Let f: [0, ) R be a continuous function such that ₀^ x^2 f(t) dt = x^6 - x^8 4 for all x 0 . The interval on which f(x) is strictly increasing is

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Let f: [0, ) R be a continuous function such that ₀^ x^2 f(t) dt = x^6 - x^8 4 for all x > 0 . The interval on which f(x) is strictly increasing is

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Correct answer: B. (0, 2)

Given equation is ₀^ x^2 f(t) dt = x^6 - x^8 4 Differentiating both sides with respect to x using the Newton-Leibniz formula, we get: f(x^2) d dx (x^2) - 0 = 6x^5 - 8x^7 4 f(x^2) 2x = 6x^5 - 2x^7 Since x > 0 , we can divide by 2x : f(x^2) = 3x^4 - x^6 Let t = x^2 . Then the function f(t) is given by: f(t) = 3t^2 - t^3 To find the interval where f(t) is strictly increasing, we compute its derivative: f'(t) = 6t - 3t^2 = 3t(2 - t) For f(t) to be strictly increasing, we must have f'(t) > 0 : 3t(2 - t) > 0 t (0, 2) Thu

Let f: [0, ) R be a continuous function such that ₀^ x^2 f(t) dt = x^6 - x^8 4 for all x > 0 . The interval on which f(x) is strictly increasing is

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