# Solve this following

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 :

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

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

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

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

Correct Option: , 3

Solution:

$f(1)=1-1-2=-2$

$f(-1)=-1-1+2=0$

$\mathrm{m}=\frac{f(1)-f(-1)}{1+1}=\frac{-2-0}{2}=-1$

$\frac{d y}{d x}=3 x^{2}-2 x-2$

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

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

$\Rightarrow(x-1)(3 x+1)=0$

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