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Question

Let f(x) be a continous function such that f(ax)+f(x)=0 for all x[0,a] then a0dx1+ef(x) is equal to

A
a
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B
a2
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C
f(a)
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D
12f(a)
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Solution

The correct option is C a2
I=a0dx1+ef(x) (say)

I=a0dx1+ef(a+0x)

adding both the equations,

2I=a01+ef(ax)+1+ef(x)(1+ef(x))(1+ef(ax))dx

=a02+ef(x)+ef(ax)1+ef(ax)+ef(x)+ef(x)+f(ax)

f(x)+f(ax)=0

2I=a02+ef(x)+ef(ax)2+ef(x)+ef(ax)dx

2I=a

I=a2

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