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WJEC C4 12 June 2015

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I think from my perspective it's that it's super unfair to do this to our year. Why our year is being targeted for C3 and C4 this year is beyond me. Every paper I've done I've gotten an A but this one I know I have failed. I was tempted to complain too.

On top of that, Michael Gove the ****** got rid of Jan exams so we have twice the workload than past years. To make the maths papers like twice as hard as well is extremely unfair. The examiners' own A Levels look like BTECs compared to exams we've had to do. Someone really should make a petition or something.

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(edited 8 years ago)

Reply 61

Djmads

Original post by fha96

Too right. Agreed.

Reply 62

wb25

When you compare this years exams to papers from 2 - 3 years + ago, the difference in difficulties is world apart. The leap is absoulutely massive.

Reply 63

Djmads

Original post by wb25

When you compare this years exams to papers from 2 - 3 years + ago, the difference in difficulties is world apart. The leap is absoulutely massive.

Exactly. You actually feel let down by the education system.

Reply 64

Mr rump

This will cheer people up https://www.youtube.com/watch?v=aZaFjAgFpUU

Reply 65

mabon96

Every ques except the partial fractions and the vectors had something that wasn't 'normal'. You cannot deny that it is clear that this year they have tried to make the papers harder. All it means is that it favours people who haven't revised (because people who have revised a lot still can't answer the questions as they are completely different-so it makes the gap between strong candidates and weaker ones lesser-the opposite of what is intended!

Original post by cnoc

On what grounds though? I mean apart from the proof question and maybe the volume thing, it wasn't that different to the usual ones.

Reply 66

TheBeardedGuy

Original post by apettah

How do you do 2b?

dy/dx = (4x^3 + 6xy) / (4y - 3x^2)

dx/dy = (4y - 3x^2) / (4x^3 + 6xy)

dx/dy = 0 (when parallel to y-axis)

0 = (4y - 3x^2)

y = (3/4)x^2

Sub into equation for curve C to get x co-ordinates, then sub x into y = (3/4)x^2. I only had one set of co-ordinates; (2,3).

(edited 8 years ago)

Reply 67

fha96

Let's hope a lot of teachers complain about it. Watch how they make up for it by making next years paper so easy. I feel like we're being made guinea pigs by the WJEC tbh

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Reply 68

steffanjones

Original post by fha96

Let's hope a lot of teachers complain about it. Watch how they make up for it by making next years paper so easy. I feel like we're being made guinea pigs by the WJEC tbh

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Definitely felt as if they were trialling their new exam techniques on us

Reply 69

fha96

Did anyone else get 1/2 for the integration by substitution bit?

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(edited 8 years ago)

Reply 70

Luke1324

Original post by fha96

Did anyone else get 1/2 for the integration by substitution bit? I just did it again at home and can't see how you'd get 1/18

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i think I got 1/18. Did you remember to change the limits? Sounds patronising but it's not intended to be :smile:

(edited 8 years ago)

Reply 71

fha96

Original post by Luke1324

i think I got 1/18. Did you remember to change the limits? Sounds patronising but it's not intended to be :smile:

Yeah I remembered that haha, I think I didn't take -1/3 out as a factor

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Reply 72

`Jts

Question 7a) was -1/3(-1/6) = 1/18

Reply 73

`Jts

Did anyone find the value of k on question 9b)?

Reply 74

Luke1324

Original post by fha96

Yeah I remembered that haha, I think I didn't take -1/3 out as a factor

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That'll probably only be a mark or two so it's not too bad. I made silly mistakes as well, like not seeing question 6 had a part B :colondollar:

Reply 75

Luke1324

Original post by `Jts

Did anyone find the value of k on question 9b)?

I did but it took longer than expected, so I assume there was a faster way of doing the question. I think I had 1/7200, but can't really remember.

Reply 76

aLittleBookWorm

Original post by `Jts

Did anyone find the value of k on question 9b)?

I didn't, and I still got to £600 - I used a simultaneous equation sort of thing and had something like this:

[1/k(800-A/800A)] / [1/k(900-A/900A)] = 3/4

the 1/k cancelled out so I had

[(800-A)/800A] / [(900-A0/900A)] = 3/4

which I used to find A

(edited 8 years ago)

Reply 77

`Jts

Original post by aLittleBookWorm

I didn't, and I still got to £600 - I used a simultaneous equation sort of thing and had something like this:

[1/k(800-A/800A)] / [1/k(900-A/900A)]

the 1/k cancelled out so I had

[(800-A)/800A] / [(900-A0/900A)]

which I used to find A

NOOOOOO, I thought you couldn't do it without k! I at least put them into simultaneous equations so that should give me a mark.

Reply 78

EatMoreThanPi

Original post by strongpar

tanx - 1 = 8tanx - 8tan^2(x)

8tan^2(x) - 7tanx - 1 = 0

(8tanx + 1)(tanx - 1) = 0

But itsnt it tanx + 1 on the LHS, so it didn't factorise :/

Reply 79

Jake2010

Ok Thanks WJEC. I am a teacher of 3 decades of experience. We spent all year on what I and my colleagues thought was preparing our students for their exams. I thought C3 was demanding but C4 was ridiculous. Elements of it would be more suited to AEA. Although I had nothing to do with the writing of the paper, I felt ashamed when I saw it. I know that I couldn't have prepared you all any better for your exams. I know that the papers (C3 and C4) have got harder elements recently to stretch A* students but this paper was in my opinion beyond the pale. I suspect many teachers feel the same way; they certainly do at my school. Keep your pecker up everyone, the grade boundaries will be much lower.