Question

For $ x \epsilon R , f (x) = | \log 2 - \sin x|$ and $g(x) = f(f(x))$, then :

• A. $g$ is not differentiable at $x = 0$
• B. $g'(0)$ = $cos(log2)$
• C. $g'(0)$ = $-cos(log2)$
• D. $g$ is differentiable at $x = 0$ and $g'(0) = -sin(log2)$

Answer 1

The Correct Option is B

To solve the problem, we need to understand the behavior and differentiability of the functions \( f(x) \) and \( g(x) = f(f(x)) \) at \( x = 0 \). Let's go through the problem step by step. 

1. Understanding the Function \( f(x) \):

The given function is \( f(x) = | \log 2 - \sin x| \). It is an absolute value function, thus it is defined and differentiable everywhere except where the expression inside the absolute value is zero.

1. Differentiability of \( f(x) \) at \( x = 0 \):

Compute \( f(x) \) at \( x = 0 \):

\( f(0) = | \log 2 - \sin 0| = |\log 2 - 0| = |\log 2| = \log 2 \)

Since \( \log 2 \neq 0 \), the function \( f(x) \) is differentiable at \( x = 0 \).

1. Finding \( g(x) = f(f(x)) \):

We have \( g(x) = f(f(x)) = f(|\log 2 - \sin x|) \).

At \( x = 0 \), we already found that \( f(0) = \log 2 \). Therefore,

\( g(0) = f(\log 2) = | \log 2 - \sin(\log 2) | \).

1. Differentiate \( g(x) \) at \( x = 0 \):

To find if \( g \) is differentiable at \( x = 0 \) and to find \( g'(0) \), we use the chain rule.

Given that \( f(x) = |\log 2 - \sin x| \), the derivative \( f'(x) \) when \( \log 2 - \sin x > 0 \) is:

\( f'(x) = -\cos x \).

Since \( \log 2 - \sin(\log 2) > 0 \) holds around \( x = 0 \),

The derivative of \( g(x) = f(f(x)) \) is given by:

\( g'(x) = f'(f(x)) \cdot f'(x) \).

Therefore, \( g'(0) = f'(\log 2) \cdot f'(0) = (-\cos(\log 2)) \cdot (-\cos 0) = \cos(\log 2) \cdot 1 = \cos(\log 2) \).

Thus, \( g'(0) = \cos(\log 2) \). Therefore, the correct answer is \(g'(0)\) = \(cos(log2)\).