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JEE Main 20163 April
Exam
JEE Main
Year
2016
Date
3 April
Subject
Mathematics
Chapter
Continuity and Differentiability

JEE Main 2016 · 3 April · Mathematics · Continuity and Differentiability

For x ∈ R, f(x) = | 2 - sin x| and g(x) = f(f(x)) , then

For x∈R,f(x)=∣log⁡2−sinx∣x \in R, f\left(x\right) = \left|\log 2 - sin x\right| and g(x)=f(f(x))g\left(x\right) = f\left(f\left(x\right)\right), then

Options

  1. Ag′(0)=−cos⁡(log⁡2){g}^{'}\left(0\right) = - \cos \left(\log 2\right)
  2. Bgisdifferentiableatis differentiable atx = 0andand{g}^{'}\left(0\right) = - \sin \left(\log 2\right)
  3. Cgg is not differentiable at x=0x = 0
  4. Dg′(0)=cos⁡(log⁡2){g}^{'}\left(0\right) = \cos \left(\log 2\right)

Correct answer

D. g′(0)=cos⁡(log⁡2){g}^{'}\left(0\right) = \cos \left(\log 2\right)

Step-by-step solution

In the neighborhood of x=0,f(x)=log⁡2−sinxx = 0, f\left(x\right) = \log 2 - sin x

∴∴ g(x)=f(f(x))=log⁡2−sin⁡(f(x))g\left(x\right) = f ( f\left(x\right) ) = \log 2 - \sin \left(f\left(x\right)\right)

=log⁡2−sin⁡(log⁡2−sinx)= \log 2 - \sin \left(\log 2 - sin x\right)

∴∴ g′(x)=−cos⁡(log⁡2−sinx)(–cosx){g}^{′}\left(x\right) = - \cos \left(\log 2 - sin x\right)\left(– cos x\right)

∴∴ ${g}^{′}\left(0\right) = \cos \left(\log 2\right)
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