MHT CET · Maths · Differentiation
For \(x \in R, f(x)=|\log 2-\sin x|\) and \(g(x)=f(f(x))\), then
- A \(g^{\prime}(0)=-\cos (\log 2)\)
- B \(g\) is not differentiable at \(x=0\).
- C \(g^{\prime}(0)=\cos (\log 2)\)
- D \(g\) is differentiable at \(x=0\) and \(g^{\prime}(0)=-\sin (\log 2)\).
Answer & Solution
Correct Answer
(C) \(g^{\prime}(0)=\cos (\log 2)\)
Step-by-step Solution
Detailed explanation
for \(x \rightarrow 0^{-}\)
\(g(x)=\log 2-\sin (\log 2-\sin x)\)
for \(x \rightarrow 0^{+}\)
\(g(x)=\log 2-\sin (\log 2-\sin x)\)
[as \(\log 2>\sin 0\) and \(x>\sin x\) ]
and \(g(x)\) is continuous at \(x=0\)
\(\begin{aligned} & \text { now } g^{\prime}(x)=0-\cos (\log 2-\sin x)(0-\cos x) \\ & \Rightarrow g^{\prime}(0)=\cos (\log 2)\end{aligned}\)
\(g(x)=\log 2-\sin (\log 2-\sin x)\)
for \(x \rightarrow 0^{+}\)
\(g(x)=\log 2-\sin (\log 2-\sin x)\)
[as \(\log 2>\sin 0\) and \(x>\sin x\) ]
and \(g(x)\) is continuous at \(x=0\)
\(\begin{aligned} & \text { now } g^{\prime}(x)=0-\cos (\log 2-\sin x)(0-\cos x) \\ & \Rightarrow g^{\prime}(0)=\cos (\log 2)\end{aligned}\)
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