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Note by Dimple Maniar
3 years, 4 months ago

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Comments

We know that f:R{3n,nZ}[1,13]f:\mathbb{R}-\{3n,n\in Z\}\rightarrow [1,13] and it is also given that (1)3p3p+3f(x)dx=3(1)\int^{3p+3}_{3p}f(x)dx=3 and (2)3q3q+3f(x)dx=39(2)\int^{3q+3}_{3q}f(x)dx=39.

(1)(1) is only possible when f(x)=1x(3p,3p+3)f(x)=1 \forall x\in (3p,3p+3) and (2)(2) is only possible when f(x)=13x(3q,3q+3)f(x)=13 \forall x\in (3q,3q+3).

As, p2+q2p^2+q^2 is odd, there are 22 cases,

  1. pp is odd and qq is even.

  2. pp is even and qq is odd.

Giving us 2\boxed{2} functions.

Then, we can easily tell that the period of f(x)f(x) is 6\boxed{6} and 0900f(x)dx=6300\int^{900}_{0}f(x)dx=\boxed{6300}.

Relation between k1k_{1} and k2k_{2} can be found only when we are specified which of the two functions is f(x)f(x).

Akshat Sharda - 3 years, 4 months ago
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