Question

The tangents to curve $y=x^3-2x^2+x-2$ which are parallel to straight line $y=x$, are

• A. $x+y=2$ and $x-y=\frac{86}{27}$
• B. $x-y=2$ and $x-y=\frac{86}{27}$
• C. $x-y=2$ and $x+y=\frac{86}{27}$
• D. $x+y=2$ and $x+y=\frac{86}{27}$

Answer 1

Answer: (B)

$x-y=2$ and $x-y=\frac{86}{27}$

Given,
$y=x^3-2x^2+x-2$

On differentiating both sides with respect to $x$, we get

$\frac{dy}{dx}=3x^2-4x+1$

and $y=x$

$\Rightarrow \frac{dy}{dx}=1$

$\therefore$ Slope of tangent will be $3x^2-4x+1$.

Since, the tangent is parallel to line $y=x$.

$\therefore 3x^2-4x+1=1$

$\Rightarrow 3x^2-4x=0$

$\Rightarrow x(3x-4)=0$

$\Rightarrow x=0,\frac{4}{3}$

When $x=0$, then $y=-2$

When $x=\frac{4}{3}$, then $y=\frac{-50}{27}$

Now, equation of tangents at point $(0,-2)$ is

$y-y_1=\frac{dy}{dx}(x-x_1)$

$\Rightarrow y+2=1(x-0)$

$\Rightarrow y+2=x$

$\Rightarrow x-y=2$ ...(i)

and equation of tangents at point $(\frac{4}{3},-\frac{50}{27})$ is

$y-y_1=\frac{dy}{dx}(x-x_1)$

$y+\frac{50}{27}=x-\frac{4}{3}$

$\Rightarrow x-y=\frac{50}{27}+\frac{4}{3}$

$\Rightarrow x-y=\frac{50+36}{27}$

$\Rightarrow x-y=\frac{86}{27}$ ....(ii)

Hence, equations (i) and (ii) are required equations of the tangents.