If $\sqrt{x+y}+\sqrt{y-x}=c$ then $\frac{d^2 y}{d x^2}$ is :
Answer & explanation
Correct answer: option 3
We are given $\sqrt{x+y}+\sqrt{y-x}=c$ . . . (1)
Also $(\sqrt{x+y})^2-(\sqrt{y-x})^2=x+y-(y-x)$
$\Rightarrow(\sqrt{x+y}+\sqrt{y-x})(\sqrt{x+y}-\sqrt{y-x})=2 x$
$\Rightarrow \sqrt{x+y}-\sqrt{y-x}=\frac{2 x}{c}$ [By (1)] . . . (2)
(1) + (2) gives $2 \sqrt{x+y}=c+\frac{2 x}{c}$
Squaring $4(x+y)=c^2+\frac{4 x^2}{c^2}+4 x$
Canceling 4x, $4 y=c^2+\frac{4 x^2}{c^2}$
$\Rightarrow 4 \frac{d y}{d x}=\frac{8 x}{c^2} \Rightarrow \frac{d^2 y}{d x^2}=\frac{2}{c^2}$
Hence (3) is correct answer.