Question

given :
f'(x+1) / f'(x+2) = 1 + ( f(x+1)/f(x+2) )
f(1)=0
and f(2)=1,
then ( f(x) / f(x+1) )= ?
a)log f(x+1)
b)log f(x)
c)(e^f(x+1)) - e
d)(e^f(x)) - 1

Answer 1

bookmark

Answer 2

@mukushla @ganeshie8 @shubhamsrg

Answer 3

got it maybe..
is it log f(x+1) ??

Answer 4

yes that correct !! how'd you get that ?

Answer 5

@UnkleRhaukus

Answer 6

@Callisto @campbell_st @hartnn

Answer 7

i just got that it cant be (b) as per hit and trial i.e. substituting x= 1 there...but x=1 matches with rest 3 options..
hmm

Answer 8

Look, this is quite reasonable to solve equation:\[\int\limits \frac{f'(x)}{f(x)} = \ln f(x)\] while \[\int\limits \frac{f'(x+1)}{f(x+1)} = \ln f(x+1)\] etc.

Answer 9

Now - to get "there" just MULTIPLY BOTH SIDES by f'(x+2) and DIVIDE BOTH SIDES by
f(x+1). Then you get to shape of formulas integrable by the above method

Answer 10

@Aryang ? You asked this , did u not?

Answer 11

@Mikael what about right side??
u get f(x+1)/ f'(x+1) for which there is no standard formula.
and what about f'(x+2)/f(x+1) ?

Answer 12

Count your "horses" once again - this is NOT the correct right side AFTER THE OPERATIONS I DESCRIBED

Answer 13

well,, how do we go about integrating f'(x+2)/f(x+1) ?

Answer 14

@amistre64 @satellite73

Answer 15

@phi

Answer 16

@shubhamsrg @shubhamsrg @shubhamsrg
@shubham.bagrecha

Answer 17

arey koi to help karo !!
i mean
please help anybody ?

Answer 18

atleast you guys bookmark it,,so that you may be in the discussion..

Answer 19

someone plz post complete solution

Mikael or Shubhamsrg

Answer 20

sorry couldnt come online whole day yesterday,,
here's my solution :

Answer 21

consider this :

d/dx ( f(x+1) / f(x+2) ) = [f'(x+1) f(x+2) - f'(x+2) f(x+1) ]/ (f(x+2) )^2
=f'(x+2) [ f'(x+1)/f'(x+2) . f(x+2) - f(x+1) ] / (f(x+2) )^2
=f'(x+2) [( 1 + ( f(x+1)/f(x+2) )) . f(x+2) - f(x+1) ] /(f(x+2))^2
=f'(x+2) [ f(x+2) + f(x+1) -f(x+1) ] / (f(x+2) )^2
d/dx ( f(x+1) / f(x+2)) = f'(x+2)/f(x+2)
=> d( f(x+1) / f(x+2) )= f'(x+2)/f(x+2) dx
integrating both sides, we have
( f(x+1) / f(x+2) ) = ln (f(x+2)) + C
for x=0, C=0
=>( f(x+1) / f(x+2) ) = ln (f(x+2))

if we substitute x-1 in place of x, we get
( f(x) / f(x+1) ) = ln(f(x+1))

hope i didnt make any fallacy somewhere!!

Answer 22

excellent @shubhamsrg !!thank you very much!!
that was really genius of you!!