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NDA 2 2025 Maths, question 76

Topic: Differential calculus

Passage · Q76 to Q77

Consider the function f(x)=x1−xf(x)=\frac{x}{1-x} (x>0,x≠1)(x>0,x\ne 1)

Q76

What is f(x)f(x+1)\frac{f(x)}{f(x+1)} equal to?

  1. (a) −f(x2)-f(x^{2})Correct answer
  2. (b) −f(x)-f(\sqrt{x})
  3. (c) f(x2)f(x^{2})
  4. (d) f(x−1)f(x-1)

Solution.

Answer: (a) −f(x2)-f(x^{2})

  1. f(x+1)=−(x+1)/x.
  2. Hence f(x)/f(x+1)=−x²/[(1−x)(x+1)]=−x²/(1−x²)=−f(x²).
  3. The given domain x>0,x≠1 keeps all needed expressions defined.