For x ∈ R, f(x) =log2-sinx and gx=ffx then

f(x)=|log 2-sin x| given that g(x)=f(f(x)) as x→0, log 2> sin x⇒|log 2-sin x|=log 2-sin x ∴f(x)=log 2-sin x ⇒g(x)=log 2-sin f(x) ∴g(x)=log 2-sin (log 2-sin x) ⇒g1(x)=- cos (log 2-sin x)(-cos x) ⇒g1(x)=cos (log 2-sin x) cos x ⇒g1(0)=cos (log 2)

For x ∈ R, f(x) =log2-sinx and gx=ffx then

f(x)=|log 2-sin x| given that g(x)=f(f(x)) as x→0, log 2> sin x⇒|log 2-sin x|=log 2-sin x ∴f(x)=log 2-sin x ⇒g(x)=log 2-sin f(x) ∴g(x)=log 2-sin (log 2-sin x) ⇒g1(x)=- cos (log 2-sin x)(-cos x) ⇒g1(x)=cos (log 2-sin x) cos x ⇒g1(0)=cos (log 2)

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For x $\in$ R, f(x) = $|\log 2 - \sin x| a n d g (x) = f (f (x))$ then

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g'(0) = cos(log2)

g is not differentiable at x = 0

g is differentiable at x = 0, g'(0) =–sin(log2)

g'(0) = –cos(log2)

answer is C.

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Detailed Solution

$f (x) = | \log 2 - \sin x | g i v e n t h a t g (x) = f (f (x)) a s x \to 0, \log 2 > \sin x \Rightarrow | \log 2 - \sin x | = \log 2 - \sin x ∴ f (x) = \log 2 - \sin x \Rightarrow g (x) = \log 2 - \sin f (x) ∴ g (x) = \log 2 - \sin (\log 2 - \sin x) \Rightarrow g^{1} (x) = - \cos (\log 2 - \sin x) (- \cos x) \Rightarrow g^{1} (x) = \cos (\log 2 - \sin x) \cos x \Rightarrow g^{1} (0) = \cos (\log 2)$