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If f(x)=tan 2 (πx/(n 2 -5n+8))+cot(n+m)πx; (n€N,m€Q),is a periodic function with 2 as the fundamental period,then m can’t belong to:

If f(x)=tan2(πx/(n2-5n+8))+cot(n+m)πx; (n€N,m€Q),is a periodic function with 2 as the fundamental period,then m can’t belong to:

 

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1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
8 years ago
Ans:
Period of tan2(ax):
\frac{\pi }{|a|}
Period of cot(ax):
\frac{\pi }{|a|}
f(x) = tan^{2}(\frac{\pi x}{n^{2}-5n+8}) + cot((n+m)\pi x)
f_{1}(x) = tan^{2}(\frac{\pi x}{n^{2}-5n+8})
Period of f1(x):
n^{2}-5n+8
f_{2}(x) = cot((n+m)\pi x)
Period of f2(x):
\frac{1}{|n+m|}
Period of f(x):
\frac{n^{2}-5n+8}{n+m} = 2
2m = n^{2}-7n+8
2m \geq \frac{-17}{4}
m \geq \frac{-17}{8}
‘m’ can’t belong to:
(-\infty , \frac{-17}{8})
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

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