Question

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

• A. $(-\infty ,\frac{-3}{5}]\bigcup [1,\infty )$
• B. $(-\infty ,\frac{-3}{5}]\bigcup (1,\infty )$
• C. $(-\infty ,\frac{3}{5})\bigcup (1,\infty )$
• D. $(-\infty ,\frac{5}{3}]\bigcup [2,\infty )$

Answer 1

The correct option is B

$(-\infty ,\frac{-3}{5}]\bigcup (1,\infty )$

The given function is y = $\frac{x^2+x+1}{x^2+x-1}$

Now, in order to find the range let’s write the above equation as x = g(y)

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

Here, y can’t be equal to 1 because as y becomes 1, the above equation becomes ‘ -2 = 0’ which is false. So y can’t be equal to 1.

Now, if y is not equal to 1 we’ll have a quadratic equation in x. To have the real roots of this equation we must have its discriminant$\ge$ 0.

So, D $\ge$ 0.

Or $(y-1{)}^2$ - 4 (y-1) (-y -1) $\ge$ 0

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

Or (5y+3) . (y-1) $\ge$ 0 ; On factorisation of the equation

This will give us, y $\le$ -$\frac{3}{5}$ or y $\ge$ 1

As we know that y can’t be equal to 1 so, the final values of y would be

y $\le$ -$\frac{3}{5}$ or y $>$1