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# WJEC C4 17th June 2016 Watch

1. I found that decent. Some tough ones on there definitely.

I reckon ~55 for an A.

Binomial I got 1/10 for the 2nd bit.

The one with Tan and cot I got 56.9 degrees, 60 something and 1 hundred and something.

For the point P at (1,9) I got 9/2 and 3/2 as values for P

Vector value P I got 10.

The diff equation I messed up part c completely. That was tough.
2. (Original post by Jack1066)
I found that decent. Some tough ones on there definitely.

I reckon ~55 for an A.

Binomial I got 1/10 for the 2nd bit.

The one with Tan and cot I got 56.9 degrees, 60 something and 1 hundred and something.

For the point P at (1,9) I got 9/2 and 3/2 as values for P

Vector value P I got 10.

The diff equation I messed up part c completely. That was tough.
For diff eqns I had time t = 10years for part c.

I found the implicit differentiation very difficult, where you had to find the two coordinates where the tangent to the line =-2

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3. (Original post by Hydeman)
How did people find it?
I really enjoyed it! Especially compared to what I was expecting. The only question I couldn't do was the second part to the implicit differentiation (where the gradient was -2)
4. (Original post by Sznsnsn)
I thought it was good apart from the second part to question 3 and 4
Easier or harder than last year?
5. (Original post by Hydeman)
How did people find it?
I found it hard but I think I've done okay in general. I used the factor theorem to solve the tan x cubic question - I think it worked out okay
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6. (Original post by Jack1066)
I found that decent. Some tough ones on there definitely.

I reckon ~55 for an A.

Binomial I got 1/10 for the 2nd bit.

The one with Tan and cot I got 56.9 degrees, 60 something and 1 hundred and something.

For the point P at (1,9) I got 9/2 and 3/2 as values for P

Vector value P I got 10.

The diff equation I messed up part c completely. That was tough.
I got these! For part c I had the time to be 10 years
7. Preferred my 2nd choice uni anyway QQ
8. (Original post by DragonRider17)
I really enjoyed it! Especially compared to what I was expecting. The only question I couldn't do was the second part to the implicit differentiation (where the gradient was -2)
Yeah, that was tough. I ran out of time, but I reckon I'll get maybe two marks on that since I set dy/dx equal to - 2, worked out x^2 in terms of y and then attempted to enter that back into the equation for C.

I also couldn't do some parts of question 4 (I now realise that you had to solve for tan x not x, so I mucked that up completely thinking I was trying to factorise to get x instead of tan x).

For the last bit on the vectors question, did you get the dot product as non-zero, and therefore said that they weren't perpendicular?
9. (Original post by Hydeman)
Yeah, that was tough. I ran out of time, but I reckon I'll get maybe two marks on that since I set dy/dx equal to - 2, worked out x^2 in terms of y and then attempted to enter that back into the equation for C.

I also couldn't do some parts of question 4 (I now realise that you had to solve for tan x not x, so I mucked that up completely thinking I was trying to factorise to get x instead of tan x).

For the last bit on the vectors question, did you get the dot product as non-zero, and therefore said that they weren't perpendicular?
Yes I did! And all my friends who sat the exam also got them non-perpednicular, so I'm very sure that's right
10. (Original post by 98matt)
I found it hard but I think I've done okay in general. I used the factor theorem to solve the tan x cubic question - I think it worked out okay
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I'm kicking myself about that now. Perfectly good marks down the drain, all because I thought I was looking for x, not tanx...

Might get like 58/59 on that. Hopefully C3 boundaries are generous...
11. The -2 gradient spat out y=0 and x^2 = -4y^3.

You ignored the x value and plugged in y=0 into original equation, that gave you x^4= 16 which is plus/minus 2 so coordinates (-2,0) (2,0)

I think that was correct.

Agreed though, tough paper but hopefully the lower boundaries push my grades up.

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12. I'm probably alone in thinking that was harder/as hard as than last year's paper...

Did anyone get weird fractional coordinates for P in question 5? I got the two values of P as 9/2 and 3/2.
13. (Original post by Hydeman)
I'm probably alone in thinking that was harder/as hard as than last year's paper...

Did anyone get weird fractional coordinates for P in question 5? I got the two values of P as 9/2 and 3/2.
Yeah I got the same. Tbh apart from the proof by contradiction question the whole paper was harder than last year I think

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The -2 gradient spat out y=0 and x^2 = -4y^3.

You ignored the x value and plugged in y=0 into original equation, that gave you x^4= 16 which is plus/minus 2 so coordinates (-2,0) (2,0)

I think that was correct.

Agreed though, tough paper but hopefully the lower boundaries push my grades up.
Did you find it easier or harder than last year's paper? :3
15. My guesses for boundaries based on this:
100ums 64-65 / 75
80ums 54/75
70ums 48/75

I think it will be slighly higher than last year's simply because we had more hard past paper qs to prepare. Similar to the difficultly of June 2013 imo

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16. (Original post by Hydeman)
Did you find it easier or harder than last year's paper? :3
I'm double posting but yeah I do.

They threw in a lot of weird follow on questions and made things tough with the powers higher than 2 so factorising etc was a *****.

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Yeah I got the same. Tbh apart from the proof by contradiction question the whole paper was harder than last year I think

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I found the proof harder as well - I ended up with x^4 - 2x^2 + 1 and couldn't get much done beyond that. Oh well. Hoping for lower boundaries as well or the A* is as good as gone.
18. (Original post by 98matt)
My guesses for boundaries based on this:
100ums 64-65 / 75
80ums 54/75
70ums 48/75

I think it will be slighly higher than last year's simply because we had more hard past paper qs to prepare. Similar to the difficultly of June 2013 imo

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I think the harder questions last year weren't all that applicable to this exam. Hopefully that means the grade boundaries stay similar to last years.

I'd guess 63 for 100%

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I think the harder questions last year weren't all that applicable to this exam. Hopefully that means the grade boundaries stay similar to last years.

I'd guess 63 for 100%

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How do you think the 8 marks for part a) of the trig question would be split? Three marks for showing and then five for the rest?
20. (Original post by Hydeman)
I found the proof harder as well - I ended up with x^4 - 2x^2 + 1 and couldn't get much done beyond that. Oh well. Hoping for lower boundaries as well or the A* is as good as gone.
I got the same but you factorise it to get x^2(x^2 - 2) + 1 is less than zero so you just have to explain that all the values are positive so it must be greater than zero hence the contradiction

Can't remember exactly what I put but I probably lost a mark somewhere

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