If \(x\sqrt{1 + y} + y\sqrt{1 + x} = 0\), where \(|x| \lt 1, |y| \lt 1\) and \(x\ne y\), then
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle(1+x)^2\frac{dy}{dx}+1=0\)
The relation can be solved for y. Moving one term across and squaring:
\(x\sqrt{1+y} = -y\sqrt{1+x}\)
\(x^2(1+y) = y^2(1+x)\)
\(x^2 - y^2 + x^2y - xy^2 = 0\)
\((x-y)(x+y) + xy(x-y) = 0\)
\((x-y)(x+y+xy) = 0\)
Since \(x \ne y\), the second factor is zero:
\(x + y + xy = 0\)
\(y(1+x) = -x\)
\(\displaystyle y = -\frac{x}{1+x}\)
By the quotient rule:
\(\displaystyle \frac{dy}{dx} = -\frac{(1+x)(1) - x(1)}{(1+x)^2} = -\frac{1}{(1+x)^2}\)
\(\displaystyle (1+x)^2\frac{dy}{dx} + 1 = 0\)
NCERT (2026-27 edition):
Continuity and Differentiability
