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} xy' = $

• 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) $
$\therefore y' = \frac{1}{2} . \frac{1}{1+a^{2}x^{2}}. a = \frac{a}{2\left(1+a^{2}x^{2}\right)}$
$ \Rightarrow y'\left(1+a^{2}x^{2}\right) = \frac{a}{2}$
$ \Rightarrow y"\left(1+a^{2}x^{2}\right) =y'\left(2a^{2}x \right) =0$
$\Rightarrow \left(1+a^{2}x^{2}\right) + y" 2a' xy' =0$