Thank you very much for uploading these useful videos I noticed that there's a small error on question 16 : option D >>> f(3) = ( 81 + 54 + 9 ) / 4 = 144 / 4 not 162/4.
1/2 = 2^-1 Useful logarithm rule to know: log_a^b(x) = 1/b log_a(x) So actually the log is -log_2(x) and now you can see where the negatives come from.
Could I ask which one of the TMUA past paper you think is the hardest? I want to save it before the exam. Thanks for your videos you're an absolute hero.
13:40 Question 10 I also came to the same conclusion that statement III was incorrect however I thought it was because a = c - b so a^2 should equal c^2 -2bc + b^2, which is not the same equation for a^2 stated in the proof. Is my thinking here valid?
It's just a very lazy bound. pi/3 is somewhere between 1 and 2 (I know much closer to 1 but I don't care) and so 2^pi/3 is between 2^1 and 2^2 and so between 2 and 4. I'm allowed to be that lazy because one of the options it turns out was fairly obviously the smallest even after finding some very lazy bounds
@@teddakkha5612 Yeah I think noticing that is fine too as a way of spotting something may have gone wrong. The key point though is that it is very hard to factorise and compare and assign values to brackets unless 1 side of the equation is 0, this has come up in TMUA a few times
at 37:03, where did the 8 and 4 come from and how do you determine from the inequality with log_1/2 8 is not greater than log_1/2 4 that the given inequality in the question is false? Also thanks a lot for your videos. They're really of help to me 😃
Hey, really appreciate these videos. The solutions for question 16 is incorrect however as (81+54+9)/4 144/4 = 36 not 40.5. My method was to differentiate C & D and spot that C becomes (4x^2+3x^2+2x+1)/2… which would not be an integer for x = 0. Assuming the first case was correct for C that means C is the right answer. However this isn’t a complete solution as the first condition may not be met for some value of x.
Yeah thanks, an error found there, fluked the answer. Going through it again, the correct (hopefully!) solution might look as follows: C is either (odd+odd+odd+odd)/2 or (even+even+even+even)/2, in either case an integer. Your counter example for C` works well to show that it is the counterexample and therefore the answer. You can actually stop there since you can only select one option. For completion, you can show that D is always an integer if the input is, by factorising it to x^2(x+1)^2/4, now if x is even then x^2 provides at least the two 2's necessary to always make this an integer and if x is odd than the x+1 does. Now D` can eventually be simplified and factored into x(x+0.5)(x+1). Now this initially looks like it won't always be an integer for integer x, because of the +0.5. But one of x or x+1 is always even, and so it multiplies out that middle bracket, which is always something.5, and makes an integer. The remaining bracket is of course an integer too. So D is not the counterexample Thanks again for pointing this out
@@MrWorzel86 oh nice! Would you like to study for it together? I can't find many resources online so it would be helpful to discuss specific questions of the papers we've been already done
solving the last one like WORDLE really helped me lol
Thank you very much for uploading these useful videos
I noticed that there's a small error on question 16 : option D >>> f(3) = ( 81 + 54 + 9 ) / 4 = 144 / 4 not 162/4.
I don't get your reasoning behind questions 18 , 1/2 to the power of 2 gives 1/4 so no negatives are involved. Could you please explain this to me?
1/2 = 2^-1
Useful logarithm rule to know: log_a^b(x) = 1/b log_a(x)
So actually the log is -log_2(x) and now you can see where the negatives come from.
Would you happen to have a video for the sequence technique used in question 15? Please link the video if you do have it, thanks a lot!
isnt q17 just asking you to negate the statement it gives you
Yeah that’s the simpler way to do it.
Could I ask which one of the TMUA past paper you think is the hardest? I want to save it before the exam. Thanks for your videos you're an absolute hero.
I'm not really sure, the Spec is definitely the easiest. Aside from, for me, they're not different enough to be noticeable
@@rtwodrew2 I see. I'll probably do it in order then. Thanks.
for q17, 3 divides 6 and 9
10:45 Question 8
You said 2^(pi/3) is approx 2. I am confused where 2^2 and the bound 4 came from?
Finding these videos really helpful :)
13:40 Question 10
I also came to the same conclusion that statement III was incorrect however I thought it was because a = c - b so a^2 should equal c^2 -2bc + b^2, which is not the same equation for a^2 stated in the proof. Is my thinking here valid?
It's just a very lazy bound. pi/3 is somewhere between 1 and 2 (I know much closer to 1 but I don't care) and so 2^pi/3 is between 2^1 and 2^2 and so between 2 and 4. I'm allowed to be that lazy because one of the options it turns out was fairly obviously the smallest even after finding some very lazy bounds
@@teddakkha5612 Yeah I think noticing that is fine too as a way of spotting something may have gone wrong. The key point though is that it is very hard to factorise and compare and assign values to brackets unless 1 side of the equation is 0, this has come up in TMUA a few times
at 37:03, where did the 8 and 4 come from and how do you determine from the inequality with log_1/2 8 is not greater than log_1/2 4 that the given inequality in the question is false?
Also thanks a lot for your videos. They're really of help to me 😃
1) They didn't come from anywhere, I made them up because I know 8
Hey, really appreciate these videos.
The solutions for question 16 is incorrect however as (81+54+9)/4 144/4 = 36 not 40.5.
My method was to differentiate C & D and spot that C becomes (4x^2+3x^2+2x+1)/2… which would not be an integer for x = 0. Assuming the first case was correct for C that means C is the right answer. However this isn’t a complete solution as the first condition may not be met for some value of x.
Yeah thanks, an error found there, fluked the answer. Going through it again, the correct (hopefully!) solution might look as follows:
C is either (odd+odd+odd+odd)/2 or (even+even+even+even)/2, in either case an integer. Your counter example for C` works well to show that it is the counterexample and therefore the answer. You can actually stop there since you can only select one option.
For completion, you can show that D is always an integer if the input is, by factorising it to x^2(x+1)^2/4, now if x is even then x^2 provides at least the two 2's necessary to always make this an integer and if x is odd than the x+1 does.
Now D` can eventually be simplified and factored into x(x+0.5)(x+1). Now this initially looks like it won't always be an integer for integer x, because of the +0.5. But one of x or x+1 is always even, and so it multiplies out that middle bracket, which is always something.5, and makes an integer. The remaining bracket is of course an integer too. So D is not the counterexample
Thanks again for pointing this out
Hey! Are you also doing the exam on oct 18?
@@aamnaz1474 Hey! Yeah I'm also sitting it then.
@@MrWorzel86 oh nice! Would you like to study for it together? I can't find many resources online so it would be helpful to discuss specific questions of the papers we've been already done
@@aamnaz1474 my comments keep getting deleted is there a way you can contact me?
Q10 (lucky guess), Q11, Q17
Could you please explain question 13 to me? I don't understand why c has to be greater than or equal to 1
Because of the definition of standard form: c x 10^n where 1
@@rtwodrew2 Right. Thank you so much
legend
Giovani Skyway