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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 $x = 0$
B. $g'(0) = \cos (\log 2)$
C. $g'(0) = - \cos (\log 2)$
D. $g$ is differentiable at $x = 0$and $g'(0) = - \sin (\log 2)$

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Last updated date: 20th Jun 2024
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Answer
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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.