# C4 Edexcel A level 22nd June 2018 Unofficial Markscheme

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bindz

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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

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

Answer:

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

Answer: (3 decimal places)

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

Answer:

a ii) Hence find ,

Answer:

b) Find .

Answer:

c) Using the substitution , find

Answer:

a) . Find in terms of x and y.

Answer:

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

Answer: ,

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 .

Answer:

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

a) Find the value of k.

Answer: (note that

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

Answer:

Given that when , solve the differential equation

Answer:

a) Find

Answer :

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.

Answer:

-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

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 ,

Answer:

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

Answer: (3 decimal places)

**Q2:**a) i) Find the values of A, B and C.

Answer:

a ii) Hence find ,

Answer:

b) Find .

Answer:

c) Using the substitution , find

Answer:

**Q3:**a) . Find in terms of x and y.

Answer:

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

Answer: ,

**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 .

Answer:

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

a) Find the value of k.

Answer: (note that

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

Answer:

**Q6:**Given that when , solve the differential equation

Answer:

**Q8:**a) Find

Answer :

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.

Answer:

-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

Post your answers here and i’ll exit

**jaspreetb**)Post your answers here and i’ll exit

1 + sqrt 2 and 1 - sqrt2

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Keira Larkin

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#3

(Original post by

Post your answers here and i’ll exit

stationary points: 1+-sqrt2

rate of change of depth=8/3pi

cos(x)=4/sqrt21

**jaspreetb**)Post your answers here and i’ll exit

stationary points: 1+-sqrt2

rate of change of depth=8/3pi

cos(x)=4/sqrt21

What was your area?

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rex2000

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rex2000

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oscar gul

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#6

MiladA

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#7

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#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 > 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

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

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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#9

Zacks121

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#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

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

Zacks121

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eja1407

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#13

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#13

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

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

(Original post by

For the last question did anyone get pi squared / 64 + pi/8 or something similar?

**p29s**)For the last question did anyone get pi squared / 64 + pi/8 or something similar?

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p29s

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#15

(Original post by

think i got pi/32 + pi cubed/64 lol

**sdfsdd**)think i got pi/32 + pi cubed/64 lol

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DrDsy

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rex2000

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#18

(Original post by

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

**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

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MiladA

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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

(Original post by

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

**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

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