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do nxt to nxt week? i dint get it :SYou know what leave the burden of solving this paper, focus on your preparations, do next to next week.
All the best.
I will be always there.
Two weeks after.do nxt to nxt week? i dint get it :S
thank u kiddo <3
yeah after 2 weeks therell be pnly p5 and p1
At first I thought that my answer is wrong; and turned out to be correct... Yay!!
i thought the sameAt first I thought that my answer is wrong; and turned out to be correct... Yay!!
Yep I also manually input the 22.something into the question, and the answers are roughly equal...i thought the same
i did this method n striked it out n did it again using laws of indices. got the same answer.
22.3 is bit abnormal for value of x for these type of questions i agree
(Y) heheYep I also manually input the 22.something into the question, and the answers are roughly equal...
(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!!
Awwww..... I got the complex number wrong for the second part... I thought I also got the differential equation wrong...
Hope so you get A*
Thanks!!!! Hope I can do statistics 2, my statistics teacher keeps giving me lots of pastpapers to practice....
Thanks!!!! Hope I can do statistics 2, my statistics teacher keeps giving me lots of pastpapers to practice....
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.
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
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.
yea even the physics notes are not his he is doing it for likes tons of them
Oh my gosh really? Where's the proof btw?
At least acknowledge the original post...
well i have seen the original post it was by someone else i am in xtremepapers from 4 yearsOh my gosh really? Where's the proof btw?
Yep I also want to get a lot of likes... But not that way.....
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