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IB Maths AA HL – Topic 2: Functions

Polynomials, factor and remainder theorems, modulus, rational functions, inequalities.

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12 questions  ·  65 marks  ·  answers included  ·  IB AA HL
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This worksheet covers:

  • Exam-style practice on HL Topic 2
  • Polynomials
  • The factor and remainder theorems
  • Sums and products of roots
  • Modulus functions
  • Rational functions
  • Inequalities

Section A is non-calculator (Paper 1 style); Sections B and C may be attempted with a GDC. Allow about 75 minutes.

Questions

Section A – Core skills

Non-calculator. Show full working.
1

Let $f(x) = 2x^3 – 5x^2 – 4x + 3$.(a) Show that $(x-3)$ is a factor of $f(x)$.(b) Hence factorise $f(x)$ completely.(c) Solve $f(x) = 0$.

Show answer
(a) $f(3) = 54 – 45 – 12 + 3 = 0$, so $(x-3)$ is a factor.(b) $f(x) = (x-3)(2x^2 + x – 1) = (x-3)(2x-1)(x+1)$(c) $x = 3,\ x = \tfrac{1}{2},\ x = -1$
[5]
2

When the polynomial $p(x) = x^4 – 3x^3 + ax + 2$ is divided by $(x-2)$ the remainder is $-4$. Find the value of $a$.

Show answer
$p(2) = 16 – 24 + 2a + 2 = 2a – 6 = -4$, so $a = 1$
[3]
3

Solve the equation $|2x – 3| = 5 – x$.

Show answer
Case $2x-3 = 5-x$: $x = \tfrac{8}{3}$. Case $2x-3 = x-5$: $x = -2$. Both satisfy $5 – x \geq 0$, so $x = -2$ or $x = \tfrac{8}{3}$.
[4]
4

Solve the inequality $\dfrac{x+1}{x-2} \leq 3$.

Show answer
$\dfrac{x+1}{x-2} – 3 = \dfrac{7-2x}{x-2} \leq 0$; critical values $x = 2$ and $x = 3.5$. Solution: $x < 2$ or $x \geq 3.5$.
[4]
5

The function $f$ is defined by $f(x) = \dfrac{3x-2}{x+1}$, $x eq -1$.(a) Write down the equations of the asymptotes of the graph of $y = f(x)$.(b) Find the coordinates of the points where the graph crosses the axes.(c) Find $f^{-1}(x)$, stating its domain.

Show answer
(a) $x = -1$ and $y = 3$(b) $\left(\tfrac{2}{3}, 0\right)$ and $(0, -2)$(c) $f^{-1}(x) = \dfrac{x+2}{3-x}$, domain $x eq 3$
[5]

Section B – Exam-style

Calculator allowed unless stated.
6

The polynomial $p(x) = x^3 + ax^2 + bx – 12$ has a factor $(x-2)$. When $p(x)$ is divided by $(x+1)$ the remainder is $-12$.(a) Find the values of $a$ and $b$.(b) Hence show that the equation $p(x) = 0$ has exactly one real root.

Show answer
(a) $p(2) = 0 \Rightarrow 2a + b = 2$; $p(-1) = -12 \Rightarrow a – b = 1$; so $a = 1,\ b = 0$.(b) $p(x) = (x-2)(x^2 + 3x + 6)$; discriminant of the quadratic is $9 – 24 < 0$, so the only real root is $x = 2$.
[6]
7

The equation $2x^3 – 7x^2 + kx + 6 = 0$ has roots $\alpha$, $\beta$ and $\gamma$.(a) Write down the value of $\alpha + \beta + \gamma$ and of $\alpha\beta\gamma$.(b) Given that one of the roots is $3$, find the value of $k$.(c) Find the other two roots, giving your answers in exact form.

Show answer
(a) $\alpha + \beta + \gamma = \tfrac{7}{2}$, $\alpha\beta\gamma = -3$(b) $54 – 63 + 3k + 6 = 0 \Rightarrow k = 1$(c) $2x^3 – 7x^2 + x + 6 = (x-3)(2x^2 – x – 2)$, so $x = \dfrac{1 \pm \sqrt{17}}{4}$
[6]
8

Let $g(x) = |x^2 – 4|$.(a) Describe the graph of $y = g(x)$, giving the coordinates of any points where it meets the axes and of any local maximum.(b) Solve the equation $|x^2 – 4| = 3x$.(c) Hence solve the inequality $|x^2 – 4| < 3x$.

Show answer
(a) Graph of $y = x^2 – 4$ with the part below the $x$-axis reflected upward: meets the $x$-axis at $(\pm 2, 0)$, local maximum at $(0, 4)$, cusps at $x = \pm 2$.(b) $x^2 – 3x – 4 = 0 \Rightarrow x = 4$ (reject $-1$); $4 – x^2 = 3x \Rightarrow x = 1$ (reject $-4$). Solutions $x = 1,\ x = 4$.(c) $1 < x < 4$
[7]
9

Solve the inequality $|x – 1| > 2|x + 2|$.

Show answer
Squaring: $(x-1)^2 > 4(x+2)^2 \Rightarrow 3x^2 + 18x + 15 < 0 \Rightarrow (x+5)(x+1) < 0$. Solution: $-5 < x < -1$.
[4]
10

Let $f(x) = \dfrac{x^2 – 3x + 3}{x – 2}$, $x eq 2$.(a) Show that $f(x)$ can be written in the form $x + a + \dfrac{b}{x-2}$ and state the values of $a$ and $b$.(b) Write down the equations of the vertical and oblique asymptotes of the graph of $y = f(x)$.(c) Find the coordinates of the stationary points of the graph.(d) Hence state the range of $f$.

Show answer
(a) $f(x) = x – 1 + \dfrac{1}{x-2}$, so $a = -1$, $b = 1$(b) $x = 2$ and $y = x – 1$(c) $f'(x) = 1 – \dfrac{1}{(x-2)^2} = 0 \Rightarrow x = 1$ or $x = 3$; local maximum $(1, -1)$, local minimum $(3, 3)$(d) $f(x) \leq -1$ or $f(x) \geq 3$
[8]

Section C – Challenge

11

The polynomial $p(x) = x^4 + ax^3 + bx^2 – 5x + 2$ has $(x-1)^2$ as a factor.(a) Explain why $p(1) = 0$ and $p'(1) = 0$.(b) Find the values of $a$ and $b$.(c) Find the number of real roots of the equation $p(x) = 0$, justifying your answer.

Show answer
(a) $x = 1$ is a repeated root, so $p(x) = (x-1)^2 q(x)$ gives $p(1) = 0$ and $p'(x) = 2(x-1)q(x) + (x-1)^2 q'(x)$ gives $p'(1) = 0$.(b) $p(1) = 0 \Rightarrow a + b = 2$; $p'(1) = 4 + 3a + 2b – 5 = 0 \Rightarrow 3a + 2b = 1$; so $a = -3$, $b = 5$.(c) $p(x) = (x-1)^2 (x^2 – x + 2)$; discriminant $1 – 8 < 0$, so $x = 1$ is the only real root (one distinct real root, of multiplicity 2).
[7]
12

(a) Find the set of values of $k$ for which $x^2 + kx + (k + 3) > 0$ for all real values of $x$.(b) Solve the inequality $\dfrac{2x}{x-1} \geq x + 1$, giving the endpoints in exact form.

Show answer
(a) $k^2 – 4(k+3) < 0 \Rightarrow (k-6)(k+2) < 0 \Rightarrow -2 < k < 6$(b) $\dfrac{2x – (x+1)(x-1)}{x-1} \geq 0 \Rightarrow \dfrac{x^2 – 2x – 1}{x – 1} \leq 0$; critical values $1 \pm \sqrt{2}$ and $x = 1$. Solution: $x \leq 1 – \sqrt{2}$ or $1 < x \leq 1 + \sqrt{2}$.
[6]

Related worksheets

More worksheets in IB Mathematics: Analysis & Approaches:

IB Maths AA SL – Topic 1: Number and AlgebraIB Maths AA SL – Topic 2: FunctionsIB Maths AA SL – Topic 3: Geometry and TrigonometryIB Maths AA SL – Topic 4: Statistics and ProbabilityIB Maths AA SL – Topic 5: CalculusIB Maths AA HL – Topic 1: Number and AlgebraIB Maths AA HL – Topic 3: Geometry and TrigonometryIB Maths AA HL – Topic 4: Statistics and Probability

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