Let the f: R → R be defined by f (x) = 2x + cos x, then f

The correct option is "is an increasing function".

To determine the nature of the function $f\left(x\right)=2x+cosx$, we need to examine its first derivative with respect to $x$. A function is strictly increasing on an interval if its first derivative is strictly positive for all values in that interval.

First, let's differentiate the function $f\left(x\right)$ with respect to $x$:
$$f^{\text{'}}\left(x\right)=\frac{d}{dx}(2x+cosx)$$
Using the sum rule and standard derivatives ($\frac{d}{dx}\left(x\right)=1$ and $\frac{d}{dx}(cosx)=-sinx$), we get:
$$f^{\text{'}}\left(x\right)=2-sinx$$

Next, we analyze the range of values that the derivative $f^{\text{'}}\left(x\right)$ can take. We know that the sine function is bounded between $-1$ and $1$ for all real numbers $x$:
$$-1\le sinx\le 1$$

Multiplying the inequality by $-1$ reverses the inequality signs:
$$-1\le -sinx\le 1$$

Now, we add $2$ to all parts of the inequality:
$$2-1\le 2-sinx\le 2+1$$
Simplifying the terms yields:
$$1\le f^{\text{'}}\left(x\right)\le 3$$

Since $f^{\text{'}}\left(x\right)\ge 1>0$ for all $x\in ℝ$, the derivative of the function is always strictly positive. Therefore, the function $f\left(x\right)$ is an increasing function on $ℝ$.