If y(x)=tan ^{-1}left(frac{sqrt{1+a^2 x^2}-1} | vedclass

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(D) Given, $y=	an ^{-1}\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}ight)$.Substitute $ax = 	an 	heta$, so $	heta = 	an ^{-1}(ax)$.Then $y = 	an ^{-1}\

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$-2$

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(D) Given, $y=	an ^{-1}\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}ight)$.Substitute $ax = 	an 	heta$, so $	heta = 	an ^{-1}(ax)$.Then $y = 	an ^{-1}\left(\frac{\sec 	heta - 1}{	an 	heta}ight) = 	an ^{-1}\left(\frac{1 - \cos 	heta}{\sin 	heta}ight) = 	an ^{-1}\left(	an \frac{	heta}{2}ight) = \frac{	heta}{2}$.Thus, $y = \frac{1}{2} 	an ^{-1}(ax)$.Differentiating with respect to $x$, we get $y^{\prime} = \frac{1}{2} \cdot \frac{a}{1+a^2 x^2}$.This implies $(1+a^2 x^2) y^{\prime} = \frac{a}{2}$.Differentiating again with respect to $x$, we get $(1+a^2 x^2) y^{\prime \prime} + (2a^2 x) y^{\prime} = 0$.Comparing this with $(1+a^2 x^2) y^{\prime \prime} + g(x) y^{\prime} = 0$, we find $g(x) = 2a^2 x$.The equation $1+a^2 x^2 + g(x) = 0$ becomes $a^2 x^2 + 2a^2 x + 1 = 0$.The sum of the roots of a quadratic equation $Ax^2 + Bx + C = 0$ is $-\frac{B}{A}$.Here, the sum of the roots is $-\frac{2a^2}{a^2} = -2$.

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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