The function $\mathrm{f}:\mathrm{R}\to \mathrm{R}$ defined by $f\left(x\right)=4+3\cos x$ is:

Given: $f\left(x\right)=4+3\cos x$
Since, $\cos \frac{\pi }{2}=\cos \left(-\frac{\pi }{2}\right)$
$\begin{array}{l}\Rightarrow 4+3\cos \frac{\pi }{2}=4+3\cos \left(-\frac{\pi }{2}\right)\\ \Rightarrow f\left(\frac{\pi }{2}\right)=f\left(-\frac{\pi }{2}\right)\end{array}$
But $\frac{\pi }{2}\ne -\frac{\pi }{2}$
So, f is not one-one.
Range of $\cos x\text{ is }[-1,1]$
$\begin{array}{l}\Rightarrow -1\le \cos x\le 1\\ \Rightarrow -3\le 3\cos x\le 3\\ \Rightarrow 1\le 4+3\cos x\le 7\\ \Rightarrow 1\le f\left(x\right)\le 7\end{array}$
So, the range of f is [1,7]
Thus, f is not onto.
Hence, f is neither one-one nor onto.