Determine Algebraically Whether the Following Function Is One-To-One $f(x)=\frac{5 x^{2}}{3 x^{2}+6}, \text { where } x>0$

Multiple Choice

Determine algebraically whether the following function is one-to-one.
$f(x) =\frac{5 x^{2}}{3 x^{2}+6}, \text { where } x>0$

$\begin{aligned}\frac{5 a^{2}}{3 a^{2}+6} &=\frac{5 b^{2}}{3 b^{2}+6} \\\frac{5 a^{2}}{3 a^{2}}+\frac{5 a^{2}}{6} &=\frac{5 b^{2}}{3 b^{2}}+\frac{5 b^{2}}{6}\end{aligned}$

A)
$\begin{aligned}\frac{5}{3}+\frac{5 a^{2}}{6} &=\frac{5}{3}+\frac{5 b^{2}}{6} ;\text { not one-to-one } \\\frac{30+5 a^{2}}{18} &=\frac{30+5 b^{2}}{18} \\ 30+5 a^{2} &=30+5 b^{2} \\5 a^{2} &=5 b^{2} \\a^{2} &=b^{2} \\\pm a &=\pm b\end{aligned}$

B)
$\begin{aligned}\frac{5 a^{2}}{3 a^{2}+6} &=\frac{5 b^{2}}{3 b^{2}+6} \\\frac{5}{3+6} &=\frac{5}{3+6} ; \text { one-to-one } \\\frac{5}{6} &=\frac{3}{6} \\a &=b\end{aligned}$


C)
$\begin{aligned}\frac{5 a^{2}}{3 a^{2}+6} &=\frac{5 b^{2}}{3 b^{2}+6} \\\frac{5 a^{2}}{3 a^{2}} &=\frac{5 b^{2}}{3 b^{2}} ; \text { one-to-one } \\\frac{5}{3} &=\frac{5}{3} \\a &=b\end{aligned}$

D)
$$\frac { 5 a ^ { 2 } } { 3 a ^ { 2 } + 6 } = \frac { 5 b ^ { 2 } } { 3 b ^ { 2 } + 6 }$$
$$\frac { 5 a ^ { 2 } } { 9 a ^ { 2 } } = \frac { 5 b ^ { 2 } } { 9 b ^ { 2 } }$$
$$\begin{aligned}\frac { 5 a } { 9 } & = \frac { 5 b } { 9 } \quad ; \text { one-to-one } \\5 a & = 5 b \\a & = b\end{aligned}$$

E)
$\begin{aligned}\frac { 5 a ^ { 2 } } { 3 a ^ { 2 } + 6 } & = \frac { 5 b ^ { 2 } } { 3 b ^ { 2 } + 6 } \\ \frac { 5 a ^ { 2 } } { 3 a ^ { 2 } } + \frac { 5 a ^ { 2 } } { 6 } & = \frac { 5 b ^ { 2 } } { 3 b ^ { 2 } } + \frac { 5 b ^ { 2 } } { 6 }\end{aligned}$
$\frac { 5 } { 3 } + \frac { 5 a ^ { 2 } } { 6 } = \frac { 5 } { 3 } + \frac { 5 b ^ { 2 } } { 6 }\ ;\text { one-to-one }$

$\begin{array} { r l c } \frac { 30 + 5 a ^ { 2 } } { 18 } & = \frac { 30 + 5 b ^ { 2 } } { 18 } \\30 + 5 a ^ { 2 } & = & 30 + 5 b ^ { 2 } \\5 a ^ { 2 } & = & 5 b ^ { 2 } \\a ^ { 2 } & = & b ^ { 2 } \\a & = & b\end{array}$