Q66 / 160 · TS EAMCET 2017 · Mathematics

Let $f:(-1,1) \rightarrow$ IR be a differentiable function with $f(0)=-1$ and $f^{\prime}(0)=1$. If $g(x)=(f(2 f(x)+2))^2$, then $g^{\prime}(0)=$

  • A
    0
  • B
    -2
  • C
    4
  • Correct answer
    D
    -4

Solution

We have, $$ \begin{aligned} & g(x)=(f(2 f(x)+2))^2 \\ & g^{\prime}(x)=2\left(f(2 f(x)+2) \cdot f^{\prime}(2 f(x)+2) f^{\prime}(x) \cdot 2\right. \\ & \Rightarrow g^{\prime}(0)=2\left(f(2 f(0)+2) \cdot f^{\prime}(2 f(0)+2) \cdot 2 f^{\prime}(0)\right. \\ & \left.\Rightarrow g^{\prime}(0)=2(f(2)-1)+2\right) \cdot f^{\prime}(2(-1)+2 \cdot 2(1) \\ & {\left[\because f(0)=-1, f^{\prime}(0)=1\right]} \\ & \Rightarrow g^{\prime}(0)=2\left[f(-2+2) \cdot f^{\prime}(-2+2)\right] \cdot 2 \\ & \Rightarrow g^{\prime}(0)=4 f(0) \cdot f^{\prime}\langle 0\rangle \\ & \Rightarrow g^{\prime}(0)=4 \times(-1)(1) \\ & =-4 \\ & \end{aligned} $$

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

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