Question

The points of the curve $y=x^3+x-2$ at which its tangent are parallel to the straight line $y=4x-1$ are

• A. $(2,7),(-2,-11)$
• B. $(0,-2),(2^{1/3},2^{1/3})$
• C. $({-2}^{1/3},{-2}^{1/3})(0,-4)$
• D. $(1,0),(-1,-4)$

Answer 1

Answer: (D)

$(1,0),(-1,-4)$

Given
$y=x^3+x-\mathrm{2...}\left(i\right)$

$y=4x-\mathrm{1....}\left(ii\right)$

Slope of tangent to the curve (i)

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

slope of tangent at point $(\alpha ,\beta )$ is

${\left|\frac{dy}{dx}\right|}_{(\alpha ,\beta )}=3{\alpha }^2+\mathrm{1...}\left(iii\right)$

Given, tangent of curve (i) is parallel to line (ii)

$\therefore$ Slope of line (ii) is $4$

$\therefore$ From Eq(iii) we get

$3{\alpha }^2+1=4\Rightarrow \alpha =\pm 1$

$\therefore (\alpha ,\beta )$ lie on curve (i)

$\beta ={(\pm 1)}^2+(\pm 1)-2\Rightarrow \beta =0,-4$

$\therefore$ Points are $(1,0)$ and $(-1,-4)$