Let F:R→R be a thrice differentiable function. Suppose that F(1)=0,F(3)=−4 and F′(x)<0 for all x∈(1/2,3). Let f(x)=xF(x) for all x∈R If ∫13 x2F′(x)dx=−12 and ∫13 x3F′′(x)dx=40, then the correct expression (s) is (are)

∫13 x3F′′(x)dx=40⇒x3F′(x)13−∫13 3x2F′(x)dx=40⇒x2F′(x)−xF(x)13−3(−12)=40⇒9f′(3)−3f(3)−f′(1)+f(1)=4⇒9f′(3)+36−f′(1)+0=4⇒9f′(3)−f′(1)+32=0⇒(C)⇒∫13 x2f′(x)dx=−12⇒x2f(x)13−∫13 2xf(x)dx=−12⇒−36−2∫13 f(x)dx=−12⇒∫13 f(x)dx=−12

Let F:R→R be a thrice differentiable function. Suppose that F(1)=0,F(3)=−4 and F′(x)<0 for all x∈(1/2,3). Let f(x)=xF(x) for all x∈R If ∫13 x2F′(x)dx=−12 and ∫13 x3F′′(x)dx=40, then the correct expression (s) is (are)

∫13 x3F′′(x)dx=40⇒x3F′(x)13−∫13 3x2F′(x)dx=40⇒x2F′(x)−xF(x)13−3(−12)=40⇒9f′(3)−3f(3)−f′(1)+f(1)=4⇒9f′(3)+36−f′(1)+0=4⇒9f′(3)−f′(1)+32=0⇒(C)⇒∫13 x2f′(x)dx=−12⇒x2f(x)13−∫13 2xf(x)dx=−12⇒−36−2∫13 f(x)dx=−12⇒∫13 f(x)dx=−12

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

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

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

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

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

answer is C.

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Detailed Solution

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