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Question:
Let $S$ be the set of all values of $x$ for which the tangent to the curve $\mathrm{y}=f(\mathrm{x})=\mathrm{x}^{3}-\mathrm{x}^{2}-2 \mathrm{x}$ at $(x, y)$ is parallel to the line segment joining the points $(1, f(1))$ and $(-1, f(-1))$, then $\mathrm{S}$ is equal to :
Correct Option: , 3
Solution:
$f(1)=1-1-2=-2$
$f(-1)=-1-1+2=0$
$\mathrm{m}=\frac{f(1)-f(-1)}{1+1}=\frac{-2-0}{2}=-1$
$\frac{d y}{d x}=3 x^{2}-2 x-2$
$3 x^{2}-2 x-2=-1$
$\Rightarrow 3 x^{2}-2 x-1=0$
$\Rightarrow(x-1)(3 x+1)=0$
$\Rightarrow x=1,-\frac{1}{3}$
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