Topic 2
[197 marks]
The diagram shows the straight line
points on L1 .
L1 . Points A(−9, − 1), M(−3, 2) and C are
1a. Find the gradient of L1 .
[2 marks]
Markscheme
2− ( −1 )
−3− ( −9 )
(M1)
Note: Award (M1) for correct substitution into the gradient formula.
=
1
2
( 36 , 0. 5)
(A1)(G2)
[2 marks]
M is the midpoint of AC.
1b. Find the coordinates of point C.
[2 marks]
Markscheme
−3 =
−9+x
2
(−6 + 9 = x) and 2 =
−1+y
2
(4 + 1 = y)
(M1)
Note: Award (M1) for correct substitution into the midpoint formula for both
coordinates.
OR
(M1)
Note: Award (M1) for a sketch showing the horizontal displacement from
to C is 6 and the vertical displacement is 3 and the coordinates at M.
M
OR
−3 + 6 = 3 and 2 + 3 = 5
(M1)
Note: Award (M1) for correct equations seen.
(3, 5)
(A1)(G1)(G1)
Note: Accept x = 3, y = 5. Award at most (M1)(A0) or (G1)(G0) if
parentheses are missing.
[2 marks]
Line L2 is perpendicular to
L1 and passes through point M.
1c. Find the equation of L2 . Give your answer in the form
, where a, b, d ∈ Z.
ax + by + d = 0 [3 marks]
Markscheme
gradient of the normal =
−2
(A1)(ft)
Note: Follow through from their gradient from part (a).
y − 2 = −2(x + 3) OR 2 = −2(−3)+c
Note: Award (M1) for correct substitution of
into straight line formula.
2x + y + 4 = 0 (accept integer multiples)
(M1)
M and their gradient of normal
(A1)(ft)(G3)
[3 marks]
The point
N(k, 4) is on L2 .
[2 marks]
1d. Find the value of k.
Markscheme
2(k)+4 + 4 = 0
(M1)
Note: Award (M1) for substitution of y = 4 into their equation of normal line
or substitution of M and (k, 4) into equation of gradient of normal.
k = −4
(A1)(ft)(G2)
Note: Follow through from part (c).
[2 marks]
1e. Find the distance between points M and
N.
[2 marks]
Markscheme
√(−4 + 3)2 + (4 − 2)2
(M1)
Note: Award (M1) for correctly substituting point
formula.
√5 (2. 24, 2. 23606 …)
(A1)(ft)
Note: Follow through from part (d).
[2 marks]
M and their N into distance
1f. Given that the length of AM is
√45, find the area of triangle ANC.
[2 marks]
Markscheme
1
2
×(2 × √45)×√5
(M1)
Note: Award (M1) for their correct substitution into area of a triangle formula.
Award (M0) for their 12 ×(√45)×√5 without any evidence of multiplication
by 2 to find length AC. Accept any other correct method to find the area.
15
(A1)(ft)(G2)
Note: Accept
from part (e).
15. 02637 … from use of a 3 sf value for √5. Follow through
[2 marks]
The coordinates of point A are (6,
Point M is the midpoint of AB.
− 7) and the coordinates of point B are (−6, 2).
L1 is the line through A and B.
The line
L2 is perpendicular to L1 and passes through M.
2. Write down, in the form
y = mx + c, the equation of L2 .
[1 mark]
Markscheme
y = 43 x −
5
2
(y = 1.33 … x − 2.5)
(A1)(ft)
(C1)
Note:
Follow through from parts (c)(i) and (a). Award (A0) if final answer is
not written in the form y = mx + c.
[1 mark]
Line L intersects the x-axis at point A and the y-axis at point B, as shown on the
diagram.
The length of line segment OB is three times the length of line segment OA, where
O is the origin.
[2 marks]
3a. Find the gradient of L.
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
−3
(A1)(A1)
Notes:
(C2)
Award (A1) for 3 and (A1) for a negative value.
Award (A1)(A0) for either
3x or −3x.
[2 marks]
Point (2,
6) lies on L.
3b. Find the equation of L in the form
y = mx + c.
[2 marks]
Markscheme
6 = −3(2) + cOR(y − 6) = −3(x − 2)
(M1)
Note:
Award (M1) for substitution of their gradient from part (a) into a
correct equation with the coordinates (2, 6) correctly substituted.
y = −3x + 12
Notes:
(a).
(A1)(ft)
(C2)
Award (A1)(ft) for their correct equation. Follow through from part
If no method seen, award (A1)(A0) for
Award (A1)(A0) for
y = −3x.
−3x + 12.
[2 marks]
[2 marks]
3c. Find the x-coordinate of point A.
Markscheme
0 = −3x + 12
Note:
(M1)
Award (M1) for substitution of
(x =) 4
(A1)(ft)
y = 0 in their equation from part (b).
(C2)
Notes:
Follow through from their equation from part (b). Do not follow
through if no method seen. Do not award the final (A1) if the value of x is
negative or zero.
[2 marks]
2 + = 10
The equation of the line
L1 is 2x + y = 10.
[2 marks]
4a. Write down
(i)
the gradient of L1 ;
(ii)
the y-intercept of L1 .
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
(i)
(ii)
−2
10
(A1)
(C1)
(A1)
(C1)
L2 is parallel to L1 and passes through the point P(0, 3).
Write down the equation of L2 .
4b. The line
[2 marks]
Markscheme
2x + y − 3 = 0
(A1)(ft)(A1)
(C2)
Notes: Award (A1)(ft) for gradient, (A1) for correct y-intercept.
The answer must be an equation.
L2 is parallel to L1 and passes through the point P(0, 3).
Find the x-coordinate of the point where L 2 crosses the x-axis.
4c. The line
[2 marks]
Markscheme
−2x + 3 = 0 or equivalent (M1)
(x =) 1.5 (A1)(ft) (C2)
Notes: Follow through from their equation in part (b). If answer given as
coordinates (1.5, 0) award at most (M1)(A0) if working seen or (A1)(A0) if
no working seen.
Consider the straight lines L1 and L2 . R is the point of intersection of these lines.
The equation of line L1 is y = ax + 5.
5a. Find the value of a.
[2 marks]
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
0 = 10a + 5
(M1)
Note: Award (M1) for correctly substituting any point from L1 into the
equation.
OR
0−5
10−0
(M1)
Note: Award (M1) for correctly substituting any two points on L1 into the
gradient formula.
5
− 10
(− 12 , −0.5)
(A1) (C2)
[2 marks]
The equation of line L2 is y = −2x + 3.
5b. Find the coordinates of R.
[2 marks]
Markscheme
(−1.33, 5.67) ((− 43 ,
(A1)(ft)(A1)(ft) (C2)
17
),
13
(−1 13 , 5 23 ) , (−1.33333 … , 5.66666 …))
Note: Award (A1) for x-coordinate and (A1) for y-coordinate. Follow through
from their part (a). Award (A1)(A0) if brackets are missing. Accept x = −1.33,
y = 5.67.
[2 marks]
5c. Line L3 is parallel to line L2 and passes through the point (2, 3).
[2 marks]
Find the equation of line L3. Give your answer in the form y = mx + c.
Markscheme
3 = −2(2) + c
(M1)
Note: Award (M1) for correctly substituting –2 and the given point into the
equation of a line.
y = −2x + 7
(A1) (C2)
Note: Award (A0) if the equation is not written in the form y = mx + c.
[2 marks]
The following diagram shows two triangles, OBC and OBA, on a set of axes. Point
C lies on the y-axis, and O is the origin.
6a. The equation of the line BC is
y = 4.
Write down the coordinates of point C.
[1 mark]
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
(0, 4)
(A1)
Notes: Accept
x = 0, y = 4.
6b. The area of triangle AOB is three times the area of triangle OBC.
Find an expression, in terms of a, for
(i)
the area of triangle OBC;
(ii)
the x-coordinate of point A.
[3 marks]
Markscheme
(i)
(ii)
2a
4x
2
(A1)
= 3 × 2a
(M1)
Note: Award (M1) for correct equation.
x = 3a
(A1)(ft)
Note: Follow through from part (d)(i).
OR
0 − 4 = − a4 (x − a)
(M1)
Note: Award (M1) for correct substitution of their gradient and the
coordinates of their point into the equation of a line.
16
a
= x−a
x=a+
16
a
(A1)(ft)
Note: Follow through from parts (b)(i) and (c)(i).
OR
4 × 0 + ax − a2 − 16 = 0
(M1)
Note: Award (M1) for correct substitution of the coordinates of A(x,
the equation of line AB.
0) into
ax − a2 − 16 = 0
x=a+
16
a
OR
x=
6c. Calculate the value of a .
(a2+16)
a
(A1)(G1)
[2 marks]
Markscheme
4(0) + a(3a) − a2 − 16 = 0
(M1)
Note: Award (M1) for correct substitution of their 3a from part (d)(ii) into the
equation of line AB.
OR
1
2 (a
+
16
)×
a
4 = 3 ( 42a )
(M1)
Note: Award (M1) for area of triangle AOB (with their substituted a + 16
and
a
4) equated to three times their area of triangle AOB.
a = 2.83 (2.82842..., 2√2, √8)
(A1)(ft)(G1)
Note: Follow through from parts (d)(i) and (d)(ii).
Professor Vinculum investigated the migration season of the Bulbul bird from their
natural wetlands to a warmer climate.
He found that during the migration season their population, P could be modelled
−t
by P = 1350 + 400(1.25) , t ≥ 0 , where t is the number of days since the start
of the migration season.
7a. Find the population of the Bulbul birds at the start of the migration
season.
[1 mark]
Markscheme
1750
A1
[1 mark]
7b. Find the population of the Bulbul birds after 5 days.
[2 marks]
Markscheme
1350 + 400(1.25)−5
= 1480
(M1)
A1
Note: Accept 1481.
[2 marks]
7c. Calculate the time taken for the population to decrease below 1400.
[2 marks]
Markscheme
1400 = 1350 + 400(1.25)−t
(M1)
9.32 (days (9.31885…) (days)) A1
[2 marks]
7d. According to this model, find the smallest possible population of Bulbul
birds during the migration season.
[1 mark]
Markscheme
1350 A1
Note: Accept 1351 as a valid interpretation of the model as P = 1350 is an
asymptote.
[1 mark]
The intensity level of sound, L measured in decibels (dB), is a function of the
sound intensity, S watts per square metre (W m −2). The intensity level is given by
the following formula.
L = 10 log10 (S × 1012 ), S ≥ 0.
8a. An orchestra has a sound intensity of 6.4 × 10−3 W m −2 . Calculate the
intensity level, L of the orchestra.
[2 marks]
Markscheme
10 log10 (6.4 × 10−3 × 1012 )
= 98.1(dB) (98.06179…)
(M1)
A1
[2 marks]
8b. A rock concert has an intensity level of 112 dB. Find the sound intensity, [2 marks]
S.
Markscheme
112 = 10 log10 (S × 1012 )
(M1)
0.158 (W m−2) (0.158489… (W m−2))
A1
[2 marks]
The braking distance of a vehicle is defined as the distance travelled from where
the brakes are applied to the point where the vehicle comes to a complete stop.
The speed, s m s−1 , and braking distance, d m , of a truck were recorded. This
information is summarized in the following table.
This information was used to create Model A, where d is a function of s,
Model A: d (s)
s ≥ 0.
= ps2 + qs, where p, q ∈ Z
At a speed of 6 m s−1 , Model A can be represented by the equation
6p + q = 2.
9a. Write down a second equation to represent Model A, when the speed is [2 marks]
10 m s−1 .
Markscheme
p(10)2 + q (10) = 60
M1
10p + q = 6 (100p + 10q = 60)
[2 marks]
A1
[2 marks]
9b. Find the values of p and q.
Markscheme
p = 1, q = −4 A1A1
Note: If p and q are both incorrect then award M1A0 for an attempt to solve
simultaneous equations.
[2 marks]
9c. Find the coordinates of the vertex of the graph of y
= d (s).
[2 marks]
Markscheme
(2, −4)
A1A1
Note: Award A1 for each correct coordinate.
Award A0A1 if parentheses are missing.
[2 marks]
9d. Using the values in the table and your answer to part (b), sketch the
[3 marks]
graph of y = d (s) for 0 ≤ s ≤ 10 and −10 ≤ d ≤ 60, clearly showing the
vertex.
Markscheme
A3
Note: Award A1 for smooth quadratic curve on labelled axes and within
correct window.
Award A1 for the curve passing through (0, 0) and (10, 60). Award A1 for
the curve passing through their vertex. Follow through from part (b).
[3 marks]
9e. Hence, identify why Model A may not be appropriate at lower speeds.
[1 mark]
Markscheme
the graph indicates there are negative stopping distances (for low speeds)
R1
Note: Award R1 for identifying that a feature of their graph results in
negative stopping distances (vertex, range of stopping distances…).
[1 mark]
Additional data was used to create Model B, a revised model for the braking
distance of a truck.
Model B: d (s)
= 0.95s2 − 3.92s
9f. Use Model B to calculate an estimate for the braking distance at a speed [2 marks]
of 20 m s−1 .
Markscheme
0.95 × 202 − 3.92 × 20
= 302 (m) (301.6 …)
(M1)
A1
[2 marks]
The actual braking distance at
20 m s−1 is 320 m.
9g. Calculate the percentage error in the estimate in part (e).
[2 marks]
Markscheme
∣∣ 301.6−320 ∣∣ × 100
320
= 5.75 (%) A1
M1
[2 marks]
9h. It is found that once a driver realizes the need to stop their vehicle, 1.6 [3 marks]
seconds will elapse, on average, before the brakes are engaged. During
this reaction time, the vehicle will continue to travel at its original speed.
A truck approaches an intersection with speed s m s−1 . The driver notices the
intersection’s traffic lights are red and they must stop the vehicle within a
distance of 330 m.
Using model B and taking reaction time into account, calculate the maximum
possible speed of the truck if it is to stop before the intersection.
Markscheme
330 = 1.6 × s + 0.95 × s2 − 3.92 × s
M1A1
Note: Award M1 for an attempt to find an expression including
stopping distance (model B) and reaction distance, equated to 330. Award A1
for a completely correct equation.
19.9 (m s−1 ) (19.8988 …)
A1
[3 marks]
Urvashi wants to model the height of a moving object. She collects the following
data showing the height, h metres, of the object at time t seconds.
She believes the height can be modeled by a quadratic function,
h (t) = at2 + bt + c, where a, b, c ∈ R.
10a. Show that 4a + 2b + c
= 34.
[1 mark]
Markscheme
t = 2, h = 34 ⇒ 34 = a22 + 2b + c
⇒ 34 = 4a + 2b + c AG
M1
[1 mark]
10b. Write down two more equations for a ,
[3 marks]
b and c.
Markscheme
attempt to substitute either (5, 38) or (7, 24)
25a + 5b + c = 38
49a + 7b + c = 24
M1
A1
A1
[3 marks]
10c. Solve this system of three equations to find the value of a ,
b and c.
[4 marks]
Markscheme
a = − 53 , b = 13, c =
44
3
M1A1A1A1
[3 marks]
Hence find
[3 marks]
10d. when the height of the object is zero.
Markscheme
− 53 t2 + 13t +
44
3
=0
M1
t = 8.8 seconds M1A1
[3 marks]
[2 marks]
10e. the maximum height of the object.
Markscheme
attempt to find maximum height, e.g. sketch of graph
h = 40.0 metres
M1
A1
[2 marks]
Elvis Presley is an extremely popular singer. Although he passed away in 1977,
many of his fans continue to pay tribute by dressing like Elvis and singing his
songs.
The number of Elvis impersonators, N(t), can be modelled by the function
N(t)= 170 × 1. 31t ,
where t, is the number of years since 1977.
11a. Write down the number of Elvis impersonators in 1977.
[1 mark]
Markscheme
170
(A1)
(C1)
[1 mark]
11b. Calculate the time taken for the number of Elvis impersonators to
reach 130 000.
[2 marks]
Markscheme
130 000 = 170 × 1. 31t
(M1)
Note: Award (M1) for equating
130 000 to the exponential function.
(t =) 24. 6 (24. 5882 … (years))
(A1)
(C2)
[2 marks]
11c. Calculate the number of Elvis impersonators when t
= 70.
[2 marks]
Markscheme
170 × 1. 3170
(M1)
Note: Award (M1) for correct substitution in the function
N(t).
2. 75 × 1010 (2. 75067 … × 1010 , 27 500 000 000, 27 506 771 343)
(A1)
(C2)
[2 marks]
11d. The world population in
2047 is projected to be 9 500 000 000 people.
Use this information to explain why the model for the number of Elvis
impersonators is unrealistic.
[1 mark]
Markscheme
The number of Elvis impersonators in
population.
(R1) (C1)
2047, is greater than the world
OR
2. 75 × 1010 > 9 500 000 000
(R1) (C1)
Note: Award (R1) for a correct comparison of their number of impersonators
with the world population. Follow through from part (c) if a reasonable
argument can be made that the model is unrealistic.
Award (R0) if the number of impersonators is not explicitly seen in part (c) or
in part (d).
[1 mark]
M-Line is a company that prints and sells custom designs on T-shirts. For each
order, they charge an initial design fee and then an additional fee for each printed
T-shirt.
M-Line charges
M euros per order. This charge is modelled by the linear function
M (x) = 5x + 40, where x is the number of T-shirts in the order.
12a. Write down the initial design fee charged for each order.
[1 mark]
Markscheme
40 (euros)
(A1) (C1)
[1 mark]
12b. Find the total amount charged for an order of 94 T-shirts.
Markscheme
(M (94) =) 5 (94) + 40
(M1)
Note: Award (M1) for correct substitution of
510 (euros)
[2 marks]
(A1) (C2)
94 into given function.
[2 marks]
EnYear is another company that prints and sells T-shirts. The price, N euros, that
they charge for an order can be modelled by the linear function N (x) = 9x,
where x is the number of T-shirts in the order.
12c. Write down the number of T-shirts in an order for which EnYear charged [1 mark]
63 euros.
Markscheme
7 (T-shirts)
(A1) (C1)
[1 mark]
12d. An order of p T-shirts will be charged the same price by both M-Line and[2 marks]
EnYear.
Find the value of p.
Markscheme
9p = 5p + 40
(M1)
Note: Award (M1) for equating the given functions. Accept a sketch showing
both functions.
(p =) 10 (T-shirts)
(A1) (C2)
[2 marks]
The graph of the quadratic function f(x)= 12 (x − 2)(x + 8) intersects the y-axis
at (0, c).
13a. Find the value of c.
[2 marks]
Markscheme
1
2 (0 −
2)(0 + 8) OR
1
2
2 (0
+ 6(0)−16) (or equivalent)
Note: Award (M1) for evaluating
(c =) − 8
(M1)
f(0).
(A1)(G2)
Note: Award (G2) if
−8 or (0, − 8) seen.
[2 marks]
The vertex of the function is (−3,
− 12. 5).
[2 marks]
13b. Write down the equation for the axis of symmetry of the graph.
Markscheme
x = −3
(A1)(A1)
Note: Award (A1) for “x = constant”, (A1) for the constant being
answer must be an equation.
−3. The
[2 marks]
The equation
f(x)= 12 has two solutions. The first solution is x = −10.
13c. Use the symmetry of the graph to show that the second solution is
x = 4.
[1 mark]
Markscheme
(−3 − −10)+ − 3
(M1)
OR
(−8 − −10)+2
(M1)
OR
−10+x
2
= −3
(M1)
OR
diagram showing axis of symmetry and given points (x-values labels, −10, −3
and 4, are sufficient) and an indication that the horizontal distances between
the axis of symmetry and the given points are 7.
(M1)
Note: Award (M1) for correct working using the symmetry between x = −10
and x = −3. Award (M0) if candidate has used x = −10 and x = 4 to show
the axis of symmetry is x = −3. Award (M0) if candidate solved f(x)= 12 or
evaluated f(−10) and f(4).
(x =) 4
(AG)
[1 mark]
[2 marks]
13d. Write down the x-intercepts of the graph.
Markscheme
−8 and 2
(A1)(A1)
Note: Accept x = −8, y = 0 and x = 2, y =
at most (A0)(A1) if parentheses are omitted.
0 or (−8, 0) and (2, 0), award
[2 marks]
= ( )
−10 ≤
≤4
13e. On graph paper, draw the graph of y = f(x) for −10 ≤ x ≤ 4 and
[4 marks]
−14 ≤ y ≤ 14. Use a scale of 1 cm to represent 1 unit on the x-axis and 1 cm to
represent 2 units on the y-axis.
Markscheme
(A1)(A1)(A1)(A1)(ft)
Note: Award (A1) for labelled axes with correct scale, correct window. Award
(A1) for the vertex, (−3, − 12. 5), in correct location.
Award (A1) for a smooth continuous curve symmetric about their vertex.
Award (A1)(ft) for the curve passing through their x and y intercepts in
correct location. Follow through from their parts (a) and (d).
If graph paper is not used:
Award at most (A0)(A0)(A1)(A1)(ft). Their graph should go through their −8
and 2 for the last (A1)(ft) to be awarded.
[4 marks]
Let
T be the tangent at x = −3.
13f. Write down the equation of T .
[2 marks]
Markscheme
y = −12. 5 OR y = 0x − 12. 5
(A1)(A1)
Note: Award (A1) for "y = constant", (A1) for the constant being −12. 5. The
answer must be an equation.
[2 marks]
13g. Draw the tangent
[1 mark]
T on your graph.
Markscheme
tangent to the graph drawn at
x = −3
(A1)(ft)
Note: Award (A1) for a horizontal straight-line tangent to curve at
approximately x = −3. Award (A0) if a ruler is not used. Follow through from
their part (e).
[1 mark]
13h. Given f(a)= 5. 5 and f'(a)= −6, state whether the function, f , is
increasing or decreasing at x = a . Give a reason for your answer.
[2 marks]
Markscheme
decreasing
(A1)
x = a) OR f'(a)< 0
(R1)
Note: Do not accept "gradient (of tangent line) is −6". Do not award (A1)
gradient (of tangent line) is negative (at
(R0).
[2 marks]
Little Green island originally had no turtles. After 55 turtles were introduced to the
island, their population is modelled by
N (t) = a × 2−t + 10, t ⩾ 0,
where a is a constant and
14a. Find the value of a .
t is the time in years since the turtles were introduced.
[2 marks]
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
55 = a × 20 + 10 (M1)
Note: Award (M1) for correct substitution of zero and 55 into the function.
45
(A1)
(C2)
[2 marks]
14b. Find the time, in years, for the population to decrease to 20 turtles.
[2 marks]
Markscheme
45 × 2−t + 10 ⩽ 20
(M1)
Note: Award (M1) for comparing correct expression involving 20 and their 45.
Accept an equation.
t = 2.17 (2.16992…)
(A1) (ft) (C2)
Note: Follow through from their part (a), but only if positive.
Answer must be in years; do not accept months for the final (A1).
[2 marks]
14c. There is a number m beyond which the turtle population will not
decrease.
[2 marks]
Find the value of m . Justify your answer.
Markscheme
m =10
(A1)
because as the number of years increases the number of turtles approaches
10
(R1) (C2)
Note: Award (R1) for a sketch with an asymptote at approximately y = 10,
OR for table with values such as 10.003 and 10.001 for t = 14 and t = 15, for
example,
OR when t approaches large numbers y approaches 10. Do not award (A1)
(R0).
[2 marks]
A potter sells x vases per month.
His monthly profit in Australian dollars (AUD) can be modelled by
1
P (x) = − x3 + 7x2 − 120, x ⩾ 0.
5
15a. Find the value of P if no vases are sold.
[1 mark]
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
(A1)
−120 (AUD)
(C1)
[1 mark]
15b. Differentiate
P (x).
[2 marks]
Markscheme
− 35 x2 + 14x
(A1)(A1)
(C2)
Note: Award (A1) for each correct term. Award at most (A1)(A0) for extra
terms seen.
[2 marks]
Let
f (x) = 2 sin (3x) + 4 for x ∈ R.
16a. The range of
f is k ≤ f (x) ≤ m. Find k and m.
[3 marks]
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
valid attempt to find range (M1)
eg
, max = 6 min = 2,
2 sin (3 × π6 ) + 4 and 2 sin (3 × π2 ) + 4 , 2 (1) + 4 and 2 (−1) + 4,
k = 2, m = 6
A1A1 N3
[3 marks]
Let
g (x) = 5f (2x).
[2 marks]
16b. Find the range of g .
Markscheme
10 ≤ y ≤ 30
A2 N2
[2 marks]
The function
g can be written in the form g (x) = 10 sin (bx) + c.
[3 marks]
16c. Find the value of b and of c.
Markscheme
evidence of substitution (may be seen in part (b))
eg
(M1)
5 (2 sin (3 (2x)) + 4) , 3 (2x)
b = 6, c = 20 (accept 10 sin (6x) + 20 )
A1A1 N3
Note: If no working shown, award N2 for one correct value.
[3 marks]
[2 marks]
16d. Find the period of g .
Markscheme
(A1)
correct working
eg
2π
b
1.04719
2π
6
(=
π
), 1.05
3
A1 N2
[2 marks]
16e. The equation
solutions.
g (x) = 12 has two solutions where π ≤ x ≤
4π
. Find both [3 marks]
3
Markscheme
valid approach
(M1)
eg
,
8
sin−1 (− 10
) , 6x = −0.927, − 0.154549, x = 0.678147
Note: Award M1 for any correct value for
domain of f .
x or 6x which lies outside the
3.81974, 4.03424
x = 3.82, x = 4.03 (do not accept answers in degrees)
[3 marks]
3
2
1
A1A1 N3
Consider the function f (x)
= x3 − 5x2 + 6x − 3 + 1x , x > 0
The function f (x) = x3 − 5x2 + 6x − 3 + 1x , x > 0, models the path of a river, as
shown on the following map, where both axes represent distance and are
measured in kilometres. On the same map, the location of a highway is defined by
−x
the function g (x) = 0.5(3)
+ 1.
The origin, O(0, 0) , is the location of the centre of a town called Orangeton.
A straight footpath, P , is built to connect the centre of Orangeton to the river at
the point where x = 12 .
17a. State the domain of P .
[2 marks]
Markscheme
0 < x < 12
(A1)(A1)
Note: Award (A1) for both endpoints correct, (A1) for correct mathematical
notation indicating an interval with two endpoints. Accept weak inequalities.
Award at most (A1)(A0) for incorrect notation such as 0 − 0.5 or a written
description of the domain with correct endpoints. Award at most (A1)(A0)
for 0 < y < 12 .
[2 marks]
Bridges are located where the highway crosses the river.
A straight road is built from the centre of Orangeton, due north, to connect the
town to the highway.
17b. Find the distance from the centre of Orangeton to the point at which
the road meets the highway.
[2 marks]
Markscheme
g (0) = 0.5(3)0 + 1
(M1)
1.5 (km) (A1)(G2)
[2 marks]
17c. This straight road crosses the highway and then carries on due north.
[2 marks]
State whether the straight road will ever cross the river. Justify your answer.
Markscheme
domain given as
x > 0 (but equation of road is x = 0)
(R1)
OR
(equation of road is x = 0) the function of the river is asymptotic to x =
(R1)
so it does not meet the river
0
(A1)
Note: Award the (R1) for a correct mathematical statement about the
equation of the river (and the equation of the road). Justification must be
based on mathematical reasoning. Do not award (R0)(A1).
[2 marks]
The amount of yeast, g grams, in a sugar solution can be modelled by the
function,
g(t) = 10 − k(c−t) for t ≥ 0
where t is the time in minutes.
The graph of g(t) is shown.
The initial amount of yeast in this solution is 2 grams.
The amount of yeast in this solution after 3 minutes is 9 grams.
18. Write down the maximum amount of yeast in this solution.
[1 mark]
Markscheme
10 (grams)
(A1) (C1)
[1 mark]
Consider the function f (x)
=
27
x2
− 16x, x ≠ 0.
19a. Sketch the graph of y = f (x), for −4 ≤ x ≤ 3 and −50 ≤ y ≤ 100.
[4 marks]
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
(A1)(A1)(A1)(A1)
Note: Award (A1) for axis labels and some indication of scale; accept y or f(x).
Use of graph paper is not required. If no scale is given, assume the given
window for zero and minimum point.
Award (A1) for smooth curve with correct general shape.
Award (A1) for x-intercept closer to y-axis than to end of sketch.
Award (A1) for correct local minimum with x-coordinate closer to y-axis than
end of sketch and y-coordinate less than half way to top of sketch.
Award at most (A1)(A0)(A1)(A1) if the sketch intersects the y-axis or if the
sketch curves away from the y-axis as x approaches zero.
[4 marks]
19b. Use your graphic display calculator to find the zero of f (x).
[1 mark]
Markscheme
1.19 (1.19055…)
(A1)
Note: Accept an answer of (1.19, 0).
Do not follow through from an incorrect sketch.
[1 mark]
19c. Use your graphic display calculator to find the coordinates of the local
minimum point.
[2 marks]
Markscheme
(−1.5, 36)
(A1)(A1)
Note: Award (A0)(A1) if parentheses are omitted.
Accept x = −1.5, y = 36.
[2 marks]
19d. Use your graphic display calculator to find the equation of the tangent [2 marks]
to the graph of y = f (x) at the point (–2, 38.75).
Give your answer in the form y = mx + c.
Markscheme
y = −9.25x + 20.3 (y = −9.25 x + 20.25)
(A1)(A1)
Note: Award (A1) for −9.25x, award (A1) for +20.25, award a maximum of
(A0)(A1) if answer is not an equation.
[2 marks]
Consider a function f (x) , for −2 ≤ x ≤ 2 . The following diagram shows the graph
of f.
20. On the grid above, sketch the graph of f −1.
[4 marks]
Markscheme
A1A1A1A1 N4
Note: Award A1 for evidence of approximately correct reflection in y = x with
correct curvature.
(y = x does not need to be explicitly seen)
Only if this mark is awarded, award marks as follows:
A1 for both correct invariant points in circles,
A1 for the three other points in circles,
A1 for correct domain.
[4 marks]
Let f(x) = ax2 − 4x − c. A horizontal line, L , intersects the graph of f at x = −1 and
x = 3.
21a. The equation of the axis of symmetry is x = p. Find p.
[2 marks]
Markscheme
METHOD 1 (using symmetry to find p)
valid approach
eg
(M1)
−1+3
2 ,
p=1
A1 N2
Note: Award no marks if they work backwards by substituting a = 2 into − 2b
a
to find p.
Do not accept
p = 2a .
METHOD 2 (calculating a first)
(i) & (ii) valid approach to calculate a
M1
eg a + 4 − c = a(32) − 4(3) − c, f(−1) = f(3)
correct working
A1
eg 8a = 16
a=2
AG N0
valid approach to find p
(M1)
4
eg − 2b ,
a
2(2)
p=1
A1 N2
[2 marks]
21b. Hence, show that a = 2.
[2 marks]
Markscheme
METHOD 1
valid approach
M1
eg − 2b , 24 (might be seen in (i)), f' (1) = 0
a
a
correct equation
A1
eg 24 = 1, 2a(1) − 4 = 0
a
a=2
AG N0
METHOD 2 (calculating a first)
(i) & (ii) valid approach to calculate a
M1
eg a + 4 − c = a(32) − 4(3) − c, f(−1) = f(3)
correct working
A1
eg 8a = 16
a=2
AG N0
[2 marks]
Consider the quadratic function f (x)
= ax2 + bx + 22.
The equation of the line of symmetry of the graph y
= f (x) isx = 1.75.
22a. Using only this information, write down an equation in terms of a and b. [1 mark]
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
1.75 =
−b
(or equivalent)
2a
(A1) (C1)
f (x) = (1.75)2 a + 1.75b or for
y = (1.75)2 a + 1.75b + 22 or for f (1.75) = (1.75)2 a + 1.75b + 22.
Note: Award (A1) for
[1 mark]
The graph intersects the x-axis at the point (−2 , 0).
22b. Using this information, write down a second equation in terms of a and
b.
[1 mark]
Markscheme
(−2)2 × a + (−2) × b + 22 = 0 (or equivalent)
Note: Award (A1) for
Award (A0) for
(A1) (C1)
(−2)2 × a + (−2) × b + 22 = 0 seen.
y = (−2)2 × a + (−2) × b + 22.
[1 mark]
22c. Hence find the value of a and of b.
[2 marks]
Markscheme
a = −2, b = 7
(A1)(ft)(A1)(ft) (C2)
Note: Follow through from parts (a) and (b).
Accept answers(s) embedded as a coordinate pair.
[2 marks]
22d. The graph intersects the x-axis at a second point, P.
Find the x-coordinate of P.
[2 marks]
Markscheme
−2 x2 + 7x + 22 = 0
(M1)
Note: Award (M1) for correct substitution of a and b into equation and setting
to zero. Follow through from part (c).
(x =) 5.5
(A1)(ft) (C2)
Note: Follow through from parts (a) and (b).
OR
x-coordinate = 1.75 + (1.75 − (−2))
(M1)
Note: Award (M1) for correct use of axis of symmetry and given intercept.
(x =) 5.5
(A1) (C2)
[2 marks]
In an experiment, a number of fruit flies are placed in a container. The population
of fruit flies, P , increases and can be modelled by the function
P (t) = 12 × 30.498t , t ⩾ 0,
where t is the number of days since the fruit flies were placed in the container.
23a. Find the number of fruit flies which were placed in the container.
[2 marks]
Markscheme
12 × 30.498×0
(M1)
Note: Award (M1) for substituting zero into the equation.
= 12
(A1) (C2)
[2 marks]
23b. Find the number of fruit flies that are in the container after 6 days.
[2 marks]
Markscheme
12 × 30.498×6
(M1)
Note: Award (M1) for substituting 6 into the equation.
320
(A1) (C2)
Note: Accept an answer of 319.756… or 319.
[2 marks]
23c. The maximum capacity of the container is 8000 fruit flies.
[2 marks]
Find the number of days until the container reaches its maximum capacity.
Markscheme
8000 = 12 × 30.498×t
(M1)
Note: Award (M1) for equating equation to 8000.
Award (M1) for a sketch of P(t) intersecting with the straight line y = 8000.
= 11.9 (11.8848…)
(A1) (C2)
Note: Accept an answer of 11 or 12.
[2 marks]
The following function models the growth of a bacteria population in an
experiment,
P(t) = A × 2t, t ≥ 0
where A is a constant and t is the time, in hours, since the experiment began.
Four hours after the experiment began, the bacteria population is 6400.
24a. Find the value of A.
[2 marks]
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
6400 = A × 24
(M1)
Note: Award (M1) for correct substitution of 4 and 6400 in equation.
(A =) 400
(A1) (C2)
[2 marks]
24b. Interpret what A represents in this context.
[1 mark]
Markscheme
the initial population OR the population at the start of experiment
(C1)
(A1)
[1 mark]
24c. Find the time since the experiment began for the bacteria population to [3 marks]
be equal to 40A.
Markscheme
40A = A × 2t OR 40 × 400 = 400 × 2t
(M1)
Note: Award (M1) for correct substitution into equation. Follow through with
their A from part (a).
40 = 2t
(M1)
Note: Award (M1) for simplifying.
5.32 (5.32192…) (hours) OR 5 hours 19.3 (19.3156…) minutes
(A1) (C3)
[3 marks]
Consider the graph of the function f (x)
=
3
x
− 2, x ≠ 0.
25a. Write down the equation of the vertical asymptote.
[2 marks]
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
x=0
(A1)(A1) (C2)
Note: Award (A1) for x = “a constant” (A1) for = 0. Award (A0)(A0) for an
answer of “0”.
[2 marks]
25b. Write down the equation of the horizontal asymptote.
[2 marks]
Markscheme
f(x) = −2 (y = −2)
(A1)(A1) (C2)
Note: Award (A1) for y = “a constant” (A1) for = −2. Award (A0)(A0) for an
answer of “−2”.
[2 marks]
25c. Calculate the value of x for which f(x) = 0 .
[2 marks]
Markscheme
3
x
−2=0
(M1)
Note: Award (M1) for equating f(x) to 0.
(x =)
3
2
(1.5)
(A1) (C2)
[2 marks]
A factory produces shirts. The cost, C, in Fijian dollars (FJD), of producing x shirts
can be modelled by
C(x) = (x − 75)2 + 100.
26a. Find the cost of producing 70 shirts.
[2 marks]
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
(70 − 75)2 + 100
(M1)
Note: Award (M1) for substituting in x = 70.
125
(A1) (C2)
[2 marks]
The cost of production should not exceed 500 FJD. To do this the factory needs to
produce at least 55 shirts and at most s shirts.
26b. Find the value of s.
[2 marks]
Markscheme
(s − 75)2 + 100 = 500
(M1)
Note: Award (M1) for equating C(x) to 500. Accept an inequality instead of =.
OR
(M1)
Note: Award (M1) for sketching correct graph(s).
(s =) 95
(A1) (C2)
[2 marks]
26c. Find the number of shirts produced when the cost of production is
lowest.
[2 marks]
Markscheme
(M1)
Note: Award (M1) for an attempt at finding the minimum point using graph.
OR
95+55
2
(M1)
Note: Award (M1) for attempting to find the mid-point between their part (b)
and 55.
OR
(C'(x) =) 2x − 150 = 0
(M1)
Note: Award (M1) for an attempt at differentiation that is correctly equated
to zero.
75
(A1) (C2)
[2 marks]
A quadratic function f is given by f(x)
(−4, 5) lie on the graph of y = f(x).
= ax2 + bx + c. The points (0, 5) and
The y-coordinate of the minimum of the graph is 3.
27. Find the value of a and of b.
[3 marks]
Markscheme
− 2ba = −2
a(−2)2 − 2b + 5 = 3 or equivalent
a(−4)2 − 4b + 5 = 5 or equivalent
2a(−2) + b = 0 or equivalent
Note:
(M1)
Award (M1) for two of the above equations.
a = 0.5 (A1)(ft)
b = 2 (A1)(ft) (C3)
Note:
Award at most (M1)(A1)(ft)(A0) if the answers are reversed.
Follow through from parts (a) and (b).
[3 marks]
Sejah placed a baking tin, that contained cake mix, in a preheated oven in order
to bake a cake. The temperature in the centre of the cake mix, T , in degrees
Celsius (°C) is given by
T (t) = 150 − a × (1.1)−t
where t is the time, in minutes, since the baking tin was placed in the oven. The
graph of T is shown in the following diagram.
28a. Write down what the value of 150 represents in the context of the
question.
[1 mark]
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
the temperature in the oven
(A1)
OR
the maximum possible temperature of the cake mix
Note:
(A1)
(C1)
Award (A0) for “the maximum temperature”.
[1 mark]
The temperature in the centre of the cake mix was 18 °C when placed in the oven.
[2 marks]
28b. Find the value of a .
Markscheme
18 = 150 − a(×1.1∘ )
(M1)
Note:
Award (M1) for correct substitution of 18 and 0. Substitution of 0
can be implied.
(a) = 132
(A1)
(C2)
[2 marks]
The baking tin is removed from the oven 15 minutes after the temperature in the
centre of the cake mix has reached 130 °C.
28c. Find the total time that the baking tin is in the oven.
[3 marks]
Markscheme
150 − 132 × 1.1−t = 130
(M1)
Note:
Award (M1) for substituting their a and equating to 130. Accept an
inequality.
Award (M1) for a sketch of the horizontal line on the graph.
t = 19.8 (19.7992 …)
Note:
(A1)(ft)
Follow through from part (b).
34.8 (minutes) (34.7992…, 34 minutes 48 seconds)
Note:
(A1)(ft)
(C3)
Award the final (A1) for adding 15 minutes to their t value.
In part (c), award (C2) for a final answer of 19.8 with no working.
[3 marks]
The size of a computer screen is the length of its diagonal. Zuzana buys a
rectangular computer screen with a size of 68 cm, a height of y cm and a width of
x cm, as shown in the diagram.
29a. Use this information to write down an equation involving x and y.
[1 mark]
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
x2 + y 2 = 682 (or 4624 or equivalent)
(A1)
(C1)
[1 mark]
The ratio between the height and the width of the screen is 3:4.
29b. Use this ratio to write down y in terms of
x.
[2 marks]
Markscheme
y
x
=
3
4
Note:
(M1)
Award (M1) for a correct equation.
y = 34 x (y = 0.75x)
(A1)
(C2)
[2 marks]
29c. Find the value of x and of y.
[3 marks]
Markscheme
2
x2 + ( 34 x) = 682 (or x2 +
9 2
x
16
= 4624 or equivalent)
(M1)
Note:
Award (M1) for correct substitution of their expression for y into
their answer to part (a). Accept correct substitution of x in terms of y.
x = 54.4 (cm), y = 40.8 (cm)
Note:
(A1)(ft)(A1)(ft)
(C3)
Follow through from parts (a) and (b) as long as
x > 0 and y > 0.
[3 marks]
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