The function f: R $\to$R defined as $f\left(x\right)=\frac{3x^2+3x-4}{3+3x-4x^2}$ is

Let $f\left(x_1\right)=f\left(x_2\right)$

$\Rightarrow \frac{3x_1^2+3x_1-4}{3+3x_1-4x_1^2}=\frac{3x_2^2+3x_2-4}{3+3x_2-4x_2^2}$

$=\frac{3\left(x_1^2-x_2^2\right)+3\left(x_1-x_2\right)}{3\left(x_1-x_2\right)-4\left(x_1^2-x_2^2\right)}$ ( using dividend rule)

$\begin{array}{l}\Rightarrow \left(3x_1^2+3x_1-4\right)\left(3\left(x_1-x_2\right)-4\left(x_1^2-x_2^2\right)\right)\\ =\left(3+3x_1-4x_1^2\right)\left[3\left(x_1^2-x_2^2\right)+3\left(x_1-x_2\right)\right]\end{array}$

On simplification we have

$7\left(x_1-x_2\right)\left(x_1+x_2-3x_1{x}_2-3\right)=0$

$\Rightarrow {\mathbf{x}}_1={\mathbf{x}}_2$ or $x_2=\frac{3-x_1}{1-3x_1}$

i.e., $f\left(x_1\right)=f\left(x_2\right)\Rightarrow x_1=x_2$ or $x_2=\frac{3-{\mathrm{x}}_1}{1-3{\mathrm{x}}_1}$

$\Rightarrow$ Function is not one to one as $x_1,\frac{3-x_1}{1-3x_1}$ are both mapped onto the same image onto:

Let $y=\frac{3x^2+3x-4}{3+3x-4x^2}$

$\Rightarrow x^2(3+4y)+x(3-3y)-4-3y=0$

If f(x) is to be onto, it has to take all real value for real x. For real x, we should have

$\begin{array}{l}(3-3y{)}^2-4(3+4y\left)\right(-4-3y)\ge 0\\ \Rightarrow 57+74y+57y^2\ge 0\\ \Rightarrow y^2+\frac{74}{57}y+1\ge 0\end{array}$

$\Rightarrow {\left(y+\frac{37}{57}\right)}^2+\frac{1880}{3249}\ge 0$ which is true.

Hence f is onto Thus f is onto but not one to one.