For x ε R, f (x)=|log2−sinx| and g(x)=f(f(x)), then :
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g is not differentiable at x=0
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g'(0)=cos(log2)
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g'(0)=-cos(log2)
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g is differentiable at x=0 and g'(0)=−sin(log2)
B.
g'(0)=cos(log2)
We have, f(x) | log 2 - sin x|
and g(x) =f(f(x)), x ε R
Note that, for x → 0 , log 2> sin x
therefore,
f(x) = log 2 - sin (f(x))
= log 2 - sin (log 2 - sin x)
Clearly (gx) is differentiable at x = 0 as sin x differentiable.
Now, g'(x) = - cos (log 2 -sin x)(-cos x)
= cos x. cos (log 2-sin x)
g'(0) = 1.cos(log 2)