Question

If $x$ is real, then the range of $\frac{x^2+2x+1}{x^2+2x+7}$ is

• A. [0, 1)
• B. $(-\infty ,0)\cup (1,\infty )$
• C. (0,1)
• D. R

Answer 1

Answer: (A)

Let $\frac{x^2+2x+1}{x^2+2x+7}=y$,
$\because y\ne 1$ ...(i)
$\Rightarrow (y-1)x^2+2(y-1)x+(7y-1)=0$
$\because x\in R$
so, $D\ge 0$
$\Rightarrow 4(y-1{)}^2-4(y-1)(7y-1)\ge 0$
$\to (y-1)[y-1-7y+1]\ge 0$
$\Rightarrow y(y-1)\le 0$ ...(ii)
$\Rightarrow y\in [0,1]$
From Eqs. (i) and (ii), we are getting
$y\in [0,1)$