Problem
Let f: [1, ∞) → R be a differentiable function such that f(1) = 1/3 and 3\( \int_{1}^{x} \)f(t)dt = xf(x) – x³/3, x ∈ [1, ∞). Let e denote the base of the natural logarithm. Then find the value of f(e).
Solution
To proceed, we’ll differentiate both sides of the given equation with respect to x.
We’ll get 3f(x) = xf'(x) + f(x) – x², which on rearranging gives f'(x) – 2f(x)/x = x.
This is a linear differential equation, where the integration factor is \( e^{\int{\frac{-2}{x}}dx} \) or 1/x².
So, the solution is given by f(x)/x² = ∫1/x dx, which gives f(x) = x²lnx + cx².
To find the value of c, we’ll use the given condition f(1) = 1/3.
On substituting x = 1 in both sides, we’ll get c = 1/3.
Finally, we’ll get f(x) = x²lnx + x²/3. Using this, f(e) = e²lne + e²/3 = 4e²/3.
Comments
A rather boring problem. Not worthy of the JEE Advanced.
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