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 - Mathematics MCQ AIEEE 2004 2004 | ExamTest.live

hardMCQAIEEE 20042004Mathematics

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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 = 3 x - 5, then the function is

Solution & Step-by-step Explanation

Given f^{''} (x) = 6 (x - 1) . Integrating once with respect to x : f^{'} (x) = 3 (x - 1)^{2} + C . The tangent at (2, 1) is y = 3 x - 5. The slope of this tangent is m = 3, which means f^{'} (2) = 3. f^{'} (2) = 3 (2 - 1)^{2} + C = 3 3 + C = 3 \Rightarrow C = 0 . So, f^{'} (x) = 3 (x - 1)^{2} . Integrating again: f (x) = (x - 1)^{3} + K . The graph passes through (2, 1), meaning f (2) = 1. f (2) = (2 - 1)^{3} + K = 1 1 + K = 1 \Rightarrow K = 0 . Thus, the function is f (x) = (x - 1)^{3} .

Try it yourself before checking the explanation above.

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 = 3 x - 5, then the function is

(x - 1)^{2}

(x - 1)^{3}

(x + 1)^{3}

(x + 1)^{2}

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