Question

If $y={\tan }^{-1}\left(\frac{\sqrt{1+a^2{x}^2}-1}{ax}\right)$
then $\left(1+a^2{x}^2\right)y^{{}^{\prime \prime }}+2a^{2}x{y}^{{}^{\prime }}$ is equal to

• A. $-2a^2$
• B. $a^2$
• C. $2a^2$
• D. 0

Answer 1

Answer: (D)

Given, $y={\tan }^{-1}\left(\frac{\sqrt{1+a^2{x}^2}-1}{ax}\right)$
Put $ax=\tan \theta$
$\therefore y={\tan }^{-1}\left(\frac{\sqrt{1+{\tan }^2\theta }-1}{\tan \theta }\right)$
$={\tan }^{-1}\left(\frac{\sec \theta -1}{\tan \theta }\right)$
$={\tan }^{-1}\left(\frac{1-\cos \theta }{\sin \theta }\right)$
$={\tan }^{-1}\left(\frac{2{\sin }^2\frac{\theta }{2}}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}\right)$
$={\tan }^{-1}\left(\tan \frac{\theta }{2}\right)$
$=\frac{\theta }{2}=\frac{1}{2}{\tan }^{-1}ax$
$\therefore y=\frac{1}{2}{\tan }^{-1}ax$
On differentiating w.r.t. $x$, we get
$y^{{}^{\prime }}=\frac{1}{2\left(1+a^2{x}^2\right)}...(i)$
Again, differentiating both side we get
$y^{{}^{\prime \prime }}=-\frac{1}{2}\frac{\left(a^{2}2x\right)}{{\left(1+a^2{x}^2\right)}^2}$
$\Rightarrow \left(1+a^2{x}^2\right)y^{{}^{\prime \prime }}=-\frac{a^{2}x}{1+a^2{x}^2}$
$=-a^{2}x\left(2y^{{}^{\prime }}\right)[\because$ from Eq.(i)]
$\Rightarrow \left(1+a^2{x}^2\right)y^{{}^{\prime \prime }}+2a^{2}x{y}^{{}^{\prime }}=0$