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Running head: MATHEMATICS 1
Mathematics
Name
Institution

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 k−10 k=e−8 k
lne−8 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

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 e−0.086 t=3
e−0.086 t = 3
231
ln (e−0.086 t )=ln ( 3
232 )
−0.086 t=ln ( 3
232 )=−4.3481
t=−4.3481
−0.086 =50.56 hrs ≅ 51 hours
Question 2
5−2 x
3 x +4 <0
5−2 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
5−2 x + + + 0 -

MATHEMATICS 4
3 x+ 4 - 0 + + +
5−2 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)2−b=x2+2 xa+(a¿ ¿2−b) ¿
x2=x2 , 2 xa=4 x ,1=(a ¿¿ 2−b) ¿
2 xa=4 x ,a=2
1=(a¿ ¿2−b)=(2¿¿ 2−b)¿ ¿
1=4−b , 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.

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)

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+ ( 1−x )=0
Using the quadratic formula,
y=−b ± √b2−4 ac
2 a =−4 ± √ 42−4(1)(1−x )
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)

MATHEMATICS 7
A sketch of y=g−1 (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

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

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 ( A−B )=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

MATHEMATICS 10
5 π
12 = ( 5 π
6 )
2
sin ( 5 π
12 )=sin ( ( 5 π
6 )
2 )
But we know, sin ( A
2 )= √ 1−cosA
2
sin ( 5 π
12 )=sin ( ( 5 π
6 )
2 )= √ 1−cos ( 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
¿ 6−2 √12+2
16 + 2+ √3
4
¿ 8−2 √ 4 (3)
16 + 2+ √ 3
4
¿ 8−4 √(3)
16 + 2+ √ 3
4

MATHEMATICS 11
¿ 8−4 √ ( 3 ) +4 ( 2+ √ 3 )
16
¿ 8−4 √3+8+ 4 √3
16 = 8+8−4 √3+4 √3
16 =16
16 =1
Question 5
Part a
Drone
30 km h−1 at a bearing of 65 °
x−component =30 sin 65=27.19 km h−1
y−component=30 cos 65=12.68 km h−1
~
d=27.19 ~i+12.68 ~j
Wind
10 km h−1 ¿ SE(that is towards NW )

MATHEMATICS 12
x−component =10 sin 45=−7.07 km h−1
y−component=10 cos 45=7.07 kmh−1
~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 h−1
tanθ= 20.12
19.75 =1.018734
θ=tan−1 1.018734=45.53 ≅ 46 °
The resultant velocity of the drone is 28.19 km h−1 at a bearing of 046 °
Question 6
f ( x )=x3 + 9
4 x2 −3 x−2
Part (a)
To get the stationary points, we determine the 1st derivative of the curve and then equate it to
zero as follows:

MATHEMATICS 13
df
dx = d
dx ( x3 + 9
4 x2 −3 x−2
) =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 x5−9−6 x3 =0 ¿
( x5−x3 )= 9
6 =1.5
x5−x3=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

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 x−2
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 x−2 is shown below.

MATHEMATICS 15

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 ( t3−12 t2+ 8t )=3t2−12(2)t1 +8
v ( t ) =3 t2−24 t +8
At t=−2
v ( 2 )=3(2)2−24 ( 2 ) +8=−28 m s−1
Part b
Acceleration=a ( t )= dV (t )
dt = d
dt ( 3 t2 −24 t+ 8 )=3 ( 2 ) t−24+ 0
a ( t )=6 t−24
Acceleration is zero at t=4 and the minimum velocity at t=4 equals
v ( 4 )=3(4 )2−24 ( 4 ) +8=−40. Hence, the velocity decreases for 0 ≤ t ≤ 4