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# CfE Higher Mathematics 2017/18 watch

1. CfE HIGHER MATHEMATICS 2017/18

Hello everyone!

Welcome to the Higher Mathematics thread. This is a thread for those who are studying Higher Mathematics for the term of 2017/18. You are welcomed to use this thread to discuss and ask questions on different aspects of the course and also suggest difference resources. Anyone who had previously studied the course would also be more than welcome to provide any help or advice as well as sharing their experience of the subject as a whole.

So who's studying this course this year?
2. Subbing, just finished my honours years of a maths degree. If anyone has any questions throughout the year feel free to PM and I'll try to get back to you ASAP.
3. Started this course in June, we've only done a tiny bit and I can already tell this year will be brutal l o l
4. (Original post by catsandhats)
Started this course in June, we've only done a tiny bit and I can already tell this year will be brutal l o l
How did everyone do?
5. honestly wishing you all the best of luck - you will need it for this
6. Just finished higher and starting AH maths. Don't panic I started the course and got so worried and ended up with a B in the mock however managed an A in the exam. My advice would be to keep up with homework/ finish exercises set in class. Do as much practice as YOU need you won't necessarily work at the same pace as everyone.

7. anyone know what i am doing wrong?
8. (Original post by NGEO)

anyone know what i am doing wrong?
edit: here's my solution (can't enter base so assume base 2)

log(2x + 1) = 3 + 2log(x)
log(2x + 1) = log(8) + log(x^2)
log(2x + 1) = log(8x^2)
2x + 1 = 8x^2
8x^2 - 2x - 1 = 0

then you'll have to use the big b^2 - 4 ac x solver thing from Nat 5 to find x value, which gives you 1/2 or -1/4. Putting it into the log equation and it shows that x = 1/2
edit: here's my solution (can't enter base so assume base 2)

log(2x + 1) = 3 + 2log(x)
log(2x + 1) = log(8) + log(x^2)
log(2x + 1) = log(8x^2)
2x + 1 = 8x^2
8x^2 - 2x - 1 = 0

then you'll have to use the big b^2 - 4 ac x solver thing from Nat 5 to find x value, which gives you 1/2 or -1/4. Putting it into the log equation and it shows that x = 1/2
Thanks man, do you know why my method was wrong?
10. (Original post by NGEO)
Thanks man, do you know why my method was wrong?
I'm not exactly sure why this is the case, but I think if you ever want to remove the logs you need to simplify both sides and have only 1 term on each side. I don't think it works otherwise.

So log(8) + log(x^2) just simplifies into the 1 term log(8x^2), allowing you to then remove the logs from each side. (note: only simplifies because both are in base 2!)

If you look at it this way:

log(8) + log(x^2) = log(8x^2)
however...
8 + x^2 ≠ 8x^2, which shows you can't remove the logs without first making them into one term
11. (Original post by cat7)
Just finished higher and starting AH maths. Don't panic I started the course and got so worried and ended up with a B in the mock however managed an A in the exam. My advice would be to keep up with homework/ finish exercises set in class. Do as much practice as YOU need you won't necessarily work at the same pace as everyone.

Hi! Well done on your spectacular result.

I'm aiming for an A in Higher this year too. I failed my National 5 prelim however I got an A in the final exam.
I'm far more confident in maths now.

I have my first unit assessment in a week or so, of such being the Expressions and Functions Unit.

Do you recommend any good books/effective techniques in succeeding in H Maths?

12. (Original post by _Matthew01_)
Hi! Well done on your spectacular result.

I'm aiming for an A in Higher this year too. I failed my National 5 prelim however I got an A in the final exam.
I'm far more confident in maths now.

I have my first unit assessment in a week or so, of such being the Expressions and Functions Unit.

Do you recommend any good books/effective techniques in succeeding in H Maths?

In terms of books any book with relevant examples is good I just used the textbook given to us by our teacher. More importantly is being confident with the material just practice as much as possible make sure you can do the topic inside out do the same questions again if needed. For now I wouldn't worry too much about the unit assessments. You do have to pass them but I found they weren't as difficult as anticipated. Apart from that maths is really just down to doing as much practice as possible. Any further questions feel free to PM me. Good luck
13. (Original post by cat7)
In terms of books any book with relevant examples is good I just used the textbook given to us by our teacher. More importantly is being confident with the material just practice as much as possible make sure you can do the topic inside out do the same questions again if needed. For now I wouldn't worry too much about the unit assessments. You do have to pass them but I found they weren't as difficult as anticipated. Apart from that maths is really just down to doing as much practice as possible. Any further questions feel free to PM me. Good luck
Thank you so much 😊
14. I'm doing the course for my second year (I didn't sit the exam last year). I have all my notes from the course last year and so far have covered straight line , quadratics and vectors. I would be more than happy to share my notes and some practice questions with solutions and think It'd be great for others to do so too. X
15. Anyone willing to help? If you could write out working would be much appreciated!

16. (Original post by gw07mcgheerachel)
Anyone willing to help? If you could write out working would be much appreciated!

A)
2x^2 + 8x - 3

2(x^2 + 4x - 3/2)
2(x^2 + 4x) - 3
2(x + 2)^2 + c

the overall constant is -3 as seen in bold

so the final part for +c is worked out by finding the total constant value currently in 2(x + 2)^2, and what c should be to equal -3
2 * (2^2) + c = -3
8 + c = -3
c = -3 - 8
c = -11

so

2(x+2)^2 - 11

B)
You can just look at it in the form we've written. You take the number in the bracket, 2, change its sign to -2, and that's the x co-ordinate. You take the number outside the bracket and keep it as it is, -11, that's y co-ordinate. No working required unless you missed out part A.

turning point at (-2,-11)

If you couldn't do part A, the alternative to finding the co-ordinates would be to get the derivative of the equation and make it equal to 0.

If you have any further questions feel free to ask.
A)
2x^2 + 8x - 3

2(x^2 + 4x - 3/2)
2(x^2 + 4x) - 3
2(x + 2)^2 + c

the overall constant is -3 as seen in bold

so the final part for +c is worked out by finding the total constant value currently in 2(x + 2)^2, and what c should be to equal -3
2 * (2^2) + c = -3
8 + c = -3
c = -3 - 8
c = -11

so

2(x+2)^2 - 11

B)
You can just look at it in the form we've written. You take the number in the bracket, 2, change its sign to -2, and that's the x co-ordinate. You take the number outside the bracket and keep it as it is, -11, that's y co-ordinate. No working required unless you missed out part A.

turning point at (-2,-11)

If you couldn't do part A, the alternative to finding the co-ordinates would be to get the derivative of the equation and make it equal to 0.

If you have any further questions feel free to ask.
Thank you so much for your help!
18. Hi guys could you check out my blog , much appreciated
https://www.thestudentroom.co.uk/sho....php?t=4998046
19. we've only done 4 topics and I've already fell behind. I can't do this!!!
20. (Original post by Beth5678)
we've only done 4 topics and I've already fell behind. I can't do this!!!
What 4 topics?

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