If the tangent at the point x 1 , y 1 on the curve y = x 3 + 3 x 2 + 5 passes through the origin, then x 1 , y 1 does NOT lie on the curve

Step-by-step solution

Given y = x 3 + 3 x 2 + 5 Now d y d x = 3 x 2 + 6 x So at point x 1 , y 1 d y d x at   x 1 , y 1 = 3 x 1 2 + 6 x 1       . . .   i Also slope of line in two point form will be y 1 - 0 x 1 - 0       . . .   ii Also x 1 , y 1 lie on curve We get y 1 = x 1 3 + 3 x 1 2 + 5       . . .   iii From equation i and ii we get 3 x 1 3 + 6 x 1 2 = y 1 (as slope is equal) Now y 1 = 3 x 1 3 + 6 x 1 2 = x 1 3 + 3 x 1 2 + 5 ⇒ 2 y 1 = 3 x 1 2 + 15 2 i.e. 2 y = 3