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lmtomciSwqh47mDgXTMln | maths | definite-integration | properties-of-definite-integration | If $$f(x) = \int\limits_0^x {t\left( {\sin x - \sin t} \right)dt\,\,\,} $$ then : | [{"identifier": "A", "content": "f'''(x) + f''(x) = sinx"}, {"identifier": "B", "content": "f'''(x) + f''(x) $$-$$ f'(x) = cosx"}, {"identifier": "C", "content": "f'''(x) + f'(x) = cosx $$-$$ 2x sinx"}, {"identifier": "D", "content": "f'''(x) $$-$$ f''(x) = cosx $$-$$ 2x sinx"}] | ["D"] | null | f(x) = $$\int_0^x {t(\sin x - \sin t).dt} $$
<br><br>= sin x$$\int_0^x {t.dt - \int_0^x {t\sin t.dt} } $$
<br><br>= $${{{x^2}} \over 2}$$ sin x +$$\left[ {t\cos t} \right]_0^x$$ + sin x
<br><br>$$ \Rightarrow $$f(x) = $${{{x^2}} \over 2}$$ sinx + xcosx + sinx
<br><br>f'(x) = $${{{x^2}} \over 2}$$ cosx + 2cos x
<br><br>f''(x) = x cos x $$-$$ $${{{x^2}} \over 2}$$ sin x $$-$$ 2sin x
<br><br>f'''(x) = cos x $$-$$ 2x sin x $$-$$ $${{{x^2}} \over 2}$$ cos x $$-$$ 2cos x
<br><br>$$\therefore\,\,\,$$ f'''(x) + f'(x) = cos x $$-$$ 2x sin x | mcq | jee-main-2018-online-16th-april-morning-slot |
18LtmLb3I2TL8sFD | maths | definite-integration | properties-of-definite-integration | The value of $$\int\limits_{ - \pi /2}^{\pi /2} {{{{{\sin }^2}x} \over {1 + {2^x}}}} dx$$ is | [{"identifier": "A", "content": "$${\\pi \\over 4}$$"}, {"identifier": "B", "content": "$${\\pi \\over 8}$$"}, {"identifier": "C", "content": "$${\\pi \\over 2}$$"}, {"identifier": "D", "content": "$${4\\pi }$$"}] | ["A"] | null | As we know,
<br><br>$$\int\limits_a^b { + \left( x \right)dx} = \int\limits_a^b {f\left( {a + b - x} \right)} dx$$
<br><br>Let $$I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\,\,{{{{\sin }^2}x} \over {1 + {2^x}}}} \,dx$$
<br><br>$$ \Rightarrow \,\,\,\,I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\,\,{{{{\sin }^2}\left( { - {\pi \over 2} + {\pi \over 2} - x} \right)} \over {1 + {2^{ - {\pi \over 2} + {\pi \over 2} - x}}}}} $$
<br><br>$$ \Rightarrow\,\,\,\, I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\sin }^2}x} \over {1 + {2^{ - x}}}}} dx$$
<br><br>$$ \Rightarrow \,\,\,\,I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{2^x}{{\sin }^2}x} \over {1 + {2^x}}}} \,\,dx$$
<br><br>$$\therefore\,\,\,$$ $$2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{{{{\sin }^2}x} \over {1 + {2^x}}} + {{{2^x}{{\sin }^2}x} \over {1 + {2^x}}}} \right)} \,\,dx$$
<br><br>$$ \Rightarrow \,\,\,\,2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^2}x\left( {{{1 + {2^x}} \over {1 + {2^x}}}} \right)} \,\,dx$$
<br><br>$$ \Rightarrow \,\,\,\,2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx$$
<br><br>$$ \Rightarrow \,\,\,\,2I = 2\int\limits_0^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx\,\,\,$$ [ as sin<sup>2</sup>
x is an even function ]
<br><br>$$ \Rightarrow \,\,\,\,I = \int\limits_0^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx\,\,\,\,$$
<br><br>[ as $$\,\,\,\,$$ $$\int\limits_0^a {f\left( a \right)\,\,dx = \int\limits_0^a {f\left( {a - x} \right)\,dx]} } $$
<br><br>So, $$\,\,\,$$ $$I = \int\limits_0^{{\pi \over 2}} {{{\sin }^2}\left( {{b \over 2} - x} \right)} \,\,dx\,\,\,\,$$
<br><br>$$\,\,\,I = \int\limits_0^{{\pi \over 2}} {{{\cos }^2}x} \,\,dx\,\,\,\,$$
<br><br>$$\therefore\,\,\,\,$$ $$\,\,\,\,2I = \int\limits_0^{{\pi \over 2}} {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)} \,\,dx\,\,\,\,$$
<br><br>$$ \Rightarrow \,\,\,\,2I = \int\limits_0^{{\pi \over 2}} \, \,\,dx\,\,\,\,$$
<br><br>$$ \Rightarrow \,\,\,\,2I = {\left[ x \right]_0}^{{\pi \over 2}}$$
<br><br>$$ \Rightarrow \,\,\,\,2I = {\pi \over 2}$$
<br><br>$$ \Rightarrow \,\,\,\,I = {\pi \over 4}$$ | mcq | jee-main-2018-offline |
sDlJBQ4AqBdys5neBbBT4 | maths | definite-integration | properties-of-definite-integration | The value of integral $$\int_{{\pi \over 4}}^{{{3\pi } \over 4}} {{x \over {1 + \sin x}}dx} $$ is : | [{"identifier": "A", "content": "$$\\pi \\sqrt 2 $$ "}, {"identifier": "B", "content": "$$\\pi \\left( {\\sqrt 2 - 1} \\right)$$"}, {"identifier": "C", "content": "$${\\pi \\over 2}\\left( {\\sqrt 2 + 1} \\right)$$"}, {"identifier": "D", "content": "$$2\\pi \\left( {\\sqrt 2 - 1} \\right)$$"}] | ["B"] | null | Let $$I = \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {{x \over {1 + \sin x}}dx} $$
<br><br>also let K = $${x \over {1 + \sin x}}$$
<br><br>Multiplying numerator and denominator by
<br>(1 $$-$$ sin x) we get;
<br><br>$$K = {{x\left( {1 - \sin x} \right)} \over {1 - {{\left( {\sin x} \right)}^2}}} = {{x(1 - \sin x)} \over {{{\left( {\cos x} \right)}^2}}}$$
<br><br>= x(1 $$-$$ sin x) sec<sup>2</sup>x
<br><br>= x sec<sup>2</sup>x $$-$$ x sin x sec<sup>2</sup>x = x sec<sup>2</sup>x $$-$$ x tan x sec x
<br><br>Now, $$I = \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {x{{\sec }^2}xdx - \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {x\sec x\,\tan \,xdx} } $$
<br><br>$$ \Rightarrow $$ I = $$\left| {x\tan x} \right|_{{\pi \over 4}}^{{{3\pi } \over 4}} - \left[ {\log \left( {\sec x} \right)} \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}$$
<br><br> - $$\left| {{x \over {\cos x}}} \right|_{{\pi \over 4}}^{{{3\pi } \over 4}} - \int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {\cos x}}} $$
<br><br>$$ \Rightarrow $$ I = $$\left| {x\tan x} \right|_{{\pi \over 4}}^{{{3\pi } \over 4}}$$ - $$\left| {{x \over {\cos x}}} \right|_{{\pi \over 4}}^{{{3\pi } \over 4}}$$
<br><br> $$ - \left[ {\log \left( {\sec x + \tan x} \right)} \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}$$
<br><br>$$ \Rightarrow $$ I = $$ - {{3\pi } \over 4} - {\pi \over 4}$$$$ + {{3\pi } \over 4}\left( {\sqrt 2 } \right)$$$$ + {\pi \over 4}\left( {\sqrt 2 } \right)$$
<br><br>$$ - \log \left( {\sqrt 2 + 1} \right)$$$$ + \log \left( {\sqrt 2 + 1} \right)$$
<br><br>$$ \Rightarrow $$ I = -$$\pi $$ + $${\sqrt 2 \pi }$$
<br><br>$$ \Rightarrow $$ I = $$\pi \left( {\sqrt 2 - 1} \right)$$ | mcq | jee-main-2018-online-15th-april-evening-slot |
nN0nJ47xujdPhmz7yV4hn | maths | definite-integration | properties-of-definite-integration | The value of the integral
<br/><br/>$$\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} x\left( {1 + \log \left( {{{2 + \sin x} \over {2 - \sin x}}} \right)} \right)dx$$ is : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$${3 \\over 4}$$"}, {"identifier": "C", "content": "$${3 \\over 8}$$ $$\\pi $$"}, {"identifier": "D", "content": "$${3 \\over 16}$$ $$\\pi $$"}] | ["C"] | null | Let
<br><br>I = $$\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} x\left( {1 + \log \left( {{{2 + \sin x} \over {2 - \sin x}}} \right)} \right)dx$$ ......(1)
<br><br>$$ \Rightarrow $$ I = $$\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}}(- x)\left( {1 + \log \left( {{{2 + \sin (- x)} \over {2 - \sin(- x)}}} \right)} \right)dx$$
<br><br>[ As $$\int\limits_a^b { f \left( x \right)dx} = \int\limits_a^b {f\left( {a + b - x} \right)} dx$$ ]
<br><br> = $$\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} x\left( {1 + \log \left( {{{2 - \sin x} \over {2 + \sin x}}} \right)} \right)dx$$
<br><br> = $$\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} x\left( {1 - \log \left( {{{2 + \sin x} \over {2 - \sin x}}} \right)} \right)dx$$ .......(2)
<br><br>Adding equation (1) and (2) we get,
<br><br>2I = 2$$\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} xdx$$
<br><br>2I = 4$$\int\limits_0^{{\pi \over 2}} {{{\sin }^4}} xdx$$
<br><br>I = 2$$\int\limits_0^{{\pi \over 2}} {{{\sin }^4}} xdx$$
<br><br> = $${{2 \times {3 \over 2} \times {1 \over 2} \times \pi } \over {2 \times 2}}$$ [ Using Gamma function ]
<br><br> = $${{3\pi } \over 8}$$ | mcq | jee-main-2018-online-15th-april-morning-slot |
bzr1Kmi7SeJ5VlDhpuDlx | maths | definite-integration | properties-of-definite-integration | If $${I_1} = \int_0^1 {{e^{ - x}}} {\cos ^2}x{\mkern 1mu} dx;$$
<br/><br/> $${I_2} = \int_0^1 {{e^{ - {x^2}}}} {\cos ^2}x{\mkern 1mu} dx$$ and
<br/><br/>$${I_3} = \int_0^1 {{e^{ - {x^3}}}} dx;$$ then | [{"identifier": "A", "content": "I<sub>2</sub> > I<sub>3</sub> > I<sub>1</sub>"}, {"identifier": "B", "content": "I<sub>2</sub> > I<sub>1</sub> > I<sub>3</sub>"}, {"identifier": "C", "content": "I<sub>3</sub> > I<sub>2</sub> > I<sub>1</sub>"}, {"identifier": "D", "content": "I<sub>3</sub> > I<sub>1</sub> > I<sub>2</sub>"}] | ["C"] | null | Given,
<br><br>$${I_1} = \int_0^1 {{e^{ - x}}} {\cos ^2}x{\mkern 1mu} dx;$$
<br><br>$${I_2} = \int_0^1 {{e^{ - {x^2}}}} {\cos ^2}x{\mkern 1mu} dx$$ and
<br><br>$${I_3} = \int_0^1 {{e^{ - {x^3}}}} dx$$
<br><br>For x $$ \in $$ (0, 1)
<br><br>$$ \Rightarrow $$ x > x<sup>2</sup> or $$-$$ x < $$-$$ x<sup>2</sup>
<br><br>and x<sup>2</sup> > x<sup>3</sup> or $$-$$ x<sup>2</sup> < $$-$$ x<sup>3</sup>
<br><br>$$ \therefore $$ $${e^{ - {x^2}}}$$ < $${e^{ - {x^3}}}$$ and $${e^{ - x}}$$ < $${e^{ - {x^2}}}$$
<br><br>$$ \Rightarrow $$ $${e^{ - x}} < {e^{ - {x^2}}} < {e^{ - {x^3}}}$$
<br><br>$$ \Rightarrow $$ $${e^{ - {x^3}}} > {e^{ - {x^2}}} > {e^{ - x}}$$
<br><br>$$ \Rightarrow $$ $${I_3} > {I_2} > {I_1}$$ | mcq | jee-main-2018-online-15th-april-evening-slot |
7dfDkLzie5qJotpAyJPsR | maths | definite-integration | properties-of-definite-integration | Let $$f(x) = \int\limits_0^x {g(t)dt} $$ where g is a non-zero even
function. If ƒ(x + 5) = g(x), then $$ \int\limits_0^x {f(t)dt} $$ equals- | [{"identifier": "A", "content": "5$$\\int\\limits_{x + 5}^5 {g(t)dt} $$"}, {"identifier": "B", "content": "$$\\int\\limits_{x + 5}^5 {g(t)dt} $$"}, {"identifier": "C", "content": "$$\\int\\limits_{5}^{x+5} {g(t)dt} $$"}, {"identifier": "D", "content": "2$$\\int\\limits_{5}^{x+5} {g(t)dt} $$"}] | ["B"] | null | $$f(x) = \int\limits_0^x {g(t)dt} $$
<br><br>$$f\left( { - x} \right) = \int\limits_0^{ - x} {g\left( t \right)} dt$$
<br><br>Put t = -v
<br><br>= $$ - \int\limits_0^x {g\left( { - v} \right)} dv$$
<br><br>= $$ - \int\limits_0^x {g\left( v \right)} d(v)$$ [ as g(v) is an even function.]
<br><br>= - f(x)
<br><br>$$ \Rightarrow $$ f(-x) = -f(x)
<br><br>$$ \therefore $$ f(x) is an odd function.
<br><br>Given ƒ(x + 5) = g(x)
<br><br>$$ \therefore $$ g(- x) = ƒ(- x + 5)
<br><br>$$ \Rightarrow $$ g(x) = - f(x - 5) [as g(x) is even and f(x) is an odd function]
<br><br>Replacing x by x + 5, we get
<br><br>f(x) = - g(x + 5) ......(1)
<br><br>Now
<br><br>$$ \int\limits_0^x {f(t)dt} $$
<br><br>= $$ - \int\limits_0^x {g\left( {t + 5} \right)} dt$$
<br><br>= $$ - \int\limits_5^{x + 5} {g\left( t \right)} dt$$
<br><br>= $$\int\limits_{x + 5}^5 {g(t)dt} $$ | mcq | jee-main-2019-online-8th-april-evening-slot |
vJn2EyBHRDhnGJNUko3rsa0w2w9jxb3cgo0 | maths | definite-integration | properties-of-definite-integration | A value of $$\alpha $$ such that
<br/>$$\int\limits_\alpha ^{\alpha + 1} {{{dx} \over {\left( {x + \alpha } \right)\left( {x + \alpha + 1} \right)}}} = {\log _e}\left( {{9 \over 8}} \right)$$ is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "- 2"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "$$-{1 \\over 2}$$"}] | ["B"] | null | $$I = \int\limits_\alpha ^{\alpha + 1} {{{dx} \over {(x + \alpha )(x + \alpha + 1)}}} $$<br><br>
Let<b> x + $$\alpha $$ = t </b> and <b>dx = dt</b><br><br>
$$I = \int\limits_{2\alpha }^{2\alpha + 1} {{{dt} \over {t(t + 1)}}} = \int\limits_{2\alpha }^{2\alpha + 1} {\left( {{1 \over t} - {1 \over {t + 1}}} \right)dx} $$<br><br>
$$ \Rightarrow \left( {\ln t - \ln (t + 1)} \right)|_{2\alpha }^{2\alpha + 1} = \ln \left( {{t \over {t + 1}}} \right)|_{2\alpha }^{2\alpha + 1}$$<br><br>
$$ \Rightarrow \ln \left( {{{2\alpha + 1} \over {2\alpha + 2}}} \right) - \ln \left( {{{2\alpha } \over {2\alpha + 1}}} \right)$$<br><br>
$$ \Rightarrow \ln \left( {{{2\alpha + 1} \over {2\alpha + 2}} \times {{2\alpha + 1} \over {2\alpha }}} \right) = \ln \left( {{9 \over 8}} \right)$$<br><br>
$$ \Rightarrow {{{{\left( {2\alpha + 1} \right)}^2}} \over {\left( {2\alpha + 2} \right)}} = {{9\alpha } \over 4}$$<br><br>
$$4(4{\alpha ^2} + 1 + 4\alpha ) = 18{\alpha ^2} + 18\alpha $$<br><br>
$$ \Rightarrow 2{\alpha ^2} - 2\alpha - 4 = 0$$<br><br>
$$ \Rightarrow {\alpha ^2} + \alpha - 2 = 0$$<br><br>
$$ \Rightarrow (\alpha + 2)(\alpha - 1) = 0$$<br><br>
$$\alpha $$ = 1 or $$\alpha $$ = -2
| mcq | jee-main-2019-online-12th-april-evening-slot |
maAWrFq0m2nmYcAJtZ3rsa0w2w9jx5cfz6r | maths | definite-integration | properties-of-definite-integration | If $$\int\limits_0^{{\pi \over 2}} {{{\cot x} \over {\cot x + \cos ecx}}} dx$$ = m($$\pi $$ + n), then m.n is equal to | [{"identifier": "A", "content": "- 1"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$ - {1 \\over 2}$$"}, {"identifier": "D", "content": "$${1 \\over 2}$$"}] | ["A"] | null | $$\int\limits_0^{\pi /2} {{{\cot x} \over {\cot x + {\mathop{\rm cosec}\nolimits} \,x}}dx = } \int\limits_0^{\pi /2} {{{\cos x} \over {\cos x + 1}}dx} $$<br><br>
$$ \Rightarrow \int\limits_0^{\pi /2} {{{2{{\cos }^2}{x \over 2} - 1} \over {2{{\cos }^2}{x \over 2}}}dx = } \int\limits_0^{\pi /2} {\left( {1 - {1 \over 2}{{\sec }^2}{x \over 2}} \right)dx} $$<br><br>
$$ \Rightarrow \left( {x - \tan {x \over 2}} \right)_0^{\pi /2} = {\pi \over 2} - 1 = {1 \over 2}\left( {\pi - 2} \right)$$<br><br>
mn = $${1 \over 2}x - 2 = - 1$$ | mcq | jee-main-2019-online-12th-april-morning-slot |
Ve0Btsv6GeVJ9Ug36y3rsa0w2w9jx25p121 | maths | definite-integration | properties-of-definite-integration | The integral $$\int\limits_{\pi /6}^{\pi /3} {{{\sec }^{2/3}}} x\cos e{c^{4/3}}xdx$$ is equal to : | [{"identifier": "A", "content": "$${3^{{5 \\over 3}}} - {3^{{1 \\over 3}}}$$"}, {"identifier": "B", "content": "$${3^{{5 \\over 6}}} - {3^{{2 \\over 3}}}$$"}, {"identifier": "C", "content": "$${3^{{4 \\over 3}}} - {3^{{1 \\over 3}}}$$"}, {"identifier": "D", "content": "$${3^{{7 \\over 6}}} - {3^{{5 \\over 6}}}$$"}] | ["D"] | null | $$\int\limits_{\pi /6}^{\pi /3} {{{\sec }^{2/3}}x\,{{{\mathop{\rm cosec}\nolimits} }^{4/3}}x\,dx} $$<br><br>
$$ = \int {{{{{\sec }^2}x} \over {{{\tan }^{4/3}}x}}dx} $$<br><br>
Let tan x = t, sec<sup>2</sup> x dx = dt<br><br>
$$ = \int {{{dt} \over {{t^{4/3}}}}} $$<br><br>
$$I = - 3\left( {{t^{ - 1/3}}} \right)$$<br><br>
$$ \Rightarrow \left( { - 3{{\left( {\tan x} \right)}^{{{ - 1} \over 3}}}} \right)_{\pi /6}^{\pi /3}$$<br><br>
$$ \Rightarrow 3\left( {{3^{{1 \over 3}}} - {1 \over {{3^{{1 \over 6}}}}}} \right)$$
= 3<sup>7/6</sup> - 3<sup>5/6</sup>
| mcq | jee-main-2019-online-10th-april-evening-slot |
g1HvVGGm3HmxOOwDTB3rsa0w2w9jwxzkfe2 | maths | definite-integration | properties-of-definite-integration | The value of $$\int\limits_0^{2\pi } {\left[ {\sin 2x\left( {1 + \cos 3x} \right)} \right]} dx$$,
<br/>where [t] denotes the greatest integer function is : | [{"identifier": "A", "content": "2$$\\pi $$"}, {"identifier": "B", "content": "$$\\pi $$"}, {"identifier": "C", "content": "-2$$\\pi $$"}, {"identifier": "D", "content": "-$$\\pi $$"}] | ["D"] | null | $$I = \int\limits_0^{2\pi } {\left[ {\sin 2x(1 + \cos 3x)} \right]} dx$$ .... (i)<br><br>
$$ \therefore\int\limits_0^a {f(x)} = \int\limits_0^a {f(a - x)} dx$$<br><br>
$$ \because I = \int\limits_0^{2\pi } {\left[ { - \sin 2x(1 + \cos 3x)} \right]} dx$$<br><br>
By (i) + (ii)<br><br>
$$ \Rightarrow 2I = \int\limits_0^{2\pi } {( - 1)dx} $$<br><br>
$$ \Rightarrow 2I = - (x)_0^{2\pi }$$<br><br>
$$ \Rightarrow $$ I = –$$\pi $$ | mcq | jee-main-2019-online-10th-april-morning-slot |
ufHaLBHc98mkwrZOC618hoxe66ijvwufwrk | maths | definite-integration | properties-of-definite-integration | The value of the integral $$\int\limits_0^1 {x{{\cot }^{ - 1}}(1 - {x^2} + {x^4})dx} $$ is :- | [{"identifier": "A", "content": "$${\\pi \\over 2} - {1 \\over 2}{\\log _e}2$$"}, {"identifier": "B", "content": "$${\\pi \\over 4} - {\\log _e}2$$"}, {"identifier": "C", "content": "$${\\pi \\over 4} - {1 \\over 2}{\\log _e}2$$"}, {"identifier": "D", "content": "$${\\pi \\over 2} - {\\log _e}2$$"}] | ["C"] | null | I = $$\int\limits_0^1 {x{{\cot }^{ - 1}}(1 - {x^2} + {x^4})dx} $$
<br><br>Let x<sup>2</sup> = t
<br><br>$$ \Rightarrow $$ 2xdx = dt
<br><br>At x = 0, t = 0
<br><br>At x = 1, t = 1
<br><br>I = $${1 \over 2}\int\limits_0^1 {{{\cot }^{ - 1}}\left( {1 - t + {t^2}} \right)dt} $$
<br><br>= $${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {{1 \over {1 - t + {t^2}}}} \right)dt} $$
<br><br>= $${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {{{t + \left( {1 - t} \right)} \over {1 - t + {t^2}}}} \right)dt} $$
<br><br>= $${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {{{t + \left( {1 - t} \right)} \over {1 - t\left( {1 - t} \right)}}} \right)dt} $$
<br><br>= $${1 \over 2}\int\limits_0^1 {\left[ {{{\tan }^{ - 1}}\left( {1 - t} \right) + {{\tan }^{ - 1}}\left( t \right)} \right]dt} $$
<br><br>= $${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} + {1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {1 - t} \right)dt} $$
<br><br>= $${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} + {1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} $$
<br><br>[ As $${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {1 - t} \right)dt} $$
<br><br>= $${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left[ {0 + 1 - \left( {1 - t} \right)} \right]dt} $$
<br><br>= $${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} $$ ]
<br><br>= $$2 \times {1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} $$
<br><br>= $$\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} $$
<br><br>Using integration by parts rule
<br><br>= $$\left[ {t.{{\tan }^{ - 1}}\left( t \right)} \right]_0^1$$$$ - \int\limits_0^1 {t.{1 \over {1 + {t^2}}}} dt$$
<br><br>= $${\pi \over 4} - {1 \over 2}\left[ {\log \left( {1 + {t^2}} \right)} \right]_0^1$$
<br><br>= $${\pi \over 4} - {1 \over 2}{\log _e}2$$ | mcq | jee-main-2019-online-9th-april-evening-slot |
AZI26JLdbb4NGYwieCd4T | maths | definite-integration | properties-of-definite-integration | The value of $$\int\limits_0^{\pi /2} {{{{{\sin }^3}x} \over {\sin x + \cos x}}dx} $$ is | [{"identifier": "A", "content": "$${{\\pi - 2} \\over 8}$$"}, {"identifier": "B", "content": "$${{\\pi - 2} \\over 4}$$"}, {"identifier": "C", "content": "$${{\\pi - 1} \\over 2}$$"}, {"identifier": "D", "content": "$${{\\pi - 1} \\over 4}$$"}] | ["D"] | null | I = $$\int\limits_0^{\pi /2} {{{{{\sin }^3}x} \over {\sin x + \cos x}}dx} $$ .....(1)
<br><br>$$ \Rightarrow $$ I = $$\int\limits_0^{{\pi \over 2}} {{{{{\cos }^3}x} \over {\sin x + \cos x}}} dx$$ ......(2)
<br><br>Adding those two
<br><br>2I = $$\int\limits_0^{{\pi \over 2}} {{{si{n^3}x + {{\cos }^3}x} \over {\sin x + \cos x}}} dx$$
<br><br>= $$\int\limits_0^{{\pi \over 2}} {{{\left( {\sin x + \cos x} \right)\left( {si{n^2}x + {{\cos }^2}x - \sin x\cos x} \right)} \over {\sin x + \cos x}}} dx$$
<br><br>= $$\int\limits_0^{{\pi \over 2}} {\left( {si{n^2}x + {{\cos }^2}x - \sin x\cos x} \right)} dx$$
<br><br>= $$\int\limits_0^{{\pi \over 2}} {\left( {1 - {{\sin 2x} \over 2}} \right)} dx$$
<br><br>= $$\left[ {x + {{\cos 2x} \over 4}} \right]_0^{{\pi \over 2}}$$
<br><br>= $$\left( {{\pi \over 2} - {1 \over 4}} \right) - \left( {{1 \over 4}} \right)$$
<br><br>$$ \therefore $$ 2I = $$\left( {{\pi \over 2} - {1 \over 2}} \right)$$
<br><br>$$ \therefore $$ I = $${{\pi - 1} \over 4}$$ | mcq | jee-main-2019-online-9th-april-morning-slot |
6w5HWVsQQcJWqT6LDMHjl | maths | definite-integration | properties-of-definite-integration | If $$f(x) = {{2 - x\cos x} \over {2 + x\cos x}}$$ and g(x) = log<sub>e</sub>x, (x > 0) then
the value of integral<br/><br/> $$\int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {g\left( {f\left( x \right)} \right)dx{\rm{ }}} $$ is | [{"identifier": "A", "content": "log<sub>e</sub>3"}, {"identifier": "B", "content": "log<sub>e</sub>2"}, {"identifier": "C", "content": "log<sub>e</sub>1"}, {"identifier": "D", "content": "log<sub>e</sub>e"}] | ["C"] | null | $$g\left( {f\left( x \right)} \right)$$ = $$\ln \left( {f\left( x \right)} \right)$$ = $$\ln \left( {{{2 - x\cos x} \over {2 + x\cos x}}} \right)$$
<br><br>$$ \therefore $$ I = $$\int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {\ln \left( {{{2 - x\cos x} \over {2 + x\cos x}}} \right)dx} $$
<br><br>( Using property $$\int\limits_{ - a}^a {f\left( x \right)} dx = \int\limits_0^a {\left( {f\left( x \right) + f\left( { - x} \right)} \right)} dx$$ )
<br><br> I = $$\int\limits_0^{{\pi \over 4}} {\left( {\ln \left( {{{2 - x\cos x} \over {2 + x\cos x}}} \right) + \ln \left( {{{2 + x\cos x} \over {2 - x\cos x}}} \right)} \right)dx} $$
<br><br>= $$\int\limits_0^{{\pi \over 4}} {\left( {\ln \left( 1 \right)} \right)dx} $$
<br><br>= 0 = log<sub>e</sub>1 | mcq | jee-main-2019-online-8th-april-morning-slot |
WJSwZSmnDltFY0gKeEQW0 | maths | definite-integration | properties-of-definite-integration | Let f and g be continuous functions on [0, a] such that f(x) = f(a – x) and g(x) + g(a – x) = 4, then $$\int\limits_0^a \, $$f(x) g(x) dx is equal to : | [{"identifier": "A", "content": "4$$\\int\\limits_0^a \\, $$f(x)dx"}, {"identifier": "B", "content": "$$-$$ 3$$\\int\\limits_0^a \\, $$f(x)dx"}, {"identifier": "C", "content": "$$\\int\\limits_0^a \\, $$f(x)dx"}, {"identifier": "D", "content": "2$$\\int\\limits_0^a \\, $$f(x)dx"}] | ["D"] | null | $${\rm I} = \int_0^a {f\left( x \right)g\left( x \right)dx} $$
<br><br>$${\rm I} = \int_0^a {f\left( {a - x} \right)g\left( {a - x} \right)dx} $$
<br><br>$${\rm I} = \int_0^a {f\left( x \right)\left( {4 - g\left( x \right.} \right)dx} $$
<br><br>$${\rm I} = 4\int_0^a {f\left( x \right)dx - {\rm I}} $$
<br><br>$$ \Rightarrow {\rm I} = 2\int_0^a {f\left( x \right)dx} $$ | mcq | jee-main-2019-online-12th-january-morning-slot |
vnub8GmPI1I4MyT3bDdFj | maths | definite-integration | properties-of-definite-integration | The integral $$\int\limits_{\pi /6}^{\pi /4} {{{dx} \over {\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}} $$ equals : | [{"identifier": "A", "content": "$${\\pi \\over {40}}$$"}, {"identifier": "B", "content": "$${1 \\over {20}}{\\tan ^{ - 1}}\\left( {{1 \\over {9\\sqrt 3 }}} \\right)$$"}, {"identifier": "C", "content": "$${1 \\over {10}}\\left( {{\\pi \\over 4} - {{\\tan }^{ - 1}}\\left( {{1 \\over {9\\sqrt 3 }}} \\right)} \\right)$$"}, {"identifier": "D", "content": "$${1 \\over 5}\\left( {{\\pi \\over 4}{{-\\tan }^{ - 1}}\\left( {{1 \\over {3\\sqrt 3 }}} \\right)} \\right)$$"}] | ["C"] | null | I $$=$$ $$\int\limits_{\pi /6}^{\pi /4} {{{dx} \over {\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}} $$
<br><br>$${\rm I} = {1 \over 2}\int\limits_{\pi /6}^{\pi /4} {{{{{\tan }^4}x{{\sec }^2}xdx} \over {\left( {1 + {{\tan }^{10}}x} \right)}}} $$ Put tan<sup>5</sup>x $$=$$ t
<br><br>$${\rm I} = {1 \over {10}}\int\limits_{{{\left( {{1 \over {\sqrt 3 }}} \right)}^5}}^1 {{{dt} \over {1 + {t^2}}}} = {1 \over {10}}\left( {{\pi \over 4} - {{\tan }^{ - 1}}{1 \over {9\sqrt 3 }}} \right)$$ | mcq | jee-main-2019-online-11th-january-evening-slot |
cSvyruRIVbuUj4gZhH0nC | maths | definite-integration | properties-of-definite-integration | The value of the integral $$\int\limits_{ - 2}^2 {{{{{\sin }^2}x} \over { \left[ {{x \over \pi }} \right] + {1 \over 2}}}} \,dx$$ (where [x] denotes the greatest integer less than or equal to x) is | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "4$$-$$ sin 4"}, {"identifier": "D", "content": "sin 4"}] | ["A"] | null | I $$=$$ $$\int\limits_{ - 2}^2 {{{{{\sin }^2}x} \over { \left[ {{x \over \pi }} \right] + {1 \over 2}}}} \,dx$$
<br><br>$${\rm I} = \int\limits_0^2 {\left( {{{{{\sin }^2}x} \over {\left[ {{x \over \pi }} \right] + {1 \over 2}}} + {{{{\sin }^2}\left( { - x} \right)} \over {\left[ { - {x \over \pi }} \right] + {1 \over 2}}}} \right)dx} $$
<br><br>$$\left( {\left[ {{x \over \pi }} \right] + \left[ { - {x \over \pi }} \right] = - 1\,\,} \right.$$ as $$\left. {\matrix{
\, \cr
\, \cr
} x \ne n\pi } \right)$$
<br><br>$${\rm I} = \int\limits_0^2 {\left( {{{{{\sin }^2}x} \over {\left[ {{x \over \pi }} \right] + {1 \over 2}}} + {{{{\sin }^2}x} \over { - 1 - \left[ {{x \over \pi }} \right] + {1 \over 2}}}} \right)dx = 0} $$ | mcq | jee-main-2019-online-11th-january-morning-slot |
ZtWfcgjHBtokIgHrSu7d4 | maths | definite-integration | properties-of-definite-integration | The value of $$\int\limits_{ - \pi /2}^{\pi /2} {{{dx} \over {\left[ x \right] + \left[ {\sin x} \right] + 4}}} ,$$ where [t] denotes the greatest integer less than or equal to t, is | [{"identifier": "A", "content": "$${1 \\over {12}}\\left( {7\\pi - 5} \\right)$$"}, {"identifier": "B", "content": "$${1 \\over {12}}\\left( {7\\pi + 5} \\right)$$"}, {"identifier": "C", "content": "$${3 \\over {10}}\\left( {4\\pi - 3} \\right)$$"}, {"identifier": "D", "content": "$${3 \\over {20}}\\left( {4\\pi - 3} \\right)$$"}] | ["D"] | null | $${\rm I} = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {\left[ x \right] + \left[ {\sin x} \right] + 4}}} $$
<br><br>$$ = \int\limits_{{{ - \pi } \over 2}}^{ - 1} {{{dx} \over { - 2 - 1 + 4}}} + \int\limits_{ - 1}^0 {{{dx} \over { - 1 - 1 + 4}}} $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \int\limits_0^1 {{{dx} \over {0 + 0 + 4}} + \int\limits_1^{{\pi \over 2}} {{{dx} \over {1 + 0 + 4}}} } $$
<br><br>$$\int\limits_{{{ - \pi } \over 2}}^{ - 1} {{{dx} \over 1} + \int\limits_{ - 1}^0 {{{dx} \over 2}} } + \int\limits_0^1 {{{dx} \over 4}} + \int\limits_1^{{\pi \over 2}} {{{dx} \over 5}} $$
<br><br>$$\left( { - 1 + {\pi \over 2}} \right) + {1 \over 2}\left( {0 + 1} \right) + {1 \over 4} + {1 \over 5}\left( {{\pi \over 2} - 1} \right)$$
<br><br>$$ - 1 + {1 \over 2} + {1 \over 4} - {1 \over 5} + {\pi \over 2} + {\pi \over {10}}$$
<br><br>$${{ - 20 + 10 + 5 - 4} \over {20}} + {{6\pi } \over {10}}$$
<br><br>$${{ - 9} \over {20}} + {{3\pi } \over 5}$$ | mcq | jee-main-2019-online-10th-january-evening-slot |
ZEOwvZM6M2qrjJ73t9XSz | maths | definite-integration | properties-of-definite-integration | Let $${\rm I} = \int\limits_a^b {\left( {{x^4} - 2{x^2}} \right)} dx.$$ If I is minimum then the ordered pair (a, b) is - | [{"identifier": "A", "content": "$$\\left( {\\sqrt 2 , - \\sqrt 2 } \\right)$$"}, {"identifier": "B", "content": "$$\\left( {0,\\sqrt 2 } \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - \\sqrt 2 ,\\sqrt 2 } \\right)$$"}, {"identifier": "D", "content": "$$\\left( { - \\sqrt 2 ,0} \\right)$$"}] | ["C"] | null | Let f(x) = x<sup>2</sup><sup></sup>(x<sup>2</sup><sup></sup> $$-$$ 2)
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266617/exam_images/eleslqkk6r7otnrfth9a.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Morning Slot Mathematics - Definite Integration Question 226 English Explanation">
<br>As long as f(x) lie below the x-axis, define integral will remain negative,
<br>so correct value of (a, b) is ($$-$$ $$\sqrt 2 $$, $$\sqrt 2 $$) for minimum of I | mcq | jee-main-2019-online-10th-january-morning-slot |
vbOxkVhe5Ag4Et5ukbMbS | maths | definite-integration | properties-of-definite-integration | If $$\int\limits_0^{{\pi \over 3}} {{{\tan \theta } \over {\sqrt {2k\,\sec \theta } }}} \,d\theta = 1 - {1 \over {\sqrt 2 }},\left( {k > 0} \right),$$ then value of k is : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2"}] | ["D"] | null | $$\int\limits_0^{{\pi \over 3}} {{{\tan \theta } \over {\sqrt {2k\,\sec \theta } }}} \,d\theta = 1 - {1 \over {\sqrt 2 }},\left( {k > 0} \right),$$
<br><br>$$ \Rightarrow \,\,\int\limits_0^{\pi /3} {{{\tan \theta } \over {\sqrt {2k\sec \theta } }}} \,d\theta = 1 - {1 \over {\sqrt 2 }}\left( {k > 0} \right)$$
<br><br>$$ \Rightarrow \,\,{1 \over {\sqrt {2k} }}\left( { - 2\sqrt {\cos \theta } } \right)_0^{\pi /3} = 1 - {1 \over {\sqrt 2 }}$$
<br><br>$$ \Rightarrow \,\,{1 \over {\sqrt {2k} }}\left( { - \sqrt 2 + 2} \right) = 1 - {1 \over {\sqrt 2 }}$$
<br><br>$$ \Rightarrow \,\,{{\sqrt 2 \left( {\sqrt 2 - 1} \right)} \over {\sqrt {2k} }} = {{\sqrt 2 - 1} \over {\sqrt 2 }}$$
<br><br>$$ \Rightarrow \,\,k = 2$$ | mcq | jee-main-2019-online-9th-january-evening-slot |
S2eHDIbrxUdmooH9dWT9i | maths | definite-integration | properties-of-definite-integration | The value of $$\int\limits_0^\pi {{{\left| {\cos x} \right|}^3}} \,dx$$ is : | [{"identifier": "A", "content": "$$4 \\over 3$$"}, {"identifier": "B", "content": "$$-$$ $$4 \\over 3$$"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$$2 \\over 3$$"}] | ["A"] | null | $$\int\limits_0^\pi {{{\left| {\cos x} \right|}^3}} \,dx$$
<br><br>The period of $$\left| {\cos x} \right|$$ = $${\pi \over 2}$$
<br><br>$$ \therefore $$ I = 2 $$\int\limits_0^{{\pi \over 2}} {{{\left| {\cos x} \right|}^3}} \,dx$$
<br><br>as in the range 0 to $${\pi \over 2}$$ $$\left| {\cos x} \right|$$ is positive.
<br><br>So, $$\left| {\cos x} \right|$$ = $$cosx$$
<br><br>$$ \therefore $$ I = 2 $$\int\limits_0^{{\pi \over 2}} {{{\cos }^3}x\,dx} $$
<br><br>= 2$$\int\limits_0^{{\pi \over 2}} {\left( {{{\cos 3x + 3\cos x} \over 4}} \right)} dx$$
<br><br>I = $${1 \over 2}\left[ {{{\sin 3x} \over 3} + 3\sin x} \right]_0^{{\pi \over 2}}$$
<br><br>I = $${1 \over 2}\left[ {{1 \over 3}\left( {{{3\pi } \over 2}} \right) + 3.\sin {\pi \over 2}} \right]$$
<br><br>I = $${1 \over 2}\left[ { - {1 \over 3} + 3} \right]$$
<br><br>= $${1 \over 2}\left( {{8 \over 3}} \right)$$
<br><br>= $${4 \over 3}$$ | mcq | jee-main-2019-online-9th-january-morning-slot |
Bi9Bsa02Lq6vcZ6XI5adz | maths | definite-integration | properties-of-definite-integration | The integral $$\int\limits_1^e {\left\{ {{{\left( {{x \over e}} \right)}^{2x}} - {{\left( {{e \over x}} \right)}^x}} \right\}} \,$$ log<sub>e</sub> x dx is equal to : | [{"identifier": "A", "content": "$$ - {1 \\over 2} + {1 \\over e} - {1 \\over {2{e^2}}}$$"}, {"identifier": "B", "content": "$${3 \\over 2} - e - {1 \\over {2{e^2}}}$$"}, {"identifier": "C", "content": "$${1 \\over 2} - e - {1 \\over {{e^2}}}$$"}, {"identifier": "D", "content": "$${3 \\over 2} - {1 \\over e} - {1 \\over {2{x^2}}}$$ "}] | ["B"] | null | $$\int\limits_1^e {\left\{ {{{\left( {{x \over e}} \right)}^{2x}} - {{\left( {{e \over x}} \right)}^x}} \right\}} \,$$
<br><br>Let $${\left( {{x \over e}} \right)^{2x}} = t,{\left( {{e \over x}} \right)^x} = v$$
<br><br>$$ = {1 \over 2}\int\limits_{{{\left( {{1 \over e}} \right)}^2}}^1 {dt + \int\limits_e^1 {dv} } $$
<br><br>$$ = {1 \over 2}\left( {1 - {1 \over {{e^2}}}} \right) + \left( {1 - e} \right) = {3 \over 2} - {1 \over {2{e^2}}} - e$$ | mcq | jee-main-2019-online-12th-january-evening-slot |
TnsaTvBW3XCkN3XElejgy2xukf3xytl4 | maths | definite-integration | properties-of-definite-integration | If the value of the integral
<br/>$$\int\limits_0^{{1 \over 2}} {{{{x^2}} \over {{{\left( {1 - {x^2}} \right)}^{{3 \over 2}}}}}} dx$$
<br/><br/>is $${k \over 6}$$, then k is equal to : | [{"identifier": "A", "content": "$$2\\sqrt 3 + \\pi $$"}, {"identifier": "B", "content": "$$3\\sqrt 2 - \\pi $$"}, {"identifier": "C", "content": "$$3\\sqrt 2 + \\pi $$"}, {"identifier": "D", "content": "$$2\\sqrt 3 - \\pi $$"}] | ["D"] | null | $$I = \int\limits_0^{{1 \over 2}} {{{{x^2}} \over {{{(1 - {x^2})}^{{3 \over 2}}}}}} dx$$<br><br>Let $$x = \sin \theta $$<br><br>$$ \Rightarrow dx = \cos \theta d\theta $$<br><br>When x = 0 then sin$$\theta $$ = 0 $$ \Rightarrow $$ $$\theta $$ = 0<br><br>When $$x = {1 \over 2}$$ then $$\sin \theta = {1 \over 2}$$ $$ \Rightarrow $$ $$\theta = {\pi \over 6}$$<br><br>$$ \therefore $$ $$I = \int\limits_0^{{\pi \over 6}} {{{{{\sin }^2}\theta } \over {{{({{\cos }^2}\theta )}^{{3 \over 2}}}}}\cos \theta d\theta } $$<br><br>$$ = \int\limits_0^{{\pi \over 6}} {{{{{\sin }^2}\theta } \over {{{\cos }^2}\theta }}d\theta } $$<br><br>$$ = \int\limits_0^{{\pi \over 6}} {{{\tan }^2}d\theta } = \int\limits_0^{{\pi \over 6}} {({{\sec }^2}\theta - 1)d\theta } $$<br><br>$$ = \left[ {\tan \theta - \theta } \right]_0^{{\pi \over 6}}$$<br><br>$$ = {1 \over {\sqrt 3 }} - {\pi \over 6}$$<br><br>Given, $${k \over 6} = {1 \over {\sqrt 3 }} - {\pi \over 6} = {{2\sqrt 3 - \pi } \over 6}$$<br><br>$$ \Rightarrow k = 2\sqrt 3 - \pi $$ | mcq | jee-main-2020-online-3rd-september-evening-slot |
U4z6G7UL3Nk0YHsoqLjgy2xukf40epci | maths | definite-integration | properties-of-definite-integration | Suppose f(x) is a polynomial of degree four,
having critical points at –1, 0, 1. If
<br/>T = {x $$ \in $$ R |
f(x) = f(0)}, then the sum of squares of all the
elements of T is : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "4"}] | ["D"] | null | Critical points = $$ - $$1, 0, 1.<br><br>$$ \therefore $$ f'(x) = a(x $$ - $$ 1)(x + 1)x<br><br>$$ \therefore $$ f(x) = a$$\left( {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right) + C$$<br><br>$$ \because $$ f(x) = f(0)<br><br>$$ - a\left( {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right) + C = C$$<br><br>$$a{{{x^2}} \over 4}\left( {{x^2} - 2} \right) = 0$$<br><br>$$ \therefore $$ x = 0, $$\sqrt 2 $$, $$ - \sqrt 2 $$<br><br>$$ \therefore $$ T = $$\left\{ {0,\sqrt 2 , - \sqrt 2 } \right\}$$<br><br>Sum of square of elements of $$T = {0^2} + {\left( {\sqrt 2 } \right)^2} + {\left( { - \sqrt 2 } \right)^2} = 4$$ | mcq | jee-main-2020-online-3rd-september-evening-slot |
nbq0G4yuH6pBXiIQ7yjgy2xukf7gl39c | maths | definite-integration | properties-of-definite-integration | Let $$f(x) = \left| {x - 2} \right|$$ and g(x) = f(f(x)), $$x \in \left[ {0,4} \right]$$. Then <br/>$$\int\limits_0^3 {\left( {g(x) - f(x)} \right)} dx$$ is equal to: | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "$${3 \\over 2}$$"}] | ["A"] | null | $$f(x) = |x - 2|$$<br><br>$$ \therefore $$ $$f(f(x)) = \left| {|x - 2| - 2} \right| = g(x)$$<br><br>$$ \Rightarrow g(x) = \left| {|x - 2| - 2} \right| = \left\{ {\matrix{
{|x - 4|} & {if\,x \ge 2} \cr
{| - x|} & {if\,x < 2} \cr
} } \right.$$<br><br>$$ \therefore $$ $$\int\limits_0^3 {(g(x) - f(x))dx} $$<br><br>$$ = \int\limits_0^3 {g(x)dx} - \int\limits_0^3 {f(x)dx} $$<br><br>$$ = \int\limits_0^2 {| - x|dx} + \int\limits_2^3 { - (x - 4)dx} - \int\limits_0^2 { - (x - 2)dx} - \int\limits_2^3 {(x - 2)dx} $$<br><br>$$ = \int\limits_0^2 {x\,dx} - \int\limits_2^3 {(x - 4)dx} + \int\limits_0^2 {(x - 2)dx} - \int\limits_2^3 {(x - 2)dx} $$<br><br>$$ = \left[ {{{{x^2}} \over 2}} \right]_0^2 - \left[ {{{{x^2}} \over 2} - 4x} \right]_2^3 + \left[ {{{{x^2}} \over 2} - 2x} \right]_0^2 - \left[ {{{{x^2}} \over 2} - 2x} \right]_2^3$$<br><br>$$ = 2 - \left\{ {{9 \over 2} - 12 - 2 + 8} \right\} + \{ 2 - 4\} - \left\{ {{9 \over 2} - 6 - 2 + 4} \right\}$$<br><br>$$ = 2 - \left\{ {{9 \over 2} - 6} \right\} - 2 - \left\{ {{9 \over 2} - 4} \right\}$$<br><br>$$ = 10 - {9 \over 2} - {9 \over 2} = {{20 - 9 - 9} \over 2} = {2 \over 2} = 1$$ | mcq | jee-main-2020-online-4th-september-morning-slot |
fCbDUKnetkr7Rql8k5jgy2xukfajri80 | maths | definite-integration | properties-of-definite-integration | The integral <br/>$$\int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{\tan }^3}x.{{\sin }^2}3x\left( {2{{\sec }^2}x.{{\sin }^2}3x + 3\tan x.\sin 6x} \right)dx} $$
<br/>is equal to: | [{"identifier": "A", "content": "$$ - {1 \\over {9}}$$"}, {"identifier": "B", "content": "$$ - {1 \\over {18}}$$"}, {"identifier": "C", "content": "$$ {7 \\over {18}}$$"}, {"identifier": "D", "content": "$${9 \\over 2}$$"}] | ["B"] | null | Given,
<br><br>I = $$\int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{\tan }^3}x.{{\sin }^2}3x\left( {2{{\sec }^2}x.{{\sin }^2}3x + 3\tan x.\sin 6x} \right)dx} $$
<br><br>$$I = \int\limits_{\pi /6}^{\pi /3} ({2.{{\tan }^3}} x{\sec ^2}x{\sin ^4}3x + 3{\tan ^4}x{\sin ^2}3x.\,2\sin 3xcos\,3x\,)dx$$<br><br>$$ = {1 \over 2}\int\limits_{\pi /6}^{\pi /3} ({4{{\tan }^3}} x{\sec ^2}x{\sin ^4}3x + 3.4{\tan ^4}x{\sin ^3}3xcos\,3x\,)dx$$<br><br>$$ = {1 \over 2}\int\limits_{\pi /6}^{\pi /3} {{d \over {dx}}\left( {{{\tan }^4}x{{\sin }^4}3x} \right)} dx$$<br><br>$$ = {1 \over 2}\left[ {{{\tan }^4}x{{\sin }^4}3x} \right]_{\pi /6}^{\pi /3}$$<br><br>$$ = {1 \over 2}\left[ {9.(0) - {1 \over 3}.{1 \over 3}(1)} \right] = - {1 \over {18}}$$ | mcq | jee-main-2020-online-4th-september-evening-slot |
P3rIEyoJ27C0Cw6PThjgy2xukfaktyu2 | maths | definite-integration | properties-of-definite-integration | Let {x} and [x] denote the fractional part of x and <br/>the greatest integer $$ \le $$ x respectively of a real<br/> number x. If $$\int_0^n {\left\{ x \right\}dx} ,\int_0^n {\left[ x \right]dx} $$ and 10(n<sup>2</sup> – n), <br/>$$\left( {n \in N,n > 1} \right)$$ are three consecutive terms of a G.P., then n is equal to_____. | [] | null | 21 | $$\int\limits_0^n {\left\{ x \right\}} dx = n\int\limits_0^1 x dx = n\left( {{{{x^2}} \over 2}} \right)_0^1 = {n \over 2}$$
<br><br>[As period of {x} = 1]
<br><br>$$\int\limits_0^n {\left[ x \right]} dx = \int\limits_0^1 0 dx + \int\limits_1^2 1 dx + ... + \int\limits_{n - 1}^n {\left( {n - 1} \right)} dx$$
<br><br>= 1 + 2 + 3 + ....+ (n - 1)
<br><br>= $${{n\left( {n - 1} \right)} \over 2}$$
<br><br>As $${n \over 2}$$, $${{n\left( {n - 1} \right)} \over 2}$$, 10(n<sup>2</sup> – n) are in GP.
<br><br>$$ \therefore $$ $${\left[ {{{n\left( {n - 1} \right)} \over 2}} \right]^2} = {n \over 2} \times 10\left( {{n^2} - n} \right)$$
<br><br>$$ \Rightarrow $$ n<sup>2</sup>
= 21n
<br><br>$$ \Rightarrow $$ n = 21 | integer | jee-main-2020-online-4th-september-evening-slot |
QwXQqjMdV5LMiJiAdJ7k9k2k5hikkq5 | maths | definite-integration | properties-of-definite-integration | If $$I = \int\limits_1^2 {{{dx} \over {\sqrt {2{x^3} - 9{x^2} + 12x + 4} }}} $$, then : | [{"identifier": "A", "content": "$${1 \\over 16} < {I^2} < {1 \\over 9}$$"}, {"identifier": "B", "content": "$${1 \\over 8} < {I^2} < {1 \\over 4}$$"}, {"identifier": "C", "content": "$${1 \\over 9} < {I^2} < {1 \\over 8}$$"}, {"identifier": "D", "content": "$${1 \\over 6} < {I^2} < {1 \\over 2}$$"}] | ["C"] | null | Let f(x) = $${1 \over {\sqrt {2{x^3} - 9{x^2} + 12x + 4} }}$$
<br><br>f'(x) = $$ - {1 \over 2}{{ - 6{x^2} - 18x + 12} \over {2{{\left( {2{x^3} - 9{x^2} + 12x + 4} \right)}^{3/2}}}}$$
<br><br>= $${{ - 6\left( {x - 1} \right)\left( {x - 2} \right)} \over {2{{\left( {2{x^3} - 9{x^2} + 12x + 4} \right)}^{3/2}}}}$$
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264874/exam_images/hg1nq2ikleda4xwios7x.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 8th January Evening Slot Mathematics - Definite Integration Question 204 English Explanation">
<br><br>f<sub>max</sub> = f(1)
<br><br>f<sub>min</sub> = f(2)
<br><br>f(1) = $${1 \over {\sqrt {2 - 9 + 12 + 4} }}$$ = $${1 \over {\sqrt 9 }}$$ = $${1 \over 3}$$
<br><br>f(2) = $${1 \over {\sqrt {16 - 36 + 24 + 4} }}$$ = $${1 \over {\sqrt 8 }}$$
<br><br>$$ \therefore $$ $${1 \over 3} < {I} <$$ $${1 \over {\sqrt 8 }}$$
<br><br>So $${1 \over 9} < {I^2} < {1 \over 8}$$ | mcq | jee-main-2020-online-8th-january-evening-slot |
ojZlauFkWetuotMrqmjgy2xukfg7467a | maths | definite-integration | properties-of-definite-integration | The value of $$\int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{1 \over {1 + {e^{\sin x}}}}dx} $$ is: | [{"identifier": "A", "content": "$$\\pi $$"}, {"identifier": "B", "content": "$${{3\\pi \\over 2}}$$"}, {"identifier": "C", "content": "$${{\\pi \\over 2}}$$"}, {"identifier": "D", "content": "$${{\\pi \\over 4}}$$"}] | ["C"] | null | I = $$\int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{1 \over {1 + {e^{\sin x}}}}dx} $$ ....(1)
<br><br>Replacing x with $$\left( {{\pi \over 2} - {\pi \over 2} + x} \right)$$, we get
<br><br>I = $$\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{dx} \over {1 + {e^{ - \sin x}}}}} $$
<br><br>= $$\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{e^{\sin x}}dx} \over {1 + {e^{\sin x}}}}} $$ .....(2)
<br><br>Adding (1) and (2), we get
<br><br>2I = $$\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {1dx} $$ = $$\left( {{\pi \over 2} - \left( { - {\pi \over 2}} \right)} \right)$$ = $$\pi $$
<br><br>$$ \Rightarrow $$ I = $${{\pi \over 2}}$$ | mcq | jee-main-2020-online-5th-september-morning-slot |
FFP4KNctrxuI4U5aFqjgy2xukfw122qb | maths | definite-integration | properties-of-definite-integration | If I<sub>1</sub> = $$\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}} dx$$ and
<br/>I<sub>2</sub> = $$\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}} dx$$ such <br/>that I<sub>2</sub>
= $$\alpha $$I<sub>1</sub>
then $$\alpha $$
equals to :
| [{"identifier": "A", "content": "$${{5051} \\over {5050}}$$"}, {"identifier": "B", "content": "$${{5050} \\over {5051}}$$"}, {"identifier": "C", "content": "$${{5050} \\over {5049}}$$"}, {"identifier": "D", "content": "$${{5049} \\over {5050}}$$"}] | ["B"] | null | I<sub>2</sub> = $$\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}} dx$$
<br><br>I<sub>2</sub> = $$\int\limits_0^1 {\left( {1 - {x^{50}}} \right){{\left( {1 - {x^{50}}} \right)}^{100}}dx} $$
<br><br>I<sub>2</sub> = $$\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}dx} - \int\limits_0^1 {{x^{50}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx} $$
<br><br>I<sub>2</sub> = I<sub>1</sub> - $$\int\limits_0^1 {x.{x^{49}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx} $$
<br><br>Now apply IBP
<br><br>I<sub>2</sub> = I<sub>1</sub> - <br>$$\left[ {x\int\limits_0^1 {{x^{49}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx} - \int {{{d\left( x \right)} \over {dx}}.\int {{{d\left( x \right)} \over {dx}}\int\limits_0^1 {{x^{49}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx} } } } \right]$$
<br><br>Let (1 – x<sup>50</sup>) = t
<br><br>$$ \Rightarrow $$ -50x<sup>49</sup>dx = dt
<br><br>I<sub>2</sub> = I<sub>1</sub> - <br>$$\left[ {x.\left( { - {1 \over {50}}} \right){{{{\left( {1 - {x^{50}}} \right)}^{101}}} \over {101}}} \right]_0^1$$ - $$ - \int\limits_0^1 {\left( { - {1 \over {50}}} \right){{{{\left( {1 - {x^{50}}} \right)}^{101}}} \over {101}}dx} $$
<br><br>= I<sub>1</sub> - 0 - $${ - {1 \over {50}}.{1 \over {101}}{I_2}}$$
<br><br>I<sub>2</sub> = I<sub>1</sub> - $${1 \over {5050}}{I_2}$$
<br><br>$$ \Rightarrow $$ I<sub>2</sub> + $${1 \over {5050}}{I_2}$$ = I<sub>1</sub>
<br><br>$$ \Rightarrow $$ $${{5051} \over {5050}}{I_2}$$ = I<sub>1</sub>
<br><br>$$ \Rightarrow $$ I<sub>2</sub> = $${{5050} \over {5051}}$$I<sub>1</sub>
<br><br>Given I<sub>2</sub>
= $$\alpha $$I<sub>1</sub>
<br><br>$$ \therefore $$ $$\alpha $$ = $${{5050} \over {5051}}$$ | mcq | jee-main-2020-online-6th-september-morning-slot |
Dh8HpGOX6THu4Upn4Gjgy2xukg38tscz | maths | definite-integration | properties-of-definite-integration | The integral $$\int\limits_1^2 {{e^x}.{x^x}\left( {2 + {{\log }_e}x} \right)} dx$$ equals : | [{"identifier": "A", "content": "e(4e + 1)\n"}, {"identifier": "B", "content": "e(2e \u2013 1)"}, {"identifier": "C", "content": "e(4e \u2013 1)"}, {"identifier": "D", "content": "4e<sup>2</sup> \u2013 1"}] | ["C"] | null | $$\int\limits_1^2 {{e^x}.{x^x}\left( {2 + {{\log }_e}x} \right)} dx$$
<br><br>= $$\int\limits_1^2 {{e^x}{x^x}\left[ {1 + \left( {1 + {{\log }_e}x} \right)} \right]} dx$$
<br><br>= $$\int\limits_1^2 {{e^x}\left[ {{x^x} + {x^x}\left( {1 + {{\log }_e}x} \right)} \right]} dx$$
<br><br>= $$\left[ {{e^x}{x^x}} \right]_1^2$$
<br><br>= e<sup>2</sup> $$ \times $$ 4 - e $$ \times $$ 1
<br><br>= 4e<sup>2</sup> - e
<br><br>= e(4e - 1)
<br><br><b>Note :</b> $$\int {{e^x}\left( {f\left( x \right) + f'\left( x \right)} \right)dx} $$ = e<sup>x</sup>f(x) + c | mcq | jee-main-2020-online-6th-september-evening-slot |
r9KdaZAY4B8jIk7oltjgy2xukezfrqar | maths | definite-integration | properties-of-definite-integration | Let [t] denote the greatest integer less than or
equal to t. <br/>Then the value of $$\int\limits_1^2 {\left| {2x - \left[ {3x} \right]} \right|dx} $$ is ______. | [] | null | 1 | $$\int\limits_1^2 {\left| {2x - \left[ {3x} \right]} \right|dx} $$ <br><br>
$$ = \int\limits_1^2 {\left| {2x - \left( {3x - \left\{ {3x} \right\}} \right)} \right|dx} $$<br><br>
$$ = \int\limits_1^2 {\left| {\left\{ {3x} \right\} - x} \right|dx} $$<br><br>
We know, 0 $$ \le $$ $$\left\{ {3x} \right\} < 1$$ and x > 1<br><br>
$$\therefore \left\{ {3x} \right\} - x < 0$$<br><br>
So, $$\left| {\left\{ {3x} \right\} - 2} \right| = - \left[ {\left\{ {3x} \right\} - x} \right]$$<br><br>
$$ = \int\limits_1^2 {\left( {x - \left\{ {3x} \right\}} \right)dx} $$<br><br>
$$ = \left[ {{{{x^2}} \over 2}} \right]_1^2 - \int\limits_{3 \times {1 \over 3}}^{6 \times {1 \over 3}} {\left\{ {3x} \right\}dx} $$ [Period of {x} = 1, so period of {3x} = $${1 \over 3}$$]<br><br>
$$ = \left( {2 - {1 \over 2}} \right) - \left( {6 - 3} \right)\int\limits_0^{{1 \over 3}} {3xdx} $$
<br><br>
$$ = {3 \over 2} - 3 \times 3\left[ {{{{x^2}} \over 2}} \right]_0^{{1 \over 3}}$$<br><br>
$$ = {3 \over 2} - {9 \over 2}\left[ {{1 \over 9} - 0} \right]$$<br><br>
$$ = {3 \over 2} - {1 \over 2}$$ = 1 | integer | jee-main-2020-online-2nd-september-evening-slot |
9sxJlaYOooElkHjpOCjgy2xukf0q9pa8 | maths | definite-integration | properties-of-definite-integration | $$\int\limits_{ - \pi }^\pi {\left| {\pi - \left| x \right|} \right|dx} $$ is equal to :
| [{"identifier": "A", "content": "$${\\pi ^2}$$"}, {"identifier": "B", "content": "2$${\\pi ^2}$$"}, {"identifier": "C", "content": "$$\\sqrt 2 {\\pi ^2}$$"}, {"identifier": "D", "content": "$${{{\\pi ^2}} \\over 2}$$"}] | ["A"] | null | $$\int\limits_{ - \pi }^\pi {\left| {\pi - \left| x \right|} \right|dx} $$
<br><br>= $$2\int\limits_0^\pi {\left| {\pi - \left| x \right|} \right|} dx$$ [As it is even function]
<br><br>= $$2\int\limits_0^\pi {\left( {\pi - x} \right)} dx$$
<br><br>= $$2\left[ {\pi x - {{{x^2}} \over 2}} \right]_0^\pi $$
<br><br>= $${\pi ^2}$$ | mcq | jee-main-2020-online-3rd-september-morning-slot |
MKf59Sb8H9TokJ3iYJ7k9k2k5e3b18i | maths | definite-integration | properties-of-definite-integration | If ƒ(a + b + 1 - x) = ƒ(x), for all x, where a and b are fixed positive real numbers, then<br/><br/>
$${1 \over {a + b}}\int_a^b {x\left( {f(x) + f(x + 1)} \right)} dx$$ is equal to:
| [{"identifier": "A", "content": "$$\\int_{a - 1}^{b - 1} {f(x+1)dx} $$"}, {"identifier": "B", "content": "$$\\int_{a + 1}^{b + 1} {f(x + 1)dx} $$"}, {"identifier": "C", "content": "$$\\int_{a - 1}^{b - 1} {f(x)dx} $$"}, {"identifier": "D", "content": "$$\\int_{a + 1}^{b + 1} {f(x)dx} $$"}] | ["A"] | null | I = $${1 \over {a + b}}\int_a^b {x\left( {f(x) + f(x + 1)} \right)} dx$$ ...(1)
<br><br>x $$ \to $$ a + b - x
<br><br>I = $${1 \over {a + b}}\int\limits_a^b {\left( {a + b - x} \right)\left( {f\left( {a + b - x} \right) + f\left( {a + b - x + 1} \right)} \right)dx} $$
<br><br>I = $${1 \over {a + b}}\int\limits_a^b {\left( {a + b - x} \right)\left( {f\left( {x + 1} \right) + f\left( x \right)} \right)dx} $$ .....(2)
<br><br>[As ƒ(x) = ƒ(a + b + 1 - x)
<br><br>$$ \Rightarrow $$ ƒ(x + 1) = ƒ(a + b - x)]
<br><br>Adding (1) and (2) we get
<br><br>2I = $${{a + b} \over {a + b}}\int\limits_a^b {\left( {f\left( {x + 1} \right) + f\left( x \right)} \right)dx} $$
<br><br>$$ \Rightarrow $$ I = $${1 \over 2}\int\limits_a^b {f\left( x \right)dx} + {1 \over 2}\int\limits_a^b {f\left( {x + 1} \right)dx} $$
<br><br>$$ \Rightarrow $$ I = $${1 \over 2}\int\limits_a^b {f\left( x \right)dx} + {1 \over 2}\int\limits_a^b {f\left( {a + b - x + 1} \right)dx} $$
<br><br>$$ \Rightarrow $$ I = $${1 \over 2}\int\limits_a^b {f\left( x \right)dx} + {1 \over 2}\int\limits_a^b {f\left( x \right)dx} $$
<br><br>$$ \Rightarrow $$ I = $$\int\limits_a^b {f\left( x \right)dx} $$
<br><br>Let x = t + 1
<br><br>$$ \therefore $$ I = $$\int\limits_{a - 1}^{b - 1} {f\left( {t + 1} \right)dt} $$
<br><br>= $$\int\limits_{a - 1}^{b - 1} {f\left( {x + 1} \right)dx} $$ | mcq | jee-main-2020-online-7th-january-morning-slot |
KYxZdnmXcvF8YGpLge7k9k2k5fia9dd | maths | definite-integration | properties-of-definite-integration | The value of $$\alpha $$ for which
<br/>$$4\alpha \int\limits_{ - 1}^2 {{e^{ - \alpha \left| x \right|}}dx} = 5$$, is: | [{"identifier": "A", "content": "$${\\log _e}2$$"}, {"identifier": "B", "content": "$${\\log _e}\\sqrt 2 $$"}, {"identifier": "C", "content": "$${\\log _e}\\left( {{4 \\over 3}} \\right)$$"}, {"identifier": "D", "content": "$${\\log _e}\\left( {{3 \\over 2}} \\right)$$"}] | ["A"] | null | $$4\alpha \int\limits_{ - 1}^2 {{e^{ - \alpha \left| x \right|}}dx} = 5$$
<br><br>$$ \Rightarrow $$ $$4\alpha \int\limits_{ - 1}^0 {{e^{\alpha x}}dx} + 4\alpha \int\limits_0^2 {{e^{ - \alpha x}}dx} $$ = 5
<br><br>$$ \Rightarrow $$ $$4\alpha \left[ {{{{e^{\alpha x}}} \over \alpha }} \right]_{ - 1}^0 + 4\alpha \left[ {{{{e^{ - \alpha x}}} \over { - \alpha }}} \right]_0^2$$ = 5
<br><br>$$ \Rightarrow $$ $$4{e^{ - 2\alpha }} + 4{e^{ - \alpha }} - 3$$ = 0
<br><br>Let e<sup>-$$\alpha $$</sup> = t
<br><br>$$ \therefore $$ 4t<sup>2</sup> + 4t - 3 = 0
<br><br>$$ \Rightarrow $$ t = $${1 \over 2}$$
<br><br>$$ \therefore $$ e<sup>-$$\alpha $$</sup> = $${1 \over 2}$$
<br><br>$$ \Rightarrow $$ $$\alpha $$ = $${\log _e}2$$ | mcq | jee-main-2020-online-7th-january-evening-slot |
jCA7Iy7zecSUIYnIOg7k9k2k5fnvr31 | maths | definite-integration | properties-of-definite-integration | If $$\theta $$<sub>1</sub>
and $$\theta $$<sub>2</sub>
be respectively the smallest and the largest values of $$\theta $$ in (0, 2$$\pi $$) - {$$\pi $$} which satisfy
the equation,
<br/>2cot<sup>2</sup>$$\theta $$ - $${5 \over {\sin \theta }}$$ + 4 = 0, then
<br/>$$\int\limits_{{\theta _1}}^{{\theta _2}} {{{\cos }^2}3\theta d\theta } $$ is equal to : | [{"identifier": "A", "content": "$${\\pi \\over 9}$$"}, {"identifier": "B", "content": "$${{2\\pi } \\over 3}$$"}, {"identifier": "C", "content": "$${{\\pi } \\over 3}$$"}, {"identifier": "D", "content": "$${\\pi \\over 3} + {1 \\over 6}$$"}] | ["C"] | null | 2cot<sup>2</sup>$$\theta $$ - $${5 \over {\sin \theta }}$$ + 4 = 0
<br><br>$$ \Rightarrow $$ $$2{{{{\cos }^2}\theta } \over {{{\sin }^2}\theta }} - {5 \over {\sin \theta }} + 4$$ = 0
<br><br>$$ \Rightarrow $$ 2sin<sup>2</sup>
$$\theta $$ – 5sin$$\theta $$ + 2 = 0
<br><br>$$ \Rightarrow $$ (2sin$$\theta $$ – 1)(sin$$\theta $$ – 2) = 0
<br><br>$$ \therefore $$ sin$$\theta $$ = $${1 \over 2}$$ [ sin$$\theta $$ = 2 not possible]
<br><br>$$ \therefore $$ $$\theta $$<sub>1</sub> = $${\pi \over 6}$$ and $$\theta $$<sub>2</sub> = $${{5\pi } \over 6}$$ as $$\theta $$<sub>1</sub> $$<$$ $$\theta $$<sub>2</sub>
<br><br>$$ \therefore $$ I = $$\int\limits_{{\pi \over 6}}^{{{5\pi } \over 6}} {{{\cos }^2}3\theta d\theta } $$
<br><br>= $$\int\limits_{{\pi \over 6}}^{{{5\pi } \over 6}} {{{1 + \cos 6\theta } \over 2}d\theta } $$
<br><br>= $${1 \over 2}\left[ {\theta + {{\sin 6\theta } \over 2}} \right]_{{\pi \over 6}}^{{{5\pi } \over 6}}$$
<br><br>= $${{\pi \over 3}}$$ | mcq | jee-main-2020-online-7th-january-evening-slot |
AHLBLmZMvlqRanb5By7k9k2k5itvbgi | maths | definite-integration | properties-of-definite-integration | If for all real triplets (a, b, c), ƒ(x) = a + bx + cx<sup>2</sup>;
then $$\int\limits_0^1 {f(x)dx} $$ is equal to : | [{"identifier": "A", "content": "$${1 \\over 6}\\left\\{ {f(0) + f(1) + 4f\\left( {{1 \\over 2}} \\right)} \\right\\}$$"}, {"identifier": "B", "content": "$$2\\left\\{ 3{f(1) + 2f\\left( {{1 \\over 2}} \\right)} \\right\\}$$"}, {"identifier": "C", "content": "$${1 \\over 3}\\left\\{ {f(0) + f\\left( {{1 \\over 2}} \\right)} \\right\\}$$"}, {"identifier": "D", "content": "$${1 \\over 2}\\left\\{ {f(1) + 3f\\left( {{1 \\over 2}} \\right)} \\right\\}$$"}] | ["A"] | null | ƒ(x) = a + bx + cx<sup>2</sup>
<br><br>$$\int\limits_0^1 {f\left( x \right)dx} $$ = $$\left[ {ax + {{b{x^2}} \over 2} + {{c{x^3}} \over 3}} \right]_0^1$$
<br><br>= $${a + {b \over 2} + {c \over 3}}$$
<br><br>= $${1 \over 6}\left[ {6a + 3b + c} \right]$$
<br><br>f(1) = a + b + c
<br><br>f(0) = a
<br><br>$$f\left( {{1 \over 2}} \right) = a + {b \over 2} + {c \over 4}$$
<br><br>By checking each option you have to find the solution.
<br><br>$${1 \over 6}\left\{ {f(0) + f(1) + 4f\left( {{1 \over 2}} \right)} \right\}$$
<br><br>= $${1 \over 6}\left[ {a + a + b + c + 4\left( {a + {b \over 2} + {c \over 4}} \right)} \right]$$
<br><br>= $${1 \over 6}\left[ {6a + 3b + c} \right]$$
<br><br>$$ \therefore $$ Option (A) is correct option. | mcq | jee-main-2020-online-9th-january-morning-slot |
xOohIAUZiGPdaMQWK1jgy2xukewsvkoq | maths | definite-integration | properties-of-definite-integration | The integral $$\int\limits_0^2 {\left| {\left| {x - 1} \right| - x} \right|dx} $$<br/> is equal to______. | [] | null | 1.50 | $$\int\limits_0^2 {\left| {\left| {x - 1} \right| - x} \right|dx} $$
<br><br>= $$\int\limits_0^1 {\left| { - \left( {x - 1} \right) - x} \right|} dx$$ + $$ + \int\limits_1^2 {\left| {\left( {x - 1} \right) - x} \right|} dx$$
<br><br>= $$\int\limits_0^1 {\left| {1 - x - x} \right|} dx + \int\limits_1^2 {dx} $$
<br><br>= $$\int\limits_0^1 {\left| {1 - 2x} \right|} dx + \int\limits_1^2 {dx} $$
<br><br>= $$\int\limits_0^{{1 \over 2}} {\left( {1 - 2x} \right)} dx + \int\limits_{{1 \over 2}}^1 {\left( {2x - 1} \right)dx} + \int\limits_1^2 {dx} $$
<br><br>= $$\left[ {x - {x^2}} \right]_0^{{1 \over 2}} + \left[ {x - {x^2}} \right]_{{1 \over 2}}^1 + \left( {2 - 1} \right)$$
<br><br>= $$\left( {{1 \over 2} - {1 \over 4}} \right) + \left[ {\left( {1 - 1} \right) - \left( {{1 \over 4} - {1 \over 2}} \right)} \right] + 1$$
<br><br>= $${3 \over 2}$$ = 1.5 | integer | jee-main-2020-online-2nd-september-morning-slot |
RKAHShP0JRx0cMT8Pe7k9k2k5ipaxrd | maths | definite-integration | properties-of-definite-integration | The value of
<br/>$$\int\limits_0^{2\pi } {{{x{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$$ is equal to : | [{"identifier": "A", "content": "4$$\\pi $$"}, {"identifier": "B", "content": "2$$\\pi $$"}, {"identifier": "C", "content": "$$\\pi $$<sup>2</sup>"}, {"identifier": "D", "content": "2$$\\pi $$<sup>2</sup>"}] | ["C"] | null | I = $$\int\limits_0^{2\pi } {{{x{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$$ .....(1)
<br><br>I = $$\int\limits_0^{2\pi } {{{\left( {2\pi - x} \right){{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$$ ......(2)
<br><br>Adding (1) and (2)
<br><br>2I = $$\int\limits_0^{2\pi } {{{2\pi {{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$$
<br><br>$$ \Rightarrow $$ I = $$2\pi \int\limits_0^\pi {{{{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$$
<br><br>$$ \Rightarrow $$ I = $$2\pi \left[ {\int\limits_0^{\pi /2} {{{{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx + \int\limits_0^{\pi /2} {{{{{\cos }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx} \right]$$
<br><br>= $${2\pi \int\limits_0^{\pi /2} 1 dx}$$
<br><br>= $$2\pi .{\pi \over 2}$$ = $${\pi ^2}$$ | mcq | jee-main-2020-online-9th-january-morning-slot |
IfPvLnsNDscWEEcX6S1klrjrrft | maths | definite-integration | properties-of-definite-integration | If $$\int\limits_{ - a}^a {\left( {\left| x \right| + \left| {x - 2} \right|} \right)} dx = 22$$, (a > 2) and [x] denotes the greatest integer $$ \le $$ x, then$$\int\limits_{ - a}^a {\left( {x + \left[ x \right]} \right)} dx$$ is equal to _________. | [] | null | 3 | $$\int\limits_{ - a}^a {\left( {\left| x \right| + \left| {x - 2} \right|} \right)} dx = 22$$
<br><br>$$ \Rightarrow $$ $$\int\limits_{ - a}^0 {( - 2x + 2)dx} + \int\limits_0^2 {(x + 2 - x)dx} + \int\limits_2^a {(2x - 2)dx} = 22$$<br><br>$$ \Rightarrow $$ $${x^2} - 2x|_0^{ - a} + 2x|_0^2 + {x^2} - 2x|_2^a = 22$$<br><br>$$ \Rightarrow $$ $${a^2} + 2a + 4 + {a^2} - 2a - (4 - 4) = 22$$<br><br>$$ \Rightarrow $$ $$2{a^2} = 18 \Rightarrow a = 3$$<br><br>$$\int\limits_3^{ - 3} {(x + [x])dx} = - \left( {\int\limits_{ - 3}^3 {(x + [x])dx} } \right) = - \left( {\int\limits_{ - 3}^3 {[x]dx} } \right)$$
<br><br>= -($$\int\limits_{ - 3}^{ - 2} {\left[ x \right]dx} + \int\limits_{ - 2}^{ - 1} {\left[ x \right]dx} + \int\limits_{ - 1}^0 {\left[ x \right]dx} $$
<br><br>+ $$\int\limits_0^1 {\left[ x \right]dx} + \int\limits_1^2 {\left[ x \right]dx} + \int\limits_2^3 {\left[ x \right]dx} $$)
<br><br>$$ = - ( - 3 - 2 - 1 + 0 + 1 + 2) = 3$$ | integer | jee-main-2021-online-24th-february-morning-slot |
fLZLj1ECpiOuil3cac1klrkgs1n | maths | definite-integration | properties-of-definite-integration | The value of the integral, $$\int\limits_1^3 {[{x^2} - 2x - 2]dx} $$, where [x] denotes the greatest integer less than or equal to x, is : | [{"identifier": "A", "content": "$$-$$ 5"}, {"identifier": "B", "content": "$$ - \\sqrt 2 - \\sqrt 3 + 1$$"}, {"identifier": "C", "content": "$$-$$ 4"}, {"identifier": "D", "content": "$$ - \\sqrt 2 - \\sqrt 3 - 1$$"}] | ["D"] | null | $$I = \int\limits_1^3 { - 3dx + \int\limits_1^3 {\left[ {{{(x - 1)}^2}} \right]dx} } $$<br><br>Put x $$-$$ 1 = t ; dx = dt<br><br>$$I = ( - 6) + \int\limits_0^2 {\left[ {{t^2}} \right]} dt$$<br><br>$$I = - 6 + \int\limits_0^1 {0dt} + \int\limits_1^{\sqrt 2 } {1dt} + \int\limits_{\sqrt 2 }^{\sqrt 3 } {2dt} + \int\limits_{\sqrt 3 }^2 {3dt} $$<br><br>$$I = - 6 + \left( {\sqrt 2 - 1} \right) + 2\sqrt 3 - 2\sqrt 2 + 6 - 3\sqrt 3 $$<br><br>$$I = - 1 - \sqrt 2 - \sqrt 3 $$ | mcq | jee-main-2021-online-24th-february-evening-slot |
oryA5enuTQBkUls5XD1klrkw9n0 | maths | definite-integration | properties-of-definite-integration | Let f(x) be a differentiable function defined on [0, 2] such that f'(x) = f'(2 $$-$$ x) for all x$$ \in $$ (0, 2), f(0) = 1 and f(2) = e<sup>2</sup>. Then the value of $$\int\limits_0^2 {f(x)} dx$$ is : | [{"identifier": "A", "content": "1 + e<sup>2</sup>"}, {"identifier": "B", "content": "2(1 + e<sup>2</sup>)"}, {"identifier": "C", "content": "1 $$-$$ e<sup>2</sup>"}, {"identifier": "D", "content": "2(1 $$-$$ e<sup>2</sup>)"}] | ["A"] | null | f'(x) = f'(2 $$-$$ x)<br><br>On integrating both side f(x) = $$-$$f(2 $$-$$ x) + c<br><br>put x = 0<br><br>f(0) + f(2) = c $$ \Rightarrow $$ c = 1 + e<sup>2</sup><br><br>$$ \Rightarrow $$ f(x) + f(2 $$-$$ x) = 1 + e<sup>2</sup> ..... (i)<br><br>$$I = \int\limits_0^2 {f(x)dx} = \int\limits_0^1 {\{ f(x) + f(2 - x)\} dx = (1 + {e^2})} $$ | mcq | jee-main-2021-online-24th-february-evening-slot |
sP3TYDhkd2SEqbv1Om1klrli04i | maths | definite-integration | properties-of-definite-integration | Let f be a twice differentiable function defined on R such that f(0) = 1, f'(0) = 2 and f'(x) $$ \ne $$ 0 for all x $$ \in $$ R. If $$\left| {\matrix{
{f(x)} & {f'(x)} \cr
{f'(x)} & {f''(x)} \cr
} } \right|$$ = 0, for all x$$ \in $$R, then the value of f(1) lies in the interval : | [{"identifier": "A", "content": "(0, 3)"}, {"identifier": "B", "content": "(9, 12)"}, {"identifier": "C", "content": "(3, 6)"}, {"identifier": "D", "content": "(6, 9)"}] | ["D"] | null | $$\left| {\matrix{
{f(x)} & {f'(x)} \cr
{f'(x)} & {f''(x)} \cr
} } \right| = 0$$<br><br>$$ \Rightarrow f(x).f''(x) - {\left( {f'(x)} \right)^2} = 0$$<br><br>Dividing by $${\left( {f(x)} \right)^2}$$, we get<br><br>$$ \Rightarrow {{f(x).f''(x) - {{\left( {f'(x)} \right)}^2}} \over {{{\left( {f(x)} \right)}^2}}} = 0$$<br><br>$$ \Rightarrow {d \over {dx}}\left( {{{f'(x)} \over {f(x)}}} \right) = 0$$<br><br>Integrating both side, <br><br>$${{f'(x)} \over {f(x)}} = c$$ (constant)<br><br>At, $$x = 0$$, $${{f'(0)} \over {f(0)}} = c$$<br><br>$$ \Rightarrow {2 \over 1} = c$$<br><br>$$ \Rightarrow c = 2$$<br><br>$$ \therefore $$ $${{f'(x)} \over {f(x)}} = 2$$<br><br>$$ \Rightarrow \int {{{f'(x)} \over {f(x)}}} dx = 2\int {dx} $$<br><br>$$ \Rightarrow \ln |f(x)|\, = 2x + c'$$<br><br>at x = 0,<br><br>$$ln|f(0)|\, = 0 + c'$$<br><br>$$ \Rightarrow 0 = 0 + c'$$<br><br>$$ \Rightarrow c' = 0$$<br><br>$$ \therefore $$ $$n|f(x)| = 2x$$<br><br>$$ \Rightarrow f(x) = {e^{2x}}$$<br><br>$$f(1) = {e^2} = {(2.71)^2} = 7.34$$<br><br>So it lie between (6, 9). | mcq | jee-main-2021-online-24th-february-evening-slot |
6mNbfsr4rNmQELaPU61kls409xr | maths | definite-integration | properties-of-definite-integration | The value of $$\int\limits_{ - 1}^1 {{x^2}{e^{[{x^3}]}}} dx$$, where [ t ] denotes the greatest integer $$ \le $$ t, is : | [{"identifier": "A", "content": "$${{e + 1} \\over 3}$$"}, {"identifier": "B", "content": "$${{e - 1} \\over {3e}}$$"}, {"identifier": "C", "content": "$${1 \\over {3e}}$$"}, {"identifier": "D", "content": "$${{e + 1} \\over {3e}}$$"}] | ["D"] | null | $$I = \int\limits_{ - 1}^1 {{x^2}{e^{[{x^3}]}}dx} $$
<br><br>Here -1 $$ \le $$ x $$ \le $$ 1 then -1 $$ \le $$ x<sup>3</sup> $$ \le $$ 1
<br><br>Integer between -1 and 1 is 0. So integration will be divided into two parts, -1 to 0 and 0 to 1.
<br><br>$$ = \int\limits_{ - 1}^0 {{x^2}{e^{[{x^3}]}}dx} + \int\limits_0^1 {{x^2}{e^{[{x^3}]}}dx} $$<br><br>$$ = \int\limits_{ - 1}^0 {{x^2}{e^{ - 1}}dx} + \int\limits_0^1 {{x^2}{e^0}dx} $$<br><br>= $${1 \over e} \times \left[ {{{{x^3}} \over 3}} \right]_{ - 1}^0 + \left[ {{{{x^3}} \over 3}} \right]_0^1$$<br><br>$$ = {1 \over e} \times \left( {0 - \left( {{{ - 1} \over 3}} \right)} \right) + {1 \over 3}$$<br><br>$$ = {1 \over {3e}} + {1 \over 3} = {{1 + e} \over {3e}}$$ | mcq | jee-main-2021-online-25th-february-morning-slot |
cIyKJCgctnZpVSvWHD1klt84ywd | maths | definite-integration | properties-of-definite-integration | If $${I_n} = \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{{\cot }^n}x\,dx} $$, then : | [{"identifier": "A", "content": "$${1 \\over {{I_2} + {I_4}}},{1 \\over {{I_3} + {I_5}}},{1 \\over {{I_4} + {I_6}}}$$ are in A.P."}, {"identifier": "B", "content": "I<sub>2</sub> + I<sub>4</sub>, I<sub>3</sub> + I<sub>5</sub>, I<sub>4</sub> + I<sub>6</sub> are in A.P."}, {"identifier": "C", "content": "$${1 \\over {{I_2} + {I_4}}},{1 \\over {{I_3} + {I_5}}},{1 \\over {{I_4} + {I_6}}}$$ are in G.P."}, {"identifier": "D", "content": "I<sub>2</sub> + I<sub>4</sub>, (I<sub>3</sub> + I<sub>5</sub>)<sup>2</sup>, I<sub>4</sub> + I<sub>6</sub> are in G.P."}] | ["A"] | null | $${I_n} = \int\limits_{\pi /4}^{\pi /2} {{{\cot }^n}xdx} = \int\limits_{\pi /4}^{\pi /2} {{{\cot }^{n - 2}}x(\cos e{c^2}x - 1)dx} $$
<br><br>= $$\int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{{\cot }^{n - 2}}x.co{{\sec }^2}} xdx - \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{{\cot }^{n - 2}}x} dx$$
<br><br>$$ = \left. {{{{{\cot }^{n - 1}}x} \over {n - 1}}} \right]_{\pi /4}^{\pi /2} - {I_{n - 2}}$$<br><br>$$ = {1 \over {n - 1}} - {I_{n - 2}}$$<br><br>$$ \Rightarrow {I_n} + {I_{n - 2}} = {1 \over {n - 1}}$$<br><br>$$ \Rightarrow {I_2} + {I_4} = {1 \over 3}$$<br><br>$${I_3} + {I_5} = {1 \over 4}$$<br><br>$${I_4} + {I_6} = {1 \over 5}$$<br><br>$$ \therefore $$ $${1 \over {{I_2} + {I_4}}},{1 \over {{I_3} + {I_5}}},{1 \over {{I_4} + {I_6}}}$$ are in A.P. | mcq | jee-main-2021-online-25th-february-evening-slot |
F2KjHonbVsNXO5wm871klt9qgz7 | maths | definite-integration | properties-of-definite-integration | The value of $$\int\limits_{ - 2}^2 {|3{x^2} - 3x - 6|dx} $$ is ___________. | [] | null | 19 | x<sup>2</sup> – x – 2 = (x - 2)(x + 1)
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263548/exam_images/bgqekrzfra42hxdnfxsj.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th February Evening Shift Mathematics - Definite Integration Question 179 English Explanation">
<br><br>$$\int\limits_{ - 2}^2 {3|{x^2} - x - 2|dx} $$<br><br>$$ = 3\int\limits_{ - 2}^2 {|{x^2} - x - 2|dx} $$<br><br>$$ = 3\left[ {\int\limits_{ - 2}^{ - 1} {\left( {{x^2} - x - 2} \right)dx} + \int\limits_{ - 1}^2 { - \left( {{x^2} - x - 2} \right)dx} } \right]$$<br><br>$$ = 3\left[ {\left. {\left( {{{{x^3}} \over 3} - {{{x^2}} \over 2} - 2x} \right)} \right|_{ - 2}^{ - 1} - \left( {{{{x^3}} \over 3} - {{{x^2}} \over 2} - 2x} \right)_{ - 1}^2} \right]$$<br><br>$$ = 3\left[ {7 - {2 \over 3}} \right]$$<br><br>= 19 | integer | jee-main-2021-online-25th-february-evening-slot |
NNxLSevDSb1pc2kNN81klugd103 | maths | definite-integration | properties-of-definite-integration | The value of $$\int\limits_{ - \pi /2}^{\pi /2} {{{{{\cos }^2}x} \over {1 + {3^x}}}} dx$$ is : | [{"identifier": "A", "content": "$$2\\pi $$"}, {"identifier": "B", "content": "$${\\pi \\over 2}$$"}, {"identifier": "C", "content": "$$4\\pi $$"}, {"identifier": "D", "content": "$${\\pi \\over 4}$$"}] | ["D"] | null | Let $$I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x} \over {1 + {3^x}}}} dx$$ .... (1)<br><br>Replace x with $$-$$x,<br><br>$$ \therefore $$ $$I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x} \over {1 + {1 \over {{3^x}}}}}} $$ .... (2)<br><br>Adding (1) and (2), we get,<br><br>$$2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x + {3^x}{{\cos }^2}x} \over {1 + {3^x}}}} dx$$<br><br>$$ = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x(1 + {3^x})} \over {1 + {3^x}}}dx} $$<br><br>$$ = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\cos }^2}x} \,dx$$<br><br>[$${{{\cos }^2}x}$$ is a even function as $$f(x) = f( - x)$$ for $${\cos ^2}x $$]
<br><br>$$= 2\int\limits_0^{{\pi \over 2}} {{{\cos }^2}x} \,dx$$<br><br>$$ = 2\int\limits_0^{{\pi \over 2}} {\left( {{{1 + \cos 2x} \over 2}} \right)} \,dx$$<br><br>$$ \Rightarrow I = {1 \over 2}\int\limits_0^{{\pi \over 2}} {(1 + \cos 2x)} \,dx$$<br><br>$$ = {1 \over 2}\left[ {x + \sin 2x} \right]_0^{{\pi \over 2}}$$<br><br>$$ = {1 \over 2}\left[ {{\pi \over 2} - 0} \right]$$<br><br>$$ = {\pi \over 4}$$ | mcq | jee-main-2021-online-26th-february-morning-slot |
xeclbysY3loXLaNaYZ1klugk4ig | maths | definite-integration | properties-of-definite-integration | The value of $$\sum\limits_{n = 1}^{100} {\int\limits_{n - 1}^n {{e^{x - [x]}}dx} } $$, where [ x ] is the greatest integer $$ \le $$ x, is : | [{"identifier": "A", "content": "100e"}, {"identifier": "B", "content": "100(e $$-$$ 1)"}, {"identifier": "C", "content": "100(1 + e)"}, {"identifier": "D", "content": "100(1 $$-$$ e)"}] | ["B"] | null | $$\sum\limits_{n = 1}^{100} {\int\limits_{n - 1}^n {{e^{x - [x]}}} dx} $$<br><br>Here, $$n - 1 \le x < n$$<br><br>$$ \therefore $$ $$[x] = n - 1$$<br><br>$$ \therefore $$ $$\int\limits_{n - 1}^n {{e^{x - (n - 1)}}} dx$$<br><br>$$ = \left[ {{e^{x - (n - 1)}}} \right]_{n - 1}^n$$<br><br>$$ = {e^1} - {e^0}$$<br><br>$$ = e - 1$$<br><br>Now, $$\sum\limits_{n = 1}^{100} {(e - 1) = 100(e - 1)} $$ | mcq | jee-main-2021-online-26th-february-morning-slot |
1gtnINHDALrDtWp8iv1kluhl7sf | maths | definite-integration | properties-of-definite-integration | The value of the integral $$\int\limits_0^\pi {|{{\sin }\,}2x|dx} $$ is ___________. | [] | null | 2 | $\begin{aligned} & \text { Let } I=\int_0^\pi|\sin 2 x| d x
\\\\ & =2 \int_0^{\pi / 2}|\sin 2 x| d x \quad[\because \sin 2 x \text { is periodic function }]
\\\\ & =2 \int_0^{\pi / 2} \sin 2 x \,d x[\sin 2 x \text { is positive in range }(0, \pi / 2)]
\\\\ & =2\left[\frac{-\cos 2 x}{2}\right]_0^{\pi / 2} \\\\ & =-[\cos \pi-\cos 0]=-(-1-1)=2 \end{aligned}$ | integer | jee-main-2021-online-26th-february-morning-slot |
VUYHG22OPDyh47l3sQ1kluwtm7l | maths | definite-integration | properties-of-definite-integration | For x > 0, if $$f(x) = \int\limits_1^x {{{{{\log }_e}t} \over {(1 + t)}}dt} $$, then $$f(e) + f\left( {{1 \over e}} \right)$$ is equal to : | [{"identifier": "A", "content": "$${1 \\over 2}$$"}, {"identifier": "B", "content": "$$-$$1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "1"}] | ["A"] | null | $$f(x) = \int_1^x {{{\ln t} \over {1 + t}}dt} $$<br><br>then $$f\left( {{1 \over x}} \right) = \int_1^{1/x} {{{\ln t} \over {1 + t}}dt} $$<br><br>Let $$t = {1 \over u} \Rightarrow dt = - {1 \over {{u^2}}}du$$<br><br>$$ \Rightarrow f\left( {{1 \over x}} \right) = \int_1^x {{{\ln {1 \over u}} \over {1 + {1 \over u}}}\left( { - {1 \over {{u^2}}}} \right)dx} $$<br><br>$$f\left( {{1 \over x}} \right) = \int_1^x {{{\ln u} \over {u(1 + u)}}} du = \int_1^x {{{\ln t} \over {t(1 + t)}}dt} $$<br><br>$$ \therefore $$ $$f(x) + f\left( {{1 \over x}} \right) = \int_1^x {\ln t\left( {{1 \over {1 + t}} + {1 \over {t(1 + t)}}} \right)} dt$$<br><br>$$ = \int_1^x {\ln t\left( {{1 \over {1 + t}} + {1 \over t} - {1 \over {t + 1}}} \right)} dt$$<br><br>$$ = \int_1^x {{{\ln t} \over t}dt = {1 \over 2}{{(\ln x)}^2}} $$<br><br>$$ \therefore $$ $$f(e) + f\left( {{1 \over e}} \right) = {1 \over 2}{(\ln e)^2} = {1 \over 2}$$ | mcq | jee-main-2021-online-26th-february-evening-slot |
r6ZrOjQx5gyiQshwUA1kluywnv0 | maths | definite-integration | properties-of-definite-integration | If $${I_{m,n}} = \int\limits_0^1 {{x^{m - 1}}{{(1 - x)}^{n - 1}}dx} $$, for m, $$n \ge 1$$, and <br/>$$\int\limits_0^1 {{{{x^{m - 1}} + {x^{n - 1}}} \over {{{(1 + x)}^{m + 1}}}}} dx = \alpha {I_{m,n}}\alpha \in R$$, then $$\alpha$$ equals ___________. | [] | null | 1 | $${I_{m,n}} = \int\limits_0^1 {{x^{m - 1}}} .{(1 - x)^{n - 1}}dx$$<br><br>Put $$x = {1 \over {y + 1}} \Rightarrow dx = {{ - 1} \over {{{(y + 1)}^2}}}dy$$<br><br>$$1 - x = {y \over {y + 1}}$$<br><br>$$ \therefore $$ $${I_{m,n}} = \int\limits_\infty ^0 {{{{y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}( - 1)dy = } \int\limits_0^\infty {{{{y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} $$ .... (i)<br><br>Similarly, $${I_{m,n}} = \int\limits_0^1 {{x^{n - 1}}.{{(1 - x)}^{m - 1}}dx} $$<br><br>$$ \Rightarrow {I_{m,n}} = \int\limits_0^\infty {{{{y^{m - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} $$ .... (ii)<br><br>From (i) & (ii)<br><br>$$2{I_{m,n}} = \int\limits_0^\infty {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} $$<br><br>$$ \Rightarrow 2{I_{m,n}} = \int\limits_0^1 {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} + \int\limits_0^\infty {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} = {I_1} + {I_2}$$<br><br>Put $$y = {1 \over z}$$ in I<sub>2</sub><br><br>$$dy = - {1 \over {{z^2}}}dz$$<br><br>$$ \Rightarrow 2{I_{m,n}} = \int\limits_0^1 {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} + \int\limits_1^0 {{{{z^{m + 1}} + {z^{n - 1}}} \over {{{(z + 1)}^{m + n}}}}( - dz)} $$<br><br>$$ \Rightarrow {I_{m,n}} = \int\limits_0^1 {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} \Rightarrow \alpha = 1$$ | integer | jee-main-2021-online-26th-february-evening-slot |
ZJfj298ec1lDoMQg4x1kmhzfc4e | maths | definite-integration | properties-of-definite-integration | Let f : R $$ \to $$ R be a continuous function such that f(x) + f(x + 1) = 2, for all x$$\in$$R. <br/><br/>If $${I_1} = \int\limits_0^8 {f(x)dx} $$ and $${I_2} = \int\limits_{ - 1}^3 {f(x)dx} $$, then the value of I<sub>1</sub> + 2I<sub>2</sub> is equal to ____________. | [] | null | 16 | $$f(x) + f(x + 1) = 2$$ .... (i)<br><br>$$x \to (x + 1)$$<br><br>$$f(x + 1) + f(x + 2) = 2$$ .... (ii)<br><br>by (i) & (ii)<br><br>$$f(x) - f(x + 2) = 0$$<br><br>$$f(x + 2) = f(x)$$<br><br>$$ \therefore $$ f(x) is periodic with T = 2<br><br>$${I_1} = \int_0^{2 \times 4} {f(x)dx} = 4\int_0^2 {f(x)dx} $$<br><br>$${I_2} = \int_{ - 1}^3 {f(x)dx} = \int_0^4 {f(x + 1)dx} = \int_0^4 {(2 - f(x))dx} $$<br><br>$$ \Rightarrow $$ $${I_2} = 8 - 2\int_0^2 {f(x)dx} $$
<br><br>$$ \Rightarrow $$ $${I_2} = 8 - $$$${{{I_1}} \over 2}$$
<br><br>$$ \Rightarrow $$ $${I_1} + 2{I_2} = 16$$ | integer | jee-main-2021-online-16th-march-morning-shift |
67uyKNlEVDc5liFIQH1kmiwkjuv | maths | definite-integration | properties-of-definite-integration | Consider the integral <br/>$$I = \int_0^{10} {{{[x]{e^{[x]}}} \over {{e^{x - 1}}}}dx} $$, <br/>where [x] denotes the greatest integer less than or equal to x. Then the value of I is equal to : | [{"identifier": "A", "content": "45 (e $$-$$ 1)"}, {"identifier": "B", "content": "45 (e + 1)"}, {"identifier": "C", "content": "9 (e + 1)"}, {"identifier": "D", "content": "9 (e $$-$$ 1)"}] | ["A"] | null | $$I = \int_0^{10} {[x]\,.\,{e^{[x] + 1 - x}}} dx$$<br><br>$$ = \int_0^1 {0\,dx} + \int_1^2 {{e^{2 - x}}dx + \int_2^3 {2\,.\,{e^{3 - x}}dx} + \int_3^4 {3.{e^{4 - x}}dx} } + ......... + \int_9^{10} {9\,{e^{10 - x}}dx} $$<br><br>$$ = - \{ (1 - e) + 2(1 - e) + 3(1 - e) + ....... + 9(1 - e)\} $$<br><br>$$ = 45(e - 1)$$ | mcq | jee-main-2021-online-16th-march-evening-shift |
r5s4VH5sw0Fo8RRTCT1kmiwph7t | maths | definite-integration | properties-of-definite-integration | Let P(x) = x<sup>2</sup> + bx + c be a quadratic polynomial with real coefficients such that $$\int_0^1 {P(x)dx} $$ = 1 and P(x) leaves remainder 5 when it is divided by (x $$-$$ 2). Then the value of 9(b + c) is equal to : | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "11"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "15"}] | ["C"] | null | $$(x - 2)Q(x) + 5 = {x^2} + bx + c$$<br><br>Put x = 2<br><br>5 = 2b + c + 4 .... (1)<br><br>$$\int_0^1 {({x^2} + bx + c)dx} = 1$$<br><br>$$ \Rightarrow {1 \over 3} + {b \over 2} + c = 1$$<br><br>$${b \over 2} + c = {2 \over 3}$$ .... (2)<br><br>Solve (1) & (2)<br><br>$$b = {2 \over 9}$$<br><br>$$c = {5 \over 9}$$<br><br>9(b + c) = 7 | mcq | jee-main-2021-online-16th-march-evening-shift |
o4pjeaV1pkGvUJWd111kmjab62m | maths | definite-integration | properties-of-definite-integration | Which of the following statements is correct for the function g($$\alpha$$) for $$\alpha$$ $$\in$$ R such that <br/><br/>$$g(\alpha ) = \int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{{{\sin }^\alpha }x} \over {{{\cos }^\alpha }x + {{\sin }^\alpha }x}}dx} $$ | [{"identifier": "A", "content": "$$g(\\alpha )$$ is a strictly increasing function"}, {"identifier": "B", "content": "$$g(\\alpha )$$ is an even function"}, {"identifier": "C", "content": "$$g(\\alpha )$$ has an inflection point at $$\\alpha$$ = $$-$$$${1 \\over 2}$$"}, {"identifier": "D", "content": "$$g(\\alpha )$$ is a strictly decreasing function"}] | ["B"] | null | $$g(\alpha ) = \int\limits_{\pi /6}^{\pi /3} {{{{{\sin }^\alpha }\left( {{\pi \over 2} - x} \right)} \over {{{\cos }^\alpha }\left( {{\pi \over 2} - x} \right)x + {{\sin }^\alpha }\left( {{\pi \over 2} - x} \right)}}dx} $$<br><br>$$ = \int\limits_{\pi /6}^{\pi /3} {{{{{\cos }^\alpha }x} \over {{{\sin }^\alpha }x + {{\cos }^\alpha }x}}dx} $$<br><br>$$ \therefore $$ $$2.g(\alpha ) = \int\limits_{\pi /6}^{\pi /3} {{{si{n^\alpha }x + {{\cos }^\alpha }x} \over {{{\sin }^\alpha }x + {{\cos }^\alpha }x}}dx} = \int\limits_{\pi /6}^{\pi /3} {dx} = {\pi \over 3} - {\pi \over 6} = {\pi \over 6}$$<br><br>$$ \Rightarrow $$ $$g(\alpha ) = {\pi \over {12}}$$ i.e. a constant function hence an even function. | mcq | jee-main-2021-online-17th-march-morning-shift |
AZS5ox5Pdy8Y10mvNF1kmjccvuv | maths | definite-integration | properties-of-definite-integration | If [ . ] represents the greatest integer function, then the value of <br/><br/><br/>$$\left| {\int\limits_0^{\sqrt {{\pi \over 2}} } {\left[ {[{x^2}] - \cos x} \right]dx} } \right|$$ is ____________. | [] | null | 1 | $$\int\limits_0^{\sqrt {{\pi \over 2}} } {\left[ {[{x^2}] - \cos x} \right]} dx$$<br><br>$$ = \int\limits_0^1 {[ - \cos x]dx} + \int\limits_1^{\sqrt {{\pi \over 2}} } {[1 - \cos x]dx} $$
<br><br>= $$\int\limits_0^1 {\left[ { - \cos x} \right]} dx + \int\limits_1^{\sqrt {{\pi \over 2}} } {1dx} + \int\limits_1^{\sqrt {{\pi \over 2}} } {\left[ { - \cos x} \right]} dx$$
<br><br>When 0 $$ \le $$ x $$ \le $$ 1 then -1 $$ \le $$ -cosx $$ \le $$ 0
<br><br>$$ \therefore $$ $${\left[ { - \cos x} \right]}$$ = -1
<br><br>When 1 $$ \le $$ x $$ \le $$ $${\sqrt {{\pi \over 2}} }$$ = 1.24 then -0.33 $$ \le $$ -cosx $$ \le $$ 0
<br><br>$$ \therefore $$ $${\left[ { - \cos x} \right]}$$ = -1 (Integer value present in the left side of -0.33)
<br><br>$$ = - \int\limits_0^1 {dx + \int\limits_1^{\sqrt {{\pi \over 2}} } {dx - \int\limits_1^{\sqrt {{\pi \over 2}} } {dx} } } $$<br><br>$$ = - (x)_0^1 = - 1$$<br><br>$$ \therefore $$ $$\left| {\int\limits_0^{\sqrt {{\pi \over 2}} } {\left[ {[{x^2}] - \cos x} \right]dx} } \right|$$ = 1 | integer | jee-main-2021-online-17th-march-morning-shift |
tBRlCNOOh5nOH5YgEh1kmkn1xxj | maths | definite-integration | properties-of-definite-integration | If the integral
<br/><br/>$$\int_0^{10} {{{[\sin 2\pi x]} \over {{e^{x - [x]}}}}} dx = \alpha {e^{ - 1}} + \beta {e^{ - {1 \over 2}}} + \gamma $$, where $$\alpha$$, $$\beta$$, $$\gamma$$ are integers and [x] denotes the greatest integer less than or equal to x, then the value of $$\alpha$$ + $$\beta$$ + $$\gamma$$ is equal to : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "25"}] | ["A"] | null | Given integral<br><br>$$\int\limits_0^{10} {{{[\sin 2\pi x]} \over {{e^{x - [x]}}}}dx = 10\int\limits_0^1 {{{[\sin 2\pi x]} \over {{e^{\{ x\} }}}}dx} } $$ (using property of definite in.)<br><br>$$ = 10\left[ {\int\limits_0^{1/2} {0.dx} + \int\limits_{1/2}^1 {{{ - 1} \over {{e^x}}}dx} } \right]$$<br><br>= $$ - 10\left[ {{{{e^{ - x}}} \over { - 1}}} \right]_{1/2}^1 = 10\left[ {{e^{ - 1}} - {e^{ - 1/2}}} \right]$$<br><br>$$ = 10{e^{ - 1}} - 10{e^{ - 1/2}}$$<br><br>comparing with the given relation,<br><br>$$\alpha$$ = 10, $$\beta$$ = $$-$$10, $$\gamma$$ = 0<br><br>$$\alpha$$ + $$\beta$$ + $$\gamma$$ = 0<br><br>Therefore, the correct answer is (A). | mcq | jee-main-2021-online-17th-march-evening-shift |
qAj5oXiZQmjkLusOyY1kmknyiwo | maths | definite-integration | properties-of-definite-integration | Let $${I_n} = \int_1^e {{x^{19}}{{(\log |x|)}^n}} dx$$, where n$$\in$$N. If (20)I<sub>10</sub> = $$\alpha$$I<sub>9</sub> + $$\beta$$I<sub>8</sub>, for natural numbers $$\alpha$$ and $$\beta$$, then $$\alpha$$ $$-$$ $$\beta$$ equals to ___________. | [] | null | 1 | $${I_n} = 2\int\limits_1^e {{x^{19}}{{(\ln x)}^n}\,.\,dx} $$<br><br>$$ = {(\ln x)^n}\,.\,\left. {{{{x^{20}}} \over {20}}} \right|_1^e -\int\limits_1^e {n{{{{(\ln x)}^{n - 1}}} \over x}{{{x^{20}}} \over {20}}dx} $$<br><br>$${I_n} = {{{e^{20}}} \over {20}} - {n \over {20}}({I_{n - 1}})$$<br><br>$$20{I_n} = {e^{20}} - n\,{I_{n - 1}}$$
<br><br>Putting n = 10, we get
<br><br>$$20{I_{10}} = ({e^{20}} - 10{I_9})$$ ...... (1)
<br><br>Putting n = 9, we get
<br><br>$$20{I_9} = {e^{20}} - 9{I_8}$$ ....... (2) <br><br>Subtracting (2) from (1), we get<br><br>$$20{I_{10}} = 10{I_9} + 9{I_8}$$
<br><br>By comparing with (20)I<sub>10</sub> = $$\alpha$$I<sub>9</sub> + $$\beta$$I<sub>8</sub>, we get
<br><br>$$\alpha$$ = 10, $$\beta$$ = 9 $$ \Rightarrow $$ $$\alpha$$ $$-$$ $$\beta$$ = 1 | integer | jee-main-2021-online-17th-march-evening-shift |
sQo3uoODmsQ775Pzlc1kmlly1cp | maths | definite-integration | properties-of-definite-integration | Let f(x) and g(x) be two functions satisfying f(x<sup>2</sup>) + g(4 $$-$$ x) = 4x<sup>3</sup> and g(4 $$-$$ x) + g(x) = 0, then the value of $$\int\limits_{ - 4}^4 {f{{(x)}^2}dx} $$ is | [] | null | 512 | $$I = 2\int\limits_0^4 {f({x^2})dx} $$ ............(1)<br><br>$$ \Rightarrow I = 2\int\limits_0^4 {f({{(4 - x)}^2})dx} $$ ..............(2)<br><br>Adding equation (1) & (2)<br><br>$$2I = 2\int\limits_0^4 {\left[ {f{{(x)}^2} + f{{(4 - x)}^2}} \right]} \,dx$$ ............(3)<br><br>Now using $$f{(x^2)} + g(4 - x) = 4{x^3}$$ ............. (4)<br><br>$$x \to 4 - x$$<br><br>$$f({(4 - x)^2}) + g(x) = 4{(4 - x)^3}$$ ..............(5)<br><br>Adding equation (4) & (5)<br><br>$$f({x^2}) + f(4 - {x^2}) + g(x) + g(4 - x) = 4({x^3} + {(4 - x)^3}]$$<br><br>$$ \Rightarrow f({x^2}) + f(4 - {x^2}) = 4({x^3} + {(4 - x)^3}]$$<br><br>Now, $$I = 4\int\limits_0^4 {\left( {{x^3} + {{(4 - x)}^3}} \right)dx = 512} $$ | integer | jee-main-2021-online-18th-march-morning-shift |
LAeG6W8QyF222p0jls1kmm38pjl | maths | definite-integration | properties-of-definite-integration | Let g(x) = $$\int_0^x {f(t)dt} $$, where f is continuous function in [ 0, 3 ] such that $${1 \over 3}$$ $$ \le $$ f(t) $$ \le $$ 1 for all t$$\in$$ [0, 1] and 0 $$ \le $$ f(t) $$ \le $$ $${1 \over 2}$$ for all t$$\in$$ (1, 3]. The largest possible interval in which g(3) lies is : | [{"identifier": "A", "content": "$$\\left[ { - 1, - {1 \\over 2}} \\right]$$"}, {"identifier": "B", "content": "$$\\left[ { - {3 \\over 2}, - 1} \\right]$$"}, {"identifier": "C", "content": "[1, 3]"}, {"identifier": "D", "content": "$$\\left[ {{1 \\over 3},2} \\right]$$"}] | ["D"] | null | Given, $g(x)=\int_0^x f(t) d t$<br/><br/>
$\therefore g(3)=\int_0^3 f(t) d t=\int_0^1 f(t) d t+\int_1^3 f(t) d t$<br/><br/>
$\Rightarrow \int_0^1 \frac{1}{3} d t+\int_1^3 0 \cdot d t \leq g(3) \leq \int_0^1 1 d t+\int_1^3 \frac{1}{2} d t$<br/><br/>
$\Rightarrow \frac{1}{3} \leq g(3) \leq 1+1$<br/><br/>
$\Rightarrow \frac{1}{3} \leq g(3) \leq 2$ | mcq | jee-main-2021-online-18th-march-evening-shift |
14bSsveDutuV4jxgMG1kmm4euz8 | maths | definite-integration | properties-of-definite-integration | Let P(x) be a real polynomial of degree 3 which vanishes at x = $$-$$3. Let P(x) have local minima at x = 1, local maxima at x = $$-$$1 and $$\int\limits_{ - 1}^1 {P(x)dx} $$ = 18, then the sum of all the coefficients of the polynomial P(x) is equal to _________. | [] | null | 8 | P'(x) = a(x + 1)(x $$-$$ 1)<br><br>$$ \therefore $$ P(x) = $${{a{x^3}} \over 3}$$ $$-$$ ax + C<br><br>P($$-$$3) = 0 (given)<br><br>$$ \Rightarrow $$ a($$-$$9 + 3) + C = 0<br><br>$$ \Rightarrow $$ 6a = C ..... (i)<br><br>Also, $$\int\limits_{ - 1}^1 {P(x)dx} = 18 $$
<br><br>$$\Rightarrow \int\limits_{ - 1}^1 {\left( {a\left( {{{{x^3}} \over 3} - x} \right) + C} \right)} dx = 18$$<br><br>$$ \Rightarrow 0 + 2C = 18 \Rightarrow C = 9$$<br><br>from (i)<br><br>$$a = {3 \over 2}$$<br><br>$$ \therefore $$ $$P(x) = {{{x^3}} \over 2} - {3 \over 2}x + 9$$<br><br>Sum of co-efficient
<br><br> = $${1 \over 2} - {3 \over 2} + 9$$ = $$-$$1 + 9 = 8 | integer | jee-main-2021-online-18th-march-evening-shift |
1krps0tcf | maths | definite-integration | properties-of-definite-integration | Let a be a positive real number such that $$\int_0^a {{e^{x - [x]}}} dx = 10e - 9$$ where [ x ] is the greatest integer less than or equal to x. Then a is equal to: | [{"identifier": "A", "content": "$$10 - {\\log _e}(1 + e)$$"}, {"identifier": "B", "content": "$$10 + {\\log _e}2$$"}, {"identifier": "C", "content": "$$10 + {\\log _e}3$$"}, {"identifier": "D", "content": "$$10 + {\\log _e}(1 + e)$$"}] | ["B"] | null | a > 0<br><br>Let $$n \le a < n + 1,n \in W$$<br><br> $$a=\matrix{
{[a]} & + & {\{ a\} } \cr
\Downarrow & {} & \Downarrow \cr
{G.I.F.} & {} & {Fractional\,part} \cr
} $$<br><br>Here [ a ] = n<br><br>Now, $$\int_0^a {{e^{x - [x]}}} dx = 10e - 9$$<br><br>$$ \Rightarrow \int\limits_0^n {{e^{\{ x\} }}dx} + \int\limits_n^a {{e^{x - [x]}}dx} = 10e - 9$$<br><br>$$\therefore$$ $$n\int\limits_0^1 {{e^x}dx} + \int\limits_n^a {{e^{x - n}}dx} = 10e - 9$$<br><br>$$ \Rightarrow n(e - 1) + ({e^{a - n}} - 1) = 10e - 9$$<br><br>$$\therefore$$ n = 0 and {a} = log<sub>e</sub> 2<br><br>So, $$a = [a] + \{ a\} = (10 + {\log _e}2)$$<br><br>$$\Rightarrow$$ Option (2) is correct. | mcq | jee-main-2021-online-20th-july-morning-shift |
1krpt3zg2 | maths | definite-integration | properties-of-definite-integration | The value of the integral $$\int\limits_{ - 1}^1 {{{\log }_e}(\sqrt {1 - x} + \sqrt {1 + x} )dx} $$ is equal to: | [{"identifier": "A", "content": "$${1 \\over 2}{\\log _e}2 + {\\pi \\over 4} - {3 \\over 2}$$"}, {"identifier": "B", "content": "$$2{\\log _e}2 + {\\pi \\over 4} - 1$$"}, {"identifier": "C", "content": "$${\\log _e}2 + {\\pi \\over 2} - 1$$"}, {"identifier": "D", "content": "$$2{\\log _e}2 + {\\pi \\over 2} - {1 \\over 2}$$"}] | ["C"] | null | $$\int\limits_{ - 1}^1 {{{\log }_e}(\sqrt 1 - x + \sqrt {1 + x} )dx} $$<br/><br/>We know, $$\int\limits_{ - a}^a {f(x)dx = 2\int\limits_0^a {f(x)dx} } $$<br/><br/>So, $$2\int\limits_0^1 {{{\log }_e}(\sqrt 1 - x + \sqrt {1 + x} )dx} $$<br/><br/>$$l = 2\int\limits_0^1 {{{\log }_e}(\sqrt 1 - x + \sqrt {1 + x} ).1dx} $$<br/><br/>$$ \Rightarrow {l \over 2} = {\log _e}(\sqrt {1 - x} + \sqrt {1 + x} ).\,x]_0^1 - \int\limits_0^1 {{{{1 \over {2\sqrt {1 + x} }} - {1 \over {2\sqrt {1 - x} }}} \over {\sqrt {1 - x} + \sqrt {1 + x} }}.\,x\,dx} $$<br/><br/>$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 - {1 \over 2}\int\limits_0^1 {\left( {{{\sqrt {1 - x} - \sqrt {1 + x} } \over {\sqrt {1 - x} + \sqrt {1 + x} }}} \right){x \over {\sqrt {1 - {x^2}} }}dx} $$<br/><br/>$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 - {1 \over 2}\int\limits_0^1 {\left( {{{(1 - x) + (1 + x) - 2\sqrt {1 - {x^2}} } \over {(1 - x) - (1 + x)}}} \right){x \over {\sqrt {1 - {x^2}} }}dx} $$<br/><br/>$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 - {1 \over 2}\int\limits_0^1 {{{2(1 - \sqrt {1 - {x^2}} )} \over { - 2x}}.{x \over {\sqrt {1 - {x^2}} }}dx} $$<br/><br/>$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 + {1 \over 2}\int\limits_0^1 {\left( {{1 \over {\sqrt {1 - {x^2}} }} - 1} \right)dx} $$<br/><br/>$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 + {1 \over 2}[{\sin ^{ - 1}}x - x]_0^1$$<br/><br/>$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 + {1 \over 2}\left( {{\pi \over 2} - 1} \right)$$<br/><br/>$$\therefore$$ $$l = {\log _e}2 + {\pi \over 2} - 1$$ | mcq | jee-main-2021-online-20th-july-morning-shift |
1krrqb9z4 | maths | definite-integration | properties-of-definite-integration | If [x] denotes the greatest integer less than or equal to x, then the value of the integral $$\int_{ - \pi /2}^{\pi /2} {[[x] - \sin x]dx} $$ is equal to : | [{"identifier": "A", "content": "$$-$$ $$\\pi$$"}, {"identifier": "B", "content": "$$\\pi$$"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "1"}] | ["A"] | null | $$I = \int_{ - \pi /2}^{\pi /2} {[[x] - \sin x]dx} $$<br><br>$$ = \int_{ - \pi /2}^{\pi /2} {\left( {[x] + [ - \sin x]} \right)dx} $$<br><br>$$ = \int_0^{\pi /2} {\left( {[x] + [ - \sin x] + [ - x] + [\sin x]} \right)} dx$$<br><br>$$ = \int_0^{\pi /2} {( - 2)dx} $$<br><br>$$ = - \pi $$ | mcq | jee-main-2021-online-20th-july-evening-shift |
1krrqkq1g | maths | definite-integration | properties-of-definite-integration | If the real part of the complex number $${(1 - \cos \theta + 2i\sin \theta )^{ - 1}}$$ is $${1 \over 5}$$ for $$\theta \in (0,\pi )$$, then the value of the integral $$\int_0^\theta {\sin x} dx$$ is equal to: | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "$$-$$1"}, {"identifier": "D", "content": "0"}] | ["A"] | null | $$z = {1 \over {1 - \cos \theta + 2i\sin \theta }}$$<br><br>$$ = {{2{{\sin }^2}{\theta \over 2} - 2i\sin \theta } \over {{{(1 - \cos \theta )}^2} + 4{{\sin }^2}\theta }}$$<br><br>$$ = {{\sin {\theta \over 2} - 2i\cos {\theta \over 2}} \over {2\sin {\theta \over 2}\left( {{{\sin }^2}{\theta \over 2} + 4{{\cos }^2}{\theta \over 2}} \right)}}$$<br><br>$${\mathop{\rm Re}\nolimits} (z) = {1 \over {2\left( {{{\sin }^2}{\theta \over 2} + 4{{\cos }^2}{\theta \over 2}} \right)}} = {1 \over 5}$$<br><br>$$\sin {{2\theta } \over 2} + 4{\cos ^2}{\theta \over 2} = {5 \over 2}$$$$1 - {\cos ^2}{\theta \over 2} + 4\cos {\theta \over 2} = {5 \over 2}$$<br><br>$$3{\cos ^2}{\theta \over 2} = {1 \over 2}$$<br><br>$${\theta \over 2} = n\pi \pm {\pi \over 4}$$<br><br>$$\theta = 2n\pi \pm {\pi \over 2}$$<br><br>$$\theta \in (0,\pi )$$<br><br>$$ \therefore $$ $$\theta = {\pi \over 2}$$<br><br>$$\int\limits_0^{{\pi \over 2}} {\sin \theta d\theta = {{[ - \cos \theta ]}_0}^{{\pi \over 2}}} $$<br><br>$$ = - (0 - 1)$$<br><br>$$ = 1$$ | mcq | jee-main-2021-online-20th-july-evening-shift |
1krru916m | maths | definite-integration | properties-of-definite-integration | Let $$g(t) = \int_{ - \pi /2}^{\pi /2} {\cos \left( {{\pi \over 4}t + f(x)} \right)} dx$$, where $$f(x) = {\log _e}\left( {x + \sqrt {{x^2} + 1} } \right),x \in R$$. Then which one of the following is correct? | [{"identifier": "A", "content": "g(1) = g(0)"}, {"identifier": "B", "content": "$$\\sqrt 2 g(1) = g(0)$$"}, {"identifier": "C", "content": "$$g(1) = \\sqrt 2 g(0)$$"}, {"identifier": "D", "content": "g(1) + g(0) = 0"}] | ["B"] | null | $$\because$$ $$f(x) = \ln \left( {x + \sqrt {{x^2} + 1} } \right)$$<br><br>$$\therefore$$ $$f(x) + f( - x) = \ln \left( {\sqrt {{x^2} + 1} + x} \right) + \ln \left( {\sqrt {{x^2} + 1} - x} \right)$$<br><br>$$\therefore$$ $$f(x) + f( - x) = 0$$ .... (i)<br><br>$$\because$$ $$g(t) = \int_{ - \pi /2}^{\pi /2} {\cos \left( {{\pi \over 4}t + f(x)} \right)} dx$$<br><br>$$ = \int_0^{\pi /2} {\left\{ {\cos \left( {{\pi \over 4}t + f(x)} \right) + \cos \left( {{\pi \over 4}t + f( - x)} \right)} \right\}} dx$$<br><br>$$ = \int_0^{\pi /2} {\left\{ {\cos \left( {{{\pi t} \over 4} + f(x)} \right) + \cos \left( {{{\pi t} \over 4} - f(x)} \right)} \right\}dx} $$<br><br>$$g(t) = 2\int_0^{\pi /2} {\cos {{\pi t} \over 4}.\cos (f(x))dx} $$<br><br>$$\therefore$$ $$g(1) = \sqrt 2 \int_0^{\pi /2} {\cos \left( {f(x)} \right)} dx$$ and <br><br>$$g(0) = 2\int_0^{\pi /2} {\cos \left( {f(x)} \right)} dx$$<br><br>$$\therefore$$ $$\sqrt 2 g(1) = g(0)$$ | mcq | jee-main-2021-online-20th-july-evening-shift |
1krtdtbei | maths | definite-integration | properties-of-definite-integration | If $$\int\limits_0^{100\pi } {{{{{\sin }^2}x} \over {{e^{\left( {{x \over \pi } - \left[ {{x \over \pi }} \right]} \right)}}}}dx = {{\alpha {\pi ^3}} \over {1 + 4{\pi ^2}}},\alpha \in R} $$ where [x] is the greatest integer less than or equal to x, then the value of $$\alpha$$ is : | [{"identifier": "A", "content": "200 (1 $$-$$ e<sup>$$-$$1</sup>)"}, {"identifier": "B", "content": "100 (1 $$-$$ e)"}, {"identifier": "C", "content": "50 (e $$-$$ 1)"}, {"identifier": "D", "content": "150 (e<sup>$$-$$1</sup> $$-$$ 1)"}] | ["A"] | null | $$I = \int_0^{100\pi } {{{{{\sin }^2}x} \over {{e^{\left( {{x \over \pi } - \left[ {{x \over \pi }} \right]} \right)}}}}dx} $$<br><br>$$\because$$ Integrand is periodic with period 1<br><br>$$\therefore$$ $$I = 100\int_0^\pi {{{{{\sin }^2}x} \over {{e^{\left\{ {{x \over \pi }} \right\}}}}}} dx$$<br><br>Let $${x \over \pi } = t \Rightarrow dx = \pi dt$$<br><br>$$ = 100\pi \int_0^1 {{{{{\sin }^2}(\pi t)dt} \over {{e^t}}}} $$<br><br>$$ = 50\pi \int_0^1 {{e^{ - t}}(1 - \cos 2\pi t)dt} $$<br><br>$$ = 50\pi \int_0^1 {{e^{ - t}}dt - 50\pi \int_0^1 {{e^{ - t}}} \cos (2\pi t)dt} $$<br><br>$$ = - 50\left[ {{e^{ - t}}} \right]_0^1 - 50\pi \left[ {{{{e^{ - t}}} \over {1 + 4{\pi ^2}}}( - \cos 2\pi t + 2\pi \sin 2\pi t)} \right]_0^1$$<br><br>$$ = - 50\pi ({e^{ - 1}} - 1) - {{50\pi } \over {1 + 4{\pi ^2}}}({e^{ - 1}}( - 1 + 0) - ( - 1 + 0))$$<br><br>$$ = - 50\pi ({e^{ - 1}} - 1) - {{50\pi (1 - {e^{ - 1}})} \over {1 + 4{\pi ^2}}}$$<br><br>$$ = {{200{\pi ^3}(1 - {e^{ - 1}})} \over {1 + 4{\pi ^2}}} = {{\alpha {\pi ^3}} \over {1 + 4{\pi ^3}}}$$ (Given)<br><br>$$\therefore$$ $$\alpha = 200(1 - {e^{ - 1}})$$ | mcq | jee-main-2021-online-22th-july-evening-shift |
1krvy4naz | maths | definite-integration | properties-of-definite-integration | The value of the definite integral $$\int\limits_{\pi /24}^{5\pi /24} {{{dx} \over {1 + \root 3 \of {\tan 2x} }}} $$ is : | [{"identifier": "A", "content": "$${\\pi \\over 3}$$"}, {"identifier": "B", "content": "$${\\pi \\over 6}$$"}, {"identifier": "C", "content": "$${\\pi \\over {12}}$$"}, {"identifier": "D", "content": "$${\\pi \\over {18}}$$"}] | ["C"] | null | Let $$I = \int\limits_{\pi /24}^{5\pi /24} {{{{{(\cos 2x)}^{1/3}}} \over {{{(\cos 2x)}^{1/3}} + {{(\sin 2x)}^{1/3}}}}dx} $$ ...... (i)<br><br>$$ \Rightarrow I = \int\limits_{\pi /2}^{5\pi /24} {{{{{\left( {\cos \left\{ {2\left( {{\pi \over 4} - x} \right)} \right\}} \right)}^{{1 \over 3}}}} \over {{{\left( {\cos \left\{ {2\left( {{\pi \over 4} - x} \right)} \right\}} \right)}^{{1 \over 3}}} + {{\left( {\sin \left\{ {2\left( {{\pi \over 4} - x} \right)} \right\}} \right)}^{{1 \over 3}}}}}} dx\left\{ {\int\limits_a^b {f(x)dx = \int\limits_a^b {f(a + b - x)dx} } } \right\}$$<br><br>So, $$I = \int\limits_{\pi /24}^{5\pi /24} {{{{{(\sin 2x)}^{1/3}}} \over {{{(\sin 2x)}^{1/3}} + {{(\sin 2x)}^{1/3}}}}dx} $$ ..... (ii)<br><br>Hence, $$2I = \int\limits_{\pi /24}^{5\pi /24} {dx} $$ [(i) + (ii)]<br><br>$$ \Rightarrow 2I = {{4\pi } \over {24}} \Rightarrow I = {\pi \over {12}}$$ | mcq | jee-main-2021-online-25th-july-morning-shift |
1krw019ap | maths | definite-integration | properties-of-definite-integration | Let $$f:[0,\infty ) \to [0,\infty )$$ be defined as $$f(x) = \int_0^x {[y]dy} $$<br/><br/>where [x] is the greatest integer less than or equal to x. Which of the following is true? | [{"identifier": "A", "content": "f is continuous at every point in $$[0,\\infty )$$ and differentiable except at the integer points."}, {"identifier": "B", "content": "f is both continuous and differentiable except at the integer points in $$[0,\\infty )$$."}, {"identifier": "C", "content": "f is continuous everywhere except at the integer points in $$[0,\\infty )$$."}, {"identifier": "D", "content": "f is differentiable at every point in $$[0,\\infty )$$."}] | ["A"] | null | $$f:[0,\infty ) \to [0,\infty ),f(x) = \int_0^x {[y]dy} $$<br><br>Let $$x = n + f,f \in (0,1)$$<br><br>So, $$f(x) = 0 + 1 + 2 + ... + (n - 1) + \int\limits_n^{n + f} {n\,dy} $$<br><br>$$f(x) = {{n(n - 1)} \over 2} + nf$$<br><br>$$ = {{[x]([x] - 1)} \over 2} + [x]\{ x\} $$<br><br>Note $$\mathop {\lim }\limits_{x \to {n^ + }} f(x) = {{n(n - 1)} \over 2},\mathop {\lim }\limits_{x \to {n^ - }} f(x) = {{(n - 1)(n - 2)} \over 2} + (n - 1)$$<br><br>$$ = {{n(n - 1)} \over 2}$$<br><br>$$f(x) = {{n(n - 1)} \over 2}(n \in {N_0})$$<br><br>so f(x) is cont. $$\forall$$ x $$\ge$$ 0 nd diff. except at integer points | mcq | jee-main-2021-online-25th-july-morning-shift |
1kryg955g | maths | definite-integration | properties-of-definite-integration | If $$\int_0^\pi {({{\sin }^3}x){e^{ - {{\sin }^2}x}}dx = \alpha - {\beta \over e}\int_0^1 {\sqrt t {e^t}dt} } $$, then $$\alpha$$ + $$\beta$$ is equal to ____________. | [] | null | 5 | $$I = 2\int_0^{\pi /2} {{{\sin }^3}x{e^{ - {{\sin }^2}x}}dx} $$<br><br>$$ = 2\int_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}dx} + \int\limits_0^{\pi /2} {\mathop {\cos x}\limits_I \,\underbrace {{e^{ - {{\sin }^2}x}}( - \sin 2x)}_{II}dx} $$$$ = 2\int\limits_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}dx} + \left[ {\cos x{e^{ - {{\sin }^2}x}}} \right]_0^{\pi /2} + \int\limits_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}} dx$$<br><br>$$ = 3\int\limits_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}dx} - 1$$<br><br>$$ = {3 \over 2}\int\limits_{ - 1}^0 {{{{e^\alpha }d\alpha } \over {\sqrt {1 + \alpha } }} - 1} $$ (Put $$-$$sin<sup>2</sup>x = t)<br><br>$$ = {3 \over {2e}}\int\limits_0^1 {{{{e^x}} \over {\sqrt x }}dx - 1} $$ (put 1 + $$\alpha$$ = x)<br><br>$$ = {3 \over {2e}}\int\limits_0^1 {\mathop {{e^x}}\limits_{II} } \mathop {{1 \over {\sqrt x }}}\limits_{II} dx - 1$$<br><br>$$ = 2 - {3 \over e}\int\limits_0^1 {{e^x}\sqrt x } dx$$<br><br>Hence, $$\alpha$$ + $$\beta$$ = 5 | integer | jee-main-2021-online-27th-july-evening-shift |
1krzlm5le | maths | definite-integration | properties-of-definite-integration | If $$f(x) = \left\{ {\matrix{
{\int\limits_0^x {\left( {5 + \left| {1 - t} \right|} \right)dt,} } & {x > 2} \cr
{5x + 1,} & {x \le 2} \cr
} } \right.$$, then | [{"identifier": "A", "content": "f(x) is not continuous at x = 2"}, {"identifier": "B", "content": "f(x) is everywhere differentiable "}, {"identifier": "C", "content": "f(x) is continuous but not differentiable at x = 2"}, {"identifier": "D", "content": "f(x) is not differentiable at x = 1"}] | ["C"] | null | $$f(x) = \int\limits_0^1 {(5 + (1 - t))dt + \int\limits_1^x {(5 + (t - 1))dt} } $$<br><br>$$ = \left. {6 - {1 \over 2} + \left( {4t + {{{t^2}} \over 2}} \right)} \right|_1^x$$<br><br>$$ = {{11} \over 2} + 4x + {{{x^2}} \over 2} - 4 - {1 \over 2}$$<br><br>$$ = {{{x^2}} \over 2} + 4x - 1$$<br><br>$$f({2^ + }) = 2 + 8 + 1 = 11$$<br><br>$$f(2) = f({2^ - }) = 5 \times 2 + 1 = 11$$<br><br>$$\Rightarrow$$ continuous at x = 2<br><br>Clearly differentiable at x = 1<br><br>Lf' (2) = 5<br><br>Rf' (2) = 6<br><br>$$\Rightarrow$$ Not differentiable at x = 2 | mcq | jee-main-2021-online-25th-july-evening-shift |
1krzmibsx | maths | definite-integration | properties-of-definite-integration | The value of the <br/><br/>integral $$\int\limits_{ - 1}^1 {\log \left( {x + \sqrt {{x^2} + 1} } \right)dx} $$ is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "$$-$$1"}, {"identifier": "D", "content": "1"}] | ["B"] | null | Let $$I = \int\limits_{ - 1}^1 {\log \left( {x + \sqrt {{x^2} + 1} } \right)dx} $$<br><br>$$\because$$ $$\log \left( {x + \sqrt {{x^2} + 1} } \right)$$ is an odd function<br><br>$$\therefore$$ I = 0 | mcq | jee-main-2021-online-25th-july-evening-shift |
1ks05rb02 | maths | definite-integration | properties-of-definite-integration | The value of the definite integral<br/><br/>$$\int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{dx} \over {(1 + {e^{x\cos x}})({{\sin }^4}x + {{\cos }^4}x)}}} $$ is equal to : | [{"identifier": "A", "content": "$$ - {\\pi \\over 2}$$"}, {"identifier": "B", "content": "$${\\pi \\over {2\\sqrt 2 }}$$"}, {"identifier": "C", "content": "$$ - {\\pi \\over 4}$$"}, {"identifier": "D", "content": "$${\\pi \\over {\\sqrt 2 }}$$"}] | ["B"] | null | $$I = \int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{dx} \over {(1 + {e^{x\cos x}})({{\sin }^4}x + {{\cos }^4}x)}}} $$ .... (1)<br><br>Using $$\int\limits_a^b {f(x)dx = \int\limits_a^b {f(a + b - x)dx} } $$<br><br>$$I = \int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{dx} \over {(1 + {e^{ - x\cos x}})({{\sin }^4}x + {{\cos }^4}x)}}} $$<br><br>Add (1) and (2)<br><br>$$2I = \int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{dx} \over {{{\sin }^4}x + {{\cos }^4}x}}} $$<br><br>$$2I = 2\int\limits_0^{{\pi \over 4}} {{{dx} \over {{{\sin }^4}x + {{\cos }^4}x}}} $$<br><br>$$I = \int\limits_0^{{\pi \over 4}} {{{\left( {1 + {1 \over {{{\tan }^2}x}}} \right){{\sec }^2}x} \over {{{\left( {\tan - {1 \over {\tan x}}} \right)}^2} + 2}}} dx$$<br><br>$$\tan x - {1 \over {\tan x}} = t$$<br><br>$$\left( {1 + {1 \over {{{\tan }^2}x}}} \right){\sec ^2}xdx = dt$$<br><br>$$I = \int\limits_{ - \infty }^0 {{{dt} \over {{t^2} + 2}}} = \left[ {{1 \over {\sqrt 2 }}{{\tan }^{ - 1}}\left( {{t \over {\sqrt 2 }}} \right)} \right]_{ - \infty }^0$$<br><br>$$I = 0 - {1 \over {\sqrt 2 }}\left( { - {\pi \over 2}} \right) = {\pi \over {2\sqrt 2 }}$$ | mcq | jee-main-2021-online-27th-july-morning-shift |
1ks0ccco5 | maths | definite-integration | properties-of-definite-integration | Let the domain of the function<br/><br/>$$f(x) = {\log _4}\left( {{{\log }_5}\left( {{{\log }_3}(18x - {x^2} - 77)} \right)} \right)$$ be (a, b). Then the value of the integral $$\int\limits_a^b {{{{{\sin }^3}x} \over {({{\sin }^3}x + {{\sin }^3}(a + b - x)}}} dx$$ is equal to _____________. | [] | null | 1 | For domain<br><br>$${\log _5}\left( {{{\log }_3}(18x - {x^2} - 77)} \right) > 0$$<br><br>$${\log _3}(18x - {x^2} - 77) > 1$$<br><br>$$18x - {x^2} - 77 > 3$$<br><br>$${x^2} - 18x + 80 < 0$$<br><br>$$x \in (8,10)$$<br><br>$$\Rightarrow$$ a = 8 and b = 10<br><br>$$I = \int\limits_a^b {{{{{\sin }^3}x} \over {{{\sin }^3}x + {{\sin }^3}(a + b - x)}}} dx$$<br><br>$$I = \int\limits_a^b {{{{{\sin }^3}x(a + b - x)} \over {{{\sin }^3}x + {{\sin }^3}(a + b - x)}}} $$<br><br>$$2I = (b - a) \Rightarrow I = {{b - a} \over 2}$$ ($$\because$$ a = 8 and b = 10)<br><br>$$I = {{10 - 8} \over 2} = 1$$ | integer | jee-main-2021-online-27th-july-morning-shift |
1ktbey39j | maths | definite-integration | properties-of-definite-integration | The value of $$\int\limits_{{{ - 1} \over {\sqrt 2 }}}^{{1 \over {\sqrt 2 }}} {{{\left( {{{\left( {{{x + 1} \over {x - 1}}} \right)}^2} + {{\left( {{{x - 1} \over {x + 1}}} \right)}^2} - 2} \right)}^{{1 \over 2}}}dx} $$ is : | [{"identifier": "A", "content": "log<sub>e</sub> 4 "}, {"identifier": "B", "content": "log<sub>e</sub> 16"}, {"identifier": "C", "content": "2log<sub>e</sub> 16"}, {"identifier": "D", "content": "4log<sub>e</sub> (3 + 2$${\\sqrt 2 }$$)"}] | ["B"] | null | <p>$$\sqrt {{{\left( {{{x + 1} \over {x - 1}}} \right)}^2} + {{\left( {{{x - 1} \over {x + 1}}} \right)}^2} - 2} $$</p>
<p>$$ = \sqrt {{{\left( {{{x + 1} \over {x - 1}} - {{x - 1} \over {x + 1}}} \right)}^2}} $$</p>
<p>$$ = \left| {{{x + 1} \over {x - 1}} - {{x - 1} \over {x + 1}}} \right|$$</p>
<p>$$ = \left| {{{{{(x + 1)}^2} - {{(x - 1)}^2}} \over {{x^2} - 1}}} \right|$$</p>
<p>$$ = \left| {{{{x^2} + 2x + 1 - {x^2} + 2x - 1} \over {{x^2} - 1}}} \right|$$</p>
<p>$$ = \left| {{{4x} \over {{x^2} - 1}}} \right|$$</p>
<p>$$\therefore$$ $$I = \int\limits_{ - {1 \over {\sqrt 2 }}}^{{1 \over {\sqrt 2 }}} {{{\left[ {{{\left( {{{x + 1} \over {x - 1}}} \right)}^2} + {{\left( {{{x - 1} \over {x + 1}}} \right)}^2} - 2} \right]}^{{1 \over 2}}}dx} $$</p>
<p>$$I = \int\limits_{{1 \over {\sqrt 2 }}}^{{1 \over {\sqrt 2 }}} {\left| {{{4x} \over {{x^2} - 1}}} \right|dx} $$</p>
<p>Let $$f(x) = \left| {{{4x} \over {{x^2} - 1}}} \right|$$</p>
<p>$$\therefore$$ $$f( - x) = \left| {{{4( - x)} \over {{{( - x)}^2} - 1}}} \right| = \left| {{{ - 4x} \over {{x^2} - 1}}} \right| = \left| {{{4x} \over {{x^2} - 1}}} \right|$$</p>
<p>$$ \Rightarrow f(x) = f( - x)$$</p>
<p>$$\therefore$$ f(x) is a even function.</p>
<p>$$\therefore$$ $$I = 2\int_0^{{1 \over {\sqrt 2 }}} {\left| {{{4x} \over {{x^2} - 1}}} \right|dx} $$</p>
<p>Using property, If f(x) is an even function then,</p>
<p>$$\int_{ - a}^a {f(x) = 2\int_0^a {f(x)dx} } $$</p>
<p>x > 0 when $$x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$$</p>
<p>$$\Rightarrow$$ x<sup>2</sup> > 0 when $$x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$$</p>
<p>$$\Rightarrow$$ x<sup>2</sup> $$-$$ 1 < 0 when $$x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$$</p>
<p>$$ \Rightarrow {1 \over {{x^2} - 1}} < 0$$ when $$x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$$</p>
<p>$$ \Rightarrow {{4x} \over {{x^2} - 1}} < 0$$ when $$x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$$</p>
<p>$$\therefore$$ $$\left| {{{4x} \over {{x^2} - 1}}} \right| = - {{4x} \over {{x^2} - 1}}$$ when $$x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$$</p>
<p>$$\therefore$$ $$I = 2\int_0^{{1 \over {\sqrt 2 }}} { - \left( {{{4x} \over {{x^2} - 1}}} \right)dx} $$</p>
<p>$$ = - 4\int_0^{{1 \over {\sqrt 2 }}} {{{2x} \over {{x^2} - 1}}dx} $$</p>
<p>$$ = - 4\left[ {{{\log }_e}\left| {{x^2} - 1} \right|} \right]_0^{{1 \over {\sqrt 2 }}}$$</p>
<p>$$ = - 4\left[ {\log \left| {{1 \over 2} - 1} \right| - \log \left| { - 1} \right|} \right]$$</p>
<p>$$ = - 4{\log _e}\left| { - {1 \over 2}} \right|$$</p>
<p>$$ = - 4{\log _e}{1 \over 2}$$</p>
<p>$$ = - 4\log _e^{{2^{ - 1}}}$$</p>
<p>$$ = 4\log _e^2$$</p>
<p>$$ = \log _e^{{2^4}}$$</p>
<p>$$ = \log _e^{16}$$</p> | mcq | jee-main-2021-online-26th-august-morning-shift |
1ktcym1ca | maths | definite-integration | properties-of-definite-integration | If the value of the integral <br/>$$\int\limits_0^5 {{{x + [x]} \over {{e^{x - [x]}}}}dx = \alpha {e^{ - 1}} + \beta } $$, where $$\alpha$$, $$\beta$$ $$\in$$ R, 5$$\alpha$$ + 6$$\beta$$ = 0, and [x] denotes the greatest integer less than or equal to x; then the value of ($$\alpha$$ + $$\beta$$)<sup>2</sup> is equal to : | [{"identifier": "A", "content": "100"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "16"}, {"identifier": "D", "content": "36"}] | ["B"] | null | <p>$$I = \int\limits_0^5 {{{x + [x]} \over {{e^{x - [x]}}}}dx = \alpha {e^{ - 1}} + \beta } $$</p>
<p>$$I = \int\limits_0^1 {{x \over {{e^x}}}dx + \int\limits_1^2 {{{x + 1} \over {{e^{x - 1}}}}dx + \int\limits_2^3 {{{x + 2} \over {{e^{x - 2}}}}dx + \int\limits_3^4 {{{x + 3} \over {{e^{x - 3}}}}dx + \int\limits_4^5 {{{x + 4} \over {{e^{x - 4}}}}dx} } } } } $$</p>
<p>Let $$I = {I_1} + {I_2} + {I_3} + {I_4} + {I_5}$$</p>
<p>Here, $${I_2} = \int\limits_1^2 {{{x + 1} \over {{e^{x - 1}}}}dx} $$ Put $$x = t + 1 \Rightarrow dx = dt$$</p>
<p>$$ = \int\limits_0^1 {{{t + 2} \over {{e^t}}}dt = \int\limits_0^1 {{t \over {{e^t}}}dt + \int\limits_0^1 {{2 \over {{e^t}}}dt} } } $$</p>
<p>$${I_2} = {I_1} + 2\int\limits_0^1 {{e^{ - t}}dt = {I_1} + 2(1 - {e^{ - 1}})} $$</p>
<p>Similarly,</p>
<p>$${I_3} = {I_1} + 4(1 - {e^{ - 1}})$$</p>
<p>$${I_4} = {I_1} + 6(1 - {e^{ - 1}})$$</p>
<p>$${I_5} = {I_1} + 8(1 - {e^{ - 1}})$$</p>
<p>$$I = {I_1} + {I_2} + {I_3} + {I_4} + {I_5} = 5{I_1} + (2 + 4 + 6 + 8)(1 - {e^{ - 1}})$$</p>
<p>$$ = 5{I_1} + 20(1 - {e^{ - 1}})$$</p>
<p>$${I_1} = \int_0^1 {x{e^{ - 1}}dx = - [{e^{ - x}}(x + 1)_0^1 = 1 - 2{e^{ - 1}}} $$</p>
<p>$$\therefore$$ $$5{I_1} + 20(1 - {e^{ - 1}}) = 5(1 - 2{e^{ - 1}}) + 20(1 - {e^{ - 1}}) = 25 - 30{e^{ - 1}}$$</p>
<p>$$\therefore$$ $$\alpha$$ = $$-$$30, $$\beta$$ = 25</p>
<p>Also it satisfy $$5\alpha + 6\beta = 0$$</p>
<p>Now, $${(\alpha + \beta )^2} = {( - 30 + 25)^2} = {( - 5)^2} = 25$$</p> | mcq | jee-main-2021-online-26th-august-evening-shift |
1ktd2e88w | maths | definite-integration | properties-of-definite-integration | The value of $$\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{{1 + {{\sin }^2}x} \over {1 + {\pi ^{\sin x}}}}} \right)} \,dx$$ is | [{"identifier": "A", "content": "$${\\pi \\over 2}$$"}, {"identifier": "B", "content": "$${{5\\pi } \\over 4}$$"}, {"identifier": "C", "content": "$${{3\\pi } \\over 4}$$"}, {"identifier": "D", "content": "$${{3\\pi } \\over 2}$$"}] | ["C"] | null | $$I = \int\limits_0^{{\pi \over 2}} {{{(1 + {{\sin }^2}x)} \over {(1 + {\pi ^{\sin x}})}} + {{{\pi ^{\sin x}}(1 + {{\sin }^2}x)} \over {(1 + {\pi ^{\sin x}})}}} dx$$<br><br>$$I = \int_0^{{\pi \over 2}} {(1 + {{\sin }^2}x)\,dx} $$<br><br>$$I = {\pi \over 2} + {\pi \over 2}.{1 \over 2} = {{3\pi } \over 4}$$ | mcq | jee-main-2021-online-26th-august-evening-shift |
1kteo6fze | maths | definite-integration | properties-of-definite-integration | $$\int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}({x^2} - 44x + 484)}}dx} $$ is equal to : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "10"}] | ["C"] | null | Let $$I = \int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}({x^2} - 44x + 484)}}dx} $$<br><br>$$I = \int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}({x^2} - 22)}}dx} $$ .... (1)<br><br>We know,<br><br>$$\int\limits_a^b {f(x)dx} = \int\limits_a^b {f(a + b - x)} \,dx$$ (king)<br><br>So, $$I = \int\limits_6^{16} {{{{{\log }_e}{{(22 - x)}^2}} \over {{{\log }_e}{{(22 - x)}^2} + {{\log }_e}{{(21 - (22 - x))}^2}}}} $$<br><br>$$I = \int\limits_6^{16} {{{{{\log }_e}{{(22 - x)}^2}} \over {{{\log }_e}{x^2} + {{\log }_e}{{(22 - x)}^2}}}dx} $$ .... (2)<br><br>(1) + (2)<br><br>$$2I = \int\limits_6^{16} {1.\,dx} = 10$$<br><br>I = 5 | mcq | jee-main-2021-online-27th-august-morning-shift |
1ktg3oaty | maths | definite-integration | properties-of-definite-integration | The value of the integral $$\int\limits_0^1 {{{\sqrt x dx} \over {(1 + x)(1 + 3x)(3 + x)}}} $$ is : | [{"identifier": "A", "content": "$${\\pi \\over 8}\\left( {1 - {{\\sqrt 3 } \\over 2}} \\right)$$"}, {"identifier": "B", "content": "$${\\pi \\over 4}\\left( {1 - {{\\sqrt 3 } \\over 6}} \\right)$$"}, {"identifier": "C", "content": "$${\\pi \\over 8}\\left( {1 - {{\\sqrt 3 } \\over 6}} \\right)$$"}, {"identifier": "D", "content": "$${\\pi \\over 4}\\left( {1 - {{\\sqrt 3 } \\over 2}} \\right)$$"}] | ["A"] | null | $$I = \int\limits_0^1 {{{\sqrt x dx} \over {(1 + x)(1 + 3x)(3 + x)}}} dx$$<br><br>Let x = t<sup>2</sup> $$\Rightarrow$$ dx = 2t . dt<br><br>$$I = \int\limits_0^1 {{{t(2t)} \over {({t^2} + 1)(1 + 3{t^2})(3 + {t^2})}}} dt$$<br><br>$$I = \int\limits_0^1 {{{(3{t^2} + 1) - ({t^2} + 1)} \over {(3{t^2} + 1)({t^2} + 1)(3 + {t^2})}}} dt$$<br><br>$$I = \int\limits_0^1 {{{dt} \over {({t^2} + 1)(3 + {t^2})}}} - \int\limits_0^1 {{{dt} \over {(1 + 3{t^2})(3 + {t^2})}}} $$<br><br>$$ = {1 \over 2}\int\limits_0^1 {{{(3 + {t^2}) - ({t^2} + 1)} \over {({t^2} + 1)(3 + {t^2})}}} dt + {1 \over 8}\int\limits_0^1 {{{(1 + 3{t^2}) - 3(3 + {t^2})} \over {(1 + 3{t^2})(3 + {t^2})}}} dt$$<br><br>$$ = {1 \over 2}\int\limits_0^1 {{{dt} \over {1 + {t^2}}} - {1 \over 2}\int\limits_0^1 {{{dt} \over {{t^2} + 3}} + {1 \over 8}\int\limits_0^1 {{{dt} \over {{t^2} + 3}} - {3 \over 8}\int\limits_0^1 {{{dt} \over {(1 + 3{t^2})}}} } } } $$<br><br>$$ = {1 \over 2}\int\limits_0^1 {{{dt} \over {{t^2} + 1}} - {3 \over 8}\int\limits_0^1 {{{dt} \over {{t^2} + 3}} - {3 \over 8}\int\limits_0^1 {{{dt} \over {1 + 3{t^2}}}} } } $$<br><br>$$ = {1 \over 2}({\tan ^{ - 1}}(t))_0^1 - {3 \over {8\sqrt 3 }}\left( {{{\tan }^{ - 1}}\left( {{t \over {\sqrt 3 }}} \right)} \right)_0^1 - {3 \over {8\sqrt 3 }}\left( {{{\tan }^{ - 1}}\left( {\sqrt 3 t} \right)} \right)_0^1$$<br><br>$$ = {1 \over 2}\left( {{\pi \over 4}} \right) - {{\sqrt 3 } \over 8}\left( {{\pi \over 6}} \right) - {{\sqrt 3 } \over 8}\left( {{\pi \over 3}} \right)$$<br><br>$$ = {\pi \over 8} - {{\sqrt 3 } \over {16}}\pi $$<br><br>$$ = {\pi \over 8}\left( {1 - {{\sqrt 3 } \over 2}} \right)$$ | mcq | jee-main-2021-online-27th-august-evening-shift |
1ktirsgas | maths | definite-integration | properties-of-definite-integration | Let [t] denote the greatest integer $$\le$$ t. Then the value of <br/><br/>$$8.\int\limits_{ - {1 \over 2}}^1 {([2x] + |x|)dx} $$ is ___________. | [] | null | 5 | $$I = \,\int\limits_{ - {1 \over 2}}^1 {([2x] + |x|)dx} $$<br><br>$$ = \int\limits_{ - 1/2}^1 {[2x]\,dx + \int\limits_{ - 1/2}^1 {|x|\,} dx} $$<br><br>$$ = 0 + \int\limits_{ - 1/2}^0 {( - x)\,} dx + \int\limits_0^1 {x\,} dx$$<br><br>$$ = \left( { - {{{x^2}} \over 2}} \right)_{ - 1/2}^0 + \left( {{{{x^2}} \over 2}} \right)_0^1$$<br><br>$$ = \left( {0 + {1 \over 8}} \right) + {1 \over 2}$$<br><br>$$ = {5 \over 8}$$<br><br>$$ \therefore $$ 8I = 5 | integer | jee-main-2021-online-31st-august-morning-shift |
1ktisti2k | maths | definite-integration | properties-of-definite-integration | If $$x\phi (x) = \int\limits_5^x {(3{t^2} - 2\phi '(t))dt} $$, x > $$-$$2, and $$\phi$$(0) = 4, then $$\phi$$(2) is __________. | [] | null | 4 | $$x\phi (x) = \int\limits_5^x {3{t^2} - 2\phi '(t)dt} $$<br><br>$$x\phi (x) = {x^3} - 125 - 2[\phi (x) - \phi (5)]$$<br><br>$$x\phi (x) = {x^3} - 125 - 2\phi (x) - 2\phi (5)$$<br><br>$$\phi (0) = 4 \Rightarrow \phi (5) = {{133} \over 2}$$<br><br>$$\phi (x) = {{{x^3} + 8} \over {x + 2}}$$<br><br>$$\phi (2) = 4$$ | integer | jee-main-2021-online-31st-august-morning-shift |
1ktka5vae | maths | definite-integration | properties-of-definite-integration | If [x] is the greatest integer $$\le$$ x, then <br/><br/>$${\pi ^2}\int\limits_0^2 {\left( {\sin {{\pi x} \over 2}} \right)(x - [x]} {)^{[x]}}dx$$ is equal to : | [{"identifier": "A", "content": "2($$\\pi$$ $$-$$ 1)"}, {"identifier": "B", "content": "4($$\\pi$$ $$-$$ 1)"}, {"identifier": "C", "content": "4($$\\pi$$ + 1)"}, {"identifier": "D", "content": "2($$\\pi$$ + 1)"}] | ["B"] | null | <p>$$I = {\pi ^2}\int_0^2 {\sin \left( {{{\pi x} \over 2}} \right){{(x - [x])}^{[x]}}dx} $$</p>
<p>$$ = {\pi ^2}\int_0^1 {\sin \left( {{{\pi x} \over 2}} \right){x^0}dx + {\pi ^2}\int_1^2 {\sin \left( {{{\pi x} \over 2}} \right)(x - 1)dx} } $$</p>
<p>$$ = {\pi ^2}\left[ {{{ - 2} \over \pi }\cos {{\pi x} \over 2}} \right]_0^1 + {\pi ^2}\left[ {(x - 1){2 \over \pi }\left( { - \cos {{\pi x} \over 2}} \right)} \right]_1^2 + {\pi ^2}\int_1^2 {{2 \over \pi }\cos {{\pi x} \over 2}dx} $$</p>
<p>$$ = \left. {{\pi ^2}\left( {{2 \over \pi }} \right) + {{2{\pi ^2}} \over \pi }(1 - 0) + 2\pi \,.\,{2 \over \pi }\left( {\sin {{\pi x} \over 2}} \right)} \right|_1^2$$</p>
<p>$$ = 2\pi + 2\pi + 4(0 - 1) = 4\pi - 4 = 4(\pi - 1)$$</p> | mcq | jee-main-2021-online-31st-august-evening-shift |
1kto5yxz5 | maths | definite-integration | properties-of-definite-integration | Let $${J_{n,m}} = \int\limits_0^{{1 \over 2}} {{{{x^n}} \over {{x^m} - 1}}dx} $$, $$\forall$$ n > m and n, m $$\in$$ N. Consider a matrix $$A = {[{a_{ij}}]_{3 \times 3}}$$ where $${a_{ij}} = \left\{ {\matrix{
{{j_{6 + i,3}} - {j_{i + 3,3}},} & {i \le j} \cr
{0,} & {i > j} \cr
} } \right.$$. Then $$\left| {adj{A^{ - 1}}} \right|$$ is : | [{"identifier": "A", "content": "(15)<sup>2</sup> $$\\times$$ 2<sup>42</sup>"}, {"identifier": "B", "content": "(15)<sup>2</sup> $$\\times$$ 2<sup>34</sup>"}, {"identifier": "C", "content": "(105)<sup>2</sup> $$\\times$$ 2<sup>38</sup>"}, {"identifier": "D", "content": "(105)<sup>2</sup> $$\\times$$ 2<sup>36</sup>"}] | ["C"] | null | A = $$\left[ {\matrix{
{a_{11}} & {a_{12}} & {a_{13}} \cr
{{a_{21}}} & {{a_{22}}} & {{a_{23}}} \cr
{{a_{31}}} & {{a_{32}}} & {{a_{33}}} \cr
} } \right]$$<br><br>$${J_{6 + i,3}} - {J_{i + 3,3}}; i \le j$$<br><br>$$ = \int_0^{{1 \over 2}} {{{{x^{6 + i}}} \over {{x^3} - 1}} - \int_0^{{1 \over 2}} {{{{x^{i + 3}}} \over {{x^3} - 1}}} } $$<br><br>$$ =\int_0^{1/2} {{{{x^{i + 3}}({x^3} - 1)} \over {{x^3} - 1}}} $$<br><br>$$ = {{{x^{3 + i + 1}}} \over {3 + i + 1}} = \left( {{{{x^{4 + i}}} \over {4 + i}}} \right)_0^{1/2}$$<br><br>$$ \therefore $$ $${a_{ij}} = {j_{6 + i,3}} - {j_{i + 3,3}} = {{{{\left( {{1 \over 2}} \right)}^{4 + i}}} \over {4 + i}}$$<br><br>$${a_{11}} = {{{{\left( {{1 \over 2}} \right)}^5}} \over 5} = {1 \over {{{5.2}^5}}}$$<br><br>$${a_{12}} = {1 \over {{{5.2}^5}}}$$<br><br>$${a_{13}} = {1 \over {{{5.2}^5}}}$$<br><br>$${a_{22}} = {1 \over {{{6.2}^6}}}$$<br><br>$${a_{23}} = {1 \over {{{6.2}^6}}}$$<br><br>$${a_{33}} = {1 \over {{{7.2}^7}}}$$<br><br>$$A = \left[ {\matrix{
{{1 \over {{{5.2}^5}}}} & {{1 \over {{{5.2}^5}}}} & {{1 \over {{{5.2}^5}}}} \cr
0 & {{1 \over {{{6.2}^6}}}} & {{1 \over {{{6.2}^6}}}} \cr
0 & 0 & {{1 \over {{{7.2}^7}}}} \cr
} } \right]$$<br><br>$$\left| A \right| = {1 \over {{{5.2}^5}}}\left[ {{1 \over {{{6.2}^6}}} \times {1 \over {{{7.2}^7}}}} \right]$$<br><br>$$\left| A \right| = {1 \over {{{210.2}^{18}}}}$$<br><br>$$\left| {adj{A^{ - 1}}} \right| = {\left| {{A^{ - 1}}} \right|^{n - 1}} = {\left| {{A^{ - 1}}} \right|^2} = {1 \over {{{\left( {\left| A \right|} \right)}^2}}}$$<br><br>$$ = {({210.2^{18}})^2}$$<br><br>= $${(105)^2} \times {2^{38}}$$ | mcq | jee-main-2021-online-1st-september-evening-shift |
1kto9vgym | maths | definite-integration | properties-of-definite-integration | The function f(x), that satisfies the condition<br/> $$f(x) = x + \int\limits_0^{\pi /2} {\sin x.\cos y\,f(y)\,dy} $$, is : | [{"identifier": "A", "content": "$$x + {2 \\over 3}(\\pi - 2)\\sin x$$"}, {"identifier": "B", "content": "$$x + (\\pi + 2)\\sin x$$"}, {"identifier": "C", "content": "$$x + {\\pi \\over 2}\\sin x$$"}, {"identifier": "D", "content": "$$x + (\\pi - 2)\\sin x$$"}] | ["D"] | null | $$f(x) = x + \int\limits_0^{\pi /2} {\sin x\cos y\,f(y)\,dy} $$<br><br>$$f(x) = x + sinx\underbrace {\int_0^{\pi /2} {\cos y\,f(y)\,dy} }_K$$<br><br>$$ \Rightarrow f(x) = x + K\sin x$$<br><br>$$ \Rightarrow f(y) = y + K\sin y$$<br><br>Now, $$K = \int_0^{\pi /2} {\mathop {y\cos y\,dy}\limits_{Apply\,IBP} } + K\int_0^{\pi /2} {\mathop {\cos y\sin y\,dy}\limits_{Put\,\sin y = t} } $$<br><br>$$K = \left( {y\sin y} \right)_0^{\pi /2} - \int_0^{\pi /2} {\sin y dy + K\int_0^1 {t\,dt} } $$<br><br>$$ \Rightarrow K = {\pi \over 2} - 1 + K\left( {{1 \over 2}} \right)$$<br><br>$$ \Rightarrow K = \pi - 2$$<br><br>So, $$f(x) = x + (\pi - 2)\sin x$$<br><br>Option (d) | mcq | jee-main-2021-online-1st-september-evening-shift |
1l544p8qe | maths | definite-integration | properties-of-definite-integration | <p>Let $$f:R \to R$$ be a function defined by :</p>
<p>$$f(x) = \left\{ {\matrix{
{\max \,\{ {t^3} - 3t\} \,t \le x} & ; & {x \le 2} \cr
{{x^2} + 2x - 6} & ; & {2 < x < 3} \cr
{[x - 3] + 9} & ; & {3 \le x \le 5} \cr
{2x + 1} & ; & {x > 5} \cr
} } \right.$$</p>
<p>where [t] is the greatest integer less than or equal to t. Let m be the number of points where f is not differentiable and $$I = \int\limits_{ - 2}^2 {f(x)\,dx} $$. Then the ordered pair (m, I) is equal to :</p> | [{"identifier": "A", "content": "$$\\left( {3,\\,{{27} \\over 4}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( {3,\\,{{23} \\over 4}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {4,\\,{{27} \\over 4}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {4,\\,{{23} \\over 4}} \\right)$$"}] | ["C"] | null | $$
\left\{\begin{array}{l}
f(x)=x^3-3 x, x \leq-1 \\\\
2,-1 < x < 2 \\\\
x^2+2 x-6,2 < x < 3 \\\\
9,3 \leq x < 4 \\\\
10,4 \leq x < 5 \\\\
11, x=5 \\\\
2 x+1, x>5
\end{array}\right.
$$<br/><br/> Clearly $\mathrm{f}(\mathrm{x})$ is not differentiable at<br/><br/>
$$
\begin{aligned}
& x=2,3,4,5 \Rightarrow m=4 \\\\
& I=\int_{-2}^{-1}\left(x^3-3 x\right) d x+\int_{-1}^2 2 \cdot d x=\frac{27}{4}
\end{aligned}
$$ | mcq | jee-main-2022-online-29th-june-morning-shift |
1l545kvee | maths | definite-integration | properties-of-definite-integration | <p>$$\int_0^5 {\cos \left( {\pi \left( {x - \left[ {{x \over 2}} \right]} \right)} \right)dx} $$,</p>
<p>where [t] denotes greatest integer less than or equal to t, is equal to:</p> | [{"identifier": "A", "content": "$$-$$3"}, {"identifier": "B", "content": "$$-$$2"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "0"}] | ["D"] | null | <p>We know,</p>
<p>$$\left[ {{x \over 2}} \right]$$ is discontinuous at 1, 2, 3, 4 ........</p>
<p>$$\therefore$$ [ x ] is discontinuous at 2, 4, 6, 8 .....</p>
<p>In between 0 to 5 it is discontinuous at 2 and 4.</p>
<p>Break the integration into 3 parts</p>
<p>(1) 0 to 2</p>
<p>(2) 2 to 4</p>
<p>(3) 4 to 5</p>
<p>$$\therefore$$ $$\int\limits_0^5 {\cos \left( {\pi \left( {x - \left[ {{x \over 2}} \right]} \right)} \right)dx} $$</p>
<p>$$ = \int\limits_0^2 {\cos \left( {\pi (x - 0)} \right)dx + \int\limits_2^4 {\cos \left( {\pi (x - 1)} \right)dx + \int\limits_4^5 {\cos \left( {\pi (x - 2)} \right)dx} } } $$</p>
<p>$$ = \int\limits_0^2 {\cos \pi x\,dx + \int\limits_2^4 {\cos (\pi x - \pi )dx + \int\limits_4^5 {\cos (\pi x - 2\pi )dx} } } $$</p>
<p>$$ = \int\limits_0^2 {\cos \pi dx - \int\limits_2^4 {\cos \pi x\,dx + \int\limits_4^5 {\cos \pi x\,dx} } } $$</p>
<p>$$ = \left[ {{{\sin \pi x} \over \pi }} \right]_0^2 - \left[ {{{\sin \pi x} \over \pi }} \right]_2^4 + \left[ {{{\sin \pi x} \over \pi }} \right]_4^5$$</p>
<p>$$ = 0 - 0 + 0$$</p>
<p>$$ = 0$$</p> | mcq | jee-main-2022-online-29th-june-morning-shift |
1l54b0obe | maths | definite-integration | properties-of-definite-integration | <p>Let f be a real valued continuous function on [0, 1] and $$f(x) = x + \int\limits_0^1 {(x - t)f(t)dt} $$.</p>
<p>Then, which of the following points (x, y) lies on the curve y = f(x) ?</p> | [{"identifier": "A", "content": "(2, 4)"}, {"identifier": "B", "content": "(1, 2)"}, {"identifier": "C", "content": "(4, 17)"}, {"identifier": "D", "content": "(6, 8)"}] | ["D"] | null | <p>Given,</p>
<p>$$f(x) = x + \int_0^1 {(x - t)f(t)dt} $$</p>
<p>$$ = x + x\int_0^1 {f(t)dt - \int_0^1 {tf(t)dt} } $$</p>
<p>$$ = x\left( {1 + \int_0^1 {f(t)dt} } \right) - \int_0^1 {tf(t)dt} $$</p>
<p>Now,</p>
<p>let $$A = 1 + \int_0^1 {f(t)dt} $$</p>
<p>and $$B = \int_0^1 {tf(t)dt} $$</p>
<p>$$\therefore$$ $$f(x) = Ax - B$$</p>
<p>$$ \Rightarrow f(t) = At - B$$</p>
<p>So, $$A = 1 + \int_0^1 {f(t)dt} $$</p>
<p>$$ = 1 + \int_0^1 {(At - B)dt} $$</p>
<p>$$ = 1 + \left[ {{{A{t^2}} \over 2} - Bt} \right]_0^1$$</p>
<p>$$ = 1 + {A \over 2} - B$$</p>
<p>$$ \Rightarrow {A \over 2} = 1 - B$$ ...... (1)</p>
<p>$$B = \int_0^1 {tf(t)dt} $$</p>
<p>$$ = \int_0^1 {t(At - B)dt} $$</p>
<p>$$ = \int_0^1 {(A{t^2} - Bt)dt} $$</p>
<p>$$ = \left[ {{{A{t^3}} \over 3} - {{B{t^2}} \over 2}} \right]_0^1$$</p>
<p>$$ = {A \over 3} - {B \over 2}$$</p>
<p>$$ \Rightarrow {{3B} \over 2} = {A \over 3}$$ ....... (2)</p>
<p>Solving equation (1) and (2) we get,</p>
<p>$$A = {{18} \over {13}}$$ and $$B = {4 \over {13}}$$</p>
<p>$$\therefore$$ $$f(x) = {{18} \over {13}}x - {4 \over {13}}$$</p>
<p>By checking all the options you can see when x = 6 we get</p>
<p>$$y = f(x) = {{18} \over {13}} \times 6 - {4 \over {13}}$$</p>
<p>$$ = {{108 - 4} \over {13}} = 8$$</p>
<p>$$\therefore$$ Point (6, 8) lies on the curve.</p> | mcq | jee-main-2022-online-29th-june-evening-shift |
1l54b41yo | maths | definite-integration | properties-of-definite-integration | <p>If $$\int\limits_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx = \int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy + \int\limits_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy + I} } } $$, then I equals</p> | [{"identifier": "A", "content": "$$\\int\\limits_0^1 {\\left( {1 + \\sqrt {1 - {y^2}} } \\right)dy} $$"}, {"identifier": "B", "content": "$$\\int\\limits_0^1 {\\left( {{{{y^2}} \\over 2} - \\sqrt {1 - {y^2}} + 1} \\right)dy} $$"}, {"identifier": "C", "content": "$$\\int\\limits_0^1 {\\left( {1 - \\sqrt {1 - {y^2}} } \\right)dy} $$"}, {"identifier": "D", "content": "$$\\int\\limits_0^1 {\\left( {{{{y^2}} \\over 2} + \\sqrt {1 - {y^2}} + 1} \\right)dy} $$"}] | ["C"] | null | <p>Given,</p>
<p>$$\int_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx = \int_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy + \int_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy + I} } } $$</p>
<p>Now,</p>
<p>$$L.H.S. = \int_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx} $$</p>
<p>$$ = \sqrt 2 \int_0^2 {{x^{1/2}}dx - \int_0^2 {\sqrt { - ({x^2} - 2x)} dx} } $$</p>
<p>$$ = \sqrt 2 \left[ {{{{x^{3/2}}} \over {{3 \over 2}}}} \right]_0^2 - \int_0^2 {\sqrt { - \left[ {{{(x - 1)}^2} - 1} \right]} dx} $$</p>
<p>$$ = \sqrt 2 \times {2 \over 3}\left[ {{2^{{3 \over 2}}}} \right] - \int_0^2 {\sqrt {1 - {{(x - 1)}^2}} dx} $$</p>
<p>[We know,</p>
<p>$$\sqrt {{a^2} - {x^2}} = {x \over 2}\sqrt {{a^2} - {x^2}} + {{{a^2}} \over 2}{\sin ^{ - 1}}\left( {{x \over a}} \right) + C$$]</p>
<p>$$ = {{2\sqrt 2 \times 2\sqrt 2 } \over 3} - \left[ {{{x - 1} \over 2}\sqrt {1 - {{(x - 1)}^2}} + {1 \over 2}{{\sin }^{ - 1}}\left( {{{x - 1} \over 1}} \right)} \right]_0^2$$</p>
<p>$$ = {8 \over 3} - \left[ {{1 \over 2}\sqrt {1 - 1} + {1 \over 2}{{\sin }^{ - 1}}(1) - \left( {\left( {{1 \over 2}} \right)\sqrt {1 - 1} + {1 \over 2}{{\sin }^{ - 1}}( - 1)} \right)} \right]$$</p>
<p>$$ = {8 \over 3} - {1 \over 2}{\sin ^{ - 1}}\left( {\sin {\pi \over 2}} \right) + {1 \over 2}{\sin ^{ - 1}}\left( {\sin {{3\pi } \over 2}} \right)$$</p>
<p>$$ = {8 \over 3} - {1 \over 2}\,.\,{\pi \over 2} + {1 \over 2}\,.\,{{3\pi } \over 2}$$</p>
<p>$$ = {8 \over 3} - {\pi \over 4} + {{3\pi } \over 4}$$</p>
<p>$$ = {8 \over 3} - {{2\pi } \over 4}$$</p>
<p>$$ = {8 \over 3} - {\pi \over 2}$$</p>
<p>Now in R.H.S.,</p>
<p>$$\int_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy} $$</p>
<p>$$ = \int_0^1 {dy - \int_0^1 {\sqrt {1 - {y^2}} - \int_0^1 {{{{y^2}} \over 2}dy} } } $$</p>
<p>$$ = \left[ y \right]_0^1 - \left[ {{y \over 2}\sqrt {1 - {y^2}} + {1 \over 2}{{\sin }^{ - 1}}\left( {{y \over 1}} \right)} \right]_0^1 - \left[ {{{{y^3}} \over 6}} \right]_0^1$$</p>
<p>$$ = 1 - \left[ {\left( {{1 \over 2}\sqrt {1 - 0} + {1 \over 2}{{\sin }^{ - 1}}(1)} \right) - \left( {0 + {{\sin }^{ - 1}}(0)} \right)} \right] - {1 \over 6}$$</p>
<p>$$ = 1 - {1 \over 2}{\sin ^{ - 1}}\left( {\sin {\pi \over 2}} \right) - {1 \over 6}$$</p>
<p>$$ = 1 - {1 \over 2} \times {\pi \over 2} - {1 \over 6}$$</p>
<p>$$ = {5 \over 6} - {\pi \over 4}$$</p>
<p>Now,</p>
<p>$$\int_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy} $$</p>
<p>$$ = \left[ {2y - {{{y^3}} \over 6}} \right]_1^2$$</p>
<p>$$ = \left[ {\left( {4 - {8 \over 6}} \right) - \left( {2 - {1 \over 6}} \right)} \right]$$</p>
<p>$$ = 4 - {8 \over 6} - 2 + {1 \over 6}$$</p>
<p>$$ = 2 - {7 \over 6}$$</p>
<p>$$ = {5 \over 6}$$</p>
<p>$$\therefore$$ $$R.H.S. = {5 \over 6} - {\pi \over 4} + {5 \over 6} + I$$</p>
<p>As $$L.H.S. = R.H.S.$$</p>
<p>$$ \Rightarrow {8 \over 3} - {\pi \over 2} = {{10} \over 6} - {\pi \over 4} + I$$</p>
<p>$$ \Rightarrow {{10} \over 6} - {8 \over 3} + I = - {\pi \over 2} + {\pi \over 4}$$</p>
<p>$$ \Rightarrow {{ - 6} \over 6} + I = - {\pi \over 4}$$</p>
<p>$$ \Rightarrow I = 1 - {\pi \over 4}$$</p>
<p>From option (c)</p>
<p>$$\int_0^1 {(1 - \sqrt {1 - {y^2}} )dy} $$</p>
<p>$$ = \left[ y \right]_0^1 - \int_0^1 {(\sqrt {1 - {y^2}} )dy} $$</p>
<p>$$ = 1 - \left[ {{y \over 2}\sqrt {1 - {y^2}} + {1 \over 2}{{\sin }^{ - 1}}\left( {{y \over 1}} \right)} \right]_0^1$$</p>
<p>$$ = 1 - \left[ {{1 \over 2}\sqrt {1 - 1} + {1 \over 2}{{\sin }^{ - 1}}(1) - \left( {0 + {1 \over 2}{{\sin }^{ - 1}}(0)} \right)} \right]$$</p>
<p>$$ = 1 - \left[ {{1 \over 2} \times {\pi \over 2} - 0} \right]$$</p>
<p>$$ = 1 - {\pi \over 4} = I$$</p>
<p>$$\therefore$$ Option (c) is the right answer.</p>
<p>Note :</p>
<p>There is no way to guess which option is correct. You have to check all the options to see which give value equal to I.</p> | mcq | jee-main-2022-online-29th-june-evening-shift |
1l55hdycb | maths | definite-integration | properties-of-definite-integration | <p>Let f : R $$\to$$ R be a differentiable function such that $$f\left( {{\pi \over 4}} \right) = \sqrt 2 ,\,f\left( {{\pi \over 2}} \right) = 0$$ and $$f'\left( {{\pi \over 2}} \right) = 1$$ and <br/><br/>let $$g(x) = \int_x^{\pi /4} {(f'(t)\sec t + \tan t\sec t\,f(t))\,dt} $$ for $$x \in \left[ {{\pi \over 4},{\pi \over 2}} \right)$$. Then $$\mathop {\lim }\limits_{x \to {{\left( {{\pi \over 2}} \right)}^ - }} g(x)$$ is equal to :</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "$$-$$3"}] | ["B"] | null | Given : $f\left(\frac{\pi}{4}\right)=\sqrt{2}, f\left(\frac{\pi}{2}\right)=0$ and $f^{\prime}\left(\frac{\pi}{2}\right)=1$
<br/><br/>
$$
\begin{aligned}
&g(x)=\int_{x}^{\frac{\pi}{4}}\left(f^{\prime}(t) \sec t+\tan t \sec t f(t)\right) d t \\\\
&=[\sec t+f(t)]_{x}^{\frac{\pi}{4}}=2-\sec x f(x)
\end{aligned}
$$
<br/><br/>
Now, $\lim \limits_{x \rightarrow \frac{\pi^{-}}{2}} g(x)=\lim \limits_{h \rightarrow 0} g\left(\frac{\pi}{2}-h\right)$
<br/><br/>
$$
=\lim \limits_{h \rightarrow 0} 2-(\operatorname{cosec} h) f\left(\frac{\pi}{2}-h\right)
$$
<br/><br/>
$=\lim \limits_{h \rightarrow 0}\left[2-\frac{f\left(\frac{\pi}{2}-h\right)}{\sin h}\right]$
<br/><br/>
$=\lim \limits_{h \rightarrow 0}\left[2+\frac{f^{\prime}\left(\frac{\pi}{2}-h\right)}{\cos h}\right]$
<br/><br/>
$$
=3
$$ | mcq | jee-main-2022-online-28th-june-evening-shift |
1l55hht4t | maths | definite-integration | properties-of-definite-integration | <p>Let f : R $$\to$$ R be a continuous function satisfying f(x) + f(x + k) = n, for all x $$\in$$ R where k > 0 and n is a positive integer. If $${I_1} = \int\limits_0^{4nk} {f(x)dx} $$ and $${I_2} = \int\limits_{ - k}^{3k} {f(x)dx} $$, then :</p> | [{"identifier": "A", "content": "$${I_1} + 2{I_2} = 4nk$$"}, {"identifier": "B", "content": "$${I_1} + 2{I_2} = 2nk$$"}, {"identifier": "C", "content": "$${I_1} + n{I_2} = 4{n^2}k$$"}, {"identifier": "D", "content": "$${I_1} + n{I_2} = 6{n^2}k$$"}] | ["C"] | null | $f: R \rightarrow R$ and $f(x)+f(x+k)=n \quad \forall x \in R$
<br/><br/>
$$
\begin{aligned}
&x \rightarrow x+k \\\\
&f(x+k)+f(x+2 k)=n \\\\
&\therefore \quad f(x+2 k)=f(x)
\end{aligned}
$$
<br/><br/>
So, period of $f(x)$ is $2 k$
<br/><br/>
$$
\begin{aligned}
&\text { Now, } I_{1}=\int_{0}^{4 n k} f(x) d x = 2 n \int_{0}^{2 k} f(x) d x \\\\
&=2 n\left[\int_{0}^{k} f(x) d x+\int_{k}^{2 k} f(x) d x\right] \\\\
&x=t+k \Rightarrow d x=d t \text { (in second integral) } \\\\
&=2 n\left[\int_{0}^{k} f(x) d x+\int_{0}^{k} f(t+k) d t\right] \\\\
&=2 n^{2} k
\end{aligned}
$$
<br/><br/>
Now, $I_2=\int_{-k}^{3 k} f(x) d x=2 \int_{0}^{2 k} f(x) d x$
<br/><br/>
$$
I_{2}=2(n k)
$$
<br/><br/>
$\therefore \quad l_{1}+n l_{2}=4 n^{2} k$ | mcq | jee-main-2022-online-28th-june-evening-shift |
1l566dpzj | maths | definite-integration | properties-of-definite-integration | <p>Let [t] denote the greatest integer less than or equal to t. Then, the value of the integral $$\int\limits_0^1 {[ - 8{x^2} + 6x - 1]dx} $$ is equal to :</p> | [{"identifier": "A", "content": "$$-$$1"}, {"identifier": "B", "content": "$${{ - 5} \\over 4}$$"}, {"identifier": "C", "content": "$${{\\sqrt {17} - 13} \\over 8}$$"}, {"identifier": "D", "content": "$${{\\sqrt {17} - 16} \\over 8}$$"}] | ["C"] | null | $\int_{0}^{1}\left[-8 x^{2}+6 x-1\right] d x$<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lc8eip72/1a7f4845-0be6-46f8-ac5d-62ddcb930a44/438942e0-8717-11ed-b3ec-0bde88094e1e/file-1lc8eip73.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lc8eip72/1a7f4845-0be6-46f8-ac5d-62ddcb930a44/438942e0-8717-11ed-b3ec-0bde88094e1e/file-1lc8eip73.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 - Definite Integration Question 123 English Explanation"><br>
$=\int_{0}^{\frac{1}{4}}(-1) d x+\int_{\frac{1}{4}}^{\frac{3}{4}} 0 d x+\int_{\frac{1}{2}}^{\frac{3}{4}}-1 d x+\int_{\frac{3}{4}}^{8}-2 d x+\int_{\frac{3+\sqrt{17}}{8}}^{1}-3 d x$<br><br> $=-\frac{1}{4}-\frac{1}{4}-2\left(\frac{3+\sqrt{17}}{8}-\frac{3}{4}\right)-3\left(1-\frac{3+\sqrt{17}}{8}\right)$<br><br> $=\frac{\sqrt{17}-13}{8}$ | mcq | jee-main-2022-online-28th-june-morning-shift |
1l56r4q3p | maths | definite-integration | properties-of-definite-integration | <p>The integral $$\int\limits_0^1 {{1 \over {{7^{\left[ {{1 \over x}} \right]}}}}dx} $$, where [ . ] denotes the greatest integer function, is equal to</p> | [{"identifier": "A", "content": "$$1 + 6{\\log _e}\\left( {{6 \\over 7}} \\right)$$"}, {"identifier": "B", "content": "$$1 - 6{\\log _e}\\left( {{6 \\over 7}} \\right)$$"}, {"identifier": "C", "content": "$${\\log _e}\\left( {{7 \\over 6}} \\right)$$"}, {"identifier": "D", "content": "$$1 - 7{\\log _e}\\left( {{6 \\over 7}} \\right)$$"}] | ["A"] | null | <p>$$\int_0^1 {{1 \over {{7^{\left[ {{1 \over x}} \right]}}}}dx} $$, let $${1 \over x} = t$$</p>
<p>$${{ - 1} \over {{x^2}}}dx = dt$$</p>
<p>$$ = \int_\infty ^1 {{1 \over { - {t^2}{7^{[t]}}}}dt = \int_1^\infty {{1 \over {{t^2}{7^{[t]}}}}dt} } $$</p>
<p>$$ = \int_1^2 {{1 \over {7{t^2}}}dt + \int_2^3 {{1 \over {{7^2}{t^2}}}dt + \,\,....} } $$</p>
<p>$$ = {1 \over 7}\left[ { - {1 \over t}} \right]_1^2 + {1 \over {{7^2}}}\left[ {{{ - 1} \over t}} \right]_2^3 + {1 \over {{7^3}}}\left[ { - {1 \over t}} \right]_2^3 + \,\,....$$</p>
<p>$$ = \sum\limits_{n = 1}^\infty {{1 \over {{7^n}}}\left( {{1 \over n} - {1 \over {n + 1}}} \right)} $$</p>
<p>$$ = \sum\limits_{n = 1}^\infty {{{{{\left( {{1 \over 7}} \right)}^n}} \over n} - 7\sum\limits_{n = 1}^\infty {{{{{\left( {{1 \over 7}} \right)}^{n + 1}}} \over {n + 1}}} } $$</p>
<p>$$ = - \log \left( {1 - {1 \over 7}} \right) + 7\log \left( {1 - {1 \over 7}} \right) + 1$$</p>
<p>$$ = 1 + 6\log {6 \over 7}$$</p> | mcq | jee-main-2022-online-27th-june-evening-shift |
1l57oc7k9 | maths | definite-integration | properties-of-definite-integration | <p>The value of the integral <br/><br/>$$\int\limits_{ - 2}^2 {{{|{x^3} + x|} \over {({e^{x|x|}} + 1)}}dx} $$ is equal to :</p> | [{"identifier": "A", "content": "5e<sup>2</sup>"}, {"identifier": "B", "content": "3e<sup>$$-$$2</sup>"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "6"}] | ["D"] | null | <p>$$I = \int\limits_{ - 2}^2 {{{|{x^3} + x|} \over {{e^{x|x|}} + 1}}dx} $$ ..... (i)</p>
<p>$$I = \int\limits_{ - 2}^2 {{{|{x^3} + x|} \over {{e^{ - x|x|}} + 1}}dx} $$ ..... (ii)</p>
<p>$$2I = \int\limits_{ - 2}^2 {|{x^3} + x|dx} $$</p>
<p>$$2I =2 \int\limits_0^2 {({x^3} + x)dx} $$</p>
<p>$$I = \int\limits_0^2 {({x^3} + x)dx} $$</p>
<p>$$ = \left. {{{{x^4}} \over 4} + {{{x^2}} \over 2}} \right]_0^2$$</p>
<p>$$ = \left( {{{16} \over 4} + {4 \over 2}} \right) - 0$$</p>
<p>$$ = 4 + 2 = 6$$</p> | mcq | jee-main-2022-online-27th-june-morning-shift |
1l58ai6jr | maths | definite-integration | properties-of-definite-integration | <p>Let f(x) = max {|x + 1|, |x + 2|, ....., |x + 5|}. Then $$\int\limits_{ - 6}^0 {f(x)dx} $$ is equal to __________.</p> | [] | null | 21 | <p>For $$\left| {x + 1} \right|$$ critical point, x + 1 = 0 $$\Rightarrow$$ x = $$-$$1</p>
<p>For $$\left| {x + 2} \right|$$ critical point, x + 2 = 0 $$\Rightarrow$$ x = $$-$$2</p>
<p>For $$\left| {x + 3} \right|$$ critical point, x + 3 = 0 $$\Rightarrow$$ x = $$-$$3</p>
<p>For $$\left| {x + 4} \right|$$ critical point, x + 4 = 0 $$\Rightarrow$$ x = $$-$$4</p>
<p>For $$\left| {x + 5} \right|$$ critical point, x + 5 = 0 $$\Rightarrow$$ x = $$-$$5</p>
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5e62gr9/169b6319-b73d-43b9-8dcb-7f912dd4450a/67482850-ffad-11ec-94de-eb2a82e25f79/file-1l5e62gra.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5e62gr9/169b6319-b73d-43b9-8dcb-7f912dd4450a/67482850-ffad-11ec-94de-eb2a82e25f79/file-1l5e62gra.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 60vh" alt="JEE Main 2022 (Online) 26th June Morning Shift Mathematics - Definite Integration Question 117 English Explanation"> </p>
<p>Here maximum function is represent by the dotted line.</p>
<p>$$\therefore$$ Point of intersection A of line y = $$-$$x $$-$$1 and y = x + 5 :</p>
<p>$$ - x - 1 = x + 5$$</p>
<p>$$ \Rightarrow 2x = - 6$$</p>
<p>$$ \Rightarrow x = - 3$$</p>
<p>$$\therefore$$ $$y = - 3 + 5 = 2$$</p>
<p>$$\therefore$$ Point $$A = ( - 3,2)$$</p>
<p>$$\therefore$$ $$\int_{ - 6}^0 {f(x)dx} $$</p>
<p>$$ = \int_{ - 6}^{ - 3} {( - x - 1)dx + \int_{ - 3}^0 {(x + 5)dx} } $$</p>
<p>$$ = \left( { - {{{x^2}} \over 2} - x} \right)_{ - 6}^{ - 3} + \left[ {{{{x^2}} \over 2} + 5x} \right]_{ - 3}^0$$</p>
<p>$$ = \left[ {\left( { - {9 \over 2} + 3} \right) - \left( { - {{36} \over 2} + 6} \right)} \right] + \left[ {0 - \left( {{9 \over 2} - 15} \right)} \right]$$</p>
<p>$$ = \left( { - {3 \over 2} + 12} \right) + {{21} \over 2}$$</p>
<p>$$ = + {{21} \over 2} + {{21} \over 2}$$</p>
<p>$$ = 21$$</p> | integer | jee-main-2022-online-26th-june-morning-shift |
1l58aq8wy | maths | definite-integration | properties-of-definite-integration | <p>The value of the integral <br/><br/>$${{48} \over {{\pi ^4}}}\int\limits_0^\pi {\left( {{{3\pi {x^2}} \over 2} - {x^3}} \right){{\sin x} \over {1 + {{\cos }^2}x}}dx} $$ is equal to __________.</p> | [] | null | 6 | <p>$$I = {{48} \over {{\pi ^4}}}\int_0^\pi {\left[ {{{\left( {{\pi \over 2} - x} \right)}^3} - {{3{\pi ^2}} \over 4}\left( {{\pi \over 2} - x} \right) + {{{\pi ^3}} \over 4}} \right]{{\sin xdx} \over {1 + {{\cos }^2}x}}} $$</p>
<p>Using $$\int_a^b {f(x)dx = \int_a^b {f(a + b - x)dx} } $$</p>
<p>$$I = {{48} \over {{\pi ^4}}}\int_0^\pi {\left[ { - {{\left( {{\pi \over 2} - x} \right)}^3} + {{3{\pi ^4}} \over 4}\left( {{\pi \over 2} - x} \right) + {{{\pi ^3}} \over 4}} \right]{{\sin xdx} \over {1 + {{\cos }^2}x}}} $$</p>
<p>Adding these two equations, we get</p>
<p>$$2I = {{48} \over {{\pi ^4}}}\int_0^\pi {{{{\pi ^3}} \over 2}\,.\,{{\sin xdx} \over {1 + {{\cos }^2}x}}} $$</p>
<p>$$ \Rightarrow I = {{12} \over \pi }\left[ { - {{\tan }^{ - 1}}(\cos x)} \right]_0^\pi = {{12} \over \pi }\,.\,{\pi \over 2} = 6$$</p> | integer | jee-main-2022-online-26th-june-morning-shift |
1l58h6qb5 | maths | definite-integration | properties-of-definite-integration | <p>The integral $${{24} \over \pi }\int_0^{\sqrt 2 } {{{(2 - {x^2})dx} \over {(2 + {x^2})\sqrt {4 + {x^4}} }}} $$ is equal to ____________.</p> | [] | null | 3 | <p>$$I = {{24} \over \pi }\int_0^{\sqrt 2 } {{{2 - {x^2}} \over {(2 + {x^2})\sqrt {4 + {x^4}} }}dx} $$</p>
<p>Let $$x = \sqrt 2 t \Rightarrow dx = \sqrt 2 dt$$</p>
<p>$$I = {{24} \over \pi }\int_0^1 {{{(2 - 2{t^2})\,.\,\sqrt 2 dt} \over {(2 + 2{t^2})\sqrt {4 + 4{t^4}} }}} $$</p>
<p>$$ = {{12\sqrt 2 } \over \pi }\int_0^1 {{{\left( {{1 \over {{t^2}}} - 1} \right)dt} \over {\left( {t + {1 \over t}} \right)\sqrt {{{\left( {t + {1 \over t}} \right)}^2} - 2} }}} $$</p>
<p>Let $$t + {1 \over t} = u$$</p>
<p>$$ \Rightarrow \left( {1 - {1 \over {{t^2}}}} \right)dt = du$$</p>
<p>$$ = {{12\sqrt 2 } \over \pi }\int_\infty ^2 {{{ - du} \over {u\sqrt {{4^2} - 2} }}} $$</p>
<p>$$ = {{12\sqrt 2 } \over \pi }\int_2^\infty {{{du} \over {{u^2}\sqrt { - {{\left( {{{\sqrt 2 } \over u}} \right)}^2}} }}} $$</p>
<p>$$ = {{12\sqrt 2 } \over \pi }\int_{{1 \over {\sqrt 2 }}}^0 {{{ - {1 \over {\sqrt 2 }}dp} \over {\sqrt {1 - {p^2}} }}} $$</p>
<p>$$ = {{12} \over \pi }\left[ {{{\sin }^{ - 1}}p} \right]_0^{{1 \over {\sqrt 2 }}}$$</p>
<p>$$ = {{12} \over \pi }\,.\,{\pi \over 4} = 3$$</p> | integer | jee-main-2022-online-26th-june-evening-shift |
1l59kj9rb | maths | definite-integration | properties-of-definite-integration | <p>If $${b_n} = \int_0^{{\pi \over 2}} {{{{{\cos }^2}nx} \over {\sin x}}dx,\,n \in N} $$, then</p> | [{"identifier": "A", "content": "$${b_3} - {b_2},\\,{b_4} - {b_3},\\,{b_5} - {b_4}$$ are in A.P. with common difference $$-$$2"}, {"identifier": "B", "content": "$${1 \\over {{b_3} - {b_2}}},{1 \\over {{b_4} - {b_3}}},{1 \\over {{b_5} - {b_4}}}$$ are in an A.P. with common difference 2 "}, {"identifier": "C", "content": "$${b_3} - {b_2},\\,{b_4} - {b_3},\\,{b_5} - {b_4}$$ are in a G.P."}, {"identifier": "D", "content": "$${1 \\over {{b_3} - {b_2}}},{1 \\over {{b_4} - {b_3}}},{1 \\over {{b_5} - {b_4}}}$$ are in an A.P. with common difference $$-$$2"}] | ["D"] | null | <p>$${b_n} - {b_{n - 1}} = \int_0^{{\pi \over 2}} {{{{{\cos }^2}nx - {{\cos }^2}(n - 1)x} \over {\sin x}}dx} $$</p>
<p>$$ = \int_0^{{\pi \over 2}} {{{ - \sin (2n - 1)x\,.\,\sin x} \over {\sin x}}dx} $$</p>
<p>$$ = \left. {{{\cos (2n - 1)x} \over {2n - 1}}} \right|_0^{\pi /2} = - {1 \over {2n - 1}}$$</p>
<p>So, $${b_3} - {b_2}$$, $${b_4} - {b_3}$$, $${b_5} - {b_4}$$ are in H.P.</p>
<p>$$ \Rightarrow {1 \over {{b_3} - {b_2}}},\,{1 \over {{b_4} - {b_3}}},\,{1 \over {{b_5} - {b_4}}}$$ are in A.P. with common difference $$-$$2.</p> | mcq | jee-main-2022-online-25th-june-evening-shift |
1l59l6mpi | maths | definite-integration | properties-of-definite-integration | <p>The value of b > 3 for which $$12\int\limits_3^b {{1 \over {({x^2} - 1)({x^2} - 4)}}dx = {{\log }_e}\left( {{{49} \over {40}}} \right)} $$, is equal to ___________.</p> | [] | null | 6 | <p>$$I = \int {{1 \over {({x^2} - 1)({x^2} - 4)}}dx = {1 \over 3}\int {\left( {{1 \over {{x^2} - 4}} - {1 \over {{x^2} - 1}}} \right)dx} } $$</p>
<p>$$ = {1 \over 3}\left( {{1 \over 4}\ln \left| {{{x - 2} \over {x + 2}}} \right| - {1 \over 2}\ln \left| {{{x - 1} \over {x + 1}}} \right|} \right) + C$$</p>
<p>$$12I = \ln \left| {{{x - 2} \over {x + 2}}} \right| + 2\ln \left| {{{x - 1} \over {x + 1}}} \right| + C$$</p>
<p>$$12\int\limits_3^b {{{dx} \over {({x^2} - 4)({x^2} - 1)}}} $$</p>
<p>$$ = \ln \left( {{{b - 2} \over {b + 2}}} \right) - 2\ln \left( {{{b - 1} \over {b + 1}}} \right) - \left( {\ln \left( {{1 \over 5}} \right) - 2\ln \left( {{1 \over 2}} \right)} \right)$$</p>
<p>$$ = \ln \left( {\left( {{{b - 2} \over {b + 2}}} \right)\,.\,{{{{(b + 1)}^2}} \over {{{(b - 1)}^2}}}} \right) - \left( {\ln {4 \over 5}} \right)$$</p>
<p>So, $${{49} \over {40}} = {{(b - 2)} \over {(b + 2)}}{{{{(b + 1)}^2}} \over {{{(b - 1)}^2}}}\,.\,{5 \over 4}$$</p>
<p>$$ \Rightarrow b = 6$$</p> | integer | jee-main-2022-online-25th-june-evening-shift |
1l5ahmkvl | maths | definite-integration | properties-of-definite-integration | <p>The value of $$\int\limits_0^\pi {{{{e^{\cos x}}\sin x} \over {(1 + {{\cos }^2}x)({e^{\cos x}} + {e^{ - \cos x}})}}dx} $$ is equal to:</p> | [{"identifier": "A", "content": "$${{{\\pi ^2}} \\over 4}$$"}, {"identifier": "B", "content": "$${{{\\pi ^2}} \\over 2}$$"}, {"identifier": "C", "content": "$${\\pi \\over 4}$$"}, {"identifier": "D", "content": "$${\\pi \\over 2}$$"}] | ["C"] | null | <p>$$\int\limits_0^\pi {{{{e^{\cos x}}\sin x} \over {\left( {1 + {{\cos }^2}x} \right)\left( {{e^{\cos x}} + {e^{ - \cos x}}} \right)}}dx} $$</p>
<p>Let $$\cos x = t$$</p>
<p>$$\sin xdx = dt$$</p>
<p>$$ = \int\limits_1^{ - 1} {{{ - {e^t}dt} \over {\left( {1 + {t^2}} \right)\left( {{e^t} + {e^{ - t}}} \right)}}} $$</p>
<p>$$I = \int\limits_{ - 1}^1 {{{{e^t}} \over {\left( {1 + {t^2}} \right)\left( {{e^t} + {e^{ - t}}} \right)}}dt} $$ ...... (i)</p>
<p>$$I = \int\limits_{ - 1}^1 {{{{e^{ - t}}} \over {\left( {1 + {t^2}} \right)\left( {{e^{ - t}} + {e^t}} \right)}}dt} $$ .... (ii)</p>
<p>Adding (i) and (ii)</p>
<p>$$2I = \int\limits_{ - 1}^1 {{{dt} \over {1 + {t^2}}}} $$</p>
<p>$$2I = \left. {{{\tan }^{ - t}}} \right|_{ - 1}^1$$</p>
<p>$$2I = {\pi \over 4} - \left( { - {\pi \over 4}} \right)$$</p>
<p>$$2I = {\pi \over 2}$$</p>
<p>$$I = {\pi \over 4}$$</p> | mcq | jee-main-2022-online-25th-june-morning-shift |
1l5b8114c | maths | definite-integration | properties-of-definite-integration | <p>The value of the integral <br/><br/>$$\int\limits_{ - \pi /2}^{\pi /2} {{{dx} \over {(1 + {e^x})({{\sin }^6}x + {{\cos }^6}x)}}} $$ is equal to</p> | [{"identifier": "A", "content": "2$$\\pi$$"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "$$\\pi$$"}, {"identifier": "D", "content": "$${\\pi \\over 2}$$"}] | ["C"] | null | <p>$$I = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {(1 + {e^x})({{\sin }^6}x + {{\cos }^6}x)}}} $$ ...... (i)</p>
<p>$$I = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {(1 + {e^{ - x}})(si{n^6}x + {{\cos }^6}x)}}} $$ ..... (ii)</p>
<p>(i) and (ii)</p>
<p>From equation (i) & (ii)</p>
<p>$$2I = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {{{\sin }^6}x + {{\cos }^6}x}}} $$</p>
<p>$$ \Rightarrow I = \int\limits_0^{{\pi \over 2}} {{{dx} \over {{{\sin }^6}x + {{\cos }^6}x}} = \int\limits_0^{{\pi \over 2}} {{{dx} \over {1 - {3 \over 4}{{\sin }^2}2x}}} } $$</p>
<p>$$ \Rightarrow I = \int\limits_0^{{\pi \over 2}} {{{4{{\sec }^2}2xdx} \over {4 + {{\tan }^2}2x}} = 2\int\limits_0^{{\pi \over 4}} {{{4{{\sec }^2}2x} \over {4 + {{\tan }^2}2x}}dx} } $$</p>
<p>when x = 0, t = 0</p>
<p>Now, $$\tan 2x = t$$</p>
<p>when $$x = {\pi \over 4},\,\,t \to \infty $$</p>
<p>$$2{\sec ^2}2x\,dx = dt$$</p>
<p>$$\therefore$$ $$I = 2\int\limits_0^\infty {{{2dt} \over {4 + {t^2}}} = 2\left( {{{\tan }^{ - 1}}{t \over 2}} \right)_0^\infty } $$</p>
<p>$$ = 2{\pi \over 2} = \pi $$</p> | mcq | jee-main-2022-online-24th-june-evening-shift |
1l5c2bxms | maths | definite-integration | properties-of-definite-integration | <p>Let $$f(\theta ) = \sin \theta + \int\limits_{ - \pi /2}^{\pi /2} {(\sin \theta + t\cos \theta )f(t)dt} $$. Then the value of $$\left| {\int_0^{\pi /2} {f(\theta )d\theta } } \right|$$ is _____________.</p> | [] | null | 1 | $f(\theta)=\sin \theta\left(1+\int_{-\pi / 2}^{\pi / 2} f(t) d t\right)+\cos \theta\left(\int_{-\pi / 2}^{\pi / 2} t f(t) d t\right)$
<br/><br/>
Clearly $f(\theta)=a \sin \theta+b \cos \theta$
<br/><br/>
Where $a=1+\int_{-\pi / 2}^{\pi / 2}(a \sin t+b \cos t) d t \Rightarrow a=1+2 b\quad\quad...(i)$
<br/><br/>
and $b=\int_{-\pi / 2}^{\pi / 2}(a t \sin t+b t \cos t) d t \Rightarrow b=2 a\quad\quad...(ii)$
<br/><br/>
from (i) and (ii) we get
<br/><br/>
$$
a=-\frac{1}{3} \text { and } b=-\frac{2}{3}
$$
<br/><br/>
So $f(\theta)=-\frac{1}{3}(\sin \theta+2 \cos \theta)$
<br/><br/>
$$
\Rightarrow\left|\int_{0}^{\pi / 2} f(\theta) d \theta\right|=\frac{1}{3}(1+2 \times 1)=1
$$ | integer | jee-main-2022-online-24th-june-morning-shift |
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