Year 12 Maths Help Thread

OP

A candid assessment of my mathematics revision:

C1: No problems
C2: No problems save perhaps a bit of recap on changing the base of a logarithm and circle geometry.
M1: No problems apart from refreshing trailer problems.
M2: No problems.
FP1: Matrix transformation (which makes no sense to me whatsoever) needs to be learnt.
S1: A lot to do:

-The effect of coding on E(X), Var(X), the standard deviation, the correlation coefficient.
-The linear interpolation formula
-Histogram problems
-Definitions and 'interpret' questions.

Reply 2863

XmayaX

2

Original post by Zacken

Why don't you pull out a factor of

3^{k+1}

in the last two terms to get

3^{k+1}(2k + 4k -1 + 4) = 3^{k+1}(6k + 3) = 3^{k+1}(3^1)(2k + 1) = 3^{k + 1 + 1}(2k+1) = 3^{k+1+ 1}(2(k+1) - 1)

Thanks, this helped a lot 😊

Posted from TSR Mobile

Reply 2864

Original post by XmayaX

Thanks, this helped a lot 😊

Posted from TSR Mobile

Glad it did. :-)

Reply 2865

LelouchViRuge

7

Original post by Palette

A candid assessment of my mathematics revision:

C1: No problems
C2: No problems save perhaps a bit of recap on changing the base of a logarithm and circle geometry.
M1: No problems apart from refreshing trailer problems.
M2: No problems.
FP1: Matrix transformation (which makes no sense to me whatsoever) needs to be learnt.
S1: A lot to do:

-The effect of coding on E(X), Var(X), the standard deviation, the correlation coefficient.
-The linear interpolation formula
-Histogram problems
-Definitions and 'interpret' questions.

Changing the base proof:

Let

Log_a b = N

From this we can deduce

a^N = b

Log_c a^N = log_c b

NLog_c a = log_c b

N = \frac {Log_c b}{Log_c a}

Hegartymaths has helpful videos on matrix transformations!

Reply 2866

Original post by LelouchViRuge

Changing the base proof:

You'll want to use \ln or \log instead of squashing the variables L, o and g together to form your logarithm.

Reply 2867

OP

Original post by LelouchViRuge

Changing the base proof:

Let

Log_a b = N

From this we can deduce

a^N = b

Log_c a^N = log_c b

NLog_c a = log_c b

N = \frac {Log_c b}{Log_c a}

Hegartymaths has helpful videos on matrix transformations!

Hopefully I should be OK with every module except S1 which is simply not designed for my brain.

Reply 2868

LelouchViRuge

7

Original post by Palette

Hopefully I should be OK with every module except S1 which is simply not designed for my brain.

Friends have told me 90% of S1 can be done with a calculator lol.

Are you doing the whole A level maths in a year?

Reply 2869

OP

Original post by LelouchViRuge

Friends have told me 90% of S1 can be done with a calculator lol.

Are you doing the whole A level maths in a year?

No, I'm doing C1 C2 M1 M2 FP1 S1.

Reply 2870

B_9710

18

Original post by LelouchViRuge

Friends have told me 90% of S1 can be done with a calculator lol.

Are you doing the whole A level maths in a year?

I think I'd have to agree with the 90%

Reply 2871

OP

Original post by B_9710

I think I'd have to agree with the 90%

I find a lot of it tedious or confusing. For example I have no idea on when to use the second z table for the normal distribution.

Reply 2872

Original post by Palette

I find a lot of it tedious or confusing. For example I have no idea on when to use the second z table for the normal distribution.

Whenever you have a normal distribution of the form

P(Z > z) = k

for special values of

k

given in the tables.

Reply 2873

samb1234

13

Original post by B_9710

I think I'd have to agree with the 90%

Nah its not that high. Theres all the nodelling stuff from chapter 1 drawing venn diagrams box plots probability questions etc etc, which while often involve the use of a calculator still require some level of understanding rather than plugging numbers in like some other parts of s1.

Reply 2874

Ayman!

15

Original post by Palette

I find a lot of it tedious or confusing. For example I have no idea on when to use the second z table for the normal distribution.

I admit - I panicked for S1 a month before the exam because I didn't do anything for it, but it turned out okay. I just did like 3 past papers before the exam and did alright in it.

Reply 2875

I'm a Peacock

1

Hello everyone, would someone please help me on a fp1 series question
I still don't understand how a base zero below the sigma effects the answer
Thank you

It is part b

Original post by I'm a Peacock

Hello everyone, would someone please help me on a fp1 series question
I still don't understand how a base zero below the sigma effects the answer
Thank you

It is part b

\displaystyle \sum_{r=1}^3 f(r) = f(1) + f(2) + f(3)

\displaystyle \sum_{r=0}^{3} f(r) = f(0) + f(1) + f(2) + f(3) = f(0) + \sum_{r=1}^n f(r)

Can you see how there is an extra term?

In general:

\displaystyle \sum_{r=1}^n f(r) = f(1) + \cdots + f(n)

- there are '

n

' terms.

\displaystyle \sum_{r=0}^n f(r) = f(0) + f(1) + \cdots + f(n) = f(0) + \sum_{r=1}^n f(r)

- there are '

n+1

' terms.

So as you can see, as long as

f(0) \neq 0

then

\sum_{r=0}^n f(r) \neq \sum_{r=1}^n f(r)

.

In this case, we have

f(r) = r^2 - 2r + 2n + 1 \Rightarrow f(0) = 2n + 1 \neq 0

.

Reply 2877

I'm a Peacock

1

Could you please provide a solution

Reply 2878

OP

This summer I may make something resembling a blog as I will be tackling STEP (and maybe MAT) questions and working through Martin Griffith's 'A Prime Puzzle'.

Reply 2879