A) \[\frac{1}{2}\]
B) \[\frac{2}{3}\]
C) \[\frac{1}{3}\]
D) \[\frac{2}{5}\]
E) \[\frac{1}{5}\]
Correct Answer: C
Solution :
\[x={{\sin }^{-1}}(3t-4{{t}^{3}})\] and \[y={{\cos }^{-1}}(\sqrt{1-{{t}^{2}}})\] Put \[t=\sin \theta \] ...(i) Then, \[x={{\sin }^{-1}}(3\sin \theta -4{{\sin }^{3}}\theta )\] \[={{\sin }^{-1}}(\sin 3\theta )=3\theta =3{{\sin }^{-1}}t\] and \[y={{\cos }^{-1}}\sqrt{1-{{\sin }^{2}}\theta }\] \[={{\cos }^{-1}}(\cos \theta )=\theta ={{\sin }^{-1}}t\] Now, \[\frac{dx}{dt}=\frac{3}{\sqrt{1-{{t}^{2}}}}\] \[\frac{dx}{dt}=\frac{1}{\sqrt{1-{{t}^{2}}}}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}\] \[\frac{dy}{dx}=\frac{1}{\sqrt{1-{{t}^{2}}}}\times \frac{\sqrt{1-{{t}^{2}}}}{3}=\frac{1}{3}\]
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