2023
TRIAL
EXAMINATION
Mathematics Extension 1
WORKED SOLUTIONS
Mathematics Extension 1 2023 Trial Examination – Worked Solutions
Section I
Question
2
Solution
Substituting 𝑥 = −1:
3(−1)4 − (−1)2 + (−1) + 8 = 9
rise
6
3
𝑚 = run = − 4 = − 2
3
3
1
𝑥
𝑥
∫ 32 +𝑥2 𝑑𝑥 = 3 × 3 tan−1 3 + 𝑐 = tan−1 3 + 𝑐
1
Answer
D
B
A
The slope field is nearly vertical at
(4, 0), above which it slopes to the
right and below which it slopes to
the left and flattens out near the
𝑦-axis.
4
5
6
7
The field is also symmetrical.
𝑃(𝑋 = 𝑟) = 𝑛𝐶𝑟 𝑝𝑟 (1 − 𝑝)𝑛−𝑟 = 𝑛𝐶𝑛−𝑟 𝑝𝑛−𝑟 (1 − 𝑝)𝑟 only holds for
𝑝 = 0.5. (If 𝑝 ≠ 0.5, the distribution is not symmetric.)
Join the o’s as one unit: 7! arrangements
Duplicate arrangements occur when the two a’s are swapped:
7!
2!
(It is not necessary to multiply or divide by 2! to account for the 2 o’s
as we are treating them as one unit and not arranging them amongst
themselves.)
When 𝑔(𝑥) is square rooted, the
portion of the graph below the 𝑥-axis
is deleted. Hence the domain of ℎ(𝑥)
is [−1, 1].
The graph of the square root will
also increase in height for points left
below 𝑦 = 1, while it is still 0 and
𝑥 = ±1 and 1 for 𝑥 = 0.
B
D
B
C
The graphs are both relatively high
above zero as 𝑥 approaches ±1, so we should not see the graph
become concave up.
The parametric curve is a left
semicircle of radius 2. At the point
(−2, 0), the tangent is vertical, hence
the gradient is undefined.
8
D
2
Mathematics Extension 1 2023 Trial Examination – Worked Solutions
For 𝑦 = 𝑒 5𝑥 ,
𝑦 ′ = 5𝑒 5𝑥
𝑦 ′′ = 25𝑒 5𝑥
Substituting into the differential equation:
(25𝑒 5𝑥 ) + 3(5𝑒 5𝑥 ) − 4(𝑒 5𝑥 ) = 25𝑒 5𝑥 + 15𝑒 5𝑥 − 4𝑒 5𝑥 ≠ 0
𝑎̰ − 𝑏̰ forms a triangle with 𝑎̰ and 𝑏̰ .
Cosine rule:
|𝑎̰ |2 + |𝑏̰ |2 − |𝑎̰ − 𝑏̰ |2 1 + 1 − |𝑎̰ − 𝑏̰ |2
cos 𝜃 =
=
2|𝑎̰ ||𝑏̰ |
2(1)(1)
2
|𝑎̰
|
2 − − 𝑏̰
1
=
> since |𝑎̰ − 𝑏̰ | < 1
2
2
1
𝜋
cos 𝜃 > 2 for 0 < 𝜃 < 3 (considering the angle between two vectors to
be 0 ≤ 𝜃 ≤ 𝜋)
9
10
C
B
Section II
Question 11
Question 11 (a)
(i) Domain of 𝑓(𝑥): [1, ∞) → range of 𝑓 −1 (𝑥): [1, ∞)
Range of 𝑓(𝑥): [0, ∞) → domain of 𝑓 −1 (𝑥): [0, ∞)
(ii)
Question 11 (b)
1
sin 𝐴 sin 𝐵 = 2 [cos(𝐴 − 𝐵) − cos(𝐴 + 𝐵)] (see Reference Sheet)
𝜋
2
1
2
7𝜋
𝐴 − 𝐵 = 12, 𝐴 + 𝐵 = 12
8𝜋
𝜋
2𝐴 = 12 → 𝐴 = 3 (adding the equations)
6𝜋
𝜋
2𝐵 = 12 → 𝐵 = 4 (subtracting the equations)
cos
𝜋
7𝜋
𝜋
𝜋
− cos = 2 sin sin
12
12
3
4
Question 11 (c)
𝑢 = √𝑥 → 𝑑𝑢 =
1
𝑑𝑥 → 2𝑑𝑢 =
1
2
𝑑𝑥
2√𝑥
√𝑥
cos(√𝑥)
∫
𝑑𝑥 = 2 ∫ cos 𝑢 𝑑𝑢 = 2 sin 𝑢 + 𝑐 = 2 sin √𝑥 + 𝑐
√𝑥
3
Mathematics Extension 1 2023 Trial Examination – Worked Solutions
Question 11 (d)
𝑐 6
(i)
𝛼𝛽 + 𝛼𝛾 + 𝛽𝛾 = = = 3
𝑎 2
𝑑
𝑏
−8
−3
(ii)
𝛼 2 𝛽𝛾 + 𝛼𝛽 2 𝛾 + 𝛼𝛽𝛾 2 = 𝛼𝛽𝛾(𝛼 + 𝛽 + 𝛾) = − × − = −
×−
=6
𝑎
𝑎
2
2
Question 11 (e)
(i) 𝑎̰ + 2𝑏̰ = [−3
] + 2[51] = [−3+2×5
] = [74]
2
2+2×1
(ii) |𝑐̰ | = √(−2)2 + (−4)2 = √20 or 2√5
(iii) 𝑐̰ ̂ = 1 [−2] = 1 [−1]
2√5 −4
√5 −2
(The answer may be presented as
(iv)
1
2
1
1
1
1 1
[ ], with a rationalised denominator
√5 2
and/or with the scalar distributed.)
𝑎̰ ⋅ 𝑐̰ = |𝑎̰ ||𝑐̰ | cos 𝜃
(−3)(−2) + (2)(−4) = √(−3)2 + (2)2 × 2√5 × cos 𝜃
−2
cos 𝜃 =
→ 𝜃 ≈ 97°
2√65
2
Question 12
Question 12 (a)
(i) At (2, 1), 𝑑𝑦 = (2)(1)2 + 2 = 4
1
(ii)
3
𝑑𝑥
𝑑𝑦
= 𝑥𝑦 2 + 𝑥 = 𝑥(𝑦 2 + 1)
𝑑𝑥
1
𝑑𝑦 = 𝑥 𝑑𝑥
2
𝑦 +1
1
∫ 2 𝑑𝑦 = ∫ 𝑥 𝑑𝑥
𝑦 +1
−1
tan
1
𝑦 = 2 𝑥2 + 𝑐
When 𝑥 = 2, 𝑦 = 1:
1
tan−1 1 = 2 (2)2 + 𝑐
𝜋
𝑐 = −2
4
1
𝜋
tan−1 𝑦 = 2 𝑥 2 + 4 − 2
1
𝜋
𝑦 = tan (2 𝑥 2 + 4 − 2)
Question 12 (b)
In Line 4, the claim that 𝑘 + 1 can be expressed as 𝑎 + 𝑏 where 0 ≤ 𝑎, 𝑏 ≤ 𝑘 + 1 is
false, as this does not hold true for 𝑘 + 1 = 1 (when 𝑘 + 1 = 1, both 𝑎 and 𝑏 must be
0). Hence, the following induction step does not hold.
Question 12 (c)
(i) Each of the 8 numbers in Set A can be paired to any of the 12 numbers in Set B
(but not more than one as this would not be a function).
There are 128 functions.
(ii) There are 12 choices for the first number from Set A, then 11 for the next, etc.
12!
There are 4! one-to-one functions.
4
2
1
1
Mathematics Extension 1 2023 Trial Examination – Worked Solutions
Question 12 (d)
(i)
(ii)
(The sketch should demonstrate understanding
of the shape obtained by rotating the area about
the 𝑥-axis.)
Find right of integration:
6 2
𝑥 =𝑥+1
25
6𝑥 2 − 25𝑥 − 25 = 0
(6𝑥 + 5)(𝑥 − 5) = 0
Right limit of integration: 𝑥 = 5.
5
𝑉 = 𝜋 ∫ [(𝑥 + 1)2 − (
0
1
3
6 2 2
𝑥 ) ] 𝑑𝑥
25
36 5
(𝑥 + 1)3 625 𝑥 5
= 𝜋[
−
]
3
5
0
1
107𝜋
= 𝜋 {[72 − 36] − [ − 0]} =
3
3
Question 12 (e)
(i) Method 1:
Using 𝑡 formula with 𝐴 = 2𝜃 (see Reference Sheet):
(2𝜃)
𝑡 = tan 2 = tan 𝜃
2
1−𝑡 2
Then cos 2𝜃 = 1+𝑡 2
1−𝑡2
1− 2
1−cos 2𝜃
√
√ 1+𝑡
=
1−𝑡2
1+cos 2𝜃
1+
1+𝑡2
1+𝑡 2 −(1−𝑡 2 )
1+𝑡 2
(multiply fraction inside radical by 1+𝑡 2)
2𝑡 2
= √1+𝑡 2 +(1−𝑡 2 ) = √ 2 = √𝑡 2 = |𝑡| = |tan 𝜃|
Method 2:
1−cos 2𝜃
1−(1−2 sin2 𝜃)
2 sin2 𝜃
=√
=√
= √tan2 𝜃 = |tan 𝜃|
1+cos 2𝜃
1+(2 cos2 𝜃−1)
2 cos2 𝜃
√
(ii)
1
5
Mathematics Extension 1 2023 Trial Examination – Worked Solutions
Question 13
Question 13 (a)
Let 𝑛 = 0. Then 𝐹0 = 1 and 𝐹2𝑛+1 − 1 = (1 + 1) − 1 = 1. So the claim is true for 𝑛 =
0.
2
Assume it is true for some 𝑛 = 𝑘. That is, assume 𝐹0 + 𝐹2 + 𝐹4 + ⋯ + 𝐹2𝑘 = 𝐹2𝑘+1 − 1.
Let 𝑛 = 𝑘 + 1. Consider 𝐹0 + 𝐹2 + 𝐹4 + ⋯ + 𝐹2𝑘 + 𝐹2(𝑘+1)
= 𝐹2𝑘+1 − 1 + 𝐹2(𝑘+1) (by the assumption step)
= 𝐹2𝑘+1 + 𝐹2𝑘+2 − 1
= 𝐹2𝑘+3 − 1 (by the recursive definition of 𝐹𝑛 )
= 𝐹2(𝑘+1)+1 − 1
Thus, the claim is true for 𝑛 ≥ 0 by the principle of mathematical induction. ∎
Question 13 (b)
(i) LHS = sin 3𝜃 = sin(𝜃 + 2𝜃) = sin 𝜃 cos 2𝜃 + cos 𝜃 sin 2𝜃 (sine angle sum ID)
= sin 𝜃 (1 − 2 sin2 𝜃) + cos 𝜃 (2 sin 𝜃 cos 𝜃) (cosine and sine double angle ID)
= sin 𝜃 − 2 sin3 𝜃 + 2 sin 𝜃 cos2 𝜃
= sin 𝜃 − 2 sin3 𝜃 + 2 sin 𝜃 (1 − sin2 𝜃) (Pythagorean ID)
= sin 𝜃 − 2 sin3 𝜃 + 2 sin 𝜃 − 2 sin3 𝜃 = 3 sin 𝜃 − 4 sin3 𝜃 = RHS ∎
(ii) 𝑥(𝑡) = ∫ sin3 𝑡 𝑑𝑡 = 1 ∫(3 sin 𝑡 − sin 3𝑡)𝑑𝑡 (by part (i))
4
1
1
= (−3 cos 𝑡 + cos 3𝑡) + 𝑐
4
3
1
1
2
2
When 𝑡 = 0, 𝑥 = 0: 0 = 4 (−3 cos 0 + 3 cos 0) + 𝑐 = − 3 + 𝑐 → 𝑐 = 3
1
1
2
𝑥(𝑡) = (−3 cos 𝑡 + cos 3𝑡) +
4
3
3
6
2
2
Mathematics Extension 1 2023 Trial Examination – Worked Solutions
Question 13 (c)
(i) Initial velocity:
(ii)
1
𝑣 √3 𝑣
𝑣̰ (0) = (𝑣0 cos 30° , 𝑣0 sin 30°) = ( 02 , 20 )
𝑎̰ (𝑡) = (0, −10)
By integrating,
𝑣̰ (𝑡) = (𝑐1 , −10𝑡 + 𝑐2 )
𝑣̰ (0) = (
𝑣0 √3 𝑣0
, )
2
2
𝑣 √3
𝑣
𝑐1 = 02 , 𝑐2 = 20
𝑣̰ (𝑡) = (
𝑣0 √3
𝑣0
, −10𝑡 + )
2
2
By integrating,
𝑣 √3
𝑣
𝑥̰ (𝑡) = ( 02 𝑡 + 𝑐3 , −5𝑡 2 + 20 𝑡 + 𝑐4 )
𝑥̰ (0) = (0, 0) (using original location of the ball as the origin)
𝑐3 = 0 and 𝑐4 = 0
𝑣 √3
𝑣
𝑥̰ (𝑡) = ( 02 𝑡, −5𝑡 2 + 20 𝑡)
The ball hits the point (25, 2.4).
Starting with horizontal component:
𝑣0 √3
50
𝑡 = 25 → 𝑡 =
2
𝑣0 √3
Substituting into vertical component:
50 2
𝑣
50
) + 20 (𝑣 3) = 2.4
3
0√
0√
12500
25
− 2 + = 2.4
3𝑣0
√3
12500
25√3
= 3 − 2.4
3𝑣02
−5 (𝑣
3𝑣02 =
12500
25√3
3 − 2.4
12500
𝑣0 = √25 3−7.2 ≈ 18.6 m/s
√
7
4
Mathematics Extension 1 2023 Trial Examination – Worked Solutions
Question 13 (d)
(i) E(𝑋) = 𝑝 = 0.52
𝑝(1−𝑝)
0.52×0.48
Var(𝑋) = 1000 = 1000 = 0.0002496
3
Normal approximation:
Let 𝜇 = E(𝑋) = 0.52 and 𝜎 = √Var(𝑋) = √0.0002496 ≈ 0.015799
𝑧- score for 𝑝̂ = 0.5:
0.50 − 0.52
𝑧=
≈ −1.27
0.015799
(ii)
According to the table on page 15, 𝑃(𝑍 ≤ 1.27) = 0.8980
Hence 𝑃(𝑍 ≤ −1.27) = 1 − 0.8980 = 0.102 = 10.2%
The sample size affects the accuracy, such that the higher the sample size, the
less likely we are to get an inaccurate result showing the losing candidate with
a majority.
2
Also, the closer the true proportion in the population is to 0.5, the less accurate
the polling will be, as it will be more likely to get an inaccurate result showing
the losing candidate with a majority.
Question 14
Question 14 (a)
For the medians to form a triangle, their vectors must sum to 0.
3
Define medians in terms of two of the original triangle vectors:
1
⃗⃗⃗⃗⃗⃗⃗
𝐴𝐴′ = ⃗⃗⃗⃗⃗
𝐴𝐵 + ⃗⃗⃗⃗⃗
𝐵𝐶
2
1
1
1
1
⃗⃗⃗⃗⃗⃗⃗′ = 𝐵𝐶
⃗⃗⃗⃗⃗ − 𝐴𝐶
⃗⃗⃗⃗⃗ = 𝐵𝐶
⃗⃗⃗⃗⃗ − (𝐴𝐵
⃗⃗⃗⃗⃗ + 𝐵𝐶
⃗⃗⃗⃗⃗ ) = 𝐵𝐶
⃗⃗⃗⃗⃗ − 𝐴𝐵
⃗⃗⃗⃗⃗
𝐵𝐵
2
2
2
2
1
⃗⃗⃗⃗⃗⃗⃗′ = −𝐵𝐶
⃗⃗⃗⃗⃗ − 𝐴𝐵
⃗⃗⃗⃗⃗
𝐶𝐶
2
1
1
1
1
⃗⃗⃗⃗⃗⃗⃗
𝐴𝐴′ + ⃗⃗⃗⃗⃗⃗⃗
𝐵𝐵′ + ⃗⃗⃗⃗⃗⃗⃗
𝐶𝐶 ′ = ⃗⃗⃗⃗⃗
𝐴𝐵 + ⃗⃗⃗⃗⃗
𝐵𝐶 + ⃗⃗⃗⃗⃗
𝐵𝐶 − ⃗⃗⃗⃗⃗
𝐴𝐵 − ⃗⃗⃗⃗⃗
𝐵𝐶 − ⃗⃗⃗⃗⃗
𝐴𝐵 = 0 ∎
2
2
2
2
Question 14 (b)
𝑓(2) = 2𝑒 2 , thus 𝑓 −1 (2𝑒 2 ) = 2.
3
𝑑
𝑔′ (𝑥) = 𝑓 −1 (𝑥) + 𝑥 𝑑𝑥 𝑓 −1 (𝑥) (product rule)
𝑑𝑥
𝑑𝑦
1
If 𝑦 = 𝑓 −1 (𝑥), then 𝑥 = 𝑓(𝑦) → 𝑑𝑦 = 𝑓 ′ (𝑦) → 𝑑𝑥 = 𝑓′ (𝑦) = 𝑓′
1
(𝑓−1 (𝑥))
𝑓 ′ (𝑥) = 𝑒 𝑥 + 𝑥𝑒 𝑥
𝑔′ (2𝑒 2 ) = 𝑓 −1 (2𝑒 2 ) + 2𝑒 2 ×
1
2𝑒 2
2𝑒 2
2 8
=
2
+
=
2
+
=2+ =
′
2
2
′
−1
2
𝑓 (2)
𝑒 + 2𝑒
3 3
𝑓 (𝑓 (2𝑒 ))
8
Mathematics Extension 1 2023 Trial Examination – Worked Solutions
Question 14 (c)
(i) Rate of rotation of minute hand:
2𝜋
𝜋
= per minute
60
1
30
Rate of rotation of hour hand:
2𝜋
60
12
𝜋
= 360 per minute
𝑑𝜃
𝜋
𝜋
11𝜋
= 360 − 30 = − 360 per minute
𝑑𝑡
(ii)
Distance of the tips of the hands (cosine rule):
𝑐 2 = 102 + 152 − 2(10)(15) cos 𝜃 = 325 − 300 cos 𝜃
3
𝑑 2
𝑑
𝑐 = (325 − 300 cos 𝜃)
𝑑𝑡
𝑑𝑡
𝑑𝑐
𝑑𝜃
2𝑐
= 300 sin 𝜃
𝑑𝑡
𝑑𝑡
2𝜋
𝜋
𝜋
At one o’clock, 𝜃 = 12 = 6 and 𝑐 = √325 − 300 cos 6 ≈ 8 mm
𝜋
11𝜋
𝑑𝑐 300 sin 6 (− 360 )
mm
=
≈ −0.9
𝑑𝑡
2(8)
min
Question 14 (d)
𝑇 = 𝐵 + 𝐴𝑒 𝑘𝑡 → 𝐴𝑒 𝑘𝑡 = 𝑇 − 𝐵
4
𝑑𝑇
= 𝑘𝐴𝑒 𝑘𝑡 = 𝑘(𝑇 − 𝐵) = 𝑘(𝑇 − 180)
𝑑𝑡
Thus, 𝐵 = 180
𝑇 = 180 + 𝐴𝑒 𝑘𝑡
When 𝑡 = 0, 𝑇 = 0:
0 = 180 + 𝐴𝑒 0
𝐴 = −180
𝑇 = 180 − 180𝑒 𝑘𝑡
When 𝑡 = 15, 𝑇 = 25:
25 = 180 − 180𝑒 15𝑘
31
= 𝑒 15𝑘
36
31
ln
= 15𝑘
36
31
ln 36
𝑘=
≈ −0.009969
15
𝑑𝑇
When 𝑇 = 100, 𝑑𝑡 = −0.009969(100 − 180) = 0.7975° C/min
9