• # question_answer Let S be the set of all values of x for which the tangent to the curve $y=f(x)={{x}^{3}}-{{x}^{2}}-2x$at $\left( x,y \right)$is parallel to the line segment joining the points $\left( 1,\left( 1 \right) \right)$and $\left( 1,\left( 1 \right) \right),$then S is equal to : [JEE Main 9-4-2019 Morning] A) $\left\{ -\frac{1}{3},-1 \right\}$                     B) $\left\{ \frac{1}{3},-1 \right\}$C) $\left\{ -\frac{1}{3},1 \right\}$D) $\left\{ \frac{1}{3},1 \right\}$

$\left( 1 \right)=112=2$ $\left( 1 \right)=11+2=0$ $m=\frac{f(1)-f(-1)}{1+1}=\frac{-2-0}{2}=-1$ $\frac{dy}{dx}=3{{x}^{2}}-2x-2$ $3{{x}^{2}}-2x-2=-1$ $\Rightarrow 3{{x}^{2}}-2x-1=0$ $\Rightarrow (x-1)(3x+1)=0$ $\Rightarrow x=1,-\frac{1}{3}$