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# For $x \in R$,$f(x) = $$\left| {\log 2 - \sin x} \right| and g(x) = f(f(x)), then:A. g is not differentiable at x = 0B. g'(0) = \cos (\log 2)C. g'(0) = - \cos (\log 2)D. g is differentiable at x = 0and g'(0) = - \sin (\log 2) Last updated date: 20th Jun 2024 Total views: 414.6k Views today: 4.14k Answer Verified 414.6k+ views Hint:We have the function f(x),g(x).According to the problem given, it is very clear that we need to find the differentiability of g(x) at x = 0 and then find the value of g'(0). So firstly we will check whether g(x) is differentiable at x = 0 and if it is, then we will find g'(0) Complete step-by-step answer: x \in R,f(x) =$$\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.