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Differential Calculus> If f(x)=tan 2 (πx/(n 2 -5n+8))+cot(n+m)πx...

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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:

vineet nimesh , 10 Years ago

Grade

1 Answers

Jitender Singh

Last Activity: 10 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

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