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

Answer: (A)

Given,
$y=x^3+x-2...(i)$
$y=4x-1...(ii)$
Slope of tangent to the curve (i)
$\frac{dy}{dx}=3x^2+1$
Slope of tangent at point $(\alpha ,\beta )$ is
${\frac{dy}{dx}|}_{(\alpha ,\beta )}-3{\alpha }^2+1...(iii)$
Given, tangent of curve (i) is parallel to line (ii).
$\therefore$ Slope of line (ii) is 4 .
$\therefore$ From Eq. (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 $(-1,-4)$