For $x \in \mathbf{R}, f(x)=|\log 2-\sin x|$ and $g(x)=f(f(x))$, then:

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

(D) g is not differentiable at $x=0$

JEE Main · 2016 · 03 Apr 2016 · Offline · Continuity and Differentiability

For

x \in R, f (x) = ∣ lo g 2 - sin x ∣

and

g (x) = f (f (x))

, then:

Answer: (A)

Near

x = 0

lo g 2 - sin x

stays positive, so

f (x) = lo g 2 - sin x

with

f (0) = lo g 2

and

f^{'} (0) = - cos 0 = - 1

. For

g (x) = f (f (x))

the chain rule gives

g^{'} (0) = f^{'} (f (0)) f^{'} (0) = f^{'} (lo g 2) \cdot (- 1)

. At

u = lo g 2 \approx 0.693 < 1

lo g 2 - sin (lo g 2)

is again positive, so

f^{'} (lo g 2) = - cos (lo g 2)

. Hence

g^{'} (0) = (- cos (lo g 2)) (- 1) = cos (lo g 2)

JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option