Hi Mr Drew, Thanks for the video. For question 15 (at 36:42), surely if we differentiate the equation, we find that we have 2 stationary points at x=0 and x=-2q. If we sub x=0 into the equation, y=2, so would the graph look slightly different to the one you have drawn? And then after we reflect the graph in the y-axis, we can say that when translated 1 unit down, the graph's second stationary point will be (0,1) so will have 3 real roots? Thank you
Yeah nicely spotted, so the turning point is actually on the y-axis. As you say, it doesn't make a difference to the answer, as a shift down of 1 from 2 is not enough to loose the roots, and we are still guaranteed to have 3
For question 17 (43:00) you said that 16 = 4^2 + 0^2 to prove every square number is a squaresum as it can written as the sum of two squares of two integers. But you forgot to explain the part about how the awkward prime factors of this squaresum should also result to an even power because if Im not mistaken you have to take into account the first statement AND the last statement in order for it to be a true squaresum (if and only if applied on both statements?) I know this may be a stupid question but how is 16 a square sum if it lacks an even amount of awkward prime factors altogether? I tried thinking through this and I may be wrong but is it because there exists an awkward prime factor for 16 (which let's call X); however, this awkward prime factor is X^0 where 0 is still of even order; therefore, the second statement is fulfilled for 16? Thanks for the help Mr Drew
in question 10 how can you be so sure that those condition are necessary for graph to have reflection symmetry in the line x =a when you have only tested out on 1 graph. is there any tips you got on doing necessary/sufficient and statement is true type of questions? i find myself struggling the most at those. Thank you.`
Q11 can be done much easier. Just choose m = 0 then your line becomes horizontal. Then choose c = -5 thus III is not true. c = 10 disproves II and c = 1/2 disproves I. Question done in no time!
Yeah if there's one thing I'll never learn, it's how to tell apart convex vs concave. I don't know what is wrong with me. I noticed this a while back but didn't think it was worth redoing the video for
@@rtwodrew2 ah okay no problem!! :) The way I find easiest to tell them apart is that "ConCAVE" graphs are shaped like the "opening of a cave", and "ConvEX" graphs are shaped like exponential curves (e.g. e^x) Thanks for the videos, really appreciate your time
I had a question, for question 16 isn’t the sequence with 3,2,1,0,-1 etc invalid as the original question mentions that of the first n terms the median will be an integer which isn’t the case for this sequence as if n is an even number the median is not an integer. Could you please clear this up
I had a question of how the tmua score is calculated. What raw score out of 40 would you need for a 7.0? And are the two papers weighted any differently or is it 50/50?
Hello Drew, just getting in the last touches before this Tuesday's exam. Thank you so much for the video. I have a question about No.10 (your favourite). I understand how what you showed proves that in order for it to be symmetrical, "I and II" are necessarily true but I don't see how they are sufficient. For example, the LHS (about a) being more steep than the RHS but equal height at those particular points. Hope you can help me please. Thank you!
I didn't do a good job on that question, these older walkthroughs will get remade before next year. All I really showed there was that conditions I and II are necessary, since if you have a symmetrical graph they must be the case, using the symmetry that you've already assumed. Looking back I didn't show sufficiency at all, which is to say, you need to show that the condition itself and alone, means you get a symmetrical graph. I is kinda straightforward. Pick a point a on the x axis, and the function is apparently the same when you move some distance x to the left or x to the right (as f(a-x)=f(a+x)). This is basically a definition of symmetry around a, and so the condition alone is enough to conclude symmetry, so it is sufficient. Condition II is harder to spot I think, f(2a-x)=f(x). Pick point a again, and then move x-a to the left, (this is the same doing a - (x-a) = 2a-x.) Then move x-a to the right, (a+(x-a)=x). But you have that f(2a-x)=f(x) and so again you are defining symmetry around a. So again sufficient
It's interesting I have a colleague who can do that kind of question absolutely instantly. There are some niche maths courses that rely heavily on that kind of logic being used, goes without saying that I never did any of them. It's probably ignorable, of the 7 paper 2s published only 3 of them have had a single question like it.
These questions become really easy when you convert 'for all' to ∀, convert there exists to '∃', and any remaining logic to f(N). Then, to find negation you NOT everything: every ∀ turns to ∃ (and vice versa), and simply negate f(N).
Hi Mr Drew,
Thanks for the video. For question 15 (at 36:42), surely if we differentiate the equation, we find that we have 2 stationary points at x=0 and x=-2q. If we sub x=0 into the equation, y=2, so would the graph look slightly different to the one you have drawn? And then after we reflect the graph in the y-axis, we can say that when translated 1 unit down, the graph's second stationary point will be (0,1) so will have 3 real roots? Thank you
Yeah nicely spotted, so the turning point is actually on the y-axis. As you say, it doesn't make a difference to the answer, as a shift down of 1 from 2 is not enough to loose the roots, and we are still guaranteed to have 3
@@rtwodrew2 Great thanks! Yep exactly - answer is still exactly the same. Thanks again
You're actually funny lol
For Q14, surely If sin x = 1/2 thus x is 150, then the triangle doesn’t exist because 150 + 30 equals 180, so we are missing an angle?
27:43 "THis is more writing than i did in gcse english" LMAO
for q14 why is 1/p
For question 17 (43:00) you said that 16 = 4^2 + 0^2 to prove every square number is a squaresum as it can written as the sum of two squares of two integers. But you forgot to explain the part about how the awkward prime factors of this squaresum should also result to an even power because if Im not mistaken you have to take into account the first statement AND the last statement in order for it to be a true squaresum (if and only if applied on both statements?)
I know this may be a stupid question but how is 16 a square sum if it lacks an even amount of awkward prime factors altogether?
I tried thinking through this and I may be wrong but is it because there exists an awkward prime factor for 16 (which let's call X); however, this awkward prime factor is X^0 where 0 is still of even order; therefore, the second statement is fulfilled for 16?
Thanks for the help Mr Drew
in question 10 how can you be so sure that those condition are necessary for graph to have reflection symmetry in the line x =a when you have only tested out on 1 graph. is there any tips you got on doing necessary/sufficient and statement is true type of questions? i find myself struggling the most at those. Thank you.`
Q11 can be done much easier. Just choose m = 0 then your line becomes horizontal. Then choose c = -5 thus III is not true. c = 10 disproves II and c = 1/2 disproves I. Question done in no time!
Also, quick question at 48:20, isn't that a concave graph? (as it looks like the opening of a cave?) Thank you
Yeah if there's one thing I'll never learn, it's how to tell apart convex vs concave. I don't know what is wrong with me. I noticed this a while back but didn't think it was worth redoing the video for
@@rtwodrew2 ah okay no problem!! :)
The way I find easiest to tell them apart is that "ConCAVE" graphs are shaped like the "opening of a cave", and "ConvEX" graphs are shaped like exponential curves (e.g. e^x)
Thanks for the videos, really appreciate your time
@@yashshah2941 The EX suffix might actually help me, I've never noticed that before. Only time will tell
Wow these explanations are so helpful, thanks a lot man
Is the video you mention at 32:10 still up? I can't seem to find it.
Sorry it was on the old channel. Reuploaded now ua-cam.com/video/nHEPC_19tL8/v-deo.html
@@rtwodrew2 Thank you very much!!
Please explain why is the square root of anything defined to be zero? and why does this mean that p must be zero thanks
since the square root of nothing will give you a negative number, the least value you can obtain from a square root is 0
I got a question on question 11, surely the third option is true as if c
I had a question, for question 16 isn’t the sequence with 3,2,1,0,-1 etc invalid as the original question mentions that of the first n terms the median will be an integer which isn’t the case for this sequence as if n is an even number the median is not an integer. Could you please clear this up
I had a question of how the tmua score is calculated. What raw score out of 40 would you need for a 7.0? And are the two papers weighted any differently or is it 50/50?
it depends on the year but 7.0 is usually 26-28 out of 40. i believe it is 50/50 as well
50/50
q18: u said the graph needs to be concave but isn't concave when second derivative is negative?
ig the graph needs to be convex, sir has made a mistake stating it to be concave, answer will be C
solving the 10th question is much easier when you use transformation of graphs
27:37 good one
Hello Drew, just getting in the last touches before this Tuesday's exam. Thank you so much for the video. I have a question about No.10 (your favourite). I understand how what you showed proves that in order for it to be symmetrical, "I and II" are necessarily true but I don't see how they are sufficient. For example, the LHS (about a) being more steep than the RHS but equal height at those particular points. Hope you can help me please. Thank you!
I didn't do a good job on that question, these older walkthroughs will get remade before next year. All I really showed there was that conditions I and II are necessary, since if you have a symmetrical graph they must be the case, using the symmetry that you've already assumed. Looking back I didn't show sufficiency at all, which is to say, you need to show that the condition itself and alone, means you get a symmetrical graph. I is kinda straightforward. Pick a point a on the x axis, and the function is apparently the same when you move some distance x to the left or x to the right (as f(a-x)=f(a+x)). This is basically a definition of symmetry around a, and so the condition alone is enough to conclude symmetry, so it is sufficient.
Condition II is harder to spot I think, f(2a-x)=f(x). Pick point a again, and then move
x-a to the left, (this is the same doing a - (x-a) = 2a-x.) Then move x-a to the right, (a+(x-a)=x). But you have that f(2a-x)=f(x) and so again you are defining symmetry around a. So again sufficient
@@rtwodrew2 Awesome! that reply cleared up a lot. Thank you very much :)
@@rtwodrew2 Hi, did you ever remake this video? Thanks for everything!
Q3, Q7, Q10, Q14, Q20
i came for r2dre2
Question 12 actually so stupid
It's interesting I have a colleague who can do that kind of question absolutely instantly. There are some niche maths courses that rely heavily on that kind of logic being used, goes without saying that I never did any of them. It's probably ignorable, of the 7 paper 2s published only 3 of them have had a single question like it.
Not really. It's a pretty simple test of elementary logic.
These questions become really easy when you convert 'for all' to ∀, convert there exists to '∃', and any remaining logic to f(N). Then, to find negation you NOT everything: every ∀ turns to ∃ (and vice versa), and simply negate f(N).