If \(\sqrt{y+x}+\sqrt{y-x}=a\), \(a \gt 1\), \(\displaystyle\frac{d^2y}{dx^2}\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{2}{a^2}\)
Instead of differentiating the roots directly, it is easier to find y as an explicit function of x. Let
\(u = \sqrt{y+x},\quad v = \sqrt{y-x}\)
Then
\(u + v = a\)
\(u^2 - v^2 = (y+x) - (y-x) = 2x\)
Since \(u^2 - v^2 = (u+v)(u-v)\):
\( \displaystyle u - v = \frac{2x}{a} \)
Adding this to \(u + v = a\):
\( \displaystyle 2u = a + \frac{2x}{a} \)
\( \displaystyle u = \frac{a}{2} + \frac{x}{a} \)
Squaring, with \(u^2 = y + x\):
\( \displaystyle y + x = \frac{a^2}{4} + x + \frac{x^2}{a^2} \)
\( \displaystyle y = \frac{a^2}{4} + \frac{x^2}{a^2} \)
Differentiating twice:
\( \displaystyle \frac{dy}{dx} = \frac{2x}{a^2} \)
\( \displaystyle \frac{d^2y}{dx^2} = \frac{2}{a^2} \)