JEE Advanced2008MathematicsApplication of DerivativesActual
Let f(x) be a non-constant twice differentiable function defined on (- , ) , such that f(x)=f(1-x) and f^ ( 1 4 )=0 . Then,
Options
- Af^ (x) vanishes atleast twice on [0,1]
- Bf^ ( 1 2 )=0
- C_ 1 2 ^ 1 2 f (x+ 1 2 ) x d x=0
- D₀^ 1 2 f(t) e^ t d t= _ 1 2 ^1 f(1-t) e^ t d t
Correct answer
A. f^ (x) vanishes atleast twice on [0,1]
Step-by-step solution
Given that, f(x)=f(1-x) On differentiating w.r.t. x , we get f^ (x)=-f^ (1-x) Let us put x= 1 2 2 f^ ( 1 2 )=0 f^ ( 1 2 )=0 Since, f^ ( 1 2 )=0 and f^ ( 1 4 )=0 f^ (x)=0 at two points in [0,1] . Now, _ -1 / 2 ^ 1 / 2 f (x+ 1 2 ) x d x=0 As, f (x+ 1 2 ) x is an odd function which is clear from the following explanation. Let g(x)=f (x+ 1 2 ) x , g(-x)=f ( 1 2 -x ) (-x)=- x f (1- ( 1 2 -x ) )=- x f ( 1 2 +x )=-g(x) Moreover, _ 1 / 2 ^1 f(1-t) e^ ( t) d t= ₀^ 1 / 2 f(u) e^ u d u where, 1-t=u .