Find the Range of the Function

Solution:

Let

\[ y=\frac{x^2-x}{x^2+2x} \]

Then,

\[ yx^2+2yx=x^2-x \]

\[ (y-1)x^2+(2y+1)x=0 \]

\[ x\left[(y-1)x+(2y+1)\right]=0 \]

Since \(x=0\) is not in domain,

\[ (y-1)x+(2y+1)=0 \]

\[ x=\frac{-(2y+1)}{y-1} \]

For \(x\) to exist,

\[ y\ne1 \]

Also,

\[ x\ne-2 \]

Putting \(x=-2\),

\[ -2=\frac{-(2y+1)}{y-1} \]

\[ 2y-2=2y+1 \]

impossible

So no extra restriction.

Hence,

\[ \boxed{\text{Range}=R-\{1\}} \]

\[ \boxed{\text{Correct Answer: (b)}} \]