IIT JEE Advanced 2008 Paper 1

We have
f(x) = f(1 – x) (1)
Substituting x =

1

2

+x , we get
f (

1

2

+x) = f (

1

2

−x)
That is,f (

1

2

+x) is an even function and sin x . f (

1

2

+x) is an odd function. Therefore,

1 2

∫

− 1 2

f(x+

1

2

) sin x = 0
Differentiating Eq. (1), we get
f ′(x) = - f ′(1- x) (2)
Substituting x = 1 / 2, we get
f′(

1

2

) = - f′(

1

2

)
⇒ f′(

1

2

) = 0
Now, substituting x = 1/4in Eq. (2), we get
f′(

1

4

) = −f′(

3

4

) = 0
⇒ f′(

1

4

) = f′(

1

2

) = f′(

3

4

)
Using Rolle’s theorem, f ″(x) = 0 has at least one solution in (

1

4

,

1

2

) and one solution in (

1 2.3

4

).
Therefore, f ′′(x) = 0 vanishes at least twice on [0, 1]. Now,

1 2

∫

0

f(t)esinπtdt
Substituting t = 1 – x, we get

1 2

∫

1

f(1−x)esinπ(1−x) (- dx) =

1

∫

1 2

f(1−x)esinπx . dx