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Question

If f(x)=ex1+ex,I1=f(a)f(a)xg{x(1x)}dx and I2=f(a)f(a)g{x(1x)}dx, then the value of I2I1 is

A
2
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B
1
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C
-1
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D
-3
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Solution

The correct option is A 2
f(a)=ea1+ea=1ex+1, f(a)=ea1+eaf(a)+f(a)=1
I1=f(a)f(a)xg{x(1x)}dx=f(a)f(a)[f(a)+f(a)x]g{[f(a)+f(a)x}{1f(a)f(a)+x}]dx
=f(a)f(a)(1x)g[(1x)x]dx=f(a)f(a)g{x(1x)}dxf(a)f(a)xg[x(1x)]dx=I2I1
2I1=I2I2I1=2

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