Question

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

• A. There is at least one $x$ in the interval $[1,\infty )$ for which $f(x+2)-f\left(x\right)<2$
• B. $\underset{x\to \infty }{lim}{f}^{\prime }\left(x\right)=1$
• C. $f(x+2)-f\left(x\right)>2$ for all $x$ in the interval $[1,\infty )$
• D. $f^{\prime }\left(x\right)$ is strictly decreasing in the interval $[1,\infty )$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $\underset{x\to \infty }{lim}{f}^{\prime }\left(x\right)=1$
C $f(x+2)-f\left(x\right)>2$ for all $x$ in the interval $[1,\infty )$
D $f^{\prime }\left(x\right)$ is strictly decreasing in the interval $[1,\infty )$

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

$f^{\prime }\left(x\right)=\frac{1}{x}\sin \frac{1}{x}+\cos \frac{1}{x}$

${lim}_{x\to \infty }{f}^{\prime }\left(x\right)=0\times \sin 0+\cos \left(0\right)$

${lim}_{x\to \infty }{f}^{\prime }\left(x\right)=1$ (option B)

Now, $xϵ[1,\infty ]\Rightarrow \frac{1}{x}ϵ[0,1]$

$f^{\prime \prime }\left(x\right)=-\frac{1}{x^3}\cos \left(\frac{1}{x}\right)<0$ (option D)

Graph of $f^{\prime }\left(x\right)$ will be (Image)

Now, in $[x,x+2],xϵ[1,\infty ],f\left(x\right)$ is continuous and Differentiable so by LMVT,

$f^{\prime }\left(x\right)=f\frac{(x+2)-f\left(x\right)}{2}$

$f^{\prime }\left(x\right)>1$

$f\frac{(x+2)-f\left(x\right)}{2}>1$

$f(x+2)-f\left(x\right)>2$

For all $xϵ[1,\infty ]$ (option C)