Question

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

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

Answer 1

The Correct Option is A

Given,
$y=x^{3}+x-2 \,...(i)$
$y=4 x-1 \, ...(ii)$
Slope of tangent to the curve (i)
$\frac{d y}{d x}=3 x^{2}+1$
Slope of tangent at point $(\alpha, \beta)$ is
$\left.\frac{d y}{d x}\right|_{(\alpha, \beta)}-3 \alpha^{2}+1\,...(iii)$
Given, tangent of curve (i) is parallel to line (ii).
$\therefore$ Slope of line (ii) is 4 .
$\therefore$ From E (iii), we get
$ 3 \alpha^{2}+1=4$
$\Rightarrow\, \alpha=\pm 1$
$\therefore\, (\alpha, \beta)$ lie on curve (i).
$\therefore \, \beta=(\pm 1)^{3}+(\pm 1)-2$
$\Rightarrow \, \beta=0,-4$
$\therefore$ Points are $(1,0)$ and $(-1,-4)$