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CUET · MATHS · PYQ PAPER 2025

If \(x^2-y^2=t-\frac{1}{t}\), and \(x^4+y^4=t^2+\frac{1}{t^2}\), then which of the following is correct?

  1. A \(x y^3 \frac{d y}{d x}+1=0\)
  2. B \(x^3 \frac{d y}{d x}+y=1\)
  3. C \(x^2 y \frac{d y}{d x}-1=0\)
  4. D \(x^3 y \frac{d y}{d x}+1=0\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(x^3 y \frac{d y}{d x}+1=0\)

Step-by-step Solution

Detailed explanation

\((x^2-y^2)^2 = (t-\frac{1}{t})^2\) \(x^4-2x^2y^2+y^4 = t^2-2+\frac{1}{t^2}\) \((x^4+y^4)-2x^2y^2 = (t^2+\frac{1}{t^2})-2\) \((t^2+\frac{1}{t^2})-2x^2y^2 = (t^2+\frac{1}{t^2})-2\) \(-2x^2y^2 = -2 \implies x^2y^2 = 1\) \(d(x^2y^2)/dx = d(1)/dx\)…
From CUET
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