Question & Answer
QUESTION

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

ANSWER Verified Verified
Hint: Find the derivative of the given curve to determine the slope of its tangent. Calculate the slope of the line segment joining the two points by using the formula: $slope=\dfrac{\Delta y}{\Delta x}$. Equate the derivative of the curve with the slope of the line segment and solve the obtained quadratic equation to get the values of $x$.

Complete step-by-step solution -
Two lines are said to be parallel when their slopes have the same value.
Now, let us come to the question. We have been provided with the curve, $y=f(x)={{x}^{3}}-{{x}^{2}}-2x$. Slope of the tangent of a given curve $y$ at any particular point is given by $\dfrac{dy}{dx}$ at that particular point. Therefore, slope of the given curve is given as,
$\begin{align}
  & \dfrac{dy}{dx}=\dfrac{d({{x}^{3}}-{{x}^{2}}-2x)}{dx} \\
 & \therefore \dfrac{dy}{dx}=3{{x}^{2}}-2x-2......................(i) \\
\end{align}$
Also, the slope of a line segment joining two points is ‘the ratio of change in $y$to the change in $x$’ or mathematically, $\dfrac{\Delta y}{\Delta x}$.
Now, $f(1)={{1}^{3}}-{{1}^{2}}-2\times 1=1-1-2=-2$, and $f(-1)={{(-1)}^{3}}-{{(-1)}^{2}}-2\times (-1)=-1-1+2=0$. Therefore, the two points are: $(1,-2)\text{ and }(-1,0)$. So, the slope of the line segment joining these two points will be,
$slope=\dfrac{\Delta y}{\Delta x}=\dfrac{-2-0}{1-(-1)}=\dfrac{-2}{2}=-1..................(ii)$
Now, for the tangent of the curve and the line segment to be parallel, their slopes must be equal.
$\begin{align}
  & \therefore 3{{x}^{2}}-2x-2=-1 \\
 & 3{{x}^{2}}-2x-2+1=0 \\
 & 3{{x}^{2}}-2x-1=0 \\
\end{align}$
Splitting the middle term we get,
$\begin{align}
  & 3{{x}^{2}}-3x+x-1=0 \\
 & 3x(x-1)+1(x-1)=0 \\
 & (3x+1)(x-1)=0 \\
 & \therefore x=1\text{ or }x=\dfrac{-1}{3} \\
\end{align}$
Hence, option (c) is the correct answer.

Note: One can note that the value of $f(1)\text{ and }f(-1)$ are determined by substituting the value of $x=1\text{ and }-1$ respectively, in the function, $y=f(x)$. You should keep in mind that, when two lines are parallel then their slopes are equal and when two lines are perpendicular then the product of their slopes is $-1$.