Question

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

• A. For at least one $x$ in the interval $[1,\infty ],f(x+2)–f\left(x\right)<2$
• B. $li{m}_{x\to \infty }f\text{'}\left(x\right)=1$
• C. For all $x$ in the interval $[1,\infty ],f(x+2)–f\left(x\right)>2$
• D. $f\left(x\right)$ is strictly increasing in the interval $[1,\infty )$
• E. $\mathrm{B},\mathrm{C}\&\mathrm{D}$

Answer 1

The correct option is E

$\mathrm{B},\mathrm{C}\&\mathrm{D}$

Explanation for the correct option:

Step1. Checking option $B$

Differentiating the function with respect to $\mathrm{x}$:

$\begin{array}{rcl}f\left(x\right)& =& x\cos \left(\frac{1}{x}\right),x\ge 1\\ \Rightarrow f\text{'}\left(x\right)& =& \frac{1}{x}\sin \left(\frac{1}{x}\right)+\cos \left(\frac{1}{x}\right)\\ \Rightarrow \underset{x\to \infty }{\lim }f\text{'}\left(x\right)& =& \underset{x\to \infty }{\lim }\left[\frac{1}{x}\sin \left(\frac{1}{x}\right)+\cos \left(\frac{1}{x}\right)\right]\\ & \Rightarrow & \underset{x\to \infty }{\lim }f\text{'}\left(x\right)=\frac{1}{\infty }\sin \left(\frac{1}{\infty }\right)+\cos \left(\frac{1}{\infty }\right)\left[\because \frac{1}{\infty }=0\right]\end{array}$

$\Rightarrow$$\underset{x\to \infty }{\lim }f\text{'}\left(x\right)=\left(0\right)\sin \left(0\right)+\cos \left(0\right)$

$\Rightarrow$$\underset{x\to \infty }{\lim }f\text{'}\left(x\right)=1$

Since, $\mathrm{f}\text{'}\left(\mathrm{x}\right)$ is exist therefore option $\left(B\right)$ is correct.

Step2. Checking option for $D$

For all $x\in [1,\infty ]$

$$\begin{gathered} \mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{1}{x}\sin \left(\frac{1}{\mathrm{x}}\right)+\cos \left(\frac{1}{\mathrm{x}}\right) \\ \mathrm{f}"\left(\mathrm{x}\right)=-\frac{1}{x^3}\cos \left(\frac{1}{\mathrm{x}}\right) \\ \mathrm{f}"\left(\mathrm{x}\right)<0 \end{gathered}$$

Since $\mathrm{f}"\left(\mathrm{x}\right)$ is negative for all $x\in [1,\infty ]$ therefore the given equation is strictly increasing & hence option $\left(D\right)$ is also correct.

Step3. Checking option $C$

As $f\text{'}\left(1\right)=\sin 1+\cos 1>1$

$\mathrm{f}\text{'}\left(\mathrm{x}\right)$is strictly decreasing and ${\lim }_{\mathrm{x}\to \infty }\mathrm{f}\text{'}\left(\mathrm{x}\right)=1$

Now, in $[x,x+2],x\in [1,\infty ],f\left(x\right)$ is continuous and differentiable.

By LMVT, $\mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{\left[\mathrm{f}\right(\mathrm{x}+2)–\mathrm{f}(\mathrm{x}\left)\right]}{2}$as $\mathrm{f}\text{'}\left(\mathrm{x}\right)>1$ for all $x\in [1,\infty ]$

$$\begin{gathered} \frac{\left[f\right(x+2)–f(x\left)\right]}{2}>1 \\ \left[f\right(x+2)–f(x\left)\right]>2, \end{gathered}$$

For all $x\in [1,\infty ]$ also exist therefore option $\left(C\right)$ is also correct.

Hence, correct option is $\left(E\right)$