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

Question

For x ϵ 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)

Tardigrade Admin · Accepted Answer

g(x)=| log e 2- sin (| log e 2- sin x|)| At x=0, g(x)= log e(2)- sin ( log e 2- sin x) ∴ g'(x)= cos ( log e(2)- sin x) × cos (x) ⇒ g'(0)= cos ( log e(2))

Q. For $x ϵ R, f (x) = ∣ lo g 2 - sin x ∣$ and $g (x) = f (f (x))$ , then :

$g$ is not differentiable at $x = 0$

$g^{'} (0)$ = $cos (l o g 2)$

$g^{'} (0)$ = $- cos (l o g 2)$

$g$ is differentiable at $x = 0$ and $g^{'} (0) = - s in (l o g 2)$

$g (x) = ∣ lo g_{e} 2 - sin (∣ lo g_{e} 2 - sin x ∣) ∣$
At $x = 0, g (x) = lo g_{e} (2) - sin (lo g_{e} 2 - sin x)$
$∴ g^{'} (x) = cos (lo g_{e} (2) - sin x) \times cos (x)$
$\Rightarrow g^{'} (0) = cos (lo g_{e} (2))$

Q. For xϵR,f(x)=∣log2−sinx∣ and g(x)=f(f(x)), then :

g is not differentiable at x=0

g′(0) = cos(log2)

g′(0) = −cos(log2)

g is differentiable at x=0 and g′(0)=−sin(log2)