Question

If $f:R-\{-1,k\}\to R-\{\alpha ,\beta \}$ is a bijective function defined by $f\left(x\right)=\frac{(2x-1)(2x^2-4px+p^3)}{(x+1)(x^2-p^{2}x+p^2)}$ where $p\ge 0,$ 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 $x^2-p^{2}x+p^2=0$
$\therefore p^4-4p^2=0\Rightarrow p^2(p^2-4)=0\Rightarrow p=0,\pm 2$
But given that $p\ge 0$
$\therefore p=0,2$

Now, for $p=0$
$f\left(x\right)=\frac{(2x-1)2x^2}{(x+1)\left(x^2\right)},$
Domain of $f$ is $R-\{-1,0\}$
$\therefore k=0$
Let $y=\frac{4(x-1)}{x+1}$
$\Rightarrow x=\frac{y+2}{4-y},(y\ne 4)$
Also, $\underset{x\to 0}{lim}f\left(x\right)=-2$
Hence, range of $f$ is $R-\{4,-2\}$
$\therefore \alpha +\beta =2$

For $p=2$
$f\left(x\right)=\frac{(2x-1)2(x-2{)}^2}{(x+1)(x-2{)}^2},$
Domain of $f$ is $R-\{-1,2\}$
$\therefore k=2$
and $\underset{x\to 2}{lim}f\left(x\right)=2$
Hence, range of $f$ is $R-\{4,2\}$
$\therefore \alpha +\beta =6$