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Question

Arrange the following function in ascending order of their fundamental period.

A
cos(3xπ2)2
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B
2cosx
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C
cos4(x+π)
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D
6cos2(x4+π)
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Solution

We know cosx has a period 2π.
Now, we know if a function f(x) has a fundamental period T and it is transformed to af(bx+c)+d, then the period of this function is T|b|.

Using this if f(x)=cosx, then 2cosx=2f(x)
Hence the period will remain the same i.e. 2π.

Now, for cos4(x+π) it can be expressed as f(bx+c), where b=4 & c=π
Hence, it's period will be 2π4=π2

Similarly, for cos(3xπ2)2 it can be expressed as: f(bx+c)+d, where b=3,c=π2 & d=2

Hence, it's period will be 2π3.

Now for the last transformation as 6cos2(x4+π), it can be written as d+af(bx+c), where d=6,a=1,b=12,c=π
Thus, it's period will be 2π1/2=4π
Thus, in ascending order of fundamental period of trigonometric functions, we can arrange them as:
cos4(x+π),cos(3xπ2)2, 2cosx,6cos2(x4+π)

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