Given function

$f(x)=\frac{x^2+2x-15}{x^2-4x+9}, \qquad x\in \mathbb R$

We must decide whether it is one-one and/or onto (taking codomain as $\mathbb R$).

First check the domain

The denominator is $x^2-4x+9=(x-2)^2+5>0 \quad \forall x\in \mathbb R$

So the denominator never vanishes, hence the domain is all real numbers.

Check whether the function is one-one

A quick way is to see whether two different inputs can give the same output.

Let us compute a few values:

$f(3)=\frac{9+6-15}{9-12+9}=\frac{0}{6}=0$

$f(-5)=\frac{25-10-15}{25+20+9}=\frac{0}{54}=0$

Since $f(3)=f(-5)=0, \qquad 3\ne -5$

the function is not one-one.

Check whether the function is onto $\mathbb R$

To find the range, let $y=\frac{x^2+2x-15}{x^2-4x+9}$

Rearranging, $y(x^2-4x+9)=x^2+2x-15$

$yx^2-4yx+9y=x^2+2x-15$

$(y-1)x^2+(-4y-2)x+(9y+15)=0$

For a real $x$ to exist for a given $y$, this quadratic in $x$ must have discriminant $\ge 0$.

So, $\Delta =(-4y-2)^2-4(y-1)(9y+15)\ge 0$

Compute: $(-4y-2)^2=16y^2+16y+4$

$4(y-1)(9y+15)=4(9y^2+6y-15)=36y^2+24y-60$

Hence $\Delta=16y^2+16y+4-(36y^2+24y-60)$ $\Delta=-20y^2-8y+64$

Therefore, $-20y^2-8y+64\ge 0$

Divide by $-4$ (reversing inequality): $5y^2+2y-16\le 0$

Solve $5y^2+2y-16=0$

$y=\frac{-2\pm\sqrt{4+320}}{10}=\frac{-2\pm 18}{10}$

So the roots are $y=-2,\quad y=\frac{8}{5}$

Since the quadratic opens upward, $5y^2+2y-16\le 0 \iff -2\le y\le \frac85$

Thus the range is $\boxed{[-2,\, 8/5]}$

This is not all of $\mathbb R$, so the function is not onto.

Conclusion

• Not one-one
• Not onto

Therefore the function is $\boxed{\text{neither one-one nor onto}}$

So the correct option is: $\boxed{\text{C}}$

Comparison with stored answer

Stored correct answer: C

Our derived answer is also C, so they agree.