Question

If $f:R-\{-1,k\}\to R-\{\alpha ,\beta \}$ is a bijective function defined by $f\left(x\right)=\frac{(2x-1)(2x^2-4px+p^3)}{(x+1)(x^2-p^{2}x+p^2)}$ for $p\ge 0,$ then which of the following statements is (are) CORRECT ?

• A. If $k\in (-1,1),$ then $\alpha +\beta =2$
• B. If $k\in (-1,1),$ then $\alpha +\beta =6$
• C. If $k\in (1,3),$ then $\alpha +\beta =4$
• D. If $k\in (1,3),$ then $\alpha +\beta =6$

Answer 1

Answer: (A), (D)

If $k\in (1,3),$ then $\alpha +\beta =6$

$f:R-\{-1,k\}\to R-\{\alpha ,\beta \}$
$f\left(x\right)=\frac{(2x-1)(2x^2-4px+p^3)}{(x+1)(x^2-p^{2}x+p^2)}$

For the domain to be $R-\{-1,k\},k$ must be a repeated root of $x^2-p^{2}x+p^2=0$
$\therefore p^4-4p^2=0\Rightarrow p^2(p^2-4)=0$
$\therefore p=0,\pm 2,$ but $p\ge 0$
$\Rightarrow p=0,2$

For $p=0,k=0$
$f\left(x\right)=\frac{(2x-1)\left(2x^2\right)}{(x+1)\left(x^2\right)}$
$\Rightarrow D_f=R-\{-1,0\},R_f=R-\{4,-2\}$
$\therefore \alpha +\beta =2$

For $p=2,k=2$
$f\left(x\right)=\frac{(2x-1)2(x-2{)}^2}{(x+1)(x-2{)}^2}$
$\Rightarrow D_f=R-\{-1,2\},R_f=R-\{4,2\}$
$\therefore \alpha +\beta =6$