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*Unofficial mark scheme* Edexcel C2 Maths 24th May 2017

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Reply 60
Oh no!!! Guess whos becoming a window cleaner after A levels, might marry a man with a level 1 in health and beauty as well >:frown: RAWR XD
Reply 61
Original post by Pedaly7
was one of the answers -3-root3 and -3+root3

i got that too
Original post by JamesHope890
When you root (3sinx -1), you get +-root2 as part of the other side. This gives two sets of two results.


I did that and I got x = arcsin(root 2)/3 + 1 and arcsin(-root 2)/3 + 1 but then one of the solutions was invalid??? Since was greater then 1?????
Original post by Strimpy
Agree with all of these except (what I assume) is your answer to 7i). The first log one

Pretty sure was 4a^3-a



Posted from TSR Mobile


I also got that m8
Anyone got working out unofficial mark scheme?
to factorise fully f(x) (the cubic equation) could you put a minus sign at the front with 3 sets of brackets or does it have to be (7-3x)
Original post by Strimpy
Agree with all of these except (what I assume) is your answer to 7i). The first log one

Pretty sure was 4a^3-a

I also got that m8

Posted from TSR Mobile


I also got that m8
Reply 67
For q10
I fin integration area from range -1/4 to 2
And then subtract area of triangle but got different answer why
Original post by Aceofall
Am I the only one who got -171585 for the sum?


Yes.
With k as 9/11 (given in b) ), you can work out the first value, a (or u1) by simply doing
7(9/11) - 5 = 8/11
ratio = -4 (use the equation you worked out for r in part a) )
therefore S10 = a(1-r^n)/1-r
= (8/11)(1-(-4)^10)/1-(-4)
= -152520
Reply 69
Original post by Brussell53
You don't find the unshaded area you do a tingle from x=1 to x=2 and calculate area of triangle and add it to the integrated area from x=1 to x=-0.25


What you got answer
Original post by Wolfram Alpha
How the hell did you get four solutions for the trig question??? I only got two and even then my answers were 31 degrees and 328 degrees???


same!
Original post by Lucas0207
Because u draw a sine graph to get 109.5


Surely doing 180-70.5 implies that there are only two angles in the triangle and is therefore wrong?

I think that 180-50-70.5 is the most sensible way to go about it.
Original post by Wolfram Alpha
How the hell did you get four solutions for the trig question??? I only got two and even then my answers were 31 degrees and 328 degrees???


I got two as well. 53.58, 122.42, 352.06 and 187.94
Reply 73
For question 4 the answer you have put is incorrect.

The constant 3 stays the same i.e. -3.

The term sqroot3 changes i.e. + sqroot3 / -sqroot3.

The answer should be sqroot3 - 3 / -sqroot3 - 3.

On another note, apparently that paper was easy. I got annihilated. So much for an A* prediction.

P.S. Anyone from RGS skipping games?
Reply 74
To be completely honest, the paper wasn't THAT hard...
Original post by reuv
i got that too


Me too m8
Reply 76
Original post by Bobrocket
You can't use Pythagoras, it wasn't a right-angled triangle.
You need to use cosine rule to work out the length BC, and then sine rule to find that sin(x) is approx. 0.987 (forgot exact value), and then you need to realise that a positive sin value can happen twice before 180 degrees, meaning that its either x or 180 - x.
Both are valid since (180 - x) + 50 < 180, meaning that the third angle WILL be positive.

EDIT:
By working out the length of BC as 12.554 (via cos. rule), you get
16*sin(50) / 12.554 = sin(x) which is approx. 0.9763
arcsin(0.9763) is approx 77.5, with the second solution being 180 - 77.5, which is 102.5

So the actual solutions for this q. are 77.5, 102.5 (both are valid)


cosine rule??? the triangle was ABC with AB=16 and AC=13 and the angle ABC was 50degrees how can you use the cosine rule if the two sides are not adjacent to the angle? isn't it just sine rule so sin(50)/13 *16=sin x ?
Original post by hey mate
to factorise fully f(x) (the cubic equation) could you put a minus sign at the front with 3 sets of brackets or does it have to be (7-3x)


It doesn't really matter, so long as it's factorised in to three linear functions and is correct (ie when multiplied out you get the unfactorised, original f(x)). Adding the - to the outside can actually carry error as you need to reverse the signs of the function as well as move it to the outside, not make it something like
(x+3)-(3x-7)(other one)
so in an exam it's probably best to leave it as you find it.
Original post by Specofranger
Heya everyone, did no one get for q2) x = 70.5 and 59.5,

Since sine rule and another angle is 180-50-70.5 = 59.5 only other possible angle. How did people get x = 109.5 out of 180 degrees??!


When you get x = 70.5 you use cast diagram to get the 180-70-5 = 109.5.

What you did, as some of my friends did too, is find the 3rd angle using that one possible value of x. It asked you to find 2 possible values of x, not one value of x and the value of the final angle.
Original post by Edgerulezs
You found the other angle in the triangle :frown: you were meant to find the 2 possible values of x so 180-70.5


Fairly certain this is wrong, if you do 180-70.5 then you are assuming that there are only two angles in the triangle which is impossible...

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