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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 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 :
Updated On: Aug 21, 2024
- (A) (x - 1)2
- (B) (x - 1)3
- (C) (x + 1)3
- (D) (x + 1)2
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The Correct Option is B
Solution and Explanation
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. ⇒ 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).
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. ⇒ 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).
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