Question

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

• A. $9f^{\prime }(3)+f^{\prime }(1)-32=0$
• B. $\underset{1}{\overset{3}{\int }}f(x)dx=12$
• C. $9f^{\prime }(3)-f^{\prime }(1)+32=0$
• D. $\underset{1}{\overset{3}{\int }}f(x)dx=-12$

Answer 1

Answer: (C), (D)

$\underset{1}{\overset{3}{\int }}{x}^3{F}^{\prime \prime }(x)dx=40{\left[x^3{F}^{\prime }(x)\right]}_1^3-\underset{1}{\overset{3}{\int }}3x^2{F}^{\prime }(x)dx=40$
$\Rightarrow {\left[x^2{f}^{\prime }(x)-xf(x)\right]}_1^3-3(-12)=40$
$\Rightarrow 9f^{\prime }(3)-3f(3)-f^{\prime }(1)+f(1)=4$
$\Rightarrow 9f^{\prime }(3)+36-f^{\prime }(1)+0=4$
$\Rightarrow 9f^{\prime }(3)-f^{\prime }(1)+32=0\Rightarrow (3)$
$\Rightarrow \underset{1}{\overset{3}{\int }}{x}^2{F}^{\prime }(x)dx=-12$
$\Rightarrow {\left[x^{2}F(x)\right]}_1^3-\underset{1}{\overset{3}{\int }}2xF(x)dx=-12$
$\Rightarrow -36-2\underset{1}{\overset{3}{\int }}f(x)dx=-12$
$\Rightarrow \underset{1}{\overset{3}{\int }}f(x)dx=-12\Rightarrow (4)$