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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 [AIEEE 2004]


A
1
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B
-3
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C
-1
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D
2
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Solution

The correct option is D 2

Given, f(x)=exx+1,I1=f(a)f(a)xg{x(1x)}dx
and I2=f(a)f(a)g{x(1x)}dx
f(a)=ea1+ea,f(a)=ea1+ea
f(a)+f(a)=1
Now, 2I1=f(a)f(a){f(a)+f(a)x}g{(1x)(x)}dx
2I1=I2I2I1=2


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