Question

Range of the function $f\left(x\right)=\frac{x^2+x+2}{x^2+x+1};xϵR$ is

• A. $(1,\infty )$
• B. $(1,\frac{11}{7})$
• C. $(1,\frac{7}{3}]$
• D. $(1,\frac{7}{5})$

Answer 1

The correct option is C

$(1,\frac{7}{3}]$

$\therefore y=\frac{x^2+x+2}{x^2+x+1},$
$y=1+\frac{1}{x^2+x+1}[i.ey>1]$ . . . (i)

$\Rightarrow$ $y{x}^2+yx+y$ = $x^2+x+2$
$\Rightarrow$ $x^2$ (y - 1) + x(y - 1) + (y - 2) = 0,

$\because$ $xϵ$ $R\Rightarrow \text{D}\ge 0$
$\Rightarrow (y-1{)}^2-4(y-1\left)\right(y-2)\ge 0$
$\Rightarrow$ $(y-1)${$(y-1)-4(y-2)$} $\ge 0$
$\Rightarrow$ $(y-1)(-3y+7)$ $\ge 0$
$\Rightarrow 1\le y\le \frac{7}{3}$ . . . (ii)
From (i) and (ii), we get
$1<y\le \frac{7}{3}$