Question

What is the range of $\frac{x^2+x+1}{x^2+x-1}$

• A. ( - , )
• B. (1 , -
• C. (- , ) (1 ,
• D. (-, ) (0,

Answer 1

The correct option is C

(- , ) (1 ,

We want to find the range of $\frac{x^2+x+1}{x^2+x-1}$. Let's call it $y$. In this case we know $x$ is a real number. We will construct a quadratic equation involving $x$ and $y$. Then we will apply the condition that the roots of the quadratic are real. Because it will be a quadratic in $x$ and we know $x$ takes only real values. So the roots will also be real (to understand this better, you can think of it the following way. We formed the quadratic by assuming $x$ takes only real values. Suppose the roots are not real. Then we get the corresponding $y$, the $y$ we get by substituting those complex values of $x$ in the expression. But we already know $x$ can take only real values. So the values of $y$ obtained thus won't form part of the range).

So we have,

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

y $x^2+yx-y=x^2+x+1$

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

y $\ne$ 1 [If y=1 , 0 $x^2$ + 0 - 1 - 1 = -2 = 0 , which is not true]

$\Rightarrow$ x is a real number and for equation (1) to have real roots $△$ $\ge$ 0 , $b^2$ - 4ac $\ge$ 0

$\Rightarrow$ $(y-1{)}^2+4(y-1\left)\right(y+1)$ $\ge$ 0

$(y-1)\times \left(\right(y-1)+(4y+4\left)\right)\ge 0$

$\Rightarrow$ $(y-1)\times (5y+3)\ge$ 0

$\Rightarrow$ y $\in$ (- $\infty$ , $\frac{-3}{5}$] $\cup$ (1 , $\infty$)