Let f:[1, ) R be a differentiable function such that the average value of f on the interval [1, x] is given by 5x f(x) - 3x^5 - 7 15(x - 1) for all x 1 . Then the value of f(2) is :

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Let f:[1, ) R be a differentiable function such that the average value of f on the interval [1, x] is given by 5x f(x) - 3x^5 - 7 15(x - 1) for all x > 1 . Then the value of f(2) is :

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Correct answer: B. 26

The average value of a function f on the interval [1, x] is given by 1 x-1 ₁^x f(t) dt . Equating this to the given expression: 1 x-1 ₁^x f(t) dt = 5x f(x) - 3x^5 - 7 15(x - 1) Multiplying both sides by 15(x-1) (since x > 1 ): 15 ₁^x f(t) dt = 5x f(x) - 3x^5 - 7 Differentiating both sides with respect to x using the Newton-Leibniz formula: 15 f(x) = 5 f(x) + 5x f'(x) - 15x^4 Rearranging the terms: 10 f(x) = 5x f'(x) - 15x^4 2 f(x) = x f'(x) - 3x^4 f'(x) - 2 x f(x) = 3x^3 This is a linear differential equation of th

Let f:[1, ) R be a differentiable function such that the average value of f on the interval [1, x] is given by 5x f(x) - 3x^5 - 7 15(x - 1) for all x > 1 . Then the value of f(2) is :

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