Question

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

• A. $\{\frac{1}{3},1\}$
• B. $\{\frac{1}{3},-1\}$
• C. $\{-\frac{1}{3},-1\}$
• D. $\{-\frac{1}{3},1\}$

Answer 1

The correct option is D $\{-\frac{1}{3},1\}$

$y=f\left(x\right)=x^3-x^2-2x\Rightarrow f^{\prime }\left(x\right)=3x^2-2x-2\cdots \left(i\right)\text{Also},f\left(1\right)=1-1-2=-2\text{and}f(-1)=-1-1+2=0\Rightarrow m=\frac{f\left(1\right)-f(-1)}{1+1}=\frac{-2-0}{2}\Rightarrow m=-1\cdots \left(ii\right)$
$From\left(i\right)and\left(ii\right)3x^2-2x-2=-1\Rightarrow 3x^2-2x-1=0\Rightarrow (3x+1)(x-1)=0\Rightarrow x=-\frac{1}{3},1$