Question

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

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

Answer 1

Answer: (D)

Given, $y=\tan ^{-1}\left(\frac{\sqrt{1+a^{2} x^{2}}-1}{a x}\right)$
Put $a x=\tan \theta$
$\therefore y=\tan ^{-1}\left(\frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta}\right)$
$=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)$
$=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)$
$=\tan ^{-1}\left(\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right)$
$=\tan ^{-1}\left(\tan \frac{\theta}{2}\right)$
$=\frac{\theta}{2}=\frac{1}{2} \tan ^{-1} ax$
$\therefore y=\frac{1}{2} \tan ^{-1} a x$
On differentiating w.r.t. $x$, we get
$y^{'}=\frac{1}{2\left(1+a^{2} x^{2}\right)}\,\,\,...(i)$
Again, differentiating both side we get
$y^{''} =-\frac{1}{2} \frac{\left(a^{2} 2 x\right)}{\left(1+a^{2} x^{2}\right)^{2}} $
$\Rightarrow \left(1+a^{2} x^{2}\right) y^{''} =-\frac{a^{2} x}{1+a^{2} x^{2}}$
$=-a^{2} x\left(2 y^{'}\right) \,\,\,[\because$ from Eq.(i)]
$\Rightarrow \left(1+a^{2} x^{2}\right) y^{''}+2 a^{2} x y^{'}=0$