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Lemur14
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Disclaimer: These are just my answers (well what I remember of them). Feel free to offer alternatives and additions to them (I will add them as soon as I can, but I have got further maths tomorrow to revise for ) I don't remember many of the marks, so feel free to add stuff!

1. Integrate \frac{2}{3}x^{3}-6x^{\frac{1}{2}}+1
Answer: \frac{1}{6}x^{4}-4x^{\frac{3}{2}}+x+c
(4 marks)
2a. Show that x^2-8x+17 is always positive
Answer (x-4)^2+1 has a vertex of (-4,1) which is >0. The coefficient is positive, so the curve opens upwards. Hence, it is always positive
b. A number plus three is squared. It is greater than the square of the original number:
-Sometimes?
-Always?
-Never?
Explain why
(n+3)^2>n^2
n^2+6n+9>n^2
6n+9>0
n>\frac{-3}{2}
Therefore, sometimes true.

3a Given A and B, find AB: A was (4i+5j) and B was (-5i-2j)
Find AB=OB-OA= -9i+3j
b) \sqrt{90}=3\sqrt{10}

4a. l1: 4y-3x=10
l2 goes through two points A(1,5) and B(-1,8)
Are these lines parallel, perpendicular or neither?
The gradients are neither equal or negative recipricols of each other, hence neither.

5a.The logs did not have the same coefficient so the substraction law could not be applied.The student did 3^2 rather than 2^3
b. Solve the log equation correctly to get 2^2=4

6a. £15 is not a sensible price, since the profit is negative (prove with P=100-6.25(15-9)^2=-125
b.80<100-6.25(x-9)^2
6.25(x-9)^2<20
(x-9)^2<3.2
-\sqrt{3.2}<x-9<\sqrt{3.2}
\frac{45-4\sqrt{5}}{5}<x<\frac{45+4\sqrt{  5}}{5}
Smallest value of x is £7.22
c.i) Maximum profit is when -6.25(x-9)^2=0
Therefore P=100
Profit= £100,000
ii)-6.25(x-9)^2=0
x-9=0
x=9
£9.00

7a. sinx=0.6
Therefore, cosx= \pm0.8
b. BC^2=5^2+10^2-2(5)(10)cosx
BC^2=125-100(-0.8)=205
BC=\sqrt{205}

8a. velocity for the minimum cost
I've written \frac{5\sqrt{330}}{11}?
b. Cost associated with that £243.16 *Pretty sure this is wrong, but it's what I've written down :dontknow:

9a. Show (x+2) was a factor of g(x)
b. Fully factorise it to (x+2)(2x-5)^2
c.i) x<=-2 OR x=\frac{5}{2}
ii)x=-1, x=\frac{4}{5}

10. Show from first principals that x^3 differentiates to 3x^2

11a. Binomial expansion of the first three terms in ascending terms of x (2-\frac{x}{16} )^9
2^9+9\times2^8\times\frac{-x}{16}+36\times2^7\times(\frac{-x}{16})^2
512-144x+18x^2
b. the first two terms of (a+bx)((2-\frac{x}{16})^9) are 128 and 36x.
Find a and b
128=512a
a=\frac{1}{4}
b=\frac{9}{64}

12a. Rearrange the trig equation
b. 16.1, 40, 80

13a q=10^{0.05}=1.122
p=10^{4.8}=63100(4sf)
b.P was the value of the painting in 1980. Q was the rate of change in the value of the painting
c. Value of painting 30 years later: £2,000,000.00

14a equation of circle: (x-3)^2+(y+5)^2=25
i) centre (3,-5)
ii) radius 5
b) Find the values of k when y=kx intersects the circle twice
Use discriminant by plugging y=kx into equation of circle
k&lt;0 OR k&gt;\frac{15}{8}

15a line equation: y=\frac{-1}{2}x+8
C=32x^{-2}+3x-8
Integrate C between 2 and 4=10
Area of triangle: \frac{1}{2}\times6\times12=36
10+36=46.(10 marks)

I've got a lesson now so I'll add more detail later if I still remember!
Last edited by Lemur14; 4 months ago
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RickHendricks
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(Original post by Lemur14)
Disclaimer: These are just my answers (well what I remember of them). Feel free to offer alternatives and additions to them (I will add them as soon as I can, but I have got further maths tomorrow to revise for ) I don't remember many of the marks, so feel free to add stuff!

1. Integrate \frac{2}{3}x^{3}-6x^{\frac{1}{2}}+1
Answer: \frac{1}{6}x^{4}-4x^{\frac{3}{2}}+x+c
(4 marks)
2a. Show that x^2-8x+17 is always positive
Answer (x-4)^2+1 has a vertex of (-4,1) which is >0. The coefficient is positive, so the curve opens upwards. Hence, it is always positive
b. A number plus three is squared. It is greater than the square of the original number:
-Sometimes?
-Always?
-Never?
Explain why
(n+3)^2&gt;n^2
n^2+6n+9&gt;n^2
6n+9&gt;0
n&gt;\frac{-3}{2}
Therefore, sometimes true.

3a Given A and B, find AB
Can't remember what A and B was, but I think AB was -9i+3j
b) \sqrt{90}=3\sqrt{10}

4a. l1: 4y-3x=10
l2 goes through two points A, and B
Are these lines parallel, perpendicular or neither?
I think I got neither, not certain

5a.??
b. 4

6a. £15 is not a sensible price, since the profit is negative (prove with P=100-6.25(15-9)^2=-125
b.80&lt;100-6.25(x-9)^2
6.25(x-9)^2&lt;20
(x-9)^2&lt;3.2
-\sqrt{3.2}&lt;x-9&lt;\sqrt{3.2}
\frac{45-4\sqrt{5}}{5}&lt;x&lt;\frac{45+4\sqrt{  5}}{5}
Smallest value of x is £7.22
c.i) Maximum profit is when -6.25(x-9)^2=0
Therefore P=100
Profit= £100,000
ii)-6.25(x-9)^2=0
x-9=0
x=9
£9.00

7a. sinx=0.6
Therefore, cosx= \pm0.8
b. BC^2=5^2+10^2-2(5)(10)cosx
BC^2=125-100(-0.8)=205
BC=\sqrt{205}

8a. velocity for the minimum cost
I've written \frac{5\sqrt{330}}{11}?
b. Cost associated with that £243.16 *Pretty sure this is wrong, but it's what I've written down :dontknow:

9a. Show (x+2) was a factor of g(x)
b. Fully factorise it to (x+2)(2x-5)^2
c.i) x<-2 OR x&gt;\frac{5}{2}
ii)x=-4 OR x=5

10. Show from first principals that x^3 differentiates to 3x^2[tex] 

 

11a. Binomial expansion of the first three terms in ascending terms of x [tex](2-\frac{x}{16})^9
2^9+9\times2^8\times\frac{-x}{16}+36\times2^7\times(\frac{-x}{16})^2
512-144x+18x^2
b. the first two terms of (a+bx)((2-\frac{x}{16})^9) are 128 and 36x.
Find a and b
128=512a
a=\frac{1}{4}
b=\frac{9}{64}

12a. Rearrange the trig equation
b. 16.1, 40, 80

13a q=10^0.05=1.122
p=10^4.8=63100(4sf)
b.??
c. Value of painting 30 years later: £1994000.00

14a equation of circle: (x-3)^2+(y+5)^2=25
i) centre (3,-5)
ii) radius 5
b) Find the values of k when y=kx intersects the circle twice
Use discriminant by plugging y=kx into equation of circle
k&lt;0 OR k&gt;\frac{39}{16} (Not sure about this one)

15a line equation: y=\frac{-1}{2}x+8
C=32x^{-2}+3x-8
Integrate C between 2 and 4=10
Area of triangle: \frac{1}{2}\times6\times12=36
10+36=46.(10 marks)

I've got a lesson now so I'll add more detail later if I still remember!
q 9(cii) was actually 4/5 and -1, since it was 2 inside the bracket, meaning a compression. You've said it's an extension.

I even put x=2x into it, and saw the roots that I go, and it was correct.

question 13c was 2,000,000 as it asked for the nearest hundred thousand, not thousand.

the logs was: P was the price of the art work in 1980, and q confused me. I just said it's the rate of change.

question 14b, the other end of the tai was 15/8
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Lemur14
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(Original post by RickHendricks)
q 9(cii) was actually 4/5 and -1, since it was 2 inside the bracket, meaning a compression. You've said it's an extension.

I even put x=2x into it, and saw the roots that I go, and it was correct.

question 13c was 2,000,000 as it asked for the nearest hundred thousand, not thousand.

the logs was: P was the price of the art work in 1980, and q confused me. I just said it's the rate of change.

question 14b, the other end of the tai was 15/8
Thanks for these
I'll add them when I get home! (currently in a further mechanics lesson so probably wouldn't be popular if I started this :lol: This is totally me looking at the mark scheme for the questions we're doing

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flowey
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Alternative answer for 2)a) is to complete the square to get (x-4)^2 +1 >0. Since the x-4 is squared, even if x is less than 4, squaring it will make it positive hence always >0.

4)a) the points were (1,5) (-1,8) i think, giving u a gradient of -3/2 which =/= 3/4 hence not parallel.
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Adnan21
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What do you expect the boundaries will be?
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RickHendricks
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(Original post by flowey)
Alternative answer for 2)a) is to complete the square to get (x-4)^2 +1 >0. Since the x-4 is squared, even if x is less than 4, squaring it will make it positive hence always >0.

4)a) the points were (1,5) (-1,8) i think, giving u a gradient of -3/2 which =/= 3/4 hence not parallel.
it was neither.
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Bradley00
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Question 3A - A was (4i+5j) and B was (-5i-2j) if my memory serves me correct
Question 4 - I got nether as the gradient of Line 1 was neither the same or the negative reciprocal of Line 2's gradient
Question 4 - I am quite sure that A was (5, -1) and B was (-1,8)
Question 8 - I think the equation was C=1500/v + 2v/11 + 60
Question 12 - The Trig equation was 4Cosx - 1 = 2(sinx)(tanx) Where x represents theta

Hopefully we can figure out question 5 now
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nyxnko_
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AHHHHHH!!! I got most of them :five:

PS. for 9cii, I'm sure it was g(2x) so x=-1 and x=5/4...
Last edited by nyxnko_; 5 months ago
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Ismaeel Ali
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Thank you so much for this, ive probably dropped like 25-40 marks, 25 if im giving method marks but 40 if im being strict, 60/100, hopefully i can get a C

Im pretty happy with myself cant lie : D
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lewisk2519
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(Original post by Lemur14)
Disclaimer: These are just my answers (well what I remember of them). Feel free to offer alternatives and additions to them (I will add them as soon as I can, but I have got further maths tomorrow to revise for ) I don't remember many of the marks, so feel free to add stuff!

1. Integrate \frac{2}{3}x^{3}-6x^{\frac{1}{2}}+1
Answer: \frac{1}{6}x^{4}-4x^{\frac{3}{2}}+x+c
(4 marks)
2a. Show that x^2-8x+17 is always positive
Answer (x-4)^2+1 has a vertex of (-4,1) which is >0. The coefficient is positive, so the curve opens upwards. Hence, it is always positive
b. A number plus three is squared. It is greater than the square of the original number:
-Sometimes?
-Always?
-Never?
Explain why
(n+3)^2&gt;n^2
n^2+6n+9&gt;n^2
6n+9&gt;0
n&gt;\frac{-3}{2}
Therefore, sometimes true.

3a Given A and B, find AB
Can't remember what A and B was, but I think AB was -9i+3j
b) \sqrt{90}=3\sqrt{10}

4a. l1: 4y-3x=10
l2 goes through two points A, and B
Are these lines parallel, perpendicular or neither?
I think I got neither, not certain

5a.??
b. 4

6a. £15 is not a sensible price, since the profit is negative (prove with P=100-6.25(15-9)^2=-125
b.80&lt;100-6.25(x-9)^2
6.25(x-9)^2&lt;20
(x-9)^2&lt;3.2
-\sqrt{3.2}&lt;x-9&lt;\sqrt{3.2}
\frac{45-4\sqrt{5}}{5}&lt;x&lt;\frac{45+4\sqrt{  5}}{5}
Smallest value of x is £7.22
c.i) Maximum profit is when -6.25(x-9)^2=0
Therefore P=100
Profit= £100,000
ii)-6.25(x-9)^2=0
x-9=0
x=9
£9.00

7a. sinx=0.6
Therefore, cosx= \pm0.8
b. BC^2=5^2+10^2-2(5)(10)cosx
BC^2=125-100(-0.8)=205
BC=\sqrt{205}

8a. velocity for the minimum cost
I've written \frac{5\sqrt{330}}{11}?
b. Cost associated with that £243.16 *Pretty sure this is wrong, but it's what I've written down :dontknow:

9a. Show (x+2) was a factor of g(x)
b. Fully factorise it to (x+2)(2x-5)^2
c.i) x<-2 OR x&gt;\frac{5}{2}
ii)x=-4 OR x=5

10. Show from first principals that x^3 differentiates to 3x^2[tex]



11a. Binomial expansion of the first three terms in ascending terms of x [tex](2-\frac{x}{16})^9
2^9+9\times2^8\times\frac{-x}{16}+36\times2^7\times(\frac{-x}{16})^2
512-144x+18x^2
b. the first two terms of (a+bx)((2-\frac{x}{16})^9) are 128 and 36x.
Find a and b
128=512a
a=\frac{1}{4}
b=\frac{9}{64}

12a. Rearrange the trig equation
b. 16.1, 40, 80

13a q=10^0.05=1.122
p=10^4.8=63100(4sf)
b.??
c. Value of painting 30 years later: £1994000.00

14a equation of circle: (x-3)^2+(y+5)^2=25
i) centre (3,-5)
ii) radius 5
b) Find the values of k when y=kx intersects the circle twice
Use discriminant by plugging y=kx into equation of circle
k&lt;0 OR k&gt;\frac{39}{16} (Not sure about this one)

15a line equation: y=\frac{-1}{2}x+8
C=32x^{-2}+3x-8
Integrate C between 2 and 4=10
Area of triangle: \frac{1}{2}\times6\times12=36
10+36=46.(10 marks)

I've got a lesson now so I'll add more detail later if I still remember!
Hopefully a lot of these are not right if not got about 10/100
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nyxnko_
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(Original post by lewisk2519)
Hopefully a lot of these are not right if not got about 10/100
I've sure you've got more than 10 :yep: You can still get method marks even if you don't get the right answer.
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haxhacks
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can someone help me with the trig identity one? i wrote a tonne but couldn't get the final answer (both parts)
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haxhacks
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(Original post by haxhacks)
can someone help me with the trig identity one? i wrote a tonne but couldn't get the final answer (both parts)

6cos^2 +costheta - 2 = sintheta x costhea

(I think)
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nyxnko_
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(Original post by haxhacks)
can someone help me with the trig identity one? i wrote a tonne but couldn't get the final answer (both parts)
sure. was it the 4cos^2\theta - cos\theta = 2sin\theta*tan\theta one?
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TarekHelmy
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final answer i think was around 16,1,40,80
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haxhacks
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(Original post by nyxnko_)
sure. was it the 4cos^2\theta - cos\theta = 2sin\theta*tan\theta one?
yeah wasnt the coefficient of cos^2thea 6? And how do you write like that
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nyxnko_
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(Original post by haxhacks)
yeah wasnt the coefficient of cos^2thea 6? And how do you write like that
I think the final answer was 6cos^2\theta - cos\theta - 2 = 0 but I can't remember the original q If you can remember it, feel free to send me a PM and I'll try and help you work through it.
It's called LaTex (pronouned La-tec) and it's just a way to write mathematical formulae on TSR.
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haxhacks
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(Original post by nyxnko_)
I think the final answer was 6cos^2\theta - cos\theta - 2 = 0 but I can't remember the original q If you can remember it, feel free to send me a PM and I'll try and help you work through it.
It's called LaTex (pronouned La-tec) and it's just a way to write mathematical formulae on TSR.
Thanks babes^10
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Anas Sheikh
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Anyone remember question 5???

Btw for question 8 i got 90. Something in speed and 93. Something for price.... I know at least two others who got this as well
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haxhacks
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(Original post by nyxnko_)
I think the final answer was 6cos^2\theta - cos\theta - 2 = 0 but I can't remember the original q If you can remember it, feel free to send me a PM and I'll try and help you work through it.
It's called LaTex (pronouned La-tec) and it's just a way to write mathematical formulae on TSR.
(Original post by haxhacks)
Thanks babes^10
I think it was a proof question but oh well
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