Which of the following functions is an injective (one-one) function in its respective domain? 

$$\begin{gathered} \left(1\right) If f\left(x\right)=2x +\sin 3x \\ \Rightarrow f\text{'}\left(x\right) =2 +3 \cos 3x =0 \\ \Rightarrow \cos 3x=\frac{-2}{3} ,which is defined \end{gathered}$$

$$\begin{gathered} and f\text{'}\text{'}\left(x\right) =- 9 \sin 3x = -3\sqrt{5}<0 \\ so f\left(x \right) has maximum value exists in its domain , so it is not one-one \end{gathered}$$

$\left(2\right) f\left(x\right) = x.\left[x\right] is not one-one \sin ce f\left(0.1\right)=0.1 \times 0 =0 and f\left(0.2\right)=0$

$\left(3\right) f\left(x\right)=\frac{2^x-1}{4^x+1}$

$\Rightarrow f\left(x\right) is continuous function and f\left(0\right)=0 and f\left(x\right) \to 0 again when x\to \infty$

      $\left(\sin ce y =\frac{{\displaystyle \frac{1}{2^x}}-{\displaystyle \frac{1}{4^x}}}{1+{\displaystyle \frac{1}{4^x}}} \to 0 as x\to \infty \right)$      $hence f\left(x \right) has a atleat one maximum exists , so it is not one-one$

$$\begin{gathered} \left(4\right) f\left(x\right) =\frac{2^x+1}{4^x-1} =\frac{2^x+1}{\left(2^x-1\right)\left(2^x+1\right)} =\frac{1}{2^x-1} \\ \Rightarrow f\text{'}\left(x\right) = -\frac{2^x \log 2}{{\left(2^x-1\right)}^2}<0 \Rightarrow f\left(x \right) is decrea\sin g function \\ so it is one-one \end{gathered}$$