Mathematics - Application of Derivative Question with Solution | TestHub

Given, $f(x)=x\cos \frac{1}{x},x\ge 1$ $$\begin{array}{rl}& \Rightarrow f^{\prime }(x)=\frac{1}{x}\sin \frac{1}{x}+\cos \frac{1}{x}\\ & \Rightarrow f^{\prime \prime }(x)=-\frac{1}{x^3}\cos \left(\frac{1}{x}\right)\end{array}$$ Now, ${\lim }_{x\to \infty }{f}^{\prime }(x)=0+1=1$ $\Rightarrow$ Option (b) is correct. Now, $x\in [1,\infty )\Rightarrow \frac{1}{x}\in (0,1]$ $\Rightarrow f^{\prime \prime }(x)<0\Rightarrow$ Option (d) is correct. As $f^{\prime }(1)=\sin 1+\cos 1>1$ $f^{\prime }(x)$ is strictly decreasing and ${\lim }_{x\to \infty }{f}^{\prime }(x)=1$ So, graph of $f^{\prime }(x)$ is shown as below.
Now, in $[x,x+2],x\in [1,\infty ),f(x)$ is continuous and differentiable so by LMVT, $$f^{\prime }(x)=\frac{f(x+2)-f(x)}{2}$$ As $f^{\prime }(x)>1$ for all $x\in [1,\infty )$ $$\begin{array}{rl}& \Rightarrow \frac{f(x+2)-f(x)}{2}>1\\ & \Rightarrow f(x+2)-f(x)>2\text{ for all }x\in [1,\infty )\end{array}$$