Question

The range of f(x) = $\frac{x^2+x+1}{x^2+x-1}$ is .

• A. $(-\infty ,\frac{3}{5}]\cup (1,\infty )$
• B. $(-\infty ,\frac{3}{5}]\cup [1,\infty )$
• C. $(-\infty ,\frac{3}{5})\cup (1,\infty )$

Answer 1

The correct option is A $(-\infty ,\frac{3}{5}]\cup (1,\infty )$

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

$\Rightarrow y(x^2+x-1)=x^2+x+1$

$\Rightarrow y{x}^2+yx-y-x^2-x-1=0$

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

This is a quadratic equation in x. Now $y\ne 1$
Also, for the value of x to be real,

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

$\Rightarrow y^2-2y+1+4(y^2-1)\ge 0$

$\Rightarrow 5y^2-2y-3\ge 0$

$\Rightarrow 5y^2-5y+3y-3\ge 0$

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

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

$\Rightarrow y\le \frac{3}{5}\text{or}y\ge 1$

Since the value of y cannot be 1, $y\le \frac{3}{5}$ or y > 1.

The range of the function is $(-\infty ,\frac{3}{5}]\cup (1,\infty )$