### Chapter 16 and 17 – Exponential Functions

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IB Math Studies – Chapter 16 and 17 – Exponential Functions – Review Questions
1.
The diagrams below are sketches of some of the following functions.
x
2
2
(i)
y=a
(ii)
y=x –a
(iii)
y=a–x
(iv) y = a – x
(v)
y=x–a
(a)
(b)
y
y
x
(c)
x
DIAGRAMS NOT
TO SCALE
(d)
y
y
x
x
Complete the table to match each sketch to the correct function.
Sketch
Function
(a)
(b)
(c)
(d)
Working:
(Total 8 marks)
1
2.
The following diagrams show the graphs of five functions.
I
y
y
II
2
1
2
1
–2 –1
–1
–2
III
1
2
–2 –1
–1
–2
x
y
2
x
1
2
x
y
IV
2
2
1
–2 –1
–1
–2
1
1
1
2
V
–2 –1
–1
–2
x
y
2
1
–2 –1
–1
–2
1
2
x
Each of the following sets represents the range of one of the functions of the graphs.
(a)
{y  y  }
(b)
{y  y  2}
(c)
{y  y > 0}
(d)
{y 1 ≤ y ≤ 2}
Write down which diagram is linked to each set.
Working:
(a) …………………………………………..
(b) …………………………………………..
(c) …………………………………………..
(d) …………………………………………..
(Total 4 marks)
2
3.
In an experiment researchers found that a specific culture of bacteria increases in number according to the formula
t
N = 150 × 2 ,
where N is the number of bacteria present and t is the number of hours since the experiment began.
Use this formula to calculate
(a)
the number of bacteria present at the start of the experiment;
(b)
the number of bacteria present after 3 hours;
(c)
the number of hours it would take for the number of bacteria to reach 19 200.
Working:
(a) …………………………………………..
(b) …………………………………………..
(c) …………………………………………..
(Total 4 marks)
4.
In an experiment it is found that a culture of bacteria triples in number every four hours.
There are 200 bacteria at the start of the experiment.
0
4
8
12
16
Hours
200
600
a
5400
16200
No. of bacteria
(a)
Find the value of a.
(1)
(b)
Calculate how many bacteria there will be after one day.
(c)
Find how long it will take for there to be two million bacteria.
(2)
(3)
Working:
(a) ...................................................
(b) ...................................................
(c) ...................................................
(Total 6 marks)
3
5.
2
2
The area, A m , of a fast growing plant is measured at noon (12:00) each day. On 7 July the area was 100 m . Every day
the plant grew by 7.5%. The formula for A is given by
t
A = 100 (1.075)
where t is the number of days after 7 July. (on 7 July, t = 0)
t
The graph of A = 100(1.075) is shown below.
A
400
300
200
100
–6
(a)
–4
–2
0
2
4
6
8
10
12
14
16
t
18
7 July
What does the graph represent when t is negative?
(2)
(b)
Use the graph to find the value of t when A = 178.
(1)
(c)
Calculate the area covered by the plant at noon on 28 July.
(3)
(Total 6 marks)
6.
The number of bacteria (y) present at any time is given by the formula:
–025t
y = 15 000e
, where t is the time in seconds and e = 2.72 correct to 3 s.f.
(a)
Calculate the values of a, b and c to the nearest hundred in the table below:
Time in seconds (t)
0
1
2
3
4
5
6
7
Amount of bacteria (y)
a 11700 9100 7100
b 4300 3300 2600
(nearest hundred)
8
c
(3)
(b)
On graph paper using 1 cm for each second on the horizontal axis and 1 cm for each thousand on the vertical
axis, draw and label the graph representing this information.
(5)
(c)
Using your graph, answer the following questions:
(i)
After how many seconds will there be 5000 bacteria? Give your answer correct to the nearest tenth of a
second.
(ii)
(iii)
How many bacteria will there be after 6.8 seconds? Give your answer correct to the nearest hundred
bacteria.
Will there be a time when there are no bacteria left? Explain your answer.
(6)
(Total 14 marks)
4
7.
Under certain conditions the number of bacteria in a particular culture doubles every 10 seconds as shown by the graph
below.
8
7
6
5
Number
of
4
bacteria
3
2
1
0
10
20
30
Time
(seconds)
(a)
Complete the table below.
Time (seconds)
Number of bacteria
0
1
10
20
30
(b)
Calculate the number of bacteria in the culture after 1 minute.
Working:
(b) ..................................................................
(Total 4 marks)
8.
x
The diagram below shows a part of the graph of y = a . The graph crosses the y-axis at the point P. The point Q (4, 16)
is on the graph.
y
Q
(4, 16)
P
Diagram not to scale
O
x
Find
(a)
the coordinates of the point P;
(b)
the value of a.
Working:
(a) ..................................................................
(b) ……………………………………..........
(Total 8 marks)
5
9.
2
x
The figure below shows the graphs of the functions y = x and y = 2 for values of x between –2 and 5.
The points of intersection of the two curves are labelled as B, C and D.
y
20
C
15
10
5
B
D A
–2
–1
1
(a)
Write down the coordinates of the point A.
(b)
Write down the coordinates of the points B and C.
(c)
Find the x-coordinate of the point D.
2
3
4
5
x
(2)
(2)
(1)
(d)
x
2
Write down, using interval notation, all values of x for which 2 ≤ x .
(3)
(Total 8 marks)
10.
A rectangle has dimensions (5 + 2x) metres and (7 – 2x) metres.
(a)
2
Show that the area, A, of the rectangle can be written as A = 35 + 4x – 4x .
(1)
(b)
2
The following is the table of values for the function A = 35 + 4x – 4x .
x
–3
–2
–1
0
1
2
3
4
A
–13
p
27
35
q
r
11
s
(i)
Calculate the values of p, q, r and s.
(ii)
On graph paper, using a scale of 1 cm for 1 unit on the x-axis and 1 cm for 5 units on the A-axis, plot the
points from your table and join them up to form a smooth curve.
(6)
(c)
Answer the following, using your graph or otherwise.
(i)
Write down the equation of the axis of symmetry of the curve,
2
(ii)
Find one value of x for a rectangle whose area is 27 m .
(iii) Using this value of x, write down the dimensions of the rectangle.
(4)
(d)
(i)
(ii)
On the same graph, draw the line with equation A = 5x + 30.
2
Hence or otherwise, solve the equation 4x + x – 5 = 0.
(3)
(Total 14 marks)
6
11.
(a)
2
A function f (x) is defined by f (x) = 2x – 10x + 60, –5  x  8.
x
–5
0
2
5
f (x)
160
a
b
60
(i)
Write down the values of a and b.
8
108
(2)
(ii)
Using the values in the above table, draw the graph of f (x) on a set of coordinate axes. Use a scale of 1 cm
to represent 1 unit on the horizontal axis and 1 cm to represent 20 units on the vertical axis.
(iii)
Show that the coordinates of the vertex of the graph are (2.5, 47.5)
(iv)
State the values of x for which the function is increasing.
(4)
(3)
(2)
(b)
A second function h (x) is defined by:
h (x),= 80, 0  x  8.
(i)
On the same axes used for part (a), draw the graph of h (x).
(2)
(ii)
Find the coordinates of the point at which f (x) = h (x).
(iii)
Find the vertical distance from the vertex of the graph
of f (x) to the line h (x).
(2)
(2)
(Total 17 marks)
12.
The following figures show the graphs of y = f (x) with f (x) chosen to be various cubic functions. The value of a is
positive.
y
y
y
B
C
A
a
a
a
a
x
a
x
a
y
y
E
D
a
a
a
(a)
(b)
(c)
x
x
a
x
In the table below, write the letter corresponding to the graph of y = f (x) in the space next to the cubic function.
(Note: one of the graphs is not represented in this table)
cubic function f (x)
graph label
3
f (x) = x + a
3
f (x) = (x – a) + a
3
f (x) = x
3
f (x) = (x – a)
State which one of the graphs represents a function that has a positive gradient for every value of x.
State how many of the graphs have the x-axis as a tangent at some point.
(Total 6 marks)
7
13.
(a)
Sketch the graph of the function f (x) =
2x  3
, for −10  x  10. Indicating clearly the axis intercepts and any
x 4
asymptotes.
(6)
(b)
Write down the equation of the vertical asymptote.
(c)
On the same diagram sketch the graph of g (x) = x + 0.5.
(d)
Using your graphical display calculator write down the coordinates of one of the points of intersection on the
graphs of f and g, giving your answer correct to five decimal places.
(e)
Write down the gradient of the line g (x) = x + 0.5.
(f)
The line L passes through the point with coordinates (−2, −3) and is perpendicular to the line g (x). Find the
equation of L.
(2)
(2)
(3)
(1)
(3)
(Total 17 marks)
14.
It is not necessary to use graph paper for this question.
3
2
(a)
Sketch the curve of the function f (x) = x − 2x + x − 3 for values of x from −2 to 4, giving the intercepts with
both axes.
(3)
(b)
On the same diagram, sketch the line y = 7 − 2x and find the coordinates of the point of intersection of the line
with the curve.
(c)
Find the value of the gradient of the curve where x =1.7.
(3)
(2)
(Total 8 marks)
15.
x
for –10 ≤ x ≤ 10.
2 x
(a)
Sketch a graph of y =
(b)
Hence write down the equations of the horizontal and vertical asymptotes.
(Total 6 marks)
16.
The functions f and g are defined by
f:x
(a)

x4
,x
x
,x0
g : x  x, x 
Sketch the graph of f for –10 ≤ x ≤ 10.
(4)
(b)
Write down the equations of the horizontal and vertical asymptotes of the function f.
(4)
(c)
Sketch the graph of g on the same axes.
(2)
(d)
Hence, or otherwise, find the solutions of
x4
= x.
x
(4)
(e)
Write down the range of function f.
(2)
(Total 16 marks)
8
17.
At the circus a clown is swinging from an elastic rope. A student decides to investigate the motion of the clown. The
x
results can be shown on the graph of the function f (x) = (0.8 )(5 sin 100x), where x is the horizontal distance in metres.
(a)
Sketch the graph of f (x) for 0 ≤ x ≤ 10 and –3 ≤ f (x) ≤ 5.
(5)
(b)
Find the coordinates of the first local maximum point.
(c)
Find the coordinates of one point where the curve cuts the y-axis.
(2)
(1)
Another clown is fired from a cannon. The clown passes through the points given in the table below:
Horizontal distance (x)
Vertical distance (y)
0.00341
0.0102
0.0238
0.0714
0.563
1.69
1.92
5.76
3.40
10.2
(d)
Find the correlation coefficient, r, and comment on the value for r.
(3)
(e)
Write down the equation of the regression line of y on x.
(f)
Sketch this line on the graph of f (x) in part (a).
(g)
Find the coordinates of one of the points where this line cuts the curve.
(2)
(1)
(2)
(Total 16 marks)
IB Math Studies – Chapter 16 and 17 – Exponential Functions – Mark Scheme
1.
(a)
(iv)
(b)
(i)
(c)
(ii)
(d)
(v)
(A2)
(C2)
(A2)
(C2)
(A2)
(C2)
(A2)
(C2)
[8]
2.
(a)
(b)
(c)
(d)
III
I
II
IV
(A1)
(A1)
(A1)
(A1)
[4]
3.
(a)
0
N = 150×2 = 150
(b)
3
N = 150×2 = 1200
(c)
19200 = 150×2
t
128 = 2
7=t
t
(A1)
(C1)
(A1)
(C1)
(M1)
(A1)
(C2)
[4]
9
4.
(a)
a = 1800
(b)
6
200 × 3 (or 16 200 × 9) = 145 800
(c)
n
6
200 × 3 = 2  10 (where n is each 4 hour interval)
Notes: Award (M1) for attempting to set up the
equation or writing a list of numbers.
n
4
3 = 10
n = 8.38 (8.383613097) correct answer only
Time = 33.5 hours
Notes: Accept 34, 35 or 36 if previous A mark
awarded
(A1)(ft) for correctly multiplying their answer by 4.
If 34, 35 or 36 seen, or 32–36 seen, award (M1)(A0)(A0).
(A1)
(C1)
(M1)(A1)
(C2)
(M1)
(A1)
(A1)(ft)
(C3)
[6]
5.
(a)
The area covered before 7 July
Note: Award (R1) for “area”, (R1) for “before” 7 July.
(R2)
2
(b)
t = 8  0.4
(A1)
1
(c)
100(1.075)
2
= 457 m
21
(M1)(A1)
(A1)
3
[6]
Note: Award (M1) for correct formula, (A1) for correct power and (A1) for correct
6.
(a)
a = 15000
b = 5500
c = 2000
(A1)
(A1)
(A1)
3
(b)
15 000
14 000
13 000
12 000
11 000
10 000
number
of
bacteria
9000
8000
7000
6000
5000
4000
3000
2000
1000
0
1
2
5
3
6
4
time (in seconds)
7
8
9
5
Note: Award (A1) for axes correctly labelled, (A1) for correct scales, (A1) for
smooth curve, (A2) for all points correctly plotted, (A1) for at least 4 points correct.
10
(c)
(i)
4.4 secs
Note: Award (M1)(A1)(ft) from graph (see (b)) or (A1) if correct and no line seen.
(M1)(A1)
(ii)
2700 bacteria (±200 bacteria)
Note: Award (M1)(A1)(ft) from graph (see (b)) or (A1) if correct and no line seen.
(M1)(A1)
(iii)
No ― theoretically, the curve never touches the horizontal axis
(or any answer to suggest that the horizontal axis is
an asymptote).
Note: Award (A1)(R1) for any time over 39 seconds with a reasonable explanation
(when the number of bacteria is less than one).
Award (A0)(R0) for a yes or no with no explanation. Do not award (A1) if (R1) is not
awarded.
(A1)(R1)
6
[14]
7.
(a)
Time (seconds)
0
10
20
30
Number of bacteria
1
2
4
8
Note: Award [½ mark] for each correct entry (round up)
(b)
N=2
(A2)
(C2)
6
(M1)
Note: Award (M1) for any correct method
= 64
(A1)
(C2)
[4]
8.
(a)
(0,1)
(b)
4
16 = a
a=2
(A2)(A2)
(C4)
(M2)
(A2)
(C4)
[8]
9.
(a)
A = (0, 1)
For parentheses
For numbers
(A1)
(A1)
2
B = (2, 4), C = (4, 16)
For 2,4
For 4,16
(A1)
(A1)
2
(c)
At D, x = –0.767
(A1)
1
(d)
x  –0.767
2x4
For inequalities
For numbers
OR
[–, –0.767]  [2, 4]
(A1)
(b)
(A1)
(A1)
(A3)
3
[8]
11
10.
(a)
A = (5 + 2x)(7 – 2x)
= 35 – 10x + 14x – 4x
2
= 35 + 4x – 4x
(b)
(i)
(M1)
2
(AG)
p = 11, q = 35, r = 27, s = –13
Note: Award (A2) for all four correct, (A1) for two or three correct.
1
(A2)
(ii)
A
45
40
35
30
25
20
15
10
5
–3
–2
–1
0 1
–5
2
3
4
5
6
x
–10
–15
(A4)
6
Notes: Award (A1) for axes with correct scales and labelling.
Award (A2) for 6, 7 or 8 points correctly plotted,
(A1) for 3, 4, or 5 points,
(A0) for 2 or fewer.
Award (A1) for a smooth curve through reasonably correct points.
(c)
(d)
(i)
Axis of symmetry is x =
1
2
(A1)
(ii)
A = 27  x = –1 or x = 2
Note: Award (A1) for one correct value of x.
(A1)
(iii)
x = –1, rectangle is (5 – 2) × (7 + 2)
i.e. 3 × 9
OR
x = 2, rectangle is (5 + 4) × (7 – 4)
i.e. 9 × 3
Notes: Award (A2) for the correct answer.
Follow through with answers for x from the candidate’s graph.
(M1)
(A1)
Line on graph.
From graph solutions are x = 1 and x = –1.3 (0.1)
(Follow through with candidate’s graph of parabola
and straight line.)
OR
Factorizing gives (x – 1)(4x + 5) = 0
 x = 1 or x = –1.25
(A1)
(A2)
(i)
(ii)
(M1)
(A1)
(M1)
(A1)
4
3
[14]
12
11.
(a)
(i)
a = 60
b = 48
(A1)
(A1)
2
(ii)
Labels and scale
all points correct
Note: At most one error (A1)(A0)
Smooth curve (not straight lines) drawn in given domain only
Note: Graphs not drawn on graph paper must be drawn
very accurately to receive marks
(iii)
x
 b  (10)

 2.5
2a
2(2)
(A1)
(A2)(ft)
(A1)(ft)
4
(M1)(A1)
(AG)
Note: (A1) is for correct substitution
OR
dy
 4 x 10
dx
4x  10 = 0, x = 2.5
(A1)
(M1)
(AG)
Note: (M1) is for setting derivative to 0.
OR
For correct symmetry argument leading to x = 2.5
(R2)
(AG)
2
f(2.5) = 2(2.5) – 10(2.5)+60=47.5
(M1)
(AG)
Notes: If graph is used and lines are drawn to the vertex from x = 2.5 and y = 47.5,
award (M1)(A0)(M1).
If vertex is indicated on graph, with or without coordinates, Award (M1) only.
If GDC is used and a sketch showing the vertex is given, award (G1) only.
(iv)
(b)
(i)
The correct answer is 2.5  x < 8 or [2.5, 8)
Accept x  2.5 or [2.5, ∞)
Notes: Allow the inequalities to be strict or non-strict
Award (A1) for 2.5 and 8 both seen.
See graph in part (a)
Domain
Note: Penalize domain only once in the question.
horizontal straight line, intercept at 80
Note: If graphs are drawn on separate axes award at most (A0)(A1).
(A2)
3
2
(A1)(ft)
(A1)
2
13
(ii)
(iii)
From GDC, x = 6.5311…
Point of intersection is (6.53, 80)
(A1)(A1)
Notes: Award (A1) for each coordinate. Award (A1)(A0) if brackets are missing.
Allow x = 6.53, y = 80
If both points are given, award appropriate marks for (6.53, 80).
If the x- coordinate is read from the graph, the method must be shown by a line
drawn on the graph or the point of intersection clearly marked. The answer can be
given to 1dp. Award (A1)(ft) for candidates intercept value  0.1.
If no method shown an answer of (6.5, 80) or 6.5, 80 receives (A0)(A1). This is not
an AP.
80 – 47.5
For subtracting appropriate values or for showing the distance on
the graph
For 32.5.
Note: Award (A0) for any other answer
2
(M1)
(A1)(G2) 2
[17]
12.
(a)
cubic function f(x)
3
f(x) = x + a
3
f(x) = (x – a) + a
3
f(x) = x
3
f(x) = (x – a)
graph label
B
(A1)
E
(A1)
C
(A1)
D
(A1) (C4)
Note: If the same letter is written in every box, award (A0) not (A1).
(b)
Graph A.
(A1)
(C1)
(c)
Two of them. (Allow “D and C”).
(A1)
(C1)
[6]
13.
(a)
20
6
y
10
x
10
–10
–10
Notes: (A1) for labels and some idea of scale.
(A1) for x-intercept seen, (A1) for y-intercept seen in
roughly the correct places (coordinates not required).
(A1) for vertical asymptote seen, (A1) for horizontal
asymptote seen in roughly the correct places (equations
of the lines not required).
(A1) for correct general shape.
(A6)
14
(b)
x = −4
(A1)(A1)(ft)
2
Note: (A1) for x =, (A1)(ft) for −4
(c)
20
2
y
10
x
10
–10
–10
Note: (A1) for correct axis intercepts,
(A1) for straight line
(d)
(–2.85078, –2.35078) OR (0.35078, 0.85078)
Notes: (A1) for x-coordinate, (A1) for y-coordinate,
(A1)(ft) for correct accuracy. Brackets required.
If brackets not used award (G1)(G0)(A1)(ft).
Accept x = −2.85078, y = −2.35078
or x = 0.35078, y = 0.85078
(e)
(f)
gradient of perpendicular = −1
Note: Can be implied in the next step
y = mx + c
−3 = −1×−2 + c
c = −5
y = −x − 5
OR
y + 3 = − (x + 2)
Note: Award (G2) for correct answer with no
working at all but (A1)(G1) if the gradient is
mentioned as −1 then correct answer with no
further working.
(A1)(A1)
(G1)(G1)(A1)(ft)
3
(A1)
1
(A1)(ft)
(M1)
(A1)(ft)(G2)
OR
(M1)(A1)(ft)(G2)
3
[17]
14.
(a)
(G3) 3
(b)
line drawn with –ve gradient and +ve y-intercept
(2.45, 2.11)
(G1)
(G1)(G1)
3
15
(c)
2
f(1.7) = 3(1.7) – 4(1.7) + 1
Note: Award (M1) for substituting in their f(x)
2.87
(M1) 2
(A1)(G2)
[8]
15.
(a)
(A4)
(C4)
Notes: Award (A1) for correct scales.
Award (A1)(A1) for two correct parts to the graph.
Award (A1) if asymptotes are shown.
(b)
Horizontal asymptote y = 1.
(A1)
(C1)
[6]
16.
(a)
y
–4
x
For x-axis from –10 to 10.
For –4 marked.
For correct shape of graph.
(A1)
(A1)
(A1)(A1)
Horizontal asymptote
y=1
Vertical asymptote
x=0
(A1)
(A1)
(A1)
(A1)
4
(c)
Line drawn on sketch
(A2)
2
(d)
(2.56, 2.56) (–1.56, –1.56)
(b)
(A1)(A1)(
4
A1)(A1)
(e)
Range y 
4
,y1
(A1)(A1)
2
[16]
16
17.
(a)
5
4
3
2
1
–1
1
2
3
4
5
6
8
9
10
–2
–3
For labels and scales.
3 maxima drawn.
2 minima drawn.
General shape
(A1)
(A1)
(A1)
(A2)
5
(b)
(0.827, 4.12)
(G2)
2
(c)
0, 1.8, 3.6, 5.4, 7.2, 9 (for any one of these answers).
(G1)
1
(d)
r=1
Perfect positive correlation.
(G2)
(R1)
3
(e)
y = 3x (accept y = 3x + 0.000274)
(G2)
2
(f)
line on graph
(A1)
1
(g)
(0, 0) or (1.16, 3.48)
(G1) (G1)
2
[16]
17
```