For question 12 rather than solving for the x value of the differential equation then plugging that back into the original equation, you can complete the square to get (x²-p²/2)² - (p^4)/4 so - (p^4)/4 must equal -9
For 17 I felt this was quite a simple approach: x^2+y^2=arcsin(1/2) => x^2+y^2=((pi/6) + 2kpi) OR ((5pi/6 + 2kpi)) => since radius is the square root of these two results, we can plot the graph of y=root(x), when the input values are equally spaced, the output values get closer and closer together, this eliminates answers B,C,D,E, since none of them have radii that get closer and closer together, therefore answer is A.
good trick for q 18 is to just consider the complex plane instead of attempting it visually. when we multiply by a complex number we rotate it so our turning point is 2+3i since rotation by complex numbers is about the origin we subtract P from our turning point and multiply by e^i3pi/2 ( = -i) so we get (2+3i - (-2+3i) ) * -i = -4i now we have to add back P to this point since we had translated it earlier so we get -2+3i -4i = -2 - i in the cartesian plane this is an x coordinate of -2 and y coordinate of -1 so therefore our new turning point is (-2,-1) and we can form our quadratic equation from here and gt B as our answer you can do this by using the rotation matrix to rotate the turning point but the translation part will still be needed
I have a much quicker method for Q14. note that for cos(pi*x) whenever x is an integer then cos(pi*x)=1 or -1. substitute in 1 you get a quadratic then apply the discriminant and then there will be 2 real roots hence 2 real solutions. sub in -1 and you get a quadratic with discriminant 0 hence only 1 solution. 2+1=3
@revilo6381tbh I did really badly and alot worse than I expected - got exam panic and my brain just couldn't do maths. For me it didn't matter too much, only one of my unis required it and it was still definitely worth doing because the preparation has helped me so much with the normal a level. It's a tough test, aslong as you're preparing as well as you reasonably can try not to get too nervous or beat yourself up about it.
for question 19, why does x have to be bigger than 90? if the first three solutions of theta are 0, 60, 90, then if x is 90, these solutions would all be within this range?
@@armaanbhartia9818Yup, lucky old me got 2 values and just used 1/2 coz i had no idea what common ratio to pick because silly me didn't read that it was a convergent series!
for question 13, i am confused because you said the integral sum could be written as 0 to 1 plus 1 to 2 plus 2 to 3 etc, but surely i should be 0 to 1 then 0 to 2 then 0 to 3 then 0 to 4 etc?
Think of the integral as the area under the curve, so the area from 0 to 3 for example, can be split into 3 separate areas being: 0 to 1 plus 1 to 2 plus 2 to 3
considering pi is 3.14... when you multiply smth>1 by x it will make the graph reduce in the x axis by that number, so when you divide pi by pi, it will factor it down to one and so on
nice video, one question on Q14, how should we hand-draw these graphs? eg. 'overalying' the left hand side graph with the right hand side, that's going to be very inaccurate surely? I guess same as always though, practice?
also for question 12 its faster to treat it as a hidden quadratic and then complete the square to find the minimum, equate it to find p^2 and then square root
I got 12 pie and was so happy that I got q20 right, then watched this and went back to depression
sameeee
FAX
For question 12 rather than solving for the x value of the differential equation then plugging that back into the original equation, you can complete the square to get (x²-p²/2)² - (p^4)/4 so - (p^4)/4 must equal -9
Thank you so much for all the TMUA content! I've got 2 weeks until my exam and these videos have helped so much over the past couple months.
For 17 I felt this was quite a simple approach: x^2+y^2=arcsin(1/2) => x^2+y^2=((pi/6) + 2kpi) OR ((5pi/6 + 2kpi)) => since radius is the square root of these two results, we can plot the graph of y=root(x), when the input values are equally spaced, the output values get closer and closer together, this eliminates answers B,C,D,E, since none of them have radii that get closer and closer together, therefore answer is A.
Thank you for this video! I'm practising for the upcoming TMUA and this has been very helpful!
Good luck, well done for preparing early
good trick for q 18 is to just consider the complex plane instead of attempting it visually.
when we multiply by a complex number we rotate it so our turning point is 2+3i
since rotation by complex numbers is about the origin we subtract P from our turning point and multiply by e^i3pi/2 ( = -i)
so we get
(2+3i - (-2+3i) ) * -i
= -4i
now we have to add back P to this point since we had translated it earlier
so we get -2+3i -4i
= -2 - i
in the cartesian plane this is an x coordinate of -2 and y coordinate of -1 so therefore our new turning point is (-2,-1) and we can form our quadratic equation from here and gt B as our answer
you can do this by using the rotation matrix to rotate the turning point but the translation part will still be needed
That’s acc smart icl
Thanks a million!!!! Treasure UA-camr!!!!
I have a much quicker method for Q14. note that for cos(pi*x) whenever x is an integer then cos(pi*x)=1 or -1. substitute in 1 you get a quadratic then apply the discriminant and then there will be 2 real roots hence 2 real solutions. sub in -1 and you get a quadratic with discriminant 0 hence only 1 solution. 2+1=3
for question 10, isnt it just easier to use the trapezium rule formula? the simplifying was much simpler in the end
i also used trapezium rule formula, but u learn it in y2 maths so maybe thats why he didnt use it here
I just did this and was absolutely certain I got 13/14 right and got 9😭😭 got the actual tmua in a week am I cooked
same
@@todismellsbenefits175 What are you applying for?
we will be fine surely!….. :-)
how did you do? nervous for my January one!
@revilo6381tbh I did really badly and alot worse than I expected - got exam panic and my brain just couldn't do maths. For me it didn't matter too much, only one of my unis required it and it was still definitely worth doing because the preparation has helped me so much with the normal a level. It's a tough test, aslong as you're preparing as well as you reasonably can try not to get too nervous or beat yourself up about it.
for question 19, why does x have to be bigger than 90? if the first three solutions of theta are 0, 60, 90, then if x is 90, these solutions would all be within this range?
How do you know that the sum of infinity is only viable for -1
it says its a convergent GP that basically means that s
@@armaanbhartia9818Yup, lucky old me got 2 values and just used 1/2 coz i had no idea what common ratio to pick because silly me didn't read that it was a convergent series!
for question 13, i am confused because you said the integral sum could be written as 0 to 1 plus 1 to 2 plus 2 to 3 etc, but surely i should be 0 to 1 then 0 to 2 then 0 to 3 then 0 to 4 etc?
Think of the integral as the area under the curve, so the area from 0 to 3 for example, can be split into 3 separate areas being: 0 to 1 plus 1 to 2 plus 2 to 3
for q14, how to draw the graph of cos (pi x)?
considering pi is 3.14... when you multiply smth>1 by x it will make the graph reduce in the x axis by that number, so when you divide pi by pi, it will factor it down to one and so on
nice video, one question
on Q14, how should we hand-draw these graphs?
eg. 'overalying' the left hand side graph with the right hand side, that's going to be very inaccurate surely?
I guess same as always though, practice?
It doesn't need to be accurate you're just counting intersections, not saying where they are.
And yes on the last bit
that last question is straight evil
also for question 12 its faster to treat it as a hidden quadratic and then complete the square to find the minimum, equate it to find p^2 and then square root
I need a little better explanation for q12 and 10 pls as i find it hard to understand
Q7, Q9, Q11, Q12, Q14, Q15, Q17, Q18, Q20
In R2Drew2 we trust