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C4 Edexcel A level 22nd June 2018 Unofficial Markscheme watch

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    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 ,

    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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    (Original post by jaspreetb)
    Post your answers here and i’ll exit
    Question 2:
    1 + sqrt 2 and 1 - sqrt2
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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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    that rate volume question had me for a bit i was so confused why they gave us 50
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    1/64pi(pi^2+4)
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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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    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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    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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    For the last question did anyone get pi squared / 64 + pi/8 or something similar?
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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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    How many marks was area of triangle?
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    3 I think
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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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    (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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    (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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    Why was I getting some weird value for binomial part b (x=-34) ??
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    What was Q coordinate?
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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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    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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    (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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