Q95 / 120 · NDA 2024 (Phase 2) · Mathematics

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

  • A
    -2
  • B
    -1
  • C
    0
  • Correct answer
    D
    2

Solution

Let $\mathrm{f}:(-1,1) \rightarrow \mathrm{R}$ be a differentiable function with $\mathrm{f}(0)$ $=-1$ and $f^{\prime}(0)=1$ $h(x)=f(2 f(n)+2)$ Differentiate both side w.r.t $x$ $\begin{aligned} & h^{\prime}(x)=f^{\prime}(2 f(x)+2) 2 f^{\prime}(x) \\ & h^{\prime}(0)=f^{\prime}(2 f(0)+2) \cdot 2 f^{\prime}(0)=f^{\prime}(-2+2) \cdot 2(1) \\ & =f^{\prime}(0) \cdot 2=(1) \cdot 2=2\left[\because f(0)=-1 \text { and } f^{\prime}(0)=1\right] \end{aligned}$

Hence, the correct answer is option: 'D'.

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