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

Answer 1

The Correct Option is B

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. $\frac{dy}{dx}= 3$⇒ 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).