Let $f: R \rightarrow R$ 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 $\int\limits_1^3 x^2 F'(x) d x=-12$ and $\int\limits_1^3 x^3 F''(x) d x=40$, then which of the following is (are) correct?
(a) $9 f'(3)+f'(1)-32=0$
(b) $\int\limits_1^3 f(x) d x=12$
(c) $9 f'(3)-f'(1)+32=0$
(d) $\int\limits_1^3 f(x) d x=-12$
Answer & explanation
Correct answer: option 3
We have, $f(x)=x F(x)$
$\int\limits_1^3 f(x) d x =\int\limits_1^3 x F(x) d x$
$=\left[\frac{x^2}{2} F(x)\right]_1^3-\frac{1}{2} \int\limits_1^3 x^2 F'(x) d x$
$=\frac{9}{2} F(3)-\frac{1}{2} F(1)-\frac{1}{2} \times(-12)$
$=\frac{9}{2}(-4)-\frac{1}{2}(0)-\frac{1}{2} \times(-12)=-18+6=-12$
It is given that
$\int\limits_1^3 x^3 F''(x) d x=40$
$\Rightarrow \left[x^3 F'(x)\right]_1^3-3 \int\limits_1^3 x^2 F'(x) d x=40$
$\Rightarrow 27 F'(3)-F'(1)-3 \times(-12)=40$ .......(i)
Now, $f(x)=x F(x)$
$\Rightarrow f'(x)=F(x)+x F'(x)$
$\Rightarrow f'(1)=F(1)+F'(1)$ and $f'(3)=F(3)+3 F'(3)$
$\Rightarrow f'(1)=0+F'(1)$ and $f'(3)=-4+3 F'(3)$
$\Rightarrow F'(1)=f'(1)$ and $F'(3)=\frac{1}{3} f'(3)+\frac{4}{3}$
Substituting these values in (i), we obtain
$9 f'(3)+36-f'(1)+36=40$
$\Rightarrow 9 f'(3)-f'(1)+32=0$
Hence, options (c) and (d) are correct.