For question 7 i subbed in x = i (root of x^2+1) and set equal to zero, this gave me -i-a+bi+c=0 so i knew b=1 and a=c, from this i was able to deduce 1,1,1 , 2,1,2 , 3,1,3 etc worked for a,b,c so 10 polynomials
on q13 option F is none of the above. can you explain how to rule out this option (with a quick method) please. just spotting the quadrants needed doesnt prove that C is the right option
Місяць тому+2
Note that the question is asking for the points on y=-x that are a distance of 1 from the origin, and it’s easy to show (or intuitively see) that such points must exist
Місяць тому+1
This comes from the fact that we need points on a perpendicular line (so the gradient is -1) that pass through the origin -> equation the centre must lie on is y=-x
Місяць тому+1
In fact, it’s easy to find the answer legit from here - the gradient is -1, so the horizontal displacement must have a magnitude of 1/sqrt 2 (draw an isosceles right triangle by taking the perpendicular from the point on the line to the x-axis)
For the first question my approach was to reverse the power rule yielding log((x^2-2x-3)^2), which showed that the function inside is positive for all real values of x, therefore I only worried about the values for which x^2-2x-3 = 1. Why is this incorrect because of course using 2log(x^2-2x-3) we can see that there are indeed negative values that could be inside the log?
Im an American student planning on doing the TMUA (Dont ask why) and what would you reccommend I learn before doing dedicated TMUA prep since the US Cirriculum is worse than the UK one and my mathmatical knowledge is currently not enough to learnt the TMUA content. Should I just learn A level maths or just look at the TMUA rubric and learn each topic individually?
The TMUA specification is the same as the AS level maths (first year of A level) specification plus A level sequences and series. So if you learn AS level maths (which you probably know a lot of anyway) plus sequences and series you should be good to go. The hard part of TMUA isn't the actual maths it's knowing how to apply your knowledge to solve the questions.
Thanks for the video - for q19 shouldn't we multiply our answer by 2? because there are two ways of adding letters, either starting with s/i then another letter, or starting with another letter then putting s/i in the second place? ( we also have the 4800 option )
thanks sir for the video. this paper felt so weird one q was really hard and the next was super easy. please can you add the video about combinations and factorials(q19)?
In question 18, If you take the arithmetic mean of the upper function f(x,y,z) Numerator. So just add the elements by opening all the brackets and apply arithmetic mean is greater than equal to geometric mean . Then somehow they all become x^3y^3z^3 on the GM side and by 1/6 as it's power it perfectly aligns with the given problem which gives answer as 6. I think it's a little bit fast than the method shown in the video.
@@gyu5164 Expanding out the brackets gives (x + y + z + xy + xz + yz) on the numerator. Now you can apply the AM-GM inequality (arithmetic mean >= geometric mean for a given sample of values) on these values to obtain the inequality: (x + y + z + xy + xz + yz)/6 >= (x^3 y^3 z^3)^(1/6) = sqrt(xyz) Rearrange this inequality to get (x + y + z + xy + xz + yz)/sqrt(xyz) >= 6, which tells us the given function is greater than or equal to 6 for any x, y, z; therefore the least value of the function is 6.
Thanks sir, please try and arrange a few more of these types of mock papers as TMUA resources are scarce and would help for the test next month.
For question 7 i subbed in x = i (root of x^2+1) and set equal to zero, this gave me -i-a+bi+c=0 so i knew b=1 and a=c, from this i was able to deduce 1,1,1 , 2,1,2 , 3,1,3 etc worked for a,b,c so 10 polynomials
Thanks for this - really helpful
on q13 option F is none of the above. can you explain how to rule out this option (with a quick method) please. just spotting the quadrants needed doesnt prove that C is the right option
Note that the question is asking for the points on y=-x that are a distance of 1 from the origin, and it’s easy to show (or intuitively see) that such points must exist
This comes from the fact that we need points on a perpendicular line (so the gradient is -1) that pass through the origin -> equation the centre must lie on is y=-x
In fact, it’s easy to find the answer legit from here - the gradient is -1, so the horizontal displacement must have a magnitude of 1/sqrt 2 (draw an isosceles right triangle by taking the perpendicular from the point on the line to the x-axis)
Thanks that's a good quick method
They took question 3 from question 21 of IC Aero/EEE Mathematics Aptitude test Sample Paper 1 by Imperial College London
good spot mate
@@seyanesiv7582 thanks mate make sure to like and subscribe for future content
For the first question my approach was to reverse the power rule yielding log((x^2-2x-3)^2), which showed that the function inside is positive for all real values of x, therefore I only worried about the values for which x^2-2x-3 = 1. Why is this incorrect because of course using 2log(x^2-2x-3) we can see that there are indeed negative values that could be inside the log?
Im an American student planning on doing the TMUA (Dont ask why) and what would you reccommend I learn before doing dedicated TMUA prep since the US Cirriculum is worse than the UK one and my mathmatical knowledge is currently not enough to learnt the TMUA content. Should I just learn A level maths or just look at the TMUA rubric and learn each topic individually?
The TMUA specification is the same as the AS level maths (first year of A level) specification plus A level sequences and series. So if you learn AS level maths (which you probably know a lot of anyway) plus sequences and series you should be good to go. The hard part of TMUA isn't the actual maths it's knowing how to apply your knowledge to solve the questions.
@@tomdavies9675Thanks alot!
Thanks for the video - for q19 shouldn't we multiply our answer by 2? because there are two ways of adding letters, either starting with s/i then another letter, or starting with another letter then putting s/i in the second place? ( we also have the 4800 option )
For Q10, couldn't you just do a u=1/2x +2 substitution instead?
I did it and still got the same answer without having to redo bounds 2x
Of course!
sir your goated you make some of it make sense, which is fucking class
i did tmua practice for 1 month but i am still stuck at 25/40, i am so depressed, what should i do, i am aiming for 33/40
Do MAT multiple choice and UKMT Senior. You might find questions that sometimes come up in future TMUA
pretty good paper
thanks sir for the video. this paper felt so weird one q was really hard and the next was super easy. please can you add the video about combinations and factorials(q19)?
Added
Amazing content!
For question 11, how can sinx = 0 be a solution as this would be 0 which does not fit in the range?
sinx is also 0 at pi
For question 5 there is an easier way
I posted a short on this
What is the easier way ?
@@Aryaa_SKfind the values where sin and cos is equal and sub it in don’t draw graphs sub in values
In question 18, If you take the arithmetic mean of the upper function f(x,y,z) Numerator. So just add the elements by opening all the brackets and apply arithmetic mean is greater than equal to geometric mean . Then somehow they all become x^3y^3z^3 on the GM side and by 1/6 as it's power it perfectly aligns with the given problem which gives answer as 6. I think it's a little bit fast than the method shown in the video.
could u elaborate further on that?
@@gyu5164 Expanding out the brackets gives (x + y + z + xy + xz + yz) on the numerator. Now you can apply the AM-GM inequality (arithmetic mean >= geometric mean for a given sample of values) on these values to obtain the inequality:
(x + y + z + xy + xz + yz)/6 >= (x^3 y^3 z^3)^(1/6) = sqrt(xyz)
Rearrange this inequality to get (x + y + z + xy + xz + yz)/sqrt(xyz) >= 6, which tells us the given function is greater than or equal to 6 for any x, y, z; therefore the least value of the function is 6.
@@aryaaskmaths thanks!!
whats a good score to get this paper, something equivalent to 7-8 tmua score?
yeah i wanna know that too
id say 14/15/16. something like that probably.
Thank you very much sir!
MOCK TUA
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