If f(x)=ex1+ex,l1=∫f(−a)f(a) xg{x(1−x)}dx .Then the value of I2/I1, is

If f(x)=ex1+ex,l1=f(a)f(a)xg{x(1x)}dx .Then the value of I2/I1, is

  1. A

    -1

  2. B

    1/2

  3. C

    2

  4. D

    1

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    Solution:

    We have, f(x)=ex1+ex 

    f(x)=ex1+ex=11+ex         f(x)+f(x)=ex1+ex+11+ex=1f(a)+f(a)=1

    Now, 

            I1=f(a)f(a)xg{x(1x)}dx                                   …(i)

    Using abf(a+bx)=abf(x) and f(a)+f(a)=1, we get

    I1=f(a)f(a)(1x)g{(1x)(1(1x))}dx I1=f(a)f(a)(1x)g{x(1x)}dx

     I1=f(a)f(a)(1x)g{x(1x)}dx                               ..(ii)

    Adding (i) and (ii), we get

    2I1=f(a)f(a)g{x(1x)}dx 2I1=I2I2I1=2

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