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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
1
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B
2
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C
2
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D
1
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Solution

The correct option is C 2
f(x)=ex1+exf(a)=ea1+eaf(a)=ea1+ea=11+ea
So,
f(a)+f(a)=ea+11+ea=1
Let
f(a)=α;f(a)=1α
Now,
I1=1ααxg(x(1x)) dx(1)
Using property,
I1=1αα(1x)g((1x)(1(1x))) dxI1=1αα(1x)g(x(1x)) dx(2)
Adding (1) and (2),m
2I1=1ααg(x(1x)) dx=I2I2I1=2

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