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NDA 2 2024 Maths, question 76

Topic: Differential calculus

Q76

If f(2x)=4x2+1f(2x)=4x^{2}+1, then for how many real values of xx will f(2x)f(2x) be the GM of f(x)f(x) and f(4x)f(4x)?

  1. (a) Four
  2. (b) Two
  3. (c) OneCorrect answer
  4. (d) None

Solution.

Answer: (c) One

  1. Replacing 2x by t gives f(t)=t²+1.
  2. The GM condition is 4x²+1=√[(x²+1)(16x²+1)].
  3. Both sides are positive, so squaring is valid.
  4. Expansion yields 16x⁴+8x²+1=16x⁴+17x²+1, hence x=0 only.
  5. It satisfies the original equality.