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8 can be done much faster by factoring out the x. 🙂

Right before the TMUA, thank yiu

watching a day to my tmua

Really do love your little ramblings between explaining things, makes it so much easier to keep focused rather than having someone continuously reciting maths/logic. Thanks a lot for the videos, they're very helpful for the TMUA 🙂

Question is quite funny considering it’s literally just like wordle. Anyone whos played the game would see the strategy immediately, i doubt a question like that would show up again

for q9 I thought the derivative of 1/x was ln(x)

oh man i was so gassed when i thought i got 12pi correct for the last one 😭

for q1, if you drew the circles out, you would've noticed that the line going through the intersections actually passes through the midpoint of the line connecting the centre of the circles and therefore you can find the gradient and midpoint of the line connecting both circles therefore find the equation for the other line.

Alternative approach to question 20: In the final expression, (a-3d)/2a = cos X, you can write the whole thing as (1/2) - (3d/2a). Since 3d>2a, 3d/2a will be greater than 1 if you divide both sides by 2a.( we know that a and d >0, so no problem with that). Thus, (1/2) - (3d/2a) can be written as (0.5)-(something bigger than 1), so cos(X)<0. X can't take any values from the first quadrant, and the only choice guarentees this condition is choice E.

Alternative approach to question 19: The x coordinate of the vertex is always equal to the artihmetic mean of the two roots(also meaning that the x coordinate of the vertex lies between the two roots) . So (p+q)/2 = (-b)/2 and we get p+q = -b (which is actually the result of the root sum formula, -b/a). And if we name the roots of the other inequality as a and b, we know that (a+b)/2= (-bc)/2. So (a+b)= -bc and we know that -b= (p+q). Finally we get a+b= cp + cq, so one of the roots is cp and the other one is cq. Since c<0 and q>p, cp is greater than cq. Thus, the answer is qc<x<pc.

For Q14, it can't be B because if the highest polynomial is ax^5, the limits have to be one tending to +infinity and the other the contrary, thus both limits can't go to -infinity which is what one local minimun and two local maxima suggests

i don’t understand question 14 😔

Solid vid mate

This is the 2016 paper

For 3, cant you also have x=4? if you have committees A,B,C,D and people x,y,z,e then A = x, z; B= x,d; C: y, d; D: y, z meets all criteria

Thank you so much for the walk-through!! super useful for students like me self-studying the mech module 2.

Would people say MAT is good for revising for TMUA (was it close to the difficulty of 2024?) I am sitting in 2 days

Well it's kinda obvious.

For Q15, is B not equivalent to D? Or is it because it’s a square root it can’t accept negative values or something like that

i dont get the bit for question 20 where u replaced 3d with 2a. my exams on thursday so any help will be appreciated

where did that a disappear to in q15

couldnt you make contrapositives for 20:14 i found that easier and still got C

In Q10 you could square both sides of each equation to get rid of modulus and find the largest solutions from both (ignoring the sign) and multiply both to get 48

For statement 1 in question 8, ac<ab<a^2, if you divide by a (a is not 0 so you can) you get c>b>a (signs change because a<0), which we are given.

Loved q5 reminds me of u-sub integration

Why can’t root A be negative? Surely if you root a number you get a positive and negative solution

For Q11 as it does not specify a minimum of terms in the series, you can just choose 2+18=20 as a series and the difference would be uneven.

for question 14 i think it should be F but should also include pi as well 11:32

for q14 surely the COMPLETE set of values should include x=pi????

Thanks G

Just as a note, q20 is really easy if you think about contrapositives, if A then B is the same as if not B then not A, so that only leaves you with C.

For q17 you know there are 2 solutions for b however 4c-5 > 0 from the second log equation, meaning c>5/4. This means b^3 >1/4 (using the third equation). the first root of b is less than 0.5, hence it's cube cannot be greater then 1/4 so surely that solution of b should be ignored and A is the correct answer?

the goat

For questions 10 literally just sub a positve function for all x

for q13, if x=-y , we have -y-(-y)y+y= -y+y^2+y, which is also bigger than 0 right ?

16 is flawed. If b can be 0 (which seems to be allowed), then a can be any real number. If b cannot be 0, then your method is correct aside from one scenario where a = 0, then b must be 0 so that option should be 0 < a <= 4/3 if b is not allowed to be 0

What would 13/20 be on a paper like this? roughly

For the last question, I applied the cosine theorem by substituting x as the lenght of PQ and get a quadratic. Then I said to myself '' If this quadratic has only one solution (meaning that the discriminant is equal to 0) then the value of p will uniquely determine PQ. And I got 4p^2-4=0 and thus p=1.

surely question 12 is a badly worded question, it is ambiguous

for the 2021 q8. Where it is said that p(a)=p(b) is sufficiant for (*). Why can you not consider the polynomial as a cubic. And therefore the point c may lie at a different turning point, greater b or less than c??

1/12 is intuitive because hexagons.

can q1 not also come up in a normal CP2 paper?

for q36 isnt condition 2 only necessary but not sufficent since a and c could equal 0

Absolutely incredible loop here

13:57 do you mind explaining whats wrong with tech?? its literally the future

Could you go through UKMT IMC 2013- 2010 please sir.

how is question 6 correct??

GOAT of frequent uploads recently

Great stuff as always

That's actually pretty smart I think I would have solved it too but it would probably take me while