Which of the following functions is injective but not surjective?

Let $f(x) = \begin{cases} x \sin \left(\frac{1}{x}\right) & \text{when } x \neq 0 \\ 1 & \text{when } x = 0 \end{cases}$ and $A = \{x \in \mathbb{R} : f(x) = 1\}$. Then,$A$ has

If the function $f:[-1,1] \rightarrow R$ is defined by $f(x) = \begin{cases} 2^x+1, & \text{for } x \in [-1,0) \\ 1, & \text{for } x=0 \\ 2^x-1, & \text{for } x \in (0,1] \end{cases}$,then in $[-1,1]$,$f(x)$ has

The mapping $f: R \to R$ defined as $f(x) = \cos x, x \in R$ is:

$A$ mapping from $\mathbb{N}$ to $\mathbb{N}$ is defined as follows: $f: \mathbb{N} \rightarrow \mathbb{N}$ where $f(n) = (n+5)^2$ for all $n \in \mathbb{N}$ (where $\mathbb{N}$ is the set of natural numbers). Then:

Let the functions $f$ and $g$ be $f: [0, \frac{\pi}{2}] \rightarrow R$ given by $f(x) = \sin x$ and $g: [0, \frac{\pi}{2}] \rightarrow R$ given by $g(x) = \cos x$,where $R$ is the set of real numbers. Consider the following statements:
Statement $(I)$: $f$ and $g$ are one-one.
Statement $(II)$: $f+g$ is one-one.
Which of the following is correct?