Question:
Let $S$ be the set of all values of $x$ for which the tangent to the curve $y=f(x)=x^{3}-x^{2}-2 x$ at $(x, y)$ is parallel to the line segment joining the points $(1, f(1))$ and $(-1, f(-1))$, then $S$ is equal to:
Correct Option: , 4
Solution:
$y=f(x)=x^{3}-x^{2}-2 x$
$\Rightarrow \frac{d y}{d x}=3 x^{2}-2 x-2$
$f(1)=1-1-2=-2, \quad f(-1)=-1-1+2=0$
Since the tangent to the curve is parallel to the line segment joining the points $(1,-2)$ and $(-1,0)$
And their slopes are equal.
$\Rightarrow 3 x^{2}-2 x-2=\frac{-2-0}{2}$
$\Rightarrow x=1, \frac{-1}{3}$
Hence, the required set $S=\left\{\frac{-1}{3}, 1\right\}$