JEE PYQ: Application Of Derivatives Question 62
Question 62 - 2019 (09 Apr Shift 1)
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:
(1) $\left{\frac{1}{3}, 1\right}$
(2) $\left{\frac{1}{3}, -1\right}$
(3) $\left{-\frac{1}{3}, -1\right}$
(4) $\left{-\frac{1}{3}, 1\right}$
Show Answer
Answer: (4)
Solution
$f(1) = -2$, $f(-1) = 0$. Slope $= \frac{-2-0}{2} = -1$. $f’(x) = 3x^2 - 2x - 2 = -1 \Rightarrow 3x^2 - 2x - 1 = 0 \Rightarrow x = 1, -\frac{1}{3}$. $S = \left{-\frac{1}{3}, 1\right}$.
BETA
