Q117 / 180 · KCET 2010 · Mathematics
If $y=\tan ^{-1} \sqrt{\mathrm{x}^{2}-1}$, then the ratio $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx} \mathrm{x}^{2}}: \frac{\mathrm{dy}}{\mathrm{dx}}$ is
- A$\frac{x\left(x^{2}-1\right)}{1+2 x^{2}}$
- Correct answerB$\frac{1-2 x^{2}}{x\left(x^{2}-1\right)}$
- C$\frac{1+2 x^{2}}{x\left(x^{2}+1\right)}$
- D$\frac{x\left(x^{2}+1\right)}{1-2 x^{2}}$
Solution
$\mathrm{y}=\tan ^{-1} \sqrt{\mathrm{x}^{2}-1}$,
Put $\left\{\begin{array}{l}\mathrm{x}=\sec \theta \\ \mathrm{dx}=\sec \theta \cdot \tan \theta \mathrm{d} \theta\end{array}\right.$
$\mathrm{y}=\tan ^{-1} \sqrt{\sec ^{2} \theta-1}=\tan ^{-1}(\tan \theta)=\theta$
$=\sec ^{-1} \mathrm{x}$
$\Rightarrow \quad \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\sec ^{-1} \mathrm{x}\right)=\frac{1}{\mathrm{x} \sqrt{\mathrm{x}^{2}-1}}$
$\begin{aligned} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}} &=\frac{1}{\mathrm{x}} \cdot \frac{-1}{2} \frac{1}{\left(\mathrm{x}^{2}-1\right)^{3 / 2}}(2 \mathrm{x})-\frac{1}{\mathrm{x}^{2}} \cdot \frac{1}{\sqrt{\mathrm{x}^{2}-1}} \\ \quad=-\frac{1}{\left(\mathrm{x}^{2}-1\right)^{3 / 2}}-\frac{1}{\mathrm{x}^{2}\left(\mathrm{x}^{2}-1\right)^{1 / 2}} \\=-\frac{1}{\mathrm{x}^{2}\left(\mathrm{x}^{2}-1\right)^{3 / 2}}\left(\mathrm{x}^{2}+\mathrm{x}^{2}-1\right) \\ \quad=-\frac{\left(2 \mathrm{x}^{2}-1\right)}{\mathrm{x}^{2}\left(\mathrm{x}^{2}-1\right)^{3 / 2}} \end{aligned}$
$$
\begin{aligned}
&\text { Now, } \frac{d^{2} y}{d x^{2}}: \frac{d y}{d x}=-\frac{\left(2 x^{2}-1\right)}{x^{2}\left(x^{2}-1\right)^{3 / 2}}: \frac{1}{x\left(x^{2}-1\right)^{1 / 2}} \\
&\frac{d^{2} y}{d x^{2}}: \frac{d y}{d x}=\left(1-2 x^{2}\right): x\left(x^{2}-1\right) \\
&\text { or } \quad\left(\frac{d^{2} y}{\frac{d x^{2}}{d y}}\right)=\frac{\left(1-2 x^{2}\right)}{x\left(x^{2}-1\right)}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.