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$$\int₀^2F(x)dx=-2$$... (1)

$$\int₀^2x^2F''(x)dx=[x^2F'(x)]₀^2-2\int₀^2xF'(x)dx$$ = 4F'(2) - 12 = 40

F'(2) = 52/4 = 13

F'(2) + $$\int₀^2F(x)dx$$ = 13 + (-2) = 11

The correct answer is Option 1: 11.

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