A) \[\left( 2,\frac{-2}{3} \right)\]
B) \[\left( 3,\frac{-3}{8} \right)\]
C) \[\left( -2,\frac{2}{3} \right)\]
D) \[\left( -\sqrt{3},\frac{\sqrt{3}}{2} \right)\]
Correct Answer: D
Solution :
Slope of normal is \[=-1\Rightarrow \]slope of tangent =1 \[f'(x)=\frac{1+{{x}^{2}}}{{{(1-{{x}^{2}})}^{2}}}=1\]i.e.\[x=0,\pm \sqrt{3}\] \[\therefore \]\[\left( -\sqrt{3},\frac{\sqrt{3}}{2} \right)\]is a pointYou need to login to perform this action.
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