Given, y=x3+x−2 ......(i) y=4x−1 ......(ii) Slope of tangent to the curve (i)
dy
dx
=3x2+1 Slope of tangent at point (α,β) is
dy
dx
|(α,β)=3α2+1 ......(iii) Given, tangent of curve (i) is parallel to line (ii) ∴ Slope of line (ii) is 4 ∴ From Eq. (iii), we get 3α2+1=4 ⇒α=±1 ∴(α,β) lie on curve (i). ∴β=(±1)3+(±1)−2 ⇒β=0,−4 ∴ Points are (1,0) and (−1,−4)