Question

If $x=\sqrt{\frac{1-t^2}{1+t^2}}$ and $y=\frac{\sqrt{1+t^2}-\sqrt{1-t^2}}{\sqrt{1+t^2}+\sqrt{1-t^2}},$ then the value of $\frac{d^{2}y}{d{x}^2}$ at $t=0$ is given by

• A. $\frac{-1}{2}$
• B. $\frac{1}{2}$
• C. $\frac{-1}{4}$
• D. $\frac{1}{4}$

Answer 1

Answer: (B)

$\frac{1}{2}$

$y=\frac{\sqrt{1+t^2}-\sqrt{1-t^2}}{\sqrt{1+t^2}+\sqrt{1-t^2}}$

Dividing both Nr' and Dr' by $\sqrt{1+t^2},$ we get

$y=\frac{1-x}{1+x}=\frac{-1-x+2}{1+x}=-1+\frac{2}{1+x}$

Differentiating w.r.t x,

$\frac{dy}{dx}=\frac{-2}{{(1+x)}^2}$

and $\frac{d^{2}y}{d{x}^2}=\frac{4}{{(1+x)}^3}=\frac{4}{8}=\frac{1}{2}.$

Because when $t=0,x=1.$