Application of Derivatives 1 Question 2
2. 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
(a) $\frac{1}{3},-1$
(b) $\frac{1}{3}, 1$
(c) $-\frac{1}{3}, 1$
(d) $-\frac{1}{3},-1$
(2019 Main, 9 April I)
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Answer:
Correct Answer: 2. (c)
Solution:
- Given curve is $y=f(x)=x^{3}-x^{2}-2 x$
So, $f(1)=1-1-2=-2$
and $f(-1)=-1-1+2=0$
Since, slope of a line passing through $\left(x _1, y _1\right)$ and
$\left(x _2, y _2\right)$ is given by $m=\tan \theta=\frac{y _2-y _1}{x _2-x _1}$
$\therefore$ Slope of line joining points $(1, f(1))$ and $(-1, f(-1))$ is $m=\frac{f(1)-f(-1)}{1-(-1)}=\frac{-2-0}{1+1}=-1$
Now, $\frac{d y}{d x}=3 x^{2}-2 x-2$
According to the question,
[differentiating Eq. (i), w.r.t. ’ $x^{2}$ ]
$$ \begin{array}{ll} & \frac{d y}{d x}=m \\ \Rightarrow & 3 x^{2}-2 x-2=-1 \Rightarrow 3 x^{2}-2 x-1=0 \\ \Rightarrow & (x-1)(3 x+1)=0 \Rightarrow x=1,-\frac{1}{3} \end{array} $$
Therefore, set $S=-\frac{1}{3}, 1$.
