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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Continuity and Differentiability

NTA Abhyas 2020 · Mathematics · Continuity and Differentiability

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

For x \in \mathbb{R}, f \left(x\right) = \left|\log 2 - sin x\right| and g(x)=g\left(x\right) =f\lleft(f\left(x\right))$, then

Options

  1. Ag′(0)=−cos⁡(log⁡2)g^{'} \left(0\right) = - \cos \left(\log 2\right)
  2. Bgisdifferentiableatis differentiable atx = 0andandg^{'} \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\lleft(f(x))∴g \left(x\right) = f ( f \left(x\right) ) = \log 2 - sin \lleft(f\left(x\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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