Question

Let $f(x)=\cos \left(\frac{\pi }{x}\right),x\ne 0$ 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)

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