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

The correct option is D $x-y=2$ and $x-y=\frac{86}{27}$

Slope of the tangent,
$m=1$
Given curve:
$y=x^3-2x^2+x-2\Rightarrow \frac{dy}{dx}=3x^2-4x+1$
So,
$\frac{dy}{dx}=3x^2-4x+1=1\Rightarrow x=0,\frac{4}{3}$

For
$x=0\Rightarrow y=-2$ and
$x=\frac{4}{3},\Rightarrow y=-\frac{50}{27}$

Tangents to the curve at the points
$(0,-2)$ and $(\frac{4}{3},-\frac{50}{27})$ are,

Therefore the equation of these tangents are
$y-(-2)=1(x-0)\Rightarrow x-y=2$

And
$y-\left(\frac{50}{27}\right)=1(x-\frac{4}{3})$
$\Rightarrow x-y=\frac{86}{27}$