A) 2
B) 4
C) 6
D) 1
Correct Answer: B
Solution :
Let\[{{y}_{1}}={{\sec }^{-1}}\frac{1}{2{{x}^{2}}-1}\]and\[{{y}_{2}}=\sqrt{1-{{x}^{2}}}\] \[\Rightarrow \] \[\frac{d{{y}_{1}}}{dx}=\frac{-2}{\sqrt{1-{{x}^{2}}}}\]and\[\frac{d{{y}_{2}}}{dx}=\frac{-x}{\sqrt{1-{{x}^{2}}}}\] \[\Rightarrow \] \[\frac{d{{y}_{1}}}{d{{y}_{2}}}=\frac{2}{x}\] \[\Rightarrow \] \[{{\left( \frac{d{{y}_{1}}}{d{{y}_{2}}} \right)}_{x=1/2}}=\frac{2}{1/2}=4\]You need to login to perform this action.
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