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If $y(x)=\tan ^{-1}\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}\right)$ and $\left(1+a^2 x^2\right) y^{\prime \prime}+g(x) y^{\prime}=0$ then, the sum of the roots of the equation $1+a^2 x^2+g(x)=0$ is
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The correct answer is:
-2
Given, $y=\tan ^{-1}\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}\right)$
$\begin{aligned} & \text { put } a x=\tan \theta \Rightarrow \theta=\tan ^{-1}(a x) \\ & y=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right) \\ & \Rightarrow \quad y=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)=\tan ^{-1} \tan \frac{\theta}{2}=\frac{\theta}{2} \\ & \Rightarrow \quad y=\frac{1}{2} \tan ^{-1}(a x) \Rightarrow y^{\prime}=\frac{1}{2} \frac{a}{1+a^2 x^2} \\ & \Rightarrow \quad\left(1+a^2 x^2\right) y^{\prime}=\frac{a}{2} \Rightarrow\left(1+a^2 x^2\right) y^{\prime \prime}+2 a^2 x y^{\prime}=0 \\ & \therefore \quad g(x)=2 a^2 x \Rightarrow 1+a^2 x^2+g(x)=0 \\ & a^2 x^2+2 a^2 x+1=0 \\ & \text { Sum of roots }=\frac{-2 a^2}{a^2}=-2\end{aligned}$
$\begin{aligned} & \text { put } a x=\tan \theta \Rightarrow \theta=\tan ^{-1}(a x) \\ & y=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right) \\ & \Rightarrow \quad y=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)=\tan ^{-1} \tan \frac{\theta}{2}=\frac{\theta}{2} \\ & \Rightarrow \quad y=\frac{1}{2} \tan ^{-1}(a x) \Rightarrow y^{\prime}=\frac{1}{2} \frac{a}{1+a^2 x^2} \\ & \Rightarrow \quad\left(1+a^2 x^2\right) y^{\prime}=\frac{a}{2} \Rightarrow\left(1+a^2 x^2\right) y^{\prime \prime}+2 a^2 x y^{\prime}=0 \\ & \therefore \quad g(x)=2 a^2 x \Rightarrow 1+a^2 x^2+g(x)=0 \\ & a^2 x^2+2 a^2 x+1=0 \\ & \text { Sum of roots }=\frac{-2 a^2}{a^2}=-2\end{aligned}$
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