$Let F : R \to R be a thrice differentiable function. Suppose that F (1) = 0, F (3) = - 4 and F^{'} (x) < 0 for all x \in (1 / 2, 3) . Let f (x) = x F (x) for all x \in R$
$If \int_{1}^{3} x^{2} F^{'} (x) d x = - 12 and \int_{1}^{3} x^{3} F^{''} (x) d x = 40, then the correct expression (s) is (are)$

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$9 f^{'} (3) + f^{'} (1) + 32 = 0$

$9 f^{'} (3) + f^{'} (1) - 32 = 0$

$\int_{1}^{3} f (x) d x = - 12$

$\int_{1}^{3} f (x) d x = 12$

answer is C.

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Detailed Solution

$\begin{array}{l} \int_{1}^{3} x^{3} F^{''} (x) d x = 40 \Rightarrow {[x^{3} F^{'} (x)]}_{1}^{3} - \int_{1}^{3} 3 x^{2} F^{'} (x) d x = 40 \\ \Rightarrow {[x^{2} F^{'} (x) - x F (x)]}_{1}^{3} - 3 (- 12) = 40 \\ \Rightarrow 9 f^{'} (3) - 3 f (3) - f^{'} (1) + f (1) = 4 \\ \Rightarrow 9 f^{'} (3) + 36 - f^{'} (1) + 0 = 4 \Rightarrow 9 f^{'} (3) - f^{'} (1) + 32 = 0 \Rightarrow (C) \\ \Rightarrow \int_{1}^{3} x 2 f^{'} (x) d x = - 12 \Rightarrow {[x^{2} f (x)]}_{1}^{3} - \int_{1}^{3} 2 x f (x) d x = - 12 \\ \Rightarrow - 36 - 2 \int_{1}^{3} f (x) d x = - 12 \Rightarrow \int_{1}^{3} f (x) d x = - 12 \end{array}$