A-Level Mathematics – Chapter 1: Algebra and Functions

Edexcel · AQA · OCR A-Level Mathematics · Updated March 2026

Algebra and Functions is the cornerstone of A-Level Mathematics. The techniques you develop here — manipulating expressions, solving equations, understanding functions — underpin every other topic in the course. This chapter covers Year 1 material on indices, surds, quadratics, and inequalities, before extending to Year 2 content on partial fractions, functions, and graph transformations.

Specification Note

Content labelled Year 2 is A2-level only (not required for AS). All other content is required for both AS and full A-Level.

1. Indices and Surds

1.1 Index Laws

The index laws (also called laws of exponents) allow you to simplify expressions involving powers. You should already know these from GCSE, but they are used extensively throughout A-Level.

Index Laws

For any base $a \neq 0$ and indices $m, n$:

$a^m \times a^n = a^{m+n}$
$a^m \div a^n = a^{m-n}$
$(a^m)^n = a^{mn}$
$a^0 = 1$
$a^{-n} = \dfrac{1}{a^n}$
$a^{1/n} = \sqrt[n]{a}$
$a^{m/n} = \left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m}$

Worked Example 1.1 — Simplifying Index Expressions

Simplify: (a) $\dfrac{x^5 \cdot x^{-2}}{x^3}$ (b) $\left(8x^6\right)^{2/3}$ (c) $\dfrac{4^{3/2} \cdot 2^{-1}}{8^{2/3}}$

a $\dfrac{x^5 \cdot x^{-2}}{x^3} = \dfrac{x^{5+(-2)}}{x^3} = \dfrac{x^3}{x^3} = x^0 = 1$

b $(8x^6)^{2/3} = 8^{2/3} \cdot (x^6)^{2/3} = \left(\sqrt[3]{8}\right)^2 \cdot x^4 = 2^2 \cdot x^4 = 4x^4$

c $4^{3/2} = (2^2)^{3/2} = 2^3 = 8$; $8^{2/3} = (2^3)^{2/3} = 2^2 = 4$

So $\dfrac{8 \cdot 2^{-1}}{4} = \dfrac{8 \cdot \frac{1}{2}}{4} = \dfrac{4}{4} = 1$

1.2 Simplifying Surds

Definition: Surd

A surd is an irrational root that cannot be simplified to a rational number. For example, $\sqrt{2}$, $\sqrt{3}$, and $\sqrt{5}$ are surds, but $\sqrt{4} = 2$ is not.

To simplify a surd $\sqrt{n}$, find the largest perfect square factor of $n$:

$$\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$$

Key rules for surds:

$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$
$\sqrt{a} \div \sqrt{b} = \sqrt{a/b}$
$(\sqrt{a})^2 = a$
$(p + q\sqrt{a})(p - q\sqrt{a}) = p^2 - q^2 a$ (difference of two squares)

1.3 Rationalising the Denominator

A fraction is in its simplest form when the denominator contains no surds. To rationalise a denominator:

If denominator is $\sqrt{a}$: multiply numerator and denominator by $\sqrt{a}$.
If denominator is $p + q\sqrt{a}$: multiply by the conjugate $p - q\sqrt{a}$.

Worked Example 1.2 — Rationalising the Denominator

Rationalise: (a) $\dfrac{5}{\sqrt{3}}$ (b) $\dfrac{4}{2 + \sqrt{5}}$

a Multiply by $\dfrac{\sqrt{3}}{\sqrt{3}}$:

$$\frac{5}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{5\sqrt{3}}{3}$$

b Multiply by the conjugate $\dfrac{2 - \sqrt{5}}{2 - \sqrt{5}}$:

$$\frac{4}{2+\sqrt{5}} \times \frac{2-\sqrt{5}}{2-\sqrt{5}} = \frac{4(2-\sqrt{5})}{4 - 5} = \frac{8 - 4\sqrt{5}}{-1} = -8 + 4\sqrt{5}$$

Practice 1A

1. Simplify $\dfrac{x^{3/2} \cdot x^{1/2}}{x^{-1}}$.

2. Write $\sqrt{180} - \sqrt{45}$ in the form $k\sqrt{5}$.

3. Rationalise and simplify $\dfrac{6}{\sqrt{3} - 1}$.

Show Solution

1. $\dfrac{x^{3/2} \cdot x^{1/2}}{x^{-1}} = \dfrac{x^2}{x^{-1}} = x^{2-(-1)} = x^3$

2. $\sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5}$; $\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}$. So $6\sqrt{5} - 3\sqrt{5} = 3\sqrt{5}$, giving $k = 3$.

3. $\dfrac{6}{\sqrt{3}-1} \times \dfrac{\sqrt{3}+1}{\sqrt{3}+1} = \dfrac{6(\sqrt{3}+1)}{3-1} = \dfrac{6(\sqrt{3}+1)}{2} = 3(\sqrt{3}+1) = 3\sqrt{3} + 3$

2. Quadratic Functions

2.1 Completing the Square

Any quadratic $ax^2 + bx + c$ can be written in completed-square (vertex) form $a(x+p)^2 + q$. This reveals the vertex of the parabola at $(-p,\, q)$.

Method: Completing the Square

For $x^2 + bx + c$: add and subtract $\left(\dfrac{b}{2}\right)^2$ inside:

$$x^2 + bx + c = \left(x + \frac{b}{2}\right)^2 - \frac{b^2}{4} + c$$

For $ax^2 + bx + c$ (with $a \neq 1$): first factor out $a$ from the $x$-terms.

Worked Example 2.1 — Completing the Square

Write $2x^2 - 8x + 3$ in the form $a(x+p)^2 + q$. Hence state the vertex of its graph.

Step 1 Factor out 2 from the $x$-terms:

$$2x^2 - 8x + 3 = 2(x^2 - 4x) + 3$$

Step 2 Complete the square inside the bracket. Half of $-4$ is $-2$, so add and subtract $(-2)^2 = 4$:

$$= 2\bigl[(x-2)^2 - 4\bigr] + 3$$

Step 3 Expand and collect:

$$= 2(x-2)^2 - 8 + 3 = 2(x-2)^2 - 5$$

So $a=2$, $p=-2$, $q=-5$. The vertex is at $(2, -5)$.

2.2 The Discriminant

For a quadratic $ax^2 + bx + c = 0$, the discriminant is $\Delta = b^2 - 4ac$. It determines the number and nature of the roots.

Nature of Roots via the Discriminant

$\Delta > 0$: two distinct real roots (graph crosses the $x$-axis twice)
$\Delta = 0$: one repeated real root (graph touches the $x$-axis once)
$\Delta < 0$: no real roots (graph does not intersect the $x$-axis)

Fig. 1 — A quadratic $y = x^2 + bx + 1$. Use the slider to vary $b$ and observe how the discriminant $b^2 - 4$ changes the number of real roots.

Worked Example 2.2 — Using the Discriminant

Find the range of values of $k$ for which $kx^2 + 4x + 1 = 0$ has two distinct real roots.

Step 1 Identify $a = k$, $b = 4$, $c = 1$.

Step 2 For two distinct roots: $\Delta > 0$:

$$b^2 - 4ac > 0 \implies 16 - 4k > 0 \implies k < 4$$

Step 3 Also, for a quadratic to exist, $k \neq 0$. So $k < 4$ and $k \neq 0$.

2.3 Sketching Parabolas

To sketch $y = ax^2 + bx + c$:

Identify whether $a > 0$ (U-shape) or $a < 0$ (∩-shape).
Find the $y$-intercept: set $x = 0$, giving $y = c$.
Find the $x$-intercepts (if they exist): solve $ax^2 + bx + c = 0$.
Find the vertex by completing the square or using $x = -\dfrac{b}{2a}$.

Practice 2A

1. Write $x^2 - 6x + 11$ in the form $(x+p)^2 + q$ and state the minimum value.

2. Show that $3x^2 - 5x + 4 = 0$ has no real solutions.

3. Find the values of $m$ for which $x^2 + mx + 9 = 0$ has equal roots.

Show Solution

1. $x^2 - 6x + 11 = (x-3)^2 - 9 + 11 = (x-3)^2 + 2$. So $p = -3$, $q = 2$. Minimum value is $2$ (when $x = 3$).

2. $\Delta = (-5)^2 - 4(3)(4) = 25 - 48 = -23 < 0$. Since $\Delta < 0$, there are no real solutions.

3. Equal roots require $\Delta = 0$: $m^2 - 4(1)(9) = 0 \Rightarrow m^2 = 36 \Rightarrow m = \pm 6$.

3. Solving Equations and Inequalities

3.1 Linear Inequalities

Solve linear inequalities just as you would equations, with one crucial rule: when multiplying or dividing by a negative number, reverse the inequality sign.

Worked Example 3.1 — Linear Inequality

Solve $3 - 2x > 7$.

$3 - 2x > 7 \implies -2x > 4 \implies x < -2$ (inequality reverses on dividing by $-2$).

3.2 Quadratic Inequalities

To solve a quadratic inequality such as $ax^2 + bx + c > 0$ or $< 0$:

Factorise (or use the quadratic formula) to find the roots $\alpha$ and $\beta$ (with $\alpha \leq \beta$).
Sketch the parabola.
Read off the solution from the sketch: for a positive leading coefficient:
- $ax^2 + bx + c > 0$: $x < \alpha$ or $x > \beta$
- $ax^2 + bx + c < 0$: $\alpha < x < \beta$

Worked Example 3.2 — Quadratic Inequality

Solve $x^2 - 5x + 6 \leq 0$.

Step 1 Factorise: $(x-2)(x-3) \leq 0$. Roots: $x = 2$ and $x = 3$.

Step 2 Sketch: parabola opening upward, crossing at $x = 2$ and $x = 3$.

Step 3 The parabola is below (or on) the $x$-axis between the roots: $2 \leq x \leq 3$.

3.3 Simultaneous Equations

Linear-Linear: Use elimination or substitution. Linear-Quadratic: Use substitution from the linear equation into the quadratic.

Worked Example 3.3 — Linear-Quadratic Simultaneous Equations

Solve simultaneously: $y = x + 1$ and $x^2 + y^2 = 13$.

Step 1 Substitute $y = x + 1$ into the circle equation:

$$x^2 + (x+1)^2 = 13$$ $$x^2 + x^2 + 2x + 1 = 13$$ $$2x^2 + 2x - 12 = 0 \implies x^2 + x - 6 = 0$$

Step 2 Factorise: $(x+3)(x-2) = 0$, so $x = -3$ or $x = 2$.

Step 3 Find corresponding $y$ values: when $x = -3$, $y = -2$; when $x = 2$, $y = 3$.

Solutions: $(-3, -2)$ and $(2, 3)$.

Practice 3A

1. Solve $2x^2 + x - 3 > 0$.

2. Solve simultaneously: $y = 2x - 1$ and $y = x^2 - 3x + 5$.

Show Solution

1. Factorise: $(2x+3)(x-1) > 0$. Roots at $x = -\frac{3}{2}$ and $x = 1$. Parabola opens upward, so solution is $x < -\frac{3}{2}$ or $x > 1$.

2. Set equal: $2x - 1 = x^2 - 3x + 5 \Rightarrow x^2 - 5x + 6 = 0 \Rightarrow (x-2)(x-3) = 0$. So $x = 2$ or $x = 3$. When $x = 2$: $y = 3$; when $x = 3$: $y = 5$. Solutions: $(2,3)$ and $(3,5)$.

4. Algebraic Fractions and Partial Fractions Year 2

4.1 Simplifying and Operating on Algebraic Fractions

Algebraic fractions follow the same rules as numerical fractions. To add or subtract, find a common denominator. To multiply, multiply numerators together and denominators together. To divide, multiply by the reciprocal.

Worked Example 4.1 — Adding Algebraic Fractions

Write $\dfrac{3}{x+1} + \dfrac{2}{x-2}$ as a single fraction.

Common denominator is $(x+1)(x-2)$:

$$\frac{3(x-2) + 2(x+1)}{(x+1)(x-2)} = \frac{3x - 6 + 2x + 2}{(x+1)(x-2)} = \frac{5x - 4}{(x+1)(x-2)}$$

4.2 Partial Fractions

Partial fractions is the reverse process: splitting a single fraction into a sum of simpler fractions. This technique is essential for integration in Year 2.

Forms of Partial Fractions

Distinct linear factors: $\dfrac{f(x)}{(ax+b)(cx+d)} \equiv \dfrac{A}{ax+b} + \dfrac{B}{cx+d}$
Repeated linear factor: $\dfrac{f(x)}{(ax+b)^2(cx+d)} \equiv \dfrac{A}{ax+b} + \dfrac{B}{(ax+b)^2} + \dfrac{C}{cx+d}$
Improper fraction (degree of numerator $\geq$ degree of denominator): perform polynomial long division first.

Worked Example 4.2 — Distinct Linear Factors

Express $\dfrac{7x - 1}{(x-1)(2x+1)}$ in partial fractions.

Step 1 Write $\dfrac{7x-1}{(x-1)(2x+1)} \equiv \dfrac{A}{x-1} + \dfrac{B}{2x+1}$.

Step 2 Multiply both sides by $(x-1)(2x+1)$:

$$7x - 1 \equiv A(2x+1) + B(x-1)$$

Step 3 Substitute convenient values of $x$:

Let $x = 1$: $7 - 1 = A(3) + 0 \Rightarrow A = 2$.

Let $x = -\frac{1}{2}$: $-\frac{7}{2} - 1 = 0 + B(-\frac{3}{2}) \Rightarrow -\frac{9}{2} = -\frac{3B}{2} \Rightarrow B = 3$.

Answer: $\dfrac{2}{x-1} + \dfrac{3}{2x+1}$.

Worked Example 4.3 — Repeated Linear Factor

Express $\dfrac{5x^2 + 3x - 2}{x^2(x+1)}$ in partial fractions.

Write: $\dfrac{A}{x} + \dfrac{B}{x^2} + \dfrac{C}{x+1}$. Multiply through: $5x^2 + 3x - 2 \equiv Ax(x+1) + B(x+1) + Cx^2$.

$x = 0$: $-2 = B$, so $B = -2$.

$x = -1$: $5 - 3 - 2 = C$, so $C = 0$.

Comparing $x^2$ terms: $5 = A + C \Rightarrow A = 5$.

Answer: $\dfrac{5}{x} - \dfrac{2}{x^2}$.

Practice 4A

1. Express $\dfrac{8x + 3}{(x+2)(x-1)}$ in partial fractions.

2. Express $\dfrac{x^2 + 1}{x(x-1)^2}$ in partial fractions. (Hint: improper? Check degrees.)

Show Solution

1. Write $\frac{A}{x+2} + \frac{B}{x-1}$. Multiply: $8x+3 = A(x-1)+B(x+2)$. Let $x=1$: $11=3B \Rightarrow B=\frac{11}{3}$. Let $x=-2$: $-13=-3A \Rightarrow A=\frac{13}{3}$. Answer: $\dfrac{13}{3(x+2)} + \dfrac{11}{3(x-1)}$.

2. Numerator degree (2) equals denominator degree (3), so it is proper. Write $\frac{A}{x}+\frac{B}{x-1}+\frac{C}{(x-1)^2}$. Multiply: $x^2+1=A(x-1)^2+Bx(x-1)+Cx$. Let $x=0$: $1=A$. Let $x=1$: $2=C$. Compare $x^2$: $1=A+B \Rightarrow B=0$. Answer: $\dfrac{1}{x} + \dfrac{2}{(x-1)^2}$.

5. Functions Year 2

5.1 Domain and Range

Definitions

The domain of a function is the set of all allowable input values ($x$-values).
The range (or image) of a function is the set of all output values ($y$-values) produced.

For example, $f(x) = \sqrt{x-2}$ requires $x - 2 \geq 0$, so the domain is $x \geq 2$, and the range is $f(x) \geq 0$.

5.2 Composite Functions

The composite function $fg(x)$ means "apply $g$ first, then $f$". That is, $fg(x) = f\!\left(g(x)\right)$.

Exam Tip

Be careful with the order: $fg$ means $f$ of $g$, so $g$ is applied first. $fg \neq gf$ in general.

Worked Example 5.1 — Composite Functions

Given $f(x) = 2x + 1$ and $g(x) = x^2 - 3$, find: (a) $fg(x)$ (b) $gf(2)$.

a $fg(x) = f\!\left(g(x)\right) = f(x^2-3) = 2(x^2-3)+1 = 2x^2 - 5$

b $f(2) = 5$, then $g(5) = 25 - 3 = 22$. So $gf(2) = 22$.

5.3 Inverse Functions

The inverse function $f^{-1}$ undoes the effect of $f$. The graph of $f^{-1}$ is the reflection of the graph of $f$ in the line $y = x$.

Condition for Inverse to Exist

A function has an inverse only if it is one-to-one (injective): each output value corresponds to exactly one input value. If a function is many-to-one on its natural domain, you must restrict the domain so it becomes one-to-one before finding the inverse.

To find $f^{-1}(x)$: write $y = f(x)$, swap $x$ and $y$, then rearrange for $y$.

Worked Example 5.2 — Finding an Inverse Function

Find $f^{-1}(x)$ where $f(x) = \dfrac{3x+2}{x-1}$, $x \neq 1$.

Step 1 Write $y = \dfrac{3x+2}{x-1}$.

Step 2 Swap $x$ and $y$: $x = \dfrac{3y+2}{y-1}$.

Step 3 Rearrange for $y$: $x(y-1) = 3y+2 \Rightarrow xy - x = 3y + 2 \Rightarrow y(x-3) = x+2 \Rightarrow y = \dfrac{x+2}{x-3}$.

So $f^{-1}(x) = \dfrac{x+2}{x-3}$, $x \neq 3$.

5.4 The Modulus Function

The modulus function $|x|$ gives the non-negative value (magnitude) of $x$:

$$|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}$$

To solve $|ax + b| = c$ (with $c > 0$): write $ax + b = c$ or $ax + b = -c$ and solve each.

To solve $|ax + b| < c$: write $-c < ax + b < c$.

Fig. 2 — The function $f(x) = 2x + 1$ (blue) and its inverse $f^{-1}(x) = \frac{x-1}{2}$ (red), reflected in the line $y = x$ (dashed green).

Practice 5A

1. $f(x) = x^2$ for $x \geq 0$ and $g(x) = 3x - 2$. Find $fg(x)$ and state its range.

2. Find the inverse of $h(x) = \dfrac{2}{x+3}$ and state the domain of $h^{-1}$.

3. Solve $|3x - 2| = 7$.

Show Solution

1. $fg(x) = f(3x-2) = (3x-2)^2$. Since $g(x) \geq 0$ requires $x \geq \frac{2}{3}$, the domain of $fg$ is $x \geq \frac{2}{3}$, so the range is $fg(x) \geq 0$.

2. Write $y = \frac{2}{x+3}$, swap: $x = \frac{2}{y+3}$, rearrange: $x(y+3)=2 \Rightarrow y = \frac{2}{x}-3 = \frac{2-3x}{x}$. So $h^{-1}(x)=\frac{2-3x}{x}$, domain: $x \neq 0$.

3. $3x-2=7 \Rightarrow x=3$ or $3x-2=-7 \Rightarrow x=-\frac{5}{3}$.

6. Transformations of Graphs Year 2

Starting from the graph of $y = f(x)$, you can produce new graphs by applying transformations. All six standard transformations are required for A-Level.

Standard Transformations Summary

$y = f(x) + a$: translation by $\begin{pmatrix}0\\a\end{pmatrix}$ (vertical shift up by $a$)
$y = f(x + a)$: translation by $\begin{pmatrix}-a\\0\end{pmatrix}$ (horizontal shift left by $a$)
$y = af(x)$: vertical stretch, scale factor $a$ (in the $y$-direction)
$y = f(ax)$: horizontal stretch, scale factor $\frac{1}{a}$ (in the $x$-direction)
$y = f(-x)$: reflection in the $y$-axis
$y = -f(x)$: reflection in the $x$-axis

Exam Tip — Common Mistakes

Students often confuse horizontal transformations. Remember: $f(x+a)$ shifts the graph LEFT by $a$ (when $a > 0$), and $f(ax)$ is a horizontal STRETCH by factor $\frac{1}{a}$, not $a$.

6.1 Combinations of Transformations

When applying multiple transformations, the order matters. For $y = af(bx + c) + d$, a safe approach is to apply them in this order: (1) horizontal shift, (2) horizontal stretch, (3) vertical stretch, (4) vertical shift. However, describing the sequence clearly is what examiners reward.

Worked Example 6.1 — Identifying Transformations

The graph of $y = f(x)$ is transformed to $y = 3f(2x) - 1$. Describe the sequence of transformations.

1 $f(x) \to f(2x)$: horizontal stretch, scale factor $\frac{1}{2}$ (parallel to the $x$-axis).

2 $f(2x) \to 3f(2x)$: vertical stretch, scale factor 3 (parallel to the $y$-axis).

3 $3f(2x) \to 3f(2x)-1$: translation by $\begin{pmatrix}0\\-1\end{pmatrix}$.

Worked Example 6.2 — Applying Transformations to Key Points

The graph of $y = f(x)$ passes through the point $(3, 4)$. Find the corresponding point on $y = f(x - 2) + 5$.

The transformation $f(x-2)+5$ is: shift right 2, then shift up 5.

$(3, 4) \to (3+2, 4+5) = (5, 9)$.

Practice 6A

1. Sketch the graph of $y = |2x - 1|$, labelling the vertex and any intercepts.

2. The curve $y = x^3$ is transformed to $y = (x+2)^3 - 3$. Describe the two transformations and state the new coordinates of the point originally at $(1, 1)$.

Show Solution

1. The vertex occurs where $2x-1=0$, i.e. at $(\frac{1}{2}, 0)$. The $y$-intercept is $|{-1}|=1$, so $(0,1)$. The graph has two straight-line portions with gradient $\pm 2$.

2. First: translation $\begin{pmatrix}-2\\0\end{pmatrix}$ (shift left by 2). Then: translation $\begin{pmatrix}0\\-3\end{pmatrix}$ (shift down by 3). The point $(1,1)$ maps to $(1-2, 1-3) = (-1, -2)$.

7. Practice Problems

The following mixed questions cover the whole chapter and are written in an exam style. Allow yourself about 90 seconds per mark as a guide.

Q1. [3 marks]

Simplify $\dfrac{\sqrt{12} + \sqrt{75}}{\sqrt{3}}$, giving your answer as an integer.

Show Solution

$\sqrt{12} = 2\sqrt{3}$, $\sqrt{75} = 5\sqrt{3}$. So numerator $= 7\sqrt{3}$.

$\dfrac{7\sqrt{3}}{\sqrt{3}} = 7$.

Q2. [4 marks]

Express $3x^2 + 12x - 1$ in the form $a(x+b)^2 + c$ and hence write down the minimum value of the expression.

Show Solution

$3x^2+12x-1 = 3(x^2+4x)-1 = 3\bigl[(x+2)^2-4\bigr]-1 = 3(x+2)^2 - 12 - 1 = 3(x+2)^2-13$.

Minimum value: $-13$ (when $x = -2$).

Q3. [3 marks]

Find the set of values of $x$ for which $x^2 - x - 12 < 0$.

Show Solution

Factorise: $(x-4)(x+3) < 0$. Roots at $x = 4$ and $x = -3$. Parabola opens upward, so the expression is negative between the roots: $-3 < x < 4$.

Q4. [4 marks]

Solve the simultaneous equations $y = 3x - 5$ and $x^2 + y^2 = 25$.

Show Solution

Substitute: $x^2 + (3x-5)^2 = 25 \Rightarrow x^2 + 9x^2 - 30x + 25 = 25 \Rightarrow 10x^2 - 30x = 0 \Rightarrow 10x(x-3)=0$.

$x = 0$: $y = -5$; $x = 3$: $y = 4$. Solutions: $(0, -5)$ and $(3, 4)$.

Q5. [5 marks] Year 2

Express $\dfrac{4x^2 + x - 1}{x(x+1)}$ in the form $A + \dfrac{B}{x} + \dfrac{C}{x+1}$.

Show Solution

Degree of numerator (2) equals degree of denominator (2), so we need to extract a constant term first.

By long division or inspection: $\frac{4x^2+x-1}{x(x+1)} = \frac{4x^2+x-1}{x^2+x}$. Dividing: $4x^2+x-1 = 4(x^2+x) + (x-1-4x) = 4(x^2+x) + (-3x-1)$.

So $\frac{4x^2+x-1}{x(x+1)} = 4 + \frac{-3x-1}{x(x+1)}$. Now decompose: $\frac{-3x-1}{x(x+1)} = \frac{B}{x}+\frac{C}{x+1}$.

$-3x-1 = B(x+1)+Cx$. $x=0$: $-1=B$. $x=-1$: $2=C(-1) \Rightarrow C=-2$. Answer: $4 - \dfrac{1}{x} - \dfrac{2}{x+1}$.

Q6. [4 marks] Year 2

Functions $f$ and $g$ are defined as $f(x) = \sqrt{x-1}$, $x \geq 1$ and $g(x) = 2x^2 + 1$, $x \in \mathbb{R}$. Find $gf(x)$ and state its range.

Show Solution

$gf(x) = g\!\left(\sqrt{x-1}\right) = 2(\sqrt{x-1})^2 + 1 = 2(x-1)+1 = 2x - 1$, domain $x \geq 1$.

When $x = 1$: $gf(1) = 1$. As $x$ increases, $gf$ increases without bound. Range: $gf(x) \geq 1$.

Q7. [5 marks] Year 2

$f(x) = \dfrac{x+3}{2x-1}$, $x \neq \frac{1}{2}$. Find $f^{-1}(x)$ and the value of $x$ for which $f(x) = f^{-1}(x)$.

Show Solution

Let $y = \frac{x+3}{2x-1}$. Swap: $x = \frac{y+3}{2y-1}$. Rearrange: $x(2y-1) = y+3 \Rightarrow 2xy - x = y + 3 \Rightarrow y(2x-1) = x+3 \Rightarrow f^{-1}(x) = \frac{x+3}{2x-1}$.

Since $f = f^{-1}$, the function is self-inverse. Any value in the domain satisfies $f(x) = f^{-1}(x)$, or: $f(x) = x \Rightarrow \frac{x+3}{2x-1} = x \Rightarrow x+3 = 2x^2-x \Rightarrow 2x^2-2x-3=0$. Using the formula: $x = \frac{2\pm\sqrt{4+24}}{4} = \frac{2\pm\sqrt{28}}{4} = \frac{1\pm\sqrt{7}}{2}$.

Q8. [3 marks]

Solve $|4x + 3| \leq 9$, giving your answer as a single inequality or set notation.

Show Solution

$-9 \leq 4x+3 \leq 9 \Rightarrow -12 \leq 4x \leq 6 \Rightarrow -3 \leq x \leq \frac{3}{2}$.

Q9. [4 marks] Year 2

The graph of $y = f(x)$ has a maximum point at $(2, 5)$. Write down the coordinates of the maximum point of: (a) $y = 2f(x-3)$ (b) $y = f(-x) + 1$.

Show Solution

(a) $f(x-3)$: shift right 3, so $(2,5) \to (5,5)$. Then $2f(x-3)$: vertical stretch factor 2, so $(5,5) \to (5, 10)$. Maximum at $(5, 10)$.

(b) $f(-x)$: reflect in $y$-axis, so $(2,5) \to (-2, 5)$. Then $+1$: shift up 1, so $(-2, 5) \to (-2, 6)$. Maximum at $(-2, 6)$.

Q10. [6 marks]

Show that $\dfrac{x^2 + 2x - 3}{x^2 - 1}$ can be written as $\dfrac{x-1+4/(x+1)}{x-1}$. Hence, or otherwise, find all values of $x$ for which $\dfrac{x^2+2x-3}{x^2-1} = 2$.

Show Solution

Note $x^2+2x-3=(x+3)(x-1)$ and $x^2-1=(x+1)(x-1)$. So the fraction equals $\frac{(x+3)(x-1)}{(x+1)(x-1)} = \frac{x+3}{x+1}$ (for $x\neq 1$).

Set $\frac{x+3}{x+1} = 2 \Rightarrow x+3 = 2x+2 \Rightarrow x=1$. But $x=1$ is excluded from the domain (denominator of original expression is zero). So there are no solutions.