# Let $S$ be the set of all values of $x$ for which the tangent to the curve $y=f\left ( x \right )=x^{3}-x^{2}-2x$ at $\left ( x,y \right )$ is parallel to the line segment joining the points $\left ( 1, f\left ( 1 \right )\right )$ and $\left ( -1,f\left ( -1 \right ) \right ),$then $S$ is equal to : Option 1)  $\left \{ \frac{1}{3}, 1\right \}$      Option 2) $\left \{ -\frac{1}{3},-1 \right \}$     Option 3) $\left \{ \frac{1}{3},-1 \right \}$ Option 4) $\left \{- \frac{1}{3},1 \right \}$

$y=f\left ( x \right )=x^{3}-x^{2}-2x$

$\frac{dy}{dx}=3x^{2}-2x-2$

Slope of line segment joining

$\left ( 1,f\left ( 1 \right ) \right )$ and $\left ( -1,f\left ( -1 \right ) \right )$ is

$m=\frac{f\left ( -1 \right )-f\left ( 1 \right )}{-1-1}$

This is equal to $\frac{dy}{dx}$

$f\left ( -1 \right )=\left ( -1 \right )^{3}-\left ( -1 \right )^{2}-2\left ( -1 \right )=0$

$f\left ( -1 \right )=\left ( 1 \right )^{3}-\left ( 1 \right )^{2}-2\left ( 1 \right )=-2$

$3x^{2}-2x-2=-1$

$3x^{2}-2x-1=0$

$x=1,x=-\frac{1}{3}$

$\left \{ 1,-\frac{1}{3} \right \}$

Option 1)

$\left \{ \frac{1}{3}, 1\right \}$

Option 2)

$\left \{ -\frac{1}{3},-1 \right \}$

Option 3)

$\left \{ \frac{1}{3},-1 \right \}$

Option 4)

$\left \{- \frac{1}{3},1 \right \}$

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