Question

If $y=\frac{2}{\sqrt{a^2-b^2}}{\tan }^{-1}\left[\sqrt{\frac{a-b}{a+b}}\tan \frac{x}{2}\right]$, then ${\frac{d^{2}y}{d{x}^2}|}_{x=\frac{\pi }{2}}=$

• A. $\frac{b}{2a^2}$
• B. $\frac{b}{a^2}$
• C. $\frac{2b}{a}$
• D. $\frac{b^2}{2a}$

Answer 1

Answer: (B)

We have,
$y=\frac{2}{\sqrt{a^2-b^2}}{\tan }^{-1}\left[\sqrt{\frac{a-b}{a+b}}\tan \frac{x}{2}\right]$
$\Rightarrow \frac{dy}{dx}=[\frac{2}{\sqrt{a^2-b^2}}\cdot \frac{1}{1+\left(\frac{a-b}{a+b}\right){\tan }^2\frac{x}{2}}$
$\cdot \sqrt{\frac{a-b}{a+b}}\cdot {\sec }^2\frac{x}{2}\cdot \frac{1}{2}]$
$=\frac{{\sec }^{2}x/2}{a+b}\frac{a+b}{(a+b)+(a-b){\tan }^{2}x/2}$
$=\frac{{\sec }^{2}x/2}{a+b+a{\tan }^2\frac{x}{2}-b{\tan }^{2}x/2}$
$=\frac{{\sec }^{2}x/2}{a\left(1+{\tan }^2\frac{x}{2}\right)+b\left(1-{\tan }^2\frac{x}{2}\right)}$
$=\frac{{\sec }^{2}x/2}{\left(1+{\tan }^2\frac{x}{2}\right)}\left[a+b\left(\frac{1-{\tan }^{2}x/2}{1+{\tan }^{2}x/2}\right)\right]$
$\frac{dy}{dx}=\frac{1}{a+b\cos x}$
$\therefore \frac{d^{2}y}{d{x}^2}=\frac{-1}{(a+b\cos x{)}^2}\cdot (-b\sin x)$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{b\sin x}{(a+b\cos x{)}^2}$
$\therefore {\frac{d^{2}y}{d{x}^2}|}_{x=\frac{\pi }{2}}=\frac{b\sin \pi /2}{{\left(a+b\cos \frac{\pi }{2}\right)}^2}=\frac{b}{a^2}$