# C4 Edexcel A level 22nd June 2018 Unofficial Markscheme

#1
Part Credz to my twin brother "chillichamps"

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 ,

Q2:

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

a ii) Hence find ,

b) Find .

c) Using the substitution , find

Q3:

a) . Find in terms of x and y.

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

Q4:

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

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

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

a) Find the value of k.

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

Q6:

Given that when , solve the differential equation

Q8:

a) Find

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

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

-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
14
3 years ago
#2
(Original post by jaspreetb)
Question 2:
1 + sqrt 2 and 1 - sqrt2
4
3 years ago
#3
(Original post by jaspreetb)
stationary points: 1+-sqrt2
rate of change of depth=8/3pi
cos(x)=4/sqrt21
I got the same cosx
0
3 years ago
#4
that rate volume question had me for a bit i was so confused why they gave us 50
7
3 years ago
#5
1/64pi(pi^2+4)
4
3 years ago
#6
i got 13+-root178/3 for y or the x (cant remember) co ords in qs 2
8
3 years ago
#7
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 &gt; 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
6
3 years ago
#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
5
3 years ago
#9
For the last question did anyone get pi squared / 64 + pi/8 or something similar?
6
3 years ago
#10
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
14
3 years ago
#11
How many marks was area of triangle?
1
3 years ago
#12
3 I think
2
3 years ago
#13
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
2
3 years ago
#14
(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
1
3 years ago
#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
1
3 years ago
#16
Why was I getting some weird value for binomial part b (x=-34) ??
11
3 years ago
#17
What was Q coordinate?
0
3 years ago
#18
(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
1
3 years ago
#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
2
3 years ago
#20
(Original post by eja1407)
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
2
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