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Question

The period of fx=10cos2πx+x-x+sin2πx is


A

1

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B

2

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C

3

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D

π

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Solution

The correct option is A

1


Explanation for the correct option:

Given function is fx=10cos2πx+x-x+sin2πx

The period of a function is that the value or interval at which the function repeats its values and is defined as the smallest constant for which fx+c=fx

Here for c=1,

fx+1=10cos2πx+1+x+1-x+1+sin2πx+1=10cos2πx+π+x-x+sin2πx+πx+1-x+1=x-x=10-cosπx2+x-x+-sinπx2cosπ+θ=-cosθ,sinπ+θ=-sinθ=10cos2πx+x-x+sin2πxfx+1=fx

fx+2=10cos2πx+2+x+2-x+2+sin2πx+2=10cos2πx+2π+x-x+sin2πx+2πx+2-x+2=x-x=10cosπx2+x-x+sinπx2cos2π+θ=cosθ,sin2π+θ=sinθ=10cos2πx+x-x+sin2πxfx+2=fx

fx+3=10cos2πx+3+x+3-x+3+sin2πx+3=10cos2πx+3π+x-x+sin2πx+3πx+3-x+3=x-x=10-cosπx2+x-x+-sinπx2cos3π+θ=-cosθ,sin3π+θ=-sinθ=10cos2πx+x-x+sin2πxfx+3=fx

From the above we can observe that c=1,2,3 satisfies the condition fx+c=fx

To determine the period of function consider the least possible value of c i.e. 1

Thus the period of the given function f(x) is 1

Hence the correct option is option (A) i.e. 1


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