A) \[\frac{1}{4}\]
B) \[\frac{3}{2}\]
C) 1
D) \[\frac{2}{3}\]
E) \[-\frac{2}{3}\]
Correct Answer: D
Solution :
\[u={{\sec }^{-1}}\left( \frac{1}{1-2{{x}^{2}}} \right),v={{\sin }^{-1}}(3x-4{{x}^{3}})\] Let \[x=\sin \theta \] \[\Rightarrow \] \[u{{\sec }^{-1}}(\sec 2\theta ),v={{\sin }^{-1}}(\sin 3\theta )\] \[\Rightarrow \] \[u=2\theta ,v=3\theta \] \[\Rightarrow \] \[u=2{{\sin }^{-1}}x,v=3{{\sin }^{-1}}x\] On differentiating w.r.t. \[\theta \]respectively \[\frac{du}{d\theta }=\frac{2}{\sqrt{1-{{x}^{2}}}},\frac{dv}{d\theta }=\frac{3}{\sqrt{1-{{x}^{2}}}}\] \[\therefore \] \[\frac{du}{d\theta }=\frac{\frac{\frac{du}{d\theta }}{dv}}{d\theta }=\frac{2}{3}\]
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