Let be a twice differentiable function such that f(2) = 1. If F(x) = xf(x) for all . and , then is equal to : [2025]

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Solution

Q.
Let $f : R \to R$ be a twice differentiable function such that f(2) = 1. If F(x) = xf(x) for all $x \in R$ . $\int_{0}^{2} x F' (x) d x = 6$ and $\int_{0}^{2} x^{2} F'' (x) d x = 40$ , then $F' (2) + \int_{0}^{2} F (x) d x$ is equal to : [2025]

1 13

2 9

3 11

4 15

Ans.
(3)

We have, $\int_{0}^{2} x F' (x) d x = 6$ ... (i)

$\Rightarrow {[x F (x)]}_{0}^{2} - \int_{0}^{2} F (x) d x = 6$

$\Rightarrow 2 F (2) - \int_{0}^{2} F (x) d x = 6$

$\Rightarrow 4 - \int_{0}^{2} F (x) d x = 6$ $[∵ F (2) = 2 f (2) = 2]$

$\Rightarrow \int_{0}^{2} F (x) d x = - 2$

Also, $\Rightarrow \int_{0}^{2} x^{2} F'' (x) d x = 40$

$\Rightarrow {[x^{2} F' (x)]}_{0}^{2} - 2 \int_{0}^{2} x F' (x) d x = 40$

$\Rightarrow 4 F' (2) - 2 (6) = 40 \Rightarrow F' (2) = 13$ [Using (i)]

Thus, $F' (2) + \int_{0}^{2} F (x) d x = 13 - 2 = 11$

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