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o4KkVYuBGGgiJsxAqi3rsa0w2w9jx2evlbl | maths | application-of-derivatives | tangent-and-normal | If the tangent to the curve $$y = {x \over {{x^2} - 3}}$$
, $$x \in \rho ,\left( {x \ne \pm \sqrt 3 } \right)$$, at a point ($$\alpha $$, $$\beta $$) $$ \ne $$ (0, 0) on it is parallel to the line
2x + 6y β 11 = 0, then : | [{"identifier": "A", "content": "| 6$$\\alpha $$ + 2$$\\beta $$ | = 9"}, {"identifier": "B", "content": "| 2$$\\alpha $$ + 6$$\\beta $$ | = 11"}, {"identifier": "C", "content": "| 2$$\\alpha $$ + 6$$\\beta $$ | = 19"}, {"identifier": "D", "content": "| 6$$\\alpha $$ + 2$$\\beta $$ | = 19"}] | ["D"] | null | $${{dy} \over {dx}}{|_{(\alpha ,\beta )}} = {{ - {\alpha ^2} - 3} \over {{{\left( {{\alpha ^2} - 3} \right)}^2}}}$$<br><br>
Given that $${{ - {\alpha ^2} - 3} \over {{{\left( {{\alpha ^2} - 3} \right)}^2}}} = - {1 \over 3}$$<br><br>
$$ \Rightarrow $$ $$\alpha $$ = 0, $$ \pm $$ 3 ($$\alpha \ne $$ 0) | mcq | jee-main-2019-online-10th-april-evening-slot |
FFjlUVpcYahbjG67Tj18hoxe66ijvwp8sn0 | maths | application-of-derivatives | tangent-and-normal | If the tangent to the curve, y = x<sup>3</sup> + ax β b at
the point (1, β5) is perpendicular to the line,
βx + y + 4 = 0, then which one of the following
points lies on the curve ? | [{"identifier": "A", "content": "(2, \u20132)"}, {"identifier": "B", "content": "(2, \u20131)"}, {"identifier": "C", "content": "(\u20132, 2)"}, {"identifier": "D", "content": "(\u20132, 1)"}] | ["A"] | null | Slope of the tangent to the curve y = x<sup>3</sup> + ax β b at point (1, β5)
<br><br>m<sub>1</sub> = $${\left. {{{dy} \over {dx}}} \right|_{\left( {1, - 5} \right)}}$$ = 3x<sup>2</sup> + a = 3 + a
<br><br>Slope of the line βx + y + 4 = 0,
<br><br>m<sub>2</sub> = 1
<br><br>As line and tangent to the curve are perpendicular to each other,
<br><br>$$ \therefore $$ m<sub>1</sub> $$ \times $$ m<sub>2</sub> = -1
<br><br>$$ \Rightarrow $$ (3 + a) $$ \times $$ 1 = -1
<br><br>$$ \Rightarrow $$ a = - 4
<br><br>$$ \therefore $$ Curve becomes y = x<sup>3</sup> - 4x β b
<br><br>This curve goes through (1, β5)
<br><br>$$ \therefore $$ -5 = 1 - 4 - b
<br><br>$$ \Rightarrow $$ b = 2
<br><br>So curve is y = x<sup>3</sup> - 4x β 2
<br><br>By checking each options you can see,
<br><br>(2, β2) lies on the curve. | mcq | jee-main-2019-online-9th-april-morning-slot |
H0eqMDxBBYA939FXRvxgg | maths | application-of-derivatives | tangent-and-normal | Let S be the set of all values of x for which the
tangent to the curve
<br/>y = Ζ(x) = x<sup>3</sup> β x<sup>2</sup> β 2x at
(x, y) is parallel to the line segment joining the
points (1, Ζ(1)) and (β1, Ζ(β1)), then S is equal
to : | [{"identifier": "A", "content": "$$\\left\\{ { {1 \\over 3}, - 1} \\right\\}$$"}, {"identifier": "B", "content": "$$\\left\\{ { - {1 \\over 3}, 1} \\right\\}$$"}, {"identifier": "C", "content": "$$\\left\\{ { - {1 \\over 3}, - 1} \\right\\}$$"}, {"identifier": "D", "content": "$$\\left\\{ { {1 \\over 3}, 1} \\right\\}$$"}] | ["B"] | null | Given Ζ(x) = x<sup>3</sup> β x<sup>2</sup> β 2x
<br><br>$$ \therefore $$ Ζ(1) = 1 β 1 β 2 = - 2
<br><br>and Ζ(-1) = -1 β 1 + 2 = 0
<br><br>So point A(1, Ζ(1)) = (1, -2)
<br><br>and point B(β1, Ζ(β1)) = (-1, 0)
<br><br>Slope of tangent at point (x, y) to the curve
<br><br> y = Ζ(x) = x<sup>3</sup> β x<sup>2</sup> β 2x is
<br><br>$${{dy} \over {dx}}$$ = 3x<sup>2</sup> - 2x - 2
<br><br>Slope of line segment joining the
points (1, -2) and (β1, 0) is
<br><br>= $${{ - 2 - 0} \over {1 - \left( { - 1} \right)}}$$
<br><br>As tangent to the curve and line segment both are parallel then slope of them are same.
<br><br>$$ \therefore $$ 3x<sup>2</sup> - 2x - 2 = $${{ - 2 - 0} \over {1 - \left( { - 1} \right)}}$$
<br><br>$$ \Rightarrow $$ 3x<sup>2</sup> - 2x - 1 = 0
<br><br>$$ \Rightarrow $$ x = $${{2 \pm \sqrt {4 - 4.3\left( { - 1} \right)} } \over {2.3}}$$
<br><br>= $${{2 \pm 4} \over 6}$$
<br><br>$$ \therefore $$ x = 1, $$ - {1 \over 3}$$
<br><br>So, S = $$\left\{ { - {1 \over 3}, 1} \right\}$$ | mcq | jee-main-2019-online-9th-april-morning-slot |
mV9YspGfsF6PJhUiFQYou | maths | application-of-derivatives | tangent-and-normal | Given that the slope of the tangent to a curve y
= y(x) at any point (x, y) is
$$2y \over x^2$$. If the curve passes through the centre of the circle x<sup>2</sup> + y<sup>2</sup> β 2x β 2y = 0, then its equation is : | [{"identifier": "A", "content": "x log<sub>e</sub>|y| = 2(x \u2013 1)"}, {"identifier": "B", "content": "x<sup>2</sup> log<sub>e</sub>|y| = \u20132(x \u2013 1)"}, {"identifier": "C", "content": "x log<sub>e</sub>|y| = x \u2013 1"}, {"identifier": "D", "content": "x log<sub>e</sub>|y| = \u20132(x \u2013 1)"}] | ["A"] | null | Slope, $${{dy} \over {dx}}$$ = $$2y \over x^2$$
<br><br>$$ \Rightarrow $$ $$\int {{{dy} \over y}} = \int {2{{dx} \over {{x^2}}}} $$
<br><br>$$ \Rightarrow $$ $${\log _e}|y| = - {2 \over x} + C$$ ....... (1)
<br><br>Center of the circle x<sup>2</sup> + y<sup>2</sup> β 2x β 2y = 0 is (1, 1)
<br><br>Equation (1) passes through point (1, 1)
<br><br>$$ \therefore $$ 0 = -2 + C
<br><br>$$ \Rightarrow $$ C = 2
<br><br>$$ \therefore $$ $${\log _e}|y| = - {2 \over x} + 2$$
<br><br>$$ \Rightarrow $$ x$${\log _e}|y| = - 2 + 2x$$
<br><br>$$ \Rightarrow $$ x$${\log _e}|y| = 2(x - 1)$$ | mcq | jee-main-2019-online-8th-april-evening-slot |
KpnzSmbznFMg3YBKF4Sio | maths | application-of-derivatives | tangent-and-normal | The tangent to the curve, y = xe<sup>x<sup>2</sup></sup> passing through the point (1, e) also passes through the point | [{"identifier": "A", "content": "$$\\left( {{4 \\over 3},2e} \\right)$$"}, {"identifier": "B", "content": "(3, 6e)"}, {"identifier": "C", "content": "(2, 3e)"}, {"identifier": "D", "content": "$$\\left( {{5 \\over 3},2e} \\right)$$"}] | ["A"] | null | y = xe<sup>x<sup>2</sup></sup>
<br><br>$${\left. {{{dy} \over {dx}}} \right|_{(1,e)}}{\left. { = \left( {e.e{x^2}.2x + {e^{{x^2}}}} \right)} \right|_{(1,e)}}$$
<br><br>$$ = 2 \cdot e + e = 3e$$
<br><br>T : y $$-$$ e = 3e (x $$-$$ 1)
<br><br>y = 3ex $$-$$ 3e + e
<br><br>y = $$\left( {3e} \right)x - 2e$$
<br><br>$$\left( {{4 \over 3},2e} \right)$$ lies on it | mcq | jee-main-2019-online-10th-january-evening-slot |
LUrxBmMKm8jszpS1sgjgy2xukg38q7pr | maths | application-of-derivatives | tangent-and-normal | If the tangent to the curve, y = f (x) = xlog<sub>e</sub> x,
<br/>(x > 0) at a point (c, f(c)) is parallel to the
line-segment<br/> joining the points (1, 0) and
(e, e), then c is equal to : | [{"identifier": "A", "content": "$${{e - 1} \\over e}$$"}, {"identifier": "B", "content": "$${e^{\\left( {{1 \\over {1 - e}}} \\right)}}$$"}, {"identifier": "C", "content": "$${e^{\\left( {{1 \\over {e - 1}}} \\right)}}$$"}, {"identifier": "D", "content": "$${1 \\over {e - 1}}$$"}] | ["C"] | null | y = f (x) = xlog<sub>e</sub> x
<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}} = $$ 1 + log<sub>e</sub> x
<br><br>$$ \Rightarrow $$ $${\left. {{{dy} \over {dx}}} \right|_{\left( {c,f\left( c \right)} \right)}}$$ = 1 + log<sub>e</sub> e = m<sub>1</sub>
<br><br>This tangent parallel to the
line-segment<br> joining the points (1, 0) and
(e, e).
<br><br>$$ \therefore $$ Slope of line-segment joining the points (1, 0) and
(e, e) = m<sub>1</sub>
<br><br>$$ \Rightarrow $$ $${{e - 0} \over {e - 1}}$$ = 1 + log<sub>e</sub> e
<br><br>$$ \Rightarrow $$ log<sub>e</sub> e = $${{e - 0} \over {e - 1}}$$ - 1 = $${1 \over {e - 1}}$$
<br><br>$$ \Rightarrow $$ c = $${e^{\left( {{1 \over {e - 1}}} \right)}}$$ | mcq | jee-main-2020-online-6th-september-evening-slot |
EkDkOZ7NbLhXskYcSejgy2xukfqfuuc9 | maths | application-of-derivatives | tangent-and-normal | If the lines x + y = a and x β y = b touch the
<br/>curve y = x<sup>2</sup>
β 3x + 2 at the points where the
curve intersects the x-axis, then $${a \over b}$$ is equal
to _______. | [] | null | 0.50 | y = x<sup>2</sup>
β 3x + 2
<br><br>$$ \Rightarrow $$ y = (x β 1)(x β 2)
<br><br>At x-axis y = 0
<br>$$ \Rightarrow $$ x = 1, 2
<br><br>So this curve intersects the x-axis
at A(1, 0) and B(2, 0).
<br><br>$${{dy} \over {dx}} = 2x - 3$$
<br><br>$${\left( {{{dy} \over {dx}}} \right)_{x = 1}} = - 1$$ and $${\left( {{{dy} \over {dx}}} \right)_{x = 2}} = 1$$
<br><br>Equation of tangent at A(1, 0) :
<br><br>y = β1(x β1)
<br><br>$$ \Rightarrow $$ x + y = 1
<br><br>and equation of tangent at B(2, 0):
<br><br>y = 1(x β 2)
<br><br>$$ \Rightarrow $$ x β y = 2
<br><br>So a = 1 and b = 2
<br><br>$$ \Rightarrow $$ $${a \over b}$$ = 0.5 | integer | jee-main-2020-online-5th-september-evening-slot |
pxlhjCCkd0Ohos8TCwjgy2xukfqb1rdv | maths | application-of-derivatives | tangent-and-normal | Which of the following points lies on the
tangent to the curve
<br/><br>x<sup>4</sup>e<sup>y</sup> + 2$$\sqrt {y + 1} $$ = 3 at the
point (1, 0)?</br> | [{"identifier": "A", "content": "(2, 2)"}, {"identifier": "B", "content": "(\u20132, 4)"}, {"identifier": "C", "content": "(2, 6)"}, {"identifier": "D", "content": "(\u20132, 6)"}] | ["D"] | null | x<sup>4</sup>e<sup>y</sup> + 2$$\sqrt {y + 1} $$ = 3
<br><br>Differentiating w.r.t. x, we get
<br><br>x<sup>4</sup>e<sup>y</sup>y' + e<sup>y</sup>4x<sup>3</sup> + $${{2y'} \over {2\sqrt {y + 1} }}$$ = 0
<br><br>At P(1, 0)
<br><br>$${y{'_P}}$$ + 4 + $${y{'_P}}$$ = 0
<br><br>$$ \Rightarrow $$ $${y{'_P}}$$ = -2
<br><br>Tangent at P(1, 0) is
<br><br>y β 0 = β 2 (x β 1)
<br><br>2x + y = 2
<br><br>Only (β2, 6) lies on it. | mcq | jee-main-2020-online-5th-september-evening-slot |
VYqe8IbBOhn1VI4F28jgy2xukewr3bph | maths | application-of-derivatives | tangent-and-normal | If the tangent to the curve y = x + sin y at a point
<br/>(a, b) is parallel to the line joining $$\left( {0,{3 \over 2}} \right)$$ and $$\left( {{1 \over 2},2} \right)$$, then : | [{"identifier": "A", "content": "b = a"}, {"identifier": "B", "content": "|b - a| = 1"}, {"identifier": "C", "content": "$$b = {\\pi \\over 2}$$ + a"}, {"identifier": "D", "content": "|a + b| = 1"}] | ["B"] | null | Slope of tangent to the curve y = x + siny
<br><br>at (a, b) is = $${{2 - {3 \over 2}} \over {{1 \over 2} - 0}}$$ = 1
<br><br>Given curve y = x + sin y
<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}} = 1 + \cos y.{{dy} \over {dx}}$$
<br><br>$$ \Rightarrow $$ (1 - cos y)$${{dy} \over {dx}}$$ = 1
<br><br>$$ \Rightarrow $$ $${\left. {{{dy} \over {dx}}} \right|_{\left( {a,b} \right)}}$$ = $${1 \over {1 + \cos b}}$$
<br><br>Now according to question, $${1 \over {1 + \cos b}} = 1$$
<br><br>$$ \Rightarrow $$ cos b = 0
<br><br>$$ \Rightarrow $$ sin b = $$ \pm $$ 1
<br><br>Point (a, b) lies on curve y = x + sin y
<br><br>$$ \therefore $$ b = a + sin b
<br><br>$$ \Rightarrow $$ |b - a| = |sin b| = 1 | mcq | jee-main-2020-online-2nd-september-morning-slot |
wcSiqoBtSVPf2VGnBMjgy2xukewra473 | maths | application-of-derivatives | tangent-and-normal | Let P(h, k) be a point on the curve
<br/>y = x<sup>2</sup>
+ 7x + 2, nearest to the line, y = 3x β 3.
<br/>Then the equation of the normal to the curve at
P is : | [{"identifier": "A", "content": "x \u2013 3y \u2013 11 = 0"}, {"identifier": "B", "content": "x \u2013 3y + 22 = 0"}, {"identifier": "C", "content": "x + 3y \u2013 62 = 0"}, {"identifier": "D", "content": "x + 3y + 26 = 0"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266899/exam_images/knvyxnlyn6etgfelas4i.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 2nd September Morning Slot Mathematics - Application of Derivatives Question 126 English Explanation">
<br><br>Let L be the common normal to parabola
<br><br>y = x<sup>2</sup>
+ 7x + 2 and line y = 3x β 3
<br><br>Slope of tangent of y = x<sup>2</sup>
+ 7x + 2 at P
<br><br>$${{{\left. {{{dy} \over {dx}}} \right|}_{P}}}$$ = 3
<br><br>$$ \Rightarrow $$ 2x + 7 = 3
<br><br>$$ \Rightarrow $$ x = -2
<br><br>$$ \therefore $$ y = -8
<br><br>So P(β2, β8)
<br><br>Normal at P : x + 3y + C = 0
<br><br>$$ \Rightarrow $$ -2 + 3(-8) + C = 0
<br><br>$$ \Rightarrow $$ C = 26
<br><br>$$ \therefore $$ Normal : x + 3y + 26 = 0 | mcq | jee-main-2020-online-2nd-september-morning-slot |
lwjOzJrH6jcRF5DUMS7k9k2k5hii08h | maths | application-of-derivatives | tangent-and-normal | The length of the perpendicular from the origin,
on the normal to the curve,<br/> x<sup>2</sup> + 2xy β 3y<sup>2</sup> = 0
at the point (2,2) is | [{"identifier": "A", "content": "$$\\sqrt 2 $$"}, {"identifier": "B", "content": "$$4\\sqrt 2 $$"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$$2\\sqrt 2 $$"}] | ["D"] | null | x<sup>2</sup> + 2xy β 3y<sup>2</sup> = 0
<br><br>Differentiate the curve
<br><br> 2x + 2y + 2xy' β 6yy' = 0
<br><br>$$ \Rightarrow $$ $$ \Rightarrow $$ x + y + xy' β 3yy' = 0
<br><br>$$ \Rightarrow $$ y'(x β 3y) = β (x + y)
<br><br>$$ \Rightarrow $$ y' = $${{x + y} \over {3y - x}}$$
<br><br>Slope of normal = $$ - {{dx} \over {dy}}$$ = $${{x - 3y} \over {x + y}}$$
<br><br>$$ \therefore $$ $${\left( { - {{dx} \over {dy}}} \right)_{\left( {2,2} \right)}}$$ = $${{2 - 6} \over {2 + 2}}$$ = -1
<br><br> Normal at (2, 2)
<br><br>y β 2 = β 1 (x β 2)
<br><br>$$ \Rightarrow $$ y + x = 4
<br><br>$$ \therefore $$ Perpendicular distance from (0,0)
<br><br>= $$\left| {{{0 + 0 - 4} \over {\sqrt 2 }}} \right|$$ = $${2\sqrt 2 }$$ | mcq | jee-main-2020-online-8th-january-evening-slot |
NLUumf2kXLiqyI6O3a7k9k2k5h0679u | maths | application-of-derivatives | tangent-and-normal | Let the normal at a point P on the curve
<br/>y<sup>2</sup> β 3x<sup>2</sup> + y + 10 = 0 intersect the y-axis at $$\left( {0,{3 \over 2}} \right)$$
. <br/>If m is the slope of the tangent at P to
the curve, then |m| is equal to | [] | null | 4 | Given curve : y<sup>2</sup>
β 3x<sup>2</sup>
+ y + 10 = 0
<br><br>$$ \Rightarrow $$ 2y$${{dy} \over {dx}}$$ - 6x + $${{dy} \over {dx}}$$ = 0
<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}}$$ = $${{6x} \over {2y + 1}}$$
<br><br>Let P be (x<sub>1</sub>, y<sub>1</sub>)
<br><br>Slope of tangent at P = $${{6{x_1}} \over {2{y_1} + 1}}$$
<br><br>$$ \therefore $$ Slope of normal at P = $$ - {{2{y_1} + 1} \over {6{x_1}}}$$
<br><br>$$ \Rightarrow $$ Equation of normal (y β y<sub>1</sub>) = $$ - \left( {{{2{y_1} + 1} \over {6{x_1}}}} \right)$$(x β x<sub>1</sub>)
<br><br>This normal passes through point $$\left( {0,{3 \over 2}} \right)$$.
<br><br>$$ \therefore $$ ($${{3 \over 2}}$$ β y<sub>1</sub>) = $$ - \left( {{{2{y_1} + 1} \over {6{x_1}}}} \right)$$(0 β x<sub>1</sub>)
<br><br>$$ \Rightarrow $$ y<sub>1</sub> = 1
<br><br>Put y<sub>1</sub> = 1 in equation of curve , then we get x<sub>1</sub> = $$ \pm $$2
<br><br>$$ \Rightarrow $$ |m| = slope of tangent = $$\left| {{{6{x_1}} \over {2{y_1} + 1}}} \right|$$ = $${{12} \over 3}$$ = 4 | integer | jee-main-2020-online-8th-january-morning-slot |
dDfWIivwafjQLW2Uenjgy2xukezarnpo | maths | application-of-derivatives | tangent-and-normal | The equation of the normal to the curve
<br/>y = (1+x)<sup>2y</sup> + cos
<sup>2</sup>(sin<sup>β1</sup>x) at x = 0 is : | [{"identifier": "A", "content": "y = 4x + 2"}, {"identifier": "B", "content": "x + 4y = 8"}, {"identifier": "C", "content": "y + 4x = 2"}, {"identifier": "D", "content": "2y + x = 4"}] | ["B"] | null | Given equation of curve
<br><br>y = (1+x)<sup>2y</sup> + cos
<sup>2</sup>(sin<sup>β1</sup>x)
<br><br>at x = 0
<br><br>$$ \Rightarrow $$ y = (1 + 0)<sup>2y</sup> + cos<sup>2</sup>(sin<sup>β1</sup>0)
<br><br>$$ \Rightarrow $$ y = 1 + 1
<br><br>$$ \Rightarrow $$ y = 2
<br><br>So we have to find the normal at (0, 2)
<br><br>Now, y = $${e^{2y\ln \left( {1 + x} \right)}} + {\cos ^2}\left( {{{\cos }^{ - 1}}\sqrt {1 - {x^2}} } \right)$$
<br><br>$$ \Rightarrow $$ y = $${e^{2y\ln \left( {1 + x} \right)}} + {\left( {\sqrt {1 - {x^2}} } \right)^2}$$
<br><br>$$ \Rightarrow $$ y = $${e^{2y\ln \left( {1 + x} \right)}} + \left( {1 - {x^2}} \right)$$
<br><br>Now differentiate w.r.t. x
<br><br>y' = $${e^{2y\ln \left( {1 + x} \right)}}\left[ {2y.\left( {{1 \over {1 + x}}} \right) + \ln \left( {1 + x} \right).2y'} \right]$$ - 2x
<br><br>Put x = 0 & y = 2
<br><br>$$ \Rightarrow $$ y' = $${e^{2y\ln \left( {1 + 0} \right)}}\left[ {2y.\left( {{1 \over {1 + 0}}} \right) + \ln \left( {1 + 0} \right).2y'} \right] - 2 \times 0$$
<br><br>$$ \Rightarrow $$ y' = e<sup>0</sup>
[4 + 0] β 0
<br><br>$$ \Rightarrow $$ y' = 4 = slope of tangent to the curve
<br><br>so slope of normal to the curve = - $${1 \over 4}$$
<br><br>Hence equation of normal at (0, 2) is
<br><br>y - 2 = - $${1 \over 4}$$(x - 0)
<br><br>$$ \Rightarrow $$ 4y β 8 = βx
<br><br>$$ \Rightarrow $$ x + 4y = 8 | mcq | jee-main-2020-online-2nd-september-evening-slot |
CkjU7HfpXnSIHE1AlR1klrgv24w | maths | application-of-derivatives | tangent-and-normal | If the tangent to the curve y = x<sup>3</sup> at the point P(t, t<sup>3</sup>) meets the curve again at Q, then the
ordinate of the point which divides PQ internally in the ratio 1 : 2 is : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "2t<sup>3</sup>"}, {"identifier": "C", "content": "-2t<sup>3</sup>"}, {"identifier": "D", "content": "-t<sup>3</sup>"}] | ["C"] | null | Given $$P(t,{t^3})$$<br><br>Let $$Q = ({t_1},t_1^3)$$<br><br>Slope of tangent at point p,<br><br>$${{dy} \over {dx}} = 3{x^2}$$<br><br>$$ \Rightarrow {\left. {{{dy} \over {dx}}} \right|_{(t,{t^3})}} = 3{t^2}$$<br><br>This slope is same as slope of line PQ.<br><br>Slope of $$PQ = {{t_1^3 - {t_3}} \over {{t_1} - t}}$$<br><br>$$ \therefore $$ $$3{t^2} = {{t_1^3 - {t_3}} \over {{t_1} - t}}$$<br><br>$$ \Rightarrow 3{t^2} = {{({t_1} - t)(t_1^2 + t{t_1} + {t^2})} \over {({t_1} - t)}}$$<br><br>$$ \Rightarrow 3{t^2} = t_1^2 + t{t_1} + {t^2}$$<br><br>$$ \Rightarrow {t_1} = - 2t$$<br><br>$$ \therefore $$ $$Q = \left( { - 2t, - 8{t^3}} \right)$$<br><br> <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267564/exam_images/huubftuoxasm3m2dm3kb.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266682/exam_images/hsrvremohnbzfysgfkae.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264701/exam_images/bxobuhrfgjuubftwrdbd.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267517/exam_images/ag9qavtioafcjzdlux7c.webp"><source media="(max-width: 1040px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266835/exam_images/ptsbfziibtdj9or4d3np.webp"><source media="(max-width: 1220px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264817/exam_images/wg2xdxagrn8ws14zj2nf.webp"><source media="(max-width: 1400px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267691/exam_images/zneg1c8tyvb5cijvffeg.webp"><source media="(max-width: 1580px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263359/exam_images/olfozptjx4ikypmsibro.webp"><source media="(max-width: 1760px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263798/exam_images/pov6jmo6ye6eriwulf1y.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267203/exam_images/eyxgnumsxyaujoadjbfq.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 24th February Morning Shift Mathematics - Application of Derivatives Question 110 English Explanation"></picture> <br><br>$$ \therefore $$ $$h = {{2t - 2t} \over 3} = 0$$<br><br>$$k = {{2{t^3} - 8{t^3}} \over 3} = - 2{t^3}$$<br><br>$$ \therefore $$ Point $$M = (0, - 2{t^3})$$ | mcq | jee-main-2021-online-24th-february-morning-slot |
WIZrag7XacI8iKDxjV1klrk8szp | maths | application-of-derivatives | tangent-and-normal | For which of the following curves, the line $$x + \sqrt 3 y = 2\sqrt 3 $$ is the tangent at the point $$\left( {{{3\sqrt 3 } \over 2},{1 \over 2}} \right)$$? | [{"identifier": "A", "content": "$$2{x^2} - 18{y^2} = 9$$"}, {"identifier": "B", "content": "$${y^2} = {1 \\over {6\\sqrt 3 }}x$$"}, {"identifier": "C", "content": "$${x^2} + 9{y^2} = 9$$"}, {"identifier": "D", "content": "$${x^2} + {y^2} = 7$$"}] | ["C"] | null | Tangent to $${x^2} + 9{y^2} = 9$$ at
<br><br>point $$\left( {{{3\sqrt 3 } \over 2},{1 \over 2}} \right)$$ is $$x\left( {{{3\sqrt 3 } \over 2}} \right) + 9y\left( {{1 \over 2}} \right) = 9$$<br><br>$$3\sqrt 3 x + 9y = 18 \Rightarrow x + \sqrt 3 y = 2\sqrt 3 $$<br><br>$$ \Rightarrow $$ option (1) is true. | mcq | jee-main-2021-online-24th-february-evening-slot |
F7Rio2iCpsDwUvu6La1klrmblum | maths | application-of-derivatives | tangent-and-normal | If the curve y = ax<sup>2</sup> + bx + c, x$$ \in $$R, passes through the point (1, 2) and the tangent line to this curve at origin is y = x, then the possible values of a, b, c are : | [{"identifier": "A", "content": "a = $$-$$ 1, b = 1, c = 1"}, {"identifier": "B", "content": "a = 1, b = 1, c = 0"}, {"identifier": "C", "content": "a = $${1 \\over 2}$$, b = $${1 \\over 2}$$, c = 1"}, {"identifier": "D", "content": "a = 1, b = 0, c = 1"}] | ["B"] | null | Given curve y = ax<sup>2</sup> + bx + c, x$$ \in $$R
<br><br>This curve passes through the point (1, 2)
<br><br>$$ \therefore $$ $$2 = a + b + c$$ ..... (i)
<br><br>Given, slope of tangent at origin is 1
<br><br>$$ \therefore $$ $${{dy} \over {dx}} = 2ax + b \Rightarrow {\left. {{{dy} \over {dx}}} \right|_{(0,0)}} = 1$$<br><br>$$ \Rightarrow b = 1 \Rightarrow a + c = 1$$<br><br>(0, 0) lie on curve<br><br>$$ \therefore $$ c = 0, a = 1 | mcq | jee-main-2021-online-24th-february-evening-slot |
s45VVInIfdRU4FBS1N1kls4lvsb | maths | application-of-derivatives | tangent-and-normal | If the curves, $${{{x^2}} \over a} + {{{y^2}} \over b} = 1$$ and $${{{x^2}} \over c} + {{{y^2}} \over d} = 1$$ intersect each other at an angle of 90$$^\circ$$, then which of the following relations is TRUE? | [{"identifier": "A", "content": "a $$-$$ c = b + d"}, {"identifier": "B", "content": "a + b = c + d "}, {"identifier": "C", "content": "$$ab = {{c + d} \\over {a + b}}$$"}, {"identifier": "D", "content": "a $$-$$ b = c $$-$$ d"}] | ["D"] | null | $${{{x^2}} \over a} + {{{y^2}} \over b} = 1$$ ..........(1)<br><br>Differentiating both sides :
<br><br>$${{2x} \over a} + {{2y} \over b}{{dy} \over {dx}} = 0 \Rightarrow {y \over b}{{dy} \over {dx}} = {{ - x} \over a}$$<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}} = {{ - bx} \over {ay}}$$ ............(2)<br><br>$${{{x^2}} \over c} + {{{y^2}} \over d} = 1$$ ........(3)<br><br>Differentiating both sides : $${{dy} \over {dx}} = {{ - dx} \over {cy}}$$ ...........(4)<br><br>$${m_1}{m_2} = - 1 \Rightarrow {{ - bx} \over {ay}} \times {{ - dx} \over {cy}} = - 1$$<br><br>$$ \Rightarrow bd{x^2} = - ac{y^2}$$ .............(5)<br><br>$$(1) - (3) \Rightarrow \left( {{1 \over a} - {1 \over c}} \right){x^2} + \left( {{1 \over b} - {1 \over d}} \right){y^2} = 0$$<br><br>$$ \Rightarrow {{c - a} \over {ac}}{x^2} + {{d - b} \over {bd}} \times \left( {{{ - bd} \over {ac}}} \right){x^2} = 0$$ (using 5)<br><br>$$ \Rightarrow (c - a) - (d - b) = 0$$<br><br>$$ \Rightarrow c - a = d - b$$<br><br>$$ \Rightarrow c - d = a - b$$ | mcq | jee-main-2021-online-25th-february-morning-slot |
Qvcb8eUQ8LwGYFPJYw1klta2a04 | maths | application-of-derivatives | tangent-and-normal | If the curves x = y<sup>4</sup> and xy = k cut at right angles, then (4k)<sup>6</sup> is equal to __________. | [] | null | 4 | $$x = {y^4}$$ and $$xy = k$$<br><br>for intersection $${y^5} = k$$ ..... (1)<br><br>Also $$x = {y^4}$$ <br><br>$$ \Rightarrow 1 = 4{y^3}{{dy} \over {dx}} \Rightarrow {{dy} \over {dx}} = {1 \over {4{y^3}}}$$<br><br>for $$xy = k \Rightarrow x = {k \over y}$$<br><br>$$ \Rightarrow 1 = - {k \over {{y^2}}}.{{dy} \over {dx}}$$<br><br>$$ \Rightarrow {{dy} \over {dx}} = {{ - {y^2}} \over k}$$<br><br>$$ \because $$ Curve cut orthogonally<br><br>$$ \Rightarrow {1 \over {4{y^3}}} \times \left( {{{ - {y^2}} \over k}} \right) = - 1$$<br><br>$$ \Rightarrow y = {1 \over {4k}}$$<br><br>$$ \therefore $$ from (1), $${y^5} = k$$<br><br>$$ \Rightarrow {1 \over {{{(4k)}^5}}} = k$$<br><br>$$ \Rightarrow 4 = {(4k)^6}$$ | integer | jee-main-2021-online-25th-february-evening-slot |
onv7vutPC56pWmNGtD1kluxctnr | maths | application-of-derivatives | tangent-and-normal | Let slope of the tangent line to a curve at any point P(x, y) be given by $${{x{y^2} + y} \over x}$$. If the curve intersects the line x + 2y = 4 at x = $$-$$2, then the value of y, for which the point (3, y) lies on the curve, is : | [{"identifier": "A", "content": "$$ - {{18} \\over {19}}$$"}, {"identifier": "B", "content": "$$ - {{4} \\over {3}}$$"}, {"identifier": "C", "content": "$${{18} \\over {35}}$$"}, {"identifier": "D", "content": "$$ - {{18} \\over {11}}$$"}] | ["A"] | null | $${{dy} \over {dx}} = {{x{y^2} + y} \over x}$$<br><br>$$ \Rightarrow {{xdy - ydx} \over {{y^2}}} = xdx$$<br>$$ \Rightarrow - d\left( {{x \over y}} \right) = d\left( {{{{x^2}} \over 2}} \right)$$<br><br>$$ \Rightarrow {{ - x} \over y} = {{{x^2}} \over 2} + C$$<br><br>Curve intersect the line x + 2y = 4 at x = $$-$$ 2<br><br>So, $$-$$ 2 + 2y = 4 $$ \Rightarrow $$ y = 3<br><br>So the curve passes through ($$-$$2, 3)<br><br>$$ \Rightarrow {2 \over 3} = 2 + C$$<br><br>$$ \Rightarrow C = {{ - 4} \over 3}$$<br><br>$$ \therefore $$ curve is $${{ - x} \over y} = {{{x^2}} \over 2} - {4 \over 3}$$<br><br>It also passes through (3, y)<br><br>$${{ - 3} \over y} = {9 \over 2} - {4 \over 3}$$<br><br>$$ \Rightarrow {{ - 3} \over y} = {{19} \over 6}$$<br><br>$$ \Rightarrow y = - {{18} \over {19}}$$ | mcq | jee-main-2021-online-26th-february-evening-slot |
kZ58r1wbH4gdw8wrHl1kluyhkmg | maths | application-of-derivatives | tangent-and-normal | Let the normals at all the points on a given curve pass through a fixed point (a, b). If the curve passes through (3, $$-$$3) and (4, $$-$$2$$\sqrt 2 $$), and given that a $$-$$ 2$$\sqrt 2 $$ b = 3, <br/>then (a<sup>2</sup> + b<sup>2</sup> + ab) is equal to __________. | [] | null | 9 | Let the equation of normal is Y $$-$$ y = $$-$$$${1 \over m}(X - x)$$, where, m = $${{dy} \over {dx}}$$<br><br>As it passes through (a, b)<br><br>$$b - y = - {1 \over m}(a - x) = - {{dx} \over {dy}}(a - x)$$<br><br>$$ \Rightarrow (b - y)dy = (x - a)dx$$<br><br>by $$ - {{{y^2}} \over 2} = {{{x^2}} \over 2} - ax + c$$ ..... (i)<br><br>It passes through (3, $$-$$3) & (4, $$-$$2$$\sqrt 2 $$)<br><br>$$ \therefore $$ $$ - 3b - {9 \over 2} = {9 \over 2} - 3a + c$$<br><br>$$ \Rightarrow - 6b - 9 = 9 - 6a + 2c$$<br><br>$$ \Rightarrow 6a - 6b - 2c = 18$$<br><br>$$ \Rightarrow 3a - 3b - c = 9$$ .... (ii)<br><br>Also, <br><br>$$ - 2\sqrt 2 b - 4 = 8 - 4a + c$$<br><br>$$4a - 2\sqrt 2 b - c = 12$$ .... (iii)<br><br>Also, $$a - 2\sqrt 2 \,b = 3$$ .... (iv) (given)<br><br>$$(ii) - (iii) \Rightarrow - a + \left( {2\sqrt 2 - 3} \right)b = - 3$$ ... (v)<br><br>$$(iv) + (v) \Rightarrow b = 0,a = 3$$<br><br>$$ \therefore $$ $${a^2} + {b^2} + ab = 9$$ | integer | jee-main-2021-online-26th-february-evening-slot |
1l54aw3dn | maths | application-of-derivatives | tangent-and-normal | <p>Let f : R $$\to$$ R be a function defined by f(x) = (x $$-$$ 3)<sup>n<sub>1</sub></sup> (x $$-$$ 5)<sup>n<sub>2</sub></sup>, n<sub>1</sub>, n<sub>2</sub> $$\in$$ N. Then, which of the following is NOT true?</p> | [{"identifier": "A", "content": "For n<sub>1</sub> = 3, n<sub>2</sub> = 4, there exists $$\\alpha$$ $$\\in$$ (3, 5) where f attains local maxima."}, {"identifier": "B", "content": "For n<sub>1</sub> = 4, n<sub>2</sub> = 3, there exists $$\\alpha$$ $$\\in$$ (3, 5) where f attains local minima."}, {"identifier": "C", "content": "For n<sub>1</sub> = 3, n<sub>2</sub> = 5, there exists $$\\alpha$$ $$\\in$$ (3, 5) where f attains local maxima."}, {"identifier": "D", "content": "For n<sub>1</sub> = 4, n<sub>2</sub> = 6, there exists $$\\alpha$$ $$\\in$$ (3, 5) where f attains local maxima."}] | ["C"] | null | <p>Given,</p>
<p>$$f(x) = {( - 3)^{{n_1}}}{(x - 5)^{{n_2}}}$$</p>
<p>Differentiating both side with respect to x, we get</p>
<p>$$f'(x) = {(x - 5)^{{n_2}}}\left( {{n_1}{{(x - 3)}^{{n_1} - 1}}} \right) + {(x - 3)^{{n_1}}}\left( {{n_2}{{(x - 5)}^{{n_2} - 1}}} \right)$$</p>
<p>$$ = {(x - 3)^{{n_1}}}{(x - 5)^{{n_2}}}\left[ {{{{n_1}} \over {x - 3}} + {{{n_2}} \over {x - 5}}} \right]$$</p>
<p>$$ = {(x - 3)^{{n_1}}}{(x - 5)^{{n_2}}}\left[ {{{{n_1}(x - 5) + {n_2}(x - 3)} \over {(x - 3)(x - 5)}}} \right]$$</p>
<p>$$ = {(x - 3)^{{n_1} - 1}}{(x - 5)^{{n_2} - 1}}[{n_1}(x - 5) + {n_2}(x - 3)]$$</p>
<h3>Option A :</h3>
<p>When n<sub>1</sub> = 3 and n<sub>2</sub> = 4 then</p>
<p>$$f'(x) = {(x - 3)^{3 - 1}}{(x - 5)^{4 - 1}}\left[ {3(x - 5) + 4(x - 3)} \right]$$</p>
<p>$$ = {(x - 3)^2}{(x - 5)^3}(7x - 27)$$</p>
<p>Critical point is at x = 3, 5 and $${{27} \over 7}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5miwu4a/b6eb5083-e519-499d-9248-a81e3fdf7ef5/e69f7da0-0445-11ed-9d9a-21bdd0f00aa4/file-1l5miwu4b.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5miwu4a/b6eb5083-e519-499d-9248-a81e3fdf7ef5/e69f7da0-0445-11ed-9d9a-21bdd0f00aa4/file-1l5miwu4b.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th June Evening Shift Mathematics - Application of Derivatives Question 77 English Explanation 1"></p>
<p>When value of f'(x) is less than $${{27} \over 5}$$, f'(x) is positive means slope of f(x) graph is positive.</p>
<p>And when value of f'(x) is greater than $${{27} \over 7}$$ but less than 5 then f'(x) is negative means slope of f(x) is negative.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5miy8jq/0355ce2a-ccfd-4014-9a4e-f11ecb74e789/0d940660-0446-11ed-9d9a-21bdd0f00aa4/file-1l5miy8jr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5miy8jq/0355ce2a-ccfd-4014-9a4e-f11ecb74e789/0d940660-0446-11ed-9d9a-21bdd0f00aa4/file-1l5miy8jr.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th June Evening Shift Mathematics - Application of Derivatives Question 77 English Explanation 2"></p>
<p>So at $${{27} \over 7}$$ between 3 and 5, f(x) attain local maxima.</p>
<p>$$\therefore$$ Option A is correct.</p>
<h3>Option B :</h3>
<p>When n<sub>1</sub> = 4 and n<sub>2</sub> = 3 then</p>
<p>$$f'(x) = {(x - 3)^{4 - 1}}{(x - 5)^{3 - 1}}\left[ {4(x - 5) + 3(x - 3)} \right]$$</p>
<p>$$ = {(x - 3)^3}{(x - 5)^2}(7x - 29)$$</p>
<p>Critical point is at x = 3, 5 and $${{29} \over 7}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5mj1eqq/48f3c678-5009-4ace-b7a5-84db65f774f0/65ca6c20-0446-11ed-9d9a-21bdd0f00aa4/file-1l5mj1eqr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5mj1eqq/48f3c678-5009-4ace-b7a5-84db65f774f0/65ca6c20-0446-11ed-9d9a-21bdd0f00aa4/file-1l5mj1eqr.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th June Evening Shift Mathematics - Application of Derivatives Question 77 English Explanation 3"></p>
<p>When f'(x) is less than $${{29} \over 7}$$ but greater than 3 then f'(x) is negative means slope of graph of f(x) is negative.</p>
<p>And when f'(x) is greater than $${{29} \over 7}$$ but less than 5 then f'(x) is positive means slope of graph of f(x) is positive.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5mj51wi/7cbdb428-88d9-49cc-9dab-6a1d9bb5b71b/cb1bec20-0446-11ed-9d9a-21bdd0f00aa4/file-1l5mj51wj.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5mj51wi/7cbdb428-88d9-49cc-9dab-6a1d9bb5b71b/cb1bec20-0446-11ed-9d9a-21bdd0f00aa4/file-1l5mj51wj.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th June Evening Shift Mathematics - Application of Derivatives Question 77 English Explanation 4"></p>
<p>So at $${{29} \over 7}$$ between 3 and 5, f(x) attain local minima.</p>
<p>$$\therefore$$ Option (B) is correct.</p>
<h3>Option C :</h3>
<p>When n<sub>1</sub> = 3 and n<sub>2</sub> = 5 then</p>
<p>$$f'(x) = {(x - 3)^{3 - 1}}{(x - 5)^{5 - 1}}\left[ {3(x - 5) + 5(x - 3)} \right]$$</p>
<p>$$ = {(x - 3)^2}{(x - 5)^4}(8x - 30)$$</p>
<p>Critical point is at x = 3, 5 and $${{30} \over 8}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5mj6a8q/2a8081a0-9fe2-49f4-95f0-647c1e4d436f/ed5c3ba0-0446-11ed-9d9a-21bdd0f00aa4/file-1l5mj6a8r.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5mj6a8q/2a8081a0-9fe2-49f4-95f0-647c1e4d436f/ed5c3ba0-0446-11ed-9d9a-21bdd0f00aa4/file-1l5mj6a8r.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th June Evening Shift Mathematics - Application of Derivatives Question 77 English Explanation 5"></p>
<p>When f'(x) is less than $${{30} \over 8}$$ but greater than 3 then f'(x) is negative means slope of graph of f(x) is negative.</p>
<p>And when f'(x) is greater than $${{30} \over 8}$$ but less than 5 then f'(x) is positive means slope of graph of f(x) is positive.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5mj7g67/ccde0452-eeab-497b-a7fe-49ce58933922/0dc02a00-0447-11ed-9d9a-21bdd0f00aa4/file-1l5mj7g68.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5mj7g67/ccde0452-eeab-497b-a7fe-49ce58933922/0dc02a00-0447-11ed-9d9a-21bdd0f00aa4/file-1l5mj7g68.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th June Evening Shift Mathematics - Application of Derivatives Question 77 English Explanation 6"></p>
<p>So, at $${{30} \over 8}$$ between 3 and 5, f(x) attain local minima.</p>
<p>$$\therefore$$ Option C is not true.</p>
<p>Similarly you can check option D also.</p> | mcq | jee-main-2022-online-29th-june-evening-shift |
1l567ppaq | maths | application-of-derivatives | tangent-and-normal | <p>Let l be a line which is normal to the curve y = 2x<sup>2</sup> + x + 2 at a point P on the curve. If the point Q(6, 4) lies on the line l and O is origin, then the area of the triangle OPQ is equal to ___________.</p> | [] | null | 13 | <p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5oc5ao6/7ae3148f-ca07-4559-92f9-0d6735c2710d/0346fe60-0545-11ed-987f-3938cfc0f7f1/file-1l5oc5ao7.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5oc5ao6/7ae3148f-ca07-4559-92f9-0d6735c2710d/0346fe60-0545-11ed-987f-3938cfc0f7f1/file-1l5oc5ao7.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 28th June Morning Shift Mathematics - Application of Derivatives Question 75 English Explanation"> </p>
<p>$${{{y_1} - 4} \over {{x_1} - 6}} = - {1 \over {4{x_1} + 1}}$$</p>
<p>$$ \Rightarrow {{2x_1^2 + {x_1} - 2} \over {{x_1} - 6}} = - {1 \over {4{x_1} + 1}}$$</p>
<p>$$ \Rightarrow 6 - {x_1} = 8x_1^3 + 6x_1^2 - 7{x_1} - 2$$</p>
<p>$$ \Rightarrow 8x_1^3 + 6x_1^2 - 6{x_1} - 8 = 0$$</p>
<p>So $${x_1} = 1 \Rightarrow {y_1} = 5$$</p>
<p>Area $$ = \left| {{1 \over 2}\left| {\matrix{
0 & 0 & 1 \cr
6 & 4 & 1 \cr
1 & 5 & 1 \cr
} } \right|} \right| = 13$$.</p> | integer | jee-main-2022-online-28th-june-morning-shift |
1l589kpc2 | maths | application-of-derivatives | tangent-and-normal | <p>Let S be the set of all the natural numbers, for which the line $${x \over a} + {y \over b} = 2$$ is a tangent to the curve $${\left( {{x \over a}} \right)^n} + {\left( {{y \over b}} \right)^n} = 2$$ at the point (a, b), ab $$\ne$$ 0. Then :</p> | [{"identifier": "A", "content": "S = $$\\phi$$"}, {"identifier": "B", "content": "n(S) = 1"}, {"identifier": "C", "content": "S = {2k : k $$\\in$$ N}"}, {"identifier": "D", "content": "S = N"}] | ["D"] | null | <p>$${\left( {{x \over a}} \right)^n} + {\left( {{y \over b}} \right)^n} = 2$$</p>
<p>Differentiating both sides with respect to x, we get</p>
<p>$$ \Rightarrow n\,.\,{\left( {{x \over a}} \right)^{n - a}}\,.\,{1 \over a} + n\,.\,{\left( {{y \over b}} \right)^{n - 1}}\,.\,{1 \over b}\,.\,{{dy} \over {dx}} = 0$$</p>
<p>$$ \Rightarrow {{dy} \over {dx}} = - {{{1 \over a}\,.\,{{\left( {{x \over a}} \right)}^{n - 1}}} \over {{1 \over b}{{\left( {{y \over b}} \right)}^{n - 1}}}}$$</p>
<p>$$ = - {b \over a}{\left( {{{xb} \over {ya}}} \right)^{n - 1}}$$</p>
<p>Now, $${{dy} \over {dx}}$$ at (a, b) $$ = - {b \over a}{\left[ {{{ab} \over {ba}}} \right]^{n - 1}} = - {b \over a}$$</p>
<p>Equation of tangent at (a, b) is,</p>
<p>$$(y - b) = - {b \over a}(x - a)$$</p>
<p>$$ \Rightarrow {{y - b} \over b} = - {{x - a} \over a}$$</p>
<p>$$ \Rightarrow {y \over b} - 1 = - {x \over a} + 1$$</p>
<p>$$ \Rightarrow {x \over a} + {y \over b} = 2$$</p>
<p>$$\therefore$$ It is tangent for all value of n.</p> | mcq | jee-main-2022-online-26th-june-morning-shift |
1l59km5jz | maths | application-of-derivatives | tangent-and-normal | <p>If the angle made by the tangent at the point (x<sub>0</sub>, y<sub>0</sub>) on the curve $$x = 12(t + \sin t\cos t)$$, $$y = 12{(1 + \sin t)^2}$$, $$0 < t < {\pi \over 2}$$, with the positive x-axis is $${\pi \over 3}$$, then y<sub>0</sub> is equal to:</p> | [{"identifier": "A", "content": "$$6\\left( {3 + 2\\sqrt 2 } \\right)$$"}, {"identifier": "B", "content": "$$3\\left( {7 + 4\\sqrt 3 } \\right)$$"}, {"identifier": "C", "content": "27"}, {"identifier": "D", "content": "48"}] | ["C"] | null | <p>$$\because$$ $${{dy} \over {dx}} = {{24(1 + \sin t)\cos t} \over {12(1 + \cos 2t)}} = {{1 + \sin t} \over {\cos t}} = \tan \left( {{\pi \over 4} + {t \over 2}} \right)$$</p>
<p>$$\because$$ $${{dy} \over {d{x_{({x_0},{y_0})}}}} = \sqrt 3 = \tan \left( {{\pi \over 4} + {t \over 2}} \right)$$</p>
<p>$$ \Rightarrow t = {\pi \over 6}$$</p>
<p>So, $${y_{0\left( {at\,t = {\pi \over 6}} \right)}} = 12{\left( {1 + \sin {\pi \over 6}} \right)^2} = 27$$</p> | mcq | jee-main-2022-online-25th-june-evening-shift |
1l5bas9mk | maths | application-of-derivatives | tangent-and-normal | <p>The slope of normal at any point (x, y), x > 0, y > 0 on the curve y = y(x) is given by $${{{x^2}} \over {xy - {x^2}{y^2} - 1}}$$. If the curve passes through the point (1, 1), then e . y(e) is equal to </p> | [{"identifier": "A", "content": "$${{1 - \\tan (1)} \\over {1 + \\tan (1)}}$$"}, {"identifier": "B", "content": "tan(1)"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "$${{1 + \\tan (1)} \\over {1 - \\tan (1)}}$$"}] | ["D"] | null | <p>$$\because$$ $$ - {{dx} \over {dy}} = {{{x^2}} \over {xy - {x^2}{y^2} - 1}}$$</p>
<p>$$\therefore$$ $${{dy} \over {dx}} = {{{x^2}{y^2} - xy + 1} \over {{x^2}}}$$</p>
<p>Let $$xy = v \Rightarrow y + x{{dy} \over {dx}} = {{dv} \over {dx}}$$</p>
<p>$$\therefore$$ $${{dv} \over {dx}} - y = {{({v^2} - v + 1)y} \over v}$$</p>
<p>$$\therefore$$ $${{dv} \over {dx}} = {{{v^2} + 1} \over x}$$</p>
<p>$$\because$$ $$y(1) = 1 \Rightarrow {\tan ^{ - 1}}(xy) = \ln x + {\tan ^{ - 1}}(1)$$</p>
<p>Put $$x = e$$ and $$y = y(e)$$ we get</p>
<p>$${\tan ^{ - 1}}(e\,.\,y(e)) = 1 + {\tan ^{ - 1}}1$$</p>
<p>$${\tan ^{ - 1}}(e\,.\,y(e)) - {\tan ^{ - 1}}1 = 1$$</p>
<p>$$\therefore$$ $$e(y(e)) = {{1 + \tan (1)} \over {1 - \tan (1)}}$$</p> | mcq | jee-main-2022-online-24th-june-evening-shift |
1l5c1al0o | maths | application-of-derivatives | tangent-and-normal | <p>If the tangent at the point (x<sub>1</sub>, y<sub>1</sub>) on the curve $$y = {x^3} + 3{x^2} + 5$$ passes through the origin, then (x<sub>1</sub>, y<sub>1</sub>) does NOT lie on the curve :</p> | [{"identifier": "A", "content": "$${x^2} + {{{y^2}} \\over {81}} = 2$$"}, {"identifier": "B", "content": "$${{{y^2}} \\over 9} - {x^2} = 8$$"}, {"identifier": "C", "content": "$$y = 4{x^2} + 5$$"}, {"identifier": "D", "content": "$${x \\over 3} - {y^2} = 2$$"}] | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5vb3wli/4dde9a3d-2417-4084-b660-1fce1c16846a/70032f50-091a-11ed-80eb-03ed789c0f4d/file-1l5vb3wlj.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5vb3wli/4dde9a3d-2417-4084-b660-1fce1c16846a/70032f50-091a-11ed-80eb-03ed789c0f4d/file-1l5vb3wlj.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 24th June Morning Shift Mathematics - Application of Derivatives Question 61 English Explanation"></p>
<p>Given curve,</p>
<p>$$y = {x^3} + 3{x^2} + 5$$</p>
<p>Slope of tangent,</p>
<p>$${{dy} \over {dx}} = 3{x^2} + 6x$$</p>
<p>Slope of line joined by (x<sub>1</sub>, y<sub>1</sub>) and (0, 0) is $$ = {{{y_0} - 0} \over {{x_1} - 0}}$$</p>
<p>$$\therefore$$ $${{{y_1}} \over {{x_1}}} = 3x_1^2 + 6{x_1}$$</p>
<p>$$ \Rightarrow {y_1} = 3x_1^3 + 6x_1^2$$ ...... (1)</p>
<p>Point (x<sub>1</sub>, y<sub>1</sub>) is on the line,</p>
<p>$$\therefore$$ $${y_1} = x_1^3 + 3x_1^2 + 5$$ ..... (2)</p>
<p>$$\therefore$$ $$3x_1^3 + 6x_1^2 = x_1^3 + 3x_1^2 + 6$$</p>
<p>$$ \Rightarrow 2x_1^3 + 3x_1^2 - 5 = 0$$</p>
<p>x = 1 satisfy the equation</p>
<p>Now, $${y_1} = x_1^3 + 3x_1^2 + 5$$</p>
<p>$$ = 1 + 3 + 5$$</p>
<p>$$ = 9$$</p>
<p>$$\therefore$$ (x<sub>1</sub>, y<sub>1</sub>) = (1, 9)</p>
<p>Option D does not satisfy point (1, 9).</p> | mcq | jee-main-2022-online-24th-june-morning-shift |
1l5c1hf56 | maths | application-of-derivatives | tangent-and-normal | <p>Let $$\lambda x - 2y = \mu $$ be a tangent to the hyperbola $${a^2}{x^2} - {y^2} = {b^2}$$. Then $${\left( {{\lambda \over a}} \right)^2} - {\left( {{\mu \over b}} \right)^2}$$ is equal to :</p> | [{"identifier": "A", "content": "$$-$$2"}, {"identifier": "B", "content": "$$-$$4"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "4"}] | ["D"] | null | $\frac{x^{2}}{\left(\frac{b^{2}}{a^{2}}\right)}-\frac{y^{2}}{b^{2}}=1$
<br/><br/>
Tangent in slope form $\Rightarrow y=m x \pm \sqrt{\frac{b^{2}}{a^{2}} m^{2}-b^{2}}$
<br/><br/>
i.e., same as $y=\frac{\lambda x}{2}-\frac{\mu}{2}$
<br/><br/>
Comparing coefficients,
<br/><br/>
$$
\begin{aligned}
&m=\frac{\lambda}{2}, \frac{b^{2}}{a^{2}} m^{2}-b^{2}=\frac{\mu^{2}}{4} \\\\
&\text { Eliminating } m, \frac{b^{2}}{a^{2}} \cdot \frac{\lambda^{2}}{4}-b^{2}=\frac{\mu^{2}}{4} \\\\
&\Rightarrow \frac{\lambda^{2}}{a^{2}}-\frac{\mu^{2}}{b^{2}}=4
\end{aligned}
$$ | mcq | jee-main-2022-online-24th-june-morning-shift |
1l6jeeqfu | maths | application-of-derivatives | tangent-and-normal | <p>Let $$M$$ and $$N$$ be the number of points on the curve $$y^{5}-9 x y+2 x=0$$, where the tangents to the curve are parallel to $$x$$-axis and $$y$$-axis, respectively. Then the value of $$M+N$$ equals ___________.</p> | [] | null | 2 | <p>Here equation of curve is</p>
<p>$${y^5} - 9xy + 2x = 0$$ ...... (i)</p>
<p>On differentiating : $$5{y^4}{{dy} \over {dx}} - 9y - 9x{{dy} \over {dx}} + 2 = 0$$</p>
<p>$$\therefore$$ $${{dy} \over {dx}} = {{9y - 2} \over {5{y^4} - 9x}}$$</p>
<p>When tangents are parallel to x-axis then $$9y - 2 = 0$$</p>
<p>$$\therefore$$ $$M = 1$$.</p>
<p>For tangent perpendicular to x-axis</p>
<p>$$5{y^4} - 9x = 0$$ ...... (ii)</p>
<p>From equation (i) and (ii) we get only one point.</p>
<p>$$\therefore$$ $$N = 1$$.</p>
<p>$$\therefore$$ $$M + N = 2$$.</p> | integer | jee-main-2022-online-27th-july-morning-shift |
1l6rfwzo1 | maths | application-of-derivatives | tangent-and-normal | <p>If the tangent to the curve $$y=x^{3}-x^{2}+x$$ at the point $$(a, b)$$ is also tangent to the curve $$y = 5{x^2} + 2x - 25$$ at the point (2, $$-$$1), then $$|2a + 9b|$$ is equal to __________.</p> | [] | null | 195 | Slope of tangent to curve $y=5 x^{2}+2 x-25$
<br/><br/>$$
=m=\left(\frac{d y}{d x}\right)_{\mathrm{at}(2,-1)}=22
$$
<br/><br/>$\therefore \quad$ Equation of tangent $: y+1=22(x-2)$
<br/><br/>$\therefore \quad y=22 x-45$.
<br/><br/>Slope of tangent to $y=x^{3}-x^{2}+x$ at point $(a, b)$
<br/><br/>$$
=3 a^{2}-2 a+1
$$
<br/><br/>$3 a^{2}-2 a+1=22$
<br/><br/>$3 a^{2}-2 a-21=0$
<br/><br/>$\therefore \quad a=3$ or $-\frac{7}{3}$
<br/><br/>Also $b=a^{3}-a^{2}+a$
<br/><br/>Then $(a, b)=(3,21)$ or $\left(-\frac{7}{3},-\frac{151}{9}\right)$.
<br/><br/>$\left(-\frac{7}{3},-\frac{151}{9}\right)$ does not satisfy the equation of tangent
<br/><br/>$\therefore \quad a=3, b=21$
<br/><br/>$\therefore|2 a+9 b|=195$ | integer | jee-main-2022-online-29th-july-evening-shift |
1ldr7heoj | maths | application-of-derivatives | tangent-and-normal | <p>The number of points on the curve $$y=54 x^{5}-135 x^{4}-70 x^{3}+180 x^{2}+210 x$$ at which the normal lines are parallel to $$x+90 y+2=0$$ is :</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "0"}] | ["C"] | null | <p>$$y'=270x^4-540x^3-210x^2+360x+210$$</p>
<p>Slope of normal $$=-\frac{1}{90}$$</p>
<p>$$\therefore$$ Slope of tangent = 90</p>
<p>$$\therefore$$ Number of normal will be number of solutions of</p>
<p>$$270x^4-540x^3-210x^2+360x+210=90$$</p>
<p>$$\Rightarrow 9x^4-18x^3-7x^2+12x+4=0$$</p>
<p>$$\therefore x=1,2,-\frac{1}{3},-\frac{2}{3}$$ are roots</p> | mcq | jee-main-2023-online-30th-january-morning-shift |
1ldsg4b6k | maths | application-of-derivatives | tangent-and-normal | <p>If the equation of the normal to the curve $$y = {{x - a} \over {(x + b)(x - 2)}}$$ at the point (1, $$-$$3) is $$x - 4y = 13$$, then the value of $$a + b$$ is equal to ___________.</p> | [] | null | 4 | <p>Given curve : $$y = {{x - a} \over {(x + b)(x - 2)}}$$ at $$(1, - 3)$$</p>
<p>$$\therefore$$ $$ - 3 = {{1 - a} \over {(1 + b)( - 1)}} \Rightarrow 3 + 3b = 1 - a$$</p>
<p>$$\beta \Rightarrow a + 3b + 2 = 0$$</p>
<p>$$y = {{x - a} \over {(x + b)(x - 2)}}$$</p>
<p>$${{dy} \over {dx}} = {{(x + b)(x - 2) - (x - a)[(x + b) + (x - 2)]} \over {{{[(x + b)(x - 2)]}^2}}}$$</p>
<p>at $$(1, - 3)\,{m_T} = {{ - (1 + b) - (1 - a)(b)} \over {{{(1 + b)}^2}}} = - 4$$</p>
<p>$$\therefore$$ $$1 + b + b - ab = 4{(1 + b)^2}$$</p>
<p>$$ \Rightarrow 1 + 2b + b(3b + 2) = 4{b^2} + 4 + 8b$$</p>
<p>$$ \Rightarrow {b^2} + 4b + 3 = 0$$</p>
<p>$$(b + 1)(b + 3) = 0$$</p>
<p>$$b = - 1,a = 1$$ but $$1 + b \ne 0$$</p>
<p>$$b = - 3,a = 7$$ $$\therefore$$ $$b \ne - 1$$</p>
<p>$$\therefore$$ $$a + b = 04$$</p> | integer | jee-main-2023-online-29th-january-evening-shift |
1lgvqx8pc | maths | application-of-derivatives | tangent-and-normal | <p>Let the quadratic curve passing through the point $$(-1,0)$$ and touching the line $$y=x$$ at $$(1,1)$$ be $$y=f(x)$$. Then the $$x$$-intercept of the normal to the curve at the point $$(\alpha, \alpha+1)$$ in the first quadrant is __________.</p> | [] | null | 11 | Let the quadratic curve be $f(x)=a x^2+b x+c$
<br/><br/>The curve passes through $(-1,0)$
<br/><br/>$0=a-b+c \Rightarrow a+c=b$ ..........(i)
<br/><br/>The curve also passes through $(1,1)$
<br/><br/>$$
\begin{gathered}
a+b+c=1 .........(ii)\\\\
2 b=1 \Rightarrow b=\frac{1}{2}
\end{gathered}
$$
<br/><br/>$$
f^{\prime}(x)=2 a x+b
$$
<br/><br/>Slope tangent of curve $=f^{\prime}(x)$ at $(1,1)=2 a+b$
<br/><br/>Slope of line $y=x$ is 1
<br/><br/>$$
\begin{aligned}
&\therefore 2 a+b =1 \\\\
&2 a+\frac{1}{2} =1 \Rightarrow a=\frac{1}{4} \\\\
&c =1-a-b \Rightarrow c=\frac{1}{4}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& f(x)=\frac{1}{4} x^2+\frac{1}{2} x+\frac{1}{4} = \frac{(x+1)^2}{4} \\\\
& f^{\prime}(x)=\frac{2(x+1)}{4}=\frac{x+1}{2}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& f(x) \text { passes through }(\alpha, \alpha+1) \\\\
& \begin{array}{l}
\therefore \alpha+1=\frac{(\alpha+1)^2}{4} \\\\
\Rightarrow \alpha+1=4 \text { or } \alpha=3
\end{array}
\end{aligned}
$$
<br/><br/>Equation of normal at $(3,4)$ is
<br/><br/>$$
y-4=-\frac{1}{2}(x-3)
$$
<br/><br/>For $x$, intercept, $y=0$
<br/><br/>$$
x-3=8 \text { or } x=11
$$
<br/><br/>Hence, the required $x$ intercept is 11 . | integer | jee-main-2023-online-10th-april-evening-shift |
1lgxsy1zh | maths | application-of-derivatives | tangent-and-normal | <p>The slope of tangent at any point (x, y) on a curve $$y=y(x)$$ is $${{{x^2} + {y^2}} \over {2xy}},x > 0$$. If $$y(2) = 0$$, then a value of $$y(8)$$ is :</p> | [{"identifier": "A", "content": "$$ - 4\\sqrt 2 $$"}, {"identifier": "B", "content": "$$2\\sqrt 3 $$"}, {"identifier": "C", "content": "$$4\\sqrt 3 $$"}, {"identifier": "D", "content": "$$ - 2\\sqrt 3 $$"}] | ["C"] | null | Let the slope of tangent at any point
<br/><br/>$(x, y)$ on a curve $y=y(x)$ is $\frac{d y}{d x}$
<br/><br/>According to the question, $\frac{d y}{d x}=\frac{x^2+y^2}{2 x y}(x>0)$
[Given]
<br/><br/>$$
\begin{aligned}
&\text { Let } y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \\\\
&\Rightarrow v+x \frac{d v}{d x}=\frac{x^2+v^2 x^2}{2 v x^2}=\frac{1+v^2}{2 v} \\\\
&\Rightarrow x \frac{d v}{d x}=\frac{1+v^2}{2 v}-v=\frac{1-v^2}{2 v}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \int \frac{2 v d v}{1-v^2}=\int \frac{d x}{x} \text { [Taking integration on both sides] } \\\\
& \Rightarrow-\ln \left|1-v^2\right|=\ln x-\ln c \\\\
& \Rightarrow \ln \left(1-v^2\right) x=\ln c \quad(c=\text { constant }) \\\\
& \Rightarrow \left(1-y^2 / x^2\right) \cdot x=c \\\\
& \Rightarrow \left(x^2-y^2\right)=c x
\end{aligned}
$$
<br/><br/>Put $y(2)=0 \Rightarrow c=2$
<br/><br/>$$
\begin{array}{rlrl}
& \therefore x^2-y^2 =2 x \\\\
& \therefore y^2 =x^2-2 x \\\\
& \Rightarrow y(x) = \pm \sqrt{x^2-2 x}
\end{array}
$$
<br/><br/>Put $x=8$, we get
<br/><br/>$$
\Rightarrow y(8)= \pm \sqrt{64-16}= \pm \sqrt{48}= \pm 4 \sqrt{3}
$$ | mcq | jee-main-2023-online-10th-april-morning-shift |
1lh2ysvt2 | maths | application-of-derivatives | tangent-and-normal | <p>Let a curve $$y=f(x), x \in(0, \infty)$$ pass through the points $$P\left(1, \frac{3}{2}\right)$$ and $$Q\left(a, \frac{1}{2}\right)$$. If the tangent at any point $$R(b, f(b))$$ to the given curve cuts the $$\mathrm{y}$$-axis at the point $$S(0, c)$$ such that $$b c=3$$, then $$(P Q)^{2}$$ is equal to __________.</p> | [] | null | 5 | Equation of tangent at $R(b, f(b))$
<br/><br/>$$
y-f(b)=f^{\prime}(b)(x-b)
$$
<br/><br/>which passes through $S(0, c)$
<br/><br/>$$
\begin{aligned}
& \therefore c-f(b)=f^{\prime}(b)(0-b) \\\\
& b f^{\prime}(b)-f(b)=-c \\\\
& \Rightarrow b f^{\prime}(b)-f(b)=\frac{-3}{b} (\because b c=3) \\\\
& \Rightarrow \frac{b f^{\prime}(b)-f(b)}{b^2}=\frac{-3}{b^3} \\\\
& \Rightarrow d\left(\frac{f(b)}{b}\right)=\frac{-3}{b^3} \\\\
& \Rightarrow \frac{f(b)}{b}=\frac{3}{2 b^2}+c
\end{aligned}
$$
<br/><br/>which passes through $P\left(1, \frac{3}{2}\right)$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \frac{3 / 2}{1}=\frac{3}{2}+c \\\\
& \Rightarrow c=0 \\\\
& \therefore f(b)=\frac{3}{2 b^2} \times b \\\\
& \Rightarrow f(b)=\frac{3}{2 b}
\end{aligned}
$$
<br/><br/>$\because$ It passes through $Q\left(a, \frac{1}{2}\right)$
<br/><br/>$$
\begin{aligned}
& \therefore \frac{1}{2}=\frac{3}{2 a} \\\\
& \Rightarrow a=3 \\\\
& \therefore P \equiv\left(1, \frac{3}{2}\right) \text { and } Q \equiv\left(3, \frac{1}{2}\right)\\\\
& \therefore (P Q)^2=(3-1)^2+\left(\frac{1}{2}-\frac{3}{2}\right)^2=4+1=5
\end{aligned}
$$ | integer | jee-main-2023-online-6th-april-evening-shift |
v8p9syKOxJlo7k47 | maths | area-under-the-curves | area-bounded-between-the-curves | The area bounded by the curves $$y = \ln x,y = \ln \left| x \right|,y = \left| {\ln {\mkern 1mu} x} \right|$$ and $$y = \left| {\ln \left| x \right|} \right|$$ is : | [{"identifier": "A", "content": "$$4$$sq. units "}, {"identifier": "B", "content": "$$6$$sq. units "}, {"identifier": "C", "content": "$$10$$sq. units"}, {"identifier": "D", "content": "none of these "}] | ["A"] | null | First we draw each curve as separate graph
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266365/exam_images/eejvpxyhkrguzqnnbwmv.webp" loading="lazy" alt="AIEEE 2002 Mathematics - Area Under The Curves Question 133 English Explanation 1">
<br><br><b>NOTE :</b> Graph of $$y = \left| {f\left( x \right)} \right|$$ can be obtained from the graph of the curve $$y = f\left( x \right)$$ by drawing the mirror image of the portion of the graph below $$x$$-axis, with respect to $$x$$-axis.
<br><br>Clearly the bounded area is as shown in the following figure.
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264182/exam_images/lbtyjmjfgihwg2ehcrky.webp" loading="lazy" alt="AIEEE 2002 Mathematics - Area Under The Curves Question 133 English Explanation 2">
<br><br>Required area $$ = 4\int\limits_0^1 {\left( { - \ln x} \right)} dx$$
<br><br>$$ = - 4\left[ {x\,\ln x + - x_0^1} \right] = 4\,\,$$ sq. units | mcq | aieee-2002 |
o1c9eAEZCudbpToS | maths | area-under-the-curves | area-bounded-between-the-curves | The area of the region bounded by the curves $$y = \left| {x - 1} \right|$$ and $$y = 3 - \left| x \right|$$ is : | [{"identifier": "A", "content": "$$6$$ sq. units"}, {"identifier": "B", "content": "$$2$$ sq. units"}, {"identifier": "C", "content": "$$3$$ sq. units"}, {"identifier": "D", "content": "$$4$$ sq. units"}] | ["D"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265258/exam_images/l9bcohz8ayjblt6nujfm.webp" loading="lazy" alt="AIEEE 2003 Mathematics - Area Under The Curves Question 132 English Explanation">
<br><br>$$A = \int\limits_{ - 1}^0 {\left\{ {\left( {3 + x} \right) - \left( { - x + 1} \right)} \right\}dx + } $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\int\limits_0^1 {\left\{ {\left( {3 - x} \right) - \left( { - x + 1} \right)} \right\}dx + } $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\int\limits_1^2 {\left\{ {\left( {3 - x} \right) - \left( {x - 1} \right)} \right\}dx} $$
<br><br>$$ = \int\limits_{ - 1}^0 {\left( {2 + 2x} \right)dx + \int\limits_0^1 {2dx + \int\limits_1^2 {\left( {4 - 2x} \right)dx} } } $$
<br><br>$$ = \left[ {2x - {x^2}} \right]_{ - 1}^0 + \left[ {2x} \right]_0^1 + \left[ {4x - {x^2}} \right]_1^2$$
<br><br>$$ = 0 - \left( { - 2 + 1} \right) + \left( {2 - 0} \right) + \left( {8 - 4} \right) - \left( {4 - 1} \right)$$
<br><br>$$ = 1 + 2 + 4 - 3 = 4$$ sq. units | mcq | aieee-2003 |
0zJaHFqpedZaI9Xm | maths | area-under-the-curves | area-bounded-between-the-curves | The area of the region bounded by the curves
<br/>$$y = \left| {x - 2} \right|,x = 1,x = 3$$ and the $$x$$-axis is : | [{"identifier": "A", "content": "$$4$$"}, {"identifier": "B", "content": "$$2$$"}, {"identifier": "C", "content": "$$3$$"}, {"identifier": "D", "content": "$$1$$"}] | ["D"] | null | The required area is shown by shaded region
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266321/exam_images/ixjlsq9plxoilqztxxsw.webp" loading="lazy" alt="AIEEE 2004 Mathematics - Area Under The Curves Question 131 English Explanation">
<br><br>$$A = \int\limits_1^3 {\left| {x - 2} \right|dx = 2\int\limits_2^3 {\left( {x - 2} \right)} } dx$$
<br><br>$$ = 2\left[ {{{{x^2}} \over 2} - 2x} \right]_2^3 = 1$$ | mcq | aieee-2004 |
0AwCj1YqAxpB6L96 | maths | area-under-the-curves | area-bounded-between-the-curves | The parabolas $${y^2} = 4x$$ and $${x^2} = 4y$$ divide the square region bounded by the lines $$x=4,$$ $$y=4$$ and the coordinate axes. If $${S_1},{S_2},{S_3}$$ are respectively the areas of these parts numbered from top to bottom ; then $${S_1},{S_2},{S_3}$$ is : | [{"identifier": "A", "content": "$$1:2:1$$"}, {"identifier": "B", "content": "$$1:2:3$$"}, {"identifier": "C", "content": "$$2:1:2$$"}, {"identifier": "D", "content": "$$1:1:1$$"}] | ["D"] | null | Intersection points of $${x^2} = 4y$$ and $${y^2} = 4x$$ are $$\left( {0,0} \right)$$ and $$\left( {4,4} \right).$$ The graph is as shown in the figure.
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267219/exam_images/qhldynplznxc2j2lyapa.webp" loading="lazy" alt="AIEEE 2005 Mathematics - Area Under The Curves Question 129 English Explanation">
<br><br>By symmetry, we observe
<br><br>$${S_1} = {S_3} = \int\limits_0^4 {ydx = \int\limits_0^4 {{{{x^2}} \over 4}dx} } $$
<br><br>$$ = \left[ {{{{x^3}} \over {12}}} \right]_0^4 = {{16} \over 3}$$ sq. unit
<br><br>Also $${S_2} = \int\limits_0^4 {\left( {2\sqrt x - {{{x^2}} \over 4}} \right)} dx$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = \left[ {{{2{x^{{3 \over 2}}}} \over {{3 \over 2}}} - {{{x^3}} \over {12}}} \right]_0^4$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = {4 \over 3} \times 8 - {{16} \over 3} = {{16} \over 3}$$
<br><br>$$\therefore$$ $${S_1}:{S_2}:{S_3} = 1:1:1$$ | mcq | aieee-2005 |
OJxelJmkjNf7amKn | maths | area-under-the-curves | area-bounded-between-the-curves | The area enclosed between the curves $${y^2} = x$$ and $$y = \left| x \right|$$ is : | [{"identifier": "A", "content": "$$1/6$$"}, {"identifier": "B", "content": "$$1/3$$"}, {"identifier": "C", "content": "$$2/3$$"}, {"identifier": "D", "content": "$$1$$"}] | ["A"] | null | The area enclosed between the curves
<br><br>$${y^2} = x$$ and $$y = \left| x \right|$$
<br><br>From the figure, area lies between
<br><br>$${y^2} = x$$ and $$y = x$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266518/exam_images/peptywbkfivgg2xjbmvn.webp" loading="lazy" alt="AIEEE 2007 Mathematics - Area Under The Curves Question 127 English Explanation">
<br><br>$$\therefore$$ Required area $$ = \int_0^1 {\left( {{y_2} - {y_1}} \right)} dx$$
<br><br>$$ = \int_0^1 {\left( {\sqrt x - x} \right)dx = \left[ {{{{x^{3/2}}} \over {3/2}} - {{{x^2}} \over 2}} \right]} _0^1$$
<br><br>$$\therefore$$ Required area $$ = {2 \over 3}\left[ {{x^{3/2}}} \right]_0^1 - {1 \over 2}\left[ {{x^2}} \right]_0^1$$
<br><br>$$ = {2 \over 3} - {1 \over 2} = {1 \over 6}$$ | mcq | aieee-2007 |
pMzAZwRFlpht8r8n | maths | area-under-the-curves | area-bounded-between-the-curves | The area of the plane region bounded by the curves $$x + 2{y^2} = 0$$ and $$\,x + 3{y^2} = 1$$ is equal to : | [{"identifier": "A", "content": "$${5 \\over 3}$$ "}, {"identifier": "B", "content": "$${1 \\over 3}$$"}, {"identifier": "C", "content": "$${2 \\over 3}$$"}, {"identifier": "D", "content": "$${4 \\over 3}$$"}] | ["D"] | null | $$x + 2{y^2} = 0 \Rightarrow {y^2} = - {x \over 2}$$
<br><br>$$\left[ {} \right.$$ Left handed parabola with vertex at $$\left( {0,0} \right)$$ $$\left. {} \right]$$
<br><br>$$x + 3{y^2} = 1 \Rightarrow {y^2} = - {1 \over 3}\left( {x - 1} \right)$$
<br><br>$$\left[ {} \right.$$ Left handed parabola with vertex at $$\left( {1,0} \right)$$ $$\left. {} \right]$$
<br><br>Solving the two equations we get the points of intersection
<br><br>as $$\left( { - 2,1} \right),\left( { - 2, - 1} \right)$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266232/exam_images/czdd2doug0cscdzgdhxu.webp" loading="lazy" alt="AIEEE 2008 Mathematics - Area Under The Curves Question 126 English Explanation">
<br><br>The required area is $$ACBDA,$$ given by
<br><br>$$ = \left| {\int\limits_{ - 1}^1 {\left( {1 - 3{y^2} - 2{y^2}} \right)dy} } \right|$$
<br><br>$$ = \left| {\left[ {y - {{5{y^3}} \over 3}} \right]_{ - 1}^1} \right|$$
<br><br>$$ = \left| {\left( {1 - {5 \over 3}} \right) - \left( { - 1 + {5 \over 3}} \right)} \right|$$
<br><br>$$ = 2 \times {2 \over 3} = {4 \over 3}\,\,$$ sq. units. | mcq | aieee-2008 |
IFwRcsOuCCngSl5t | maths | area-under-the-curves | area-bounded-between-the-curves | The area of the region bounded by the parabola $${\left( {y - 2} \right)^2} = x - 1,$$ the tangent of the parabola at the point $$(2, 3)$$ and the $$x$$-axis is : | [{"identifier": "A", "content": "$$6$$"}, {"identifier": "B", "content": "$$9$$"}, {"identifier": "C", "content": "$$12$$"}, {"identifier": "D", "content": "$$3$$"}] | ["B"] | null | The given parabola is $${\left( {y - 2} \right)^2} = x - 1$$
<br><br>Vertex $$\left( {1,2} \right)$$ and it meets $$x$$-axis at $$\left( {5,0} \right)$$
<br><br>Also it gives $${y^2} - 4y - x + 5 = 0$$
<br><br>So, that equation of tangent to the parabola at $$\left( {2,3} \right)$$ is
<br><br>$$y.3 - 2\left( {y + 3} \right) - {1 \over 2}\left( {x + 2} \right) + 5 = 0$$
<br><br>or $$x - 2y + 4 = 0$$
<br><br>which meets $$x$$-axis at $$\left( { - 4,0} \right).$$
<br><br>In the figure shaded area is the required area.
<br><br>Let us draw $$PD$$ perpendicular to $$y$$-axis.
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264492/exam_images/dhvwqvaj51muxjrzua8j.webp" loading="lazy" alt="AIEEE 2009 Mathematics - Area Under The Curves Question 125 English Explanation">
<br><br>Then required area
<br><br>$$ = Ar\,\,\Delta BOA + Ar\,\,\left( {OCPD} \right)\, - \,Ar\left( {\Delta APD} \right)$$
<br><br>$$ = {1 \over 2} \times 4 \times 2 + \int_0^3 {xdy - {1 \over 2}} \times 2 \times 1$$
<br><br>$$ = 3 + \int_0^3 {{{\left( {y - 2} \right)}^2} + 1\,dy} $$
<br><br>$$ = 3 + \left[ {{{{{\left( {y - 2} \right)}^3}} \over 3} + y} \right]_0^3$$
<br><br>$$ = 3 + \left[ {{1 \over 3} + 3 + {8 \over 3}} \right] = 3 + 6 = 9$$ Sq. units | mcq | aieee-2009 |
AX7qp2uZbAWLtltG | maths | area-under-the-curves | area-bounded-between-the-curves | The area bounded by the curves $$y = \cos x$$ and $$y = \sin x$$ between the ordinates $$x=0$$ and $$x = {{3\pi } \over 2}$$ is | [{"identifier": "A", "content": "$$4\\sqrt 2 + 2$$ "}, {"identifier": "B", "content": "$$4\\sqrt 2 - 1$$"}, {"identifier": "C", "content": "$$4\\sqrt 2 + 1$$"}, {"identifier": "D", "content": "$$4\\sqrt 2 - 2$$"}] | ["D"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267132/exam_images/qwadcct8cbggvfesgjha.webp" loading="lazy" alt="AIEEE 2010 Mathematics - Area Under The Curves Question 124 English Explanation">
<br><br>$$\therefore$$ Required area
<br><br>$$ = \left[ {\int\limits_0^{{\pi \over 4}} {\left( {\cos \,x - \sin x} \right)dx + } } \right.$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\int\limits_{{\pi \over 4}}^{{5\pi \over 4}} {\left( {\sin x - \cos x} \right)dx + } $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\int\limits_{{5\pi \over 4}}^{{3\pi \over 2}} {\left( {\cos x - \sin x} \right)dx } $$
<br><br>$$ = 4\sqrt 2 - 2$$ | mcq | aieee-2010 |
6sHxXhoKXtc1O9cA | maths | area-under-the-curves | area-bounded-between-the-curves | The area of the region enclosed by the curves $$y = x,x = e,y = {1 \over x}$$ and the positive $$x$$-axis is : | [{"identifier": "A", "content": "$$1$$ square unit"}, {"identifier": "B", "content": "$${3 \\over 2}$$ square units"}, {"identifier": "C", "content": "$${5 \\over 2}$$ square units"}, {"identifier": "D", "content": "$${1 \\over 2}$$ square unit"}] | ["B"] | null | Area of required region $$AOCB$$
<br><br>$$ = \int\limits_0^1 {xdx} + \int\limits_1^e {{1 \over x}dx = {1 \over 2}} + 1 = {3 \over 2}$$ sq. units
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267579/exam_images/y3qu2ps3iqhwrvq5xytr.webp" loading="lazy" alt="AIEEE 2011 Mathematics - Area Under The Curves Question 123 English Explanation">
<br><br> | mcq | aieee-2011 |
SVlnixqVTARTxEc5 | maths | area-under-the-curves | area-bounded-between-the-curves | The area between the parabolas $${x^2} = {y \over 4}$$ and $${x^2} = 9y$$ and the straight line $$y=2$$ is : | [{"identifier": "A", "content": "$$20\\sqrt 2 $$ "}, {"identifier": "B", "content": "$${{10\\sqrt 2 } \\over 3}$$ "}, {"identifier": "C", "content": "$${{20\\sqrt 2 } \\over 3}$$"}, {"identifier": "D", "content": "$$10\\sqrt 2 $$"}] | ["C"] | null | Given curves $${x^2} = {y \over 4}$$ and $${x^2} = 9y$$ are the parabolas whose equations can be written as $$y = 4{x^2}$$ and $$y = {1 \over 9}{x^2}.$$
<br><br>Also, given $$y=2.$$
<br><br>Now, shaded portion shows the required area which is symmetric.
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263338/exam_images/lx0pkl74ucmclccrmyhl.webp" loading="lazy" alt="AIEEE 2012 Mathematics - Area Under The Curves Question 122 English Explanation">
<br><br>$$\therefore$$ Area $$ = 2\int\limits_0^2 {\left( {\sqrt {9y} - \sqrt {{y \over 4}} } \right)} dy$$
<br><br>Area $$ = 2\int\limits_0^2 {\left( {3\sqrt y - {{\sqrt y } \over 2}} \right)} dy$$
<br><br>$$ = 2\left[ {{2 \over 3} \times 3.{y^{{3 \over 2}}} - {1 \over 2} \times {2 \over 3}.{y^{{3 \over 2}}}} \right]_0^2$$
<br><br>$$ = 2\left[ {2{y^{{3 \over 2}}} - {1 \over 3}{y^{{3 \over 2}}}} \right] = \left. {2 \times {5 \over 3}{y^{{3 \over 2}}}} \right|_0^2$$
<br><br>$$ = 2.{5 \over 3}2\sqrt 2 = {{20\sqrt 2 } \over 3}$$ | mcq | aieee-2012 |
2vnQTfUpqrB6aQZg | maths | area-under-the-curves | area-bounded-between-the-curves | The area (in square units) bounded by the curves $$y = \sqrt {x,} $$ $$2y - x + 3 = 0,$$ $$x$$-axis, and lying in the first quadrant is : | [{"identifier": "A", "content": "$$9$$ "}, {"identifier": "B", "content": "$$36$$ "}, {"identifier": "C", "content": "$$18$$"}, {"identifier": "D", "content": "$${{27} \\over 4}$$ "}] | ["A"] | null | Given curves are
<br><br>$$y = \sqrt x $$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>and $$2y - x + 3 = 0\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
<br><br>On solving both we get $$y=-1,3$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266589/exam_images/p8cxbhqzhcyvdc09huxd.webp" loading="lazy" alt="JEE Main 2013 (Offline) Mathematics - Area Under The Curves Question 118 English Explanation">
<br><br>Required area $$ = \int\limits_0^3 {\left\{ {\left( {2y + 3} \right) - {y^2}} \right\}} dy$$
<br><br>$$\left. {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {y^2} + 3y - {{{y^3}} \over 3}} \right|_0^3 = 9.$$ | mcq | jee-main-2013-offline |
DI501ESEKnVjEWBw | maths | area-under-the-curves | area-bounded-between-the-curves | The area of the region described by
<br/>$$A = \left\{ {\left( {x,y} \right):{x^2} + {y^2} \le 1} \right.$$ and $$\left. {{y^2} \le 1 - x} \right\}$$ is : | [{"identifier": "A", "content": "$${\\pi \\over 2} - {2 \\over 3}$$ "}, {"identifier": "B", "content": "$${\\pi \\over 2} + {2 \\over 3}$$"}, {"identifier": "C", "content": "$${\\pi \\over 2} + {4 \\over 3}$$"}, {"identifier": "D", "content": "$${\\pi \\over 2} - {4 \\over 3}$$"}] | ["C"] | null | Given curves are $${x^2} + {y^2} = 1$$ and $${y^2} = 1 - x.$$
<br><br>Intersection points are $$x = 0,1$$
<br><br>Area of shaded portion is the required area.
<br><br>So, Required Area $$=$$ Area of semi-circle $$+$$ Area bounded by parabola
<br><br>$$ = {{\pi {r^2}} \over 2} + 2\int\limits_0^1 {\sqrt {1 - x} dx} $$
<br><br>$$ = {\pi \over 2} + 2\int\limits_0^1 {\sqrt {1 - x} \,dx} $$
<br><br>( As radius of circle $$=1$$ )
<br><br>$$ = {\pi \over 2} + 2\left[ {{{{{\left( {1 - x} \right)}^{{\raise0.5ex\hbox{$\scriptstyle 3$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}} \over { - {\raise0.5ex\hbox{$\scriptstyle 3$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}} \right]_0^1$$
<br><br>$$ = {\pi \over 2} - {4 \over 3}\left( { - 1} \right) = {\pi \over 2} + {4 \over 3}$$ Sq. unit | mcq | jee-main-2014-offline |
XTxcDJGWpMyOANNI | maths | area-under-the-curves | area-bounded-between-the-curves | The area (in sq. units) of the region described by
<br/><br>$$\left\{ {\left( {x,y} \right):{y^2} \le 2x} \right.$$ and $$\left. {y \ge 4x - 1} \right\}$$ is :</br> | [{"identifier": "A", "content": "$${{15} \\over {64}}$$ "}, {"identifier": "B", "content": "$${{9} \\over {32}}$$"}, {"identifier": "C", "content": "$${{7} \\over {32}}$$"}, {"identifier": "D", "content": "$${{5} \\over {64}}$$"}] | ["B"] | null | Required area
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l91nf3yh/b4e886d1-7f41-4b72-ab7a-245bf94417c9/ef2e8480-47fb-11ed-8757-0f869593f41f/file-1l91nf3yi.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l91nf3yh/b4e886d1-7f41-4b72-ab7a-245bf94417c9/ef2e8480-47fb-11ed-8757-0f869593f41f/file-1l91nf3yi.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2015 (Offline) Mathematics - Area Under The Curves Question 120 English Explanation"><br>$$ = $$ Area of $$ABCD$$ $$-$$ $$ar$$ $$(ABOCD)$$
<br><br $$="\int\limits_{" -="" 1="" 2}^1="" {{{y="" +="" 1}="" \over="" 4}}="" dy="" \int\limits_{="" {{{{y^2}}="" 2}dy}="" <br=""><br>$$ = {1 \over 4}\left[ {{{{y^2}} \over 2} + y} \right]_{ - 1/2}^1\,\, - {1 \over 2}\left[ {{{{y^3}} \over 3}} \right]_{ - 1}^1$$
<br><br>$$ = {1 \over 4}\left[ {{3 \over 2} + {3 \over 8}} \right] - {9 \over {48}}$$
<br><br>$$ = {{15} \over {32}} - {9 \over {48}} = {{27} \over {96}} = {9 \over {32}}$$ | mcq | jee-main-2015-offline |
pgV7X2hCYOLs7nkDCGAqG | maths | area-under-the-curves | area-bounded-between-the-curves | The area (in sq. units) of the region described by
<br/><br/>A= {(x, y) $$\left| {} \right.$$y$$ \ge $$ x<sup>2</sup> $$-$$ 5x + 4, x + y $$ \ge $$ 1, y $$ \le $$ 0} is : | [{"identifier": "A", "content": "$${7 \\over 2}$$ "}, {"identifier": "B", "content": "$${{19} \\over 6}$$ "}, {"identifier": "C", "content": "$${{13} \\over 6}$$"}, {"identifier": "D", "content": "$${{17} \\over 6}$$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265211/exam_images/xedy87gs6fsevx9ehium.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2016 (Online) 9th April Morning Slot Mathematics - Area Under The Curves Question 112 English Explanation">
<br><br>Required Area
<br><br>= A<sub>1</sub> + A<sub>2</sub>
<br><br>= $$\left| {\int\limits_1^3 {\left( {1 - x} \right)} dx} \right| + \left| {\int\limits_3^4 {\left( {{x^2} - 5x + 4} \right)dx} } \right|$$
<br><br>= $$\left| {\left[ {x - {{{x^2}} \over 2}} \right]_1^3} \right| + \left| {\left[ {{{{x^3}} \over 3} - {5 \over 2}{x^2} + 4x} \right]_3^4} \right|$$
<br><br>= $$\left| {\left[ {\left( {3 - {9 \over 2}} \right) - \left( {1 - {1 \over 2}} \right)} \right]} \right| + \left| {\left[ {\left( {{{64} \over 3} - 40 + 16} \right) - \left( {9 - {{45} \over 2} + 12} \right)} \right]} \right|$$
<br><br>= $$\left| {\left( {2 - 4} \right)} \right| + \left| {\left( {{{ - 8} \over 3} + {3 \over 2}} \right)} \right|$$
<br><br>= 2 + $${7 \over 6}$$
<br><br>= $${{19} \over 6}$$ sq. unit. | mcq | jee-main-2016-online-9th-april-morning-slot |
UNVOXbPQzfaJPPXv | maths | area-under-the-curves | area-bounded-between-the-curves | The area (in sq. units) of the region $$\left\{ {\left( {x,y} \right):{y^2} \ge 2x\,\,\,and\,\,\,{x^2} + {y^2} \le 4x,x \ge 0,y \ge 0} \right\}$$ is : | [{"identifier": "A", "content": "$$\\pi - {{4\\sqrt 2 } \\over 3}$$ "}, {"identifier": "B", "content": "$${\\pi \\over 2} - {{2\\sqrt 2 } \\over 3}$$ "}, {"identifier": "C", "content": "$$\\pi - {4 \\over 3}$$ "}, {"identifier": "D", "content": "$$\\pi - {8 \\over 3}$$"}] | ["D"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264764/exam_images/zyqeh2enbhsl6lnxeebb.webp" loading="lazy" alt="JEE Main 2016 (Offline) Mathematics - Area Under The Curves Question 121 English Explanation">
<br><br>Points of intersection of the two curves are $$\left( {0,0} \right),\left( {2,2} \right)$$ and $$\left( {2, - 2} \right)$$
<br><br>Area $$=$$ Area $$(OAB)-$$ area under parabola ($$0$$ to $$2$$ )
<br><br>$$ = {{\pi \times {{\left( 2 \right)}^2}} \over 4} - \int\limits_0^2 {\sqrt 2 \sqrt x } \,dx$$
<br><br>$$ = \pi - {8 \over 3}$$ | mcq | jee-main-2016-offline |
d3UyzVS5F5KKpkXlftdxY | maths | area-under-the-curves | area-bounded-between-the-curves | The area (in sq. units) of the smaller portion enclosed between the curves, x<sup>2</sup> + y<sup>2</sup> = 4 and y<sup>2</sup> = 3x, is : | [{"identifier": "A", "content": "$${1 \\over {2\\sqrt 3 }} + {\\pi \\over 3}$$ "}, {"identifier": "B", "content": "$${1 \\over {\\sqrt 3 }} + {{2\\pi } \\over 3}$$"}, {"identifier": "C", "content": "$${1 \\over {2\\sqrt 3 }} + {{2\\pi } \\over 3}$$"}, {"identifier": "D", "content": "$${1 \\over {\\sqrt 3 }} + {{4\\pi } \\over 3}$$"}] | ["D"] | null | <p>The given equation $${x^2} + {y^2} = 4$$ is equation of circle of radius 2 centred at origin and equation $${y^2} = 3x$$ is the equation of parabola.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l3bio1x7/04a063ba-ee8e-4754-b8f8-b564026e9f6e/e47771a0-d69f-11ec-a354-cb09522b689a/file-1l3bio1x8.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l3bio1x7/04a063ba-ee8e-4754-b8f8-b564026e9f6e/e47771a0-d69f-11ec-a354-cb09522b689a/file-1l3bio1x8.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2017 (Online) 8th April Morning Slot Mathematics - Area Under The Curves Question 113 English Explanation"></p>
<p>$${x^2} + {y^2} = 4$$ ..... (1)</p>
<p>$${y^2} = 3x$$ ..... (2)</p>
<p>Substituting Eq. (2) in Eq. (1), we get</p>
<p>$${x^2} + 3x - 4 = 0$$</p><p></p>
<p>$$ \Rightarrow {x^2} + 4x - x - 4 = 0$$</p>
<p>$$ \Rightarrow x(x + 4) - 1(x + 4) = 0$$</p>
<p>$$ \Rightarrow (x - 1)(x + 4) = 0$$</p>
<p>$$ \Rightarrow (x - 1) = 0$$ and $$(x + 4) = 0$$</p>
<p>Therefore, x = 1, $$-$$4. Considering x = 1, then from Eq. (2), we get $$y = \sqrt 3 , - \sqrt 3 $$.</p>
<p>Therefore $$(1,\sqrt 3 )$$ and $$(1, - \sqrt 3 )$$ are the points of intersection of parabola and circle.</p>
<p>The required area (A) is the area of the shaded region shown in the figure. Therefore,</p>
<p>$$A = 2\left[ {\int\limits_0^1 {{y_2}dx + \int\limits_1^2 {{y_1}dx} } } \right]$$</p>
<p>From Eq. (1), we get $${y_1} = \sqrt {4 - {x^2}} $$</p>
<p>From Eq. (2), we get $${y_2} = \sqrt {3x} $$</p>
<p>Therefore, $$A = 2\left[ {\int\limits_0^1 {\sqrt {3x} dx + \int\limits_1^2 {\sqrt {4 - {x^2}} dx} } } \right]$$</p>
<p>$$ = 2\left[ {\int\limits_0^1 {\sqrt 3 {x^{1/2}}dx + \int\limits_1^2 {\sqrt {{2^2} - {x^2}} dx} } } \right]$$</p>
<p>Using standard integral : $$\int {{x^n}dx = {{{x^{n - 1}}} \over {n + 1}}} $$, we have</p>
<p>$$\int {\sqrt {{a^2} - {x^2}} dx = {x \over 2}\sqrt {{a^2} - {x^2}} + {{{a^2}} \over 2}{{\tan }^{ - 1}}{x \over {\sqrt {{a^2} - {x^2}} }}} $$</p>
<p>Therefore,</p>
<p>$$A = 2\left[ {\left. {\left( {\sqrt 3 {{{x^{3/2}}} \over {3/2}}} \right)} \right|_0^1 + \left. {\left( {{x \over 2}\sqrt {4 - {x^2}} + {4 \over 2}{{\tan }^{ - 1}}{x \over {\sqrt {4 - {x^2}} }}} \right)} \right|_1^2} \right]$$</p>
<p>$$ = 2\left[ {\sqrt 3 .{1 \over {3/2}} - 0 + {2 \over 2}\sqrt {4 - 4} + 2{{\tan }^{ - 1}}{2 \over {\sqrt {4 - 4} }} - {1 \over 2}\sqrt {4 - 1} - 2{{\tan }^{ - 1}}{1 \over {\sqrt {4 - 1} }}} \right]$$</p>
<p>$$ = 2\left[ {{{2\sqrt 3 } \over 3} + {{\tan }^{ - 1}}(\infty ) - {{\sqrt 3 } \over 2} - 2{{\tan }^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)} \right]$$</p>
<p>Now, $$\tan {\pi \over 2} = \infty $$ and $$\tan {\pi \over 6} = {1 \over {\sqrt 3 }}$$. Therefore, the area of the smaller portion enclosed between the two curves is obtained as follows:</p>
<p>$$A = 2\left[ {{{2\sqrt 3 } \over 3} - {{\sqrt 3 } \over 2} + 2{{\tan }^{ - 1}}\left( {\tan {\pi \over 2}} \right) - 2{{\tan }^{ - 1}}\left( {\tan {\pi \over 6}} \right)} \right]$$</p>
<p>$$ = 2\left[ {\sqrt 3 {1 \over 6} + 2{\pi \over 2} - 2{\pi \over 6}} \right]$$</p>
<p>$$ = 2\left[ {\sqrt 3 + {1 \over 6} + 2{{2\pi } \over 6}} \right] = 2\left[ {{{\sqrt 3 } \over 6} + {{4\pi } \over 6}} \right]$$</p>
<p>$$ = {{\sqrt 3 } \over 3} + {{4\pi } \over 3} = \left( {{1 \over {\sqrt 3 }} + {{4\pi } \over 3}} \right)$$ sq. units</p> | mcq | jee-main-2017-online-8th-april-morning-slot |
FEfb6CgsB2oAvcfP | maths | area-under-the-curves | area-bounded-between-the-curves | The area (in sq. units) of the region
<br/><br/>$$\left\{ {\left( {x,y} \right):x \ge 0,x + y \le 3,{x^2} \le 4y\,and\,y \le 1 + \sqrt x } \right\}$$ is | [{"identifier": "A", "content": "$${3 \\over 2}$$"}, {"identifier": "B", "content": "$${7 \\over 3}$$"}, {"identifier": "C", "content": "$${5 \\over 2}$$"}, {"identifier": "D", "content": "$${59 \\over 12}$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266336/exam_images/dfewqownwnzl1h0lfyw7.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2017 (Offline) Mathematics - Area Under The Curves Question 116 English Explanation">
<br>Area of shaded region
<br><br>= $$\int\limits_0^1 {\left( {1 + \sqrt x } \right)dx} + \int\limits_1^2 {\left( {3 - x} \right)dx} - \int\limits_0^2 {{{{x^2}} \over 4}dx} $$
<br><br>= $$\left[ x \right]_0^1 + \left[ {{{{x^{{3 \over 2}}}} \over {{3 \over 2}}}} \right]_0^1$$ + $$3\left[ x \right]_1^2 - \left[ {{{{x^2}} \over 2}} \right]_1^2 - \left[ {{{{x^3}} \over {12}}} \right]_0^2$$
<br><br>= $${5 \over 2}$$ sq. unit | mcq | jee-main-2017-offline |
ThczWi1ecPYlB5lQwPItg | maths | area-under-the-curves | area-bounded-between-the-curves | The area (in sq. units) of the region
<br/><br/>{x $$ \in $$ <b>R</b> : x $$ \ge $$ 0, y $$ \ge $$ 0, y $$ \ge $$ x $$-$$ 2 Β <i>and</i> y $$ \le $$ $$\sqrt x $$}, is : | [{"identifier": "A", "content": "$${{13} \\over 3}$$"}, {"identifier": "B", "content": "$${{8} \\over 3}$$"}, {"identifier": "C", "content": "$${{10} \\over 3}$$"}, {"identifier": "D", "content": "$${{5} \\over 3}$$"}] | ["C"] | null | y = $$\sqrt x $$
<br><br>y = x $$-$$ 2
<br><br>$$\therefore\,\,\,$$ $$\sqrt x $$ = x $$-$$ 2
<br><br>$$ \Rightarrow $$$$\,\,\,$$ x = x<sup>2</sup> $$-$$ 4x + 4
<br><br>x<sup>2</sup> $$-$$ 5x + 4 = 0
<br><br>x<sup>2</sup> $$-$$ 4x $$-$$ x + 4 = 0
<br><br>$$ \Rightarrow $$$$\,\,\,$$ x(x $$-$$ 4) $$-$$ (x $$-$$ 4) = 0
<br><br>$$ \Rightarrow $$$$\,\,\,$$ (x $$-$$ 4) (x $$-$$ 1) = 0
<br><br>$$\therefore\,\,\,$$ x = 4, 1
<br><br>and y = 2, $$-$$ 1
<br><br>$$\therefore\,\,\,$$ Their point of intersection (4, 2) and (1, $$-$$ 1)
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263924/exam_images/cx2cz6kvnuizfpmqmwgs.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2018 (Online) 15th April Morning Slot Mathematics - Area Under The Curves Question 115 English Explanation">
<br><br>Required area is shown in the shaded figure.
<br><br>$$\therefore\,\,\,$$ Required area
<br><br>= $$\int\limits_0^2 {\sqrt x \,dx + \int\limits_2^4 {\left( {\sqrt x - x + 2} \right)\,dx} } $$
<br><br>= $$\int\limits_0^4 {\sqrt x \,dx + \int\limits_2^4 {\left( {2 - x} \right)\,dx} } $$
<br><br>= $$\left[ {{2 \over 3}{x^{{3 \over 2}}}} \right]_0^4 + \left[ {2x - {{{x^2}} \over 2}} \right]_2^4$$
<br><br>= $${2 \over 3}\left( 8 \right)$$ + 2$$\left( {4 - 2} \right)$$ $$-$$ $${1 \over 2}$$ (16 $$-$$ 4)
<br><br>= $${{16} \over 3}$$ + 4 $$-$$ 6
<br><br>= $${{10} \over 3}$$ | mcq | jee-main-2018-online-15th-april-morning-slot |
8L5Ej5Etn8PQ5O4I0IQ6b | maths | area-under-the-curves | area-bounded-between-the-curves | If the area of the region bounded by the curves, $$y = {x^2},y = {1 \over x}$$ and the lines y = 0 and x= t (t >1) is 1 sq. unit, then t is equal to : | [{"identifier": "A", "content": "$${e^{{3 \\over 2}}}$$"}, {"identifier": "B", "content": "$${4 \\over 3}$$"}, {"identifier": "C", "content": "$${3 \\over 2}$$"}, {"identifier": "D", "content": "$${e^{{2 \\over 3}}}$$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265062/exam_images/b58nqrkplxkhfj2zdyvj.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2018 (Online) 16th April Morning Slot Mathematics - Area Under The Curves Question 114 English Explanation">
<br><br>Point of intersection of y = x<sup>2</sup> and y = $${1 \over x}$$.
<br><br>put y = $${1 \over x}$$ in y = x<sup>2</sup>, then we get,
<br><br>$${1 \over x} = {x^2}$$
<br><br>$$ \Rightarrow $$ $$\,\,\,$$ x<sup>3</sup> $$-$$ 1 = 0
<br><br>$$ \Rightarrow $$ $$\,\,\,$$ x = 1
<br><br>$$\therefore\,\,\,$$ y = 1
<br><br>$$\therefore\,\,\,$$ point B = (1, 1)
<br><br>Area of region ABCDA
<br><br>= $$\int\limits_0^1 {{x^2}} $$ dx + $$\int\limits_1^t {{1 \over x}} $$ dx
<br><br>$$=$$ $$\left[ {{{{x^3}} \over 3}} \right]_0^1$$ + $$\left[ {\ell n\,x} \right]_1^t$$
<br><br>= $${1 \over 3}$$ + $$\ell n\,t$$ $$-$$ $$\ell n$$ 1
<br><br>= $${1 \over 3}$$ + $$\ell n\,t$$ [ as $$\ell n$$ 1 = 0]
<br><br>given this Area = 1 sq unit.
<br><br>$$\therefore\,\,\,$$ $${1 \over 3}$$ + $$\ell n\,t$$ = 1
<br><br>$$ \Rightarrow $$ $$\ell n\,t$$ = $${2 \over 3}$$
<br><br>$$ \Rightarrow $$ t = e$${^{{2 \over 3}}}$$ | mcq | jee-main-2018-online-16th-april-morning-slot |
1YN4VE8zviTQTiOCB718hoxe66ijvwoz4p3 | maths | area-under-the-curves | area-bounded-between-the-curves | The area (in sq. units) of the region
<br/><br/>A = {(x, y) : x<sup>2</sup> $$ \le $$ y $$ \le $$ x + 2} is | [{"identifier": "A", "content": "$${{31 \\over 6}}$$"}, {"identifier": "B", "content": "$${{10 \\over 3}}$$"}, {"identifier": "C", "content": "$${{13 \\over 6}}$$"}, {"identifier": "D", "content": "$${{9 \\over 2}}$$"}] | ["D"] | null | Parabola : x<sup>2</sup> = y
<br><br>Straight line : y = x + 2
<br><br>$$ \therefore $$ x<sup>2</sup> = x + 2
<br><br>$$ \Rightarrow $$ x<sup>2</sup> - x - 2 = 0
<br><br>x = -1, 2
<br><br>$$ \therefore $$ y = 1, 4
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264695/exam_images/e5vhjwjekrzf7uf2prs5.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264428/exam_images/yibgkyg5oc2qlpiazwxv.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266090/exam_images/qqdptvmz6bwhqedphfk6.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th April Morning Slot Mathematics - Area Under The Curves Question 103 English Explanation"></picture>
<br><br>Required area = $$\int\limits_{ - 1}^2 {\left[ {\left( {x + 2} \right) - {x^2}} \right]} dx$$
<br><br>= $$\left[ {{{{x^2}} \over 2} + 2x - {{{x^3}} \over 3}} \right]_{ - 1}^2$$
<br><br>= $$\left( {2 + 4 - {8 \over 3}} \right) - \left( {{1 \over 2} - 2 + {1 \over 3}} \right)$$
<br><br>= $${{9 \over 2}}$$ | mcq | jee-main-2019-online-9th-april-morning-slot |
olQ2CuKJqbEOapJsuh3rsa0w2w9jxaoline | maths | area-under-the-curves | area-bounded-between-the-curves | If the area (in sq. units) bounded by the parabola y<sup>2</sup>
= 4$$\lambda $$x and the line y = $$\lambda $$x, $$\lambda $$ > 0, is $${1 \over 9}$$
, then $$\lambda $$ is equal to : | [{"identifier": "A", "content": "$$4\\sqrt 3 $$"}, {"identifier": "B", "content": "2$$\\sqrt 6 $$"}, {"identifier": "C", "content": "48"}, {"identifier": "D", "content": "24"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266672/exam_images/z4km1mkoieljdfzlmsq9.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th April Evening Slot Mathematics - Area Under The Curves Question 99 English Explanation">
y<sup>2</sup> = 4$$\lambda $$x and y = $$\lambda x$$<br><br>
If $$\lambda $$ > 0 then<br><br>
Hence $$\int\limits_0^{4/\lambda } {(2\sqrt \lambda \sqrt x } - \lambda x)dx = {1 \over 9}$$<br><br>
$$ \Rightarrow $$ $${\left( {{{2\sqrt \lambda {x^{3/2}}} \over {3/2}} - {{\lambda {x^2}} \over 2}} \right)^{{4 \over \lambda }}} = {1 \over 9}$$<br><br>
$$ \Rightarrow $$ $${4 \over 3}\sqrt \lambda {8 \over {{\lambda ^{3/2}}}} - \lambda {8 \over {{\lambda ^2}}} = {1 \over 9}$$<br><br>
$$ \Rightarrow $$ $${{32} \over {3\lambda }} - {8 \over \lambda } = {1 \over 9}$$<br><br>
$$ \Rightarrow $$ $$ {8 \over 3\lambda } = {1 \over 9}$$<br><br>
$$ \Rightarrow $$ $$\lambda $$ = 24 | mcq | jee-main-2019-online-12th-april-evening-slot |
VgYeNczCUuvuSa2x6S3rsa0w2w9jx61y3q1 | maths | area-under-the-curves | area-bounded-between-the-curves | If the area (in sq. units) of the region {(x, y) : y<sup>2</sup>
$$ \le $$ 4x, x + y $$ \le $$ 1, x $$ \ge $$ 0, y $$ \ge $$ 0} is a $$\sqrt 2 $$ + b, then a β b is equal
to : | [{"identifier": "A", "content": "$${8 \\over 3}$$"}, {"identifier": "B", "content": "$$ - {2 \\over 3}$$"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "$${{10} \\over 3}$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267762/exam_images/pgiki0cdc9cjw0ntf4lc.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th April Morning Slot Mathematics - Area Under The Curves Question 100 English Explanation">
Let P be the point common to x + y = 1 and y<sup>2</sup> = 4x<br><br>
Then y<sup>2</sup> = 4(1-y) $$ \Rightarrow $$ y<sup>2</sup> - 4y - 4 = 0 <br><br>
$$ \Rightarrow y = {{ - 4 \pm \sqrt {16 + 16} } \over 2}$$<br><br>
$$ \Rightarrow y = - 2 + 2\sqrt 2 $$<br><br>
Then P is (3 - $$ 2\sqrt 2$$, -2 + $$ 2\sqrt 2$$).<br><br>
Hence shaded area = Area of region (OPN) + area of ($$\Delta OPQ$$)<br><br>
$$ \Rightarrow {\int\limits_0^{3 - 2\sqrt 2 } {2\sqrt {xdx} + {1 \over 2}\left[ {1 - (3 - 2\sqrt 2 )} \right]} ^2}$$<br><br>
$$ \Rightarrow {2 \over 3}2\left( {\sqrt 2 - 1} \right)\left( {3 - 2\sqrt 2 } \right) + {1 \over 2}{\left[ {2(\sqrt 2 - 1)} \right]^2}$$<br><br>
$$ \Rightarrow {4 \over 3}\left\{ { - 7 + 5\sqrt 2 } \right\} + 2\left( {3 - 2\sqrt 2 } \right)$$<br><br>
$$ \Rightarrow \left( {{{20} \over 3} - 4} \right)\sqrt 2 + 6 - {{28} \over 3}$$<br><br>
$$ \Rightarrow {8 \over 3}\sqrt 2 - {{10} \over 3}$$<br><br>
Then a = $${8 \over 3}$$ and b = $${-10 \over 3}$$, so a - b = 6 | mcq | jee-main-2019-online-12th-april-morning-slot |
I1nmB5bUtKNBoasg383rsa0w2w9jx25emr2 | maths | area-under-the-curves | area-bounded-between-the-curves | The area (in sq.units) of the region bounded by the curves y = 2<sup>x</sup>
and y = |x + 1|, in the first quadrant is :
| [{"identifier": "A", "content": "$${1 \\over 2}$$"}, {"identifier": "B", "content": "$${3 \\over 2}$$"}, {"identifier": "C", "content": "$${3 \\over 2} - {1 \\over {\\log _e^2}}$$"}, {"identifier": "D", "content": "$$\\log _e^2 + {3 \\over 2}$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263379/exam_images/pwnxyxw58zdpddygd57j.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th April Evening Slot Mathematics - Area Under The Curves Question 101 English Explanation"><br><br>
Required area = $$\int\limits_0^1 {((x + 1) - {2^x})dx} $$<br><br>
$$ \Rightarrow \left( {{{{x^2}} \over 2} + x - {{{2^x}} \over {{{\log }_e}2}}} \right)_0^1$$<br><br>
$$ \Rightarrow {3 \over 2} - {1 \over {{{\log }_e}2}}$$ | mcq | jee-main-2019-online-10th-april-evening-slot |
TsupCJrGClDjxoGzUl18hoxe66ijvww5mu1 | maths | area-under-the-curves | area-bounded-between-the-curves | The area (in sq. units) of the region<br/>
A = {(x, y) : $${{y{}^2} \over 2}$$ $$ \le $$ x $$ \le $$ y + 4} is :- | [{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "18"}, {"identifier": "C", "content": "$${{53} \\over 3}$$"}, {"identifier": "D", "content": "16"}] | ["B"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265986/exam_images/g7vhs2kmokzw9dgrvimz.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264472/exam_images/wbpeghezagympwrgiinu.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264662/exam_images/oejbpdgknl2hbpnnncph.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th April Evening Slot Mathematics - Area Under The Curves Question 102 English Explanation"></picture>
<br><br>y<sup>2</sup> = 2x ...........(1)
<br><br>and x = y + 4 .............(2)
<br><br>Solving (1) and (2)
<br><br>(x - 4)<sup>2</sup> = 2x
<br><br>$$ \Rightarrow $$ x<sup>2</sup> - 10x + 16 = 0
<br><br>$$ \Rightarrow $$ x = 8, 2 and y = 4, -2
<br><br>Integrating in y direction from A to B
<br><br>Required area (A) = $$\int\limits_{ - 2}^4 {\left( {y + 4 - {{{y^2}} \over 2}} \right)dy} $$
<br><br>= $$\left[ {{{{y^2}} \over 2} + 4y - {{{y^3}} \over 6}} \right]_{ - 2}^4$$
<br><br>= $${\left( {{{16} \over 2} + 16 - {{64} \over 6}} \right)}$$ - $${\left( {{4 \over 2} - 8 + {8 \over 6}} \right)}$$
<br><br>= 30 - 12 sq. unit
<br><br>= 18 sq. unit | mcq | jee-main-2019-online-9th-april-evening-slot |
1dTbUwB1nv6yAQrJbYghx | maths | area-under-the-curves | area-bounded-between-the-curves | The area (in sq. units) of the region bounded by the parabola, y = x<sup>2</sup> + 2 and the lines, y = x + 1, x = 0 and x = 3, is | [{"identifier": "A", "content": "$${{15} \\over 4}$$"}, {"identifier": "B", "content": "$${{15} \\over 2}$$"}, {"identifier": "C", "content": "$${{21} \\over 2}$$"}, {"identifier": "D", "content": "$${{17} \\over 4}$$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264477/exam_images/utoesgjenyhl0zmw0ys8.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Morning Slot Mathematics - Area Under The Curves Question 106 English Explanation">
<br><br>Required area
<br><br>$$ = \int\limits_0^3 {\left( {{x^2} + 2} \right)dx - {1 \over 2}.5.3 = 9 + 6 - {{15} \over 2}} {}$$
<br><br>$$ = {{15} \over 2}$$ | mcq | jee-main-2019-online-12th-january-morning-slot |
mdpcBn9fdbD5jQ5ut30AV | maths | area-under-the-curves | area-bounded-between-the-curves | The area (in sq. units) in the first quadrant bounded by the parabola, y = x<sup>2</sup> + 1, the tangent to it at the point (2, 5) and the coordinate axes is : | [{"identifier": "A", "content": "$${8 \\over 3}$$"}, {"identifier": "B", "content": "$${{14} \\over 3}$$"}, {"identifier": "C", "content": "$${{187} \\over {24}}$$"}, {"identifier": "D", "content": "$${{37} \\over {24}}$$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264058/exam_images/uhew9xrggiezdnledix0.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 11th January Evening Slot Mathematics - Area Under The Curves Question 107 English Explanation">
<br><br>Area $$ = \int\limits_0^2 {\left( {{x^2} + 1} \right)dx - {1 \over 2}} \left( {{5 \over 4}} \right)\left( 5 \right) = {{37} \over {24}}$$ | mcq | jee-main-2019-online-11th-january-evening-slot |
jYiGJS0vOPirByaUJpCZg | maths | area-under-the-curves | area-bounded-between-the-curves | The area (in sq. units) of the region bounded by the curve x<sup>2</sup> = 4y and the straight line x = 4y β 2 is :
| [{"identifier": "A", "content": "$${3 \\over 4}$$"}, {"identifier": "B", "content": "$${5 \\over 4}$$"}, {"identifier": "C", "content": "$${7 \\over 8}$$"}, {"identifier": "D", "content": "$${9 \\over 8}$$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263553/exam_images/ajducliqewdrjwwiwzo7.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 11th January Morning Slot Mathematics - Area Under The Curves Question 108 English Explanation">
<br>x = 4y $$-$$ 2 & x<sup>2</sup> = 4y
<br><br>$$ \Rightarrow $$ x<sup>2</sup> = x + 2 $$ \Rightarrow $$ x<sup>2</sup> $$-$$ x $$-$$ 2 = 0
<br><br>x = 2, $$-$$ 1
<br><br>So, $$\int\limits_{ - 1}^2 {\left( {{{x + 2} \over 4} - {{{x^2}} \over 4}} \right)\,dx = {9 \over 8}} $$ | mcq | jee-main-2019-online-11th-january-morning-slot |
Ln6s5OIuo2KWZAUu98dpB | maths | area-under-the-curves | area-bounded-between-the-curves | If the area enclosed between the curves y = kx<sup>2</sup> and x = ky<sup>2</sup>, (k > 0), is 1 square unit. Then k is - | [{"identifier": "A", "content": "$$\\sqrt 3 $$"}, {"identifier": "B", "content": "$${{\\sqrt 3 } \\over 2}$$"}, {"identifier": "C", "content": "$${2 \\over {\\sqrt 3 }}$$"}, {"identifier": "D", "content": "$${1 \\over {\\sqrt 3 }}$$"}] | ["D"] | null | Area bounded by
<br><br>y<sup>2</sup> = 4ax & x<sup>2</sup> = 4by, a, b $$ \ne $$ 0
<br><br>is $$\left| {{{16ab} \over 3}} \right|$$
<br><br>by using formula :
<br><br>4a $$=$$ $${1 \over k} = 4b,k > 0$$
<br><br>Area $$ = \left| {{{16.{1 \over {4k}}.{1 \over {4k}}} \over 3}} \right| = 1$$
<br><br>$$ \Rightarrow $$ k<sup>2</sup> $$ = {1 \over 3}$$
<br><br>$$ \Rightarrow $$ k $$ = {1 \over {\sqrt 3 }}$$ | mcq | jee-main-2019-online-10th-january-morning-slot |
uQTkAboyY20GxQ1XzsGfr | maths | area-under-the-curves | area-bounded-between-the-curves | The area (in sq. units) bounded by the parabolae y = x<sup>2</sup> β 1, the tangent at the point (2, 3) to it and the y-axis is : | [{"identifier": "A", "content": "$$56\\over3$$"}, {"identifier": "B", "content": "$$32\\over3$$"}, {"identifier": "C", "content": "$$8\\over3$$"}, {"identifier": "D", "content": "$$14\\over3$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265696/exam_images/atq9a0prxhuwrfqaru4w.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th January Morning Slot Mathematics - Area Under The Curves Question 111 English Explanation">
<br><br>Equation of tangent at (2, 3) on the parabola y = x<sup>2</sup> $$-$$ 1 is
<br><br>$${{y + 3} \over 2} = 2x - 1$$
<br><br>$$ \Rightarrow $$ y + 3 = 4x $$-$$ 2
<br><br>$$ \Rightarrow $$ y = 4x $$-$$ 5
<br><br>When x = 0 then for the tangent y = $$-$$ 5
<br><br>$$ \therefore $$ Tangent cuts x y axis at (0, $$-$$ 5) point.
<br><br>$$ \therefore $$ Area of the bounded region is
<br><br>= $$\int\limits_{ - 5}^3 {{{y + 5} \over 4}} \,\,\,dy - \int\limits_{ - 1}^3 {\sqrt {y + 1} } \,\,\,dy$$
<br><br>= $${1 \over 4}\left[ {{{{y^2}} \over 2} + 5y} \right]_{ - 5}^3 - \left[ {{2 \over 3} \times {{\left( {y + 1} \right)}^{{3 \over 2}}}} \right]_{ - 1}^3$$
<br><br>$${1 \over 4}\left[ {\left( {{9 \over 2} + 15} \right) - \left( {{{25} \over 2} - 25} \right)} \right] - {2 \over 3}{\left( 4 \right)^{{3 \over 2}}}$$
<br><br>= $${1 \over 4}\left[ {{{93} \over 2} + {{25} \over 2}} \right] - {2 \over 3} \times 8$$
<br><br>= $${1 \over 4} \times {{64} \over 2} - {{16} \over 3}$$
<br><br>= $$8 - {{16} \over 3}$$
<br><br>= $${8 \over 3}$$ | mcq | jee-main-2019-online-9th-january-morning-slot |
vc73pE7QbP2xsKjFnTjgy2xukg395b9n | maths | area-under-the-curves | area-bounded-between-the-curves | The area (in sq. units) of the region enclosed
<br/>by the curves y = x<sup>2</sup> β 1 and y = 1 β x<sup>2</sup> is equal to : | [{"identifier": "A", "content": "$${8 \\over 3}$$"}, {"identifier": "B", "content": "$${4 \\over 3}$$"}, {"identifier": "C", "content": "$${7 \\over 2}$$"}, {"identifier": "D", "content": "$${{16} \\over 3}$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267732/exam_images/mtmqm6fur1hjtylyhwhz.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 6th September Evening Slot Mathematics - Area Under The Curves Question 88 English Explanation">
<br><br>A = $$\int\limits_{ - 1}^1 {\left( {\left( {1 - {x^2}} \right) - \left( {{x^2} - 1} \right)} \right)dx} $$
<br><br>= $$\int\limits_{ - 1}^1 {\left( {2 - 2{x^2}} \right)dx} $$
<br><br>= $$4\int\limits_0^1 {\left( {1 - {x^2}} \right)dx} $$
<br><br>= $$4\left( {x - {{{x^3}} \over 3}} \right)_0^1$$
<br><br>= $$4\left( {{2 \over 3}} \right)$$ = $${8 \over 3}$$ | mcq | jee-main-2020-online-6th-september-evening-slot |
7aBmgaMUOgZkTjOAaljgy2xukfuuu3jm | maths | area-under-the-curves | area-bounded-between-the-curves | The area (in sq. units) of the region <br/>A = {(x, y) : |x| + |y| $$ \le $$ 1, 2y<sup>2</sup> $$ \ge $$ |x|} | [{"identifier": "A", "content": "$${1 \\over 6}$$"}, {"identifier": "B", "content": "$${5 \\over 6}$$"}, {"identifier": "C", "content": "$${1 \\over 3}$$"}, {"identifier": "D", "content": "$${7 \\over 6}$$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264818/exam_images/s3vrvzluxopimmp3ygw5.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 6th September Morning Slot Mathematics - Area Under The Curves Question 89 English Explanation">
For point of intersection
<br><br>x + y = 1 $$ \Rightarrow $$ x = 1 β y
<br><br>y<sup>2</sup> = $${x \over 2}$$ $$ \Rightarrow $$ 2y<sup>2</sup>
= x
<br><br>2y<sup>2</sup>
= 1 β y $$ \Rightarrow $$ 2y<sup>2</sup>
+ y β 1 = 0
<br><br>$$ \Rightarrow $$ (2y β 1) (y + 1) = 0
<br><br>$$ \Rightarrow $$ y = $${1 \over 2}$$ or -1
<br><br>Total area = $$4\int\limits_0^{{1 \over 2}} {\left[ {\left( {1 - x} \right) - \left( {\sqrt {{x \over 2}} } \right)} \right]} dx$$
<br><br>= $$4\left[ {x - {{{x^2}} \over 2} - {1 \over {\sqrt 2 }}{{{x^{3/2}}} \over {3/2}}} \right]_0^{{1 \over 2}}$$
<br><br>= $$4\left[ {{1 \over 2} - {1 \over 8} - {{\sqrt 2 } \over 3}{{\left( {{1 \over 2}} \right)}^{3/2}}} \right]$$
<br><br>= 4 $$ \times $$ $${5 \over {24}}$$ = $${5 \over 6}$$ | mcq | jee-main-2020-online-6th-september-morning-slot |
TjLSpk6Qv1h0WchkY2jgy2xukfqfasd9 | maths | area-under-the-curves | area-bounded-between-the-curves | The area (in sq. units) of the region
<br/><br>A = {(x, y) : (x β 1)[x] $$ \le $$ y $$ \le $$ 2$$\sqrt x $$, 0 $$ \le $$ x $$ \le $$ 2}, where [t]
<br/><br>denotes the greatest integer function, is :</br></br> | [{"identifier": "A", "content": "$${8 \\over 3}\\sqrt 2 - 1$$"}, {"identifier": "B", "content": "$${4 \\over 3}\\sqrt 2 + 1$$"}, {"identifier": "C", "content": "$${8 \\over 3}\\sqrt 2 - {1 \\over 2}$$"}, {"identifier": "D", "content": "$${4 \\over 3}\\sqrt 2 - {1 \\over 2}$$"}] | ["C"] | null | y = (x β 1)[x] = $$\left\{ {\matrix{
{0,} & {0 \le x < 1} \cr
{x - 1,} & {1 \le x < 2} \cr
{2,} & {x = 2} \cr
} } \right.$$
<br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264666/exam_images/koeynuqfjc3pdhjdcnwf.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 5th September Evening Slot Mathematics - Area Under The Curves Question 90 English Explanation">
<br><br>A = $$\int\limits_0^2 {2\sqrt x } dx - {1 \over 2}.1.1$$
<br><br>= 2$$\left[ {{{{x^{3/2}}} \over {{3 \over 2}}}} \right]_0^2$$ - $${1 \over 2}$$
<br><br>= $${{8\sqrt 2 } \over 3} - {1 \over 2}$$ | mcq | jee-main-2020-online-5th-september-evening-slot |
oliGjHCWAC079qXRzK7k9k2k5higv7o | maths | area-under-the-curves | area-bounded-between-the-curves | The area (in sq. units) of the region<br/>
<br>{(x,y) $$ \in $$ R<sup>2</sup> : x<sup>2</sup> $$ \le $$ y $$ \le $$ 3 β 2x}, is :</br> | [{"identifier": "A", "content": "$${{34} \\over 3}$$"}, {"identifier": "B", "content": "$${{29} \\over 3}$$"}, {"identifier": "C", "content": "$${{31} \\over 3}$$"}, {"identifier": "D", "content": "$${{32} \\over 3}$$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265543/exam_images/g6ehrq5oh7fip48nkm6y.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 8th January Evening Slot Mathematics - Area Under The Curves Question 95 English Explanation">
<br><br>x<sup>2</sup> $$ \le $$ y $$ \le $$ β 2x + 3
<br><br>$$ \Rightarrow $$ x<sup>2</sup>
= β 2x + 3
<br><br>$$ \Rightarrow $$ x<sup>2</sup>
+ 2x β 3 = 0
<br><br>$$ \Rightarrow $$ (x + 3) (x β 1) = 0
<br><br>$$ \Rightarrow $$ x = β 3, x = 1
<br><br>Area = $$\int\limits_{ - 3}^1 {\left( { - 2x + 3 - {x^2}} \right)dx} $$
<br><br>= $$\left( { - {x^2} + 3x - {{{x^3}} \over 3}} \right)_{ - 3}^1$$
<br><br>= $$\left( { - 1 + 3 - {1 \over 3} + 9 + 9 - 9} \right)$$
<br><br>= 11 - $${{1 \over 3}}$$
<br><br>= $${{{32} \over 3}}$$ sq. unit | mcq | jee-main-2020-online-8th-january-evening-slot |
gHNdC9wBr4VGCvWafkjgy2xukf0p4kl4 | maths | area-under-the-curves | area-bounded-between-the-curves | The area (in sq. units) of the region
<br/><br/>{ (x, y) : 0 $$ \le $$ y $$ \le $$ x<sup>2</sup> + 1, 0 $$ \le $$ y $$ \le $$ x + 1,
<br/><br> $${1 \over 2}$$ $$ \le $$ x $$ \le $$ 2 } is :</br> | [{"identifier": "A", "content": "$${{79} \\over {16}}$$"}, {"identifier": "B", "content": "$${{79} \\over {24}}$$"}, {"identifier": "C", "content": "$${{23} \\over {6}}$$"}, {"identifier": "D", "content": "$${{23} \\over {16}}$$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267143/exam_images/d1iaq40mqsj3mmajo5mu.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 3rd September Morning Slot Mathematics - Area Under The Curves Question 91 English Explanation">
$$A = \int\limits_{{1 \over 2}}^1 {({x^2} + 1)dx + } $$$$\int\limits_1^2 {} $$$${(x + 1)dx}$$<br><br>= $${\left[ {{{{x^3}} \over 3} + x} \right]_{{1 \over 2}}^1 + \left[ {{{{x^2}} \over 2} + x} \right]_1^2}$$<br><br>$${ = \left( {{4 \over 3} - {{13} \over {24}}} \right) + \left( {4 - {3 \over 2}} \right)}$$<br><br>$${ = {{19} \over {24}} + {5 \over 2}}$$<br><br>$${ = {{79} \over {24}}}$$ | mcq | jee-main-2020-online-3rd-september-morning-slot |
vVmxpIPbu31oaPTkNFjgy2xukezery5f | maths | area-under-the-curves | area-bounded-between-the-curves | Consider a region R = {(x, y) $$ \in $$ R : x<sup>2</sup> $$ \le $$ y $$ \le $$ 2x}.
if a line y = $$\alpha $$ divides the area of region R into
two equal parts, then which of the following is
true? | [{"identifier": "A", "content": "3$$\\alpha $$<sup>2</sup> - 8$$\\alpha $$ + 8 = 0"}, {"identifier": "B", "content": "$$\\alpha $$<sup>3</sup> - 6$$\\alpha $$<sup>3/2</sup> - 16 = 0"}, {"identifier": "C", "content": "3$$\\alpha $$<sup>2</sup> - 8$$\\alpha $$<sup>3/2</sup> + 8 = 0"}, {"identifier": "D", "content": "$$\\alpha $$<sup>3</sup> - 6$$\\alpha $$<sup>2</sup> + 16 = 0"}] | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264598/exam_images/wusjy1qensakgeqp8glu.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267704/exam_images/ngz0ndqarqosp7kzmh1p.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265887/exam_images/mvphisn5wpvy2c8cysdw.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263709/exam_images/njtahhijwsodoo0btdmn.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 2nd September Evening Slot Mathematics - Area Under The Curves Question 92 English Explanation"></picture>
<br><br> y $$ \ge $$ x<sup>2</sup> $$ \Rightarrow $$ upper region of y = x<sup>2</sup>
<br><br>y $$ \le $$ 2x $$ \Rightarrow $$ lower region of y = 2x
<br><br> According to question, area of OABC = 2 $$ \times $$ area of OAC
<br><br>$$ \Rightarrow $$ $$\int\limits^{4}_{0} \left( \sqrt{y} -\frac{y}{2} \right) dy$$ = 2$$\int\limits^{\alpha }_{0} \left( \sqrt{y} -\frac{y}{2} \right) dy$$<br><br>
$$\Rightarrow \left[ {{2 \over 3}{y^{{3 \over 2}}} - {{{y^2}} \over 4}} \right]_0^4 = 2\left[ {{2 \over 3}{y^{{3 \over 2}}} - {{{y^2}} \over 4}} \right]_0^\alpha $$<br><br>
$$ \Rightarrow {{16} \over 3} - 4 = 2\left[ {{2 \over 3}{{\left( \alpha \right)}^{{3 \over 2}}} - {{{\alpha ^2}} \over 4}} \right]$$<br><br>
$$ \Rightarrow {4 \over 3} = 2\left[ {{2 \over 3}{\alpha ^{{3 \over 2}}} - {{{\alpha ^2}} \over 4}} \right]$$<br><br>
$$ \Rightarrow {2 \over 3} = {2 \over 3}{\alpha ^{{3 \over 2}}} - {{{\alpha ^2}} \over 4}$$<br><br>
$$ \Rightarrow 8 = 8{\alpha ^{{3 \over 2}}} - 3{\alpha ^2}$$<br><br>
$$ \Rightarrow 3{\alpha ^2} - 8{\alpha ^{{3 \over 2}}} + 8 = 0$$ | mcq | jee-main-2020-online-2nd-september-evening-slot |
P3CagPVm1jMSVgFFUg7k9k2k5kgx0xy | maths | area-under-the-curves | area-bounded-between-the-curves | Given : $$f(x) = \left\{ {\matrix{
{x\,\,\,\,\,,} & {0 \le x < {1 \over 2}} \cr
{{1 \over 2}\,\,\,\,,} & {x = {1 \over 2}} \cr
{1 - x\,\,\,,} & {{1 \over 2} < x \le 1} \cr
} } \right.$$<br/><br/>
and $$g(x) = \left( {x - {1 \over 2}} \right)^2,x \in R$$ <br/><br/>Then the area
(in sq. units) of the region bounded by the
curves, y = Ζ(x) and y = g(x) between the lines,
2x = 1 and 2x = $$\sqrt 3 $$, is : | [{"identifier": "A", "content": "$${1 \\over 2} + {{\\sqrt 3 } \\over 4}$$"}, {"identifier": "B", "content": "$${1 \\over 2} - {{\\sqrt 3 } \\over 4}$$"}, {"identifier": "C", "content": "$${1 \\over 3} + {{\\sqrt 3 } \\over 4}$$"}, {"identifier": "D", "content": "$${{\\sqrt 3 } \\over 4} - {1 \\over 3}$$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264106/exam_images/d94wa1tsnhgimjfnfsfb.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 9th January Evening Slot Mathematics - Area Under The Curves Question 94 English Explanation">
<br><br>Required area = Area of trepezium ABCD β
Area of parabola between x = $${1 \over 2}$$ and x = $${{\sqrt 3 } \over 2}$$
<br><br>= Area of trepezium ABCD - $$\int\limits_{{1 \over 2}}^{{{\sqrt 3 } \over 2}} {{{\left( {x - {1 \over 2}} \right)}^2}dx} $$
<br><br>= $${1 \over 2}\left( {{{\sqrt 3 } \over 2} - {1 \over 2}} \right)\left( {{1 \over 2} + 1 - {{\sqrt 3 } \over 2}} \right)$$ - $${1 \over 3}\left[ {{{\left( {x - {1 \over 2}} \right)}^3}} \right]_{{1 \over 2}}^{{{\sqrt 3 } \over 2}}$$
<br><br>= $${1 \over 2}\left( {{{\sqrt 3 - 1} \over 2}} \right)\left( {{{3 - \sqrt 3 } \over 2}} \right)$$ $$ - {1 \over 3}\left[ {{{\left( {{{\sqrt 3 - 1} \over 2}} \right)}^3} - 0} \right]$$
<br><br>= $${{\sqrt 3 } \over 4} - {1 \over 3}$$ | mcq | jee-main-2020-online-9th-january-evening-slot |
97vagoGtOJjPTsqqEL7k9k2k5gryhb0 | maths | area-under-the-curves | area-bounded-between-the-curves | For a > 0, let the curves C<sub>1</sub> : y<sup>2</sup> = ax and
C<sub>2</sub> : x<sup>2</sup> = ay intersect at origin O and a point P.
Let the line x = b (0 < b < a) intersect the chord
OP and the x-axis at points Q and R,
respectively. If the line x = b bisects the area
bounded by the curves, C<sub>1</sub> and C<sub>2</sub>, and the area of
<br/>$$\Delta $$OQR = $${1 \over 2}$$, then 'a' satisfies the equation : | [{"identifier": "A", "content": "x<sup>6</sup> \u2013 12x<sup>3</sup> + 4 = 0"}, {"identifier": "B", "content": "x<sup>6</sup> \u2013 12x<sup>3</sup> \u2013 4 = 0"}, {"identifier": "C", "content": "x<sup>6</sup> + 6x<sup>3</sup> \u2013 4 = 0"}, {"identifier": "D", "content": "x<sup>6</sup> \u2013 6x<sup>3</sup> + 4 = 0"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264200/exam_images/yig7dqvbpytiruqt2hnf.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 8th January Morning Slot Mathematics - Area Under The Curves Question 96 English Explanation">
C<sub>1</sub> : y<sup>2</sup>
= ax, C<sub>2</sub> : x<sup>2</sup>
= ay (a > 0)
<br><br>P : intersection point of y<sup>2</sup>
= ax and x<sup>2</sup>
= ay
<br><br>$$ \Rightarrow $$ x<sup>4</sup>
= a<sup>2</sup>
y<sup>2</sup>
<br><br>$$ \Rightarrow $$ x<sup>4</sup>
= a<sup>2</sup>ax
<br><br>$$ \Rightarrow $$ x = a, y = a
<br><br>$$ \therefore $$ Point P : (a, a)
<br><br>Line OP : y = x
<br><br>$$ \Rightarrow $$ Point Q = (b, b)
<br><br>Area $$\Delta $$OQR = $${1 \over 2}$$
<br><br>$$ \Rightarrow $$ $${1 \over 2} \times b \times b$$ = $${1 \over 2}$$
<br><br>$$ \Rightarrow $$ b = 1
<br><br>As line x = b bisect the area between curve
<br><br>$$ \therefore $$ $${1 \over 2}\int\limits_0^a {\left( {\sqrt {ax} - {{{x^2}} \over a}} \right)} dx$$ = $$\int\limits_0^1 {\left( {\sqrt {ax} - {{{x^2}} \over a}} \right)} dx$$
<br><br>$$ \Rightarrow $$ $$\left[ {{1 \over 2}\sqrt a {x^{{3 \over 2}}} \times {2 \over 3} - {1 \over 2}{{{x^3}} \over {3a}}} \right]_0^a$$ = $$\left[ {\sqrt a {x^{{3 \over 2}}} \times {2 \over 3} - {{{x^3}} \over {3a}}} \right]_0^1$$
<br><br>$$ \Rightarrow $$ $${{{a^2}} \over 3} - {1 \over 2}{{{a^2}} \over 3}$$ - 0 = $${{2\sqrt a } \over 3} - {1 \over {3a}}$$ - 0
<br><br>$$ \Rightarrow $$ $${{{a^2}} \over 2} + {1 \over a} = 2\sqrt a $$
<br><br>$$ \Rightarrow $$ $${a^3} + 2 = 4a\sqrt a $$
<br><br>Squareing both sides, we get
<br><br>$$ \Rightarrow $$ $${a^6} + 4{a^3} + 4 = 16{a^3}$$
<br><br>$$ \Rightarrow $$ $${a^6} - 12{a^3} + 4 = $$ 0
<br><br> Hence $$a$$ satisfy x<sup>6</sup> β 12x<sup>3</sup> + 4 = 0. | mcq | jee-main-2020-online-8th-january-morning-slot |
ev69psbeho98LVk3LE7k9k2k5fjbh1k | maths | area-under-the-curves | area-bounded-between-the-curves | The area (in sq. units) of the region
<br/>{(x, y) $$ \in $$ R<sup>2</sup> | 4x<sup>2</sup> $$ \le $$ y $$ \le $$ 8x + 12} is : | [{"identifier": "A", "content": "$${{125} \\over 3}$$"}, {"identifier": "B", "content": "$${{128} \\over 3}$$"}, {"identifier": "C", "content": "$${{127} \\over 3}$$"}, {"identifier": "D", "content": "$${{124} \\over 3}$$"}] | ["B"] | null | For point of intersection
4x<sup>2</sup>
= 8x + 12
<br><br>$$ \Rightarrow $$ x<sup>2</sup> - 2x - 3 = 0
<br>$$ \Rightarrow $$ x = β1, 3
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263719/exam_images/casyif6hyghpfwzxghwt.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 7th January Evening Slot Mathematics - Area Under The Curves Question 97 English Explanation">
<br><br>Required area = area of the shaded region
<br><br>= $$\int\limits_{ - 1}^3 {\left( {8x + 12 - 4{x^2}} \right)dx} $$
<br><br>= $$4\left[ {2.{{{x^2}} \over 2} + 3x - {{{x^3}} \over 3}} \right]_{ - 1}^3$$
<br><br>= (36 + 36 β 36) β (4 β 12 + $${4 \over 3}$$)
<br><br>= $${{128} \over 3}$$ | mcq | jee-main-2020-online-7th-january-evening-slot |
mrazloIgO7RLHAf87O7k9k2k5e2xp0r | maths | area-under-the-curves | area-bounded-between-the-curves | The area of the region, enclosed by the circle x<sup>2</sup> + y<sup>2</sup> = 2 which is not common to the region bounded by the parabola y<sup>2</sup> = x and the straight line y = x, is: | [{"identifier": "A", "content": "$${1 \\over 6}\\left( {24\\pi - 1} \\right)$$"}, {"identifier": "B", "content": "$${1 \\over 3}\\left( {12\\pi - 1} \\right)$$"}, {"identifier": "C", "content": "$${1 \\over 3}\\left( {6\\pi - 1} \\right)$$"}, {"identifier": "D", "content": "$${1 \\over 6}\\left( {12\\pi - 1} \\right)$$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263874/exam_images/ilutwsifhzc063d2ocbw.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 7th January Morning Slot Mathematics - Area Under The Curves Question 98 English Explanation">
<br><br>Required area = Area of circle β $$\int\limits_0^1 {\left( {\sqrt x - x} \right)dx} $$
<br><br>= $$\pi $$r<sup>2</sup> - $$\left( {{{2{x^{3/2}}} \over 3} - {{{x^2}} \over 2}} \right)_0^1$$
<br><br>= $$\pi {\left( {\sqrt 2 } \right)^2}$$ - $${1 \over 6}$$
<br><br>= $${1 \over 6}\left( {12\pi - 1} \right)$$ | mcq | jee-main-2020-online-7th-january-morning-slot |
d6xFyNgEj2CmZ6SHmgjgy2xukewn1vf0 | maths | area-under-the-curves | area-bounded-between-the-curves | Area (in sq. units) of the region outside
<br/><br>$${{\left| x \right|} \over 2} + {{\left| y \right|} \over 3} = 1$$ and inside the ellipse $${{{x^2}} \over 4} + {{{y^2}} \over 9} = 1$$ is :</br> | [{"identifier": "A", "content": "$$6\\left( {4 - \\pi } \\right)$$"}, {"identifier": "B", "content": "$$3\\left( {4 - \\pi } \\right)$$"}, {"identifier": "C", "content": "$$6\\left( {\\pi - 2} \\right)$$"}, {"identifier": "D", "content": "$$3\\left( {\\pi - 2} \\right)$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266337/exam_images/yov949hdcyfww9kn13ai.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 2nd September Morning Slot Mathematics - Area Under The Curves Question 93 English Explanation">
<br><br>Area of Ellipse = $$\pi $$ab = 6$$\pi $$
<br><br>$$ \therefore $$ Required area = Area of ellipse
β 4 (Area of triangle OAB)
<br><br>= 6$$\pi $$ - $$4\left( {{1 \over 2} \times 2 \times 3} \right)$$
<br><br>= 6$$\pi $$ - 12
<br><br>= $$6\left( {\pi - 2} \right)$$ sq.units | mcq | jee-main-2020-online-2nd-september-morning-slot |
xq9dux4OasWkcREEcJ1klrggz5j | maths | area-under-the-curves | area-bounded-between-the-curves | The area (in sq. units) of the part of the circle x<sup>2</sup> + y<sup>2</sup> = 36, which is outside the parabola
y<sup>2</sup> = 9x, is : | [{"identifier": "A", "content": "$$12\\pi - 3\\sqrt 3 $$"}, {"identifier": "B", "content": "$$24\\pi + 3\\sqrt 3 $$"}, {"identifier": "C", "content": "$$24\\pi - 3\\sqrt 3 $$"}, {"identifier": "D", "content": "$$12\\pi + 3\\sqrt 3 $$"}] | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264209/exam_images/f914oa66zhlhb39t3fuf.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266556/exam_images/dee6z6pgqdahguovrp24.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264224/exam_images/bfwply8vstrjg4u5plqy.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265525/exam_images/ivv6ztsoxu8fgepeagns.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266907/exam_images/sbavak08yt3gbpyt9h8l.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 24th February Morning Shift Mathematics - Area Under The Curves Question 87 English Explanation"></picture> <br>$${x^2} + {y^2} = 36$$ and $${y^2} = 9x$$<br><br>$$ \therefore $$ $${x^2} + 9x - 36 = 0$$<br><br>$$ \Rightarrow x = 3, - 12$$<br><br>Required Area, <br><br>$$A = \pi {{{(6)}^2}} - 2\left[ {{A_1} + {A_2}} \right]$$<br><br>$$A = \pi {{{(6)}^2} - 2\left[ {\int_0^3 {\sqrt {9x} dx + \int_3^6 {\sqrt {36 - {x^2}} } dx} } \right]} $$<br><br>$$ = 36\pi - $$$$2\left[ {\left[ {3 \times {2 \over 3}{x^{{3 \over 2}}}} \right]_0^3 + \left[ {{x \over 2}\sqrt {36 - {x^2}} + {{36} \over 2}{{\sin }^{ - 1}}{x \over 6}} \right]_3^6} \right]$$
<br><br>$$ = 36\pi - $$$$2\left[ {\left[ {2{x^{{3 \over 2}}}} \right]_0^3 + \left[ {{x \over 2}\sqrt {36 - {x^2}} + 18{{\sin }^{ - 1}}{x \over 6}} \right]_3^6} \right]$$
<br><br>$$ = 36\pi - $$$$2\left[ {\left[ {6\sqrt 3 - 0} \right] + \left[ {\left( {0 + 18{{\sin }^{ - 1}}{6 \over 6}} \right) - \left( {{3 \over 2} \times 3\sqrt 3 + 18{{\sin }^{ - 1}}{3 \over 6}} \right)} \right]} \right]$$
<br><br>$$ = 36\pi - $$$$2\left[ {\left[ {6\sqrt 3 } \right] + \left[ {\left( {{{18\pi } \over 2}} \right) - \left( {{3 \over 2} \times 3\sqrt 3 + {{18\pi } \over 6}} \right)} \right]} \right]$$
<br><br>$$ = 36\pi - $$ $$\left( {12\sqrt 3 + 18\pi - 9\sqrt 3 - 6\pi } \right)$$
<br><br>= $$24\pi - 3\sqrt 3 $$
<br><br><b>Note :</b>
<br>(i) $$\int {\sqrt {{x^2} + {a^2}} dx = {1 \over 2}\left[ {x\sqrt {{x^2} + {a^2}} + {a^2}\log |x + \sqrt {{x^2} + {a^2}} |} \right]} + C$$<br><br>(ii) $$\int {\sqrt {{a^2} - {x^2}} dx = {1 \over 2}\left[ {x\sqrt {{a^2} - {x^2}} + {a^2}{{\sin }^{ - 1}}\left( {{x \over a}} \right)} \right]} + C$$<br><br>(iii) $$\int {\sqrt {{x^2} - {a^2}} dx = {1 \over 2}\left[ {x\sqrt {{x^2} - {a^2}} - {a^2}\log |x + \sqrt {{x^2} - {a^2}} |} \right]} + C$$ | mcq | jee-main-2021-online-24th-february-morning-slot |
yPCA6PkvYr7HYrsAwf1klrmdwhc | maths | area-under-the-curves | area-bounded-between-the-curves | The area of the region : $$R = \{ (x,y):5{x^2} \le y \le 2{x^2} + 9\} $$ is : | [{"identifier": "A", "content": "$$6\\sqrt 3 $$ square units"}, {"identifier": "B", "content": "$$12\\sqrt 3 $$ square units"}, {"identifier": "C", "content": "$$11\\sqrt 3 $$ square units"}, {"identifier": "D", "content": "$$9\\sqrt 3 $$ square units"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264729/exam_images/jm9chvj4tml7xrg2bqui.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 24th February Evening Shift Mathematics - Area Under The Curves Question 86 English Explanation">
<br>Required area<br><br>$$ = 2\int\limits_0^{\sqrt 3 } {\left( {2{x^2} + 9 - 5{x^2}} \right)} dx$$<br><br>$$ = 2\int\limits_0^{\sqrt 3 } {\left( {9 - 3{x^2}} \right)dx} $$<br><br>$$ = 2|\,9x - {x^3}\,|_0^{\sqrt 3 } = 12\sqrt 3 $$ | mcq | jee-main-2021-online-24th-february-evening-slot |
YJVsU602GxrlTeWim91kls5gdoe | maths | area-under-the-curves | area-bounded-between-the-curves | The graphs of sine and cosine functions, intersect each other at a number of points and between two consecutive points of intersection, the two graphs enclose the same area A. Then A<sup>4</sup> is equal to __________. | [] | null | 64 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264150/exam_images/scswst5jdqmnovfjgtnf.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th February Morning Shift Mathematics - Area Under The Curves Question 85 English Explanation"><br><br>$$A = \int\limits_{{\pi \over 4}}^{{{5\pi } \over 4}} {(\sin x - \cos x)dx} $$
<br><br>$$= [ - \cos x - \sin x]_{\pi /4}^{5\pi /4}$$<br><br>$$ = - \left[ {\left( {\cos {{5\pi } \over 4} + \sin {5\pi \over 4}} \right) - \left( {\cos {\pi \over 4} + \sin {\pi \over 4}} \right)} \right]$$<br><br>$$ = - \left[ {\left( { - {1 \over {\sqrt 2 }} - {1 \over {\sqrt 2 }}} \right) - \left( {{1 \over {\sqrt 2 }} + {1 \over {\sqrt 2 }}} \right)} \right]$$<br><br>$$ = {4 \over {\sqrt 2 }} = 2\sqrt 2 $$<br><br>$$ \Rightarrow {A^4} = {\left( {2\sqrt 2 } \right)^4} = 64$$ | integer | jee-main-2021-online-25th-february-morning-slot |
YTNLpgdL2WAyCcWQVi1kluhnqzk | maths | area-under-the-curves | area-bounded-between-the-curves | The area bounded by the lines y = || x $$-$$ 1 | $$-$$ 2 | is ___________. | [] | null | 8 | Question is incomplete it should be area bounded
by y = || x $$-$$ 1 | $$-$$ 2 | and y = 2. | integer | jee-main-2021-online-26th-february-morning-slot |
cffEOVLEI7ehYNp5lZ1kmko9iza | maths | area-under-the-curves | area-bounded-between-the-curves | Let f : [$$-$$3, 1] $$ \to $$ R be given as <br/><br/>$$f(x) = \left\{ \matrix{
\min \,\{ (x + 6),{x^2}\}, - 3 \le x \le 0 \hfill \cr
\max \,\{ \sqrt x ,{x^2}\} ,\,0 \le x \le 1. \hfill \cr} \right.$$<br/><br/>If the area bounded by y = f(x) and x-axis is A, then the value of 6A is equal to ___________. | [] | null | 41 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265792/exam_images/adjt0f0zaxeqso6stpct.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 17th March Evening Shift Mathematics - Area Under The Curves Question 82 English Explanation">
<br>Area is $$\int\limits_{ - 3}^{ - 2} {(x + 6)dx + \int\limits_{ - 2}^0 {{x^2}dx + \int\limits_0^1 {\sqrt {x}dx = A} } } $$<br><br>$$ = {7 \over 2} + \left[ {{{{x^3}} \over 3}} \right]_{ - 2}^0 + \left[ {{2 \over 3}{x^{3/2}}} \right]_0^1$$<br><br>$$ = {7 \over 2} + {8 \over 3} + {2 \over 3} = {{41} \over 6}$$<br><br>So, 6A = 41 | integer | jee-main-2021-online-17th-march-evening-shift |
1krq14l9w | maths | area-under-the-curves | area-bounded-between-the-curves | Let T be the tangent to the ellipse E : x<sup>2</sup> + 4y<sup>2</sup> = 5 at the point P(1, 1). If the area of the region bounded by the tangent T, ellipse E, lines x = 1 and x = $$\sqrt 5 $$ is $$\alpha$$$$\sqrt 5 $$ + $$\beta$$ + $$\gamma$$ cos<sup>$$-$$1</sup>$$\left( {{1 \over {\sqrt 5 }}} \right)$$, then |$$\alpha$$ + $$\beta$$ + $$\gamma$$| is equal to ______________. | [] | null | 1.25 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263552/exam_images/hlnbvgwmqwwtyylb5tjw.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 20th July Morning Shift Mathematics - Area Under The Curves Question 80 English Explanation"><br>E : x<sup>2</sup> + 4y<sup>2</sup> = 5<br><br>Tangent at P : x + 4y = 5<br><br>Required area<br><br>$$ = \int\limits_1^{\sqrt 5 } {\left( {{{5 - x} \over 4} - {{\sqrt {5 - {x^2}} } \over 2}} \right)dx} $$<br><br>$$ = \left[ {{{5x} \over 4} - {{{x^2}} \over 8} - {x \over 4}\sqrt {5 - {x^2}} - {5 \over 2}{{\sin }^{ - 1}}{x \over {\sqrt 5 }}} \right]_1^{\sqrt 5 }$$<br><br>$$ = {5 \over 4}\sqrt 5 - {5 \over 4} - {5 \over 4}{\cos ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)$$<br><br>If we assume $$\alpha$$, $$\beta$$, $$\gamma$$, $$\in$$ Q (Not given in question) then $$\alpha$$ = $${5 \over 4}$$, $$\beta$$ = $$-$$$${5 \over 4}$$ & $$\gamma$$ = $$-$$$${5 \over 4}$$<br><br>|$$\alpha$$ + $$\beta$$ + $$\gamma$$| = 1.25 | integer | jee-main-2021-online-20th-july-morning-shift |
1krub9ptn | maths | area-under-the-curves | area-bounded-between-the-curves | The area (in sq. units) of the region bounded by the curves x<sup>2</sup> + 2y $$-$$ 1 = 0, y<sup>2</sup> + 4x $$-$$ 4 = 0 and y<sup>2</sup> $$-$$ 4x $$-$$ 4 = 0, in the upper half plane is _______________. | [] | null | 2 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266774/exam_images/u1ozidkzbjd8lcnvbnev.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 22th July Evening Shift Mathematics - Area Under The Curves Question 79 English Explanation"><br>Required area (shaded)<br><br>$$ = 2\left[ {\int\limits_0^2 {\left( {{{4 - {y^2}} \over 4}} \right)dy - \int\limits_0^1 {\left( {{{1 - {x^2}} \over 2}} \right)dx} } } \right]$$<br><br>$$ = 2\left[ {{4 \over 3} - {1 \over 3}} \right] = (2)$$ | integer | jee-main-2021-online-22th-july-evening-shift |
1krvzemnw | maths | area-under-the-curves | area-bounded-between-the-curves | The area (in sq. units) of the region, given by the set $$\{ (x,y) \in R \times R|x \ge 0,2{x^2} \le y \le 4 - 2x\} $$ is : | [{"identifier": "A", "content": "$${8 \\over 3}$$"}, {"identifier": "B", "content": "$${{17} \\over 3}$$"}, {"identifier": "C", "content": "$${{13} \\over 3}$$"}, {"identifier": "D", "content": "$${7 \\over 3}$$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264240/exam_images/tijy7o0iderdha3gedzq.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th July Morning Shift Mathematics - Area Under The Curves Question 78 English Explanation"><br>Required area = $$\left. {\int\limits_0^1 {\left( {4 - 2x - 2{x^2}} \right)dx = 4x - {x^2} - {{2{x^3}} \over 3}} } \right|_0^1$$<br><br>$$ = 4 - 1 - {2 \over 3} = {7 \over 3}$$ | mcq | jee-main-2021-online-25th-july-morning-shift |
1krxjjxgo | maths | area-under-the-curves | area-bounded-between-the-curves | The area of the region bounded by y $$-$$ x = 2 and x<sup>2</sup> = y is equal to : | [{"identifier": "A", "content": "$${{16} \\over 3}$$"}, {"identifier": "B", "content": "$${{2} \\over 3}$$"}, {"identifier": "C", "content": "$${{9} \\over 2}$$"}, {"identifier": "D", "content": "$${{4} \\over 3}$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264963/exam_images/u8k9qqbaojrvwvtkepl5.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Evening Shift Mathematics - Area Under The Curves Question 77 English Explanation"><br><br>y $$-$$ x = 2, x<sup>2</sup> = y<br><br>Now, x<sup>2</sup> = 2 + x<br><br>$$\Rightarrow$$ x<sup>2</sup> $$-$$ x $$-$$ 2 = 0<br><br>$$\Rightarrow$$ (x + 1)(x $$-$$ 2) = 0<br><br>Area = $$\int\limits_{ - 1}^2 {(2 + x - {x^2})} $$<br><br>$$ = \left| {2x + {{{x^2}} \over 2} - {{{x^3}} \over 3}} \right|_{ - 1}^2$$<br><br>$$ = \left( {4 + 2 - {8 \over 3}} \right) - \left( { - 2 + {1 \over 2} + {1 \over 3}} \right)$$<br><br>$$ = 6 - 3 + 2 - {1 \over 2} = {9 \over 2}$$ | mcq | jee-main-2021-online-27th-july-evening-shift |
1ks06ssp5 | maths | area-under-the-curves | area-bounded-between-the-curves | If the area of the bounded region <br/>$$R = \left\{ {(x,y):\max \{ 0,{{\log }_e}x\} \le y \le {2^x},{1 \over 2} \le x \le 2} \right\}$$ is , <br/>$$\alpha {({\log _e}2)^{ - 1}} + \beta ({\log _e}2) + \gamma $$, then the value of $${(\alpha + \beta - 2\lambda )^2}$$ is equal to : | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "1"}] | ["B"] | null | $$R = \left\{ {(x,y):\max \{ 0,{{\log }_e}x\} \le y \le {2^x},{1 \over 2} \le x \le 2} \right\}$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264045/exam_images/gmfzldxp4t8nl2p4kdgf.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Morning Shift Mathematics - Area Under The Curves Question 76 English Explanation"><br><br>$$\int\limits_{{1 \over 2}}^2 {{2^x}dx} - \int\limits_1^2 {\ln xdx} $$<br><br>$$ \Rightarrow \left[ {{{{2^x}} \over {\ln 2}}} \right]_{1/2}^2 - [x\ln x - x]_1^2$$<br><br>$$ \Rightarrow {{({2^2}) - {2^{1/2}}} \over {{{\log }_e}2}} - (2\ln 2 - 1)$$<br><br>$$ \Rightarrow {{\left( {{2^2} - \sqrt 2 } \right)} \over {{{\log }_e}2}} - 2\ln 2 + 1$$<br><br>$$\therefore$$ $$\alpha = {2^2} - \sqrt 2 $$, $$\beta = - 2$$, $$\gamma = 1$$<br><br>$$ \therefore {(\alpha + \beta + 2\gamma )^2}$$<br><br>$$ = {({2^2} - \sqrt 2 - 2 - 2)^2}$$<br><br>$$ = {(\sqrt 2 )^2} = 2$$ | mcq | jee-main-2021-online-27th-july-morning-shift |
1ktbilndb | maths | area-under-the-curves | area-bounded-between-the-curves | The area of the region $$S = \{ (x,y):3{x^2} \le 4y \le 6x + 24\} $$ is ____________. | [] | null | 27 | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267515/exam_images/gnxx2fmvzigd5eqlwpbs.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265574/exam_images/sa1cxk6zsgh0gri62bbr.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265595/exam_images/tuxmxd7kcn0tmyn0de3f.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266757/exam_images/yc56rqbdajzn1a2loepd.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th August Morning Shift Mathematics - Area Under The Curves Question 75 English Explanation"></picture> <br><br>For A & B<br><br>3x<sup>2</sup> = 6x + 24 $$\Rightarrow$$ x<sup>2</sup> $$-$$ 2x $$-$$ 8 = 0<br><br>$$\Rightarrow$$ x = $$-$$2, 4<br><br>Area $$ = \int\limits_{ - 2}^4 {\left( {{3 \over 2}x + 6 - {3 \over 4}{x^2}} \right)dx} $$<br><br>$$ = \left[ {{{3{x^2}} \over 4} + 6x - {{{x^3}} \over 4}} \right]_{ - 2}^4 = 27$$ | integer | jee-main-2021-online-26th-august-morning-shift |
1ktd2yfdk | maths | area-under-the-curves | area-bounded-between-the-curves | Let a and b respectively be the points of local maximum and local minimum of the function f(x) = 2x<sup>3</sup> $$-$$ 3x<sup>2</sup> $$-$$ 12x. If A is the total area of the region bounded by y = f(x), the x-axis and the lines x = a and x = b, then 4A is equal to ______________. | [] | null | 114 | f'(x) = 6x<sup>2</sup> $$-$$ 6x $$-$$ 12 = 6(x $$-$$ 2) (x + 1)<br><br>Point = (2, $$-$$20) & ($$-$$1, 7)<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263659/exam_images/k3ibititsvzikcqrsdja.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th August Evening Shift Mathematics - Area Under The Curves Question 74 English Explanation"><br>$$A = \int\limits_{ - 1}^0 {(2{x^3} - 3{x^2} - 12x)dx + \int\limits_0^2 {(12x + 3{x^2} - 2{x^3})\,dx} } $$<br><br>$$A = \left( {{{{x^4}} \over 2} - {x^3} - 6{x^2}} \right)_{ - 1}^0 + \left( {6{x^2} + {x^3} - {{{x^4}} \over 2}} \right)_0^2$$<br><br>4A = 114 | integer | jee-main-2021-online-26th-august-evening-shift |
1ktg2vo7y | maths | area-under-the-curves | area-bounded-between-the-curves | The area of the region bounded by the parabola (y $$-$$ 2)<sup>2</sup> = (x $$-$$ 1), the tangent to it at the point whose ordinate is 3 and the x-axis is : | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "6"}] | ["A"] | null | y = 3 $$\Rightarrow$$ x = 2<br><br>Point is (2, 3)<br><br>Diff. w.r.t x<br><br>2 (y $$-$$ 2) y' = 1<br><br>$$\Rightarrow$$ $$y' = {1 \over {2(y - 2)}}$$<br><br>$$ \Rightarrow y{'_{(2,3)}} = {1 \over 2}$$<br><br>$$ \Rightarrow {{y - 3} \over {x - 2}} = {1 \over 2} \Rightarrow x - 2y + 4 = 0$$<br><br>Area $$ = \int\limits_0^3 {\left( {{{(y - 2)}^2} + 1 - (2y - 4)} \right)} \,dy$$<br><br>= 9 sq. units<br><br> <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265260/exam_images/etwbivpocxnqqwklpzbs.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264663/exam_images/g2jng2fhz6nhcng1xnnf.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263996/exam_images/er6uqxcqynx3kojs74k9.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th August Evening Shift Mathematics - Area Under The Curves Question 73 English Explanation"></picture> | mcq | jee-main-2021-online-27th-august-evening-shift |
1kto652vo | maths | area-under-the-curves | area-bounded-between-the-curves | The area, enclosed by the curves $$y = \sin x + \cos x$$ and $$y = \left| {\cos x - \sin x} \right|$$ and the lines $$x = 0,x = {\pi \over 2}$$, is : | [{"identifier": "A", "content": "$$2\\sqrt 2 (\\sqrt 2 - 1)$$"}, {"identifier": "B", "content": "$$2(\\sqrt 2 + 1)$$"}, {"identifier": "C", "content": "$$4(\\sqrt 2 - 1)$$"}, {"identifier": "D", "content": "$$2\\sqrt 2 (\\sqrt 2 + 1)$$"}] | ["A"] | null | $$A = \int_0^{{\pi \over 2}} {\left( {(\sin x + \cos x) - \left| {\cos x - \sin x} \right|} \right)\,dx} $$<br><br>$$A = \int_0^{{\pi \over 2}} {\left( {(\sin x + \cos x) - (\cos x - \sin x)} \right)\,dx} + \int_{{\pi \over 4}}^{{\pi \over 2}} {\left( {(\sin x + \cos x) - (\sin x - \cos x)} \right)\,dx} $$<br><br>$$A = 2\int_0^{{\pi \over 2}} {\sin x\,dx + 2\int_{{\pi \over 4}}^{{\pi \over 2}} {\cos x\,dx} } $$<br><br>$$A = - 2\left( {{1 \over {\sqrt 2 }} - 1} \right) + \left( {1 - {1 \over {\sqrt 2 }}} \right)$$<br><br>$$A = 4 - 2\sqrt 2 = 2\sqrt 2 (\sqrt 2 - 1)$$<br><br>Option (a) | mcq | jee-main-2021-online-1st-september-evening-shift |
1l544ulfw | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area enclosed by y<sup>2</sup> = 8x and y = $$\sqrt2$$ x that lies outside the triangle formed by y = $$\sqrt2$$ x, x = 1, y = 2$$\sqrt2$$, is equal to:</p> | [{"identifier": "A", "content": "$${{16\\sqrt 2 } \\over 6}$$"}, {"identifier": "B", "content": "$${{11\\sqrt 2 } \\over 6}$$"}, {"identifier": "C", "content": "$${{13\\sqrt 2 } \\over 6}$$"}, {"identifier": "D", "content": "$${{5\\sqrt 2 } \\over 6}$$"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5niit1g/6da14760-cb2e-4c6b-b5f6-e80807f5e0ce/291c7440-04d1-11ed-93b8-936002ac8631/file-1l5niit1h.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5niit1g/6da14760-cb2e-4c6b-b5f6-e80807f5e0ce/291c7440-04d1-11ed-93b8-936002ac8631/file-1l5niit1h.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th June Morning Shift Mathematics - Area Under The Curves Question 70 English Explanation"></p>
<p>$$A\left( {2,2\sqrt 2 } \right),B\left( {1,2\sqrt 2 } \right),C\left( {1,\sqrt 2 } \right)$$</p>
<p>Area $$ = \int\limits_0^{4\sqrt 2 } {\left( {{y \over {\sqrt 2 }} - {{{y^2}} \over 8}} \right)dy - } $$ area ($$\Delta$$BAC)</p>
<p>$$ = \left[ {{{{y^2}} \over {2\sqrt 2 }} - {{{y^3}} \over {24}}} \right]_0^{4\sqrt 2 } - {1 \over 2} \times AB \times BC$$</p>
<p>$$ = 8\sqrt 2 - {{32 \times 4\sqrt 2 } \over {24}} - {1 \over 2} \times 1 \times \sqrt 2 $$</p>
<p>$$ = 8\sqrt 2 - {{16\sqrt 2 } \over 3} - {{\sqrt 2 } \over 2}$$</p>
<p>$$ = {{\sqrt 2 } \over 6}(48 - 32 - 3) = {{13\sqrt 2 } \over 6}$$</p> | mcq | jee-main-2022-online-29th-june-morning-shift |
1l54tvpeu | maths | area-under-the-curves | area-bounded-between-the-curves | <p>For real numbers a, b (a > b > 0), let</p>
<p>Area $$\left\{ {(x,y):{x^2} + {y^2} \le {a^2}\,and\,{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} \ge 1} \right\} = 30\pi $$ </p>
<p>and</p>
<p>Area $$\left\{ {(x,y):{x^2} + {y^2} \le {b^2}\,and\,{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} \le 1} \right\} = 18\pi $$</p>
<p>Then, the value of (a $$-$$ b)<sup>2</sup> is equal to ___________.</p> | [] | null | 12 | $x^{2}+y^{2} \leq a^{2}$ is interior of circle
<br><br>
and $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \geq 1$ is exterior of ellipse<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lc5rg14e/3ec2057b-efed-4095-9156-11e5f06f21d7/760d25d0-85a3-11ed-b688-91788a069596/file-1lc5rg14f.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lc5rg14e/3ec2057b-efed-4095-9156-11e5f06f21d7/760d25d0-85a3-11ed-b688-91788a069596/file-1lc5rg14f.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th June Evening Shift Mathematics - Area Under The Curves Question 69 English Explanation"><br>
$\therefore$ Area $=\pi a^{2}-\pi a b=30 \pi$
<br><br>
Similarly $x^{2}+y^{2} \geq b^{2}$ and $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leq 1$ gives
<br><br>
$\pi a b-\pi b^{2}=18 \pi$
<br><br>
By (1) and (2), $\frac{a}{b}=\frac{5}{3} \Rightarrow a=\frac{5 b}{3}$
<br><br>
$\Rightarrow \pi \cdot \frac{25 b^{2}}{9}-\pi \cdot \frac{5 b^{2}}{3}=30 \pi$
<br><br>
$\Rightarrow\left(\frac{25}{9}-\frac{5}{3}\right) b^{2}=30$
<br><br>
$\Rightarrow \frac{10}{9} b^{2}=30 \Rightarrow b^{2}=27$
<br><br>
and $a^{2}=\frac{25}{9} \cdot 27=75$
<br><br>
$(a-b)^{2}=(5 \sqrt{3}-3 \sqrt{3})^{2}=3 \cdot 4=12$ | integer | jee-main-2022-online-29th-june-evening-shift |
1l566hjjl | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region S = {(x, y) : y<sup>2</sup> $$\le$$ 8x, y $$\ge$$ $$\sqrt2$$x, x $$\ge$$ 1} is</p> | [{"identifier": "A", "content": "$${{13\\sqrt 2 } \\over 6}$$"}, {"identifier": "B", "content": "$${{11\\sqrt 2 } \\over 6}$$"}, {"identifier": "C", "content": "$${{5\\sqrt 2 } \\over 6}$$"}, {"identifier": "D", "content": "$${{19\\sqrt 2 } \\over 6}$$"}] | ["B"] | null | <p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5obth1c/3674f2a1-bb5f-42c5-bcda-0e644fc0ece5/ba7c4f10-0543-11ed-987f-3938cfc0f7f1/file-1l5obth1d.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5obth1c/3674f2a1-bb5f-42c5-bcda-0e644fc0ece5/ba7c4f10-0543-11ed-987f-3938cfc0f7f1/file-1l5obth1d.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 28th June Morning Shift Mathematics - Area Under The Curves Question 67 English Explanation"> </p>
<p>Required area</p>
<p>$$ = \int\limits_1^4 {\left( {\sqrt {8x} - \sqrt 2 x} \right)dx} $$</p>
<p>$$ = \left. {{{2\sqrt 8 } \over 3}{x^{{3 \over 2}}} - {{{x^2}} \over {\sqrt 2 }}} \right|_1^4$$</p>
<p>$$ = {{16\sqrt 3 } \over 3} - {{16} \over {\sqrt 2 }} - {{2\sqrt 8 } \over 3} + {1 \over {\sqrt 2 }}$$</p>
<p>$$ = {{11\sqrt 2 } \over 6}$$ sq. units</p> | mcq | jee-main-2022-online-28th-june-morning-shift |
1l56u5mze | maths | area-under-the-curves | area-bounded-between-the-curves | <p>If the area of the region $$\left\{ {(x,y):{x^{{2 \over 3}}} + {y^{{2 \over 3}}} \le 1,\,x + y \ge 0,\,y \ge 0} \right\}$$ is A, then $${{256A} \over \pi }$$ is equal to __________.</p> | [] | null | 36 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5p7lae4/469ec989-bde2-4216-be88-8dbb75d383c8/fb02f1c0-05bf-11ed-8617-d71e6444d1a0/file-1l5p7lae5.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5p7lae4/469ec989-bde2-4216-be88-8dbb75d383c8/fb02f1c0-05bf-11ed-8617-d71e6444d1a0/file-1l5p7lae5.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 27th June Evening Shift Mathematics - Area Under The Curves Question 66 English Explanation"></p>
<p>$$\therefore$$ Area of shaded region</p>
<p>$$ = \int\limits_{ - {{\left( {{1 \over 2}} \right)}^{{3 \over 2}}}}^0 {\left( {{{\left( {1 - {x^{{2 \over 3}}}} \right)}^{{3 \over 2}}} + x} \right)dx + \int\limits_0^1 {{{\left( {1 - {x^{{2 \over 3}}}} \right)}^{{3 \over 2}}}dx} } $$</p>
<p>$$ = \int\limits_{ - {{\left( {{1 \over 2}} \right)}^{{3 \over 2}}}}^0 {{{\left( {1 - {x^{{2 \over 3}}}} \right)}^{{3 \over 2}}}dx + \int\limits_{ - {{\left( {{1 \over 2}} \right)}^{{3 \over 2}}}}^0 {xdx} } $$</p>
<p>Let $$x = {\sin ^3}\theta $$</p>
<p>$$\therefore$$ $$dx = 3{\sin ^2}\theta \cos \theta d\theta $$</p>
<p>$$ = \int\limits_{ - {\pi \over 4}}^{{\pi \over 2}} {3{{\sin }^2}\theta {{\cos }^4}\theta d\theta + \left( {0 + {1 \over {16}}} \right)} $$</p>
<p>$$ = {{9\pi } \over {64}} + {1 \over {16}} - {1 \over {16}} = {{36\pi } \over {256}} = A$$</p>
<p>$$\therefore$$ $${{256A} \over \pi } = 36$$</p> | integer | jee-main-2022-online-27th-june-evening-shift |
1l57p3c3w | maths | area-under-the-curves | area-bounded-between-the-curves | Let
<p>$${A_1} = \left\{ {(x,y):|x| \le {y^2},|x| + 2y \le 8} \right\}$$ and</p>
<p>$${A_2} = \left\{ {(x,y):|x| + |y| \le k} \right\}$$. If 27 (Area A<sub>1</sub>) = 5 (Area A<sub>2</sub>), then k is equal to :</p> | [] | null | 6 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5q9zfw8/6cdd28b7-2353-4e56-94fa-08f93a22752d/1f7f8b80-0656-11ed-903e-c9687588b3f3/file-1l5q9zfw9.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5q9zfw8/6cdd28b7-2353-4e56-94fa-08f93a22752d/1f7f8b80-0656-11ed-903e-c9687588b3f3/file-1l5q9zfw9.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 27th June Morning Shift Mathematics - Area Under The Curves Question 65 English Explanation 1"></p>
<p>Required area (above x-axis)</p>
<p>$${A_1} = 2\int\limits_0^4 {\left( {{{8 - x} \over 2} - \sqrt x } \right)dx} $$</p>
<p>$$ = 2\left( {16 - {{16} \over 4} - {8 \over {3/2}}} \right) = {{40} \over 3}$$</p>
<p>and $${A_2} = 4\left( {{1 \over 2}\,.\,{k^2}} \right) = 2{k^2}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5qa0x8z/b3f515b6-b4dd-4a2e-b38c-8edeeca6de75/48b68a30-0656-11ed-903e-c9687588b3f3/file-1l5qa0x90.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5qa0x8z/b3f515b6-b4dd-4a2e-b38c-8edeeca6de75/48b68a30-0656-11ed-903e-c9687588b3f3/file-1l5qa0x90.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 27th June Morning Shift Mathematics - Area Under The Curves Question 65 English Explanation 2"></p>
<p>$$\therefore$$ $$27\,.\,{{40} \over 3} = 5\,.\,(2{k^2})$$</p>
<p>$$\Rightarrow$$ k = 6</p> | integer | jee-main-2022-online-27th-june-morning-shift |
1l589ntbl | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area bounded by the curve y = |x<sup>2</sup> $$-$$ 9| and the line y = 3 is :</p> | [{"identifier": "A", "content": "$$4(2\\sqrt 3 + \\sqrt 6 - 4)$$"}, {"identifier": "B", "content": "$$4(4\\sqrt 3 + \\sqrt 6 - 4)$$"}, {"identifier": "C", "content": "$$8(4\\sqrt 3 + 3\\sqrt 6 - 9)$$"}, {"identifier": "D", "content": "$$8(4\\sqrt 3 + \\sqrt 6 - 9)$$"}] | ["D"] | null | <p>$$y = 3$$ and $$y = |{x^2} - 9|$$</p>
<p>Intersect in first quadrant at $$x = \sqrt 6 $$ and $$x = \sqrt {12} $$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5rh7hff/4ef2bdc8-3288-4017-9611-a67305104639/27b58cb0-06ff-11ed-b821-f5ba0940c0a2/file-1l5rh7hfg.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5rh7hff/4ef2bdc8-3288-4017-9611-a67305104639/27b58cb0-06ff-11ed-b821-f5ba0940c0a2/file-1l5rh7hfg.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 26th June Morning Shift Mathematics - Area Under The Curves Question 64 English Explanation"></p>
<p>Required area</p>
<p>$$ = 2\left[ {{2 \over 3}\left( {6 \times \sqrt 6 } \right) + \int\limits_{\sqrt 6 }^3 {\left( {3 - \left( {9 - {x^2}} \right)} \right)dx + \int\limits_3^{\sqrt {12} } {\left( {3 - \left( {{x^2} - 9} \right)} \right)dx} } } \right]$$</p>
<p>$$ = 2\left[ {4\sqrt 6 + \left. {\left( {{{{x^3}} \over 3} - 6x} \right)} \right|_{\sqrt 6 }^3 + \left. {\left( {12x - {{{x^3}} \over 3}} \right)} \right|_3^{\sqrt {12} }} \right]$$</p>
<p>$$ = 2\left[ {4\sqrt 6 + \left( {4\sqrt 6 - 9} \right) + \left( {8\sqrt {12} - 27} \right)} \right]$$</p>
<p>$$ = 2\left[ {8\sqrt 6 + 16\sqrt 3 - 36} \right] = 8\left[ {2\sqrt 6 + 4\sqrt 3 - 9} \right]$$</p> | mcq | jee-main-2022-online-26th-june-morning-shift |
1l58f80fj | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region bounded by y<sup>2</sup> = 8x and y<sup>2</sup> = 16(3 $$-$$ x) is equal to:</p> | [{"identifier": "A", "content": "$${{32} \\over 3}$$"}, {"identifier": "B", "content": "$${{40} \\over 3}$$"}, {"identifier": "C", "content": "16"}, {"identifier": "D", "content": "19"}] | ["C"] | null | <p>$${c_1}:{y^2} = 8x$$</p>
<p>$${c_2}:{y^2} = 16(3 - x)$$</p>
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5qp2am3/c5e50fc7-f237-4ff6-b8a1-b5d6a9b181f3/17c870b0-0691-11ed-93bf-f57702a71509/file-1l5qp2am4.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5qp2am3/c5e50fc7-f237-4ff6-b8a1-b5d6a9b181f3/17c870b0-0691-11ed-93bf-f57702a71509/file-1l5qp2am4.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 26th June Evening Shift Mathematics - Area Under The Curves Question 63 English Explanation"></p>
<p>Solving c<sub>1</sub> and c<sub>2</sub></p>
<p>$$48 - 16x = 8x$$</p>
<p>$$x = 2$$</p>
<p>$$\therefore$$ $$y = \pm \,4$$</p>
<p>$$\therefore$$ Area of shaded region</p>
<p>$$ = 2\int\limits_0^4 {\left\{ {\left( {{{48 - {y^2}} \over {16}}} \right) - \left( {{{{y^2}} \over 8}} \right)} \right\}dy} $$</p>
<p>$$ = {1 \over 8}\left[ {48y - {y^3}} \right]_0^4 = 16$$</p> | mcq | jee-main-2022-online-26th-june-evening-shift |
1l59ju9ac | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region enclosed between the parabolas y<sup>2</sup> = 2x $$-$$ 1 and y<sup>2</sup> = 4x $$-$$ 3 is</p> | [{"identifier": "A", "content": "$${1 \\over {3}}$$"}, {"identifier": "B", "content": "$${1 \\over {6}}$$"}, {"identifier": "C", "content": "$${2 \\over {3}}$$"}, {"identifier": "D", "content": "$${3 \\over {4}}$$"}] | ["A"] | null | <p>Area of the shaded region</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5th819e/347b9698-98c3-43fc-baeb-9ed9884fb187/c8540420-0818-11ed-98aa-f9038709a939/file-1l5th819f.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5th819e/347b9698-98c3-43fc-baeb-9ed9884fb187/c8540420-0818-11ed-98aa-f9038709a939/file-1l5th819f.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th June Evening Shift Mathematics - Area Under The Curves Question 62 English Explanation"></p>
<p>$$ = 2\int_0^1 {\left( {{{{y^2} + 3} \over 4} - {{{y^2} + 1} \over 2}} \right)dy} $$</p>
<p>$$ = 2\int_0^1 {\left( {{1 \over 4} - {{{y^2}} \over 4}} \right)dy} $$</p>
<p>$$ = 2\left[ {{1 \over 4} - {1 \over {12}}} \right] = {1 \over 3}$$</p> | mcq | jee-main-2022-online-25th-june-evening-shift |
1l5bb4eq9 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area (in sq. units) of the region enclosed between the parabola y<sup>2</sup> = 2x and the line x + y = 4 is __________.</p> | [] | null | 18 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5v6294y/2f8c41d0-8e8c-4db4-b12e-8da89b735e16/b46f9a20-0906-11ed-a790-b11fa70c8a36/file-1l5v6294z.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5v6294y/2f8c41d0-8e8c-4db4-b12e-8da89b735e16/b46f9a20-0906-11ed-a790-b11fa70c8a36/file-1l5v6294z.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 24th June Evening Shift Mathematics - Area Under The Curves Question 61 English Explanation"></p>
<p>The required area $$ = \int_{ - 4}^2 {\left( {4 - y - {{{y^2}} \over 2}} \right)dy} $$</p>
<p>$$ = \left[ {4y - {{{y^2}} \over 2} - {{{y^3}} \over 6}} \right]_{ - 4}^2$$</p>
<p>$$ = 18$$ square units</p> | integer | jee-main-2022-online-24th-june-evening-shift |
1l5c2geb9 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let S be the region bounded by the curves y = x<sup>3</sup> and y<sup>2</sup> = x. The curve y = 2|x| divides S into two regions of areas R<sub>1</sub>, R<sub>2</sub>. If max {R<sub>1</sub>, R<sub>2</sub>} = R<sub>2</sub>, then $${{{R_2}} \over {{R_1}}}$$ is equal to ______________.</p> | [] | null | 19 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l8lc1nby/e7dc3cf8-3c01-44b7-9f93-5093bb7a6fcf/ea0230d0-3f02-11ed-8d74-051dc2e154aa/file-1l8lc1nbz.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l8lc1nby/e7dc3cf8-3c01-44b7-9f93-5093bb7a6fcf/ea0230d0-3f02-11ed-8d74-051dc2e154aa/file-1l8lc1nbz.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 24th June Morning Shift Mathematics - Area Under The Curves Question 60 English Explanation"><br>$C_{1}: y=x^{3}$<br><br> $C_{2}: y^{2}=x$
<br><br>
and $C_{3}=y=2|x|$
<br><br>
$C_{1}$ and $C_{2}$ intersect at $(1,1)$
<br><br>
$C_{2}$ and $C_{3}$ intersect at $\left(\frac{1}{4}, \frac{1}{2}\right)$
<br><br>
Clearly $R_{1}=\int_{0}^{1 / 4}(\sqrt{x}-2 x) d x=\frac{2}{3}\left(\frac{1}{8}\right)-\frac{1}{16}=\frac{1}{48}$
<br><br>
and $R_{1}+R_{2}=\int_{0}^{1}\left(\sqrt{x}-x^{3}\right) d x=\frac{2}{3}-\frac{1}{4}=\frac{5}{12}$
<br><br>
So, $\frac{R_{1}+R_{2}}{R_{1}}=\frac{5 / 12}{1 / 48} \Rightarrow 1+\frac{R_{2}}{R_{1}}=20$
<br><br>
$\Rightarrow \frac{R_{2}}{R_{1}}=19$ | integer | jee-main-2022-online-24th-june-morning-shift |
1l5w1a9en | maths | area-under-the-curves | area-bounded-between-the-curves | <p>If for some $$\alpha$$ > 0, the area of the region $$\{ (x,y):|x + \alpha | \le y \le 2 - |x|\} $$ is equal to $${3 \over 2}$$, then the area of the region $$\{ (x,y):0 \le y \le x + 2\alpha ,\,|x| \le 1\} $$ is equal to ____________.</p> | [] | null | 4 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l65usgec/bd118792-9d01-40fb-b8a3-8435f69bbadb/3ead1c40-0ee7-11ed-a7de-eff776fdb55c/file-1l65usged.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l65usgec/bd118792-9d01-40fb-b8a3-8435f69bbadb/3ead1c40-0ee7-11ed-a7de-eff776fdb55c/file-1l65usged.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 30th June Morning Shift Mathematics - Area Under The Curves Question 59 English Explanation 1"></p>
<p>Point A is the intersection of y = x + $$\alpha$$ and y = 2 $$-$$ x lines.</p>
<p>$$\therefore$$ $$y = 2 - y + \alpha $$</p>
<p>$$ \Rightarrow 2y = 2 + \alpha $$</p>
<p>$$ \Rightarrow y = {{2 + \alpha } \over 2}$$</p>
<p>and $$x = 2 - {{2 + \alpha } \over 2} = {{2 - \alpha } \over 2}$$</p>
<p>$$\therefore$$ Point $$A = \left( {{{2 - \alpha } \over 2},{{2 + \alpha } \over 2}} \right)$$</p>
<p>$$\therefore$$ AN = y-coordinate of point $$A = {{2 + \alpha } \over 2}$$</p>
<p>Point B is the intersection of $$y = 2 + x$$ and $$y = - x - \alpha $$ lines.</p>
<p>$$\therefore$$ $$y = 2 - y - \alpha $$</p>
<p>$$ \Rightarrow 2y = 2 - \alpha $$</p><p>
</p><p>$$ \Rightarrow y = {{2 - \alpha } \over 2}$$</p>
<p>$$\therefore$$ $$x = {{2 - \alpha } \over 2} - 2 = {{2 - \alpha - 4} \over 2} = {{ - \alpha - 2} \over 2}$$</p>
<p>$$\therefore$$ Point $$B = \left( {{{ - \alpha - 2} \over 2},{{2 - \alpha } \over 2}} \right)$$</p>
<p>$$\therefore$$ $$BM = y - $$coordinate of point $$B = {{2 - \alpha } \over 2}$$</p>
<p>Area of the common region $$BRAE$$</p>
<p>$$ = \Delta CDE - \left( {\Delta BCR + \Delta ARD} \right)$$</p>
<p>$$ = {1 \over 2} \times 4 \times 2 - \left( {{1 \over 2}( - \alpha + 2) \times BM + {1 \over 2} \times (2 + \alpha ) \times AN} \right)$$</p>
<p>$$ = 4 - \left( {{1 \over 2} \times (2 - \alpha ) \times {{(2 - \alpha )} \over 2} + {1 \over 2} \times (2 + \alpha ) \times {{2 + \alpha } \over 2}} \right)$$</p>
<p>$$ = 4 - \left[ {{{{{(2 - \alpha )}^2}} \over 4} + {{{{(2 + \alpha )}^2}} \over 4}} \right]$$</p>
<p>Given, $$4 - \left[ {{{{{(2 - \alpha )}^2}} \over 4} + {{{{(2 + \alpha )}^2}} \over 4}} \right] = {3 \over 2}$$</p>
<p>$$ \Rightarrow {{{{(2 - \alpha )}^2}} \over 4} + {{{{(2 + \alpha )}^2}} \over 4} = {5 \over 2}$$</p>
<p>$$ \Rightarrow {(2 - \alpha )^2} + {(2 + \alpha )^2} = 10$$</p>
<p>$$ \Rightarrow 4 + {\alpha ^2} - 4\alpha + 4 + {\alpha ^2} + 4\alpha = 10$$</p>
<p>$$ \Rightarrow 2{\alpha ^2} + 8 = 10$$</p>
<p>$$ \Rightarrow 2{\alpha ^2} = 2$$</p>
<p>$$ \Rightarrow {\alpha ^2} = 1$$</p>
<p>$$ \Rightarrow \alpha = \, \pm \,1$$</p>
<p>Given that $$\alpha > 0$$ so accepted value of $$\alpha = + \,1$$.</p>
<p>Now, $$0 \le y \le x + 2\alpha $$ and $$|x| \le 1$$</p>
<p>$$ \Rightarrow 0 \le y \le x + 2$$ and $$ - 1 \le x \le 1$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l65uu2u5/3cd72241-ee42-4674-8663-b746043011c3/6bd1a8d0-0ee7-11ed-a7de-eff776fdb55c/file-1l65uu2u6.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l65uu2u5/3cd72241-ee42-4674-8663-b746043011c3/6bd1a8d0-0ee7-11ed-a7de-eff776fdb55c/file-1l65uu2u6.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 30th June Morning Shift Mathematics - Area Under The Curves Question 59 English Explanation 2"></p>
<p>Area of $$ABCD = {1 \over 2}(1 + 3) \times (1 - ( - 1))$$</p>
<p>$$ = {1 \over 2} \times 4 \times 2$$</p>
<p>$$ = 4$$ sq. unit</p> | integer | jee-main-2022-online-30th-june-morning-shift |
1l6dvcsqp | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region given by</p>
<p>$$A=\left\{(x, y): x^{2} \leq y \leq \min \{x+2,4-3 x\}\right\}$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{31}{8}$$"}, {"identifier": "B", "content": "$$\\frac{17}{6}$$"}, {"identifier": "C", "content": "$$\\frac{19}{6}$$"}, {"identifier": "D", "content": "$$\\frac{27}{8}$$"}] | ["B"] | null | $A=\left\{(x, y): x^{2} \leq y \leq \min \{x+2,4-3 x\}\right.$<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l97rmv56/3d3d2137-f038-407b-8cca-4933b5be4cbf/1f426190-4b59-11ed-bfde-e1cb3fafe700/file-1l97rmv57.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l97rmv56/3d3d2137-f038-407b-8cca-4933b5be4cbf/1f426190-4b59-11ed-bfde-e1cb3fafe700/file-1l97rmv57.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th July Morning Shift Mathematics - Area Under The Curves Question 58 English Explanation"><br><br>
So area of required region
<br><br>
$$
\begin{aligned}
&A=\int_{-1}^{\frac{1}{2}}\left(x+2-x^{2}\right) d x+\int_{\frac{1}{2}}^{1}\left(4-3 x-x^{2}\right) d x \\\\
&=\left[\frac{x^{2}}{2}+2 x-\frac{x^{3}}{3}\right]_{-1}^{\frac{1}{2}}+\left[4 x-\frac{3 x^{2}}{2}-\frac{x^{3}}{3}\right]_{\frac{1}{2}}^{1} \\\\
&=\left(\frac{1}{8}+1-\frac{1}{24}\right)-\left(\frac{1}{2}-2+\frac{1}{3}\right)+\left(4-\frac{3}{2}-\frac{1}{3}\right)-\left(2-\frac{3}{8}-\frac{1}{24}\right) \\\\
&=\frac{17}{6}
\end{aligned}
$$ | mcq | jee-main-2022-online-25th-july-morning-shift |
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