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: (D)

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

Answer 2

Given,
$y=x^{3}-2 x^{2}+x-2$
On differentiating both sides w.r.t. $x$, we get
$\frac{d y}{d x}=3 x^{2}-4 x+1$
and $y=x$
$\Rightarrow \frac{d y}{d x} =1$
$\therefore$ Slope of tangent will be $3 x^{2}-4 x+1$
Since, the tangent is parallel to line $y=x$.
$\therefore 3 x^{2}-4 x+1=1$
$\Rightarrow 3 x^{2}-4 x=0$
$\Rightarrow x(3 x-4) =0$
$\Rightarrow x =0, \frac{4}{3}$
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
$y-y_{1}= \frac{d y}{d x}\left(x-x_{1}\right)$
$\Rightarrow y+2=1(x-0)$
$\Rightarrow y+2=x$
$\Rightarrow x-y=2$ ...(i)
and equation of tangents at point $\left(\frac{4}{3},-\frac{50}{27}\right)$ is
$y -y_{1} =\frac{d y}{d x}\left(x-x_{1}\right)$
$y +\frac{50}{27} =x-\frac{4}{3}$
$\Rightarrow x-y =\frac{50}{27}+\frac{4}{3}$
$\Rightarrow x-y =\frac{50+36}{27}$
$\Rightarrow x-y =\frac{86}{27}$ ...(ii)
Hence, Eqs. (i) and (ii) are required equations of the tangents.