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

(x-1)²

(x-1)³

(x+1)³

(x+1)²

Solution

(x-1)³

f''(x) = 6(x - 1) f'(x) = 3(x - 1)2 + c ........ (i)
At the point (2, 1) the tangent to graph is y = 3x - 5 Slope of tangent = 3
∴ f'(2) = 3(2 - 1)2 + c = 3
3 + c = 3
⇒ c = 0
∴ From equation (i) f'(x) = 3(x - 1)²
f'(x) = 3(x - 1)2 f(x) = (x - 1)3 + k ...... (ii)
Since graph passes through (2, 1)
∴ 1 = (2 - 1)² + k k = 0
∴ Equation of function is f(x) = (x - 1)³

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Continuity And Differentiability

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

(x-1)²

(x-1)³

(x+1)³

(x+1)²

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For Daily Free Study Material Join wiredfaculty

Continuity And Differentiability

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 (x-1)2 (x-1)3 (x+1)3 (x+1)2

Some More Questions From Continuity and Differentiability Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

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

(x-1)²

(x-1)³

(x+1)³

(x+1)²