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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Differentiation

NDA 2024 · Mathematics · Differentiation

Paragraph: Let f:(-1,1) arrow R be a differentiable function with...

Paragraph: Let f:(−1,1)→Rf:(-1,1) \rightarrow \mathrm{R} be a differentiable function with f(0)=−1f(0)=-1 and f′(0)=1f^{\prime}(0)=1. Let h(x)=f(2f(x)+2)h(x)=f(2 f(x)+2) and g(x)=(h(x))2g(x)=(h(x))^2.
Question: What is g′(0)g^{\prime}(0) equal to?

Options

  1. A-4
  2. B-2
  3. C0
  4. D4

Correct answer

A. -4

Step-by-step solution

Let f:(−1,1)→R\mathrm{f}:(-1,1) \rightarrow \mathrm{R} be a differentiable function with f(0)=−1andf′(0)=1g(n)=(h(x))2\mathrm{f}(0)=-1 and f^{\prime}(0)=1\mathrm{g}(\mathrm{n})=(\mathrm{h}(\mathrm{x}))^2 Differentiate both side w.r.t x\mathrmg′(x)=2(h(x))⋅\mathrmh′(x)x\mathrmg^{\prime}(\mathrm{x})=2(\mathrm{ h}(\mathrm{x})) \cdot \mathrmh^{\prime}(\mathrm{x})g′(0)=2(h(0))⋅h′(0)[∵h′(0)=2][∵f′(0)=−1]=2f(2+10)+2)⋅2=2f(2(−1)+2)=2f(0)⋅2=(−2)⋅2=−4.\begin{aligned} & g^{\prime}(0)=2(h(0)) \cdot h^{\prime}(0) \\ & {\left[\because h^{\prime}(0)=2\right]} \\ & {\left[\because f^{\prime}(0)=-1\right]} \\ & =2 f(2+10)+2) \cdot 2 \\ & =2 f(2(-1)+2)=2 f(0) \cdot 2=(-2) \cdot 2=-4.\end{aligned}$$
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