CHAPTER 11 REVIEW SOLUTIONS
1.
(a) y = 1.25 – a0 1.25 – 1
(M1)
= 0.25
(A1) (C2)
Note: Award (M1)(A1) for (0, 0.25).
(b) 1 = 1.25 – a–2
(M1)
a =2
(A1) (C2)
(c) y = 1.25
(A1)(A1) (C2)
Note: Award (A1) for y = “a constant”, (A1) for 1.25.
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2.
(a)
(b)
(i)
(A1)(A1)(A1)
Notes: Award (A1) for labels and scale on y-axis.
Award (A1) for smooth increasing curve in the given domain.
Award (A1) for asymptote implied (gradient → 0).
(ii) (0, 2) (accept x = 0, y = 2)
(A1) (C4)
Note: If incorrect domain used and both (0, 2) and (–0.737, 0) seen
award (A1)(ft).
line passing through (0, 5), parallel to x-axis and not intersecting
their graph.
(A1) (C1)
zero
(A1) (C1)
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3.
(a)
p+q=6
0.5p + q = 4
(b)
p = 4, q = 2
(c)
y=2
(A1)
(A1) (C2)
Note: Accept correct equivalent forms of the equations.
(A1)(A1)(ft) (C2)
Notes: If both answers are incorrect, award (M1) for attempt at solving
simultaneous equations.
(A1)(A1)(ft) (C2)
Notes: Award (A1) for “y = a constant”, (A1)(ft) for 2. Follow through
from their value for q as long as their constant is greater than 2 and less
than 6.
An equation must be seen for any marks to be awarded.
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IB Questionbank Mathematical Studies 3rd edition
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4.
(a)
(b)
f(–2) = 2 × 3–2
(M1)
2
=
(0.222)
(A1)
9
f(5) = 2 × 35
= 486
(A1)
2
2


Range
≤ f(x) ≤ 486 OR  , 486
(A1) (C4)
9
9


Note: Award (M1) for correct substitution of –2 or 5 into f(x),
(A1)(A1)for each correct end point.
x
2 × 3 = 162
(M1)
x=4
(A1) (C2)
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5.
(a)
(b)
(c)
a = 1800
(A1) (C1)
6
200 × 3 (or 16 200 × 9) = 145 800
(M1)(A1) (C2)
n
6
200 × 3 = 2  10 (where n is each 4 hour interval)
(M1)
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
(A1)
Time = 33.5 hours
(A1)(ft)
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).
(C3)
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6.
(a)
(b)
(c)
f(0) = a0 + b = 6
(M1)
b=5
(A1)
f(1) = a1 + 5 = 9  a = 4
(A1) (C3)
Note: (M1) for attempt at solving any appropriate equations
(simultaneously).
(A1)(A1) for each correct answer.
f(2) = 21
(A1)(ft) (C1)
Note: Follow through from their f(x)
4c + 5 = 5.5
(M1)
Note: Correct substitution in their f(x)
1
c= 
(A1)(ft) (C2)
2
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7.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
B
D
A
E
C
F
(A1)
(A1)
(A1)
(A1)
(A1)
(A1) (C6)
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IB Questionbank Mathematical Studies 3rd edition
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8.
(a)
(b)
(i)
(ii)
(A4) (C4)
Note: Award (A1) for some indication of scale on the y-axis. Award (A1)
for at least one asymptote drawn. Award (A1) for each of the two
(smooth) branches. The left hand branch must pass through 0. One
branch should be above the horizontal asymptote and the other below but
if the asymptote is not drawn, then there should be little or no overlap in
heights of the branches. If this condition is not fulfilled, award (A1)(A0)
for the curve.
Horizontal asymptote y = 3
(A1)
Vertical asymptote x = 1
(A1)(ft) (C2)
Equations for x and y must be seen, (ft) if reversed.
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9.
(a)
(A4) (C4)
(b)
Notes: Award (A1) for correct scales.
Award (A1)(A1) for two correct parts to the graph.
Award (A1) if asymptotes are shown.
Horizontal asymptote y = 1.
(A1) (C1)
(c)
Vertical asymptote x = –2.
(A1) (C1)
(a)
(b)
(c)
(x – 2)(x – 4)
(A1)(A1)
x = 2, x = 4
(A1)(ft)(A1)(ft)
x = 0.807, x = 6.19
(A1)(A1)
Note: Award maximum of (A0)(A1) if coordinate pairs given.
OR
(M1) for an attempt to solve x2 – 7x + 5 = 0 via formula with
correct values substituted.
(M1)
7  29
x=
(A1)
2
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10.
(C2)
(C2)
(C2)
(C2)
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IB Questionbank Mathematical Studies 3rd edition
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11.
(a)
With the given domain, the correct answer is
(A1)
(A1)(ft)
(A1)(ft) (C3)
(b)
(c)
Notes: Award (A1) for a neat window complying reasonably with the
requirements.
The window must clearly have used x values from –3 to 3 and y values at
least from 0 to 1. Axes labels are not essential. Some indication of scale
must be present but this need not be a formal scale, eg tick marks, a
single number on each axis or coordinates of the intersection are all
adequate.
Award (A1) for each curve correct and correctly labelled with f and g or
the expressions for f and g. Can follow through both curves, for example
if curves are incomplete due to a poor window, and penalize only once if
both curve labels are missing. Examiners should familiarize themselves
1
with the graph of 2  1 as this is expected to appear in error. With the
x
correct window, this graph will not be seen at all, but with a larger y
interval it might look a little like the correct graph except that it would
have asymptotes at x = 0 and y = 1. Award (A0) for this curve.
One solution.
(A1)(ft) (C1)
Solution occurs at the point of intersection of the curves,
where x = 0.569840
0.570.
(M1)(A1) (C2)
Notes: The (M1) can also be awarded for the intersection point indicated on the
sketch.
(0.57 is an (AP))
If a coordinate pair is given as the answer and the x value is correct with
no method mentioned, award (C1) or if the method is mentioned, award
(M1)(A0).
1
Can follow through if curve 2  1 is drawn, answer to (c) is then 1.75.
x
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12.
(a)
(b)
A( –1.79, 0.789) and B(1.14, 2.70)
(C2) (C2)
Notes: Award (C2) for each pair of coordinates obtained from the GDC
Award (A1)(A2)(ft) if bracket is not used.
–1.79  x  1.14
(A1)(ft)(A1)(ft) (C2)
Note: Award (A1) for both numbers, (A1) for correct inequalities.
[6]
IB Questionbank Mathematical Studies 3rd edition
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