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For $x\epsilon R$ , $f(x)=\mid \log2-\sin x\mid$ and $g(x)=f((f(x)), then$

A) g is not differentiable at x=0

$g'(0)=\cos (\log 2)$

$g'(0)=-\cos (\log 2)$

D) g is differentiable at x=0 and g'(0)=-sin(log 2)

Option B

We have $f(x)=\mid\log2-\sin x$ and $g(x)=f(f(x)),x\epsilon R$

Note that , for $x\rightarrow0, \log2>\sin x$

$f(x)=\log 2-\sin x$

$\Rightarrow g(x)=\log 2-\sin(f(x))$

Clearly g(x) is differentiable at x=0 as sin x is diffenerntiable

Now $g'(x)=-\cos (\log 2-\sin x)(-\cos x)$

= $\cos x.\cos(\log2-\sin x)$

$\Rightarrow g'(0)=1.\cos(\log 2)$

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