[ugk4life]

how are we supposed to lay out the iterative questions where we have to find the roots?

i know how to do them, just not sure of how to present them

[lamalas600]

(Original post by ugk4life)
how are we supposed to lay out the iterative questions where we have to find the roots?

i know how to do them, just not sure of how to present them

Set them up like this
x0= Starting value
x1= 2nd value
x2= 3rd value

etc....

Until you have two answers the same for the given decimal place(in question), for each iterate you should write about 6 sig figs

[1234girl]

CAN SOMEONE PLEASE GIVE ME THE SOLUTION TO 9iii OF JUNE 2010?????

[19hannah94]

(Original post by lamalas600)
You know that it's mass is initially 40 grams, and 2years later 31.4 grams, and it's an exponential.

Therefore you know at time=0 (initially) it's 40 grams therefore M(40)= there x must be equal to 40.

Then you have M=40e^kt (time and k=constant)
Your given 2 years later it's 31.4... 31.4=40e^2k

Can you do the rest?

Thank you so much! I had something like this in mind but was using a less logical value for k. Thanks!

[ugk4life]

(Original post by lamalas600)
Set them up like this
x0= Starting value
x1= 2nd value
x2= 3rd value

etc....

Until you have two answers the same for the given decimal place(in question), for each iterate you should write about 6 sig figs

(Original post by lamalas600)
Set them up like this
x0= Starting value
x1= 2nd value
x2= 3rd value

etc....

Until you have two answers the same for the given decimal place(in question), for each iterate you should write about 6 sig figs

cool thanks.

[19hannah94]

(Original post by 1234girl)
CAN SOMEONE PLEASE GIVE ME THE SOLUTION TO 9iii OF JUNE 2010?????

Put the equations equal to each other and equate coefficients of either b or a i think and rearrange.

[Master S P]

Anybody have the Jan 2012 MARKSCHEME ?

[lamalas600]

(Original post by Master S P)
Anybody have the Jan 2012 MARKSCHEME ?

PM me your email

[kontemptXD]

Can any one explain how to do Q7(ii) part b from the June 2009 paper please?

[19hannah94]

(Original post by kontemptXD)
Can any one explain how to do Q7(ii) part b from the June 2009 paper please?

Relate it to the 8sin(theta)-6cos(theta) and then take values of x and y to plug into the equation given in part (b) i.e. 1 or -1

[kontemptXD]

(Original post by 19hannah94)
Relate it to the 8sin(theta)-6cos(theta) and then take values of x and y to plug into the equation given in part (b) i.e. 1 or -1

I took a similar approach and ended up with 40 - 20 but the mark scheme answer is 60. The examiners report on the question hasn't really helped me to comprehend why. Can you explain further?

[lamalas600]

(Original post by kontemptXD)
I took a similar approach and ended up with 40 - 20 but the mark scheme answer is 60. The examiners report on the question hasn't really helped me to comprehend why. Can you explain further?

You should get a double negative 40 - -20 = 60

[ugk4life]

hey can someone explain june 11 qu 6 please

[lamalas600]

(Original post by ugk4life)
hey can someone explain june 11 qu 6 please

It's a long question... the x=10/3 bit or the find area bit

[Sweetcorn_1]

(Original post by lamalas600)
From the first part you know that
Let's do some rearranging..
It is given in the question that
Plug that into

Thanks a bunch.

[ugk4life]

(Original post by lamalas600)
It's a long question... the x=10/3 bit or the find area bit

no worries, seen it explained on an older thread.

the x=10/3 was confusing me though

[Aimanos]

Hey could someone please explain Jan 2012 question 8iii please? Much appreciated...

[Axellia]

Does anyone know what is usually the lowest mark for 90ums+

[andy180594]

(Original post by Aimanos)
Hey could someone please explain Jan 2012 question 8iii please? Much appreciated...

I am assuming you have done 8i and 8ii?
You should have 1 - 8sin^2xcos^2x for 8i and (2- route 3)/16 for 8ii.

Cos4x = 2cos^2 2x - 1 = 1 - 8sin^2xcos^2x.
cos4x + 1 = 2cos^2 2x (substitutes for first part)
re arranging some of the equations you can get 8sin^2xcos^2x = 1-cos4x ( /3 then substitute for second part)

So,

8/3 sin^2xcos^2x = (1-cos4x)/3

plug in (1-cos4x) where 2cos^2 2x is and (1-cos4x)/3
where 8/3 sin^2xcos^2x and your equation should look like 1 - cos4x - (1-cos4x)/3

get the same common denominator so you get 3/3 - 3cos4x/3 - (1-cos4x)/3

rearrange to get (4cos4x + 2)/3

max cos4x can ever be is 1 and min cos4x can ever be is -1,

So max and min points are 2 and -2/3.

I have probably made heavy work of this questions but In short, Your substituting cos4x into the equation where you can by rearranging what you did in 8i.

Hope this helps

[Fward]

not confident at all, but only retaking to try gain A*