Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Differentiation question

2011 · Shift 1 · Q33
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Advanced
  3. /Mathematics
  4. /Differentiation
  5. /2011 · Shift 1 · Q33

Differentiation question

2011 · Shift 1 · Q33

JEE AdvancedMathematicsDifferentiationNumerical+4 / −1
Let f(θ)=sin⁡(tan⁡−1(sin⁡θcos⁡2θ)),f\left( \theta \right) = \sin \left( {{{\tan }^{ - 1}}\left( {{{\sin \theta } \over {\sqrt {\cos 2\theta } }}} \right)} \right),f(θ)=sin(tan−1(cos2θ​sinθ​)), where −π4<θ<π4.- {\pi \over 4} \lt \theta \lt {\pi \over 4}.−4π​<θ<4π​. Then the value of dd(tan⁡θ)(f(θ)){d \over {d\left( {\tan \theta } \right)}}\left( {f\left( \theta \right)} \right)d(tanθ)d​(f(θ)) is
Numerical answer
View written solutionFree

Correct answer: 1

  1. Given function

We have

f(θ)=sin⁡(tan⁡−1(sin⁡θcos⁡2θ)),f(\theta)=\sin\left(\tan^{-1}\left(\frac{\sin\theta}{\sqrt{\cos 2\theta}}\right)\right),f(θ)=sin(tan−1(cos2θ​sinθ​)),

with

π4<θ<π4.\frac{\pi}{4}<\theta<\frac{\pi}{4}.4π​<θ<4π​.

We need to find

dd(tan⁡θ)(f(θ)).\frac{d}{d(\tan\theta)}\bigl(f(\theta)\bigr).d(tanθ)d​(f(θ)).
  1. Simplify f(θ)f(\theta)f(θ) using an identity

Let

x=sin⁡θcos⁡2θ.x=\frac{\sin\theta}{\sqrt{\cos 2\theta}}.x=cos2θ​sinθ​.

Then

f(θ)=sin⁡(tan⁡−1x).f(\theta)=\sin(\tan^{-1}x).f(θ)=sin(tan−1x).

Now use the standard identity:

sin⁡(tan⁡−1x)=x1+x2.\sin(\tan^{-1}x)=\frac{x}{\sqrt{1+x^2}}.sin(tan−1x)=1+x2​x​.

So,

f(θ)=sin⁡θcos⁡2θ1+sin⁡2θcos⁡2θ.f(\theta)=\frac{\dfrac{\sin\theta}{\sqrt{\cos 2\theta}}}{\sqrt{1+\dfrac{\sin^2\theta}{\cos 2\theta}}}.f(θ)=1+cos2θsin2θ​​cos2θ​sinθ​​.

Simplify the denominator:

1+sin⁡2θcos⁡2θ=cos⁡2θ+sin⁡2θcos⁡2θ.1+\frac{\sin^2\theta}{\cos 2\theta} =\frac{\cos 2\theta+\sin^2\theta}{\cos 2\theta}.1+cos2θsin2θ​=cos2θcos2θ+sin2θ​.

Since

cos⁡2θ=cos⁡2θ−sin⁡2θ,\cos 2\theta=\cos^2\theta-\sin^2\theta,cos2θ=cos2θ−sin2θ,

we get

cos⁡2θ+sin⁡2θ=cos⁡2θ.\cos 2\theta+\sin^2\theta=\cos^2\theta.cos2θ+sin2θ=cos2θ.

Hence

1+sin⁡2θcos⁡2θ=cos⁡2θcos⁡2θ.1+\frac{\sin^2\theta}{\cos 2\theta}=\frac{\cos^2\theta}{\cos 2\theta}.1+cos2θsin2θ​=cos2θcos2θ​.

Therefore,

1+sin⁡2θcos⁡2θ=cos⁡2θcos⁡2θ=cos⁡θcos⁡2θ.\sqrt{1+\frac{\sin^2\theta}{\cos 2\theta}} =\sqrt{\frac{\cos^2\theta}{\cos 2\theta}} =\frac{\cos\theta}{\sqrt{\cos 2\theta}}.1+cos2θsin2θ​​=cos2θcos2θ​​=cos2θ​cosθ​.

Here we used that in the interval −π4<θ<π4-\frac\pi4<\theta<\frac\pi4−4π​<θ<4π​, we have cos⁡θ>0\cos\theta>0cosθ>0.

So,

f(θ)=sin⁡θcos⁡2θcos⁡θcos⁡2θ=sin⁡θcos⁡θ=tan⁡θ.f(\theta)=\frac{\dfrac{\sin\theta}{\sqrt{\cos 2\theta}}}{\dfrac{\cos\theta}{\sqrt{\cos 2\theta}}} =\frac{\sin\theta}{\cos\theta}=\tan\theta.f(θ)=cos2θ​cosθ​cos2θ​sinθ​​=cosθsinθ​=tanθ.

Thus,

f(θ)=tan⁡θ.f(\theta)=\tan\theta.f(θ)=tanθ.
  1. Differentiate with respect to tan⁡θ\tan\thetatanθ

Since

f(θ)=tan⁡θ,f(\theta)=\tan\theta,f(θ)=tanθ,

we directly get

dd(tan⁡θ)(f(θ))=dd(tan⁡θ)(tan⁡θ)=1.\frac{d}{d(\tan\theta)}\bigl(f(\theta)\bigr)=\frac{d}{d(\tan\theta)}(\tan\theta)=1.d(tanθ)d​(f(θ))=d(tanθ)d​(tanθ)=1.
  1. Final answer
1\boxed{1}1​
PreviousNext

More from Differentiation

  • Consider the functions defined implicitly by the equation y3−3y+x=0 on various intervals in the real line. If x∈(−∞,−2)∪(2,∞), the equation implicitly defines a unique real valued differentiable function y=f(x). If x∈(−2,2)…2008 · MCQ
  • Let f and g be real valued functions defined on interval (−1,1) such that g′′(x) is continuous, g(0)e0.g′(0)=0, g′′(0)e0, and f(x)=g(x)sinx…2008 · MCQ
  • Consider the function f:(−∞,∞)→(−∞,∞) defined by f(x)=x2+ax+1x2−ax+1​,0<a<2.Which of the following is true?2008 · MCQ
  • Let g(x)=logf(x), where f(x) is a twice differentiable positive function on (0, ∞) such that f(x+1)=xf(x). Then for N = 1, 2, 3, ..., g′′(N+21​)−g′′(21​)=2008 · MCQ
  • Let S be the set of all twice differentiable functions f from R to R such that dx2d2f​(x)>0 for all x∈(−1,1). For f∈S, let Xf​ be the number of points x∈(−1,1) for which…2023 · Multiple correct
  • Let f:R→R,g:R→R and h:R→R be differentiable functions such that f(x)=x3+3x+2,g(f(x))=x and h(g(g(x)))=x…2016 · Multiple correct
  • Let F:R→R be a thrice differentiable function. Suppose that F(1)=0,F(3)=−4 and F′(x)<0 for all x∈(21​,3). Let f(x)=xF(x)…2015 · Multiple correct
  • Let f:[0,2]→R be a function which is continuous on [0,2] and is differentiable on (0,2) with f(0)=1. Let F(x)=0∫x2​f(t​)dt for x∈[0,2]…2014 · MCQ