Question

The expression $\frac{x^2+2x+1}{x^2+2x+7}$ lies in the interval : $\left(xϵR\right)$.

• A. $[0,-1]$
• B. $(-\infty ,0]U[1,\infty )$
• C. $[0,1)$
• D. None of the above.

Answer 1

Answer: (C)

$[0,1)$

Let $y=\frac{x^2+2x+1}{x^2+2x+7}$

$(y-1)x^2+2(y-1)x+(7y-1)=0$

Given that $xϵR$

$\therefore △=4{(y-1)}^2-4(y-1)(7y-1)\ge 0$

$(y-1)(y-1-7y+1)\ge 0$

$y(y-1)\le 0$

$\therefore yϵ[0,1]$

But $y\ne$ ($\because$ co efficient of $x^2\ne 0$)

$\therefore yϵ[0,1)$

$\therefore \frac{x^2+2x+1}{x^2+2x+7}ϵ[0,1)$