A) 0
B) 1/2
C) 1/3
D) None of these
Correct Answer: A
Solution :
\[{{\sec }^{-1}}\frac{1}{(2{{x}^{2}}-1)}=2{{\cos }^{-1}}x\] \[\therefore y=2{{\cos }^{-1}}x,\ \ z=\sqrt{1+3x}\] \[\frac{dy}{dz}=\frac{dy}{dx}\div \frac{dz}{dx}=-\frac{2}{\sqrt{1-{{x}^{2}}}}.\frac{2\sqrt{1+3x}}{3}=0,\,\left( \text{at}\ \ \ x=-\frac{1}{3} \right)\].
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