SQA Higher Mathematics - Paper 1 Non-calculator - 13th May 2024 [Exam Chat]

Reply 20

8

Original post

by ilovelaerning

however, the second last question on paper 2 was confusing it said, sin2x-sinsquaredx=0 so when I expanded it and took out the common factor I got sinx(4cosx-sinx)=0 so obv sinx=180 but for the second part I wrote no solutions

Yeah same I got (4cosx-sinx) too but was not sure how to finish it so I did the sinx=180 and stuff like that

Reply 21

Ahudndinshsjendn

6

Original post

by Ornella1017

Yeah same I got (4cosx-sinx) too but was not sure how to finish it so I did the sinx=180 and stuff like that

i just took sin^2 = 4sinxcosx - then rooted the other side so sinx = 2sinxcosx and was able to put in brackets - idk if thats what your meant to do tho

Reply 22

ahad017

5

Original post

by Ornella1017

Yeah same I got (4cosx-sinx) too but was not sure how to finish it so I did the sinx=180 and stuff like that

I used the wave function on the second bracket and got that to be sqrt(17)sin(x-75.96). So I did sinx = 0 and sqrt(17)sin(x-75.96) = 0 so I got 0°, 75.96°, 180°, and 255.96° as the solutions. Probbalbly not the correct and easiest way to do it but I had no clue what else to do. Desmos suggests these are the right answers though so fingers crossed ig.

Reply 23

ahad017

5

Original post

by Ahudndinshsjendn

i just took sin^2 = 4sinxcosx - then rooted the other side so sinx = 2sinxcosx and was able to put in brackets - idk if thats what your meant to do tho

Sounds like what your meant to do especially since it's five marks

Reply 24

Ornella1017

8

Original post

by ahad017

I used the wave function on the second bracket and got that to be sqrt(17)sin(x-75.96). So I did sinx = 0 and sqrt(17)sin(x-75.96) = 0 so I got 0°, 75.96°, 180°, and 255.96° as the solutions. Probbalbly not the correct and easiest way to do it but I had no clue what else to do. Desmos suggests these are the right answers though so fingers crossed ig.

I got 0 and 180 in my solution but idk if I can possibly get a mark cause I remember getting those values but my answer has 0,180,360 and something else idek what I did tbh

Reply 25

Ornella1017

8

Original post

by zlongstaff777

I got that too don’t worry! I found the papers really challenging too so hopefully the grade boundaries reflect how difficult it was!

Do grade boundaries change much for higher maths?

Reply 26

Ahudndinshsjendn

6

Original post

by Ornella1017

Do grade boundaries change much for higher maths?

lets hope so

Reply 27

CossLne

19

Original post

by Ornella1017

Do grade boundaries change much for higher maths?

Not usually. They changed drastically in 2015 after so many would have failed but usually there isnt much it.
Does anyone know if the paper has been uploaded like the n5 ones were?

Reply 28

Roarms271

1

Original post

by ahad017

I used the wave function on the second bracket and got that to be sqrt(17)sin(x-75.96). So I did sinx = 0 and sqrt(17)sin(x-75.96) = 0 so I got 0°, 75.96°, 180°, and 255.96° as the solutions. Probbalbly not the correct and easiest way to do it but I had no clue what else to do. Desmos suggests these are the right answers though so fingers crossed ig.

Yeah well an easier way to do it that I found in the exam was to solve for sinx = 0 and 4cosx - sin x = 0, sinx = 0 being x = 0, 180, 360 and 4cosx - sinx = 0 being factorised to tanx = 4 (sinx/cosx = tanx) and solve for x. I didn't understand what you meant when you used sqrt(17). As a sidenote, did anyone else get the last circle question right under that question? The one with the perpendicular bisector where you had to find the equation of the circle. I approached it by using an isosceles triangle for the points on the horizontal lines but I'd be glad to hear how you approached it.

Reply 29

Maryam x

Original post

by Ornella1017

Yeah same I got (4cosx-sinx) too but was not sure how to finish it so I did the sinx=180 and stuff like that

If you go to Youtube- Clelland Maths did both papers on there

Reply 30

ahad017

5

Original post

by Roarms271

Yeah well an easier way to do it that I found in the exam was to solve for sinx = 0 and 4cosx - sin x = 0, sinx = 0 being x = 0, 180, 360 and 4cosx - sinx = 0 being factorised to tanx = 4 (sinx/cosx = tanx) and solve for x. I didn't understand what you meant when you used sqrt(17). As a sidenote, did anyone else get the last circle question right under that question? The one with the perpendicular bisector where you had to find the equation of the circle. I approached it by using an isosceles triangle for the points on the horizontal lines but I'd be glad to hear how you approached it.

Yeah you did it exactly as Mr Clelland did it (probably the intended way). I half panicked in the exam and decided to give my method a shot and it worked out thankfully hopefully I should get the credit for it. The root 17 was the value of k I got when I did the wave function for 4xosx - sinx. For the circle one I did the same as part a) basically got the perpendicular bisector of QR which was x = 9 I believe and the point of intersection between that and the equation from part a) is the centre of the circle. Used the distance formula from P (or Q or R) to C to get the radius. If I remember it was (x - 9)^2 + (y - 7)^2 = 40 or something like that. Links to Mr Clelland's videos:

Paper 1: https://youtu.be/Vkp2t9gy3DA?si=wES6dvCoPNaRU_K_

Paper 2: https://youtu.be/rhI1Qw2DjkE?si=0TGV_OxugsmW4xLx

Reply 31

eva.wright

1

Original post

by Mills15

How did everyone find the paper, I thought they were both awful, I was on course for a B but I don’t think i got that.
What did everyone get for U2 and the limit on the recurrence relation on Paper 1, I got U2 = 16 and the limit = 15 but that can’t be possible but I couldn’t find any mistake in my working.