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NDA 2 2024 Maths, question 95

Topic: Differential calculus

Passage · Q95 to Q96

Let f:(−1,1)→Rf:(-1,1)\to R be a differentiable function with f(0)=−1f(0)=-1 and f′(0)=1f'(0)=1. Let h(x)=f(2f(x)+2)h(x)=f(2f(x)+2) and g(x)=(h(x))2g(x)=(h(x))^{2}.

Q95

What is h′(0)h'(0) equal to?

  1. (a) −2-2
  2. (b) −1-1
  3. (c) 00
  4. (d) 22Correct answer

Solution.

Answer: (d) 22

  1. The inner argument at 0 is 2f(0)+2=0.
  2. By the chain rule h′(0)=f′(0)×2f′(0)=1×2×1=2.
  3. Differentiability ensures the composition is defined in a sufficiently small neighbourhood of 0.