C4 Edexcel A level 22nd June 2018 Unofficial Markscheme

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bindz
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#1
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#1
Part Credz to my twin brother "chillichamps"


Post your answers here and i’ll edit
q1)if 1+3=6, then what is the integral of 1/4x+5x^1/3? rosemary cayenne pepper +c
Q1:

a) Use the bionmial series to find the expansion of Image, Image

Answer: Image

b) Use your answer to part (a) to estimate the value of Image. Give your answer to 3 decimal places.

Answer: Image (3 decimal places)

Q2:

a) i) Image Find the values of A, B and C.

Answer: Image

a ii) Hence find Image, Image

Answer: Image

b) Find Image.

Answer: Image

c) Using the substitution Image, find Image

Answer: Image

Q3:

a) Image. Find Image in terms of x and y.

Answer: Image

b) Hence find the x coordinates of the two points where Image

Answer: Image, Image

Q4:

a) Show that Image. We could use the triangle to say that Image, therefore Image.

b) Assume the rate of increase of the volume of water in the cone is Image. Find the rate of change of the depth of water, leaving your answer in terms of Image.

Answer: Image

Q5:
Image, Image. The point P (k, 2) is a point of the curve with these parametric equations.

a) Find the value of k.

Answer: Image (note that Image

b) Find the equation of the tangent at P,. Give your answer in the form Image, where p and q are exact real constants.

Answer: Image

Q6:

Given that Image when Image, solve the differential equation Image

Answer: Image

Q8:

a) Find Image

Answer : Image

A sketch of C, Image is shown above, as is the finite region R between Image and Image. This region is rotated through Image radians about the x-axis to form a solid of revolution.

b) Find the exact value of the volume of the solid generated.

Answer: Image


-stationary points: 1+-sqrt2
-rate of change of depth=8/3pi
-integral= 3/8lnu(5+4x^2/3)
-vol of revolution=0.68 (not exact i know) included pi^3 and pi
-binomial expansion=17.6- let x=0.1 and then times by 10

k=6-pi/2
-lambda=-8/5
cos(x)=4/sqrt21
area=12sqrt5
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Keira Larkin
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#2
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#2
(Original post by jaspreetb)
Post your answers here and i’ll exit
Question 2:
1 + sqrt 2 and 1 - sqrt2
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Keira Larkin
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(Original post by jaspreetb)
Post your answers here and i’ll exit
stationary points: 1+-sqrt2
rate of change of depth=8/3pi
cos(x)=4/sqrt21
I got the same cosx
What was your area?
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rex2000
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that rate volume question had me for a bit i was so confused why they gave us 50
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rex2000
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#5
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#5
1/64pi(pi^2+4)
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oscar gul
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#6
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i got 13+-root178/3 for y or the x (cant remember) co ords in qs 2
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MiladA
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#7
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Q(4,17/2,11/2) ?
Volume = (Pi/64)(Pi^2 + 4)
Binomial 2 - 9x/8 - 81x^2/64
Also x = 0.1 for part b
Y = 6/(2 - tan(2x)) differential equation
K = 6 - Pi/2 as K > 0 (t was equal to -Pi/2)
Area = 12root5
Integral sub was (3/8)*ln(4x^2/3 + 5)
Rates question was 3/(8*Pi)
1 +- root 2 for x coordinates for dy/dx =0
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iwouldn'tworry
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#8
might misremember some of these but I'll try:

1) 2 - 9/4 x -81/64 x^2

2) 1+/-sqrt2

vectors question:
B: (1,1,4)
area of triangle =12sqrt5
cos of angle = 4/sqrt21

definitely got 8/3pi for something

k=6-pi/2

might post again if I remember anything else
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p29s
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For the last question did anyone get pi squared / 64 + pi/8 or something similar?
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Zacks121
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1a) 2-9/4x-81/64x^2 b) 17.623
2a) (4-2x-y)/(x+2y-5) b) 1+√2 1-√2
3a) a=-2 b=6 c=1 b) ln(x+3/2x+1) - 3/2x+1 c) 1/3e^3x+3/2e^2x+3e^x+X d) 3/8ln(4x^3/2+5)
4a) show b) 8/3π
5a)k=(π+8)/2 b) y=4x-2(π-9)
6) y=6/(2+tan2x)
7a) (1,1,4) b) cosx=4√21/21 c) 12√5 d) r=(9,1,8)+k(4,-6,2) e) Q=(4,8.5,5.5)
8a) 1/4xsin4x+1/16cos4x b) π^3/64+π/16
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Keira Larkin
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#11
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#11
How many marks was area of triangle?
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Zacks121
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#12
3 I think
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eja1407
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Some of my answers:
Root 310: 17.623
Y= -6/(tan2x+1)
K=6-pi/2
Y= -4x -2pi - 26
Costheta = 4root21/21
Area = 12root5
Q(-1,16,3) - although friends disagree
Pi/16(Pi^2/4 +1) oe
U integral: 3/8 ln(4x^2/3+5) +c
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СУКА БЛЯТЬ
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#14
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(Original post by p29s)
For the last question did anyone get pi squared / 64 + pi/8 or something similar?
think i got pi/32 + pi cubed/64 lol
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p29s
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#15
(Original post by sdfsdd)
think i got pi/32 + pi cubed/64 lol
I might’ve got that tbh I just remember the 64, found that question so hard though aha
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DrDsy
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#16
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#16
Why was I getting some weird value for binomial part b (x=-34) ??
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Keira Larkin
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#17
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What was Q coordinate?
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rex2000
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(Original post by Zacks121)
1a) 2-9/4x-81/64x^2 b) 17.623
2a) (4-2x-y)/(x+2y-5) b) 1+√2 1-√2
3a) a=-2 b=6 c=-1 b) ln(x+3/2x+1) - 3/2x+1 c) 1/3e^3x+3/2e^2x+3e^x+X d) 3/8ln(4x^3/2+5)
4a) show b) 8/3π
5a)k=(π+8)/2 b) y=4x-2(π-9)
6) y=6/(2+tan2x)
7a) (1,1,4) b) cosx=4√21/21 c) 12√5 d) r=(9,1,8)+k(4,-6,2) e) Q=(4,8.5,5.5)
8a) 1/4xsin4x+1/16cos4x b) π^3/64+π/16
i agree with most but wasnt a=2 and c=1
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MiladA
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#19
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#19
Binomial expansion is approximate for small values of x. In this case |x| < 4/9 so -4/9 < x < 4/9. X was 0.1 btw
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Fogstaz
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#20
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#20
(Original post by eja1407)
Some of my answers:
Root 310: 17.623
Y= -6/(tan2x+1)
K=6-pi/2
Y= -4x -2pi - 26
Costheta = 4root21/21
Area = 12root5
Q(-1,16,3) - although friends disagree
Pi/16(Pi^2/4 +1) oe
U integral: 3/8 ln(4x^2+5) +c
Did you use a parallelogram for vectors? I got somewhat the same answer
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