For the function $f(x)=x \cos \frac{1}{x}, x \geq 1$,

• A

  for atleast one $x$ in the interval $[1, \infty)$, $f(x+2)-f(x) < 2$
• Correct answer
  B

  $\lim _{x \rightarrow \infty} f^{\prime}(x)=1$
• Correct answer
  C

  for all $x$ in the interval $[1, \infty)$, $f(x+2)-f(x)>2$
• Correct answer
  D

  $f^{\prime}(x)$ is strictly decreasing in the interval $[1, \infty)$

Solution

Given, $f(x)=x \cos \frac{1}{x}, x \geq 1$ $$ \begin{aligned} & \Rightarrow \quad f^{\prime}(x)=\frac{1}{x} \sin \frac{1}{x}+\cos \frac{1}{x} \\ & \Rightarrow \quad f^{\prime \prime}(x)=-\frac{1}{x^3} \cos \left(\frac{1}{x}\right) \end{aligned} $$ Now, $\lim _{x \rightarrow \infty} f^{\prime}(x)=0+1=1$ $\Rightarrow$ Option (b) is correct. Now, $\quad x \in[1, \infty) \Rightarrow \frac{1}{x} \in(0,1]$ $\Rightarrow f^{\prime \prime}(x) < 0 \Rightarrow$ Option (d) is correct. As $f^{\prime}(1)=\sin 1+\cos 1>1$ $f^{\prime}(x)$ is strictly decreasing and $\lim _{x \rightarrow \infty} f^{\prime}(x)=1$ 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 $x \in[1, \infty)$ $$ \begin{aligned} & \Rightarrow \quad \frac{f(x+2)-f(x)}{2}>1 \\ & \Rightarrow f(x+2)-f(x)>2 \quad \text { for all } x \in[1, \infty) \end{aligned} $$

Hence, the correct answer is options: 'B', 'C', 'D'.