For Q18, once you know that B has 5 guaranteed solutions you already know that has the most possible solutions because f(x)=g(x) is just a degree 5 polynomial (after you get it all on one side) which has a maximum number of 5 solutions
16 ultra neat solution: you know x = +- cos theta. Turn them into +- sin theta by replacing every x^2 in the original equation to (1-x^2). Expand, get B in 20 seconds
For Q. 16 I believe you can just substitute Cos(#) into the x values, convert them all into 1-sin^2(#) and expand and simplify. Then if you let sin(#) = x, it cancels down to equation B and the solutions will be sin(#). I have no idea if this solution is mathematically correct but it gives the right answer so /shrug
For Q14 just to share: A nice reasoning why an equilateral triangle will have the largest area is because since PQ is a fixed line, we can treat that as the base of a triangle, since area = 0.5 * base * height, moving R will reduce the height, therefore the largest height will be when the perpendicular bisector of PQ goes through the height of the triangle.
for q14, we can use the fact that area of a triangle is base x perpendicular height I think. knowing that PQ is fixed, we can maximise the perpendicular height by making the line OR coincident with the perpendicular bisector of PQ. then we have 3 identical triangles with area 9sqrt3 then multiply that by 3 no?
Hi mr drew For question 9 I need to know if my attempt at it was a fluke or could be classified as a valid solution: for f(x) - g(x) = 2sinx I drew 2sinx and found the least value of this graph to be -2 so wanted to see if f(x) could go to this value too I realised this was possible because if f(x) was -2 and g(x) was 0 -> -2 - 0 = -2 In the equation f(x)g(x) = cos^2x I substituted f(x) as -2 and realised g(x) had to be negative or 0 too to give positive values for cos^x which was also possible e.g -2 * -1/2 or -2 * 0 which were all on the cos^2x therefore thats how I found out the most minimum value that f(x) could be was -2
Q16 is discussed as a PS problem in Further Maths chapter 1. Since cos(theta)=sqrt(1-sin^2(theta)), Let sin be y and cos be x; substitute this x in the given biquadratic and the resulting biquadratic in y is your answer. You were pretty much completely there...
For Q18, once you know that B has 5 guaranteed solutions you already know that has the most possible solutions because f(x)=g(x) is just a degree 5 polynomial (after you get it all on one side) which has a maximum number of 5 solutions
Broski but the time pressure and dat
😂😂😂he has no idea about what that means
Farx
literally like some of his solutions near 4 minutes just to write up
Q 17 can also be solved using sine rule, you just limit the other sin value between 30
Doesn't this assume the triangle is drawn to scale, how do you know without looking at their picture of this information?
Q20, for the case a
16 ultra neat solution: you know x = +- cos theta. Turn them into +- sin theta by replacing every x^2 in the original equation to (1-x^2). Expand, get B in 20 seconds
For Q. 16 I believe you can just substitute Cos(#) into the x values, convert them all into 1-sin^2(#) and expand and simplify. Then if you let sin(#) = x, it cancels down to equation B and the solutions will be sin(#). I have no idea if this solution is mathematically correct but it gives the right answer so /shrug
i omor
average proof fan vs average "seems to work" enjoyer
It really worked. Nice solution
For Q14 just to share: A nice reasoning why an equilateral triangle will have the largest area is because since PQ is a fixed line, we can treat that as the base of a triangle, since area = 0.5 * base * height, moving R will reduce the height, therefore the largest height will be when the perpendicular bisector of PQ goes through the height of the triangle.
for q14, we can use the fact that area of a triangle is base x perpendicular height I think. knowing that PQ is fixed, we can maximise the perpendicular height by making the line OR coincident with the perpendicular bisector of PQ. then we have 3 identical triangles with area 9sqrt3 then multiply that by 3 no?
Thanks for the video! Test in 2 days 😬 I find paper 1 quite a lot harder than paper 2…
Any tips for P2. P1 is ok but p2 is so hard
Amazing explanation, thank you!
Hi mr drew
For question 9 I need to know if my attempt at it was a fluke or could be classified as a valid solution:
for f(x) - g(x) = 2sinx I drew 2sinx and found the least value of this graph to be -2 so wanted to see if f(x) could go to this value too
I realised this was possible because if f(x) was -2 and g(x) was 0 -> -2 - 0 = -2
In the equation f(x)g(x) = cos^2x I substituted f(x) as -2 and realised g(x) had to be negative or 0 too to give positive values for cos^x which was also possible e.g -2 * -1/2 or -2 * 0 which were all on the cos^2x
therefore thats how I found out the most minimum value that f(x) could be was -2
Principal Solution!
Q16 is discussed as a PS problem in Further Maths chapter 1. Since cos(theta)=sqrt(1-sin^2(theta)), Let sin be y and cos be x; substitute this x in the given biquadratic and the resulting biquadratic in y is your answer. You were pretty much completely there...
Did anyone else try Q16 with pair sum and product? I mean it wasn't that bad using that method
For 17, what will happen if x is 3 it doesn’t have 2 triangles
do i need to know the triangle conrguency rules for tmua
yuh
8:25 make r2 you say..... is that a ref?
for q11 you could do an arithmetic sequence as well :)
Yeah sum of n being n(n+1)/2 isn't actually in the spec so I should have used sum of an arithmetic formula
For 14 wouldn’t it be x^2 + y^2 =36
yes
principle solution lol
Q9, Q10, Q13, Q14, Q15, Q17, Q18 fluke, Q19, Q20
mandem this paper was hard you know
unsubscribing
I got 1-9, 11, 12, 14 right and rest wrong
if it's not intrusive may I ask what your tmua score eventually was?🥲
@@UniCorn-m6j 5.2 😂
@@heeseunglee880 What do you think made you do so much worse than in your practices?
@@flippergail9946 mhm idk maybe I wasn’t too smart after all 😂 I was confident when I walked out the exam but I got a 4 in paper 1 and 6.5 in paper 2
@@heeseunglee880 You went from a 6.8 for paper 1 in a practice to a 4?!