If f(x)=ex1+ex,I1=f(a)∫f(−a)xg(x(1−x))dx and I2=f(a)∫f(−a)g(x(1−x))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 C2 f(x)=ex1+exf(a)=ea1+eaf(−a)=e−a1+e−a=11+ea So, f(−a)+f(a)=ea+11+ea=1 Let f(−a)=α;f(a)=1−α Now, I1=1−α∫αxg(x(1−x))dx⋯(1) Using property, ⇒I1=1−α∫α(1−x)g((1−x)(1−(1−x)))dx⇒I1=1−α∫α(1−x)g(x(1−x))dx⋯(2) Adding (1) and (2),m ⇒2I1=1−α∫αg(x(1−x))dx=I2⇒I2I1=2