Question 11

Let $$f : R\rightarrow R$$ be a twice differentiable function such that $$f(2)=1$$. If $$F(x)=xf(x)$$ for all $$x \in R$$, $$\int_{0}^{2}x F'(x)dx=6$$ and $$\int_{0}^{2}x^{2}F''(x)dx=40$$, then $$F'(2)+\int_{0}^{2}F(x)dx$$ is equal to :

F(x) = xf(x). F'(x) = f(x)+xf'(x). F''(x) = 2f'(x)+xf''(x).

∫₀² xF'(x)dx = [xF(x)]₀² - ∫₀²F(x)dx = 2F(2) - ∫₀²F(x)dx = 2·2·f(2) - ∫₀²F(x)dx = 4 - ∫₀²F(x)dx = 6

So ∫₀²F(x)dx = -2 ... (1)

∫₀² x²F''(x)dx = [x²F'(x)]₀² - 2∫₀²xF'(x)dx = 4F'(2) - 12 = 40

F'(2) = 52/4 = 13

F'(2) + ∫₀²F(x)dx = 13 + (-2) = 11

The correct answer is Option 1: 11.

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