Limit Continuity and Differentiability 7 Question 11
11. For $x \in R, f(x)=|\log 2-\sin x|$ and $g(x)=f(f(x))$, then
(a) $g$ is not differentiable at $x=0$
(2016 Main)
(b) $g^{\prime}(0)=\cos (\log 2)$
(c) $g^{\prime}(0)=-\cos (\log 2)$
(d) $g$ is differentiable at $x=0$ and $g^{\prime}(0)=-\sin (\log 2)$
Show Answer
Answer:
Correct Answer: 11. (d)
Solution:
- Here, $g$ is the inverse of $f(x)$. $\Rightarrow$
$$ f \circ g(x)=x $$
On differentiating w.r.t. $x$, we get
$$ \begin{gathered} f^{\prime}{g(x)} \times g^{\prime}(x)=1 \Rightarrow g^{\prime}(x)=\frac{1}{f^{\prime}(g(x))} \\ =\frac{1}{\frac{1}{1+{g(x)}^{5}}} \quad \because f^{\prime}(x)=\frac{1}{1+x^{5}} \\ \Rightarrow \quad g^{\prime}(x)=1+{g(x)}^{5} \end{gathered} $$