Question

$y={\tan }^{-1}\left(\frac{\sqrt{1+a^2{x}^2}-1}{ax}\right)\Rightarrow \left(1+a^2{x}^2\right)y"+2a^{2}x{y}^{\prime }=$

• A. $-2a^2$
• B. $a^2$
• C. $2a^3$
• D. 0

Answer 1

Answer: (D)

Put $ax=\tan \theta$
$\therefore y={\tan }^{-1}\left(\frac{\sqrt{1+{\tan }^2\theta }-1}{\tan \theta }\right)$
$={\tan }^{-1}\left(\frac{\sec \theta -1}{\tan \theta }\right)={\tan }^{-1}\left(\frac{1-\cos \theta }{\sin \theta }\right)$
$={\tan }^{-1}\left(\frac{2{\sin }^2\theta /2}{2\sin \frac{\theta }{2}\cos \theta /2}\right)=\frac{\theta }{2}$
$={\tan }^{-1}\left(\tan \frac{\theta }{2}\right)=\frac{\theta }{2}=\frac{1}{2}{\tan }^{-1}\left(ax\right)$
$\therefore y^{\prime }=\frac{1}{2}.\frac{1}{1+a^2{x}^2}.a=\frac{a}{2\left(1+a^2{x}^2\right)}$
$\Rightarrow y^{\prime }\left(1+a^2{x}^2\right)=\frac{a}{2}$
$\Rightarrow y"\left(1+a^2{x}^2\right)=y^{\prime }\left(2a^{2}x\right)=0$
$\Rightarrow \left(1+a^2{x}^2\right)+y"2a^{\prime }x{y}^{\prime }=0$