A) \[\frac{2}{3}\]
B) \[\frac{3}{2}\]
C) \[\frac{1}{2}\]
D) \[1\]
E) \[0\]
Correct Answer: A
Solution :
Let \[y={{\sin }^{-1}}(2x\sqrt{1-{{x}^{2}}})\] ?. (i) and \[z={{\sin }^{-1}}(3x-4{{x}^{3}})\] ...(ii) Now,\[x=cos\theta \]putting in Eq. (i), we get \[y={{\sin }^{-1}}(2\cos \theta \sqrt{1-{{\cos }^{2}}\theta })\] \[={{\sin }^{-1}}\,(2\cos \,\theta \,\sin \theta )\] \[={{\sin }^{-1}}(\sin 2\theta )\] \[\Rightarrow \] \[y=2\theta \] \[\Rightarrow \] \[y=2{{\cos }^{-1}}x\] Differentiating it w.r.t.\[\theta \], we get \[\frac{dz}{d\theta }=3\] ...(iii) Also, putting\[x=sin\theta \]in Eq. (ii), we get \[z={{\sin }^{-1}}(3\sin \theta -4{{\sin }^{3}}\theta )={{\sin }^{-1}}(\sin 3\theta )\] \[\therefore \] \[z=3\theta \] Differentiating it w.r.t.\[\theta \], we get \[\frac{dz}{d\theta }=3\] ...(iv) Now, \[\frac{dy}{dz}=\frac{dy}{d\theta }.\frac{d\theta }{dz}\] \[=2.\frac{1}{3}=\frac{2}{3}\] \[\therefore \] \[\frac{d({{\sin }^{-1}}2x\sqrt{1-{{x}^{2}}})}{d({{\sin }^{-1}}(3x-4{{x}^{3}}))}=\frac{2}{3}\]
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