Find the equation of the tangent to the curve y = x + 4 x 2 which is parallel to x axis.

Let the point of contact be x 1 , y 1 Given : slope of tangent at x 1 , y 1 is 0 because tangent is parallel to x axis. Let the slope of tangent at x 1 , y 1 be m ⇒ m = 0 . . . . . . . . . . 1 Now given curve is y = x + 4 x 2 Differentiating both side with respect to x d y d x = 1 + 4 - 2 · x - 2 - 1 [ d d x x n = n x n - 1 ] ⇒ d y d x = 1 - 8 x 3 At point x 1 , y 1 d y d x = 1 - 8 x 1 3 ⇒ m = 1 - 8 x 1 3 . . . . . . . . . . . . 2 On comparing equations 1 and 2 1 - 8 x 1 3 = 0 ⇒ x 1 3 = 8 ⇒ x 1 = 2 Also point x 1 , y 1 lies on the curve y = x + 4 x 2 ⇒ y 1 = x 1 + 4 x 1 2 Put x 1 = 2 y 1 = 2 + 4 4 = 3 Hence, point of contact is 2 , 3 Now, we know that, equation of tangent at x 1 , y 1 having slope m is y - y 1 = m x - x 1 On putting m = 0 and 2 , 3 and x 1 , y 1 Equation of tangent is, y - 3 = 0 x - 2 ⇒ y = 3 Hence, equation of tangent is y = 3 .

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Understanding ISC Mathematics Class 12 Volume 1>Chapter 7 - Applications of Derivatives>EXERCISE 7.2

M L Aggarwal Solutions for Chapter: Applications of Derivatives, Exercise 2: EXERCISE 7.2

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Applications of Derivatives, Exercise 2: EXERCISE 7.2

Attempt the practice questions on Chapter 7: Applications of Derivatives, Exercise 2: EXERCISE 7.2 with hints and solutions to strengthen your understanding. Understanding ISC Mathematics Class 12 Volume 1 solutions are prepared by Experienced Embibe Experts.