Question

Let $f\left(x\right)$ be an even function such that $f\left(x\right)+f(x-3)=x(x-3)+1.$ Then $\underset{0}{\overset{3}{\int }}\frac{f\left(x\right)dx}{(x^2-3x+1)}$ is

• A. $-\frac{2}{3}$
• B. $-\frac{3}{2}$
• C. $\frac{2}{3}$
• D. $\frac{3}{2}$

Answer 1

The correct option is D $\frac{3}{2}$

Let $I=\underset{0}{\overset{3}{\int }}\frac{f\left(x\right)dx}{\left(x\right(x-3)+1)}\dots \left(1\right)$
$\Rightarrow I=\underset{0}{\overset{3}{\int }}\frac{f(3-x)}{\left(\right(3-x\left)\right(-x)+1)}=\underset{0}{\overset{3}{\int }}\frac{f(3-x)}{\left(\right(x\left)\right(x-3)+1)}\dots \left(2\right)$

Adding $\left(1\right)+\left(2\right),$
$2I=\underset{0}{\overset{3}{\int }}\frac{f\left(x\right)+f(3-x)}{\left(x\right(x-3)+1)}dx$
Given, $f\left(x\right)+f(x-3)=x(x-3)+1$
Replace $x$ by $3-x$
$f(3-x)+f(-x)=f(3-x)+f\left(x\right)=x(x-3)+1$
$\{\because f\left(x\right)$ is an even function $\}$

Now, $2I=\underset{0}{\overset{3}{\int }}\frac{x(x-3)+1}{x(x-3)+1}dx$
$\Rightarrow 2I=3\Rightarrow I=\frac{3}{2}$