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

Let $y=f\left(x\right)=\frac{x^2+x+2}{x^2+x+1};xϵR$
$\therefore y=\frac{x^2+x+2}{x^2+x+1},$
$y=1+\frac{1}{x^2+x+1}[i.ey>1]$ . . . (i)
$\Rightarrow y{x}^2+yx+y=x^2+x+2\Rightarrow x^2(y-1)+x(y-1)+(y-2)=0,\because xϵR\Rightarrow \text{D}\ge 0\Rightarrow (y-1{)}^2-4(y-1\left)\right(y-2)\ge 0\Rightarrow (y-1\left)\right\{(y-1)-4(y-2)\}\ge 0\Rightarrow (y-1\left)\right(-3y+7)\ge 0$
$\Rightarrow 1\le y\le \frac{7}{3}$ . . . (ii)
From (i) and (ii), we get
$1<y\le \frac{7}{3}$