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Differentiation

JEE Mathematics · 66 questions · Page 4 of 7 · Click an option or "Show Solution" to reveal answer

Q31
Let f(x)=cos⁡(2tan⁡−1sin⁡(cot⁡−11−xx))f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right), 0 < x < 1. Then :
A (1−x)2f′(x)−2(f(x))2=0{(1 - x)^2}f'(x) - 2{(f(x))^2} = 0
B (1+x)2f′(x)+2(f(x))2=0{(1 + x)^2}f'(x) + 2{(f(x))^2} = 0
C (1−x)2f′(x)+2(f(x))2=0{(1 - x)^2}f'(x) + 2{(f(x))^2} = 0
D (1+x)2f′(x)−2(f(x))2=0{(1 + x)^2}f'(x) - 2{(f(x))^2} = 0
Correct Answer
Option C
Solution
f(x)=cos⁡(2tan⁡−1sin⁡(cot⁡−11−xx))f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)
cot⁡−11−xx=sin⁡−1x{\cot ^{ - 1}}\sqrt {{{1 - x} \over x}} = {\sin ^{ - 1}}\sqrt x

or

f(x)=cos⁡(2tan⁡−1x)f(x) = \cos (2{\tan ^{ - 1}}\sqrt x )
=cos⁡tan⁡−1(2x1−x)= \cos {\tan ^{ - 1}}\left( {{{2\sqrt x } \over {1 - x}}} \right)
f(x)=1−x1+xf(x) = {{1 - x} \over {1 + x}}

Now,

f′(x)=−2(1+x)2f'(x) = {{ - 2} \over {{{(1 + x)}^2}}}

or

f′(x)(1−x)2=−2(1−x1+x)2f'(x){(1 - x)^2} = - 2{\left( {{{1 - x} \over {1 + x}}} \right)^2}

or

(1−x)2f′(x)+2(f(x))2=0{(1 - x)^2}f'(x) + 2{(f(x))^2} = 0

.

Q32
If y(x)=cot⁡−1(1+sin⁡x+1−sin⁡x1+sin⁡x−1−sin⁡x),x∈(π2,π)y(x) = {\cot ^{ - 1}}\left( {{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} } \over {\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right),x \in \left( {{\pi \over 2},\pi } \right), then dydx{{dy} \over {dx}} at x=5π6x = {{5\pi } \over 6} is :
A −12 - {1 \over 2}
B −-1
C 12{1 \over 2}
D 0
Correct Answer
Option A
Solution

We have,

y(x)=cot⁡−1(1+sin⁡x+1−sin⁡x1+sin⁡x−1−sin⁡x)y(x)=\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)
=cot⁡−1∣cos⁡x2+sin⁡x2∣+∣cos⁡x2−sin⁡x2∣∣cos⁡x2+sin⁡x2∣−∣cos⁡x2−sin⁡x2∣=\cot ^{-1} \frac{\left|\cos \frac{x}{2}+\sin \frac{x}{2}\right|+\left|\cos \frac{x}{2}-\sin \frac{x}{2}\right|}{\left|\cos \frac{x}{2}+\sin \frac{x}{2}\right|-\left|\cos \frac{x}{2}-\sin \frac{x}{2}\right|}
[ as cos⁡2x2+sin⁡2x2=1\left[\text{ as } \cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}=1\right.

and sin⁡x=2sin⁡x2cos⁡x2]\left.\sin x=2 \sin \dfrac{x}{2} \cos \dfrac{x}{2}\right]

=cot⁡−1(cos⁡x2+sin⁡x2+sin⁡x2−cos⁡x2cos⁡x2+sin⁡x2−sin⁡x2+cos⁡x2)∀x∈(π2,π)\begin{aligned} & =\cot ^{-1}\left(\frac{\cos \frac{x}{2}+\sin \frac{x}{2}+\sin \frac{x}{2}-\cos \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}-\sin \frac{x}{2}+\cos \frac{x}{2}}\right) \forall x \in\left(\frac{\pi}{2}, \pi\right) \\ & \end{aligned}
=cot⁡−1(sin⁡x2cos⁡x2)=cot⁡−1(tan⁡x2)=\cot ^{-1}\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right)=\cot ^{-1}\left(\tan \frac{x}{2}\right)
=π2−tan⁡−1(tan⁡x2)=\frac{\pi}{2}-\tan ^{-1}\left(\tan \frac{x}{2}\right)
∴y′(x)=π2−x2⇒dydx=y′(x)=−12=y′(5π6)\begin{aligned} & \therefore y^{\prime}(x)=\frac{\pi}{2}-\frac{x}{2} \\\\ & \Rightarrow \frac{d y}{d x}=y^{\prime}(x)=-\frac{1}{2}=y^{\prime}\left(\frac{5 \pi}{6}\right) \end{aligned}
Q33
If $${\cos ^{ - 1}}\left( {{y \over 2}} \right) = {\log _e}{\left( {{x \over 5}} \right)^5},\,|y|
A x2y′′+xy′−25y=0{x^2}y'' + xy' - 25y = 0
B x2y′′−xy′−25y=0{x^2}y'' - xy' - 25y = 0
C x2y′′−xy′+25y=0{x^2}y'' - xy' + 25y = 0
D x2y′′+xy′+25y=0{x^2}y'' + xy' + 25y = 0
Correct Answer
Option D
Solution
cos⁡−1(y2)=log⁡e(x5)5         ∣y∣Differentiatingonbothside{\cos ^{ - 1}}\left( {{y \over 2}} \right) = {\log _e}{\left( {{x \over 5}} \right)^5}\,\,\,\,\,\,\,\,\,|y| Differentiating on both side

- {1 \over {\sqrt {1 - {{\left( {{y \over 2}} \right)}^2}} }} \times {{y'} \over 2} = {5 \over {{x \over 5}}} \times {1 \over 5}

{{ - xy'} \over 2} = 5\sqrt {1 - {{\left( {{y \over 2}} \right)}^2}}

SquareonbothsideSquare on both side

{{{x^2}y{'^2}} \over 4} = 25\left( {{{4 - {y^2}} \over 4}} \right)

DiffonbothsideDiff on both side

2xy{'^2} + 2y'y''{x^2} = - 25 \times 2yy'

xy' + y''{x^2} + 25y = 0$$

Q34
Let f : R →\to R be defined as f(x)=x3+x−5f(x) = {x^3} + x - 5. If g(x) is a function such that f(g(x))=x,∀′x′∈Rf(g(x)) = x,\forall 'x' \in R, then g'(63) is equal to ________________.
A 149{1 \over {49}}
B 349{3 \over {49}}
C 4349{43 \over {49}}
D 9149{91 \over {49}}
Correct Answer
Option A
Solution
f(x)=3x2+1f(x) = 3{x^2} + 1

f'(x) is bijective function and

f(g(x))=x⇒g(x)f(g(x)) = x \Rightarrow g(x)

is inverse of f(x)

g(f(x))=xg(f(x)) = x
g′(f(x)) . f′(x)=1g'(f(x))\,.\,f'(x) = 1
g′(f(x))=13x2+1g'(f(x)) = {1 \over {3{x^2} + 1}}

Put x = 4 we get

g′(63)=149g'(63) = {1 \over {49}}
Q35
If $$y = {\tan ^{ - 1}}\left( {\sec {x^3} - \tan {x^3}} \right),{\pi \over 2}
A xy′′+2y′=0xy'' + 2y' = 0
B x2y′′−6y+3π2=0{x^2}y'' - 6y + {{3\pi } \over 2} = 0
C x2y′′−6y+3π=0{x^2}y'' - 6y + 3\pi = 0
D xy′′−4y′=0xy'' - 4y' = 0
Correct Answer
Option B
Solution

Let

x3=θ⇒θ2∈(π4, 3π4){x^3} = \theta \Rightarrow {\theta \over 2} \in \left( {{\pi \over 4},\,{{3\pi } \over 4}} \right)

∴\therefore

y=tan⁡−1(sec⁡θ−tan⁡θ)y = {\tan ^{ - 1}}(\sec \theta - \tan \theta )
=tan⁡−1(1−sin⁡θcos⁡θ)= {\tan ^{ - 1}}\left( {{{1 - \sin \theta } \over {\cos \theta }}} \right)

∴\therefore

y=π4−θ2y = {\pi \over 4} - {\theta \over 2}
y=π4−x32y = {\pi \over 4} - {{{x^3}} \over 2}

∴\therefore

y′=−3x22y' = {{ - 3{x^2}} \over 2}
y′′=−3xy'' = - 3x

∴\therefore

x2y′′−6y+3π2=0{x^2}y'' - 6y + {{3\pi } \over 2} = 0
Q36
The value of log⁡e2ddx(log⁡cos⁡xcosec⁡x)\log _{e} 2 \dfrac{d}{d x}\left(\log _{\cos x} \operatorname{cosec} x\right) at x=π4x=\dfrac{\pi}{4} is
A −22-2 \sqrt{2}
B 222 \sqrt{2}
C −4-4
D 4
Correct Answer
Option D
Solution

Let

f(x)=log⁡cos⁡xcos⁡ec xf(x) = {\log _{\cos x}}\cos ec\,x
=log⁡cos⁡ec xlog⁡cos⁡x= {{\log \cos ec\,x} \over {\log \cos x}}
⇒f′(x)=log⁡cos⁡x . sin⁡x . (−cos⁡ec xcot⁡x−log⁡cos⁡ec x . 1cos⁡x . −sin⁡x)(log⁡cos⁡x)2\Rightarrow f'(x) = {{\log \cos x\,.\,\sin x\,.\,\left( { - \cos ec\,x\cot x - \log \cos ec\,x\,.\,{1 \over {\cos x}}\,.\, - \sin x} \right)} \over {{{(\log \cos x)}^2}}}

at

x=π4x = {\pi \over 4}
f′(π4)=−log⁡(12)+log⁡2(log⁡12)2=2log⁡2f'\left( {{\pi \over 4}} \right) = {{ - \log \left( {{1 \over {\sqrt 2 }}} \right) + \log \sqrt 2 } \over {{{\left( {\log {1 \over {\sqrt 2 }}} \right)}^2}}} = {2 \over {\log \sqrt 2 }}

∴\therefore

log⁡e2f′(x){\log _e}2f'(x)

at

x=π4=4x = {\pi \over 4} = 4
Q37
Let x(t)=22cos⁡tsin⁡2tx(t)=2 \sqrt{2} \cos t \sqrt{\sin 2 t} and y(t)=22sin⁡tsin⁡2t,t∈(0,π2)y(t)=2 \sqrt{2} \sin t \sqrt{\sin 2 t}, t \in\left(0, \dfrac{\pi}{2}\right). Then 1+(dydx)2d2ydx2\dfrac{1+\left(\dfrac{d y}{d x}\right)^{2}}{\dfrac{d^{2} y}{d x^{2}}} at t=π4t=\dfrac{\pi}{4} is equal to :
A −223\dfrac{-2 \sqrt{2}}{3}
B 23\dfrac{2}{3}
C 13\dfrac{1}{3}
D −23 \dfrac{-2}{3}
Correct Answer
Option D
Solution
x=22cos⁡tsin⁡2t, y=22sin⁡tsin⁡2tx = 2\sqrt 2 \cos t\sqrt {\sin 2t} ,\,y = 2\sqrt 2 \sin t\sqrt {\sin 2t}

∴\therefore

dxdt=22cos⁡3tsin⁡2t, dydt=22sin⁡3tsin⁡2t{{dx} \over {dt}} = {{2\sqrt 2 \cos 3t} \over {\sqrt {\sin 2t} }},\,{{dy} \over {dt}} = {{2\sqrt 2 \sin 3t} \over {\sqrt {\sin 2t} }}

∴\therefore

dydx=tan⁡3t, (at t=π4, dydx=−1){{dy} \over {dx}} = \tan 3t,\,\left( {\mathrm{at}\,t = {\pi \over 4},\,{{dy} \over {dx}} = - 1} \right)

and

d2ydx2=3sec⁡23t . dtdx=3sec⁡23t . sin⁡2t22cos⁡3t{{{d^2}y} \over {d{x^2}}} = 3{\sec ^2}3t\,.\,{{dt} \over {dx}} = {{3{{\sec }^2}3t\,.\,\sqrt {\sin 2t} } \over {2\sqrt 2 \cos 3t}}
(At t=π4, d2ydx2=−3)\left( {\mathrm{At}\,t = {\pi \over 4},\,{{{d^2}y} \over {d{x^2}}} = - 3} \right)

∴\therefore

1+(dydx)2d2ydx2=2−3=−23{{1 + {{\left( {{{dy} \over {dx}}} \right)}^2}} \over {{{{d^2}y} \over {d{x^2}}}}} = {2 \over { - 3}} = {{ - 2} \over 3}
Q38
If y(x)=xx,x>0y(x)=x^{x},x > 0, then y′′(2)−2y′(2)y''(2)-2y'(2) is equal to
A 4(log⁡e2)2+24(\log_{e}2)^{2}+2
B 8log⁡e2−28\log_{e}2-2
C 4log⁡e2+24\log_{e}2+2
D 4(log⁡e2)2−24(\log_{e}2)^{2}-2
Correct Answer
Option D
Solution

y=xxy′=xx(1+ln⁡x)y′′=xx(1+ln⁡x)2+xxxf′′(2)−2f′(2)=(4(1+ln⁡2)2+2)−(2)(4(1+ln⁡2))=4(1+(ln⁡2)2)+2−8=4(ln⁡2)2−2\begin{aligned} & y=x^x \\\\ & y^{\prime}=x^x(1+\ln x) \\\\ & y^{\prime \prime}=x^x(1+\ln x)^2+\dfrac{x^x}{x} \\\\ & f^{\prime \prime}(2)-2 f^{\prime}(2)=\left(4(1+\ln 2)^2+2\right)-(2)(4(1+\ln 2)) \\\\ & =4\left(1+(\ln 2)^2\right)+2-8 \\\\ & =4(\ln 2)^2-2 \\\\ & \end{aligned}

Q39
Let f(x)=2x+tan⁡−1xf(x) = 2x + {\tan ^{ - 1}}x and g(x)=log⁡e(1+x2+x),x∈[0,3]g(x) = {\log _e}(\sqrt {1 + {x^2}} + x),x \in [0,3]. Then
A there exists x^∈[0,3]\widehat x \in [0,3] such that f′(x^)<g′(x^)f'(\widehat x) < g'(\widehat x)
B there exist 0<x1<x2<30 < {x_1} < {x_2} < 3 such that f(x)<g(x),∀x∈(x1,x2)f(x) < g(x),\forall x \in ({x_1},{x_2})
C min⁡f′(x)=1+max⁡g′(x)\min f'(x) = 1 + \max g'(x)
D max⁡f(x)>max⁡g(x)\max f(x) > \max g(x)
Correct Answer
Option D
Solution
f′(x)=2+11+x2,g′(x)=1x2+1f′′(x)=−2x(1+x2)2<0g′′(x)=−12(x2+1)−3/2⋅2x<0f′(x)∣min⁡=f′(3)=2+110=2110g′(x)∣max⁡=g′(0)=1f′(x)∣max⁡=f(3)=2+tan⁡−13g(x)∣max⁡=g(3)=ln⁡(3+10)<ln⁡<7<2\begin{aligned} & f^{\prime}(x)=2+\frac{1}{1+x^2}, g^{\prime}(x)=\frac{1}{\sqrt{x^2+1}} \\\\ & f^{\prime \prime}(x)=-\frac{2 x}{\left(1+x^2\right)^2}<0 \\\\ & g^{\prime \prime}(x)=-\frac{1}{2}\left(x^2+1\right)^{-3 / 2} \cdot 2 x<0 \\\\ & \left.f^{\prime}(x)\right|_{\min }=f^{\prime}(3)=2+\frac{1}{10}=\frac{21}{10} \\\\ & \left.g^{\prime}(x)\right|_{\max }=g^{\prime}(0)=1 \\\\ & \left.f^{\prime}(x)\right|_{\max }=f(3)=2+\tan ^{-1} 3 \\\\ & \left.g(x)\right|_{\max }=g(3)=\ln (3+\sqrt{10})<\ln <7<2 \end{aligned}
Q40
Let y=f(x)=sin⁡3(π3(cos⁡(π32(−4x3+5x2+1)32)))y=f(x)=\sin ^{3}\left(\dfrac{\pi}{3}\left(\cos \left(\dfrac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{\dfrac{3}{2}}\right)\right)\right). Then, at x = 1,
A 2y′+3π2y=02 y^{\prime}+\sqrt{3} \pi^{2} y=0
B y′+3π2y=0y^{\prime}+3 \pi^{2} y=0
C 2y′−3π2y=0\sqrt{2} y^{\prime}-3 \pi^{2} y=0
D 2y′+3π2y=02 y^{\prime}+3 \pi^{2} y=0
Correct Answer
Option D
Solution

f(x)=sin⁡3(π3cos⁡(π32(−4x3+5x2+1)3/2))f(x)=\sin ^{3}\left(\dfrac{\pi}{3} \cos \left(\dfrac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right)

f′(x)=3sin⁡2(π3cos⁡(π32(−4x3+5x2+1)3/2))cos⁡(π3cos⁡(π32(−4x3+5x2+1)3/2))π3(−sin⁡(π32(−4x3+5x2+1)3/2))π3232(−4x3+5x3+1)1/2(−12x2+10x)\begin{aligned} & f^{\prime}(x)=3 \sin ^{2}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right) \\\\ & \cos \left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right) \\\\ & \frac{\pi}{3}\left(-\sin \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right) \\\\ & \frac{\pi}{3 \sqrt{2}} \frac{3}{2}\left(-4 x^{3}+5 x^{3}+1\right)^{1 / 2}\left(-12 x^{2}+10 x\right) \end{aligned}

f′(1)=3π216f^{\prime}(1)=\dfrac{3 \pi^{2}}{16}

f(1)=sin⁡3(π3cos⁡(π3222))=sin⁡3(−π6)=−18∴2f′(1)+3π2f(1)=0\begin{aligned} & f(1)=\sin ^{3}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}} 2 \sqrt{2}\right)\right) \\\\ &=\sin ^{3}\left(-\frac{\pi}{6}\right)=\frac{-1}{8} \\\\ & \therefore 2 f^{\prime}(1)+3 \pi^{2} f(1)=0 \end{aligned}
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