# Pick a question.Watch

This discussion is closed.
#1
Hey, I haven't got the markscheme to this, and I plan to do it later, (maybe around 10.00pm), so just incase I get stuck, could people pick a question and answer it (thorough workings out!)? (I'm afraid that you lot won't be online by the time I come back from the hospital visiting my brother). When answering a question, please let me know which, the question number is the name of each file. Thanks!
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#2
Question 6 is here, it's in two parts.
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#3
Erm......... hello? people?
0
13 years ago
#4
(Original post by [email protected])
Erm......... hello? people?
hello.
I'm not onto that maths yet, sorry.
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13 years ago
#5
for number 6) this is what i got:

y=x^2sin(1/2x)

u=x^2 and du/dx=2x
dv/dx=sin(1/2x) and v=-2cos(1/2x)

-2x^2cos(1/2x) + INT 4xcos(1/2x)

u=4x du/dx=4
dv/dx=cos(1/2x) v=2sin(1/2x)

8xsin(1/2x) - INT 8sin91/2x)
8xsin(1/2x) + 16cos(1/2x)

integral of y= -2x^2cos(1/2x) + 8xsin(1/2x) + 16cos(1/2x)
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13 years ago
#6
Did you know that (your) attachment 1 and attachment 2 are the same file/question ?

Q3)
l1: r1 = (11,5,6) + λ(4,2,4) = (11+4λ, 5+2λ, 6+4λ)
l2: r2 = (24,4,13) + μ(7,1,5) = (24+7μ, 4+μ, 13+5μ)

a)
let P be a point on r1 for some value of λ
let Q be a point on r2 for some value of μ

Assume P = Q, then
(11+4λ, 5+2λ, 6+4λ) = (24+7μ, 4+μ, 13+5μ)

equating coeffts of i,j,k,

11+4λ = 24+7μ
5+2λ = 4+μ
6+4λ = 13+5μ

which gives λ = -2, μ = -3
==================
There is a valid solution, of real values for λ and μ, hence P = Q.

Therfefore the lines l1 and l2 intersect
=========================

b)
P = (11+4λ, 5+2λ, 6+4λ)
P = (11 - 8, 5 - 4, 6 - 8)
P = (3, 1, -2)
=========

c)
let V1 = (4,2,4)
let V2 = (7,1,5)

V1.V2 = 28 + 2 + 20 = 50
|V1| = √(16 + 4 + 16) = √36 = 6
|V2| = √(49 + 1 + 25) = √75 = 5√3

cosθ = V1.V2 / (|V1||V2|)
cosθ = 50 / (6*5√3) = 10√3/(6*3)
cosθ = (5/9)√3
===========
0
13 years ago
#7
Q4)
x = cost, y = sin(2t), 0 <= t < 2pi

a)
dx/dt = -sint, dy/dt = 2cos(2t)
dy/dx = dy/dt / dx/dt
dy/dx = (2cos(2t)) / (-sint)
dy/dx = -2cos(2t)/sint
=================

b)
when dy/dx = 0,
cos(2t) = 0
2t = pi/2, 3pi/2, 5pi/2, 7pi/2
t = pi/4, 3pi/4, 5pi/4, 7pi.4
====================

c)
x = cost, y = sin(2t)
y = 2sint.cost
y = 2√(1 - sin²t).cost
y = 2√(1 - x²).x
y = 2x√(1 - x²)
===========

d)
for pi <= t < 2pi

y = -2x√(1 - x²)
============
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13 years ago
#8
Q1)

a) A(1+2x) + B(1-x) = 1+14x

x=1, 3A = 15, A=5

x=-1/2, 3/2B = -6, B=-4

5/(1-x) - 4/(1+2x)
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#9
Guys, question 5, I can't do part D and E. Rep will be given one day at a time
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13 years ago
#10
1.) a.) f(x) = (1 + 14x)/[(1 - x)(1 + 2x)] = [A(1 + 2x) + B(1 - x)]/[1 - x)(1 + 2x)]

-> A(1 + 2x) + B(1 - x) = 1 + 14x
x = 1: 3A = 15 -> A = 5
x = -1/2: 3B/2 = -6 -> B = -4

-> f(x) = 5/(1 - x) - 4/(1 + 2x)

b.) Int. (1/3 to 1/6) f(x) dx. = Int. (1/3 to 1/6) 5/(1 - x) - 4(1 + 2x) dx. = [5ln|1 - x| - 2ln|1 + 2x|] (Limits 1/3 & 1/6) = [5ln|2/3| - 2ln|5/3|] - [5ln|5/6| - 2ln|4/3|] = 5[ln|2/3| - ln|5/6|] + 2[ln|4/3| - ln|5/3|] = 7ln(4/5) = ln(16384/78125)

c.) f(x) = (1 + 14x)(1 - x)^(-1).(1 + 2x)(-1) = (1 + 14x)[1 + x + x^2 + x^3][1 - 2x + 4x^2 - 8x^3] = (1 + 15x + 15x^2 + 15x^3...)(1 - 2x + 4x^2 - 8x^3...) = 1 - 2x + 4x^2 - 8x^3 + 15x - 30x^2 + 60x^3 + 15x^2 - 30x^3 + 15x^3... = 1 + 13x - 11x^2 + 37x^3...

2.) L1: r = [11, 5 , 6] + Lamba[4, 2, 4]
L2: [24, 4, 13] + mu[7, 1, 5]

a.) At POI: [11 + 4Lambda, 5 + 2Lambda, 6 + 4Lambda] = [24 + 7mu, 4 + mu, 13 + 5mu]

-> 11 + 4Lambda = 24 + 7mu -> Lambda = (13 + 7mu)/4 * (1)
-> 5 + 2Lambda = 4 + mu -> Lambda = (mu - 1)/2 * (2)
-> 6 + 4Lambda = 13 + 5mu -> 4Lambda = 7 + 5mu -> Lambda = (7 + 5mu)/4

Equate (1) and (2): (13 + 7mu)/4 = (mu - 1)/2 -> 26 + 14mu = 4mu - 4 -> 10mu = -30 -> mu = -3

Lamda = (7 - 15)/4 = -8/4 = -2

Check solutions with (3): -2 = (7 - 15)/4 = -2 -> True.

-> Equations are consistent -> POI exists.

b.) At POI, Lambda = -2:
-> POI = [11, 5, 6] + (-2)[4, 2, 4] = [11, 5, 6] + [-8, -4, -8] = [3, 1, -2]
-> POI: (3, 1 , -2).

c.) Let direction vector of L1 be: d1 = [4, 2, 4]
Let direction vector of L2 be: d2 = [7, 1, 5]

d1.d2 = 28 + 2 + 20 = 50

50 = |d1|*|d2|*cos(theta)
-> 50 = Sqrt[(4^2 + 2^2 + 4^2)(7^2 + 1^2 + 5^2)]*cos(theta)
-> 50 = Sqrt(2700)*cos(theta) = 30Sqrt(3)*cos(theta)
-> cos(theta) = 50/[30Sqrt(3)]
-> cos(theta) = 50Sqrt(3)/90
-> cos(theta) = 5Sqrt(3)/9
-> k = 5/9

3.) a.) x = cost, y = sin2t.

dx/dt = -sint, dy/dt = 2cos2t

dy/dx = 2cos2t * (-1)/sint = (-2cos2t)/sint

b.) dy/dx = 0
-> -2cos2t = 0
-> cos2t = 0
-> 2t = Pi/2, 3Pi/2, 5Pi/2, 7Pi/2
-> t = Pi/4, 3Pi/4, 5Pi/4, 7Pi/4

c.) When tangent is parallel to x-axis -> dy/dx = 0.
-> Hence at such points: t = Pi/4, 3Pi/4, 5Pi/4, 7Pi/4.

t = Pi/4: x = Sqrt(2)/2, y = 1 -> Point: [Sqrt(2)/2 , 1]
t = 3Pi/4: x = -Sqrt(2)/2, y = -1 -> Point: [-Sqrt(2)/2, -1]
t = 5Pi/4: x = -Sqrt(2)/2, y = 1 -> Point: [-Sqrt(2)/2, 1]
t = 7Pi/4: x = Sqrt(2)/2, y = -1 -> Point: [Sqrt(2)/2, -1]

d.) x = cost, y = sin(2t) = 2sintcost
-> y = 2sint.cost = 2xsint
-> y = 2x[√(1 - cos^2t)]
y = 2x.Sqrt(1 - x^2)

e.) y = -2x.Sqrt(1 - x^2)

4.) a.) dM/dt = -kM
b.) Don't understand notation here, sorry.
c.) 10/(10M - 1) dM = -k dt
-> ln|10M - 1| = c - kt
t = 0, M = 10: c = ln99
t = 10, M = 8.85: ln87.5 = ln99 - 10k -> 10k = ln[99/(175/2)] = ln(198/175) -> k = (1/10)ln(198/175)

Therefore: ln|10M - 1| = ln99 - (t/10)ln(198/175) = ln99 - ln[(198/175)^(t/10)] = ln{99/[(198/175)^(t/10)]}
-> 10M - 1 = 99/[(198/175)^(t/10)]

When t = 15: 10M - 1 = 99/[(198/175)^(1.5)]
-> 10M - 1 = 82.261
-> 10M = 83.261
-> M = 8.33 g (3.S.F)

5.) a.) R = Int. (2Pi to Pi) x^2.sin(x/2) dx.
Let u = x^2 -> du/dx = 2x
Let dv/dx = sin(x/2) -> v = -2cos(x/2)

R = [-2x^2.cos(x/2)] (Limits 2Pi and Pi) + 2 Int (2Pi to Pi) 2x.cos(x/2) dx.

To find Int. 2x.cos(x/2) dx.
Let u = 2x -> du/dx = 2
Let dv/dx = cos(x/2) -> v = 2sin(x/2)

-> Int. 2x.cos(x/2) dx. = 4xsin(x/2) - 4Int. sin(x/2) dx. = 4xsin(x/2) + 8cos(x/2)

Hence: R = [-2x^2.cos(x/2)] (Limits 2Pi and Pi) + [8xsin(x/2) + 16cos(x/2)] (Limits 2 Pi and Pi) = 2(4Pi^2) + [-16Pi - 8Pi] = 8Pi^2 - 24Pi = 8Pi(Pi - 3)

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13 years ago
#11
4 b)

which is the form of DE given in part a)
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