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.
@kausarlolz
2 жыл бұрын
solving the last one like WORDLE really helped me lol
@misto.13
Жыл бұрын
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?
@pawelpow
Ай бұрын
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.
@LiamLI-yh5dp
3 ай бұрын
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!
@roshanxd4499
17 күн бұрын
isnt q17 just asking you to negate the statement it gives you
@Everything-mh4bp
5 күн бұрын
Yeah that’s the simpler way to do it.
@shepherd3974
2 жыл бұрын
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.
@rtwodrew2
2 жыл бұрын
I'm not really sure, the Spec is definitely the easiest. Aside from, for me, they're not different enough to be noticeable
@shepherd3974
2 жыл бұрын
@@rtwodrew2 I see. I'll probably do it in order then. Thanks.
@teddakkha5612
2 жыл бұрын
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 :)
@teddakkha5612
2 жыл бұрын
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?
@rtwodrew2
2 жыл бұрын
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
@rtwodrew2
2 жыл бұрын
@@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
@kene4906
2 жыл бұрын
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 😃
@rtwodrew2
2 жыл бұрын
1) They didn't come from anywhere, I made them up because I know 8
@jagzey
2 ай бұрын
for q17, 3 divides 6 and 9
@MrWorzel86
2 жыл бұрын
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.
@rtwodrew2
2 жыл бұрын
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
@aamnaz1474
2 жыл бұрын
Hey! Are you also doing the exam on oct 18?
@MrWorzel86
2 жыл бұрын
@@aamnaz1474 Hey! Yeah I'm also sitting it then.
@aamnaz1474
2 жыл бұрын
@@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
@MrWorzel86
2 жыл бұрын
@@aamnaz1474 my comments keep getting deleted is there a way you can contact me?
@user-ky5ev6sk1q
Жыл бұрын
Q10 (lucky guess), Q11, Q17
@wcook5806
2 жыл бұрын
legend
@aamnaz1474
2 жыл бұрын
Could you please explain question 13 to me? I don't understand why c has to be greater than or equal to 1
@rtwodrew2
2 жыл бұрын
Because of the definition of standard form: c x 10^n where 1
Пікірлер: 33