Question

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

• A. for atleast one $x$ in interval $[1, \infty), f(x+2)-f(x)<2$
• B. $\displaystyle\lim _{x \rightarrow \infty} f'(x)=1$
• C. for all $x$ in the interval $[1, \infty), f(x+2)-f(x)>2$
• D. $f'( x )$ is strictly decreasing in the interval $[1, \infty)$

Answer 1

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

For $f(x)=x \cos \left(\frac{1}{x}\right), x \geq 1$
$f'(x)=\cos \left(\frac{1}{x}\right)+\frac{1}{x} \sin \left(\frac{1}{x}\right) \rightarrow 1$ for $x \rightarrow \infty$
also $f''(x)=\frac{1}{x^{2}} \sin \left(\frac{1}{x}\right)-\frac{1}{x^{2}} \sin \left(\frac{1}{x}\right)-\frac{1}{x^{3}} \cos \left(\frac{1}{x}\right)$
$=-\frac{1}{x^{3}} \cos \left(\frac{1}{x}\right)<0 $ for $x \geq 1$
$\Rightarrow f'(x)$ is decreasing for $[1, \infty)$
$\Rightarrow f'(x+2)< f'(x) $
Also, $\displaystyle\lim _{x \rightarrow \infty} f(x+2)-f(x)$
$=\displaystyle\lim _{x \rightarrow \infty}\left[(x+2) \cos \frac{1}{x+2}-x \cos \frac{1}{x}\right]=2$
$\therefore f(x+2)-f(x)>2 \forall x \geq 1$