Question

The expression $\frac{x^2-2x+p^2}{x^2+2x+p^2}$ lies between (if $p<-1$ )

• A. $[\frac{p+1}{p-1},\frac{p-1}{p+1}]$
• B. $[\frac{p-1}{p+1},1]$
• C. $[p^2-1,p^2+1]$
• D. $[p,p^2]$

Answer 1

The correct option is A $[\frac{p+1}{p-1},\frac{p-1}{p+1}]$

Let $\frac{x^2-2x+p^2}{x^2+2x+p^2}=y\Rightarrow x^2(y-1)+2x(y+1)+p^2(y-1)=0$
.
For real $x$, $D$ of the above equation should be greater than or equal to zero.
$4{(y+1)}^2-4{(y-1)}^2{p}^2\ge 0\Rightarrow {(y+1)}^2-{(y-1)}^2{p}^2\ge 0\Rightarrow [y+1+(y-1)p][y+1-(y-1)p]\ge 0\Rightarrow [y(1+p)-(p-1)][y(1-p)+p+1]\ge 0$
.
Now dividing by $(1+p)(1-p)$, we get
$[y-\frac{p-1}{p+1}][y-\frac{p+1}{p-1}]\le 0\Rightarrow y\in [\frac{p+1}{p-1},\frac{p-1}{p+1}]$ ...(Inequality changes $\because (1+p)(1-p)<0$)