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Properties of Functions

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

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Question 1
SL • Paper 1
Easy
Non Calculator
SL • Paper 1
Easy
Non Calculator

The points A(−2,5)A(-2,5) and B(4,−1)B(4,-1) lie on a straight line. The line LL is perpendicular to ABAB and passes through the midpoint of ABAB.

A

the gradient of ABAB.

[2]
B

the equation of LL, giving your answer in the form y=mx+cy=mx+c.

[3]
C

the coordinates of the xx-intercept of LL.

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

Let f(x)=5−2x+1f(x)=\sqrt{5-2x}+1.

A

the largest possible domain of ff.

[2]
B

the range of ff.

[1]
C

f(x)=4f(x)=4.

[2]
Question 3
SL • Paper 1
Easy
Non Calculator
SL • Paper 1
Easy
Non Calculator

Let f(x)=2x−3f(x)=2x-3 and g(x)=x2+1g(x)=x^2+1, for x∈Rx\in\mathbb{R}.

A

(f∘g)(x)(f\circ g)(x).

[2]
B

(g∘f)(x)(g\circ f)(x).

[2]
C

whether f∘g=g∘ff\circ g=g\circ f. your answer.

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

The total cost, CC dollars, of a guided walking tour for nn people is modelled by a linear function. For 88 people the cost is 146,andfor146, and for 15peoplethecostispeople the cost is237.

A

a model for CC in terms of nn.

[3]
B

Another company models its cost by D=9.5n+105D=9.5n+105. the number of people for which the two companies have the same cost.

[2]

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Question 5
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

The graphs of y=x2−4x+1y=x^2-4x+1 and y=−x+3y=-x+3 intersect at the points PP and QQ.

A

Form an equation in xx whose solutions give the xx-coordinates of PP and QQ.

[1]
B

this equation for xx.

[2]
C

the coordinates of PP and QQ.

[2]
Question 6
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

The function ff is defined by f(x)=−(x+1)2+4f(x)=-(x+1)^2+4, for −4≤x≤2-4\le x\le 2.

A

the xx-intercepts of the graph of ff.

[2]
B

the vertex and the maximum value of ff.

[2]
C

the graph of ff, showing the endpoints of the domain.

[2]
Question 7
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

Consider the equation

e2x−5ex+6=0e^{2x}-5e^x+6=0
A

Use the substitution u=exu=e^x to write the equation as a quadratic equation in uu.

[2]
B

the original equation for xx.

[2]
Question 8
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The function ff is defined by f(x)=(x−1)2+3f(x)=(x-1)^2+3, for x≥1x\ge 1.

A

the range of ff.

[1]
B

f−1(x)f^{-1}(x), stating its domain.

[3]
C

f−1(x)=5f^{-1}(x)=5.

[1]

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Question 9
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let

f(x)=sin⁡x+asin⁡2x+bcos⁡xf(x)=\sin x+a\sin 2x+b\cos x

where a,b∈Ra,b\in\mathbb{R} and x∈Rx\in\mathbb{R}. The function ff is odd, and f(π6)=1f\left(\dfrac{\pi}{6}\right)=1.

A

the value of bb.

[2]
B

the value of aa.

[2]
C

the value of f(−π6)f\left(-\dfrac{\pi}{6}\right).

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

The function ff is defined by

f(x)=7−2x+1f(x)=\sqrt{7-2x}+1

where xx is in the largest possible real domain.

A

the domain and range of ff.

[2]
B

an expression for f−1(x)f^{-1}(x), stating its domain.

[3]
Question 11
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A machine produces a component whose mass after tt minutes is modelled by

m(t)=1.5t+20,t≥0m(t)=1.5t+20,\quad t\geq 0

The energy, EE joules, required to heat a component of mass mm grams is modelled by

E(m)=0.04m2+12E(m)=0.04m^2+12
A

an expression for (E∘m)(t)(E\circ m)(t).

[2]
B

the energy required when t=10t=10.

[1]
C

the time at which the energy required is 8080 joules.

[2]
Question 12
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The function ff is defined by

f(x)=(x−2)2−5,x≥2f(x)=(x-2)^2-5,\quad x\geq 2
A

the range of ff.

[1]
B

an expression for f−1(x)f^{-1}(x), stating its domain.

[3]
C

f−1(x)=6.5f^{-1}(x)=6.5.

[1]

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Question 13
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The function ff is defined for x∈Rx\in\mathbb{R} by

f(x)=asin⁡x+bcos⁡x+cf(x)=a\sin x+b\cos x+c

It is known that ff is an even function, f(0)=5f(0)=5 and f(π)=1f(\pi)=1.

A

the values of aa, bb and cc.

[4]
B

the range of ff.

[1]
Question 14
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The function ff is defined by

f(x)=x2−6x+5,x≤3f(x)=x^2-6x+5,\quad x\leq 3
A

why ff has an inverse function.

[1]
B

f−1(x)f^{-1}(x), stating its domain.

[3]
C

f−1(x)=−2f^{-1}(x)=-2.

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

Let f(x)=x5+ax3+bx2+cxf(x)=x^5+ax^3+bx^2+cx, where a,b,c∈Ra,b,c\in\mathbb{R}. The function ff is odd. Also, f(1)=0f(1)=0 and f(2)=18f(2)=18.

A

the value of bb.

[2]
B

the values of aa and cc.

[3]
Question 16
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let f(x)=mx+cf(x)=mx+c, where m,c∈Rm,c\in\mathbb{R}. The function ff is self-inverse, ff is not the identity function, and f(3)=1f(3)=1.

A

m=−1m=-1.

[2]
B

cc and f(x)f(x).

[2]
C

the fixed point of ff.

[1]

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Question 17
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The function ff is defined by

f(x)=2x+3x−2,x≠2f(x)=\frac{2x+3}{x-2},\quad x\ne 2
A

f−1(x)f^{-1}(x), stating its domain.

[3]
B

whether ff is self-inverse.

[1]
C

the fixed points of ff.

[2]
Question 18
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The function ff is defined by f(x)=x2−4x+1f(x)=x^2-4x+1, for x≤2x\le 2.

A

Write f(x)f(x) in the form (x−h)2+k(x-h)^2+k and the range of ff.

[2]
B

f−1(x)f^{-1}(x), stating its domain.

[3]
C

f−1(6)f^{-1}(6).

[1]
Question 19
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The function ff is defined by

f(x)=x3−4x+1,−3≤x≤3f(x)=x^3-4x+1,\quad -3\leq x\leq 3
A

Using your GDC, the zeros of ff.

[3]
B

the coordinates of the local maximum and the local minimum of the graph of y=f(x)y=f(x).

[3]
Question 20
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

Two quantities, AA and BB, are modelled by

A(t)=8e−0.18t+2,B(t)=0.5t+3A(t)=8e^{-0.18t}+2,\quad B(t)=0.5t+3

where tt is measured in hours and 0≤t≤200\leq t\leq 20.

A

A(0)−B(0)A(0)-B(0).

[1]
B

Using your GDC, the value of tt for which A(t)=B(t)A(t)=B(t). Give your answer correct to 2 decimal places.

[3]
C

the time interval during which A(t)>B(t)A(t)>B(t). Give the endpoint correct to 2 decimal places.

[1]

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Question 21
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The function ff is defined by

f(x)=2x−5x−2f(x)=\frac{2x-5}{x-2}
A

the domain and range of ff.

[2]
B

algebraically that ff is self-inverse.

[3]
Question 22
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The function hh is defined for x∈Rx\in\mathbb{R} by

h(x)=asin⁡(2x)+bcos⁡x+cxh(x)=a\sin(2x)+b\cos x+cx

It is known that hh is an odd function, h(π2)=1h\left(\frac{\pi}{2}\right)=1 and h(π6)=5h\left(\frac{\pi}{6}\right)=5.

A

the value of bb.

[1]
B

the value of cc.

[2]
C

the value of aa.

[2]
D

h(−π6)h\left(-\frac{\pi}{6}\right).

[1]
Question 23
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

A linear function ff is defined by f(x)=ax+bf(x)=ax+b, where a,b∈Ra,b\in\mathbb{R}. It is known that f(2)=7f(2)=7 and f(6)=−1f(6)=-1.

A

the function ff.

I.

the values of aa and bb.

[3]
II.

f(x)f(x).

[1]
B

The inverse function is denoted by f−1f^{-1}.

I.

f−1(x)f^{-1}(x).

[2]
II.

f(x)=f−1(x)f(x)=f^{-1}(x).

[2]
C

Let PP be the point where the graphs of y=f(x)y=f(x) and y=f−1(x)y=f^{-1}(x) intersect. the coordinates of PP and why PP lies on the line y=xy=x.

[2]
Question 24
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

The function ff is defined by f(x)=3−2x+4f(x)=3-\sqrt{2x+4}, for x≥−2x\geq -2.

A

Consider the domain and range of ff.

I.

the domain of ff.

[1]
II.

the range of ff.

[1]
B

the inverse function.

I.

an expression for f−1(x)f^{-1}(x).

[3]
II.

the domain and range of f−1f^{-1}.

[2]
C

equations involving ff.

I.

f(x)=1f(x)=1.

[2]
II.

f(x)=xf(x)=x.

[2]

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Question 25
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

The function ff is defined by
f(x)=x+2f(x)=x+2 for −3≤x≤1-3\leq x\leq 1, and f(x)=2x+1f(x)=2x+1 for 1<x≤41<x\leq 4.

A

Consider the values and range of ff.

AI.

f(−3)f(-3).

[1]
AII.

f(4)f(4).

[1]
AIII.

the range of ff.

[2]
B

The function ff is one-to-one.

BI.

f−1(x)f^{-1}(x) as a piecewise function.

[3]
BII.

the domain of f−1f^{-1}.

[1]
C

Use your expression for f−1f^{-1} to f−1(x)=2f^{-1}(x)=2.

[2]
Question 26
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

Let f(x)=x−1f(x)=x-1, where x∈Rx\in\mathbb{R}, and let g(x)=x+3g(x)=\sqrt{x+3}, where x≥−3x\geq -3.

A

the two composite functions and their domains.

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

(f∘g)(x)=(g∘f)(x)(f\circ g)(x)=(g\circ f)(x).

[3]
C

Consider the inverse of ff.

CI.

f−1(x)f^{-1}(x).

[1]
CII.

whether f∘gf\circ g and g∘fg\circ f are the same function. your answer.

[1]
Question 27
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A straight cable is to be installed above a curved roof. The cable passes through the points A(0,18)A(0,18) and B(60,42)B(60,42), where coordinates are measured in metres. The height of the roof above the same horizontal axis is modelled by

r(x)=10+8sin⁡(0.06x),0≤x≤60r(x)=10+8\sin(0.06x),\quad 0\le x\le 60

The support strut is a straight line that passes through the point C(45,62)C(45,62) and is perpendicular to the cable.

A
I.

the gradient of the cable.

[1]
II.

an equation for the height, L(x)L(x), of the cable.

[2]
B

A support strut is perpendicular to the cable and passes through the point C(45,62)C(45,62).

I.

an equation of the support strut.

[2]
II.

the coordinates of the point where the support strut meets the cable.

[2]
C
I.

a function d(x)d(x) for the vertical distance from the roof to the cable.

[2]
II.

Using your GDC, the minimum vertical distance between the roof and the cable.

[2]
Question 28
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The function ff is defined by

f(x)=4−18−3xf(x)=4-\sqrt{18-3x}

where xx is in the largest possible real domain.

A
I.

the domain of ff.

[1]
II.

the range of ff.

[2]
B
I.

an expression for f−1(x)f^{-1}(x), stating its domain.

[3]
II.

f−1(1)f^{-1}(1).

[1]
C

the fixed point of ff, giving your answer to three significant figures.

[3]

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Question 29
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A heating process is modelled by the function

T(t)=18+12(1−e−0.25t),0≤t≤12T(t)=18+12(1-e^{-0.25t}),\quad 0\le t\le 12

where TT is the temperature in degrees Celsius after tt minutes. The quality score of the product at temperature TT is modelled by

Q(T)=100−0.5(T−28)2Q(T)=100-0.5(T-28)^2
A
I.

an expression for (Q∘T)(t)(Q\circ T)(t).

[2]
II.

the quality score after 66 minutes.

[2]
B
I.

the time at which the quality score is a maximum.

[2]
II.

the maximum quality score.

[2]
C

the interval of time for which the quality score is at least 9898.

[2]
Question 30
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The function ff is defined for x∈Rx\in\mathbb{R} by

f(x)=x+x2+9f(x)=x+\sqrt{x^2+9}

The function is one-to-one.

A

the range of ff.

[1]
B

f−1(x)f^{-1}(x), stating its domain.

[3]
C

f(x)=10f(x)=10.

[1]
Question 31
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

For a real constant aa, the function ff is defined by

f(x)=ax+63x+2,x≠−23f(x)=\frac{ax+6}{3x+2},\quad x\neq -\frac{2}{3}
A

the value of aa for which ff is self-inverse.

[4]
B

For this value of aa, the domain and range of ff.

[2]
Question 32
HL • Paper 3
Medium
Calculator Permitted
HL • Paper 3
Medium
Calculator Permitted

For a∈Ra\in\mathbb{R}, define the function fa:R→Rf_a:\mathbb{R}\to\mathbb{R} by fa(x)=a−xf_a(x)=a-x. Also define f0(x)=−xf_0(x)=-x and, for n∈Z+n\in\mathbb{Z}^+, let TanT_a^n denote applying TaT_a repeatedly nn times, where Ta=fa∘f0T_a=f_a\circ f_0.

A
I.

faf_a is self-inverse.

[2]
II.

the fixed point of faf_a.

[2]
B
I.

Ta(x)T_a(x) and Tan(x)T_a^n(x).

[3]
II.

Tan(x)=fa(x)T_a^n(x)=f_a(x), giving your answer in terms of aa and nn.

[3]

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Question 33
HL • Paper 3
Medium
Calculator Permitted
HL • Paper 3
Medium
Calculator Permitted

A function ff is defined on a domain symmetric about 00. Define

E(x)=f(x)+f(−x)2,O(x)=f(x)−f(−x)2E(x)=\frac{f(x)+f(-x)}{2},\qquad O(x)=\frac{f(x)-f(-x)}{2}

The functions EE and OO are called the even and odd parts of ff.

A
I.

EE is an even function.

[2]
II.

OO is an odd function.

[2]
B

For f(x)=exf(x)=e^x, E(x)E(x) and O(x)O(x).

[2]
C
I.

O(x)=3O(x)=3 exactly.

[2]
II.

why EE has no inverse function on R\mathbb{R}, but has an inverse function if its domain is restricted to x≥0x\ge 0.

[2]
Question 34
HL • Paper 3
Medium
Calculator Permitted
HL • Paper 3
Medium
Calculator Permitted

Two linear functions are defined by f(x)=mx+cf(x)=mx+c and g(x)=px+qg(x)=px+q, where m,p,c,q∈Rm,p,c,q\in\mathbb{R}. This question investigates when the order of two linear calibrations matters.

A
I.

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

[2]
II.

f∘g=g∘ff\circ g=g\circ f if and only if c(p−1)=q(m−1)c(p-1)=q(m-1).

[2]
B

Now let f(x)=2x+cf(x)=2x+c and g(x)=px+3g(x)=px+3, where p≠1p\ne 1. If f∘g=g∘ff\circ g=g\circ f, express cc in terms of pp.

[2]
C
I.

Suppose also that gg is self-inverse. pp and cc.

[3]
II.

the common fixed point of ff and gg for these values.

[1]
Question 35
HL • Paper 3
Medium
Calculator Permitted
HL • Paper 3
Medium
Calculator Permitted

Let a>0a>0. Define

f(x)=x+a,x≥−af(x)=\sqrt{x+a},\qquad x\ge -a

and

g(x)=x2−a,x∈Rg(x)=x^2-a,\qquad x\in\mathbb{R}

The two functions appear to undo each other, but only in one order without further restriction.

A
I.

the domain and range of ff.

[2]
II.

(g∘f)(x)=x(g\circ f)(x)=x for all xx in the domain of ff.

[2]
B

(f∘g)(x)(f\circ g)(x) and its largest possible domain.

[2]
C
I.

Restrict the domain of gg so that g−1=fg^{-1}=f.

[2]
II.

why (f∘g)(x)=x(f\circ g)(x)=x only on this restricted domain.

[2]
Question 36
HL • Paper 3
Medium
Calculator Permitted
HL • Paper 3
Medium
Calculator Permitted

Let a>0a>0 and define

f(x)=a2−x2f(x)=\sqrt{a^2-x^2}

The largest possible real domain is used unless otherwise stated.

A
I.

the domain and range of ff.

[2]
II.

ff is even, and why it has no inverse function on its largest domain.

[2]
B

Restrict the domain to 0≤x≤a0\le x\le a. f−1(x)f^{-1}(x), stating its domain.

[3]
C

On the restricted domain 0≤x≤a0\le x\le a, the intersection of y=f(x)y=f(x) and y=x2y=\frac{x}{2}.

[2]

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Question 37
HL • Paper 3
Medium
Calculator Permitted
HL • Paper 3
Medium
Calculator Permitted

A function on a finite set can be represented by a mapping table. Let S={1,2,3,4}S=\{1,2,3,4\}. Functions f:S→Sf:S\to S and g:S→Sg:S\to S are given by the table.

x

f(x)

g(x)

1

3

2

2

4

1

3

1

4

4

2

3

A
I.

that ff is self-inverse.

[2]
II.

whether gg is self-inverse.

[1]
B

Let h=f∘gh=f\circ g. Copy and complete the mapping table for hh.

[3]
C
I.

whether hh is self-inverse.

[2]
II.

the fixed points of hh.

[1]
Question 38
SL • Paper 1
Hard
Non Calculator
SL • Paper 1
Hard
Non Calculator

Let f(x)=x2−2x−3f(x)=x^2-2x-3 and g(x)=x+1g(x)=x+1, for x∈Rx\in\mathbb{R}. The function dd is defined by d(x)=f(x)−g(x)d(x)=f(x)-g(x).

A

Consider the difference function dd.

I.

d(x)d(x).

[1]
II.

d(x)=0d(x)=0.

[3]
B

the coordinates of the intersection points of the graphs of y=f(x)y=f(x) and y=g(x)y=g(x).

[2]
C

Analyse the vertical separation of the two graphs.

I.

Write d(x)d(x) in completed-square form and its minimum value.

[2]
II.

the value of xx for which g(x)−f(x)g(x)-f(x) is greatest, and this greatest value.

[2]
Question 39
SL • Paper 1
Hard
Non Calculator
SL • Paper 1
Hard
Non Calculator

Consider the function hh defined by h(x)=e2x−6ex+8h(x)=e^{2x}-6e^x+8, for x∈Rx\in\mathbb{R}.

A

Use the substitution u=exu=e^x.

I.

Using the substitution u=exu=e^x, write h(x)=0h(x)=0 as a quadratic equation in uu.

[1]
II.

this quadratic equation for uu.

[2]
B

h(x)=0h(x)=0.

[2]
C

the inequality h(x)≤0h(x)\leq 0.

[3]
D

The equation h(x)=0h(x)=0 can be interpreted as the intersection of the graphs y=exy=e^x and y=6−8e−xy=6-8e^{-x}. the coordinates of these intersection points.

[2]
Question 40
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let f(x)=ax4+bx3+cx2+dx+ef(x)=ax^4+bx^3+cx^2+dx+e, where a,b,c,d,e∈Ra,b,c,d,e\in\mathbb{R}. The function ff is even. It is also known that f(0)=3f(0)=3, f(1)=0f(1)=0 and f(2)=15f(2)=15.

A

Part (a): the constants in ff.

I.

Use the fact that ff is even to the values of bb and dd.

[2]
II.

aa, cc and ee.

[4]
B

Part (b): Define the function by g(x)=xf(x)g(x)=xf(x).

I.

gg is an odd function.

[2]
II.

all zeros of gg.

[2]

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Question 41
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

For a real constant aa, the function ff is defined by f(x)=(x−a)2+1f(x)=(x-a)^2+1, for x≥ax\geq a. It is known that f(5)=10f(5)=10.

A

the value of aa and the range of ff.

I.

a=2a=2.

[3]
II.

the range of ff.

[1]
B

Use a=2a=2 for this part. If you did not obtain a=2a=2, use f(x)=(x−2)2+1f(x)=(x-2)^2+1, x≥2x\geq 2.

I.

f−1(x)f^{-1}(x), stating its domain.

[3]
II.

f−1(x)=xf^{-1}(x)=x.

[3]
Question 42
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let h(x)=x3−4x2+x+6h(x)=x^3-4x^2+x+6, for x∈Rx\in\mathbb{R}. Define
E(x)=h(x)+h(−x)2E(x)=\dfrac{h(x)+h(-x)}{2} and O(x)=h(x)−h(−x)2O(x)=\dfrac{h(x)-h(-x)}{2}.

A

E(x)E(x) and O(x)O(x).

I.

h(−x)h(-x).

[1]
II.

E(x)E(x) and O(x)O(x).

[3]
B

Consider the symmetry of EE and OO.

I.

EE is even and OO is odd.

[2]
II.

that h(x)=E(x)+O(x)h(x)=E(x)+O(x).

[1]
C

h(x)=h(−x)h(x)=h(-x).

[2]
D

the zeros of EE and their symmetry.

[2]
Question 43
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let f(x)=x−1f(x)=\sqrt{x-1}, where x≥1x\geq 1, and let g(x)=3x−2g(x)=3x-2, where x∈Rx\in\mathbb{R}. Define h=f∘gh=f\circ g.

A

the function hh.

I.

an expression for h(x)h(x).

[1]
II.

the domain and range of hh.

[2]
B

h−1(x)h^{-1}(x) directly, stating its domain.

[3]
C

Use the inverse functions of ff and gg.

I.

f−1(x)f^{-1}(x) and g−1(x)g^{-1}(x).

[2]
II.

that h−1=g−1∘f−1h^{-1}=g^{-1}\circ f^{-1}.

[1]
D

evaluate h−1(3)h^{-1}(3) and its meaning in terms of hh.

[1]
Question 44
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

The height, yy metres, of a small drone above level ground after xx seconds is modelled by

y=f(x)=x3−6x2+9x+2,0≤x≤5y=f(x)=x^3-6x^2+9x+2,\quad 0\le x\le 5

A warning signal is triggered whenever the drone is above 44 metres.

A
I.

the coordinates of the local maximum point of the graph of ff.

[2]
II.

the coordinates of the local minimum point of the graph of ff.

[2]
B

the times at which the warning signal changes .

[3]
C
I.

the intervals of time during which the warning signal is on.

[2]
II.

the total area between the graph of ff and the line y=4y=4 over the intervals when the warning signal is on.

[3]

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Question 45
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

Consider the function

f(x)=x3−3x+2f(x)=x^3-3x+2

The domain of ff is first taken to be −1≤x≤1-1\le x\le 1.

A
I.

the range of ff on this domain.

[2]
II.

why ff has an inverse function on this domain.

[1]
B
I.

the domain and range of f−1f^{-1}.

[2]
II.

Using your GDC, f−1(1.5)f^{-1}(1.5).

[2]
C
I.

the graphs of y=f(x)y=f(x) and y=f−1(x)y=f^{-1}(x) on the same axes.

[2]
II.

the point of intersection of the graphs of y=f(x)y=f(x) and y=f−1(x)y=f^{-1}(x).

[3]
Question 46
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

Let

h(x)=cos⁡x+0.2x,0≤x≤10h(x)=\cos x+0.2x,\quad 0\le x\le 10

The graph of y=h(x)y=h(x) is used to equations and inequalities.

A
I.

Using your GDC, all zeros of hh.

[2]
II.

the solution of h(x)<0h(x)<0.

[2]
B
I.

the minimum value of h(x)h(x) on 0≤x≤100\le x\le 10.

[2]
II.

why there are no zeros of hh for 5<x≤105<x\le 10.

[1]
C

the area enclosed by the graph of hh and the xx-axis.

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

The function ff is defined for x∈Rx\in\mathbb{R} by

f(x)=asin⁡(2x)+bcos⁡x+ccos⁡(2x)+df(x)=a\sin(2x)+b\cos x+c\cos(2x)+d

where a,b,c,d∈Ra,b,c,d\in\mathbb{R}. It is known that ff is even, f(0)=5f(0)=5, f(π)=1f(\pi)=1 and f(π3)=1f\left(\frac{\pi}{3}\right)=1.

A
I.

a=0a=0.

[2]
II.

the values of bb, cc and dd.

[3]
B

For the rest of the question, take f(x)=2cos⁡x+2cos⁡(2x)+1f(x)=2\cos x+2\cos(2x)+1.

I.

the range of ff.

[3]
II.

all solutions of f(x)=1f(x)=1 for 0≤x≤2π0\le x\le 2\pi.

[1]
C

The domain of ff is restricted to 0≤x≤π0\le x\le \pi. why this restricted function does not have an inverse function.

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

The function ff is defined by

f(x)=x3−3x2+2f(x)=x^3-3x^2+2

First consider ff on the domain x∈Rx\in\mathbb{R}.

A
I.

why ff does not have an inverse function on R\mathbb{R}.

[2]
II.

The domain is now restricted to x≥2x\ge2. the range of this restricted function.

[2]
B

Let gg be the inverse function of the restriction of ff to the domain x≥2x\ge2.

I.

the domain and range of gg.

[2]
II.

Using your GDC, g(10)g(10).

[2]
C

the gradient of the graph of y=g(x)y=g(x) at the point where x=10x=10.

[3]

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Question 49
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The function ff is defined by

f(x)=x4−4x2f(x)=x^4-4x^2
A
I.

ff is even.

[2]
II.

why ff does not have an inverse function on R\mathbb{R}.

[2]
B

For parts (b) and (c), restrict the domain of ff to x≥2x\ge\sqrt{2}.

I.

the range of this restricted function.

[2]
II.

f−1(x)f^{-1}(x), stating its domain.

[3]
C

the exact value of f−1(12)f^{-1}(12) and your answer by composition.

[3]
Question 50
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For real constants pp and qq, consider

f(x)=px+qx−p,x≠pf(x)=\frac{px+q}{x-p},\qquad x\ne p

You may assume q≠−p2q\ne -p^2.

A
I.

f−1(x)f^{-1}(x).

[3]
II.

the role of the condition q≠−p2q\ne -p^2.

[2]
B

For p=2p=2 and q=5q=5, the equations of the asymptotes and the intercepts with the axes.

[3]
C
I.

For p=2p=2 and q=5q=5, the fixed points of ff.

[2]
II.

geometrically why the graph is symmetric in the line y=xy=x.

[2]
Question 51
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Let a>0a>0. The function ff is defined by

f(x)=x2−2ax,x≥af(x)=x^2-2ax,\qquad x\ge a

This is one branch of a parabola.

Object

Equation

Domain / condition

Branch of ff

y=x2−2axy=x^2-2ax

a>0a>0, x≥ax\ge a

Line

y=xy=x

All real xx

A
I.

the vertex of the graph and the range of ff.

[3]
II.

f−1(x)f^{-1}(x), stating its domain.

[3]
B

, in terms of aa, the intersection point of the graph of ff and the line y=xy=x.

[2]
C

Given that the minimum value of ff is −9-9, the intersection point of the graph of ff and the line y=xy=x.

[2]
Question 52
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For a real parameter kk, define

hk(x)=e−x−x−k,x≥0h_k(x)=e^{-x}-x-k,\qquad x\ge 0

The equation hk(x)=0h_k(x)=0 may be interpreted as the intersection of y=e−xy=e^{-x} and y=x+ky=x+k on the restricted domain x≥0x\ge 0.

A
I.

hkh_k is strictly decreasing on x≥0x\ge 0.

[2]
II.

the values of kk for which hk(x)=0h_k(x)=0 has a solution on x≥0x\ge 0.

[3]
B
I.

Use your GDC to e−x=x+0.2e^{-x}=x+0.2 for x≥0x\ge 0.

[2]
II.

the value of kk for which the solution is x=0.7x=0.7.

[3]

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Question 53
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For a≠0a\ne 0, define

fa(x)=ax,x≠0f_a(x)=\frac{a}{x},\qquad x\ne 0

This question investigates a simple family of self-inverse functions and their intersections with straight lines through the origin.

A
I.

that faf_a is self-inverse.

[2]
II.

whether faf_a is odd, even or neither.

[2]
B

Let m≠0m\ne 0. the condition on aa and mm for the graphs of y=fa(x)y=f_a(x) and y=mxy=mx to have real intersections.

[3]
C
I.

For a=12a=12 and m=3m=3, the intersection points.

[2]
II.

the fixed points of f12f_{12}.

[1]
Question 54
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For a∈Ra\in\mathbb{R}, define

fa(x)=(x−a)3+a,x∈Rf_a(x)=(x-a)^3+a,\qquad x\in\mathbb{R}

The graph is a cubic translated so that its centre of rotational symmetry is (a,a)(a,a).

A
I.

fa−1(x)f_a^{-1}(x).

[2]
II.

why faf_a is one-to-one on R\mathbb{R}.

[1]
B

the fixed points of faf_a, giving your answers as coordinates.

[3]
C
I.

the value of aa for which the three fixed points have xx-coordinates with product 66.

[2]
II.

For this value of aa, use your GDC to confirm the intersections of y=fa(x)y=f_a(x) and y=xy=x.

[2]
Question 55
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

For k∈Rk\in\mathbb{R}, define Fk(x)=kx+1x−kF_k(x)=\dfrac{kx+1}{x-k}, where x≠kx\neq k.

A

Consider the case k=2k=2.

I.

the domain of F2F_2.

[1]
II.

22 is not in the range of F2F_2.

[1]
B

Use composition to examine F2F_2.

I.

(F2∘F2)(x)=x(F_2\circ F_2)(x)=x.

[3]
II.

F2−1(x)F_2^{-1}(x).

[1]
C

the fixed points of F2F_2.

[3]
D

FkF_k is self-inverse for every real value of kk.

[3]
Question 56
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

The function ff is defined by f(x)=ln⁡(1+x1−x)f(x)=\ln\left(\dfrac{1+x}{1-x}\right), for −1<x<1-1<x<1.

A

Consider the symmetry of ff.

I.

ff is an odd function.

[3]
II.

the symmetry of the graph of y=f(x)y=f(x).

[1]
B

the inverse function.

I.

f−1(x)f^{-1}(x).

[4]
II.

the domain and range of f−1f^{-1}.

[2]
C

Use the inverse function.

I.

f(x)=ln⁡3f(x)=\ln 3.

[1]
II.

f−1f^{-1} is an odd function.

[1]

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Question 57
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

For a real parameter aa, define

fa(x)=2x+aax−2,x≠2a, a≠0f_a(x)=\frac{2x+a}{ax-2},\quad x\ne \frac{2}{a},\ a\ne0
A
I.

the domain of faf_a.

[1]
II.

algebraically that faf_a is self-inverse.

[3]
B

It is given that fa(1)=5f_a(1)=5.

I.

Using the given condition fa(1)=5f_a(1)=5, the value of aa.

[2]
II.

fa−1(5)f_a^{-1}(5).

[1]
C

For a=3a=3, the fixed points of faf_a and the set of values of xx for which fa(x)>xf_a(x)>x.

[5]
Question 58
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Consider the function

h(x)=2cos⁡x+sin⁡2x,−π≤x≤πh(x)=2\cos x+\sin^2 x,\quad -\pi\le x\le \pi
A
I.

hh is an even function.

[2]
II.

why hh does not have an inverse function on [−π,π][-\pi,\pi].

[1]
B

Let gg be the restriction of hh to the domain 0≤x≤π0\le x\le\pi.

I.

the range of the restricted function.

[2]
II.

an expression for h−1(x)h^{-1}(x), stating its domain.

[3]
C
I.

g−1(0)g^{-1}(0).

[2]
II.

Using your GDC, the intersections of the graphs of y=h(x)y=h(x) and y=g−1(x)y=g^{-1}(x).

[2]
Question 59
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The function ff is defined for all real xx by

f(x)=x3+4xf(x)=x^3+4x

Let g=f−1g=f^{-1}.

A
I.

ff is odd.

[2]
II.

why ff has an inverse function.

[2]
B
I.

Using your GDC, g(20)g(20).

[2]
II.

g(−20)g(-20), giving a reason.

[2]
C

f(f(x))=10f(f(x))=10.

[4]
Question 60
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For a∈Ra\in\mathbb{R}, define

fa(x)=cos⁡x+acos⁡2x,x∈Rf_a(x)=\cos x+a\cos 2x,\qquad x\in\mathbb{R}

The graph is periodic. This question investigates its symmetry and range by using u=cos⁡xu=\cos x.

A
I.

faf_a is even.

[1]
II.

Using u=cos⁡xu=\cos x, express fa(x)f_a(x) as a quadratic in uu.

[2]
B

For a=1a=1, the range of f1f_1.

[4]
C
I.

the value of aa for which the quadratic in uu has its vertex at u=−12u=-\frac{1}{2}.

[2]
II.

For this value of aa, fa(x)=0f_a(x)=0 for 0≤x<2π0\le x<2\pi.

[3]

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Polynomials

Quadratics