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Question

Let f(x)=ax (a>0) be written as f(x)=f1(x)+f2(x), where f1(x) is an even function and f2(x) is an odd function. Then f1(x+y)+f1(x−y) equals :

A
2f1(x)f1(y)
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B
2f1(x+y)f1(xy)
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C
2f1(x)f2(y)
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D
2f1(x+y)f2(xy)
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Solution

The correct option is A 2f1(x)f1(y)
Every function f(x) can be represented as
f(x)=f(x)+f(x)2+f(x)f(x)2,
where f(x)+f(x)2 is even function and f(x)f(x)2 is odd function.

f(x)=ax
So, f1(x)=ax+ax2

f1(x+y)+f1(xy)
=ax+y+a(x+y)2+axy+a(xy)2
=ax(ay+ay)+ax(ay+ay)2
=(ax+ax)(ay+ay)2
=2f1(x)f1(y)

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