Let $f\left(x\right)=\cos 2\pi x+x-\left[x\right]$, $\left[·\right]$ denotes the greatest integer function. Then, number of points in $[0,10]$ at which $f\left(x\right)$ assumes its local maximum value, is

Consider,

$f\left(x\right)=\cos 2\pi x-\left(x\right)\dots \dots [x\in (0,10\left)\right]$

$\because f$ is a periodic function of period $1$ .

$f\left(x\right)=\cos 2\pi x+x\dots \dots [n\in (0,1\left)\right]$

This function is continuous and differentiable in the interval $[0,1]$

$$\begin{gathered} f\left(0\right)=\cos \left(0\right)+0 \\ f\left(0\right)=1 \\ f\left(1\right)=\cos 2\pi +1 \\ f\left(1\right)=2 \end{gathered}$$

Differentiate function $f\left(x\right)$,

$$\begin{gathered} f\text{'}\left(x\right)=-\sin 2\pi x\left(2\pi \right)+1\text{ for all }x\in {}^9\cup _{k=0}(k,k+1) \\ f\text{'}\left(x\right)=1-2\pi \times \sin \left(2\pi x\right) \end{gathered}$$

Consider, $f^{\text{'}}\left(x\right)=0$

$\sin 2\pi x=\frac{1}{2\pi }$

Consider, differentiated equation $\sin 2\pi x=\frac{1}{2\pi }$ has two solutions in each of the sub-intervals $(k,k+1)$ were $k=1,2,3,\dots ,9$ since point of maxima and minima occur alternatively.

Therefore, there are$10$  points of local maxima and $10$ points of local minima.

Hence, there are $10$ points in the local maxima.

Therefore, the correct answer is option $\left(2\right)$.