1is the ans 1/2 ??
am getting this one only... [3]
If 0∫x f(t)dt = x + x∫1 t f(t)dt , Find f(1).
I am getting two answers for this!! Pls. try it & explain!!
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10 Answers
1me also gettin jus one ans...
diffrentiate by newton leibnitz...
f(x)=1+(-xf(x))
(1+x)f(x)=1
put x=1
ans 1/2
62If 0∫x f(t)dt = x + x∫1 t f(t)dt , Find f(1).
0∫x f(t)dt = x + 0∫1 t f(t)dt - 0∫x t f(t)dt
thus, 0∫x (1+t)f(t)dt = x + 0∫1 t f(t)dt
Take derivative wrt x
(1+x)f(x)=1
f(x)=1/(1+x)
Tx sky for the hint ;)
62Why lebinitz!!!
I hate that thing when it comes in a JEE question... This does not need that theorem! :(
1sry bhaiya haame to min.. amt of theorams ate hain unhi ka use karna padta hai...
62arrey sid but i din use the theorem at all..
If you look closely u din either.. (only lebinitz made u belive that you did!)
If g(x)=a∫x f(t)dt
then isnt g'(x)=f(x)??
That is all that i have used...
RHS is a+ constant!
11I am really sorry for the trouble!!! I understood where I was making a mistake.I'll tell you what went wrong.This is how I was getting 2 answers :
a) f(x)=1-xf(x) ⇒ f(x)=1/(1+x) ∴ f(0)=1
b) 0∫x f(x-t)dt = x + x∫1 (1+x-t) f(1+x-t)dt f(0)=1-f(1) ⇒ f(1)=1-f(0)
∴From a) only, we get f(1)=1/(1+1)=1/2
& from b) & a), we get f(1)=1-1=0
" BUT ".....the big mistake lies in part b).
Actually, (1+x-t)f(1+x-t) cannot be differntiated as I have done.I had taken 'x' to be a const. while differentiating...which is WRONG!!!
Again,sorry for the trouble!!!It's my mistake!!