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Desklib is an online library for study material with solved assignments, essays, dissertations, etc. It offers a wide range of study materials for various subjects and courses. This article covers topics such as mathematics, functions, graphs, and velocity.
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Running head: MATHEMATICS 1
Mathematics
Name
Institution
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MATHEMATICS 2
Question 1
Part a
f ( t )= A ekt (t 1)
Whent=2 hours , f (t)=195
When t=10 hours , f (t)=98
195= A e2 t (i)
98=A e10t (ii)
Dividing equation ( i )by equation ( ii ) we obtain
195
98 = A e2k
A e10 k
195
98 =e2 k10 k=e8 k
lne8 k =ln ( 195
98 )
8 k =ln ( 195
98 )
k =ln ( 195
98 ) ÷ ( 8 ) =0.086
Substituting t into equation 1 we get
195= A e2 k
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MATHEMATICS 3
A=195
e2 k = 195
e2(0.086) =231.599
A=232 ,k =0.086 (¿ 3 sf )
Part b
giventhat f ( t )=3=A ekt
232 e0.086 t=3
e0.086 t = 3
231
ln (e0.086 t )=ln ( 3
232 )
0.086 t=ln ( 3
232 )=4.3481
t=4.3481
0.086 =50.56 hrs 51 hours
Question 2
52 x
3 x +4 <0
52 x=0 , x= 5
2
3 x+ 4=0 , x=4
3
x 4
3 x=4
3
4
3 < x < 5
2 x= 5
2 x > 5
2
52 x + + + 0 -
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MATHEMATICS 4
3 x+ 4 - 0 + + +
52 x
3 x +4
- undefined + 0 -
From the table, we chose the ranges that satisfy the required conditions. That is,
x 4
3 x > 5
2
Question 3
Part a(i)
f ( x)=x2 + 4 x+1
f ( x )=(x +a)2b=x2+2 xa+(a¿ ¿2b) ¿
x2=x2 , 2 xa=4 x ,1=(a ¿¿ 2b) ¿
2 xa=4 x ,a=2
1=(a¿ ¿2b)=(2¿¿ 2b)¿ ¿
1=4b , b=3
f ( x )=(x +2)2 3
Part a(ii)
The graph of f ( x)=x2 + 4 x+1can be obtained from f ( x)=x2 by a vertical stretch of 4 followed
by a vertical translation of +1 unit.
Part a(iii)
The image set of f ( x ) is undefined since we don’t have the range of x.
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MATHEMATICS 5
Part b(i)
A graph of g ( x )=x2+ 4 x +1 for (2 x 2)
( x ) -2 -1 0 1 2
g ( x ) -3 -2 1 6 13
The image set of g ( x ) is g (2 x 2 ) ={3,13 }
Part b(ii)
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MATHEMATICS 6
Inverse of g ( x )=x2+4 x +1 for (2 x 2)
y=x2 +4 x+1
Interchanging the variables
x= y2 +4 y +1
y2 +4 y+ ( 1x )=0
Using the quadratic formula,
y=b ± b24 ac
2 a =4 ± 424(1)(1x )
2(1)
y=4 ± 12+4 x
2 =4 ± 4( 3+x)
2 =4 ± 2 (3+x )
2 =2± x+3
g ' ( x )=2+ x+ 3 or g ' ( x )=2 x +3
The domain of the function is x 3. That is, (2 x 2)
The image set of g '( x ) is
g' (2 )=2+ 2+3=1, g' (2 )=2+ 2+3=0.236, g' ( x ) =[1 ,0.236 ]
Or
g' (2 )=2 2+3=3
g' (2 )=2 2+3=4.236
g' ( x ) =[3 ,4.236]
Part b(iii)
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MATHEMATICS 7
A sketch of y=g1 (x)
( x ) g ' ( x ) ¿ g ' ( x )
-2 -1 -3
-1 -0.586 -3.414
0 -0.268 -3.732
1 0 -4
2 0.236 -4.236
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MATHEMATICS 8
Question 4
Part a
cos ( 5 π
12 ), 3 π
4 , π
3
3 π
4 = 3 π × 3
4 ×3 = 9 π
12
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MATHEMATICS 9
π
3 = π ×4
3 × 4 = 4 π
12
sin ( 3 π
4 )=sin ( 9 π
12 )= 2
2
cos ( 3 π
4 )=cos ( 9 π
12 )= 2
2
5 π
12 = 9 π
12 4 π
12
sin ( 4 π
12 )= 3
2
cos ( 4 π
12 )=1
2
Using the identity
cos ( AB )=cosAcosB+ sinAsinB
cos ( 5 π
12 ) =cos ( 9 π
12 ) cos ( 4 π
12 ) +sin ( 9 π
12 ) sin ( 4 π
12 )
¿ 2
2 ( 1
2 )+ 2
2 ( 3
2 )= 6
4 2
4 = 6 2
4
Part b
sin ( 5 π
12 )using half angle identity for sine} ¿ cos ( 5 π
6 )
cos ( 5 π
6 )= 3
2
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MATHEMATICS 10
5 π
12 = ( 5 π
6 )
2
sin ( 5 π
12 )=sin ( ( 5 π
6 )
2 )
But we know, sin ( A
2 )= 1cosA
2
sin ( 5 π
12 )=sin ( ( 5 π
6 )
2 )= 1cos ( 5 π
6 )
2 = 1 ( 3
2 )
2 = ( 1+ 3
2 )
2 = 2+ 3
2
Part c
cos ( 5 π
12 )= 6 2
4
sin ( 5 π
12 )= 2+ 3
2
cos2 θ+ sin2 θ=1
cos2
(5 π
12 )+sin2
( 5 π
12 )= ( 6 2
4 )2
+ ( 2+ 3
2 )2
¿ 62 12+2
16 + 2+ 3
4
¿ 82 4 (3)
16 + 2+ 3
4
¿ 84 (3)
16 + 2+ 3
4
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MATHEMATICS 11
¿ 84 ( 3 ) +4 ( 2+ 3 )
16
¿ 84 3+8+ 4 3
16 = 8+84 3+4 3
16 =16
16 =1
Question 5
Part a
Drone
30 km h1 at a bearing of 65 °
xcomponent =30 sin 65=27.19 km h1
ycomponent=30 cos 65=12.68 km h1
~
d=27.19 ~i+12.68 ~j
Wind
10 km h1 ¿ SE(that is towards NW )
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MATHEMATICS 12
xcomponent =10 sin 45=7.07 km h1
ycomponent=10 cos 45=7.07 kmh1
~w=7.07 ~
i+ 7.07 ~
j
Part b
Resultant velocity ¿ ~
d +~w
¿ ( 27.19 ~i+ 12.68 ~j ) + (7.07 ~i+ 7.07 ~j )
¿ 20.12 ~
i+19.75 ~
j
Part c
magnitude= 20.122 +19.752=28.19 km h1
tanθ= 20.12
19.75 =1.018734
θ=tan1 1.018734=45.53 46 °
The resultant velocity of the drone is 28.19 km h1 at a bearing of 046 °
Question 6
f ( x )=x3 + 9
4 x2 3 x2
Part (a)
To get the stationary points, we determine the 1st derivative of the curve and then equate it to
zero as follows:
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MATHEMATICS 13
df
dx = d
dx ( x3 + 9
4 x2 3 x2
) =3 x2 + 9 ( 2 )
4 x3 3
¿ 3 x2 9
2 x3 3
3 x2 9
2 x3 3=0
We multiply the equation by 2 x3 and simplify to get
(3 x ¿¿ 2 9
2 x3 3=0)2 x3 =6 x596 x3 =0 ¿
( x5x3 )= 9
6 =1.5
x5x3=1.5
Solving the above equation we obtain, x=1.29826
At x=1.29826 , f ( x )=1.298263 + 9
4 ×1.298262 3 ( 1.29826 )2=2.3717
Hence, the stationary point is ( 1.29826 ,2.3717 )
Part b
To determine whether the stationary point is a maximum or a minimum we choose values on
either sides of the turning point and evaluate the derivatives as summarized in the table below.
df
dx =3 x2 9
2 x3 3
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MATHEMATICS 14
When x=1 x=1.29826 when x=2
df
dx
4.5(negative) 0 8.4375( positive)
Sign ¿ ¿ ¿
Therefore, we can see that ( 1.29826 ,2.3717 )is a minimum point.
Part c
f ( x )=x3 + 9
4 x2 3 x2
x -2 -1.5 -1 0 0.5 1 1.298 2
f ( x ) -3.438 0.125 2.25 undefined 0.5625 -1.75 -2.372 0.563
f ( x ) has no y-intercept.
A graph of f ( x )=x3 + 9
4 x2 3 x2 is shown below.
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MATHEMATICS 15
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MATHEMATICS 16
Part d
In the interval [-3,2] f ( x ) has the least value at x=3. That is,
f ( 3 ) =(3)3+ 9
4 (3)2 3 ( 3 ) 2=19.75
However, the largest value approaches infinity (+ ) as x approaches zero.
Question 7
s ( t ) =t3 12t2 +8 t(t 0)
Part a
Velocity=v (t )= ds ( t )
dt = d
dt ( t312 t2+ 8t )=3t212(2)t1 +8
v ( t ) =3 t224 t +8
At t=2
v ( 2 )=3(2)224 ( 2 ) +8=28 m s1
Part b
Acceleration=a ( t )= dV (t )
dt = d
dt ( 3 t2 24 t+ 8 )=3 ( 2 ) t24+ 0
a ( t )=6 t24
Acceleration is zero at t=4 and the minimum velocity at t=4 equals
v ( 4 )=3(4 )224 ( 4 ) +8=40. Hence, the velocity decreases for 0 t 4
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