1f ' (x) = \sin \left(\frac{\pi }{x} \right) .\left(\frac{\Pi }{x^{2}} \right)
for f(x) to be increasing,
2n\Pi <\left(\frac{\Pi }{x} \right)<\left(2n+1)\Pi \right
hence x \epsilon \left(\frac{1}{2n+1},\frac{1}{2n} \right)
f\left(x \right)=\cos \left(\frac{\Pi }{x} \right)
then prove that f(x) is increasing in the interval \left(\frac{1}{2n+1} ,\frac{1}{2n}\right) n\epsilon N
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