The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Edexcel A2 C4 Mathematics June 2016 - Official Thread

Scroll to see replies

Original post by Supermanxxxxxx
Screen Shot 2016-06-23 at 14.41.06.png
Attachment not found


can someone explain how your meant to know to put an a in front as didnt think you had to


Well we can't perform partial fractions straight away since we require the degree of the numerator to be less than the degree of the denominator. Here both degrees are 2, so we will get that constant A A out the front!

So first you'll need to expand the bottom and do some algebraic division before doing partial fractions as follows:

[br]9x2+20x−10=3(x+2)(3x−1)+5x−4[br] [br]9x^2 + 20x - 10 = 3(x+2)(3x-1) + 5x - 4[br]

So A=3 A = 3
(edited 7 years ago)
Original post by Euclidean
V=π∫ab(y12−y22)dx \displaystyle V = \pi\int_a^b{(y_1^2-y_2^2)}dx

This integral should work. I'd always do it separately though.


Thanks. Just realised I can't read 'about y' :facepalm:
What is the hardest C4 paper?
Original post by Polyrogue
Do we have to know how to integrate a^x like we do for differentiating? Thanks!


Well if you can differentiate it you can automatically integrate it!

[br]∫ax dx=1ln⁡(a)ax+c[br][br]\int a^x \, \mathrm d x = \frac1{\ln(a)} a^x + c[br]
Original post by TercioOfParma
What is the hardest C4 paper?


The C4 paper with the lowest grade boundaries since 2009 is June 2013 (R), but there isn't really one that stands out, like June 2013 for C3.
Original post by TercioOfParma
What is the hardest C4 paper?


The hardest C4 paper that I've done is probably the January 2014 IAL C4 paper.
Original post by kdevitto


LOL i remember doing this question in my real exam (I did A-Levels in 2014) and just not having a clue for this one.

Basically you know that the integral of 1/cosec2ycosecy = integral of sin2ysiny (as cosec(Y) = 1/sin(Y))

So you can use the double angle formulae (Sin2Y = 2SinYCosY) and rearrange to get the integral of 2Sin^2YCosY. -----> I don't know how to use LATEX form but I'm not saying "2 sin squared 2Y" I'm saying "2 sin squared Y" just to be clear

Proceed from there. Think about how the Cos and Sin are related, and you'll know what the best integration method is.
(edited 7 years ago)
Original post by TercioOfParma
What is the hardest C4 paper?


Since the spec changed in 2009, the UK paper with the lowest boundaries is June 2014.

But the January 2014 IAL paper is probably the hardest one. June 14 for the most wasn't hard, just had 1-2 exceptionally nasty questions worth a lot of marks.
Why is the 90UMS boundary so much higher for C4 than C3. C4 is so much harder and takes longer, one silly slip up and you can lose 7 marks and drop to an A?

EDIT: do you need over 90UMS in C3 and C4 for and A*, or an average, or what ?
Original post by ccharlie97
Not really - is it to do with volume of revolution? Because that has y2 y^2 as the integrand so it wouldn't matter whether the function lied below or above the curve!

What is this left/right question you refer to?


So lets just look at this question

I have to find the area of the shaded region, so I integrate to find it. Why does the area underneath the x-axis not cancel with the area above the x-axis when I integrate?
Original post by ccharlie97
That won't work here as it's about a different axis of rotation!


I thought the OP was talking about the general case of a revolution around a co-ordinate axis, I was mistaken :tongue:
Original post by jfahdgjdfshg
Why is the 90UMS boundary so much higher for C4 than C3. C4 is so much harder and takes longer, one silly slip up and you can lose 7 marks and drop to an A?

EDIT: do you need over 90UMS in C3 and C4 for and A*, or an average, or what ?

On average between the two
thanks - but can you answer why the boundary for C4 is so muchigher?
Original post by jfahdgjdfshg
Why is the 90UMS boundary so much higher for C4 than C3. C4 is so much harder and takes longer, one silly slip up and you can lose 7 marks and drop to an A?

EDIT: do you need over 90UMS in C3 and C4 for and A*, or an average, or what ?


Original post by jfahdgjdfshg
thanks - but can you answer why the boundary for C4 is so muchigher?


C4 is somewhat easier.
Really? Maybe I havent done enough practice but my whole class is struggling much more with C4, and all the questions involve loads of algebra
Original post by jfahdgjdfshg
Really? Maybe I havent done enough practice but my whole class is struggling much more with C4, and all the questions involve loads of algebra


Each to their own I guess. I think nationally C3 tends to have a lower mark distribution.
Original post by ccharlie97
Well we can't perform partial fractions straight away since we require the degree of the numerator to be less than the degree of the denominator. Here both degrees are 2, so we will get that constant A A out the front!

So first you'll need to expand the bottom and do some algebraic division before doing partial fractions as follows:

[br]9x2+20x−10=3(x+2)(3x−1)+5x−4[br] [br]9x^2 + 20x - 10 = 3(x+2)(3x-1) + 5x - 4[br]

So A=3 A = 3


Can you tell me how you can easily spot whether you have to use long division or not? I know that if the degrees were unequal (for example x^4 on the top and x^3 on the bottom) you'd have to use it, but in this case the order was 2 for both the numerator and denominator surely?
Original post by Music With Rocks
So lets just look at this question

I have to find the area of the shaded region, so I integrate to find it. Why does the area underneath the x-axis not cancel with the area above the x-axis when I integrate?


It will cancel - what is the question? If you tell me or give me the curve I'll be able to understand your question properly :smile:
kk ty for replying - is it true you need an average of 90UMS between C3 and C4 for the A*?
Original post by Don Pedro K.
Can you tell me how you can easily spot whether you have to use long division or not? I know that if the degrees were unequal (for example x^4 on the top and x^3 on the bottom) you'd have to use it, but in this case the order was 2 for both the numerator and denominator surely?


Think about it like this:

If you do partial fractions (after you multiply by the denominator), you need to have at least the degree of the LHS on the RHS. In other words, you can't have a LHS of degree 3 and a RHS of degree < 3.

If you have equal degrees, you won't be able to get that degree on the RHS unless you long divide first or add a constant to the end of your unknown PF terms.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you