{2 \mathrm{Y}^{\prime}(\mathrm{x})}+1$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoxkuuf/8206ce56-06e1-4b71-9f96-f1cdb4f19baf/cd6cd370-ccf1-11ee-a330-494dca5e9a63/file-6y3zli1lsoxkuug.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsoxkuuf/8206ce56-06e1-4b71-9f96-f1cdb4f19baf/cd6cd370-ccf1-11ee-a330-494dca5e9a63/file-6y3zli1lsoxkuug.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 30th January Evening Shift Mathematics - Differential Equations Question 22 English Explanation"></p>
<p>$$\Rightarrow\left(-y+x Y^{\prime}(x)\right)\left(y-x Y^{\prime}(x)\right)=-y^2+2 Y^{\prime}(x)$$</p>
<p>$$\begin{aligned}
-y^2+x y Y^{\prime}(x)+x y Y^{\prime}(x) & -x^2\left[Y^{\prime}(x)\right]^2 \\
= & -y^2+2 Y^{\prime}(x)
\end{aligned}$$</p>
<p>$$\begin{aligned}
& 2 x y-x^2 Y^{\prime}(x)=2 \\
& \frac{d y}{d x}=\frac{2 x y-2}{x^2} \\
& \frac{d y}{d x}-\frac{2}{x} y=\frac{-2}{x^2} \\
& \text { I.F. }=e^{-2 \ln x}=\frac{1}{x^2} \\
& y \cdot \frac{1}{x^2}=\frac{2}{3} x^{-3}+c \\
& \text { Put } x=1, y=1 \\
& 1=\frac{2}{3}+c \Rightarrow c=\frac{1}{3} \\
& Y=\frac{2}{3} \cdot \frac{1}{X}+\frac{1}{3} X^2 \\
\Rightarrow \quad & 12 Y(2)=\frac{5}{3} \times 12=20
\end{aligned}$$</p> | integer | jee-main-2024-online-30th-january-evening-shift |
1lsgaiiyb | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=y(x)$$ be the solution of the differential equation $$\sec x \mathrm{~d} y+\{2(1-x) \tan x+x(2-x)\} \mathrm{d} x=0$$ such that $$y(0)=2$$. Then $$y(2)$$ is equal to:</p> | [{"identifier": "A", "content": "$$2\\{\\sin (2)+1\\}$$\n"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "$$2\\{1-\\sin (2)\\}$$"}] | ["B"] | null | <p>$$\frac{d y}{d x}=2(x-1) \sin x+\left(x^2-2 x\right) \cos x$$</p>
<p>Now both side integrate</p>
<p>$$\begin{aligned}
& y(x)=\int 2(x-1) \sin x d x+\left[\left(x^2-2 x\right)(\sin x)-\int(2 x-2) \sin x d x\right] \\
& y(x)=\left(x^2-2 x\right) \sin x+\lambda \\
& y(0)=0+\lambda \Rightarrow 2=\lambda \\
& y(x)=\left(x^2-2 x\right) \sin x+2 \\
& y(2)=2
\end{aligned}$$</p> | mcq | jee-main-2024-online-30th-january-morning-shift |
lv0vxcw0 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>If the solution $$y=y(x)$$ of the differential equation $$(x^4+2 x^3+3 x^2+2 x+2) \mathrm{d} y-(2 x^2+2 x+3) \mathrm{d} x=0$$ satisfies $$y(-1)=-\frac{\pi}{4}$$, then $$y(0)$$ is equal to :</p> | [{"identifier": "A", "content": "$$-\\frac{\\pi}{12}$$\n"}, {"identifier": "B", "content": "$$\\frac{\\pi}{2}$$\n"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$$\\frac{\\pi}{4}$$"}] | ["D"] | null | <p>$$\begin{aligned}
& \left(x^4+2 x^3+3 x^2+2 x+2\right) d y-\left(2 x^2+2 x+3\right) d x=0 \\
& \int d y=\int\left(\frac{2 x^2+2 x+3}{x^4+2 x^3+3 x^2+2 x+2} d x\right. \\
& \int d y=\int \frac{1}{x^2+1} d x+\int \frac{}{x^2+2 x+2} d x \\
& y=\tan ^{-1}(x)+\tan ^{-1}(1+x)+C \\
& y(-1)=\tan ^{-1}(-1)+\tan ^{-1}(1-1)+C \\
& y(-1)=-\frac{\pi}{4}+C=\left(\frac{-\pi}{4}\right)-\{\text { given }\} \\
& \Rightarrow C=0 \\
& \text { So, } y(x)=\tan ^{-1}(x)+\tan ^{-1}(1+x) \\
& y(0)=\tan ^{-1}(0)+\tan ^{-1}(1+0) \\
& y(0)=\frac{\pi}{4}
\end{aligned}$$</p> | mcq | jee-main-2024-online-4th-april-morning-shift |
lv2erury | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=y(x)$$ be the solution of the differential equation $$(x^2+4)^2 d y+(2 x^3 y+8 x y-2) d x=0$$. If $$y(0)=0$$, then $$y(2)$$ is equal to</p> | [{"identifier": "A", "content": "$$2 \\pi$$\n"}, {"identifier": "B", "content": "$$\\frac{\\pi}{8}$$\n"}, {"identifier": "C", "content": "$$\\frac{\\pi}{16}$$\n"}, {"identifier": "D", "content": "$$\\frac{\\pi}{32}$$"}] | ["D"] | null | <p>$$\frac{d y}{d x}+\frac{y\left(2 x^3+8 x\right)}{\left(x^2+4\right)^2}=\frac{2}{\left(x^2+4\right)^2}$$</p>
<p>$$\mathrm{IF}=e^{\int \frac{2 x^3+8 x}{\left(x^2+4\right)^2} d x}$$</p>
<p>$$\text { Let }\left(x^2+4\right)^2=t \quad \Rightarrow 2\left(x^2+4\right)(2 x) d x=d t$$</p>
<p>$$\begin{gathered}
=e^{\int \frac{d t}{2 t}}=e^{\log \sqrt{t}}=\sqrt{t}=\left(x^2+4\right) \\
\therefore \quad y\left(x^2+4\right)=\int \frac{2}{x^2+4}+c \\
\Rightarrow y\left(x^2+4\right)=\tan ^{-1}\left(\frac{x}{2}\right)+c \\
y(0)=0
\end{gathered}$$</p>
<p>$$\begin{aligned}
& \Rightarrow 0=0+c \quad \Rightarrow c=0 \\
& \text { put } x=2 \\
& y(8)=\frac{\pi}{4} \quad \Rightarrow \quad y=\frac{\pi}{32} \\
\end{aligned}$$</p> | mcq | jee-main-2024-online-4th-april-evening-shift |
lv2erh30 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=y(x)$$ be the solution of the differential equation $$(x+y+2)^2 d x=d y, y(0)=-2$$. Let the maximum and minimum values of the function $$y=y(x)$$ in $$\left[0, \frac{\pi}{3}\right]$$ be $$\alpha$$ and $$\beta$$, respectively. If $$(3 \alpha+\pi)^2+\beta^2=\gamma+\delta \sqrt{3}, \gamma, \delta \in \mathbb{Z}$$, then $$\gamma+\delta$$ equals _________.</p> | [] | null | 31 | <p>$$\begin{aligned}
& \frac{d y}{d x}=(x+y+z)^2 \\
& \text { Put } x+y+z=t \\
& \Rightarrow 1+\frac{d y}{d x}=\frac{d t}{d x} \\
& \text { Given DE } \Rightarrow \frac{d t}{d x}-1=t^2 \\
& \Rightarrow \frac{d t}{1+t^2}=d x \Rightarrow \tan ^{-1} t=x+c \\
& \Rightarrow x+y+z=\tan (x+c) \\
& \Rightarrow y(x)=\tan (x+c)-x-2 \\
& \because y(0)=-2 \Rightarrow-2=\tan c-0-2 \\
& \qquad \Rightarrow c=0 \\
& \Rightarrow y(x)=\tan x-x-2 \\
& \frac{d y}{d x}=\sec ^2 x-1 \geq 0
\end{aligned}$$</p>
<p>$$\Rightarrow y(x)$$ is increasing if $$x \in\left(0, \frac{\pi}{3}\right)$$</p>
<p>$$\begin{aligned}
& \Rightarrow \alpha=y\left(\frac{\pi}{3}\right), \beta=y(0) \\
& \Rightarrow \alpha=-\frac{\pi}{3}-2+\sqrt{3} \text { and } \beta=-2
\end{aligned}$$</p>
<p>Now, $$(3 \alpha+\pi)^2+\beta^2=(6+3 \sqrt{3})^2+(-2)^2$$</p>
<p>$$=67-36 \sqrt{3}=y+\delta \sqrt{3}$$.</p>
<p>$$\Rightarrow \gamma+\delta=31$$</p> | integer | jee-main-2024-online-4th-april-evening-shift |
lv3ve44l | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=y(x)$$ be the solution curve of the differential equation $$\sec y \frac{\mathrm{d} y}{\mathrm{~d} x}+2 x \sin y=x^3 \cos y, y(1)=0$$. Then $$y(\sqrt{3})$$ is equal to:</p> | [{"identifier": "A", "content": "$$\\frac{\\pi}{6}$$\n"}, {"identifier": "B", "content": "$$\\frac{\\pi}{12}$$\n"}, {"identifier": "C", "content": "$$\\frac{\\pi}{3}$$\n"}, {"identifier": "D", "content": "$$\\frac{\\pi}{4}$$"}] | ["D"] | null | <p>$$\begin{aligned}
& \sec y \frac{d y}{d x}+2 x \sin y=x^3 \cos y \\
& \Rightarrow \sec ^2 y \frac{d y}{d x}+2 x \tan y=x^3
\end{aligned}$$</p>
<p>Let $$z=\tan y$$</p>
<p>$$\begin{aligned}
& \frac{d z}{d x}=\sec ^2 y \frac{d y}{d x} \\
& \Rightarrow \frac{d z}{d x}+2 x z=x^3
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \text { I.F. }=e^{x^2} \\
& \Rightarrow z . e^{x^2}=\int e^{x^2} \cdot x^3 d x+c
\end{aligned}$$</p>
<p>$$\Rightarrow \tan y \cdot e^{x^2}=\frac{1}{2}\left(x^2 e^{x^2}-e^{x^2}\right)+c$$</p>
<p>$$\Rightarrow \tan (0) \cdot e=\frac{1}{2}(1 \cdot e-e)+c$$</p>
<p>$$\Rightarrow c=0$$</p>
<p>$$\Rightarrow \tan y=\frac{x^2-1}{2}$$</p>
<p>$$f(x)=\tan ^{-1}\left(\frac{x^2-1}{2}\right) \Rightarrow f(\sqrt{3})=\frac{\pi}{4}$$</p> | mcq | jee-main-2024-online-8th-april-evening-shift |
lv3ve63m | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$\alpha|x|=|y| \mathrm{e}^{x y-\beta}, \alpha, \beta \in \mathbf{N}$$ be the solution of the differential equation $$x \mathrm{~d} y-y \mathrm{~d} x+x y(x \mathrm{~d} y+y \mathrm{~d} x)=0,y(1)=2$$. Then $$\alpha+\beta$$ is equal to ________</p> | [] | null | 4 | <p>$$\begin{aligned}
& \alpha|x|=|y| e^{x y-\beta} \\
& \frac{x d y-y d x}{y^2}+\frac{x y(x d y+y d x)}{y^2}=0 \\
& -d\left(\frac{x}{y}\right)+\frac{x}{y} d(x y)=0
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \int d(x y)=\int \frac{d\left(\frac{x}{y}\right)}{\frac{x}{y}} \\
& x y=\ln \left|\frac{x}{y}\right|+\ln c \\
& x y=\ln \left(\left|\frac{x}{y}\right| \cdot c\right) \\
& \because y(1)=2 \\
& 2=\ln \left|\frac{1}{2}\right| c \Rightarrow c=2 e^2 \\
& \therefore \quad \operatorname{solution} x y=\ln \left(\left|\frac{x}{y}\right| \cdot 2 e^2\right) \\
& e^{x y}=\frac{|x|}{|y|} \cdot 2 e^2 \\
& 2|x|=|y| e^{x y-2} \\
& \Rightarrow \alpha=2, \beta=2, \alpha+\beta=4
\end{aligned}$$</p> | integer | jee-main-2024-online-8th-april-evening-shift |
lv5grwhw | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Let $$y=y(x)$$ be the solution of the differential equation $$(1+y^2) e^{\tan x} d x+\cos ^2 x(1+e^{2 \tan x}) d y=0, y(0)=1$$. Then $$y\left(\frac{\pi}{4}\right)$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{1}{e^2}$$\n"}, {"identifier": "B", "content": "$$\\frac{2}{e^2}$$\n"}, {"identifier": "C", "content": "$$\\frac{2}{e}$$\n"}, {"identifier": "D", "content": "$$\\frac{1}{e}$$"}] | ["D"] | null | <p>$$\begin{aligned}
& \left(1+y^2\right) e^{\tan x} d x+\cos ^2 x\left(1+e^{2 \tan x}\right) d y=0 \\
& \frac{d y}{1+y^2}=-\frac{e^{\tan x} \cdot \sec ^2 x d x}{1+e^{2 \tan x}} \\
& \int \frac{d y}{1+y^2}=-\int \frac{e^{\tan x} \cdot \sec ^2 x d x}{1+e^{2 \tan x}}
\end{aligned}$$</p>
<p>Let $$e^{\tan x}=t$$</p>
<p>$$\begin{aligned}
& e^{\tan x} \cdot \sec ^2 x d x=d t \\
& \int \frac{d y}{1+y^2}=-\int \frac{d t}{1+t^2}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \tan ^{-1} y=-\tan ^{-1} t+c \\
& \tan ^{-1} y=-\tan ^{-1}\left(e^{\tan x}\right)+c \\
& \text { at } x=0, y=1 \\
& \tan ^{-1}(1)=-\tan ^{-1}(1)+c \\
& \frac{\pi}{4}=-\frac{\pi}{4}+c \\
& c=\frac{\pi}{2} \\
& \tan ^{-1} y=-\tan ^{-1}\left(e^{\tan x}\right)+\frac{\pi}{2}
\end{aligned}$$</p>
<p>Now, at $$x=\frac{\pi}{4}$$</p>
<p>$$\begin{aligned}
& \tan ^{-1} y=-\tan ^{-1}(e)+\frac{\pi}{2} \\
& \tan ^{-1} y=\cot ^{-1} e=\tan ^{-1} \frac{1}{e} \\
& \Rightarrow y=\frac{1}{e}
\end{aligned}$$</p> | mcq | jee-main-2024-online-8th-april-morning-shift |
lvb294oh | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>Suppose the solution of the differential equation $$\frac{d y}{d x}=\frac{(2+\alpha) x-\beta y+2}{\beta x-2 \alpha y-(\beta \gamma-4 \alpha)}$$ represents a circle passing through origin. Then the radius of this circle is :</p> | [{"identifier": "A", "content": "$$\\sqrt{17}$$\n"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "$$\\frac{\\sqrt{17}}{2}$$\n"}, {"identifier": "D", "content": "$$\\frac{1}{2}$$"}] | ["C"] | null | <p>$$\begin{aligned}
& \frac{d y}{d x}=\frac{(2+\alpha) x-\beta y+2}{\beta x-2 \alpha y-(\beta \gamma-4 \alpha)} \\
& \beta x d y-2 \alpha y d y-(\beta \gamma-4 \alpha) d y \\
& =2 x d x+\alpha x d x-\beta y d x+2 d x \\
& \beta \int(x d y+y d y)-\alpha y^2-(\beta \gamma-4 x) y=x^2+\frac{\alpha x^2}{2}+2 x \\
& \beta x y-\alpha y^2-(\beta \gamma-4 \alpha) y=x^2+\frac{\alpha x^2}{2}+2 x \\
& \left(1+\frac{\alpha}{2}\right) x^2+\alpha y^2-\beta x y+2 x+(\beta \gamma-4 \alpha) y=0 \\
& \because \text { this represents circle passing through origin } \\
& \Rightarrow \beta=0 \text { and } 1+\frac{\alpha}{2}=\alpha \\
& \Rightarrow \alpha=2
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \therefore C: 2 x^2+2 y^2+2 x-8 y=0 \\
& x^2+y^2+x-4 y=0 \\
& \text { Radius }=\sqrt{\frac{1}{4}+4-0} \\
& \quad=\frac{\sqrt{17}}{2}
\end{aligned}$$</p> | mcq | jee-main-2024-online-6th-april-evening-shift |
lvb29503 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | <p>If the solution $$y(x)$$ of the given differential equation $$\left(e^y+1\right) \cos x \mathrm{~d} x+\mathrm{e}^y \sin x \mathrm{~d} y=0$$ passes through the point $$\left(\frac{\pi}{2}, 0\right)$$, then the value of $$e^{y\left(\frac{\pi}{6}\right)}$$ is equal to _________.</p> | [] | null | 3 | <p>Given the differential equation</p>
<p>$$\left(e^y + 1\right) \cos x \, dx + e^y \sin x \, dy = 0,$$</p>
<p>we aim to find the value of $ e^{y\left(\frac{\pi}{6}\right)} $ given that the solution $ y(x) $ passes through the point $\left(\frac{\pi}{2}, 0\right)$.</p>
<p>First, we recognize that the differential equation can be rearranged as:</p>
<p>$$d\left(e^y \sin x\right) + \cos x \, dx = 0.$$</p>
<p>Integrating this expression, we obtain:</p>
<p>$$ e^y \sin x + \sin x = C,$$</p>
<p>where $ C $ is a constant. Given that the solution passes through the point $\left(\frac{\pi}{2}, 0\right)$, we substitute these values into the equation to find $ C $:</p>
<p>$$ e^0 \sin\left(\frac{\pi}{2}\right) + \sin\left(\frac{\pi}{2}\right) = C \Rightarrow 1 + 1 = C \Rightarrow C = 2.$$</p>
<p>Thus, the equation simplifies to:</p>
<p>$$e^y \sin x + \sin x = 2.$$</p>
<p>We now need to determine the value of $ e^{y\left(\frac{\pi}{6}\right)} $. Substituting $ x = \frac{\pi}{6} $ into the equation, we get:</p>
<p>$$ \left(e^{y\left(\frac{\pi}{6}\right)} + 1\right) \cdot \sin\left(\frac{\pi}{6}\right) = 2.$$</p>
<p>Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$, the equation becomes:</p>
<p>$$ \left(e^{y\left(\frac{\pi}{6}\right)} + 1\right) \cdot \frac{1}{2} = 2 \Rightarrow e^{y\left(\frac{\pi}{6}\right)} + 1 = 4 \Rightarrow e^{y\left(\frac{\pi}{6}\right)} = 3.$$</p>
<p>Therefore, the value of $ e^{y\left(\frac{\pi}{6}\right)} $ is $ 3 $.</p> | integer | jee-main-2024-online-6th-april-evening-shift |
FFUHPxJb4rDcgpkbqKevN | maths | differentiation | differentiation-of-a-function-with-respect-to-another-function | If $$x = \sqrt {{2^{\cos e{c^{ - 1}}}}} $$ and $$y = \sqrt {{2^{se{c^{ - 1}}t}}} \,\,\left( {\left| t \right| \ge 1} \right),$$ then $${{dy} \over {dx}}$$ is equal to : | [{"identifier": "A", "content": "$${y \\over x}$$ "}, {"identifier": "B", "content": "$${x \\over y}$$"}, {"identifier": "C", "content": "$$-$$ $${y \\over x}$$"}, {"identifier": "D", "content": "$$-$$ $${x \\over y}$$"}] | ["C"] | null | x = $$\sqrt {{2^{\cos e{c^{ - 1}}t}}} $$
<br><br>$$\therefore\,\,\,\,$$ $${{dx} \over {dt}}$$ = $${1 \over {2\sqrt {{2^{\cos e{c^{ - 1}}t}}} }}$$ $$ \times $$ ($${2^{\cos e{c^{ - 1}}t}}\,.\,\log 2$$) $$ \times $$ $${{ - 1} \over {t\sqrt {{t^2} - 1} }}$$
<br><br>$${{dy} \over {dt}}$$ = $${1 \over {2\sqrt {{2^{{{\sec }^{ - 1}}t}}} }}$$ $$ \times $$ $$\left( {{2^{{{\sec }^{ - 1}}t}}\log 2} \right)$$ $$ \times $$ $${1 \over {t\sqrt {{t^2} - 1} }}$$
<br><br>$$\therefore\,\,\,\,$$ $${{dy} \over {dx}}$$
<br><br>= $${{{{dy} \over {dt}}} \over {{{dx} \over {dt}}}}$$
<br><br>= $${{ - \sqrt {{2^{\cos e{c^{ - 1}}t}}} } \over {\sqrt {{2^{\cos e{c^{ - 1}}t}}} }}$$ $$ \times $$ $${{{2^{{{\sec }^{ - 1}}t}}} \over {{2^{\cos e{c^{ - 1}}t}}}}$$
<br><br>= $$ - \sqrt {{{{2^{{{\sec }^{ - 1}}t}}} \over {{2^{\cos e{c^{ - 1}}t}}}}} $$
<br><br>= $$-$$ $${y \over x}$$ | mcq | jee-main-2018-online-16th-april-morning-slot |
udGIdarrHkh9036m2n3rsa0w2w9jxb3mi6x | maths | differentiation | differentiation-of-a-function-with-respect-to-another-function | The derivative of $${\tan ^{ - 1}}\left( {{{\sin x - \cos x} \over {\sin x + \cos x}}} \right)$$, with respect to $${x \over 2}$$
, where $$\left( {x \in \left( {0,{\pi \over 2}} \right)} \right)$$ is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "$${2 \\over 3}$$"}, {"identifier": "D", "content": "$${1 \\over 2}$$"}] | ["B"] | null | $$y = {\tan ^{ - 1}}\left( {{{\tan x - 1} \over {\tan x + 1}}} \right)$$<br><br>
$$ \Rightarrow $$ $$ - {\tan ^{ - 1}}\left( {{{1 - \tan x} \over {1 + \tan x}}} \right)$$<br><br>
$$ \Rightarrow $$ $$ - {\tan ^{ - 1}}\left( {\tan \left( {{\pi \over 4} - x} \right)} \right)$$<br><br>
$$ \Rightarrow $$ $$ - \left( {{\pi \over 4} - x} \right)$$<br><br>
$$ \Rightarrow $$ $${{dy} \over {dx}} = 1$$<br><br>
Now if differentiation of $${x \over 2}$$ w.r.t is $${1 \over 2}$$<br><br>
$$ \Rightarrow $$ So differentiation of y w.r.t $${x \over 2}$$ is $$1 \over {1 \over 2}$$ = 2 | mcq | jee-main-2019-online-12th-april-evening-slot |
s3JCuNr7c41ktbzUE9jgy2xukfqf5mcc | maths | differentiation | differentiation-of-a-function-with-respect-to-another-function | The derivative of
<br/>$${\tan ^{ - 1}}\left( {{{\sqrt {1 + {x^2}} - 1} \over x}} \right)$$ with<br/> respect to $${\tan ^{ - 1}}\left( {{{2x\sqrt {1 - {x^2}} } \over {1 - 2{x^2}}}} \right)$$ at x = $${1 \over 2}$$ is : | [{"identifier": "A", "content": "$${{2\\sqrt 3 } \\over 3}$$"}, {"identifier": "B", "content": "$${{2\\sqrt 3 } \\over 5}$$"}, {"identifier": "C", "content": "$${{\\sqrt 3 } \\over {10}}$$"}, {"identifier": "D", "content": "$${{\\sqrt 3 } \\over {12}}$$"}] | ["C"] | null | Let f = $${\tan ^{ - 1}}\left( {{{\sqrt {1 + {x^2}} - 1} \over x}} \right)$$
<br><br>Put x = tan $$\theta $$ $$ \Rightarrow $$ $$\theta $$ = tan<sup>–1</sup> x
<br><br>f = $${\tan ^{ - 1}}\left( {{{\sec \theta - 1} \over {\tan \theta }}} \right)$$
<br><br>$$ \Rightarrow $$ f = $${\tan ^{ - 1}}\left( {{{1 - \cos \theta } \over {\sin \theta }}} \right)$$ = $${\theta \over 2}$$
<br><br>$$ \Rightarrow $$ f = $${{{{\tan }^{ - 1}}x} \over 2}$$
<br><br>$$ \therefore $$ $${{df} \over {dx}}$$ = $${1 \over {2\left( {1 + {x^2}} \right)}}$$ ....(1)
<br><br>Let g = $${\tan ^{ - 1}}\left( {{{2x\sqrt {1 - {x^2}} } \over {1 - 2{x^2}}}} \right)$$
<br><br>Put x = sin $$\theta $$ $$ \Rightarrow $$ $$\theta $$ = sin<sup>–1</sup>
x
<br><br>$$ \Rightarrow $$ g = $${\tan ^{ - 1}}\left( {{{2\sin \theta \cos \theta } \over {1 - 2{{\sin }^2}\theta }}} \right)$$
<br><br>$$ \Rightarrow $$ g = tan<sup>–1</sup>
(tan 2$$\theta $$) = 2$$\theta $$
<br><br>$$ \Rightarrow $$ g = 2sin<sup>-1</sup> x
<br><br>$$ \Rightarrow $$ $${{dg} \over {dx}}$$ = $${2 \over {\sqrt {1 - {x^2}} }}$$ ...(2)
<br><br>Using (i) and (ii),
<br><br>$$ \therefore $$ $${{df} \over {dg}}$$ = $${1 \over {2\left( {1 + {x^2}} \right)}}{{\sqrt {1 - {x^2}} } \over 2}$$
<br><br>At x = $${1 \over 2}$$, $${\left( {{{df} \over {dg}}} \right)_{x = {1 \over 2}}}$$ = $${{\sqrt 3 } \over {10}}$$ | mcq | jee-main-2020-online-5th-september-evening-slot |
h3iEjCw8iLxihq9W | maths | differentiation | differentiation-of-composite-function | Let $$f:\left( { - 1,1} \right) \to R$$ be a differentiable function with $$f\left( 0 \right) = - 1$$ and $$f'\left( 0 \right) = 1$$. Let $$g\left( x \right) = {\left[ {f\left( {2f\left( x \right) + 2} \right)} \right]^2}$$. Then $$g'\left( 0 \right) = $$ | [{"identifier": "A", "content": "$$-4$$ "}, {"identifier": "B", "content": "$$0$$ "}, {"identifier": "C", "content": "$$-2$$ "}, {"identifier": "D", "content": "$$4$$ "}] | ["A"] | null | $$g'\left( x \right) = 2\left( {f\left( {2f\left( x \right) + 2} \right)} \right)$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{d \over {dx}}\left( {f\left( {2f\left( x \right) + 2} \right)} \right)} \right)$$
<br><br>$$ = 2f\left( {2f\left( x \right) + 2} \right)f'\left( {2f\left( x \right)} \right)$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left. 2 \right).\left( {2f'\left( x \right)} \right)$$
<br><br>$$ \Rightarrow g'\left( 0 \right) = 2f\left( {2f\left( 0 \right) + 2} \right).$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f'\left( {2f\left( 0 \right) + 2} \right).2f'\left( 0 \right)$$
<br><br>$$ = 4f\left( 0 \right){\left( {f'\left( 0 \right)} \right)^2}$$
<br><br>$$ = 4\left( { - 1} \right){\left( 1 \right)^2} = - 4$$ | mcq | aieee-2010 |
mgG2EkWdjrRPzc2r | maths | differentiation | differentiation-of-composite-function | If $$g$$ is the inverse of a function $$f$$ and $$f'\left( x \right) = {1 \over {1 + {x^5}}},$$ then $$g'\left( x \right)$$ is equal to: | [{"identifier": "A", "content": "$${1 \\over {1 + {{\\left\\{ {g\\left( x \\right)} \\right\\}}^5}}}$$ "}, {"identifier": "B", "content": "$$1 + {\\left\\{ {g\\left( x \\right)} \\right\\}^5}$$ "}, {"identifier": "C", "content": "$$1 + {x^5}$$ "}, {"identifier": "D", "content": "$$5{x^4}$$ "}] | ["B"] | null | Since $$f(x)$$ and $$g(x)$$ are inverse of each other
<br><br>$$\therefore$$ $$g'\left( {f\left( x \right)} \right) = {1 \over {f'\left( x \right)}}$$
<br><br>$$ \Rightarrow g'\left( {f\left( x \right)} \right) = 1 + {x^5}$$
<br><br>$$\left( \, \right.$$ As $$\,f'\left( x \right) = {1 \over {1 + {x^5}}}$$ $$\left. \, \right)$$
<br><br>Here $$x=g(y)$$
<br><br>$$\therefore$$ $$g'\left( y \right) = 1 + \left\{ {g\left( y \right)} \right\}$$
<br><br>$$ \Rightarrow g'\left( x \right) = 1 + \left\{ {g\left( x \right)} \right\}$$ | mcq | jee-main-2014-offline |
nJGKt5gcnsbeag2W6jFE6 | maths | differentiation | differentiation-of-composite-function | If ƒ(1) = 1, ƒ'(1) = 3, then the derivative of
ƒ(ƒ(ƒ(x))) + (ƒ(x))<sup>2
</sup> at x = 1 is : | [{"identifier": "A", "content": "33"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "15"}] | ["A"] | null | Given ƒ(1) = 1, ƒ'(1) = 3
<br><br>Let y = ƒ(ƒ(ƒ(x))) + (ƒ(x))<sup>2
</sup>
<br><br>On differentiating both sides with respect to x we get,
<br><br>$${{dy} \over {dx}}$$ = ƒ'(ƒ(ƒ(x))).ƒ'(ƒ(x)).ƒ'(x) + 2ƒ(x).ƒ'(x)
<br><br>Now at x = 1,
<br><br>$${{dy} \over {dx}}$$ = ƒ'(ƒ(ƒ(1))).ƒ'(ƒ(1)).ƒ'(1) + 2ƒ(1).ƒ'(1)
<br><br>= ƒ'(ƒ(1)).ƒ'(1).ƒ'(1) + 2.1.ƒ'(1)
<br><br>= ƒ'(1).ƒ'(1).ƒ'(1) + 2.1.ƒ'(1)
<br><br>= 3$$ \times $$3$$ \times $$3 + 2$$ \times $$3
<br><br>= 33 | mcq | jee-main-2019-online-8th-april-evening-slot |
HrO6IQwQJN5Eo4N4zJ3rsa0w2w9jx220fw8 | maths | differentiation | differentiation-of-composite-function | Let f(x) = log<sub>e</sub>(sin x), (0 < x < $$\pi $$) and g(x) = sin<sup>–1</sup>
(e<sup>–x</sup>
), (x $$ \ge $$ 0). If $$\alpha $$ is a positive real number such that
a = (fog)'($$\alpha $$) and b = (fog)($$\alpha $$), then : | [{"identifier": "A", "content": "a$$\\alpha $$<sup>2</sup> + b$$\\alpha $$ - a = -2$$\\alpha $$<sup>2</sup>"}, {"identifier": "B", "content": "a$$\\alpha $$<sup>2</sup> + b$$\\alpha $$ + a = 0"}, {"identifier": "C", "content": "a$$\\alpha $$<sup>2</sup> - b$$\\alpha $$ - a = 0"}, {"identifier": "D", "content": "a$$\\alpha $$<sup>2</sup> - b$$\\alpha $$ - a = 1"}] | ["D"] | null | f(x) = ln(sin x), g(x) = sin<sup>–1</sup> (e<sup>–x</sup>)<br><br>
f(g(x)) = ln(sin(sin<sup>–1</sup> e<sup>–x</sup>)) = -x<br><br>
f(g($$\alpha $$)) = – $$\alpha $$ = b<br><br>
As f(g(x)) = – x <br><br>
$$ \therefore $$ (f(g(x)))' = – 1<br><br>
$$ \Rightarrow $$ (f(g($$\alpha $$)))' = – 1 = a<br><br>
$$ \therefore $$ b = – $$\alpha $$,
a = – 1<br><br>
$$ \therefore $$ a$$\alpha $$<sup>2</sup> - b$$\alpha $$ - a = - $$\alpha $$<sup>2</sup> + $$\alpha $$<sup>2</sup> + 1 = 1 | mcq | jee-main-2019-online-10th-april-evening-slot |
1l5ai1at2 | maths | differentiation | differentiation-of-composite-function | <p>Let f : R $$\to$$ R be defined as $$f(x) = {x^3} + x - 5$$. If g(x) is a function such that $$f(g(x)) = x,\forall 'x' \in R$$, then g'(63) is equal to ________________.</p> | [{"identifier": "A", "content": "$${1 \\over {49}}$$"}, {"identifier": "B", "content": "$${3 \\over {49}}$$"}, {"identifier": "C", "content": "$${43 \\over {49}}$$"}, {"identifier": "D", "content": "$${91 \\over {49}}$$"}] | ["A"] | null | <p>$$f(x) = 3{x^2} + 1$$</p>
<p>f'(x) is bijective function</p>
<p>and $$f(g(x)) = x \Rightarrow g(x)$$ is inverse of f(x)</p>
<p>$$g(f(x)) = x$$</p>
<p>$$g'(f(x))\,.\,f'(x) = 1$$</p>
<p>$$g'(f(x)) = {1 \over {3{x^2} + 1}}$$</p>
<p>Put x = 4 we get</p>
<p>$$g'(63) = {1 \over {49}}$$</p> | mcq | jee-main-2022-online-25th-june-morning-shift |
1ldr85iqr | maths | differentiation | differentiation-of-composite-function | <p>Let $$f^{1}(x)=\frac{3 x+2}{2 x+3}, x \in \mathbf{R}-\left\{\frac{-3}{2}\right\}$$ For $$\mathrm{n} \geq 2$$, define $$f^{\mathrm{n}}(x)=f^{1} \mathrm{o} f^{\mathrm{n}-1}(x)$$. If $$f^{5}(x)=\frac{\mathrm{a} x+\mathrm{b}}{\mathrm{b} x+\mathrm{a}}, \operatorname{gcd}(\mathrm{a}, \mathrm{b})=1$$, then $$\mathrm{a}+\mathrm{b}$$ is equal to ____________.</p> | [] | null | 3125 | <p>$$f'(x) = {{3x + 2} \over {2x + 3}}x \in R - \left\{ { - {3 \over 2}} \right\}$$</p>
<p>$${f^5}(x) = {f_o}{f_o}{f_o}{f_o}f(x)$$</p>
<p>$${f_o}f(x) = {{13x + 12} \over {12x + 13}}$$</p>
<p>$${f_o}{f_o}{f_o}{f_o}f(x) = {{1563x + 1562} \over {1562x + 1563}}$$</p>
<p>$$ \equiv {{ax + b} \over {bx + a}}$$</p>
<p>$$\therefore$$ $$a = 1563,b = 1562$$</p>
<p>$$ = 3125$$</p> | integer | jee-main-2023-online-30th-january-morning-shift |
lv0vxcdq | maths | differentiation | differentiation-of-composite-function | <p>Let $$f(x)=x^5+2 \mathrm{e}^{x / 4}$$ for all $$x \in \mathbf{R}$$. Consider a function $$g(x)$$ such that $$(g \circ f)(x)=x$$ for all $$x \in \mathbf{R}$$. Then the value of $$8 g^{\prime}(2)$$ is :</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "16"}, {"identifier": "D", "content": "8"}] | ["C"] | null | <p>Given that $$(g \circ f)(x) = x$$ for all $$x \in \mathbf{R}$$. This means $$g(f(x)) = x$$ for all $$x \in \mathbf{R}$$. Differentiating both sides with respect to $$x$$, we get:
<p>$$g'(f(x)) \cdot f'(x) = 1$$</p>
</p>
<p>Now, we want to find the value of $$8g'(2)$$. To do this, we need to find a value of $$x$$ such that $$f(x) = 2$$. Let's solve for $$x$$:
<p>$$x^5 + 2e^{x/4} = 2$$</p>
</p>
<p>By inspection, we see that $$x = 0$$ is a solution. Therefore, $$f(0) = 2$$. Now, we can substitute this into our differentiated equation:
<p>$$g'(f(0)) \cdot f'(0) = 1$$</p>
<p>$$g'(2) \cdot f'(0) = 1$$</p>
</p>
<p>Let's find $$f'(0)$$:
<p>$$f'(x) = 5x^4 + \frac{1}{2}e^{x/4}$$</p>
<p>$$f'(0) = \frac{1}{2}$$</p>
</p>
<p>Substituting this back into our equation:
<p>$$g'(2) \cdot \frac{1}{2} = 1$$</p>
<p>$$g'(2) = 2$$</p>
</p>
<p>Finally, we can calculate $$8g'(2)$$:
<p>$$8g'(2) = 8 \cdot 2 = \boxed{16}$$</p></p>
<p></p> | mcq | jee-main-2024-online-4th-april-morning-shift |
lvb294td | maths | differentiation | differentiation-of-composite-function | <p>Suppose for a differentiable function $$h, h(0)=0, h(1)=1$$ and $$h^{\prime}(0)=h^{\prime}(1)=2$$. If $$g(x)=h\left(\mathrm{e}^x\right) \mathrm{e}^{h(x)}$$, then $$g^{\prime}(0)$$ is equal to:</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "8"}] | ["A"] | null | <p>To determine $$g^{\prime}(0)$$, we start by applying the chain rule and product rule to find the derivative of the given function $$g(x) = h\left(\mathrm{e}^x\right) \mathrm{e}^{h(x)}$$.</p>
<p>The product rule states that if we have two functions $$u(x)$$ and $$v(x)$$, then the derivative of their product is given by:</p>
<p>$$ (u(x)v(x))' = u'(x)v(x) + u(x)v'(x) $$</p>
<p>Let's denote $$u(x) = h(\mathrm{e}^x)$$ and $$v(x) = \mathrm{e}^{h(x)}$$.</p>
<p>First, we need to find $$u'(x)$$ and $$v'(x)$$. Using the chain rule, we find:</p>
<p>$$ u'(x) = \frac{d}{dx}h(\mathrm{e}^x) = h'(\mathrm{e}^x) \cdot \frac{d}{dx}(\mathrm{e}^x) = h'(\mathrm{e}^x) \cdot \mathrm{e}^x $$</p>
<p>Now, the derivative of $$v(x)$$ is:</p>
<p>$$ v'(x) = \frac{d}{dx}(\mathrm{e}^{h(x)}) = \mathrm{e}^{h(x)} \cdot h'(x) $$</p>
<p>Using the product rule, we get the derivative of $$g(x)$$:</p>
<p>$$ g'(x) = u'(x) v(x) + u(x) v'(x) $$</p>
<p>Substituting $$u(x)$$, $$v(x)$$, $$u'(x)$$, and $$v'(x)$$ into the above expression, we get:</p>
<p>$$ g'(x) = \left( h'(\mathrm{e}^x) \mathrm{e}^x \right) \mathrm{e}^{h(x)} + \left( h(\mathrm{e}^x) \right) \left( \mathrm{e}^{h(x)} h'(x) \right) $$</p>
<p>Next, we need to evaluate this at $$x = 0$$:</p>
<p>First, we know that:</p>
<p>$$ h(0) = 0 $$</p>
<p>$$ h(1) = 1 $$</p>
<p>$$ h'(0) = 2 $$</p>
<p>$$ h'(1) = 2 $$</p>
<p>Substituting $$x = 0$$ into the expressions, we get:</p>
<p>$$ u(0) = h(\mathrm{e}^0) = h(1) = 1 $$</p>
<p>$$ v(0) = \mathrm{e}^{h(0)} = \mathrm{e}^0 = 1 $$</p>
<p>$$ u'(0) = h'(\mathrm{e}^0) \mathrm{e}^0 = h'(1) \cdot 1 = 2 $$</p>
<p>$$ v'(0) = \mathrm{e}^{h(0)} h'(0) = \mathrm{e}^0 \cdot 2 = 2 $$</p>
<p>Therefore, evaluating $$g'(0)$$:</p>
<p>$$ g'(0) = \left( u'(0) v(0) \right) + \left( u(0) v'(0) \right) $$</p>
<p>$$ g'(0) = \left( 2 \cdot 1 \right) + \left( 1 \cdot 2 \right) $$</p>
<p>$$ g'(0) = 2 + 2 = 4 $$</p>
<p>Thus, the value of $$g^{\prime}(0)$$ is 4, which corresponds to Option A.</p>
<p>The correct answer is <strong>Option A: 4</strong>.</p> | mcq | jee-main-2024-online-6th-april-evening-shift |
W9kJiVVFBaQQJEim | maths | differentiation | differentiation-of-implicit-function | If $$y = {\left( {x + \sqrt {1 + {x^2}} } \right)^n},$$ then $$\left( {1 + {x^2}} \right){{{d^2}y} \over {d{x^2}}} + x{{dy} \over {dx}}$$ is | [{"identifier": "A", "content": "$${n^2}y$$ "}, {"identifier": "B", "content": "$$-{n^2}y$$"}, {"identifier": "C", "content": "$$-y$$ "}, {"identifier": "D", "content": "$$2{x^2}y$$ "}] | ["A"] | null | $$y = {\left( {x + \sqrt {1 + {x^2}} } \right)^n}$$
<br><br>$${{dy} \over {dx}} = n{\left( {x + \sqrt {1 + {x^2}} } \right)^{n - 1}}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {1 + {1 \over 2}{{\left( {1 + {x^2}} \right)}^{ - 1/2}}.2x} \right);$$
<br><br>$${{dy} \over {dx}} = n{\left( {x + \sqrt {1 + {x^2}} } \right)^{n - 1}}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\left( {\sqrt {1 + {x^2}} + x} \right)} \over {\sqrt {1 + {x^2}} }}$$
<br><br>$$ = {{n{{\left( {\sqrt {1 + {x^2}} + x} \right)}^n}} \over {\sqrt {1 + {x^2}} }}$$
<br><br>or $$\sqrt {1 + {x^2}} {{dy} \over {dx}} = ny$$
<br><br>or $$\sqrt {1 + {x^2}} {y_1} = ny$$
<br><br>$$\left( {{y_1} = {{dy} \over {dx}}} \right)$$
<br><br>Squaring, $$\left( {1 + {x^2}} \right){y_1}^2 = {n^2}{y^2}$$
<br><br>Differentiating, $$\left( {1 + {x^2}} \right)2{y_1}{y_2} + {y_1}^2.2x$$
<br><br>$$ = {n^2}.2y{y_1}$$
<br><br>or $$\left( {1 + {x^2}} \right){y_2} + x{y_1} = {n^2}y$$
| mcq | aieee-2002 |
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