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

Answer 1

Answer: (A), (C)

The correct options are
A $f\left(x\right)$ increases in the interval $(\frac{1}{2k+1},\frac{1}{2k})$
C $f\left(x\right)$ decreases in the interval $(\frac{1}{2k+2},\frac{1}{2k+1})$

$f\left(x\right)=\cos \left(\frac{\pi }{x}\right)$
$f^{\prime }\left(x\right)=-\sin \left(\frac{\pi }{x}\right)(-\frac{\pi }{x^2})=\frac{\pi }{x^2}\sin \left(\frac{\pi }{x}\right)$

For increasing function $f^{\prime }\left(x\right)>0$
$\Rightarrow \sin \left(\frac{\pi }{x}\right)>0\Rightarrow \left(2k\pi \right)<\frac{\pi }{x}<(2k+1)\pi \Rightarrow \frac{1}{2k}>x>\frac{1}{2k+1}$

For decreasing function $f^{\prime }\left(x\right)<0$
$\Rightarrow \sin \left(\frac{\pi }{x}\right)<0\Rightarrow (2k+1)\pi <\frac{\pi }{x}<(2k+2)\pi \Rightarrow \frac{1}{2k+1}>x>\frac{1}{2k+2}$