Question

If the equation $\frac{x-1}{x^2+2}=\frac{3-x}{x+1}$ has odd number of solution(s), then interval of $x$ is

• A. $(0,1)$
• B. $(1,2)$
• C. $(2,3)$
• D. None of the above

Answer 1

The correct option is C $(2,3)$

Let $f\left(x\right)=\frac{x-1}{x^2+2}-\frac{3-x}{x+1}$

Case $1:f\left(0\right)=-\frac{7}{2}$ and $f\left(1\right)=0-1<0$
$\Rightarrow f\left(x\right)=0$ has either two solutions or no solution.

Case $2:f\left(1\right)<0$ and $f\left(2\right)=\frac{1}{6}-\frac{1}{3}<0$
$\Rightarrow f\left(x\right)=0$ has either two solutions or no solution.

Case $3:f\left(2\right)<0$ and $f\left(3\right)=\frac{2}{11}>0$
$\Rightarrow f\left(x\right)=0$ has atleast one solution.

Case $4:f\left(3\right)>0$ and $f\left(4\right)=\frac{3}{18}+\frac{1}{3}>0$
$\Rightarrow$ No solution
$\therefore f\left(x\right)=0$ has atleast one solution in $(2,3)$