Question

$\huge\sf\fbox\blue{Question}$

For x∈R, f(x) = |log2 – sinx| and g(x) = f(f(x)), then:

(1) g is differentiable at x = 0 and g'(0) = –sin(log2)

(2) g is not differentiable at x = 0

(3) g'(0) = cos(log2)

(4) g'(0) = –cos(log2)​

Answer 1

$\large\underline\mathfrak\red{Correct \: Option \: :- }$

(3) g'(0) = cos(log2)

$\large\underline\mathfrak\red{Explanation \: :- }$

$\sf{g(x) = f(f(x))}$

$\sf{g'(x) = f'(f(x)f'(x)}$

$\sf{g'(0) = f'(f(0)f'(0)}$

x tending to 0, log2 - sinx

$\sf{f(x) = log2 - \sin x }$

$\sf{f'(x) = - \cos x }$

$\sf{f'(0) = -1 }$

$\sf{f'(log2) = - cos (log2) }$

$\sf{g'(0) = (-cos (log2) )(-1) = cos(log2) }$

Hence Option (3) is correct answer

Answer 2

Answer:

CorrectOption:−

(3) g'(0) = cos(log2)

\large\underline\mathfrak\red{Explanation \: :- }

Explanation:−

\sf{g(x) = f(f(x))}g(x)=f(f(x))

\sf{g'(x) = f'(f(x)f'(x)}g

′

(x)=f

′

(f(x)f

′

(x)

\sf{g'(0) = f'(f(0)f'(0)}g

′

(0)=f

′

(f(0)f

′

(0)

x tending to 0, log2 - sinx

\sf{f(x) = log2 - \sin x }f(x)=log2−sinx

\sf{f'(x) = - \cos x }f

′

(x)=−cosx

\sf{f'(0) = -1 }f

′

(0)=−1

\sf{f'(log2) = - cos (log2) }f

′

(log2)=−cos(log2)

\sf{g'(0) = (-cos (log2) )(-1) = cos(log2) }g

′

(0)=(−cos(log2))(−1)=cos(log2)

Hence Option (3) is correct answer