CCEA C2 Exam Thread (6th June)

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Reply 60

apen610

Original post by HugeBicepLAD

I emailed ccea haha

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hahaha, it's the only way

Reply 61

Frank the Tankk

Original post by apen610

I sure hope so, the lower the better. But I thought last years was a bit harder?

I'm not sure, can't remember what it was like but a lot of people came out in my school making stupid mistakes and such. So while it mightn't have been as hard I reckon there was scope to make more mistakes

Reply 62

GhostlyFish

Original post by apen610

Yeah that seemed to be an issue with this paper, make a small error or miss something and you're pretty much ****ed for the whole question...

And frankly, I didn't have enough time to go back and check any of my work, so the likelihood of me making one of those little mistakes is almost inevitable. Oh, the sweet embrace of failure...

Glad to hear the answer for 10 was 4root3 etc, though. Couple of my friends got that. Good for them :biggrin:

Reply 63

Henrymac

I changed the base at the start of question 8 and got completely random answers. Does anyone have any idea if I can get any marks out of the 10?

Reply 64

HugeBicepLAD

Original post by Henrymac

I changed the base at the start of question 8 and got completely random answers. Does anyone have any idea if I can get any marks out of the 10?

I dropped it

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Reply 65

HugeBicepLAD

Original post by Henrymac

I changed the base at the start of question 8 and got completely random answers. Does anyone have any idea if I can get any marks out of the 10?

Yeah you'll get marks don't worry
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(edited 9 years ago)

Reply 66

HugeBicepLAD

Original post by apen610

hahaha, it's the only way

Had to be done

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Reply 67

Hollienikita

Original post by apen610

yes

can you post it please??

Reply 68

Cameron225

Seemed to be a fairly easy paper, only question that I couldn't work out was 4b (kite and circle thing) and 7a (4-sinx=6cos^2x)

Reply 69

d22

For the kite, I found out the centre and radius. When I found out the centre, I used the distance formula to find out the length of AD. That line will split the kite into two right angled triangle. Used the pythagoras theorem then to find out the other side and simply multiplied 1/2 × radius × length. Then multiplied that by two.

7a) The 6cos^2 can be changed into 6(1-sin^2) and it should form a quadratic equation when multiplied out and rearranged.

Reply 70

borysek01

The paper was okay. C1 was MUCH MUCH more harder than this. I just wonder, how did you guys even manage to get an answer for the kite? All I did was to complete the square, and from the result find the radius and the centre coordinates. Then I found the gradient of AC (line directly through the middle) and the equation for AC. I then said that the line perpendicular to that (from B to D) is the negative reciprocal and this is where I stopped. Did I get any marks here?

I completely skipped the curved length surface area one. No clue. Apart from that, it was okay. I think I scraped my A. Was surprising how easy the end of the paper was.

Reply 71

borysek01

Original post by Frank the Tankk

x = 4root(3)/3

y = 4root(3)

There was two set of answers for this question. Square roots of number give either a positive or a negative result.

Reply 72

d22

Original post by borysek01

There was two set of answers for this question. Square roots of number give either a positive or a negative result.

I think you can't have a negative number because if you put it into the calculator, it comes up error. I tested it during the exam lol.

Reply 73

borysek01

Original post by Mr Tall

That was so hard. Answers i got: 1i) 7/36 ii)didnt get it iii) converges. b)-189 2. -16/x +3/4x^3/4 +9/2x^2 -7x +c 3.i) 9.82 km ii) 81.3 iii) 10.1km^2 4. Showed it okay to get pi r l b) 37.9 5.ai) 0.047 ii) 250 b) couldnt prove it fml 6i) A(4,16) ii) 42(2/3) units squared 7.a) 41.8, 138,210,330 b) proved it 8. X=-/+4/3 root3, y=+/-4root3

This is the same answer I got for 8.

Reply 74

sarah-1234

What was question 7a about?

Reply 75

Hollienikita

okay let theta=y for ease here

for 7A I subbed 1-sin^2y in for cos^y and ended up with 6sin^y - siny - 2=0
then I let x=sin^2y and got 6x^2-x-2=0 then I worked out that x=-2/3 or 1/2
I subbed siny back in so siny = -2/3 or 1/2 then I worked out that Y= -41.8 or 30

so I drew cast diagram from 0-360 and put the angle 41.8 in the cos and tan quadrants giving me 221.8 and 318.2 then drew another one for 30 and put the angle in the all and sin quadrants giving me 30 and 150

so I ended up with theta = 41.8, 221.8, 30 and 150

I know that I went wrong there but can someone explain how and tell me how many marks (if any) i'll get for that??

Reply 76

d22

Original post by Hollienikita

I got Siny=2/3 and -1/2. You might have got the signs mixed up. You probably lost 1 or 2 marks but I'm not quite sure how much they would take. I'm sure it won't that much. Two at most probably.

Reply 77

Hollienikita

Original post by d22

I got Siny=2/3 and -1/2. You might have got the signs mixed up. You probably lost 1 or 2 marks but I'm not quite sure how much they would take. I'm sure it won't that much. Two at most probably.

ahh okay! thank god it wasn't something big! thank you!!

Reply 78

KatiePhilips1996

1ai) 7/36
ii) n+1/n^2
iii) converges

b) -189

2) -16x^-1 + 3/4(x^4/3) + 9/2(x)^2 -7x +c

3i) 3.13km
ii) 81.3
iii) 10.0km^2

4a) Set surface area of cone equal to area of sector but for the area of sector I use L as the radius. This gave me an equation for the angle in terms of r pie and l
ii) I sub the previous answer into area of sector and it proved the pie L R thing..
b) I firstly found the centre and radius. I then found the distance of the line which cut the kite in 2 (cant remember letters). I then used Pythagoras theorem and used this figure and the radius to get the are. I got 37.95

5ai) 0.047
ii) 250

b) proved it

6i) (4,16)
ii) 42 and 2/3 units^2

7a) 41.8, 138, 210, 330

b) proved it

8) x= 4root3 / 3
y= 4root3

Over all i was happy with the paper. I definitely think C1 was harder!