C2 Maths AS aqa 2016 (unofficial mark scheme new)Watch

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2 years ago
#281
Anybody, for the sector question i thought you could write it as 19.9 = 2 multiplide by area of sector? isnt this right im sure it is, since the sector multiplied would get you area of triangle?
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2 years ago
#282
(Original post by beanigger)
UPDATED THE MARK SCHEME SORRY FOR THE MISTAKES
i know the marks dont add up to 75 but i cant recall all the correct marks for each question
if there is anything else wrong please let me know
thanks
hey where the updated version ? and well mistakes ok cant expect to b perfect did better than i would have
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2 years ago
#283
(Original post by Alexgood123)
I believe that question 5a) said to expand the binomial equation and show it in its simplest form which cant you divide the expansion in your answer by 8 to show it in its simplest form.
It just said expand this in the form " " with p q and r being integers or something.
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2 years ago
#284
(Original post by SunDun111)
Anybody, for the sector question i thought you could write it as 19.9 = 2 multiplide by area of sector? isnt this right im sure it is, since the sector multiplied would get you area of triangle?
Thats correct; because it is area of sector=area of triangle-area of sector. So you rearrange and get area of sector x 2=area of triangle. theta x r^2=19.9
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2 years ago
#285
(Original post by Parhomus)
Thats correct; because it is area of sector=area of triangle-area of sector. So you rearrange and get area of sector x 2=area of triangle. theta x r^2=19.9
I didnt show the rearrangement in my answer though because of time restrictions but thanks
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2 years ago
#286
(Original post by SunDun111)
I didnt show the rearrangement in my answer though because of time restrictions but thanks
Yeah, I considered doing it but I wasn't sure, so I showed it anyways. I think you get the marks for the working regardless because it is implied.
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2 years ago
#287
(Original post by Porkieee.ee)
18/2(56+17(-2))
9(56-34)
9(22)
=198
No, if you are going to do it that way, your first term is u4, which is 22, not 28. If you use this, you get the answer 90, which is the same as above.
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2 years ago
#288
(Original post by beanigger)
Unoffical Mark scheme for C2 AQA 2016

It would help if you could correct the amount of marks in each question if it is wrong please

Questions:

1)a) integrate something to get -36x-1 + ax2 /2 [3]
b) find the value of a when the answer to 1)a) is equal to 16 with limits of 3 to 1, a = - 2 [2]

2)a) Draw the graph of (0.2)x and state where itcrosses axis (0,1)[2]
b) use logarithms to solve (0.2)x = 4 , x = -0.861 [2]
c) describe transformation of (0.2)x onto (5)x reflection in y axis [1]

3)a) differentiate something to get 3x-1/2 -1 [2]
b) find y co-ordinate of the maximum point M M(9,6) [3]
c) equation of normal to curve at P, P was (4,5) y=-2x+13 [2]
d) the normal to curve at P is translated by (k,0) find value of k, k = 5.5 [3]

4)a) show that a+10d=8 [2]
b)sum to 2nd and 3rd term is 50, work out the 12th term of the series D=-2 and A=28, u12 = 6 [3]
c) work out the sum of the 21st term minus sum of 3rd term = 90 [3]

5)a) show that thetha (in radians) was 0.568 [2]
b) work out area of triangle 19.9 [2]
c) work out perimeter of shaded area, first piece of info given -> area of sector = area of shaded area

you had to realise that the area of sector = area of triangle - area of sector
re-arrange to get 2(area of sector) = 19.9 and solve for r to get 5.829
perimeter of shaded = (9-r) + (8-r) + 5 + (length of arc = r*theta)
perimeter of shaded = 13.8 [6]

6)a)trapezium rule to get 65.6 [4]
b) describe transformation of graph y=(x2 +9)1/2 onto y = 5 + (x2 +9)1/2
Translation by (0, 5) [2]
c) describe transformation of graph y=(x2 +9)1/2 onto y=3(x2 +1)1/2 Stretch by scale factor 1/3 in x direction [2]

7)a) expand (1+2x)5 work out p q r, i think p = -10, q= 40 r = - 80 (correct me if im wrong) [4]
b)find the coefficient of x10 in the expansion of (1+2x)^5 * (2+x)^7
it was something like -1648 [5]

8)a)find the value of tan(x) tan(x) = -5/4 [2]
b) values of tanx in the range of 0 to 360 degrees 45, 129, 225, 309 [3]
b) re-arrange 16+9sin2x / 5 - 3cosx into the form p + qcosx and find the min value = 2, this is when x is equal to pi [4]

9)re-arrange something into the form (c)1/2 / d^2 to get 3y y = 1/2 ( m - 12n) [4]
b) find the one and only solution of something, x = 5/2 ( you should get two equal roots i think) [4]
3c was 3 marks, not 2
Q4 a,b,c was 3, 4 and 3 marks
Q5 and Q6 are the wrong way around - Q5 is the trapezium rule a4, b2, c2 marks and Q6 the triangle question - a3, b2,c6 marks
Q7 a, b are 3 and 5 marks

The rest of the marks are correct
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2 years ago
#289
(Original post by SunDun111)
Anybody, for the sector question i thought you could write it as 19.9 = 2 multiplide by area of sector? isnt this right im sure it is, since the sector multiplied would get you area of triangle?
Yeah that'd work, as it's just area of sector = area of triangle / 2 rearranged
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2 years ago
#290
(Original post by Alexgood123)
I believe that question 5a) said to expand the binomial equation and show it in its simplest form which cant you divide the expansion in your answer by 8 to show it in its simplest form.
It wasn't an equation, just the expansion of (1-2x)^5. It didn't ask for it in its simplest form, but as it's not an equation, you can't divide by 8 anyway
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2 years ago
#291
(Original post by Eisobdxhsonw)
For question 6 part C I got stretch in the X by a scale factor of 1/9

Ill show you how I got it
First I rearranged the y=(x^2+9)^1/2 to make it easier to compare

Y=(x^2+9)^1/2
Y=(9(x^2/9 +1))^1/2
Y=3(x^2/9+1)^1/2

Comparing that with y=3(x^2+1)^1/2 you would have to stretch it in the X by a scale factor of 1/9???

I know this doesn't look too nice with all the powers and divides but if you write it down it makes more sense

I was just wondering if anyone could confirm my method or tell me where I went wrong as I believe this could have been one of those transformations with multiple answers??
I might be wrong, but aren't you doing this the wrong way around? We want to transform (x^2 +9)^1/2 into 3(x^2+1)^1/2. For it to become 3(x^2+1)^1/2, it must first become (9(x^2 + 1)^1/2 = (9x^2+9)^1/2 = ((3x)^2+9)^1/2. This means x has transformed into 3x, so it is a stretch parallel to the x axis, sf 1/3)
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2 years ago
#292
(Original post by Bosssman)
Yeah that'd work, as it's just area of sector = area of triangle / 2 rearranged
kinda nervous cos i didnt right that statement down tho, but i did show it in my working i think
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Thread starter 2 years ago
#293
(Original post by smithj)
3c was 3 marks, not 2
Q4 a,b,c was 3, 4 and 3 marks
Q5 and Q6 are the wrong way around - Q5 is the trapezium rule a4, b2, c2 marks and Q6 the triangle question - a3, b2,c6 marks
Q7 a, b are 3 and 5 marks

The rest of the marks are correct
thanks, its been updated
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2 years ago
#294
(Original post by beanigger)
Unoffical Mark scheme for C2 AQA 2016

7)a) expand (1+2x)5 work out p q r, i think p = -10, q= 40 r = - 80 (correct me if im wrong) [3]
b)find the coefficient of x10 in the expansion of (1+2x)^5 * (2+x)^7
it was something like -1648 [5]
Not sure if I read it wrong but I thought it was (2-x)^7 rather than positive
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Thread starter 2 years ago
#295
(Original post by KiranG23)
Not sure if I read it wrong but I thought it was (2-x)^7 rather than positive
yeah ur right my bad ill change it
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2 years ago
#296
If i'm lucky i got 63, but probably go a less so i'm worried I wont get an A, what are the grade boundary predictions? Hopefully i can pull up my grade with stats but thats usually my worst one!
0
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