For the function $f\left(x\right)=x\cos \frac{1}{x},x\ge 1$ which one of the following is incorrect?

we have,

$f\left(x\right)=x\cos \frac{1}{x},x\ge 1$

$\Rightarrow f^{\mathrm{\prime }}\left(x\right)=\cos \frac{1}{x}+\frac{1}{x}\sin \frac{1}{x}$and $f^{\prime \prime }\left(x\right)=-\frac{1}{x^3}\cos \frac{1}{x}$

$\Rightarrow \underset{x\to \mathrm{\infty }}{lim} f^{\mathrm{\prime }}\left(x\right)=1$ and $f^{\prime \prime }\left(x\right)<0$for all $x\ge 1$

$\Rightarrow f^{\mathrm{\prime }}\left(x\right)$ is strictly decreasing in the interval $(1 ,\infty )$

Let $g\left(x\right)=f(x+2)-f\left(x\right)$ . Then 

$g^{\mathrm{\prime }}\left(x\right)=f^{\mathrm{\prime }}(x+2)-f^{\mathrm{\prime }}\left(x\right)<0$      [$\therefore f\text{'}\left(x\right)$) is strictly decreasing]

$\Rightarrow g\left(x\right)$ is strictly decreasing on  $(1 ,\infty )$

Now, 

$\begin{array}{l}\underset{x\to \mathrm{\infty }}{lim} g\left(x\right)=\underset{x\to \mathrm{\infty }}{lim} f(x+2)-f\left(x\right)\\ =\underset{x\to \mathrm{\infty }}{lim} (x+2)\cos \frac{1}{x+2}-x\cos \frac{1}{x}\end{array}$ 

$\begin{array}{l}=\underset{x\to \mathrm{\infty }}{lim} x\left(\cos \frac{1}{x+2}-\cos \frac{1}{x}\right)+2\cos \frac{1}{x+2}\\ =\underset{x\to \mathrm{\infty }}{lim} 2x\sin \frac{x+1}{x^2+2x}\sin \frac{1}{x^2+2x}+2\cos \frac{1}{x+2}\\ =2\underset{x\to \mathrm{\infty }}{lim} \frac{\sin \left(\frac{x+1}{x^2+\frac{1}{2x}}\right)}{\left(\frac{x+1}{x^2}+\frac{1}{2x}\right)}\times \frac{\sin \left(\frac{1}{x^2+2x}\right)}{\left(\frac{1}{x^2+2x}\right)}\times \frac{x(x+1)}{{\left(x^2+2x\right)}^2}+2\cos \frac{1}{x}\\ =2\times 1\times 1\times 0+2=2\end{array}$

As $g\left(x\right)$ is decreasing on $[1, \infty ]$ and $\underset{x\to \mathrm{\infty }}{lim} g\left(x\right)=2$

$\therefore g\left(x\right)>2$for all $x\in [1,\mathrm{\infty })$

$\Rightarrow f(x+2)-f\left(x\right)>2$ for all $x\in [1,\mathrm{\infty })$

Hence, option (a) is incorrect