Question

Let $f(x) = \cos \left( \frac{\pi}{x} \right) , x \neq 0 $ then assuming $k$ as an integer,

• A. f(x) increases in the interval $\left(\frac{1}{2k + 1} , \frac{1}{2k} \right)$
• B. f(x) decreases in the interval $\left(\frac{1}{2k + 1} , \frac{1}{2k} \right)$
• C. f(x) decreases in the interval $\left(\frac{1}{2k + 1} , \frac{1}{2k + 2 } \right)$
• D. f(x) increases in the interval $\left(\frac{1}{2k + 1} , \frac{1}{2k + 2 } \right)$

Answer 1

Answer: (A), (C)

$f(x)=\cos \left(\frac{\pi}{x}\right)$
$\Rightarrow f'(x)=-\sin \left(\frac{\pi}{x}\right)\left(\frac{-\pi}{x^{2}}\right)=\frac{\pi}{x^{2}} \sin \frac{\pi}{x}$
For increasing function, $f'(x) > 0$
$\Rightarrow \sin \left(\frac{\pi}{x}\right) > 0 $
$\Rightarrow 2 k \pi < \frac{\pi}{x} < (2 k+1) \pi$
$\Rightarrow \frac{1}{2 k} > x > \frac{1}{2 k+1}$
For decreasing function, $f'(x) < 0$
$\Rightarrow \sin \left(\frac{\pi}{x}\right) < 0$
$\Rightarrow \frac{\pi}{x} \in[(2 k+1) \pi,(2 k+2) \pi] $
$\Rightarrow x \in\left(\frac{1}{2 k+2}, \frac{1}{2 k+1}\right)$