Mr M's OCR FSMQ Additional Maths Answers Thread

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BTW, for Q13) ii), what does derive the formula mean? What are the solutions? Does it mean find t?

And also please can someone confirm for Q 4) ii) Is the answer 19/144 or 19/1296. Because Mr M's answer is 19/1296 but I keep getting 19/144. Please can someone explain why I'm wrong. I did 1-[P(x].

Reply 42

misskimono

Mr M

No this one is correct. (8, 2) is the centre of the circle. You do not find the sums of the squares of the coordinates of the centre of the circle in order to find the radius.

Can you show me how you got his then? Because we were taught x squared + y squared = r squared.

Reply 43

Mr M

Study Forum Helper

hearmecry

can u go through 10 and tan question they were so hard!!

Question 10 isn't hard (possibly just unfamiliar).

You are told y = k (x-2) (x-4).

You are given the point (0,4). Just substitute it in to find k.

Then you are given y = c (x-2) (x-2) (x-4).

Shove the same point in to find c.

Then try to put the point (3, 0.25) into both equations to see which one is best.

Which one is the tan question? Question 7?

You just needed to cross-multiply to prove part (i) and then use the identity tan x = sin x / cos x to do part (ii).

You get here:

\tan \theta + \frac{1}{\tan \theta} = 4

Multiply through by tan theta to form a quadratic. Rearrange to make it equal to zero and then use the quadratic formula or complete the square.

You obtain:

\tan \theta = 2 \pm \sqrt 3

and then solve in the interval.

Reply 44

misskimono

Ivo

BTW, for Q13) ii), what does derive the formula mean? What are the solutions? Does it mean find t?

I did not know what that meant at all. It was like when I first saw constant retardation (or something similar) on one of the kinematics questions on a past paper.

Reply 45

Mr M

Study Forum Helper

misskimono

Can you show me how you got his then? Because we were taught x squared + y squared = r squared.

That is only true if the centre of the circle is the origin.

P (1,3) and Q (15,1) were on the ends of a diameter. The centre of a circle is the MIDPOINT of a diameter so it is M (8, 2).

Now use Pythagoras to find the distance PM or MQ.

Now write the equation in the form

(x-8)^2 + (y-2)^2 = (5 \sqrt 2)^2

and expand it out.

Reply 46

Edwin Okli

You did it between 0 and 5? I was really worried about that question because I didn't know if I was supposed to integrate between time in minutes or seconds..

s = \displaystyle \int_0^5 (60t^4 - 600t^3 + 1500t^2) dt

\Rightarrow = \left[ 12t^5 - 150t^4 + 500t^3 \right]_0^5

\Rightarrow = (37500 - 93750 + 62500) - 0

\Rightarrow = 6250m

Ha, I was bored so I gave it a go. Now I realise that was a page ago..

Reply 47

Mr M

Study Forum Helper

Ivo

And also please can someone confirm for Q 4) ii) Is the answer 19/144 or 19/1296. Because Mr M's answer is 19/1296 but I keep getting 19/144. Please can someone explain why I'm wrong. I did 1-[P(x].

That is correct. When I altered the answer, I obviously didn't change it properly. It is fixed now.

Reply 48

Mr M

Study Forum Helper

Edwin Okli

You did it between 0 and 5? I was really worried about that question because I didn't know if I was supposed to integrate between time in minutes or seconds..

s = \displaystyle \int_0^5 (60t^4 - 600t^3 + 1500t^2) dt

\Rightarrow = \left[ 12t^5 - 150t^4 + 500t^3 \right]_0^5

\Rightarrow = (37500 - 93750 + 62500) - 0

\Rightarrow = 6250m

Ha, I was bored so I gave it a go. Now I realise that was a page ago..

The question says t is measured in minutes.

Reply 49

jusjes

Wow that SUPER failed. Don't want to think about it untill results day so i can explain away the awful result. Oops

Reply 50

Mr M

Study Forum Helper

Ivo

BTW, for Q13) ii), what does derive the formula mean? What are the solutions? Does it mean find t?

Ali makes 72/t components per hour.

Beth makes 72/(t+2) components per hour.

The difference between the number Ali makes and the number Beth makes is 3 per hour.

\frac{72}{t} - \frac{72}{t+2} = 3

72t + 144 - 72t = 3t (t+2)

3t^2 + 6t - 144 = 0

t^2 + 2t - 48 = 0

(t+8)(t-6)=0

etc

Reply 51

mrmanps

Mr M

I don't know. This is the first year I have taught this qualification so I don't have a huge amount of experience to draw on.

My first year teaching it next year :biggrin:

Reply 52

Mr M

Study Forum Helper

mrmanps

My first year teaching it next year :biggrin:

How are you doing it? I did it one lesson per week after school for about 20 weeks for those who did GCSE in November early entry and got A*. It was a rush to cover everything in time. I have some Year 10s that did it too - they haven't even taken GCSE yet!

Reply 53

Stoney15

Hey, thanks for the thread. It's great.
Do you think you could attach the question paper as any other file type, only my computer is having trouble opening it, and other .pdf files?

Reply 54

sassy :)

ahh this is great! i got way more answers than i thought i had. :biggrin:

well relieved.
anyone any ideas on what the grade boundaries will be on this paper?

Reply 55

misskimono

jusjes

Wow that SUPER failed. Don't want to think about it untill results day so i can explain away the awful result. Oops

I feel the same. But I think i will be in a better position for AS maths next year. My algebra skills have improved massively. My least favourite units were co-ordinate geometry and inequalities (linear programming :s-smilie:

) Integreation, differentiation, polynomials and kinematics were the best. I've checked with Mr M and I have full marks for questions on those topics. :woo:

Reply 56

mastermind_107

misskimono

My problem is not how you worked it out. It's how you got (8,2) because y2-y1= 3-1/2 =1 x2-x1= 15-1/2 = 7. therefore isn't it (7,1)

y2 and y1 refer to the respective y co-ordinates of the two points.

So the y co-ordinate of the midpoint the average of the two y co-ordinates; the x co-ordinate of the midpoint is the average of the two x co-ordinates.

So Mr M is right. :yep:

Reply 57

Ivo

misskimono

My problem is not how you worked it out. It's how you got (8,2) because y2-y1= 3-1/2 =1 x2-x1= 15-1/2 = 7. therefore isn't it (7,1)

No, some theory you are getting muddled. Check back the textbook, formula is:

(x1+x2)/2; (y1+y2)/2

Also think, if one co-ordinate is (1,3) and (15,1), how can midpoint y-co-ordinate be 1?