For the function f(x)=x cos 1/x, x ≥ 1, which one of the following is…

We have,

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

$\Rightarrow f'(x)=\cos \frac{1}{x}+\frac{1}{x} \sin \frac{1}{x}$ and $f''(x)=-\frac{1}{x^3} \cos \frac{1}{x}$

$\Rightarrow \lim\limits_{x \rightarrow \infty} f'(x)=1$ and f''(x) < 0  for all $x \geq 1$

⇒ f'(x) is strictly decreasing in the interval $[1, \infty)$

Let g(x) = f(x + 2) - f(x). Then,

$g'(x)=f'(x+2)-f'(x)<0$             [∵ f'(x) is strictly decreasing]

⇒ g(x) is strictly decreasing on $[1, \infty)$

Now,

$\lim\limits_{x \rightarrow \infty} g(x)=\lim\limits_{x \rightarrow \infty} f(x+2)-f(x)$

$=\lim\limits_{x \rightarrow \infty}(x+2) \cos \frac{1}{x+2}-x \cos \frac{1}{x}$

$=\lim\limits_{x \rightarrow \infty} x\left(\cos \frac{1}{x+2}-\cos \frac{1}{x}\right)+2 \cos \frac{1}{x+2}$

$=\lim\limits_{x \rightarrow \infty} 2 x \sin \frac{x+1}{x^2+2 x} \sin \frac{1}{x^2+2 x}+2 \cos \frac{1}{x+2}$

$=2 \lim\limits_{x \rightarrow \infty} \frac{\sin \left(\frac{x+1}{x^2+2 x}\right)}{\left(\frac{x+1}{x^2+2 x}\right)} \times \frac{\sin \left(\frac{1}{x^2+2 x}\right)}{\left(\frac{1}{x^2+2 x}\right)} \times \frac{x(x+1)}{\left(x^2+2 x\right)^2}+2 \cos \frac{1}{x+2}$

$=2 \times 1 \times 1 \times 0+2=2$

As g(x) is decreasing on $[1, \infty)$ and $\lim\limits_{x \rightarrow \infty} g(x)=2$

∴  $g(x)>2$ for all $x \in[1, \infty)$

$\Rightarrow f(x+2)-f(x)>2$ for all $x \in[1, \infty)$.

Hence, option (a) is incorrect.