Question

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

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

Answer 1

Answer: (A), (C)

$f\left(x\right)$ increases in the interval $(\frac{1}{2k+1},\frac{1}{2k})$
$f\left(x\right)$ increases in the interval $(\frac{1}{2k+1},\frac{1}{2k+1})$

$f\left(x\right)=cos\left(\frac{\pi }{x}\right)f^{\prime }\left(x\right)=-sin\left(\frac{\pi }{x}\right)\left(\frac{-\pi }{x}\right)=\frac{\pi }{x^2}sin\frac{\pi }{x}whichis>0$
for increasing function f'(x) > 0
$\Rightarrow sin\frac{\pi }{x}>0$
$2k\pi <\frac{\pi }{x}<(2k+1)\pi$
$\frac{1}{2k}>x>\frac{1}{2k+1}$
for decreasing function f '(x) < 0
$\Rightarrow sin\frac{\pi }{x}<0$
$\Rightarrow \frac{\pi }{x}$ is lying between $\left(\right(2k+1)\pi ,(2k+2\left)\pi \right)$
$\Rightarrow x$ belongs to $(\frac{1}{2k+2},\frac{1}{2k+1})$
Therefore Answer must be $AC$