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Properties of Functions

Practice exam-style IB Math AI questions for Properties of Functions, aligned with the syllabus and grouped by topic.

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Question 1
SL • Paper 1
Easy
Calculator Permitted
SL • Paper 1
Easy
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The real-valued function ff is defined by

f(x)=12−3x+2f(x)=\sqrt{12-3x}+2
A

f(−1)f(-1).

[1]
B

the domain of ff.

[2]
C

the range of ff.

[1]
Question 2
SL • Paper 1
Easy
Calculator Permitted
SL • Paper 1
Easy
Calculator Permitted

A temperature of FF degrees Fahrenheit is converted to CC degrees Celsius using

C(F)=59(F−32)C(F)=\frac{5}{9}(F-32)
A

the Celsius temperature corresponding to 95∘F95^\circ\text{F}.

[1]
B

the Fahrenheit temperature corresponding to 18∘C18^\circ\text{C}.

[2]
C

how the calculation in part (b) is related to the inverse of CC.

[1]
Question 3
HL • Paper 1
Easy
Calculator Permitted
HL • Paper 1
Easy
Calculator Permitted

A delivery route has length mm miles, where 0≤m≤3000\leq m\leq 300. The function

f(m)=1.609mf(m)=1.609m

converts the distance to kilometres. The delivery charge for a route of kk kilometres is modelled by

g(k)=0.12k+4g(k)=0.12k+4

where g(k)g(k) is measured in dollars.

A

an expression for (g∘f)(m)(g\circ f)(m).

[2]
B

the delivery charge for a route of 8585 miles.

[1]
C

the domain and range of g∘fg\circ f in this context.

[2]
Question 4
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

The function gg is defined by

g(x)=x3−4x2−x+4,−2≤x≤5g(x)=x^3-4x^2-x+4,\qquad -2\leq x\leq 5
A

the xx-intercepts of the graph of gg.

[2]
B

the coordinates of the local minimum point of gg.

[2]
C

the maximum value of gg on the given domain.

[1]

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Question 5
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
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For a particular product, the demand price D(q)D(q) and supply price S(q)S(q), in dollars per item, are modelled by

D(q)=80e−0.08q+10D(q)=80e^{-0.08q}+10

and

S(q)=2.5q+8S(q)=2.5q+8

where qq is the number of hundreds of items produced. Market equilibrium occurs when the demand price equals the supply price.

A

the point of intersection of the graphs of DD and SS.

[3]
B

the first coordinate of this point in context.

[1]
Question 6
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

The functions ff and gg are defined by

f(x)=x+2,g(x)=1x−1f(x)=\sqrt{x+2},\qquad g(x)=\frac{1}{x-1}

The difference function dd is defined by d(x)=f(x)−g(x)d(x)=f(x)-g(x).

A

an expression for d(x)d(x).

[1]
B

the domain of dd.

[2]
C

d(2)d(2).

[1]
Question 7
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

A continuous function hh has domain −3≤x≤5-3\leq x\leq 5. Its graph has the following features:

  • endpoints (−3,2)(-3,2) and (5,−12)(5,-\frac{1}{2});
  • xx-intercepts (−2,0)(-2,0) and (2,0)(2,0);
  • a local minimum at (−2,0)(-2,0);
  • a local maximum at (0,4)(0,4);
  • a local minimum at (4,−1)(4,-1).

The graph has no other turning points or intercepts.

A

the graph of hh on the axes provided. all intercepts and turning points.

[4]
B

, the range of hh.

[2]
Question 8
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

The function hh is defined by

h(x)=2x+3x−4,x≠4h(x)=\frac{2x+3}{x-4},\qquad x\neq 4
A

the coordinates of the xx-intercept and the yy-intercept of the graph of hh.

[2]
B

the equations of the vertical and horizontal asymptotes of the graph of hh.

[2]
C

the range of hh.

[1]

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Question 9
HL • Paper 1
Medium
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HL • Paper 1
Medium
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The function qq is defined by

q(x)=(x−2)2+5q(x)=(x-2)^2+5

with domain x≥2x\geq 2.

A

q−1(x)q^{-1}(x) and its domain.

[3]
B

q−1(30)q^{-1}(30).

[1]
C

If the domain of qq were instead restricted to x≤2x\leq2, the resulting inverse function and its value at 3030.

[2]
Question 10
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A temperature sensor converts a temperature TT, in degrees Celsius, into a voltage V(T)V(T), in volts, using

V(T)=0.02T+1.5V(T)=0.02T+1.5

where −20≤T≤80-20\leq T\leq80. A digital converter then changes a voltage vv into a reading R(v)R(v) using

R(v)=409.6vR(v)=409.6v
A

an expression for the digital reading as a function of temperature.

[2]
B

A digital reading of 860860 is recorded. the corresponding temperature.

[2]
Question 11
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

The height of a weather balloon above the ground, h(t)h(t) metres, is modelled for 0≤t≤140\leq t\leq 14 by

h(t)=−0.04t3+0.48t2+1.2t+2h(t)=-0.04t^3+0.48t^2+1.2t+2

where tt is the time in minutes after its release.

A

when the balloon reaches its maximum height.

[2]
B

the maximum height of the balloon.

[1]
C

the height of the balloon at the end of the modelled interval.

[1]
D

the range of hh over the modelled interval.

[2]
Question 12
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The function ff is defined by

f(x)=3x−2x+4,x≠−4f(x)=\frac{3x-2}{x+4},\qquad x\neq -4
A

f−1(x)f^{-1}(x).

[3]
B

the domain of f−1f^{-1}.

[1]
C

that (f−1∘f)(x)=x(f^{-1}\circ f)(x)=x on the domain of ff.

[2]

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Question 13
HL • Paper 1
Medium
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HL • Paper 1
Medium
Calculator Permitted

The functions ff and gg are defined by

f(x)=2x+1,x∈Rf(x)=2x+1,\qquad x\in\mathbb{R}

and

g(x)=x2,x≥0g(x)=x^2,\qquad x\geq0
A

expressions for (f∘g)(x)(f\circ g)(x) and (g∘f)(x)(g\circ f)(x).

[2]
B

(f∘g)(x)=(g∘f)(x)(f\circ g)(x)=(g\circ f)(x), taking account of the domains of both composite functions.

[3]
Question 14
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The function ff is defined by

f(x)=ln⁡(x−1)+3f(x)=\ln(x-1)+3

with domain x>1x>1.

A

f−1(x)f^{-1}(x).

[3]
B

the domain and range of f−1f^{-1}.

[2]
C

that (f∘f−1)(x)=x(f\circ f^{-1})(x)=x.

[1]
Question 15
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The linear function ff is defined by

f(x)=ax+bf(x)=ax+b

where a≠0a\neq0. Its inverse is

f−1(x)=0.25x−3f^{-1}(x)=0.25x-3
A

the values of aa and bb.

[3]
B

both inverse composition identities for these functions.

[2]
Question 16
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
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The function pp is defined by

p(x)=3−(x+1)2p(x)=3-(x+1)^2

with domain x≤−1x\leq-1.

A

why the stated domain allows pp to have an inverse function.

[1]
B

p−1(x)p^{-1}(x) and its domain.

[3]
C

, p(x)=−13p(x)=-13 on the stated domain.

[2]

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Question 17
SL • Paper 2
Medium
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SL • Paper 2
Medium
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The depth of water, f(t)f(t) metres, in a drainage channel is modelled by

f(t)=36−4t+1f(t)=\sqrt{36-4t}+1

where tt is the time in hours after a gate is opened and 0≤t≤90\leq t\leq9. A second channel has depth g(t)=0.5t+2g(t)=0.5t+2 metres over the same interval.

A
I.

the initial depth and the depth after 99 hours in the first channel.

[2]
II.

the domain and range of ff in this context.

[2]
B

when the two channels have the same depth, and this common depth.

[2]
C

the length of time for which the depth in the first channel is at least 44 metres.

[2]
Question 18
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The concentration of a nutrient in a tank is modelled by

p(t)=8+12tt+3,0≤t≤30p(t)=8+\frac{12t}{t+3},\qquad 0\leq t\leq30

where p(t)p(t) is measured in milligrams per litre and tt is measured in minutes.

A
I.

p(0)p(0) and p(30)p(30).

[2]
II.

the equation of the horizontal asymptote of the unrestricted graph of pp.

[2]
B

the time at which the concentration first reaches 1717 milligrams per litre.

[2]
C

why the model never predicts a concentration of 2020 milligrams per litre at a finite time.

[2]
Question 19
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The strength of a wireless signal at horizontal distance dd metres from a transmitter is modelled by

S(d)=100d2+4,0≤d≤20S(d)=\frac{100}{d^2+4},\qquad 0\leq d\leq20

where S(d)S(d) is measured in arbitrary signal units.

A
I.

the maximum and minimum values of SS on the stated domain.

[2]
II.

the range of SS in this context.

[2]
B

S−1(2)S^{-1}(2) and your answer in context.

[2]
C

why S−1S^{-1} exists when SS has the stated domain, but would not exist if the domain were −20≤d≤20-20\leq d\leq20.

[2]
Question 20
HL • Paper 3
Medium
Calculator Permitted
HL • Paper 3
Medium
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A monitoring system records the time separation from an event at t=4t=4 seconds using

x(t)=∣t−4∣x(t)=|t-4|

where 0≤t≤100\leq t\leq10. A detector converts this separation into a signal

q(u)=3u+2q(u)=3u+2

A
I.

the signal function S=q∘xS=q\circ x and its range.

[2]
II.

why SS has no inverse function on 0≤t≤100\leq t\leq10.

[2]
B
I.

Restrict the domain to 4≤t≤104\leq t\leq10. the inverse of the restricted signal function.

[2]
II.

A signal of 1414 is recorded during the restricted interval. the time of the recording.

[2]
C

all possible recording times for a signal of 1414 when the original domain 0≤t≤100\leq t\leq10 is used, and the difference from part (b).

[2]

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Question 21
SL • Paper 2
Hard
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SL • Paper 2
Hard
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Two functions are defined for 0≤x≤80\leq x\leq8 by

f(x)=ln⁡(x+2),g(x)=0.4xf(x)=\ln(x+2),\qquad g(x)=0.4x

The sum and difference functions are s(x)=f(x)+g(x)s(x)=f(x)+g(x) and d(x)=f(x)−g(x)d(x)=f(x)-g(x).

A
I.

expressions for s(x)s(x) and d(x)d(x).

[2]
II.

the domain of each of the two new functions.

[2]
B

Use your GDC to the maximum value of dd and the value of xx at which it occurs.

[2]
C

the point, other than any endpoint, at which the graphs of ff and gg intersect.

[2]
Question 22
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
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During a laboratory trial, a response index is modelled by

r(t)=te−0.4t,0≤t≤12r(t)=te^{-0.4t},\qquad 0\leq t\leq12

where tt is measured in hours.

A
I.

Use your GDC to the coordinates of the maximum point of the graph of rr.

[2]
II.

the range of rr on the stated domain.

[2]
B

the two times at which the response index is 0.2500.250.

[2]
C

the length of time for which the response index exceeds 0.2500.250. If you did not obtain the times in part (b), use t=0.280t=0.280 and t=8.94t=8.94.

[2]
Question 23
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
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The function qq is defined by

q(x)=x4−8x2+5q(x)=x^4-8x^2+5

for −3≤x≤3-3\leq x\leq3.

A
I.

Use your GDC to all the zeros of qq on the stated domain.

[2]
II.

the coordinates of all turning points of qq on the stated domain.

[2]
B

why the graph of qq is symmetric about the yy-axis.

[2]
C

the graph of qq on the stated domain, labelling the turning points and endpoint values.

[2]
Question 24
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

The functions ff and gg are defined by

f(x)=3x+1x−2,g(x)=0.5x+4f(x)=\frac{3x+1}{x-2},\qquad g(x)=0.5x+4
A
I.

the equations of the vertical and horizontal asymptotes of the graph of ff.

[2]
II.

the coordinates of the intercepts of the graph of ff with the coordinate axes.

[2]
B

Use your GDC to the points of intersection of the graphs of ff and gg.

[2]
C

why a GDC window showing only −4≤x≤4-4\leq x\leq4 would lead to an incomplete conclusion about the intersections.

[2]

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Question 25
SL • Paper 2
Hard
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SL • Paper 2
Hard
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The vertical displacement of a floating sensor is modelled for 0≤t≤120\leq t\leq12 by

h(t)=5+3cos⁡(πt6)+0.2th(t)=5+3\cos\left(\frac{\pi t}{6}\right)+0.2t

where h(t)h(t) is measured in metres and tt in hours.

A
I.

h(0)h(0) and h(12)h(12).

[2]
II.

Use your GDC to the local maximum and local minimum points in the interior of the stated domain.

[2]
B

the absolute maximum value of hh on the stated domain and why it is not the value at the local maximum.

[2]
C

the range of hh on the stated domain.

[2]
Question 26
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A sensor converts a concentration cc into a voltage using f(c)=0.8c+1f(c)=0.8c+1. A processor converts a voltage vv into an index using g(v)=50ln⁡vg(v)=50\ln v. The concentration satisfies 0≤c≤100\leq c\leq10, and the overall index is H(c)=(g∘f)(c)H(c)=(g\circ f)(c).

A
I.

an expression for H(c)H(c).

[2]
II.

the domain and range of HH in this context.

[2]
B

H−1(x)H^{-1}(x).

[2]
C

An index of 8080 is recorded. the corresponding concentration and your result using composition.

[2]
Question 27
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The function qq is defined by

q(x)=2(x−1)2−5q(x)=2(x-1)^2-5

with domain x≥1x\geq1. The function rr is defined by r(x)=3x−2r(x)=3x-2.

A
I.

why the stated domain allows qq to have an inverse.

[2]
II.

q−1(x)q^{-1}(x) and its domain.

[2]
B

an expression for (q−1∘r)(x)(q^{-1}\circ r)(x) and its domain.

[2]
C

(q−1∘r)(x)=5(q^{-1}\circ r)(x)=5.

[2]
Question 28
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The function pp is defined by

p(x)=5x+2x+3p(x)=\frac{5x+2}{x+3}

with domain x>−3x>-3.

A
I.

the range of pp.

[2]
II.

p−1(x)p^{-1}(x) and its domain.

[2]
B

algebraically that (p−1∘p)(x)=x(p^{-1}\circ p)(x)=x for x>−3x>-3.

[2]
C

p−1(4.2)p^{-1}(4.2) and the result as a solution of an equation involving pp.

[2]

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Question 29
HL • Paper 3
Hard
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HL • Paper 3
Hard
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A water-quality instrument measures pollutant concentration xx, in milligrams per litre, where 0≤x≤400\leq x\leq40. Its sensor voltage is modelled by

s(x)=0.8x+1+0.2s(x)=0.8\sqrt{x+1}+0.2

A processor converts a voltage vv into a displayed reading using

p(v)=120ln⁡vp(v)=120\ln v

A
I.

the range of ss in this context.

[2]
II.

an expression for the displayed reading R(x)=(p∘s)(x)R(x)=(p\circ s)(x) and its contextual domain.

[2]
B
I.

the displayed reading when x=15x=15.

[2]
II.

R−1(r)R^{-1}(r), expressing the pollutant concentration in terms of a displayed reading rr.

[2]
C

A displayed reading of 200200 is observed. whether this reading is consistent with the stated concentration interval. your answer.

[2]
Question 30
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A digital image stores a brightness value xx in the interval −1≤x≤1-1\leq x\leq1. A normalization function and a function are defined by

f(x)=x+12,g(u)=u2f(x)=\frac{x+1}{2},\qquad g(u)=u^2

where 0≤u≤10\leq u\leq1. A final function h(v)=255vh(v)=255v converts the result to a display level.

A
I.

an expression for D(x)=(h∘g∘f)(x)D(x)=(h\circ g\circ f)(x).

[2]
II.

the domain and range of DD in this context.

[2]
B
I.

D−1(y)D^{-1}(y) for 0≤y≤2550\leq y\leq255.

[2]
II.

A pixel has display level 180180. its original brightness value.

[2]
C

If were applied before normalization, the intermediate output would be (f∘g)(x)(f\circ g)(x). Since gg is defined for inputs in [0,1][0,1], the two orders on their common domain x∈[0,1]x\in[0,1]. the two orders give the same intermediate output only when x=1x=1.

[2]
Question 31
HL • Paper 3
Hard
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HL • Paper 3
Hard
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On the interval −1≤x≤8-1\leq x\leq8, two functions are defined by

f(x)=x+1,g(x)=x−1f(x)=\sqrt{x+1},\qquad g(x)=x-1

The sum and difference functions are s=f+gs=f+g and d=f−gd=f-g.

A
I.

expressions for s(x)s(x) and d(x)d(x).

[2]
II.

the domain common to ss and dd.

[1]
B

the point at which the graphs of ff and gg intersect.

[2]
C
I.

the coordinates of the maximum point of dd on the stated interval.

[2]
II.

the graph of dd, labelling its endpoints, maximum point and zero.

[3]
III.

, the range of dd on the stated interval.

[1]
Question 32
HL • Paper 3
Hard
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HL • Paper 3
Hard
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The vertical position of a robotic arm is modelled by

h(t)=(t−6)2+4h(t)=(t-6)^2+4

where 0≤t≤120\leq t\leq12 and tt is measured in seconds. A camera records the transformed value

R(t)=ln⁡(h(t))R(t)=\ln(h(t))

A
I.

the range of hh.

[1]
II.

why hh does not have an inverse function on 0≤t≤120\leq t\leq12.

[2]
B
I.

The arm is observed only during its return motion, 6≤t≤126\leq t\leq12. the inverse of the restricted function hh.

[2]
II.

the domain of this inverse.

[1]
C

During the return motion, the camera records R(t)=ln⁡13R(t)=\ln13. tt.

[2]
D

the inverse of RR during the return motion and its domain.

[2]

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Question 33
HL • Paper 3
Hard
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HL • Paper 3
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A medical device uses a patient's mass ww, in kilograms, where 30≤w≤15030\leq w\leq150. It first calculates a body-size index

m(w)=w+1080m(w)=\sqrt{\frac{w+10}{80}}

and then calculates a dose, in milligrams, using

d(u)=240u+20d(u)=240u+20

A

Dose function

I.

the composite dose function D=d∘mD=d\circ m.

[2]
II.

the range of DD in this context.

[2]
B

Inverse dose function

I.

Let rr denote the displayed dose in milligrams. D−1(r)D^{-1}(r), including its domain and the units of the output.

[2]
II.

A dose of 300300 milligrams is displayed. the corresponding patient mass.

[2]
C

why the composition m∘dm\circ d does not represent the operation of this medical device.

[2]
Question 34
HL • Paper 3
Hard
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HL • Paper 3
Hard
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A traveller exchanges an amount xx using Bank A. After its fee, the amount received is modelled by

f(x)=0.92x−4f(x)=0.92x-4

The traveller later exchanges the full received amount back using Bank B, which applies

g(y)=1.06y−3g(y)=1.06y-3

All monetary amounts are measured in the same reference currency for comparison.

A
I.

the round-trip function T=g∘fT=g\circ f.

[2]
II.

the amount returned from an original amount of 500500.

[2]
B

an expression for the round-trip loss L(x)=x−T(x)L(x)=x-T(x), and the loss when x=500x=500.

[2]
C
I.

f−1(y)f^{-1}(y).

[2]
II.

why gg is not the inverse of ff.

[2]
Question 35
HL • Paper 3
Hard
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HL • Paper 3
Hard
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A continuous, strictly increasing function ff has domain [1,7][1,7] and range [−2,5][-2,5]. Its graph passes through (1,−2)(1,-2), (3,0)(3,0), (4,1)(4,1) and (7,5)(7,5).

A

On the same axes, the graph of f−1f^{-1}. the images of the four given points.

[3]
B
I.

f−1(1)f^{-1}(1).

[1]
II.

f−1(5)f^{-1}(5).

[1]
C

the domain and range of f−1f^{-1}.

[2]
D

Evaluate (f−1∘f)(3)(f^{-1}\circ f)(3) and (f∘f−1)(−2)(f\circ f^{-1})(-2).

[2]
E

The graph is extended by adding the point (8,4)(8,4) while remaining continuous. why the extended function cannot have an inverse.

[1]
Question 36
HL • Paper 2
Hard
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HL • Paper 2
Hard
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The functions ff and gg are defined by

f(x)=x−2,g(x)=ln⁡(7−x)f(x)=\sqrt{x-2},\qquad g(x)=\ln(7-x)
A
I.

an expression for (f∘g)(x)(f\circ g)(x).

[2]
II.

the domain of f∘gf\circ g.

[2]
B

(g∘f)(x)(g\circ f)(x) and its domain.

[2]
C

(f∘g)(x)=1(f\circ g)(x)=1, giving your answer to three significant figures.

[2]

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Question 37
HL • Paper 2
Hard
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HL • Paper 2
Hard
Calculator Permitted

A calibration system applies two functions in sequence. The first is A(x)=3x−7A(x)=3x-7 and the second is B(y)=ey/20B(y)=e^{y/20}. The overall calibration is C=B∘AC=B\circ A.

A
I.

an expression for C(x)C(x).

[2]
II.

A−1(x)A^{-1}(x) and B−1(x)B^{-1}(x).

[2]
B

C−1(x)C^{-1}(x), making clear the order in which the inverse functions are applied.

[2]
C

the input that produces a calibrated output of 4.54.5, and the result.

[2]
Question 38
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The function pp is defined by

p(x)=x3+x+1p(x)=x^3+x+1

for x∈Rx\in\mathbb R. The function qq is defined by q(x)=2x−3q(x)=2x-3.

A
I.

Use a graph to why pp has an inverse function.

[2]
II.

Use your GDC to p−1(20)p^{-1}(20), giving your answer to three significant figures.

[2]
B

Without finding an algebraic formula for p−1p^{-1}, (p−1∘q)(x)=2(p^{-1}\circ q)(x)=2.

[2]
C

why (p−1∘p)(x)=x(p^{-1}\circ p)(x)=x for every real xx.

[2]
Question 39
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The functions ff and gg are defined by

f(x)=4e0.3x−2f(x)=4e^{0.3x}-2

for x∈Rx\in\mathbb R, and

g(x)=x+2g(x)=\sqrt{x+2}

for x≥−2x\geq-2.

A
I.

f−1(x)f^{-1}(x) and its domain.

[2]
II.

and simplify (g∘f)(x)(g\circ f)(x).

[2]
B

an expression for (f−1∘g)(x)(f^{-1}\circ g)(x) and its domain.

[2]
C

(g∘f)(x)=10(g\circ f)(x)=10.

[2]
Question 40
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For x≥2x\geq2, the function ff is defined by

f(x)=x+4xf(x)=x+\frac{4}{x}

A

(a)

I.

the minimum value and range of ff.

[2]
II.

why ff has an inverse function on the stated domain.

[2]
B

(b)

I.

f−1(x)f^{-1}(x) and its domain.

[3]
II.

that (f−1∘f)(x)=x(f^{-1}\circ f)(x)=x for x≥2x\geq2.

[2]
C

The family fa(x)=x+axf_a(x)=x+\frac{a}{x} is restricted to x≥ax\geq\sqrt a, where a>0a>0. , fa−1(x)f_a^{-1}(x) and its domain.

[2]

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Question 41
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For a real parameter aa, define

fa(x)=ax+1x−a,x≠af_a(x)=\frac{ax+1}{x-a},\qquad x\neq a

A
I.

the equations of the vertical and horizontal asymptotes of the graph of faf_a.

[2]
II.

the domain and range of faf_a.

[2]
B
I.

fa−1(x)=fa(x)f_a^{-1}(x)=f_a(x).

[2]
II.

, the value of (fa∘fa)(x)(f_a\circ f_a)(x) for x≠ax\neq a.

[2]
C

the coordinates of the two points where the graph of faf_a intersects the line y=xy=x.

[2]
D

The horizontal distance between these two intersection points is 44. the possible values of aa.

[2]
Question 42
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The functions ff and gg are defined by

f(x)=9−x,g(x)=x2+1f(x)=\sqrt{9-x},\qquad g(x)=x^2+1

using their largest possible real domains.

The graph is shown over a finite plotting window; any curve endpoints visible within that window are plotting truncations, unless forced by the stated domain.

A
I.

(f∘g)(x)(f\circ g)(x) and its domain.

[2]
II.

(g∘f)(x)(g\circ f)(x) and its domain.

[2]
B

why simplifying (g∘f)(x)(g\circ f)(x) to 10−x10-x does not allow its domain to be changed to all real numbers.

[2]
C
I.

f−1(x)f^{-1}(x) and its domain.

[2]
II.

that (f∘f−1)(x)=x(f\circ f^{-1})(x)=x on the domain of f−1f^{-1}.

[2]
D

(f∘g)(x)=2(f\circ g)(x)=2 on its domain.

[2]
Question 43
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The function ff is defined by

f(x)=∣2x−6∣+1f(x)=|2x-6|+1

for all real xx.

A
I.

the range of ff.

[1]
II.

why ff has no inverse on R\mathbb R.

[2]
B
I.

Restrict the domain to x≥3x\geq3. the inverse function rr.

[2]
II.

Restrict the domain instead to x≤3x\leq3. the inverse function ll.

[2]
C

(r∘f)(x)(r\circ f)(x) separately for x≥3x\geq3 and x<3x<3.

[2]
D

geometrically why (r∘f)(x)=6−x(r\circ f)(x)=6-x when x<3x<3.

[2]
Question 44
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For k>0k>0, define

fk(x)=kx+1,x≥0f_k(x)=\frac{k}{x+1},\qquad x\geq0

xx

fk(x)f_k(x)

0

kk

1

k2\frac{k}{2}

2

k3\frac{k}{3}

A
I.

the range of fkf_k and why fkf_k has an inverse.

[2]
II.

fk−1(x)f_k^{-1}(x) and its domain.

[2]
B

the positive fixed point of fkf_k in terms of kk.

[2]
C
I.

an expression for (fk∘fk)(x)(f_k\circ f_k)(x).

[1]
II.

the non-negative solutions of (fk∘fk)(x)=x(f_k\circ f_k)(x)=x are precisely the fixed points of fkf_k.

[2]
D

kk if the fixed point is x=2x=2.

[2]

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Question 45
HL • Paper 3
Hard
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HL • Paper 3
Hard
Calculator Permitted

An unfamiliar function is defined by

h(x)=x+e−x,x∈Rh(x)=x+e^{-x},\qquad x\in\mathbb R

A
I.

Use graphing technology to the coordinates of the minimum point of hh.

[2]
II.

the range of hh.

[2]
B

why hh has no inverse function on R\mathbb R, and a domain restriction that retains the point (0,1)(0,1) and makes hh one-to-one.

[2]
C
I.

h(x)=2h(x)=2 on the restricted domain x≥0x\geq0.

[1]
II.

h(x)=2h(x)=2 on the restricted domain x≤0x\leq0.

[1]
D

the number of real solutions of h(x)=ch(x)=c for each of the cases c<1c<1, c=1c=1 and c>1c>1.

[2]
Question 46
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The concentration of a medicine tt hours after administration is modelled by

f(t)=100e−0.2t,t≥0f(t)=100e^{-0.2t},\qquad t\geq0

A response score is calculated from concentration cc using

g(c)=cc+20,c>0g(c)=\frac{c}{c+20},\qquad c>0

The response as a function of time is h=g∘fh=g\circ f.

A
I.

and simplify an expression for h(t)h(t).

[2]
II.

the domain and range of hh in this context.

[2]
B

h−1(r)h^{-1}(r), stating its domain.

[2]
C

when the response score first falls to 0.6000.600.

[2]
Question 47
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The family of curves

y=hk(x)=x3−3x+ky=h_k(x)=x^3-3x+k

is compared with the line y=xy=x, where kk is a real parameter. Intersections satisfy

x3−4x+k=0x^3-4x+k=0

A
I.

For k=0k=0, all points of intersection.

[2]
II.

For k=2k=2, the three points of intersection, giving coordinates to three significant figures.

[2]
B

Let Fk(x)=x3−4x+kF_k(x)=x^3-4x+k. the coordinates of the local maximum and local minimum of FkF_k in terms of kk.

[4]
C
I.

, the interval of values of kk for which the cubic and the line have three distinct intersections.

[2]
II.

the intersections when k=1633k=\frac{16}{3\sqrt3}.

[2]
D

why changing the graphing window could lead to an incorrect conclusion about the number of intersections.

[2]
Question 48
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Two reversible coding steps are represented by the affine functions

f(x)=1.5x+2,g(x)=0.8x−5f(x)=1.5x+2, \qquad g(x)=0.8x-5
A
I.

(f∘g)(x)(f\circ g)(x).

[2]
II.

(g∘f)(x)(g\circ f)(x).

[2]
B

why there is no value of xx for which the two coding orders give the same output.

[2]
C

Let F(x)=ax+bF(x)=ax+b and G(x)=cx+dG(x)=cx+d, where a≠0a\neq0 and c≠0c\neq0. FF and GG commute if and only if

d(a−1)=b(c−1)d(a-1)=b(c-1)

[3]
D
I.

f−1(x)f^{-1}(x) and g−1(x)g^{-1}(x).

[2]
II.

, in the correct order, the composition that reverses f∘gf\circ g.

[1]

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Linear Equations & Graphs

Transformations