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Mathematics: Post your doubts here!

TimBluesWin

TimBluesWin

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Thought blocker said:
9709_w14_ms_32 By Thought blocker.
*Copyrights Violation*
Click to expand...

Awwww..... I got the complex number wrong for the second part... I thought I also got the differential equation wrong... :( (I think I got x=0.5 instead of 100/57)... As I said before, I also got wrong with the trapezium rule (I mistakenly thought the area was above the graph!!), and also the 10i.... (10ii I got same sets of answers...). Im fairly confident with the "show" type questions, I nailed them mostly... Wish I can get an A* though!!
 
Thought blocker

Thought blocker

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TimBluesWin said:
Awwww..... I got the complex number wrong for the second part... I thought I also got the differential equation wrong... :( (I think I got x=0.5 instead of 100/57)... As I said before, I also got wrong with the trapezium rule (I mistakenly thought the area was above the graph!!), and also the 10i.... (10ii I got same sets of answers...). Im fairly confident with the "show" type questions, I nailed them mostly... Wish I can get an A* though!!
Click to expand...
(y) Hope so you get A* ;)
 
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Meleko

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Thought blocker said:
9709_w14_ms_32 By Thought blocker.

1)
NOTE using n in powers instead of x as x isn't available in power form.
eⁿ = 3ⁿ ⁻ ² So what is the value of n?
eⁿ = 3ⁿ/3² = 3ⁿ/9
9 = 3ⁿ/eⁿ
9 = (3/e)ⁿ
n = log base (3/e) of 9 = 22.3.

2)
i)
h = [(b - a) / n]
h = [{(2Π/3) - (Π/6)}/3) = Π/6

f(x) = cosec(x)
x₀ = Π/6
x₁ = 2Π/6 = Π/3
x₂= 3Π/6 = Π/2
x₃ = 4Π/6 = 2Π/3
y₀ = cosec(Π/6) = 1/(sin(Π/6) = 2
y₁ = cosec(Π/3) = 1/sin(Π/3) = (2√3/3)
y₂ = cosec(Π/2) = 1/sin(Π/2) = 1
y₃ = cosec(2Π/3) = 1/sin(2Π/3) = (2√3/3)

(2Π/3 → Π/6)∫(cosec(x).dx = (1/2 times 1Π/6) times {2 + 2 ((2√3/3) + 1) + (2√3/3)} = 1.95 (2dp)

ii)
Overestimate as the curve for y = cosec(x) is below the lines forming trapezium.
View attachment 48476

3)
p(x) = ax³ + bx² + x + 3 --- 1.
(3x + 1) factor = p(-1/3) = 0
p(x) divided by (x - 2) and remainder r(x) = 21. (x - 2) factor = p(2) = 21.
Substitute the factor of (3x - 1) in equation 1.
p(-1/3) = a(-1/3)³ + b(-1/3)² + (-1/3) + 3 = 0
= (-1/27)a + (1/9)b + (-1/3) + 3 = 0
Multiply the p(x) with 27
= -a + 3b -9 + 81 = 0
= 3b - a = -72
Arranged :¬
p(-1/3) = a - 3b = 72

Now substitute the factor of (x - 2) in equation 1 with the remainder 21.
a(2)³ + b(2)² + 2 + 3 = 21
8a + 4b = 16
Divide equation by 4.
2a + b = 16
Get the value of b and substitute is in the a - 3b = 72 equation.
So, b = 4 - 2a
Substituting:
a - 3(4 - 2a) = 72
a - 12 + 6a = 72
a = 12
As we solved for b, b = 4 - 2a Substitute the value of a as 12 we got.
b = 4 - 24 = -20

Hence Values of a and b are 12 and -20 respectively.

4)
Given ; x = 1/cos³t, y = tan³t for 0 ≤ t ≤ Π/2
i)
dy/dx = (dy/dt)/(dx/dt)
So in order to solve this we need to find dy/dt and dx/dt as follows :
Let a = cos(t), hence x = a⁻³
So, da/dt = -sin(t)
And, dx/da = -3a⁻⁴ = -3/cos⁴t
Hence we get the value of dx/dt to be,
dx/dt = 3sin(t)/cos⁴t
Now,
y = (sin(t)/cos(t))³
dy/dt = 3sec²(t).tan²(t)
Let b = sin(t)/cos(t)
Hence b = y³
So, db/dt = sec²(t)
And, dy/db = 3b²
Hence we get the value of dy/dt to be,
dy/dt = 3sec²(t) times tan²(t)

So dy/dx =[(3 times 1/cos²(t) times (sin²(t)/cos²(t))]/(3sin(t)/cos⁴t) Solving it we get it equal to sin(t).

ii)
Equation of a tangent at a point with parameter t. Given ; y₁= tan³(t), x₁ = 1/cos³(t)
Use the formula for the equation that is y - y₁ = m(x - x₁)
So,
y - tan³(t) = m(x - 1/cos³(t))
m - dy/dx = sin(t)
y - (sin³(t)/cos³(t)) = (x)sin(t) - (sin(t)/cos³(t))
y = (x)sin(t) - (sin(t)/cos³(t)) + (sin(t).sin²(t)/cos³(t))
y = (x)sin(t) - (sin(t)/cos³(t)).(1 - sin²(t)) -> cos²(t)
y = (x)sin(t) - (sin(t)/cos(t))
y = (x)sin(t) - tan(t)

5)
Given ; w = (z + i)/(iz + 2)
i)
As z = 1 + i
w = (1 + 2i)/(i + i² +2) = (1 + 2i)/(i + 1) = (1 + 2i - i - 2i²)/(1 - i²)
w = (3 + i)/2
w = 3/2 + (1/2)i
ii)
If w = z and real part of z is < zero
z = (z + i)/(iz + 2)
Hence,
iz² + 2z = z + i
z²i + z(2 - 1) - i = 0
z²i + 2z - z - i = 0 (Here 2z - z = z)

Discriminant = b² - 4ac = 1² - 4(i)(-i) = 1 - 4 = -3
Now use z = (-b ± √ (b² - 4ac)/2a) [Here b² - 4ac = -3]
z₁,₂ = (-1 ± √-3)/2i = (-1 ± √3i)/2i
z₁,₂ = (-i ± √3)/-2
So,
z₁ = (√3 - i)/ -2 = -√3/2 + (1/2)i
z₂ = (-√3 - i)/ -2 = √3/2 + (1/2)i

As we know real part of z is < zero, z₁ is accepted.
Hence z = -√3/2 + (1/2)i

6)
Given (a → 1)∫(ln(2x)).dx = 1 and a > 1.
i)
I(x) = ∫(ln(2x)).dx = ∫1.ln(2x).dx
Let u = ln(2x)
So, du/dx = (1/2x).2 = 1/x
Now, dv/dx = 1 as v = x
∫u.(dv/dx).dx = uv - ∫v.(du/dx).dx
I(x) = (x)ln(2x) - ∫1.(1/x).dx = (x)ln(2x) - ∫1.dx
∫ln(2x).dx = (x)ln(2x) - x + c
(a → 1)∫ln(2x).dx = [(x)ln(2x) - x]a → 1 = (a)ln(2a) - a - ln(2) + 1
I(x) = (a)ln(2a) - a - ln(2) + 1
So, (a)ln(2a) = a + ln(2)
ln(2a) =(1 + (ln(2)/a)
2a = e^((1 + (ln(2)/a))
a = 1/2 times e^((1 + (ln(2)/a))

ii)
an ₊ ₁ = 1/2 times e^((1 + (ln(2)/an))
a₀ = 2
a₁ = 1.9221
a₂ = 1.9493
a₃ = 1.9395
a₄ = 1.94340
a₅ = 1.9418
a₆ = 1.9422
a₇ = 1.9420
a₈ = 1.9421
a₉ = 1.9421
a₁₀= 1.9421

Hence x = 1.94 (2dp)

7)
Given ; dR/dx = R((1/x) - 0.57) and when x = 0.5, R = 16.8
i)
∫(1/R).dR = ∫((1/x) - 0.57). dx
lnR = ln(x) - 0.57(x) + c
ln16.8 = ln0.5 - (0.57 times 0.5) + c
c = ln(16.8/0.5) + 0.285 = ln(33.6) + 0.285
lnR = ln(x) - 0.57(x) + ln33.6 + 0.285
ln(R/33.6x) = 0.285 - 0.57x
R/33.6x = e^(0.285 - 0.57(x))
R = 33.6(x) times e^(0.285 - 0.57(x))

ii)
dR/dx = 0
33.6(x).e^(0.285 - 0.57(x)) . ((1/x) - 0.57) = 0
1/x = 0.57
x = 100/57 <---------- Maximum value of R is when x has this value.
maxR = 33.6 times 100/57 times e^(0.285 - 0.57 times 1/0.57)
R = 33.6 times 100/57 times e^(0.285 - 1)
R = 28.8 (3sf)

Now just easy questions left. ;)

8)
i)
sin(2θ + θ) = L.H.S
= sin2θcosθ + cos2θsinθ
= (2sinθcosθ)cosθ + (cos²θ-sin²θ)sinθ
= 2sinθcos²θ + sinθcos²θ - sin³θ
= 3sinθ.cos²θ - sin³θ
= 3sinθ.(1 - sin²θ) - sin³θ
= 3sinθ - 3sin³θ - sin³θ
Hence, 3sinθ - 4sin³θ = R.H.S
Sin3θ = 3sinθ - 4sin³θ

ii)
If x = 2sinθ/√3
x√3 = 2sinθ
sinθ = (√3x/2)
sin3θ = 3(√3/2)x - 4(√3/2)³x³ = (3√3/2)x - 6√3x³
x³ - x + (√3/6) = 0
(2sinθ/√3)³ - (2sinθ/√3) + √3/6 = 0
(8sin³θ/3√3) - (2sinθ/√3) + 1/2√3 = 0
(8sin³θ/3) - 2sinθ + 1/2 = 0
Multiplying 3 to both sides
8sin³θ - 6sinθ + 3/2 = 0
Divide both sides by 2
4sin³θ - 3sinθ = (-3/4)
-sin3θ = -3/4
sin3θ = 3/4

iii)
x³ - x + (√3/6) ≡ sin3θ = 3/4
I : 3θ = sin⁻¹(3/4) Hence, θ₁ = (sin⁻¹(3/4))/3
II : 3θ = Π - sin⁻¹(3/4) Hence, θ₂ = Π/3 - sin⁻¹(3/4)
III : 3θ = sin⁻¹(3/4) ± 2Π Hence, 3θ = sin⁻¹(3/4) + Π ± 2Π

x₁ = 0.322 (3sf)
x₂ = 0.799 (3sf)

x ∈ {0.322, 0.799} to 3sf

9)
i)
f(x) = ((x² - 8x + 9)/(1 - x) (2 - x)²) = (A/(1 - x)) + (B/(2 - x)) + (C/(2 - x)²)
f(x) = x² - 8x + 9 = A(2 - x)² + B(1 - x)(2 - x) + C(1 - x)
If x = 1, 1 - 8 + 9 = 2 = A
If x = 2, 4 - 16 + 9 = 3 = C
If x = 0, 9 = 4A + 2B + C hence B = -1
So partial form is f(x) = (2/(1 - x)) - (1/(2 - x) + (3/(2 - x)²))

ii)
f(x) = 2(1 - x)⁻¹ - 1(2 - x)⁻¹ + 3( 2 - x)⁻²
(1 - x)⁻¹ = 1 + (-1)(-x) + ((-1)(-1-1)/2!).(-x)² + ... = 1 + x + x²+...
(2 - x)⁻¹ = 2⁻¹.(1 - (x/2)⁻¹ = (1/2)(1 + (x/2) + (x²/4)+...)
(2 - x)⁻² = 2⁻².(1 - (x/2)⁻² = (1/4).(1 + x + ((-2)(-3)/2!).(x²/4)+...)
f(x) = 2(1 + x + x²+...) - (1/2)(1 + (x/2) + (x²/4)+...) + (3/4)(1 +x + 3/4)
f(x) = 2 + 2x + 2x² - (1/2) - (x/4) - (x²/8) + (3/4) + ( 3x/4) + (9x²/16)
f(x) = (9/4) + (5x/2) + (39x²/16)

Now the most easiest one from the paper comes. :D

10)

Given;
l : r = 4i - 9j + 9k + λ(-2i +j -2k) = View attachment 48491

OA = 3i + 8j + 5k = View attachment 48492

i)
Let P be the foot of perpendicular and B be the point on l such that OB = 4i - 9j + 9k = View attachment 48493

P on l is OP and OP = View attachment 48494

AP will be AO + OP = View attachment 48495

As AP is perpendicular to l = View attachment 48496
Solving it to get λ value.
-2 + 4 λ - 17 + λ - 8 + 4 λ = 0
λ = 3
Substitute in AP = View attachment 48497
AP = √((-5)² + (-14)² + (-2)²)
AP = √225 = 15 units.

ii)
P and B on l = P and B on plane.
Get OP by substituting value of λ = View attachment 48498
OB = View attachment 48492

ax + by - 3z + 1 = 0
From OP
-2a - 6b - 3(3)+ 1 = 0
-2a - 6b = 8
a + 3b = -4 ---> 1.
From OB
4a - 9b -3(9) + 1 = 0
4a - 9b = 26 ---> 2.

Now 1 ∩ 2 that is simultaneous turn to get values of a and b
a + 3b = -4 should be multiplied with -4
so, -4a - 12b = 16 and 4a - 9b = 26
-4a and 4a gets cut so -21b = 42 so b = -2 substitute in equation 2, 4a + 24 = 26 so a = 2
a = 2 and b = -2.

Finally it ends, tell me if I have mistakes.
By - Thought blocker
Click to expand...

Dear Thought blocker... I think it took you a lot of time to type the solution.. did you also solve the paper? or you just copied from someone else? You added a few more words I must admit.. Next time don't take credits for something that is not yours.. you even made your small pics.. wasn't easier to put the original ones and give the link to the website?? justpastpapers.com tells you anything?
tell me if I have mistakes
Click to expand...
Yes you have a few, mistyping nothing else, but the results are ok.. but you made another big mistake... you copied without permission.. and guess what?! this is not the first time - you also copied paper 12 and you made a mistake at the quadratic sketch... Just pay attention and don't take credits for something that does not belong to you.
ps: the 2 links in your signature have a wrong code.. the code for the exam is 9709 not 9702...
 
Last edited:
TimBluesWin

TimBluesWin

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Meleko said:
Dear Thought blocker... I think it took you a lot of time to type the solution.. did you also solve the paper? or you just copied from someone else? You added a few more words I must admit.. Next time don't take credits for something that is not yours.. you even made your small pics.. wasn't easier to put the original ones and give the link to the website?? justpastpapers.com tells you anything?
Yes you have a few, mistyping nothing else, but the results are ok.. but you made another big mistake... you copied without permission.. and guess what?! this is not the first time - you also copied paper 12 and you made a mistake at the quadratic sketch... Just pay attention and don't take credits for something that does not belong to you.
ps: the 2 links in your signature have a wrong code.. the code for the exam is 9707 not 9702...
Click to expand...

Oh my goshh.... I thought he's a really really smart guy.... :O
PS: Math is 9709 btw....
 
Thought blocker

Thought blocker

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Meleko said:
Dear Thought blocker... I think it took you a lot of time to type the solution.. did you also solve the paper? or you just copied from someone else? You added a few more words I must admit.. Next time don't take credits for something that is not yours.. you even made your small pics.. wasn't easier to put the original ones and give the link to the website?? justpastpapers.com tells you anything?
Yes you have a few, mistyping nothing else, but the results are ok.. but you made another big mistake... you copied without permission.. and guess what?! this is not the first time - you also copied paper 12 and you made a mistake at the quadratic sketch... Just pay attention and don't take credits for something that does not belong to you.
ps: the 2 links in your signature have a wrong code.. the code for the exam is 9707 not 9702...
Click to expand...
(y) Thank you.
 
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