Question

Let $F:R\to R$ be a thrice differentiable function. Supose that $F\left(1\right)=0,F\left(3\right)=–4$ and $F^{\prime }\left(x\right)<0$ for all $x\in (1/2,3)$. Let $f\left(x\right)=xF\left(x\right)$ for all $x\in R$.

If $\underset{1}{\overset{3}{\int }}{x}^2{F}^{\prime }\left(x\right)dx=-12$ and $\underset{1}{\overset{3}{\int }}{x}^3{F}^{\prime \prime }\left(x\right)dx=40$,then the correct expression(s) is(are)

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

Answer 1

Answer: (C), (D)

$\underset{1}{\overset{3}{\int }}f\left(x\right)dx=-12$

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

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

$\because \left[F^{\prime }\right(x)=\frac{f^{\prime }\left(x\right)-F\left(x\right)}{x}]$
$\Rightarrow 9\left[f^{\prime }\right(3)-F(3\left)\right]-\left[f^{\prime }\right(1)-F(1\left)\right]=4\Rightarrow 9f^{\prime }\left(3\right)-9F\left(3\right)-f^{\prime }\left(1\right)=4\Rightarrow 9f^{\prime }\left(3\right)-f^{\prime }\left(1\right)+32=0$