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**Hint:**We have the function $f(x),g(x)$.According to the problem given, it is very clear that we need to find the differentiability of $g(x)$ at $x = 0$ and then find the value of $g'(0)$. So firstly we will check whether $g(x)$ is differentiable at $x = 0$ and if it is, then we will find $g'(0)$

**Complete step-by-step answer:**

$x \in R$,$f(x) = $$\left| {\log 2 - \sin x} \right|$ $ - - - - - (1)$

$g(x) = f(f(x))$$ - - - - - (2)$

It is very clear from the options that we need to find the differentiability of $g(x)$ at $x = 0$

We know that if $g(x)$ is continuous at $x = 0$, when

$g(0) = \mathop {\lim }\limits_{x \to 0} g(x)$

So firstly,

$g(0) = f(f(0))$

Using (1),

$g(0) = \left| {\log 2 - \sin 0} \right|$

As $\sin 0 = 0$, so

$g(0) = f\left| {\log 2} \right|$

$g(0) = \left| {\log 2 - \sin (\log 2)} \right|$$ - - - - - (3)$

Now taking

$\mathop {\lim }\limits_{x \to 0} g(x) = \mathop {\lim }\limits_{x \to 0} f(f(x))$

$ = \mathop {\lim }\limits_{x \to 0} $$f(\left| {\log 2 - \sin x} \right|)$

Using (1),

$ = \mathop {\lim }\limits_{x \to 0} $$\left| {\log 2 - \sin (\log 2)} \right|$

Therefore:

$\mathop {\lim }\limits_{x \to 0} g(x)$$ = \left| {\log 2 - \sin (\log 2)} \right|$$ - - - - - (4)$

Now from (3) and (4)

$g(0) = \mathop {\lim }\limits_{x \to 0} g(x)$

So $g(x)$ is continuous at $x = 0$, hence differentiable also.

Now we will find $g'(0)$

Differentiating (2) and using the chain rule,

$g'(0)$$ = f'(f(x)).f'(x)$

Using $x = 0$

$ = f'(f(0)).f'(0)$

Now differentiating (1) with respect to $x$

$f'(x) = - \cos x - - - - - (6)$

Put $x = 0$ in (1) and (6)

$f(0) = \left| {\log 2} \right|$ and $f'(0) = - \cos 0 = - 1$

Now use (5) by putting these values in it,

$g'(0)$$ = f'(\left| {\log 2} \right|).( - 1) - - - - - (7)$

Now putting $x = \left| {\log 2} \right|$in (6)

$f'(\left| {\log 2} \right|) = - \cos (\left| {\log 2} \right|)$

Now putting this in (7)

$g'(0)$$ = - \cos (\left| {\log 2} \right|).( - 1) = \cos \left( {\left| {\log 2} \right|} \right)$

**So, the correct answer is “Option B”.**

**Note:**Here in the last step we have written that $\left| {\log 2} \right| = \log 2$ because $\log 2$ is the positive number.in this question, we have also taken the derivative of $f$ which means that it is differentiable. So in this way we can easily solve the problem.

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