Question

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

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

Answer 1

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

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