Question

Determine the range of the function f(x) defined as:
$f\left(x\right)=\frac{x^2+x+2}{x^2+x+1},xϵR$

• 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.e. y>1] ....(i)
$\Rightarrow y{x}^2+yx+y=x^2+x+2$
$\Rightarrow x^2(y-1)+x(y-1)+(y-2)=0,\forall xϵR$
Since, x is real, $D\ge 0$
$\Rightarrow (y-1{)}^2-4(y-1\left)\right(y-2)\ge 0\Rightarrow (y-1\left)\right\{(y-1)-4(y-2)\}\ge 0\Rightarrow (y-1\left)\right(-3y+7)\ge 0\Rightarrow 1\le y\le \frac{7}{3}$ ....(iii)
From Eqs. (i) and (ii), $Rangeϵ(1,\frac{7}{3}]$