4 Math Questions
Top_A
solutions.docxQuestion 1.
a) (i)
(4+i2).(1+i3)
4(1+i3)+i2(1+i3)
4+i12+i2-6
=-2+i14
(ii)
(4+i2)/(1+i3)
numerator
4(1-i3)+i2(1-i3)
4-i12+i2-(i)26
4-i10+6
10-i10
Denominator
1(1-i3)+i3(1-i3)
1-i3+i3-9(i)2
1+9=10
Quotient=
=1-i
b)
r=/z/=√(-2)2+142
=√200
Argument of z
Tanθ=b/a
=14/-2
=-7
Θ=tan-1(-7)
=-81.87ᵒ×πᶜ/180ᵒ
=πᶜ….4th quadrant
ϴ1=- πᶜ+πᶜ
ϴ2= πᶜ….2nd quadrant(principal arg)
Therefore,arg z= πᶜ+2πᶜn where n=intergers
Z==√200[cos ( πᶜ)+isin( πᶜ)]
c)
(4+i2) to polar form
R=/z/=√42+22
=√20
Arg z1
Tanθ1=2/4
Θ=tan-11/2
=26.6ᵒ×πᶜ/180ᵒ
=πᶜ….1st quadrant
Arg z1= πᶜ+2πᶜn…..where n =interger
Z1=√20[cos(πᶜ)+isin(πᶜ)]
(1+i3) to polar form
R=/z/=√12+32=√10
Argz2
Tanθ2=3/1
Θ=tan-13
=71.56ᵒ×πᶜ/180ᵒ
=πᶜ
Therefore z2=πᶜ+2πᶜn….n=integers
Z2=√10[cos(πᶜ)+isin(πᶜ)]
(i)
Product in polar representation
Z1.z2=√20[cos(πᶜ)+isin(πᶜ)].√10[cos(πᶜ)+isin(πᶜ)]
=√20.√10[cos(πᶜ+πᶜ)+isin(πᶜ+πᶜ)]
=10√2[cos(πᶜ)+isin(πᶜ)]
(ii)
Quotient in polar representation
Z1/z2=[cos(πᶜ - πᶜ)+isin(πᶜ - πᶜ)]
= [cos(πᶜ )+isin((πᶜ)]
=√2[cos(πᶜ )-isin((πᶜ)]
d(i)
10√2[cos(πᶜ)+isin(πᶜ)]
=10√2[-0.139+i0.99]
=-1.968+i14
(ii)
√2[cos(πᶜ )-isin((πᶜ)]
=√2[0.803-i0.719]
=1.135-i1.017
Qn 9
Z1=2+j10
Z2=j14
(i)
z=z1+z2 in series
=2+j10+j14
=(2+0)+j(10+14)
=2+j24
(ii)
Z=z1+z2 in parallel
zE=
z1.z2=(2+j10).(0+ j14)
=2(0+j14)+j10(0+j14)
=0+j28+0+140(j)2
=j28-140
=-140+j28
zE=×
numerator
=-140(2-j24)+j28(2-j24)
=-280+j3360+j56-672(j)2
=-280+j3416+672
=392+j3416
Denominator
zE=2(2-j24)+j24(2-j24)
=4-j48+j48-576j2
=4+576
=580
zE=
=
=0.676+j5.890
Question 2
The following data within the working area consists of measurements of resistor values from a production line.
For the sample as a set of ungrouped data, calculate (using at least 2 decimal places):
1. arithmetic mean
1. standard deviation
1. variance
The following data consists of measurements of resistor values from a production line:
|
51.4 |
54.1 |
53.7 |
55.4 |
53.1 |
53.5 |
54.0 |
|
56.0 |
53.0 |
55.3 |
55.0 |
52.8 |
55.9 |
52.8 |
|
50.5 |
54.2 |
56.2 |
55.6 |
52.7 |
56.1 |
52.1 |
|
54.2 |
50.2 |
54.7 |
56.2 |
55.6 |
52.7 |
52.1 |
|
56.1 |
54.2 |
50.2 |
54.7 |
55.1 |
54.8 |
56.5 |
|
55.8 |
55.3 |
54.5 |
57.0 |
56.0 |
53.9 |
57.3 |
|
55.3 |
54.4 |
49.6 |
54.1 |
51.6 |
53.2 |
54.6 |
|
56.4 |
53.9 |
50.9 |
54.0 |
51.8 |
56.1 |
53.2 |
|
54.6 |
56.4 |
53.9 |
50.9 |
54.0 |
51.8 |
56.1 |
|
‘n’ No. |
Value (X) |
|
Mean |
|
(X – Mean) |
(X-Mean)2 |
|
1 |
51.4 |
|
58.0413 |
|
0.115873016 |
0.013426556 |
|
2 |
56 |
|
58.0413 |
|
2.015873016 |
4.063744016 |
|
3 |
50.5 |
|
58.0413 |
|
1.715873016 |
2.944220207 |
|
4 |
54.2 |
|
58.0413 |
|
1.215873016 |
1.478347191 |
|
5 |
56.1 |
|
58.0413 |
|
2.315873016 |
5.363267826 |
|
6 |
55.8 |
|
58.0413 |
|
0.515873016 |
0.266124969 |
|
7 |
55.3 |
|
58.0413 |
|
0.015873016 |
0.000251953 |
|
8 |
56.4 |
|
58.0413 |
|
-1.084126984 |
1.175331318 |
|
9 |
54.6 |
|
58.0413 |
|
0.115873016 |
0.013426556 |
|
10 |
54.1 |
|
58.0413 |
|
-3.884126984 |
15.08644243 |
|
11 |
53 |
|
58.0413 |
|
0.115873016 |
0.013426556 |
|
12 |
54.2 |
|
58.0413 |
|
1.215873016 |
1.478347191 |
|
13 |
50.2 |
|
58.0413 |
|
0.315873016 |
0.099775762 |
|
14 |
54.2 |
|
58.0413 |
|
-0.184126984 |
0.033902746 |
|
15 |
55.3 |
|
58.0413 |
|
2.315873016 |
5.363267826 |
|
16 |
54.4 |
|
58.0413 |
|
-0.384126984 |
0.14755354 |
|
17 |
53.9 |
|
58.0413 |
|
0.115873016 |
0.013426556 |
|
18 |
56.4 |
|
58.0413 |
|
2.015873016 |
4.063744016 |
|
19 |
53.7 |
|
58.0413 |
|
1.715873016 |
2.944220207 |
|
20 |
55.3 |
|
58.0413 |
|
1.215873016 |
1.478347191 |
|
21 |
56.2 |
|
58.0413 |
|
2.115873016 |
4.476918619 |
|
22 |
54.7 |
|
58.0413 |
|
0.615873016 |
0.379299572 |
|
23 |
50.2 |
|
58.0413 |
|
-3.884126984 |
15.08644243 |
|
24 |
54.5 |
|
58.0413 |
|
0.415873016 |
0.172950365 |
|
25 |
49.6 |
|
58.0413 |
|
-4.484126984 |
20.10739481 |
|
26 |
50.9 |
|
58.0413 |
|
-3.184126984 |
10.13866465 |
|
27 |
53.9 |
|
58.0413 |
|
-0.184126984 |
0.033902746 |
|
28 |
55.4 |
|
58.0413 |
|
1.315873016 |
1.731521794 |
|
29 |
55 |
|
58.0413 |
|
0.915873016 |
0.838823381 |
|
30 |
55.6 |
|
58.0413 |
|
1.515873016 |
2.297871 |
|
31 |
56.2 |
|
58.0413 |
|
2.115873016 |
4.476918619 |
|
32 |
54.7 |
|
58.0413 |
|
0.615873016 |
0.379299572 |
|
33 |
57 |
|
58.0413 |
|
2.915873016 |
8.502315445 |
|
34 |
54.1 |
|
58.0413 |
|
0.015873016 |
0.000251953 |
|
35 |
54 |
|
58.0413 |
|
-0.084126984 |
0.007077349 |
|
36 |
50.9 |
|
58.0413 |
|
-3.184126984 |
10.13866465 |
|
37 |
53.1 |
|
58.0413 |
|
-0.984126984 |
0.968505921 |
|
38 |
52.8 |
|
58.0413 |
|
-1.284126984 |
1.648982111 |
|
39 |
52.7 |
|
58.0413 |
|
-1.384126984 |
1.915807508 |
|
40 |
55.6 |
|
58.0413 |
|
1.515873016 |
2.297871 |
|
41 |
55.1 |
|
58.0413 |
|
1.015873016 |
1.031997984 |
|
|
42 |
56 |
|
58.0413 |
|
1.915873016 |
3.670569413 |
|
|
43 |
51.6 |
|
58.0413 |
|
-2.484126984 |
6.170886873 |
|
|
44 |
51.8 |
|
58.0413 |
|
-2.284126984 |
5.21723608 |
|
|
45 |
54 |
|
58.0413 |
|
-0.084126984 |
0.007077349 |
|
|
46 |
53.5 |
|
58.0413 |
|
-0.584126984 |
0.341204334 |
|
|
47 |
55.9 |
|
58.0413 |
|
1.815873016 |
3.29739481 |
|
|
48 |
56.1 |
|
58.0413 |
|
2.015873016 |
4.063744016 |
|
|
49 |
52.7 |
|
58.0413 |
|
-1.384126984 |
1.915807508 |
|
|
50 |
54.8 |
|
58.0413 |
|
0.715873016 |
0.512474175 |
|
|
51 |
53.9 |
|
58.0413 |
|
-0.184126984 |
0.033902746 |
|
|
52 |
53.2 |
|
58.0413 |
|
-0.884126984 |
0.781680524 |
|
|
53 |
56.1 |
|
58.0413 |
|
2.015873016 |
4.063744016 |
|
|
54 |
51.8 |
|
58.0413 |
|
-2.284126984 |
5.21723608 |
|
|
55 |
54 |
|
58.0413 |
|
-0.084126984 |
0.007077349 |
|
|
56 |
52.8 |
|
58.0413 |
|
-1.284126984 |
1.648982111 |
|
|
57 |
52.1 |
|
58.0413 |
|
-1.984126984 |
3.936759889 |
|
|
58 |
52.1 |
|
58.0413 |
|
-1.984126984 |
3.936759889 |
|
|
59 |
56.5 |
|
58.0413 |
|
2.415873016 |
5.836442429 |
|
|
60 |
57.3 |
|
58.0413 |
|
3.215873016 |
10.34183925 |
|
|
61 |
54.6 |
|
58.0413 |
|
0.515873016 |
0.266124969 |
|
|
62 |
53.2 |
|
58.0413 |
|
-0.88413 |
0.781681 |
|
|
63 |
56.1 |
|
58.0413 |
|
2.015873 |
4.063744 |
|
|
3407.3 |
= Total X |
58.0413 |
54.95645
|
215.4841
|
|||
|
Mean = Total X / ‘n’(max) = |
|
|
|
||||
|
‘n’ max – 1 (Y) = |
|
|
|||||
|
Total (X-Mean)2 ……. let’s call this (Z) |
|
||||||
|
Variance |
|
Variance = (Total(X-Mean)2) / (‘n’ max -1) = Z / Y = 3.920998
|
|||||
|
Standard Deviation |
|
Standard Deviation = square-root of Variance
= 1.980151
|
Question 3
Using the data in Q2, we now need to arrange the data into groups so that we can “tally” the data accordingly.
|
No. |
Data Range |
Gap |
Freq. (F) |
Mid-Point (X) |
(F x X) |
Gap x F (Bar Area) |
(X – Mean) |
(X-Mean)2 |
(X-Mean)2 x F |
|
1 |
49.5 – 50.5 |
1 |
5 |
50 |
250 |
5 |
-4.84127 |
23.4379 |
117.1895 |
|
2 |
50.5 – 51.5 |
1 |
3 |
51 |
153 |
3 |
-3.84127 |
14.75536 |
44.26607 |
|
3 |
51.5 – 52.5 |
1 |
5 |
52 |
260 |
5 |
-2.84127 |
8.072815 |
40.36408 |
|
4 |
52.5 – 53.5 |
1 |
9 |
53 |
477 |
9 |
-1.84127 |
3.390275 |
30.51248 |
|
5 |
53.5 – 54.5 |
1 |
14 |
54 |
756 |
14 |
-0.84127 |
0.707735 |
9.908293 |
|
6 |
54.5 – 55.5 |
1 |
11 |
55 |
605 |
11 |
0.15873 |
0.025195 |
0.277147 |
|
7 |
55.5 – 56.5 |
1 |
15 |
56 |
840 |
15 |
1.15873 |
1.342655 |
20.13983 |
|
8 |
56.5 – 57.5 |
1 |
2 |
57 |
114 |
2 |
2.15873 |
4.660115 |
9.32023 |
|
9 |
|
|
|
|
|
|
|
|
|
|
10 |
|
|
|
|
|
|
|
|
|
|
11 |
|
|
|
|
|
|
|
|
|
|
Total F…Let’s call this (Y) |
63 |
3455 |
54.84127 |
271.97759 |
|||||
|
Total (F x X) |
|
|
|
||||||
|
|
|
|
|||||||
|
Mean = (Total (F x X)) / (Y) |
|
|
|||||||
|
Total (X-Mean)2 x F ……. let’s call this (Z) |
|
|
Variance |
|
Variance = (Total(X-Mean)2 x f) / (Total F) = Z / Y 4.95936 |
|
Standard Deviation |
|
Standard Deviation = square-root of Variance
=2.2269
|
Question 4
i)
Amplitude is the height of the wave
Periodic time is the time taken to complete one complete oscillation
Frequency refers to the number of oscillations made per second of time
In the above, for time T, the wave oscillates once, hence has a frequency of one.
V = 310 Sin (285t + 0.65)
X y
|
0.2 |
0 |
|
0.6 |
-310 |
|
0.9 |
0 |
|
1.2 |
310 |
ii)The wave form is leading
iii) The phase angle in degrees =0.65
iv)The amplitude is 310
v)Periodic time T is
wt=2π0
t=2850/(360*πc)
=1.58πr0
T =2αc /1.581 αc
=1.262 seconds
vi)Frequency
f=1/t
=1/1.263
=0.792hz
With t=1.263
Divide by 4 threfore we have
t=0,A=310
t=0.3, A=0
t=0.6, A=-310
t=0.9,A=0
t=1.2,A=310
hence the graph
Question 6
Differentiate:
1. y = (3x2 – 2x)7 ans:7(3x2-2x) 6(6x-2)
1. y = 6x3 .sin4x ans:18x2sin4x+24x3cos4x
(C) y=5e^6x/x-8
Let u=5e^6x and t=6x
dt/dx=6
du/dt=5e^t
.: du/dx=du/dt*dt/dx
=5e^t.6
Let v=x-8
Dv/dx=1
By quotient rule
Dy/dx=(Vdu/dx-Udv/dx)/V^2
={(x-8).30e^6x – 5e^6x.1 }/(x-8)(x-8)
/ (x-8)^2
Question 7
(a)
(i) ∫ (4cos3ϴ+sin6 ϴ)dϴ
=4∫cos3ϴdϴ+∫sin6ϴdϴ
For 4∫cos3ϴdϴ
Let u-3ϴ
1/3du=dϴ
Substituting for dϴ
4∫cos u.1/3du
=4/3(sin u)+c….eqn(i)
For ∫sin6ϴdϴ
Let u-6ϴ
1/6du=dϴ
Substituting or dϴ
∫sin u.1/6du
=1/6∫(sin u)du
=1/6(-cos u)+c
=-1/6cos 6ϴ +c ….eqn(ii)
Adding eqn (i) and (ii)
=
(ii) ∫(2+cos0.83)dϴ
=2∫dϴ+∫cos0.83ϴ dϴ
For 2∫dϴ
=2ϴ+c…eqn(i)
For ∫cos0.83ϴ dϴ
=1/0.83sin0.83+c…eqn(ii)
Adding eqn (i) and (ii)
C(I) y=3x^2+6
Values of x and y
When x=0 ; y=3(0)^2+6 =6
x=1 ; y=3(1)^2+6 =9
x=2; y=3(2)^2+6 =18
x=3 ; y=3(3)^2+6 =33
x=4 ; y=3(4)^2+6 =54
NB Using these values draw the required graph pliz
(ii)
A =∫
=
=3[1/3 +c
=3[64/3-1/3]
=63….(i)
6
=6[x+c]14
=6[4-1]
=18…eqn(ii)
Adding (i) and (ii)
=63+18= 81 sq units
Question 8
y=axn
Introducing logs
log y= log(axn )
loga+logxn
loga+nlogx this is the straight line form
Taking a tablex:a,0,1,2,…
Since y intercept=0,the equation becomes:
logy=nlogx+0
log1=nlog1
-0.3=0.3n
n=-1
0.9=loga+0.3(-1)
loga=1.2
gradient= n=-1
intersept= loga=1.2
sketching the graph:
|
n |
x |
y |
n logx |
antilog |
|
1 |
2 |
8 |
0.301 |
1.9999 |
|
2 |
2.5 |
6.4 |
0.7959 |
6 |
|
3 |
3 |
5.3 |
1.4314 |
27.00023 |
|
4 |
3.5 |
4.6 |
2.1763 |
150.072 |
|
5 |
4 |
4 |
3.0103 |
1024.0001 |
Qn 9
Z1=2+j10
Z2=j14
(i)
z=z1+z2 in series
=2+j10+j14
=(2+0)+j(10+14)
=2+j24
(ii)
Z=z1+z2 in parallel
zE=
z1.z2=(2+j10).(0+ j14)
=2(0+j14)+j10(0+j14)
=0+j28+0+140(j)2
=j28-140
=-140+j28
zE=×
numerator
=-140(2-j24)+j28(2-j24)
=-280+j3360+j56-672(j)2
=-280+j3416+672
=392+j3416
Denominator
zE=2(2-j24)+j24(2-j24)
=4-j48+j48-576j2
=4+576
=580
zE=
=
=0.676+j5.890
0.2 0.60000000000000064 0.9 1.2 0 -310 0 310 0 0.30000000000000032 0.60000000000000064 0.9 1.2 1.5 310 0 -310 0 310 0 Y-Values 0.70000000000000062 1.8 2.6 2.7 3.2 0.81
)
x
-
x
(
=
s
Deviation
dard
tan
S
1
-
n
)
x
-
x
(
s
Variance
2
2
2
-
å
å
=
n
1
)
x
-
x
(
=
s
Deviation
dard
tan
S
1
-
n
)
x
-
x
(
s
Variance
2
2
2
-
å
å
=
n
Variance s
x
-
x
)
n
S
dard Devia
tion s
=
x
-
x
)
2
2
2
=
å
å
(
tan
(
n
