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Question

If f(a+b+1x)=f(x) x, where a and b are fixed positive real numbers, then 1(a+b)bax(f(x)+f(x+1))dx is equal to

A
b1a1f(x)dx
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B
b+1a+1f(x+1)dx
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C
b1a1f(x+1)dx
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D
b+1a+1f(x)dx
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Solution

The correct option is C b1a1f(x+1)dx
f(a+b+1x)=f(x) ...(1)
xx+1
f(a+bx)=f(x+1) ...(2)
I=1a+bbax(f(x)+f(x+1))dx ...(3)

From (1) and (2)
I=1a+bba(a+bx)(f(x+1)+f(x))dx ...(4)

Adding (3) and (4)
2I=ba(f(x)+f(x+1))dx2I=baf(x+1)dx+baf(x)dx2I=baf(a+bx+1)dx+baf(x)dx2I=2baf(x)dxI=baf(x)dx
Let x=t+1,dx=dt
I=b1a1f(t+1)dtI=b1a1f(x+1)dx

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