A) - 4
B) 4
C) -1
D) None of these
Correct Answer: A
Solution :
Let \[y=\cos e{{c}^{-1}}\left( \frac{1}{2{{x}^{2}}-1} \right)\] and \[z=\sqrt{1-{{x}^{2}}}\] Now, \[y={{\sin }^{-1}}(2{{x}^{2}}-1)=\frac{\pi }{2}-{{\cos }^{1}}(2{{x}^{2}}-1)\] Put \[x=\cos \theta ,\] \[y=\frac{\pi }{2}-{{\cos }^{-1}}(cos2\theta )=\frac{\pi }{2}-2.{{\cos }^{-1}}x\] \[\Rightarrow \] \[\frac{dy}{dx}=+\frac{2}{\sqrt{1-{{x}^{2}}}}\] ?(i) and \[z=\sqrt{1-{{x}^{2}}},\frac{dz}{dx}=\frac{-x}{\sqrt{1-{{x}^{2}}}}\] \[\therefore \] \[\frac{dy}{dz}=\frac{dy}{dx}\times \frac{dx}{dz}=\frac{2}{\sqrt{1-{{x}^{2}}}}\times \left( -\frac{\sqrt{1-{{x}^{2}}}}{x} \right)\] \[\Rightarrow \] \[\frac{dy}{dz}=-\frac{2}{x}\] \[{{\left( \frac{dy}{dx} \right)}_{at(x=1/2)}}=-4\]
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