Question

A function y = f(x) has a second order derivative f''(x) = 6(x - 1). If its graph passes through the point (2, 1) and at that point the tangent to the graph is y = 3x -5. then the function is :

• A. (A) (x - 1)2
• B. (B) (x - 1)3
• C. (C) (x + 1)3
• D. (D) (x + 1)2

Answer 1

Answer: (B)

(B) (x - 1)3

Answer 2

Explanation:
Given :A function y = f(x) such that f''(x) = 6(x - 1)....... (A)The graph of f (x) passes through the point (2, 1) at which the tangent to the graph is y = 3x - 5.We have to find the value of f (x). We shall use the concept of tangents and normals in this question.From eqn (A), we havef'(x) = 3(x -1)2 +C ........ (i)At the point (2, 1) the tangent to the graph is given byy = 3x - 5Slope of tangent i.e. dydx= 3⇒ f'(2) = 3Therefore, f'(2) = 3(2 - 1)2 +C = 3 + C = 0.From equation (i), we getf'(x) = 3(x - 1)2⇒ f'(x) = 3(x - 1)2⇒ f(x) = (x - 1)3 + k ..........(ii)Since, graph of f(x) passes through (2, 1), therefore1 = (2 - 1)3 + k⇒ k = 0Equation of funciton isf(x) = (x - 1)3Hence, the correct option is (B).