Let S be the set of all values of x for which the tangent to the curve
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:

  1. (1) $\left\{\frac{1}{3}, 1\right\}$

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

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

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


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

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