If f'(x) = x^2/2 - kx + 1 , such that f(0) = 0 and f(3) = 15 . Find the value of k . - Mathematics MCQ Mathematics Mock Test - 1 2026 | ExamTest.live

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If f^{'} (x) = \frac{x ^{2}}{2} - k x + 1, such that f (0) = 0 and f (3) = 15 . Find the value of k .

Solution & Step-by-step Explanation

Integrate f^{'} (x) to find f (x) : f (x) = \int (\frac{x ^{2}}{2} - k x + 1) d x f (x) = \frac{x ^{3}}{6} - \frac{k x ^{2}}{2} + x + C Given f (0) = 0 : 0 = 0 - 0 + 0 + C ⟹ C = 0 So, f (x) = \frac{x ^{3}}{6} - \frac{k x ^{2}}{2} + x .Given f (3) = 15 : 15 = \frac{3 ^{3}}{6} - \frac{k ( 3 ^{2} )}{2} + 3 15 = \frac{27}{6} - \frac{9 k}{2} + 3 15 = 4.5 - 4.5 k + 3 15 = 7.5 - 4.5 k 4.5 k = 7.5 - 15 = - 7.5 k = - \frac{7.5}{4.5} = - \frac{75}{45} = - \frac{5}{3}

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If f^{'} (x) = \frac{x ^{2}}{2} - k x + 1, such that f (0) = 0 and f (3) = 15 . Find the value of k .

5/3

3/5

- 5/3

- 3/5

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