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?
- A \(x y^3 \frac{d y}{d x}+1=0\)
- B \(x^3 \frac{d y}{d x}+y=1\)
- C \(x^2 y \frac{d y}{d x}-1=0\)
- D \(x^3 y \frac{d y}{d x}+1=0\)
Answer & Solution
Correct Answer
(D) \(x^3 y \frac{d y}{d x}+1=0\)
Step-by-step Solution
Detailed explanation
\((x^2-y^2)^2 = (t-\frac{1}{t})^2\) \(x^4-2x^2y^2+y^4 = t^2-2+\frac{1}{t^2}\) \((x^4+y^4)-2x^2y^2 = (t^2+\frac{1}{t^2})-2\) \((t^2+\frac{1}{t^2})-2x^2y^2 = (t^2+\frac{1}{t^2})-2\) \(-2x^2y^2 = -2 \implies x^2y^2 = 1\) \(d(x^2y^2)/dx = d(1)/dx\)…
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