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NDA 1 2024 Maths, question 86

Topic: Differential calculus

Passage · Q85 to Q86

Let f(x)f(x) and g(x)g(x) be two functions such that g(x)=x−1xg(x)=x-\frac{1}{x} and f∘g(x)=x3−1x3f\circ g(x)=x^{3}-\frac{1}{x^{3}}.

Q86

What is f′′(x)f''(x) equal to ?

  1. (a) −2x3-\frac{2}{x^{3}}
  2. (b) 2x+2x32x+\frac{2}{x^{3}}
  3. (c) 6x+36x+3
  4. (d) 6x6xCorrect answer

Solution.

Answer: (d) 6x6x

  1. The identity (x − 1/x)³ + 3(x − 1/x) = x³ − 1/x³ gives f(t) = t³ + 3t.
  2. Therefore f′(t) = 3t² + 3 and f″(t) = 6t; at argument x the answer is 6x.