A) \[\frac{x({{x}^{2}}-1)}{1+2{{x}^{2}}}\]
B) \[\frac{1-2{{x}^{2}}}{x({{x}^{2}}-1)}\]
C) \[\frac{1+2{{x}^{2}}}{x({{x}^{2}}+1)}\]
D) \[\frac{x({{x}^{2}}+1)}{1-2{{x}^{2}}}\]
Correct Answer: B
Solution :
\[y={{\tan }^{-1}}\sqrt{{{x}^{2}}-1},\] Put\[\left\{ \begin{matrix} x=\sec \theta \\ dx=\sec \theta .\tan \theta d\theta \\ \end{matrix} \right.\] \[y={{\tan }^{-1}}\sqrt{{{\sec }^{2}}\theta -1}={{\tan }^{-1}}(tan\theta )=\theta \] \[={{\sec }^{-1}}x\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{d}{dx}({{\sec }^{-1}}x)=\frac{1}{x\sqrt{{{x}^{2}}-1}}\] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{1}{x}.\frac{-1}{2}=\frac{1}{{{({{x}^{2}}-1)}^{3/2}}}(2x)-\frac{1}{{{x}^{2}}}.\frac{1}{\sqrt{{{x}^{2}}-1}}\] \[=-\frac{1}{{{({{x}^{2}}-1)}^{3/2}}}-\frac{1}{{{x}^{2}}{{({{x}^{2}}-1)}^{3/2}}}\] \[=-\frac{1}{{{x}^{2}}{{({{x}^{2}}-1)}^{3/2}}}({{x}^{2}}+{{x}^{2}}-1)\] \[=-\frac{(2{{x}^{3}}-1)}{{{x}^{2}}{{({{x}^{2}}-1)}^{3/2}}}\] Now,\[\frac{{{d}^{2}}y}{d{{x}^{2}}}:\frac{dy}{dx}=-\frac{(2{{x}^{2}}-1)}{{{x}^{2}}{{({{x}^{2}}-1)}^{3/2}}}:\frac{1}{x{{({{x}^{2}}-1)}^{1/2}}}\] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}:\frac{dy}{dx}=(1-2{{x}^{2}}):x({{x}^{2}}-1)\] Or \[\left( \frac{\frac{{{d}^{2}}x}{d{{x}^{2}}}}{\frac{dy}{dx}} \right)=\frac{(1-2{{x}^{2}})}{x({{x}^{2}}-1)}\]
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