CUET · MATHS · PYQ PAPER 2025
If \(x \sqrt{1+y}+y \sqrt{1+x}=0\), where \(|x|<1,|y|<1\) and \(x \neq y\), then
- A \(\frac{d y}{d x}=\sqrt{\frac{1+y}{1+x}}\)
- B \((1+x)^2 \frac{d y}{d x}+1=0\)
- C \(\left(1+x^2\right) \frac{d y}{d x}+1=0\)
- D \((1+x)^2 \frac{d y}{d x}=1\)
Answer & Solution
Correct Answer
(B) \((1+x)^2 \frac{d y}{d x}+1=0\)
Step-by-step Solution
Detailed explanation
\(x \sqrt{1+y} = -y \sqrt{1+x}\) \(x^2(1+y) = y^2(1+x)\) \(x^2 - y^2 = y^2x - x^2y\) \((x-y)(x+y) = -xy(x-y)\) \(x+y = -xy\) \(y(1+x) = -x\) \(y = \frac{-x}{1+x}\) \(\frac{dy}{dx} = \frac{-(1)(1+x) - (-x)(1)}{(1+x)^2}\) \(\frac{dy}{dx} = \frac{-1}{(1+x)^2}\)…
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