Question

Find the points on the curve $y=x^3-2x^2-x$ at which the tangent lines are parallel to the line $y=3x-2$

• A. $(2,2),\left(-\frac{2}{3},-\frac{14}{27}\right)$
• B. $(2,-2),\left(-\frac{2}{3},-\frac{14}{37}\right)$
• C. $(2,-2),\left(-\frac{2}{3},-\frac{14}{27}\right)$
• D. None of the above:

Answer 1

Answer: (C)

Let $(x_1,y_1)$ be the required point. We have the given curve $y=x^3-2x^2-x$ ...(i) $\frac{dy}{dx}=3x^2-4x-1$ ${\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=3x_1^2-4x_1-1$ This is the slope of the tangent to the curve but the tangent is parallel to the line $y=3x-2$ $\therefore$ $3x_1^2-4x_1-1=3$ $\Rightarrow$ $3x_1^2-4x_1-4=0$ $\Rightarrow$ $(x_1-2)(3x_1+2)=0$ $\Rightarrow$ $x_1=2,-\frac{2}{3}$ Since, the point $(x_1,y_1)$ lies on the curve (i), $\therefore$ At $x_1=2y_1=2^3-2{(2)}^2-2=-2$ at $x_1=-\frac{2}{3},y_1={\left(-\frac{2}{3}\right)}^3-2{\left(-\frac{2}{3}\right)}^2+\frac{2}{3}=-\frac{14}{27}$ Hence, the required points are $(2,-2)$ and $\left(-\frac{2}{3},-\frac{14}{27}\right)$