Mathematics

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Date of submission
Question 1
The model is: –
gt=Aekt (t≥0)Finding A and K constants
42=Ae3k ………….i
1195=Ae10k………ii
Making A subject of the formulae we have: –
A=42e3kReplacing the A in equation ii
1195=42e3ke1OK1195 = 42e7K 28.4524=e7k ln28.4524=7kK =ln⁡(28.4524)7K = 0.478
From the value of k we can calculate A
A = 42e3*0.478A =10.011
Increase after every 2 years
Model = 10.011e0.478tPopulation increase after 2 years is: –
10.011e0.478*2 = 26.041
Increase = 26.041Question 2
-3×2-6x+9≥0 x2+2x-3≥0(x-1)(x+3)
1819274104775088582412382500123825112585500123824621030104775135255-∞ -3 1 ∞
x-1 – – 0 +
x+3 – 0 + +
f(x) + 0 – 0 +
the interval is -∞,-3U[1,∞)Question 3
fx=x2+6x+8Completing square
x2+6x+622+8-622=x+32-1F(x)= x+32-1Explain how the graph of f can be obtained: –
The graph can be obtained by transforming y x by x times along x axis. This will lead to obtaining x^2 transformations.
gx=x2+6x+8 (-3≤x≤0)Sketch the graph of g, using equal scales on the axes.

Find the inverse function g−1, specifying its rule, domain and
image set.
gx=x2+6x+8 (-3≤x≤0) y=x2+6x+8 (-3≤x≤0) x=y2+6y+8 0=x2+6x+8-x x=-b±b2-4ac2a -6±62-4(8-x)2 =-6±4+4×2=-3±1+x g-=-3-1+x , (x≥0)Add a sketch of y = g−1(x) to the graph that you produced in
part (b)(i). Mark the coordinates of the endpoints of the graph
of g−1.

Question 4
Given angle A = 36oSide b = 10
Side c = 12
Use this information to calculate the area of triangle ABC, giving your
answer to two decimal places.
Area of triangle = 0.5 AB sin C
= 0.5 * 10 * 12 sin (36)
= 35.27
(i). Use the cosine rule to find the length of side a, to two decimals
places.
Using cosine rule we have
a2=b2+c2-2bc cos⁡(a) =102+122-2×120cos⁡(36)= 49.8359
Length of A = 7.06
ii) Using only the lengths of the sides of the triangle, find the area of
triangle ABC, giving your answer to two decimal places, and check
that it is the same as the area that you found in part (a).
(Hint: look up Heron’s formula in Unit 4.)
area=S(S-A)(S-B)(S-C) S=12+10+7.062 S=14.53 area=14.53(14.53-7.06)(14.53-10)(14.53-12)Area = 35.27
The answers are equal
Without using the cosine rule again, find the remaining angles B and C,
giving your answers to the nearest degree.
Solution
Area of triangle = 0.5 AB sin C
Angle A= 36
Angle B
35.27=0.5(7.06 * 12) SIN B
0.832625 = sin B
B = 56
Angle C
35.

27= 0.5(7.06*10) SIN C
C = 88
QUESTION 5
find the exact value of cot Ø
sin = opposite/hypotenuse
sin =1741using Pythagoras theorem, we have
AD2=412-172Ad =37.31
cot Ø =adjacent/opposite
= 37.31/17
= 2.195
11π12=22×32+-22×12=6-24-
Use the exact value of cos(11π/6) and the half-angle identity for cosine to find the exact value of cos(11π/12).
2cos2t=1+32=2cot2t-1Therefore, cos11π12= -2+32Question 6
Finding the coordinates of the intersecting points
Use Maxima to plot the parabola y = 4_ 3x _ 2×2 and the circle
x2 + y2 + 2x _ 2y _ 7 = 0 on the same graph, and to find the coordinates of
the points of intersection between the parabola and the circle.
solutionUsing graphical tool to plot we have: –

The co-ordinates are (-2.922,-4.307) and (1.255,-2.915)
Question 7
Express the velocity b of the bat relative to the air and the velocity w
of the wind in component form, giving the numerical values in ms−1 to
two decimal places.
b= U + w
b = 0.6944j – 0.3936i + 1.75j – 1.75ib = (0.6944 + 1.75)j + (-0.3936 – 1.75)ib = 2.4444j – 2.1436i
Express the resultant velocity v of the bat in component form, giving numerical values in ms−1 to two decimal places.
U = 4 cos (-80°)j + 4 sin (-80°)iU = 4 (0.1736)j + 4 (-0.9840)i
U = 0.6944j – 0.3936i
Hence find the magnitude and direction of the resultant velocity v of
the bat, giving the magnitude in ms−1 to two decimal places and the
direction as a bearing to the nearest degree.
solutionTan θ = -2.1436 / 2.444
Question 8
Find the stationary points of f, and give their exact x- and
y-coordinates.
Solution
fx=-x3-32×2+6x+6 dydx=-3×2-3x+6Therefore, stationary points are
0=-3×2-3x+6(x+2)(x-1)
Stationary points are: –
(-2,8) and (1,19/2)
Use the first derivative test to classify the stationary points that you
found in part (a).
using the first derivative done above, point (-2,8) is the local maximum while (1,19/2) is local minimum.
c
Question 9
Find expressions for the velocity v(t) and the acceleration a(t) of the
object at time t.
solution
the equation is: –
st=4t3-21t2+18t+5Velocity is the first derivative of displacement
dsdt=12t2-42t+18 velocity=12t2-42t+18Acceleration is the 2nd derivative of displacement
d2sdt=24t-42 acceleration=24t-42Find the times when the object is momentarily at rest, i.e. the
Velocity v(t) equals zero.
Solution
0=12t2-42t+18T1=3 and T2 = 0.5
At what time does the object have its minimum velocity?
Solution
At lowest velocity, acceleration =0
0=24t-4224t=42
t = 42/42
= 1.75