AI Functions Summer [61 marks]
1. [Maximum mark: 4] 23N.1.SL.TZ1.6
Consider the following function:
h(x) =
2
+
1
2
, for x > 1.
√ x−1
(a) Find h−1(1). [2]
Markscheme
setting h(x) = 1 (M1)
(h
−1
(1) =)17 A1
[2 marks]
(b) Find the domain of h−1(x). [2]
Markscheme
x >
1
2
A1A1
Note: Award A1 for 1
2
seen, A1 for completely correct answer.
[2 marks]
2. [Maximum mark: 7] 21M.1.AHL.TZ2.2
A function is defined by f (x) = 2 −
12
x+5
for −7 ≤ x ≤ 7, x ≠ −5.
(a) Find the range of f . [3]
Markscheme
(f (−7) =) 8 and (f (7) =) 1 (A1)
range is f (x) ≤ 1, f (x) ≥ 8 A1A1
Note: Award at most A1A1A0 if strict inequalities are used.
[3 marks]
(b) Find an expression for the inverse function f (x). The
−1
domain is not required. [3]
Markscheme
interchanging x, y at any stage (A1)
12
y = 2 −
x+5
12
= 2 − y
x+5
12
2−y
= x + 5 (A1)
12
− 5 = x
2−y
12 2+5x
( f
−1
(x) =)
2−x
− 5 (=
2−x
) A1
[3 marks]
(c) Write down the range of f
−1
(x). [1]
Markscheme
range is −7 ≤ f
−1
(x) ≤ 7, f
−1
(x) ≠ −5 A1
[1 mark]
3. [Maximum mark: 6] 21N.1.SL.TZ0.4
Dilara is designing a kite ABCD on a set of coordinate axes in which one unit
represents 10 cm.
The coordinates of A, B and C are (2, 0), (0, 4) and (4, 6) respectively.
Point D lies on the x-axis. [AC] is perpendicular to [BD]. This information is
shown in the following diagram.
(a) Find the gradient of the line through A and C. [2]
Markscheme
6−0
m =
4−2
= 3 (M1)A1
[2 marks]
(b) Write down the gradient of the line through B and D. [1]
Markscheme
(m =) −
1
3
(−0. 333, − 0. 333333 …) A1
[1 mark]
(c) Find the equation of the line through B and D. Give your
answer in the form ax + by + d = 0, where a, b and d are
integers. [2]
Markscheme
an equation of line with a correct intercept and either of their gradients
from (a) or (b) (M1)
e.g. y = −
1
3
x + 4 OR y − 4 = −
1
3
(x − 0)
Note: Award (M1) for substituting either of their gradients from parts (a) or
(b) and point B or (3, 3) into equation of a line.
x + 3y − 12 = 0 or any integer multiple A1
[2 marks]
(d) Write down the x-coordinate of point D. [1]
Markscheme
(x =) 12 A1
[1 mark]
4. [Maximum mark: 16] 23M.2.AHL.TZ2.3
A particular park consists of a rectangular garden, of area A m2, and a concrete
path surrounding it. The park has a total area of 1200 m2.
The width of the path at the north and south side of the park is 2 m.
The width of the path at the west and east side of the park is 1. 5 m.
The length of the park (along the north and south sides) is x metres,
3 < x < 300.
diagram not to scale
(a) Show that A = 1212 − 4x −
3600
. [5]
x
Markscheme
Note: In methods 1 and 2, full marks are available for candidates who work
with a dummy variable, e.g. y, that represents the width of the park and
hence is equal to 1200
x
. The substitution to express an answer in only x
may come as late as the final line.
METHOD 1 (finding dimensions of garden)
1200
(width of park =) x
(A1)
1200
(length of garden =) x − 3, (width of garden =) x
− 4 (A1)(A1)
A = (x − 3) × (
1200
x
− 4) A1
= 1200 − 4x −
3600
x
+ 12 A1
= 1212 − 4x −
3600
x
AG
METHOD 2 (subtracting the area of the path)
width of park = 1200
x
(A1)
attempt to cut path into 4 (or 8) pieces (M1)
four (or eight) areas of the path expressed in terms of x (A1)
1200 1200
A = 1200 − 2x − 2x − 1. 5( − 4) − 1. 5( − 4)
x x
A1
correct manipulation leading to given result A1
1800 1800
= 1212 − 4x − −
x x
= 1212 − 4x −
3600
x
AG
Note: To award (M1)(A1) without a diagram the division of the park must be
clear.
[5 marks]
(b) Find the possible dimensions of the park if the area of the
garden is 800 m2. [4]
Markscheme
setting 1212 − 4x − 3600
x
= 800 (accept a sketch) (M1)
x = 9. 64 (9. 64011 …) (m) OR
x = 9. 34 (93. 3598 …) (m) A1
(width =) 124 (124. 479 …) (m) A1
(width =) 12. 9 (12. 8534 …) (m) A1
Note: To award the final A1 both values of x and both values of the width
must be seen. Accept 12. 8 for second value of width from candidate
dividing 1200 by 3 sf value of 93. 4.
[4 marks]
(c) Find an expression for
dA
. [3]
dx
Markscheme
dA 3600
(
dx
=) − 4 +
x
2 OR −4 + 3600x
−2
A1A1A1
Note: Award A1 for −4, A1 for +3600, and A1 for x−2 or x2 in
denominator.
[3 marks]
(d) Use your answer from part (c) to find the value of x that will
maximize the area of the garden. [2]
Markscheme
dA dA
setting their dx equal to 0 OR sketch of their dx
with x-intercept
highlighted M1
(x =) 30 (m) A1
Note: To award A1FT the candidate’s value of x must be within the domain
given in the problem (3 < x < 300).
[2 marks]
(e) Find the maximum possible area of the garden. [2]
Markscheme
EITHER
evidence of using GDC to find maximum of graph of
3600
A = 1212 − 4x −
x
(M1)
OR
substitution of their x into A (M1)
OR
dividing 1200 by their x to find width of park and subtracting 3 from their x
and 4 from the width to find park dimensions (M1)
Note: For the last two methods, only follow through if 3 < their x < 300
THEN
(A =) 972 (m )
2
A1
[2 marks]
5. [Maximum mark: 18] 22M.2.SL.TZ2.3
The Voronoi diagram below shows four supermarkets represented by points with
coordinates A(0, 0), B(6, 0), C(0, 6) and D(2, 2). The vertices X, Y, Z
are also shown. All distances are measured in kilometres.
(a) Find the midpoint of [BD]. [2]
Markscheme
2+6 2+0
(
2
,
2
) (M1)
(4, 1) A1
Note: Award A0 if parentheses are omitted in the final answer.
[2 marks]
(b) Find the equation of (XZ). [4]
Markscheme
attempt to substitute values into gradient formula (M1)
(
0−2
6−2
=) −
1
2
(A1)
therefore the gradient of perpendicular bisector is 2 (M1)
so y − 1 = 2(x − 4) (y = 2x − 7) A1
[4 marks]
The equation of (XY) is y = 2 − x and the equation of (YZ) is
y = 0. 5x + 3. 5.
(c) Find the coordinates of X. [3]
Markscheme
identifying the correct equations to use: (M1)
y = 2 − x and y = 2x − 7
evidence of solving their correct equations or of finding intersection point
graphically (M1)
(3, − 1) A1
Note: Accept an answer expressed as “x = 3, y = −1”.
[3 marks]
The coordinates of Y are (−1, 3) and the coordinates of Z are (7, 7).
(d) Determine the exact length of [YZ]. [2]
Markscheme
attempt to use distance formula (M1)
2 2
YZ = √ (7 − (−1)) + (7 − 3)
= √ 80 (4√ 5) A1
[2 marks]
(e) Given that the exact length of [XY] is √32, find the size of
XŶZ in degrees. [4]
Markscheme
METHOD 1 (cosine rule)
length of XZ is √80 (4√ 5, 8. 94427 …) (A1)
Note: Accept 8. 94 and 8. 9.
attempt to substitute into cosine rule (M1)
cos XŶZ =
80+32−80
(= 0. 316227 …) (A1)
2×√ 80√ 32
Note: Award A1 for correct substitution of XZ, YZ, √32 values in the cos
rule. Exact values do not need to be used in the substitution.
(XŶZ =) 71. 6° (71. 5650 …°) A1
Note: Last A1 mark may be lost if prematurely rounded values of XZ, YZ
and/or XY are used.
METHOD 2 (splitting isosceles triangle in half)
length of XZ is √80 (4√ 5, 8. 94427 …) (A1)
Note: Accept 8. 94 and 8. 9.
√ 32
required angle is cos−1( ) (M1)(A1)
2√ 80
√ 32
Note: Award A1 for correct substitution of XZ (or YZ), 2 values in the
cos rule. Exact values do not need to be used in the substitution.
(XŶZ =) 71. 6° (71. 5650 …°) A1
Note: Last A1 mark may be lost if prematurely rounded values of XZ, YZ
and/or XY are used.
[4 marks]
(f ) Hence find the area of triangle XYZ. [2]
Markscheme
1 1
(area =) 2
√ 80√ 32 sin 71. 5650 … OR (area =) 2
√ 32√ 72
(M1)
A1
2
= 24 km
[2 marks]
A town planner believes that the larger the area of the Voronoi cell XYZ, the
more people will shop at supermarket D.
(g) State one criticism of this interpretation. [1]
Markscheme
Any sensible answer such as:
There might be factors other than proximity which influence shopping
choices.
A larger area does not necessarily result in an increase in population.
The supermarkets might be specialized / have a particular clientele who
visit even if other shops are closer.
Transport links might not be represented by Euclidean distances.
etc. R1
[1 mark]
6. [Maximum mark: 4] 21N.1.AHL.TZ0.10
The graph of y = f (x) is given on the following set of axes. The graph passes
through the points (−2, 6) and (0, 1), and has a horizontal asymptote at
y = 0.
Let g(x) = 2f (x − 2) + 4.
(a) Find g(0). [2]
Markscheme
g(0) = 16 M1A1
[2 marks]
(b) On the same set of axes draw the graph of y = g(x), showing
any intercepts and asymptotes. [2]
Markscheme
y-asymptote (y = 4) A1
concave up decreasing curve and passing through (0, 16) A1
[2 marks]
7. [Maximum mark: 6] 20N.2.SL.TZ0.S_1
Consider the function f (x) = x
2
+ x +
50
x
, x ≠ 0.
(a) Find f (1). [2]
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
attempt to substitute x = 1 (M1)
eg f (1), 1
2
+ 1 +
50
52 (exact) A1 N2
[2 marks]
(b) Solve f (x) = 0. [2]
Markscheme
−4. 04932
−4. 05 A2 N2
[2 marks]
(c) The graph of f has a local minimum at point A.
Find the coordinates of A. [2]
Markscheme
(2. 76649, 28. 4934)
A(2. 77, 28. 5) A1A1 N2
[2 marks]
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