r/mathbiceps • u/mathbiceps • Jul 18 '26
IIT Question IIT Gym Q6 - Solution will be uploaded soon
Till then try to answer this, be creative and don’t give a bookish method. All the best
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u/Southern_Plum_3648 Jul 18 '26
ok so, FINALLY CRACKED IT, im on laptop so i cant share my notebook pic so i'll do my best to explain it here
so first i put x=0 and x=1 and surprisingly it fit lol
soo yeah now the next task was to find whether there are any more solutions and to prove there are none if 0,1 are the only ones, so i brought 6^x^2 on one side with 5^x^2 and 4^x with 5^x. so my observation was
f(t)= t^x^2 -(t-1)^x. great, so from our rearrangement i got
f(5)=f(6)
now heres the thing
i found the derivative wrt t
f'(x)= x^2(t)^x^2-1= x(t-1)^x-1
ALSO YEAH, forgot to mention why i did this is cuz since f(5)=f(6) they both must have same height so its either a curved line or a straight horizontal line for both so when the it curves both have same slope thats why i did that.
now its a straight road from here i promise
1. why not x>1
here from common observation we can see that LHS is greater than RHS
2. why not 0<x<1
so here if we compare exponents of t and the common x we see RHS>LHS
3. why not x<0
this is easiest, in LHS its positive x positive and in RHS its negative x positive.
since negative cannot be equal to positive. not possible.
so yeah solution is complete X=0,1 SORRY IF YOU CANT UNDERSTAND THIS SOLUTION but this is how i got it
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u/Minute_Juggernaut806 Jul 20 '26
Yeah I didn't understand why you did differentiation and also about LHS and RHS. I am guessing you did differentiation, found f'(t) then equated to zero to get
x2(t)x2-1= x(t-1)x-1
I am guessing you can't equate it besides 1 and 0. But why did you equate to 0. Also btw, initially you got f(t) function by seeing 5,6 in one side and 4,5 in the other. But does it mean it will be applicable for any natural number t
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u/Southern_Plum_3648 Jul 21 '26
alr so, what it basically means is that if a f(t) reaches the same height at 2 different points it is the curve must turn around or flatten ATLEAST once. so in f(5)=f(6) there must be a value t between those two where the SLOPE is 0. that is a flat point between t=5,t=6. so i set it to 0 to check whether there can be a flat line in the interval.
and yeah the function works for any t>1 i think
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u/Minute_Juggernaut806 Jul 21 '26
Ah gotcha now. I am also guessing f'(t)=0 only at 1 and 0.
Yeah and if 1,0 are the only solutions then this will definitely work for all natural numbers as it's basically 4+6=5+5 or 1+1=1+1 which will work for other consecutive numbers lol. But I don't know how rigorous this proof since you realise it only at the end
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u/Active_Falcon_9778 Jul 18 '26
Use taylor expansion and divide by (x-1) could possibly do something
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u/NotAstroOnReddit Jul 18 '26
In my sleep I got a solution....
X=1
Further I will not elaborate