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NDA 1 2025 Maths, question 75

Topic: Differential calculus

Passage · Q75 to Q76

Let (x+y)p+q=xpyq(x + y)^{p + q} = x^{p}y^{q}, where pp, qq are positive integers.

Q75

The derivative of yy with respect to xx

  1. (a) depends on pp only
  2. (b) depends on qq only
  3. (c) depends on both pp and qq
  4. (d) is independent of both pp and qqCorrect answer

Solution.

Answer: (d) is independent of both pp and qq

  1. Take logs: (p+q)log⁡(x+y)=plog⁡x+qlog⁡y(p + q)\log(x + y) = p\log x + q\log y, then differentiate: (p+q)(1+y′)x+y=px+qy′y\frac{(p + q)(1 + y')}{x + y} = \frac{p}{x} + \frac{qy'}{y}.
  2. Rearranging gives (py−qx)(xy′−y)=0(py - qx)(xy' - y) = 0, and the factor py−qxpy - qx cannot be zero on a real solution.
  3. So y′=y/xy' = y/x, which contains neither pp nor qq.

Trap: The curve itself depends on pp and qq; the option means that the formula for dy/dxdy/dx has no pp or qq in it.