Clastify logo
Subjects
Features
Review
HOT
Tutoring

Modulus & Inequalities

Practice exam-style IB Math AA questions for Modulus & Inequalities, aligned with the syllabus and grouped by topic.

Verified by Karim
Verified by Karim
Paper
Difficulty
Status
Level

This content can't be printed from the browser. Use Export PDF with Pro for an offline copy.

Question 1
HL • Paper 1
Easy
Non Calculator
HL • Paper 1
Easy
Non Calculator

Consider the inequality ∣3x−2∣<x+4|3x-2|<x+4.

A

the necessary condition on x+4x+4.

[1]
B

the inequality.

[4]
Question 2
HL • Paper 2
Easy
Calculator Permitted
HL • Paper 2
Easy
Calculator Permitted

For a function ff, it is known that

f(X)≥0if and only if−4≤X≤0 or X≥3f(X)\geq0 \quad \text{if and only if} \quad -4\leq X\leq0 \text{ or } X\geq3

Let h(x)=f(2x−5)h(x)=f(2x-5).

A

the values of xx corresponding to X=−4X=-4, X=0X=0 and X=3X=3 under the transformation X=2x−5X=2x-5.

[2]
B

h(x)≥0h(x)\geq0.

[2]
Question 3
HL • Paper 2
Easy
Calculator Permitted
HL • Paper 2
Easy
Calculator Permitted

Let f(x)=x2−3x−4f(x)=x^2-3x-4, where x∈Rx\in\mathbb{R}.

A

an expression for f(∣x∣)f(|x|).

[1]
B

f(∣x∣)<0f(|x|)<0.

[3]
Question 4
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let f(x)=x2+1f(x)=x^2+1 and g(x)=x3+x2−3x+3g(x)=x^3+x^2-3x+3, where x∈Rx\in\mathbb{R}.

A

g(x)−f(x)=(x+2)(x−1)2g(x)-f(x)=(x+2)(x-1)^2.

[2]
B

g(x)≥f(x)g(x)\geq f(x).

[3]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 5
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Consider the equation ∣x+2∣+∣2x−1∣=7|x+2|+|2x-1|=7.

A

the critical values of xx at which the modulus expressions change form.

[1]
B

the equation.

[4]
Question 6
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let f(x)=x2−4x+3f(x)=x^2-4x+3, where x∈Rx\in\mathbb{R}.

A

the graph of y=∣f(x)∣y=|f(x)|, clearly showing the intercepts and the coordinates of any turning points.

[3]
B

∣f(x)∣=5|f(x)|=5.

[3]
Question 7
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let f(x)=x3−3xf(x)=x^3-3x and let h(x)=f(2x−1)h(x)=f(2x-1).

A

the zeros of hh.

[2]
B

h(x)≥0h(x)\geq0.

[3]
Question 8
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let f(x)=x2−2x−3f(x)=x^2-2x-3, where x∈Rx\in\mathbb{R}.

A

Rewrite [f(x)]2≤9[f(x)]^2\leq9 as a double inequality involving f(x)f(x).

[1]
B

[f(x)]2≤9[f(x)]^2\leq9.

[4]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 9
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Consider the inequality ∣x−2∣≥x2−4|x-2|\geq x^2-4.

A

the inequality for x<2x<2.

[2]
B

the inequality for x≥2x\geq2.

[2]
C

the solution set.

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

Let f(x)=x2+1f(x)=x^2+1 and g(x)=x3−x2−5x+7g(x)=x^3-x^2-5x+7, for x∈Rx\in\mathbb{R}.

A

g(x)−f(x)=(x+2)(x−1)(x−3)g(x)-f(x)=(x+2)(x-1)(x-3).

[2]
B

g(x)≥f(x)g(x)\geq f(x).

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

Let f(x)=(x+3)(x−1)(x−4)f(x)=(x+3)(x-1)(x-4), where x∈Rx\in\mathbb{R}.

A

∣f(x)∣=f(x)|f(x)|=f(x).

[3]
B

the domain of the function h(x)=1f(x)h(x)=\dfrac{1}{f(x)} and the equations of its vertical asymptotes.

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

A

the equation ∣2x−5∣+∣x+1∣=8|2x-5|+|x+1|=8.

[5]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 13
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A

the equation ∣x2−4∣=x+2|x^2-4|=x+2.

[5]
Question 14
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The graph of y=∣x2−4∣y=|x^2-4| is intersected by the horizontal line y=ky=k, where k∈Rk\in\mathbb{R}.

A

the coordinates of the local maximum and the local minima of y=∣x2−4∣y=|x^2-4|.

[2]
B

the range of values of kk for which the equation ∣x2−4∣=k|x^2-4|=k has exactly four distinct real solutions.

[2]
Question 15
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The function ff is defined by f(x)=1+2x−1f(x)=1+\dfrac{2}{x-1}, for x≠1x\neq1. Let h(x)=f(∣x∣)h(x)=f(|x|).

A

Write h(x)h(x) as a piecewise function and its domain.

[3]
B

the equations of the asymptotes of y=h(x)y=h(x) and h(x)=3h(x)=3.

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

Let f(x)=x2−4f(x)=x^2-4, where x∈Rx\in\mathbb{R}.

A

the graph of y=1f(x)y=\dfrac{1}{f(x)}, showing all asymptotes and the yy-intercept.

[4]
B

1f(x)≥15\dfrac{1}{f(x)}\geq\dfrac15.

[3]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 17
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Consider the inequality

∣x+1x−2∣<2\left|\frac{x+1}{x-2}\right|<2

where x≠2x\neq2.

A

the values of xx for which ∣x+1x−2∣=2\left|\dfrac{x+1}{x-2}\right|=2.

[2]
B

the inequality.

[3]
Question 18
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let f(x)=x2−2x−3f(x)=x^2-2x-3, where x∈Rx\in\mathbb{R}.

A

the graph of y=f(∣x∣)y=f(|x|), showing the intercepts and the coordinates of any local maxima or minima.

[4]
B

why f(∣x∣)f(|x|) is an even function.

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

Consider the inequality

∣3xarccos⁡x∣>1|3x\arccos x|>1
A

Use a GDC to the inequality for −1≤x≤1-1\leq x\leq1.

[5]
Question 20
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A

the inequality ∣x−4∣≤2∣x+1∣−3|x-4|\leq2|x+1|-3.

[5]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 21
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Let f(x)=(x−2)e−xf(x)=(x-2)e^{-x}, x∈Rx\in\mathbb{R}.

Function

Domain

f(x)=(x−2)e−xf(x)=(x-2)e^{-x}

xinmathbbRx\\in\\mathbb{R}

A

the equation of the vertical asymptote of the graph of y=1f(x)y=\dfrac{1}{f(x)}.

[1]
B

Use a GDC to ∣f(x)∣<0.1|f(x)|<0.1.

[4]
Question 22
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Let f(x)=ln⁡x−x3f(x)=\ln x-\dfrac{x}{3}, where x>0x>0.

A

Use a GDC to ∣f(x)∣≤0.5|f(x)|\leq0.5.

[5]
Question 23
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A

the inequality x+2x−1≥x\dfrac{x+2}{x-1}\geq x.

[6]
Question 24
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let f(X)=(X−3)(X−2)(X+2)f(X)=(X-3)(X-2)(X+2), where X∈RX\in\mathbb{R}. A second function is defined by h(x)=f(1−2x)h(x)=f(1-2x), where x∈Rx\in\mathbb{R}.

A
I.

the intervals for which f(X)≤0f(X)\leq0.

[3]
II.

h(x)≤0h(x)\leq0.

[1]
B
I.

h(x)=−2(x+1)(2x+1)(2x−3)h(x)=-2(x+1)(2x+1)(2x-3).

[3]
II.

h(x)>0h(x)>0. If you did not obtain the result in part (b)(i), use h(x)=−2(x+1)(2x+1)(2x−3)h(x)=-2(x+1)(2x+1)(2x-3).

[2]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 25
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Consider the function F(x)=∣x+4∣+2∣x−1∣−(x+10)F(x)=|x+4|+2|x-1|-(x+10), where x∈Rx\in\mathbb{R}.

A
I.

the critical values of xx at which the modulus expressions change form.

[1]
II.

Express F(x)F(x) without modulus signs on each of the intervals x<−4x<-4, −4≤x<1-4\leq x<1 and x≥1x\geq1.

[3]
B
I.

the equation ∣x+4∣+2∣x−1∣=x+10|x+4|+2|x-1|=x+10.

[3]
II.

the inequality ∣x+4∣+2∣x−1∣≤x+10|x+4|+2|x-1|\leq x+10. If you did not obtain the equation solutions in part (b)(i), use x=−2x=-2 and x=4x=4 as boundary points.

[2]
III.

the length of the solution interval found in part (b)(ii).

[1]
Question 26
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let F(x)=∣x−2∣+∣x+2∣F(x)=|x-2|+|x+2| and L(x)=x+5L(x)=x+5, where x∈Rx\in\mathbb{R}.

A
I.

Express F(x)F(x) as a piecewise function.

[3]
II.

the points of intersection of y=F(x)y=F(x) and y=L(x)y=L(x).

[1]
B
I.

F(x)≤L(x)F(x)\leq L(x).

[2]
II.

the exact area of the finite region enclosed by the graphs of y=F(x)y=F(x) and y=L(x)y=L(x). If you did not obtain the intersection points in part (a)(ii), use x=−1x=-1 and x=5x=5.

[4]
Question 27
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let p∈Rp\in\mathbb{R}. Consider the equation ∣x−1∣=px+3|x-1|=px+3.

A

For x<1x<1, the solution in terms of pp and when it is valid.

[2]
B

For x≥1x\geq1, the solution in terms of pp and when it is valid.

[2]
C

the values of pp for which the equation has exactly one solution.

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

Let f(x)=x2−5x+4f(x)=x^2-5x+4, for x∈Rx\in\mathbb{R}.

A

Consider the graph of y=∣f(x)∣y=|f(x)|.

I.

the coordinates of the local maximum and local minima of y=∣f(x)∣y=|f(x)|.

[3]
II.

∣f(x)∣≤5|f(x)|\leq5.

[3]
B

Define hh by h(x)=∣f(2x−3)∣h(x)=|f(2x-3)|.

I.

the values of xx for which h(x)=0h(x)=0.

[2]
II.

h(x)≤5h(x)\leq5.

[2]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 29
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Let f(x)=x2−4x−5f(x)=\dfrac{x^2-4}{x-5}, where x≠5x\neq5. Define h(x)=1f(∣x∣)h(x)=\dfrac{1}{f(|x|)}.

A

the domain and asymptotes of hh.

I.

the domain of hh.

[2]
II.

h(x)=∣x∣−5x2−4h(x)=\dfrac{|x|-5}{x^2-4}, and the equations of all vertical asymptotes.

[3]
B

h(x)≥0h(x)\geq0.

[4]
Question 30
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Let f(x)=x2−2x+1f(x)=x^2-2x+1 and g(x)=x3+2x2−7x+4g(x)=x^3+2x^2-7x+4, for x∈Rx\in\mathbb{R}.

A

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

I.

g(x)−f(x)=(x+3)(x−1)2g(x)-f(x)=(x+3)(x-1)^2.

[2]
II.

g(x)≥f(x)g(x)\geq f(x).

[3]
B

Let F(x)=f(2x+1)F(x)=f(2x+1) and G(x)=g(2x+1)G(x)=g(2x+1).

I.

a factorised expression for G(x)−F(x)G(x)-F(x).

[2]
II.

G(x)<F(x)G(x)<F(x).

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

Let A=x2−2x−8A=x^2-2x-8 and B=x+2B=x+2.

A

Consider the equation ∣A∣=∣B∣|A|=|B|.

I.

x=−2x=-2, x=3x=3 and x=5x=5 satisfy ∣x2−2x−8∣=∣x+2∣|x^2-2x-8|=|x+2|.

[3]
II.

why x=−2x=-2 is a repeated boundary point when solving inequalities involving ∣A∣|A| and ∣B∣|B|.

[2]
B

∣x2−2x−8∣≥∣x+2∣|x^2-2x-8|\geq|x+2|.

[4]
Question 32
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Let f(X)=(X+2)(X−3)f(X)=(X+2)(X-3) and define h(x)=∣f(3−2x)∣h(x)=|f(3-2x)|, for x∈Rx\in\mathbb{R}.

A

the transformation from y=f(X)y=f(X) to y=h(x)y=h(x).

I.

h(x)=∣4x2−10x∣h(x)=|4x^2-10x|.

[2]
II.

the coordinates of the local maximum of y=h(x)y=h(x) between its two zeros.

[3]
B

h(x)≤6h(x)\leq6.

[3]
C

The horizontal line y=ky=k, where k>0k>0, intersects the graph of y=h(x)y=h(x) at exactly four distinct points. the possible values of kk.

[2]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 33
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

The function ff is defined by f(x)=x−1x+2f(x)=\dfrac{x-1}{x+2}, for x≠−2x\neq-2.

A
I.

the equations of the asymptotes of y=f(x)y=f(x) and the intercepts with the axes.

[3]
II.

y=∣f(x)∣y=|f(x)|, clearly showing any asymptotes and intercepts.

[2]
B
I.

an expression for 1f(x)\dfrac{1}{f(x)} and its domain.

[2]
II.

∣f(x)∣≤12|f(x)|\leq\dfrac12.

[4]
Question 34
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let f(x)=x2−6x+5f(x)=x^2-6x+5, where x∈Rx\in\mathbb{R}, and let h(x)=f(∣x∣)h(x)=f(|x|).

A
I.

Write h(x)h(x) in terms of ∣x∣|x| and its zeros.

[2]
II.

the coordinates of the local maximum and local minima of y=h(x)y=h(x).

[3]
B
I.

h(x)≤0h(x)\leq0.

[2]
II.

the domain and the vertical asymptotes of y=1h(x)y=\dfrac{1}{h(x)}.

[2]
III.

1h(x)<0\dfrac{1}{h(x)}<0. If you did not obtain the zeros in part (a)(i), use x=−5,−1,1,5x=-5,-1,1,5.

[1]
Question 35
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let f(x)=x2−4xf(x)=x^2-4x, where x∈Rx\in\mathbb{R}, and let q(x)=[f(x)]2q(x)=[f(x)]^2.

A
I.

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

[2]
II.

the coordinates of the local maxima and local minima of y=q(x)y=q(x).

[2]
B
I.

q(x)≤16q(x)\leq16.

[4]
II.

the number of distinct real solutions of q(x)=kq(x)=k, for each possible range of k∈Rk\in\mathbb{R}.

[3]
Question 36
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let f(X)=(X+4)(X−2)(X−5)f(X)=(X+4)(X-2)(X-5), where X∈RX\in\mathbb{R}. The function hh is defined by h(x)=f(3−2x)h(x)=f(3-2x).

A
I.

f(X)≥0f(X)\geq0.

[3]
II.

h(x)≥0h(x)\geq0.

[1]
B
I.

h(x)=−2(x+1)(2x−1)(2x−7)h(x)=-2(x+1)(2x-1)(2x-7).

[3]
II.

∣h(x)∣=−h(x)|h(x)|=-h(x). If you did not obtain the expression in part (b)(i), use h(x)=−2(x+1)(2x−1)(2x−7)h(x)=-2(x+1)(2x-1)(2x-7).

[2]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 37
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

For a≥0a\geq0, define Fa(x)=∣x−a∣+∣x+a∣F_a(x)=|x-a|+|x+a|, where x∈Rx\in\mathbb{R}.

A
I.

Fa(x)=2aF_a(x)=2a for ∣x∣≤a|x|\leq a and Fa(x)=2∣x∣F_a(x)=2|x| for ∣x∣≥a|x|\geq a.

[3]
II.

For a=3a=3, Fa(x)=8F_a(x)=8.

[1]
B
I.

the number of distinct real solutions of Fa(x)=8F_a(x)=8 for each possible value of a≥0a\geq0.

[3]
II.

Fa(x)≤8F_a(x)\leq8 in terms of aa.

[3]
Question 38
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let f(t)=(t+1)(t−2)(t−4)f(t)=(t+1)(t-2)(t-4), where t∈Rt\in\mathbb{R}. Define p(x)=f(∣x∣)p(x)=f(|x|).

A
I.

the zeros of pp.

[2]
II.

why pp is an even function.

[2]
B
I.

p(x)≥0p(x)\geq0.

[3]
II.

The function ss is defined by s(x)=1p(x)s(x)=\dfrac{1}{p(x)}. the domain of ss and s(x)<0s(x)<0. If you did not obtain the zeros in part (a)(i), use x=−4,−2,2,4x=-4,-2,2,4.

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

Let f(x)=ln⁡(x+3)f(x)=\ln(x+3), for x>−3x>-3, and let g(x)=0.15x2−0.7g(x)=0.15x^2-0.7. Define h(x)=g(x)−f(x)h(x)=g(x)-f(x).

A

Use a GDC to the intersections of the two curves.

I.

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

[3]
II.

g(x)≥f(x)g(x)\geq f(x).

[2]
B

Let h(x)=g(x)−f(x)h(x)=g(x)-f(x).

I.

the minimum value of h(x)h(x) for x>−3x>-3.

[3]
II.

The graph of y=g(x)y=g(x) is translated vertically upwards by kk units. the least value of kk for which the translated graph is on or above the graph of y=f(x)y=f(x) for all x>−3x>-3.

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

Consider the equation and inequality involving two modulus expressions
∣x+1∣+∣2x−5∣|x+1|+|2x-5| and the quadratic expression x2−2x^2-2.

A

the equation ∣x+1∣+∣2x−5∣=x2−2|x+1|+|2x-5|=x^2-2.

I.

the critical values of xx and the three corresponding intervals that must be considered.

[2]
II.

the equation ∣x+1∣+∣2x−5∣=x2−2|x+1|+|2x-5|=x^2-2, giving exact answers.

[4]
B

the inequality ∣x+1∣+∣2x−5∣≤x2−2|x+1|+|2x-5|\leq x^2-2.

[4]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 41
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Let f(x)=x3−4xf(x)=x^3-4x, for x∈Rx\in\mathbb{R}.

A

Consider the equation [f(x)]2=9[f(x)]^2=9.

I.

the solutions of f(x)=3f(x)=3 are x=−1x=-1 and x=1±132x=\dfrac{1\pm\sqrt{13}}{2}.

[3]
II.

the solutions of f(x)=−3f(x)=-3.

[3]
B

[f(x)]2≤9[f(x)]^2\leq9.

[5]
Question 42
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Let f(x)=ln⁡xf(x)=\ln x and g(x)=2−x2g(x)=2-\dfrac{x}{2}, for x>0x>0. Also define d(x)=g(x)−f(x)d(x)=g(x)-f(x).

A

Use a GDC to the two functions.

I.

the xx-coordinate of the point of intersection of y=f(x)y=f(x) and y=g(x)y=g(x).

[2]
II.

g(x)≥f(x)g(x)\geq f(x).

[2]
B

Let d(x)=g(x)−f(x)d(x)=g(x)-f(x).

I.

why dd is strictly decreasing on x>0x>0.

[2]
II.

∣g(x)−f(x)∣≤1|g(x)-f(x)|\leq1.

[4]
Question 43
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A sensor has signed error e(t)=0.4t2−3t+2e(t)=0.4t^2-3t+2, where tt is measured in hours and 0≤t≤80\leq t\leq8. The absolute error is modelled by E(t)=∣e(t)∣E(t)=|e(t)|.

A

the model for the absolute error.

I.

the times at which the signed error is zero.

[2]
II.

the maximum value of E(t)E(t) on 0≤t≤80\leq t\leq8.

[3]
B

The sensor is considered to be within tolerance when E(t)≤1.5E(t)\leq1.5.

I.

the time intervals during which the sensor is within tolerance.

[4]
II.

the total length of time for which the sensor is within tolerance.

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

For a>0a>0, define Da(x)=∣x−a∣+∣x+a∣D_a(x)=|x-a|+|x+a|. This question investigates the shape of y=Da(x)y=D_a(x) and inequalities involving a sloping line.

A
I.

Write Da(x)D_a(x) as a piecewise-defined function.

[3]
II.

y=D3(x)y=D_3(x), showing the coordinates of the two corner points.

[2]
B

D3(x)≤x+6D_3(x)\leq x+6.

[4]
C

For 0<m<20<m<2, Da(x)≤mx+2aD_a(x)\leq mx+2a in terms of aa and mm.

[5]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 45
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For a real parameter tt, define ht(X)=(X+2)(X−t)(X−4)h_t(X)=(X+2)(X-t)(X-4) and Ht(x)=ht(2x−1)H_t(x)=h_t(2x-1). This question compares the sign of a cubic before and after a horizontal transformation.

Object

Description

Role

ht(X)h_t(X)

A cubic depending on tt

Object studied

h1(X)h_1(X)

A special case of ht(X)h_t(X)

Used in part (a)

Ht(x)H_t(x)

A transformed cubic

Used in parts (b) and (c)

Transformation

A linear change of variable

Relates XX and xx

A

h1(X)≥0h_1(X)\geq0.

[3]
B
I.

the three roots of Ht(x)=0H_t(x)=0.

[2]
II.

H3(x)≥0H_3(x)\geq0.

[3]
C

the values of tt for which Ht(x)≥0H_t(x)\geq0 for every xx in the interval 0≤x≤20\leq x\leq2.

[4]
Question 46
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Let f(x)=x2−2x−8f(x)=x^2-2x-8, where x∈Rx\in\mathbb{R}. This question compares the graphs of y=f(x)y=f(x), y=∣f(x)∣y=|f(x)| and y=[f(x)]2y=[f(x)]^2.

A

the zeros of ff and the coordinates of its vertex.

[3]
B

Using the supplied graph, all intercepts and local extrema of y=∣f(x)∣y=|f(x)|, how it is obtained from y=f(x)y=f(x), and its end behaviour.

[4]
C

[f(x)]2≤16[f(x)]^2\leq16.

[5]
D

∣f(x)∣≥[f(x)]2|f(x)|\geq [f(x)]^2.

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

Define N(x)=∣∣x−2∣−3∣N(x)=\big||x-2|-3\big|. This question investigates a nested modulus function.

A

Write N(x)N(x) as a piecewise-defined function.

[4]
B

Using the graph, the range of N(x)N(x) and the intervals on which N(x)N(x) is increasing and decreasing.

[3]
C

N(x)≤2N(x)\leq2.

[3]
D

the number of real solutions of N(x)=cN(x)=c for all real values of cc.

[4]
Question 48
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Let S(x)=∣x+1∣+2∣x−4∣S(x)=|x+1|+2|x-4|. This question investigates inequalities involving a weighted sum of distances on a number line.

A

Write S(x)S(x) as a piecewise-defined function.

[3]
B

S(x)≤9S(x)\leq9.

[3]
C
I.

For p>0p>0, define Tp(x)=∣x∣+p∣x−6∣T_p(x)=|x|+p|x-6|. Write Tp(x)T_p(x) as a piecewise-defined function.

[3]
II.

the value or values of xx for which Tp(x)T_p(x) is a minimum, considering separately 0<p<10<p<1, p=1p=1 and p>1p>1.

[3]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 49
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Let F(x)=(x+4)(x−2)2F(x)=(x+4)(x-2)^2. This question investigates inequalities involving a cubic with a repeated root and transformations of its input.

A
I.

F(x)≥0F(x)\geq0.

[3]
II.

∣F(x)∣=F(x)|F(x)|=F(x).

[2]
B

F(3x−1)<0F(3x-1)<0.

[3]
C

1F(∣x∣)>0\frac{1}{F(|x|)}>0.

[4]
Question 50
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let f(x)=(x−2)2−1f(x)=(x-2)^2-1, where x∈Rx\in\mathbb{R}. The function rr is defined by r(x)=1f(x)r(x)=\dfrac{1}{f(x)}.

A
AI.

the zeros of ff and the vertical asymptotes of y=r(x)y=r(x).

[2]
AII.

the coordinates of the turning point of y=f(x)y=f(x) and the horizontal asymptote of y=r(x)y=r(x).

[2]
B
BI.

the values of xx for which r(x)=1r(x)=1.

[2]
BII.

r(x)≥1r(x)\geq1. If you did not obtain the values in part (b)(i), use x=2−2x=2-\sqrt2 and x=2+2x=2+\sqrt2.

[4]
Question 51
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

For k>2k>2, the graphs of y=∣x−1∣y=|x-1| and y=k−∣x+1∣y=k-|x+1| enclose a finite region.

A
I.

the xx-coordinates of the points of intersection are x=−k2x=-\dfrac{k}{2} and x=k2x=\dfrac{k}{2}.

[3]
II.

For k=4k=4, the area of the enclosed region.

[1]
B
I.

the area AA of the enclosed region is A=(k−2)22+2(k−2)A=\dfrac{(k-2)^2}{2}+2(k-2).

[5]
II.

Given that the area of the enclosed region is 2424, kk. If you did not obtain the formula in part (b)(i), use A=(k−2)22+2(k−2)A=\dfrac{(k-2)^2}{2}+2(k-2).

[3]
Question 52
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let R(x)=∣x−3x+1∣R(x)=\left|\dfrac{x-3}{x+1}\right|, where x∈Rx\in\mathbb{R} and x≠−1x\neq-1.

A
I.

R(x)=2R(x)=2.

[3]
II.

R(x)>2R(x)>2. If you did not obtain the values in part (a)(i), use x=−5x=-5 and x=13x=\dfrac13 as boundary points.

[2]
B
I.

For m∈Rm\in\mathbb{R}, R(x)=mR(x)=m in terms of mm, where possible.

[3]
II.

the number of distinct real solutions of R(x)=mR(x)=m for all m∈Rm\in\mathbb{R}.

[2]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 53
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Let f(x)=1x−1f(x)=\dfrac{1}{x-1} and g(x)=x+2x+4g(x)=\dfrac{x+2}{x+4}, where x≠1,−4x\neq1,-4.

A

ff and gg algebraically.

I.

g(x)−f(x)=x2−6(x+4)(x−1)g(x)-f(x)=\dfrac{x^2-6}{(x+4)(x-1)}.

[2]
II.

g(x)≥f(x)g(x)\geq f(x).

[4]
B

∣g(x)−f(x)∣≤1|g(x)-f(x)|\leq1.

[6]
Question 54
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Let p(x)=∣2sin⁡x−1∣p(x)=|2\sin x-1| and q(x)=1+cos⁡xq(x)=1+\cos x, for 0≤x≤2π0\leq x\leq2\pi.

A

the intersection points of the two graphs.

I.

For the case 2sin⁡x−1≥02\sin x-1\geq0, p(x)=q(x)p(x)=q(x) on 0≤x≤2π0\leq x\leq2\pi.

[4]
II.

For the case 2sin⁡x−1<02\sin x-1<0, p(x)=q(x)p(x)=q(x) on 0≤x≤2π0\leq x\leq2\pi.

[3]
B

∣2sin⁡x−1∣≤1+cos⁡x|2\sin x-1|\leq1+ \cos x for 0≤x≤2π0\leq x\leq2\pi.

[4]
Question 55
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Let q(x)=∣xln⁡(x+2)∣q(x)=|x\ln(x+2)|, where x>−2x>-2. This question uses both graphical and numerical methods to a modulus inequality.

A
I.

the domain of qq.

[1]
II.

Use a GDC to q(x)=1q(x)=1.

[4]
B

q(x)≤1q(x)\leq1.

[3]
C

the smallest value of cc such that the solution of q(x)≤cq(x)\leq c contains the whole interval [−1,1][-1,1].

[4]
Question 56
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Let f(x)=x2−5x+6f(x)=x^2-5x+6 and define R(x)=1f(x)R(x)=\frac{1}{f(x)}. This question investigates inequalities involving a reciprocal graph.

A

the domain of RR and the equations of its vertical asymptotes.

[2]
B

R(x)≥12R(x)\geq\frac12.

[4]
C
I.

For k>0k>0, the solutions of R(x)=kR(x)=k are x=5±1+4k2x=\frac{5\pm\sqrt{1+\frac4k}}{2}.

[2]
II.

R(x)≥kR(x)\geq k for k>0k>0.

[2]
D

the value of kk for which the total length of the two intervals in part (c)(ii) is 55.

[3]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview
Question 57
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Let f(x)=x3−5x+4f(x)=x^3-5x+4. This question compares y=f(∣x∣)y=f(|x|) and y=∣f(x)∣y=|f(x)|.

A

the zeros of ff and f(x)≥0f(x)\geq0.

[4]
B

f(∣x∣)≥0f(|x|)\geq0.

[4]
C

∣f(x)∣=f(∣x∣)|f(x)|=f(|x|).

[5]
Question 58
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Let p(x)=∣e−x−x∣p(x)=|e^{-x}-x|. This question uses technology to a modulus inequality and then studies the effect of a horizontal transformation.

A
I.

Use a GDC to p(x)=0.5p(x)=0.5.

[3]
II.

p(x)≤0.5p(x)\leq0.5.

[2]
B

Let r(x)=p(2x−1)r(x)=p(2x-1). r(x)≤0.5r(x)\leq0.5. If you did not obtain the values in part (a), use 0.2662490.266249 and 0.9046740.904674.

[3]
C

Suppose the solution of p(X)≤0.5p(X)\leq0.5 is α≤X≤β\alpha\leq X\leq\beta. For a>0a>0, b∈Rb\in\mathbb{R}, the solution of p(ax+b)≤0.5p(ax+b)\leq0.5.

[4]
Question 59
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For a∈Ra\in\mathbb{R}, let CaC_a be the graph y=∣x−a∣y=|x-a| and let PP be the parabola y=x2−1y=x^2-1. This question investigates their intersections.

A
I.

For a=0a=0, ∣x∣=x2−1|x|=x^2-1.

[3]
II.

∣x∣≤x2−1|x|\leq x^2-1.

[3]
B

intersections of CaC_a and PP must satisfy x2−x+a−1=0x^2-x+a-1=0 when x≥ax\geq a, and x2+x−a−1=0x^2+x-a-1=0 when x<ax<a.

[2]
C

that CaC_a and PP have exactly two points of intersection for every real value of aa.

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

Let A(X)=∣X−3∣A(X)=|X-3| and B(X)=ln⁡(X+4)B(X)=\ln(X+4), where X>−4X>-4. This question investigates the inequality A(X)≤B(X)A(X)\leq B(X) and then applies an inside transformation.

A
I.

Use a GDC to ∣X−3∣=ln⁡(X+4)|X-3|=\ln(X+4).

[3]
II.

∣X−3∣≤ln⁡(X+4)|X-3|\leq\ln(X+4).

[3]
B

∣2x−4∣≤ln⁡(2x+3)|2x-4|\leq\ln(2x+3). Give your answer to 2 decimal places. If you did not obtain the values in part (a), use 1.3271.327 and 5.2225.222 (or use the unrounded GDC values).

[3]
C

Suppose A(X)≤B(X)A(X)\leq B(X) has solution α≤X≤β\alpha\leq X\leq\beta. For a<0a<0, the solution of A(ax+b)≤B(ax+b)A(ax+b)\leq B(ax+b) in terms of aa, bb, α\alpha and β\beta.

[3]

Build a Practice Exam — Math AA HL

Test your knowledge with a custom paper containing questions from selected topics.

Practice exam question preview
Practice exams dashboard preview

Exponent-Log Functions

Polynomials