For x∈R;f(x)=|loge2−sinx| and g(x)=f(f(x)), then

∴ f(f(x))=|loge2−sin|log22−sinx||g(x) is differentiable at x=0g′(x)=−cos⁡(log⁡2−sin⁡x)(−cos⁡x)=cos⁡x⋅cos⁡(log⁡2−sin⁡x)g′(0)=1⋅cos⁡(log⁡2)∴ g'(0)=cos(loge2)

For x∈R;f(x)=|loge2−sinx| and g(x)=f(f(x)), then

∴ f(f(x))=|loge2−sin|log22−sinx||g(x) is differentiable at x=0g′(x)=−cos⁡(log⁡2−sin⁡x)(−cos⁡x)=cos⁡x⋅cos⁡(log⁡2−sin⁡x)g′(0)=1⋅cos⁡(log⁡2)∴ g'(0)=cos(loge2)

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For $x \in R; f (x) = | \log_{e} 2 - \sin x |$ and $g (x) = f (f (x)),$ then

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$g' (0) = \cos (\log_{e} 2)$

$g' (0) = - \cos (\log_{e} 2)$

g is differentiable at $x = 0$ and $g' (0) = - \sin (\log_{e} 2)$

g is not differentiable at $x = 0$

answer is A.

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Detailed Solution

$∴ f (f (x)) = | \log_{e} 2 - \sin | \log_{2} 2 - \sin x | |$
$g (x)$ is differentiable at $x = 0$

$\begin{array}{l} g^{'} (x) = - \cos (\log 2 - \sin x) (- \cos x) \\ = \cos x \cdot \cos (\log 2 - \sin x) \\ g^{'} (0) = 1 \cdot \cos (\log 2) \end{array}$