Algebra readiness

Recognise the number systems

Classify a number by whether it can be written as a ratio of integers. A decimal approximation alone does not establish that a number is rational.

The revision starts with the integers $\mathbb Z=\{\ldots,-2,-1,0,1,2,3,4,\ldots\}$ and positive integers $\mathbb Z^+=\{1,2,3,4,\ldots\}$. Rational numbers have the form

$$\mathbb Q=\left\{\frac pq:p,q\in\mathbb Z,\ q\ne0\right\}.$$

The source examples are $-\frac12$, $\frac56$, $\frac83$, $2122$, $12.6$, $-99.4$ and $-0.875$. Integers and finite decimals can also be written as such fractions.

The real numbers $\mathbb R$ are all numbers on the number line; $\mathbb R^+=\{x\in\mathbb R:x>0\}$. An irrational number is a real number that is not rational. The printed examples are $\sqrt3\approx1.73205\ldots$, $\pi\approx3.14159\ldots$, $e\approx2.71828\ldots$ and $\ln2\approx0.693147\ldots$.

Complete a square without changing the polynomial

Halve the coefficient of x, square it, then add and subtract that same amount. Expand your result to check that you have preserved the polynomial.

A quadratic polynomial is $ax^2+bx+c$ with constant coefficients and $a\ne0$. Completing the square writes it as $a(x+p)^2+q$. The source identity is

$$x^2+nx=\left(x+\frac n2\right)^2-\left(\frac n2\right)^2.$$

For example, $x^2+4x=(x+\frac42)^2-(\frac42)^2=(x+2)^2-4$. When the leading coefficient is not 1, first factor it from the variable terms, then complete the square inside the bracket; every correction is still multiplied by that coefficient. Example 1 retains both printed calculations. Continue with Tutorial Question 2.

Derive the quadratic formula

The formula comes from completing the square. Follow the rearrangement in two stages: isolate a square, then take both square-root signs.

Starting from $ax^2+bx+c=0$, with $a\ne0$, the printed derivation successively gives

$$\begin{aligned} a\left(x^2+\frac ba x\right)+c&=0,\\ a\left(x^2+\frac ba x\right)&=-c,\\ x^2+\frac ba x&=-\frac ca,\\ \left(x+\frac b{2a}\right)^2-\left(\frac b{2a}\right)^2&=-\frac ca,\\ \left(x+\frac b{2a}\right)^2&=-\frac ca+\frac{b^2}{4a^2}. \end{aligned}$$

Taking both square-root signs and subtracting $b/(2a)$ gives

$$\begin{aligned} x+\frac b{2a}&=\pm\sqrt{\frac{b^2-4ac}{4a^2}},\\ x&=-\frac b{2a}\pm\sqrt{\frac{b^2-4ac}{4a^2}},\\ x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. \end{aligned}$$

For real roots the radicand must be nonnegative. The final ± expression represents the same pair for either sign of a; it does not assert $\sqrt{a^2}=a$ for negative a.

The quantity $D=b^2-4ac$ is the discriminant. Example 2 uses this formula with a parameter. Continue with Tutorial Question 3.

Read roots and strict signs from a graph

Ask two questions: which way does the parabola open, and how many times does it meet the x-axis? The leading coefficient answers the first; the discriminant answers the second.

For real coefficients and $a\ne0$, $D>0$ gives two real distinct roots; $D=0$ gives two equal real roots (one distinct root); $D<0$ gives no real roots. The six source sketches combine each case with a>0 or a<0. An upward parabola crosses, touches or stays above the x-axis; a downward parabola crosses, touches or stays below it.

The exploration lets you compare all six cases one at a time. Its labelled equations are illustrative examples, not coordinates prescribed by the source; the original schematic remains linked. Real roots exist exactly when $D\ge0$, equivalently when the graph has a point on the x-axis, including tangency.

A quadratic is strictly positive for every real x exactly when $a>0$ and $D<0$. It is strictly negative for every real x exactly when $a<0$ and $D<0$. Both strict-sign conditions need a negative discriminant: D=0 touches zero and fails strict positivity or negativity. Examples 3–4 retain the source calculations. Continue with Tutorial Question 4 and Tutorial Question 5.

Use the factor theorem and five identities

Test a proposed linear factor by substituting its root. For x+a, the test input is −a; the sign inside the factor changes.

A polynomial of degree n has form $a_0+a_1x+a_2x^2+a_3x^3+\cdots+a_nx^n$, with $a_n\ne0$. The factor theorem says

$$(x-a)\text{ is a factor of }f(x)\quad\Longleftrightarrow\quad f(a)=0.$$

A value a with $f(a)=0$ is called a root. Example 5 tests x+1 by substituting −1. The source lists all five useful factorisation identities:

$$\begin{aligned} a^2+2ab+b^2&=(a+b)^2,\\ a^2-2ab+b^2&=(a-b)^2,\\ a^2-b^2&=(a+b)(a-b),\\ a^3+b^3&=(a+b)(a^2-ab+b^2),\\ a^3-b^3&=(a-b)(a^2+ab+b^2). \end{aligned}$$

Continue with Tutorial Question 6.

Simplify surds while respecting the radical

Look for a square factor inside the radical before multiplying out. To rationalise a two-term denominator, multiply numerator and denominator by its conjugate.

The source revisits rational numbers as a/b with integer a,b and $b\ne0$, using $1$, $0.5$ and $1/3$. Irrational examples are $\sqrt2$, $-\sqrt5$ and $\pi$. A surd expresses an irrational number using a radical/root symbol. For positive a,b, the three printed properties are

$$\sqrt a\sqrt b=\sqrt{ab},\qquad \frac{\sqrt a}{\sqrt b}=\sqrt{\frac ab},\qquad \sqrt a\sqrt a=a.$$

The source warns against distributing a square root over addition or subtraction: $\sqrt{a+b}$ is not generally $\sqrt a+\sqrt b$, and $\sqrt{a-b}$ is not generally $\sqrt a-\sqrt b$. Where subtraction is used over the reals, its radicand must also be nonnegative. Example 6 retains the printed product and conjugate rationalisation. Continue with Tutorial Question 7.

Keep the index laws and their domains in view

Identify the bases before choosing a law. Multiplying powers with the same base adds exponents; raising a power to another power multiplies them.

For a positive integer n, $a^n$ means n factors of a multiplied together; a is the base and n the index or exponent. The source groups its nine laws as follows.

Same base:

$$a^ma^n=a^{m+n},\qquad\frac{a^m}{a^n}=a^{m-n},\qquad(a^m)^n=a^{mn}.$$

Same index:

$$a^mb^m=(ab)^m,\qquad\frac{a^n}{b^n}=\left(\frac ab\right)^n.$$

Zero, negative and fractional indices:

$$\begin{aligned} a^0&=1,\\ a^{-m}&=\frac1{a^m},\\ a^{1/m}&=\sqrt[m]a\quad(m\in\mathbb Z^+),\\ a^{m/n}&=\sqrt[n]{a^m}=(\sqrt[n]a)^m\quad(n\in\mathbb Z^+). \end{aligned}$$

The source lists these compactly without all domain qualifications. For unrestricted real exponents, use positive bases. Denominators and negative powers require nonzero bases; the zero-power rule above excludes zero. A real even root requires a nonnegative radicand.

For negative bases, rational exponents need a specified real-root convention and cannot be manipulated as unrestricted real exponents. Example 7(b) uses thirds: interpret them via the real cube root, with $x\ne0$. Continue with Tutorial Question 8.

Solve exponential equations with equal bases or a substitution

First try to express both sides using the same base. If several powers of one exponential appear, substitute for that exponential and retain its positive range.

For $a>0$ and $a\ne1$, the exponential is one-to-one: $a^x=a^n$ implies $x=n$. The source graphs show an increasing curve for a>1 and a decreasing curve for 0<a<1. Both pass through (0,1), stay positive and approach the x-axis without reaching zero. Their domain is all real numbers and range is positive real numbers.

Example 8(a) is the same printed equation as Tutorial Question 9. The lecture working writes $9=3^2$ and $27=3^3$:

$$\begin{aligned} (3^2)^{2x+1}-(3^3)^x&=0,\\ 3^{4x+2}-3^{3x}&=0,\\ 3^{4x+2}&=3^{3x},\\ 4x+2&=3x,\\ x&=-2. \end{aligned}$$

The tutorial solution also gives a factorisation argument. Example 8(b) introduces $y=2^x>0$ and rejects a negative quadratic candidate; the new exercise keeps all its printed stages.

Translate between index and logarithmic forms

Read a logarithm as an exponent: “what power of the base gives this positive number?” Keep the base unchanged when switching forms.

For $a>0$, $a\ne1$ and $x>0$,

$$a^y=x\quad\Longleftrightarrow\quad y=\log_ax.$$

The source calls base 10 the common logarithm: $10^y=x$ gives $y=\log_{10}x=\lg x$. Base e gives the natural logarithm: $e^y=x$ gives $y=\log_ex=\ln x$. Example 9 uses the definition directly. Continue with Tutorial Question 10.

The two source logarithm sketches both pass through (1,0), are defined only for positive x and approach the y-axis as a vertical asymptote. For a>1 the logarithm increases; for 0<a<1 it decreases. Its range is all real numbers. The exploration compares both exponential and logarithm curves using bases 2 and 1/2. These are illustrative choices within the two source base ranges; the original four-case sketch remains linked.

Apply all five logarithm laws

Check that each original logarithm has a positive argument before combining terms. Solve the transformed equation, then check candidates in the original domain.

For positive m,n and valid base a ($a>0$, $a\ne1$), the printed laws are

$$\begin{aligned} \log_a(mn)&=\log_am+\log_an,\\ \log_a\left(\frac mn\right)&=\log_am-\log_an,\\ \log_a(m^n)&=n\log_am,\\ a^{\log_am}&=m,\\ \log_ab&=\frac{\log_cb}{\log_ca}. \end{aligned}$$

For the power rule, a positive m also permits any real exponent n. Change of base additionally needs $b>0$, $c>0$ and $c\ne1$; its denominator is nonzero because $a\ne1$. The source remarks give $\log_a1=0$ and $\log_aa=1$. Example 10 changes base 9 to base 3; Example 11 takes natural logarithms and applies the power rule to differing exponential bases. Continue with Tutorial Question 11.

Distinguish proper and improper rational functions

Compare the numerator and denominator degrees before decomposing. Equal degrees require division too.

The source defines a polynomial as $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_2x^2+a_1x+a_0$, with real coefficients, $a_n\ne0$ and nonnegative integer degree n. A rational function is P(x)/Q(x), where P,Q are polynomials and the denominator is nonzero at each allowed input.

It is proper when $\deg P<\deg Q$, as in the source example

$$\frac{4x^2+5x-1}{3x^3+2x-4}.$$

It is improper when $\deg P\ge\deg Q$. Both printed examples are retained:

$$\frac{3x^2+2x-1}{5x-4},\qquad \frac{3x^4+2x-1}{(x+1)(x+3)(x-1)^2}.$$

In the latter denominator the degrees add to 4, so equal degrees still mean improper. Divide first to write an improper expression as a polynomial plus a proper rational remainder; the polynomial part must not be lost.

Choose the right partial-fraction form

Build the complete template before calculating coefficients. A missing repeated-factor term cannot be repaired by choosing different coefficient values.

The source’s four-step procedure is: divide first if improper; factorise the denominator as far as possible; assign the correct terms for its factor types; then solve their unknown coefficients.

Case 1: distinct linear factors.

$$\frac{px+q}{(ax+b)(cx+d)}=\frac A{ax+b}+\frac B{cx+d}.$$

The displayed example form is

$$\frac{x+1}{(x-2)(x+3)}=\frac A{x-2}+\frac B{x+3}.$$

Case 2: a repeated linear factor. Include every power up to its multiplicity:

$$\frac{px^2+qx+r}{(ax+b)(cx+d)^2} =\frac A{ax+b}+\frac B{cx+d}+\frac C{(cx+d)^2}.$$

Both printed example forms are

$$\begin{aligned} \frac{x+1}{(x-2)(x+3)^2}&=\frac A{x-2}+\frac B{x+3}+\frac C{(x+3)^2},\\ \frac{x+1}{(x-2)x^2}&=\frac A{x-2}+\frac Bx+\frac C{x^2}. \end{aligned}$$

Case 3: a non-repeated irreducible quadratic. Use a linear numerator over the quadratic:

$$\frac{px^2+qx+r}{(ax+b)(x^2+c^2)} =\frac A{ax+b}+\frac{Bx+C}{x^2+c^2}.$$

Here c must be nonzero for $x^2+c^2$ to be irreducible over the reals. The two printed example forms are

$$\begin{aligned} \frac{x+1}{(x-2)(x^2+3)}&=\frac A{x-2}+\frac{Bx+C}{x^2+3},\\ \frac{x+1}{(x-2)(x^2+3x+4)}&=\frac A{x-2}+\frac{Bx+C}{x^2+3x+4}. \end{aligned}$$

The latter quadratic has discriminant −7; it also has no real linear factors. The three template cases are labelled “in MF27” in the source; that is a retained source reference. Linear-factor templates assume nonzero leading coefficients and distinct roots between different factors, while inputs must avoid every denominator zero. No coefficient values are printed for these illustrative templates; Examples 12–15 calculate coefficients in full.

Solve coefficients without requiring cover-up knowledge

Clear denominators first, then use convenient inputs or compare like powers. Check the final decomposition by recombining its fractions.

The source permits substitution, comparing coefficients, or the optional direct cover-up rule if previously taught. Substitution and coefficient comparison suffice. First clear denominators to obtain a polynomial identity. It is legitimate to substitute a denominator zero into that identity; the original rational function is still undefined there.

Example 12 isolates each of three distinct linear-factor coefficients. Example 13 includes both powers of a repeated factor and uses a third convenient input for the remaining coefficient. Example 14 finds its simple linear-factor coefficient by cover-up, then compares x² and constant coefficients for the quadratic numerator. Its source callout reassures students that substitution is an equally valid method.

Direct cover-up isolates simple linear-factor coefficients and the highest power of a repeated linear factor; it does not directly give the other repeated-factor coefficients or all coefficients over a quadratic. These restrictions preserve the source remark without making cover-up compulsory.

Example 15 retains the complete division with a zero x² placeholder, then decomposes a positive remainder numerator before restoring the minus sign around the whole remainder. Continue with the existing, distinct Tutorial Question 12, which asks for three different decompositions.

Use radians for circular measure

Choose whether the question asks for a length or an area. Both formulas use the same fraction of a full turn, but they scale differently with radius.

The source gives $180^\circ=\pi$ radians. An arc PQ of length s subtends angle θ at the centre O of a circle of radius r. For θ in radians,

$$\frac{\text{arc length}}{\text{circumference}}=\frac\theta{2\pi},\qquad \frac{s}{2\pi r}=\frac\theta{2\pi}\ \Longrightarrow\ s=r\theta.$$

For sector POQ of area A, the corresponding area fraction gives

$$\frac{\text{sector area}}{\text{circle area}}=\frac\theta{2\pi},\qquad \frac A{\pi r^2}=\frac\theta{2\pi}\ \Longrightarrow\ A=\frac12r^2\theta.$$

The two source circle diagrams distinguish the labelled arc length s from the sector area. Both retain O,P,Q,r and the central angle θ. Convert degrees to radians before using either formula.

Match triangle sides to the chosen angle

Mark the angle first, then name its opposite and adjacent sides. Those names depend on the chosen angle; the hypotenuse stays opposite the right angle.

In the source right triangle ABC, B is the right angle and the acute angle θ is at C. Thus AB is opposite, BC adjacent and AC the hypotenuse:

$$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{AB}{AC},\quad \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{BC}{AC},\quad \tan\theta=\frac{\text{opposite}}{\text{adjacent}}=\frac{AB}{BC}=\frac{\sin\theta}{\cos\theta}.$$

The reciprocal ratios are $\operatorname{cosec}\theta=1/\sin\theta$, $\sec\theta=1/\cos\theta$ and $\cot\theta=1/\tan\theta=\cos\theta/\sin\theta$, wherever the corresponding denominators are nonzero. The cotangent ratio cosθ/sinθ also defines cotangent when tanθ itself is undefined but sinθ is nonzero; the reciprocal-of-tangent equality needs both expressions defined.

Use these five exact special-angle cases. Tangent is undefined at π/2 because cosine is zero:

0

sinθ 0

cosθ 1

tanθ 0

$\pi/6$ (30°)

sinθ $1/2$

cosθ $\sqrt3/2$

tanθ $\sqrt3/3$

$\pi/4$ (45°)

sinθ $\sqrt2/2$

cosθ $\sqrt2/2$

tanθ 1

$\pi/3$ (60°)

sinθ $\sqrt3/2$

cosθ $1/2$

tanθ $\sqrt3$

$\pi/2$ (90°)

sinθ 1

cosθ 0

tanθ Undefined

Complementary-angle properties are

$$\sin(\pi/2-\theta)=\cos\theta,\quad \cos(\pi/2-\theta)=\sin\theta,\quad \tan(\pi/2-\theta)=\cot\theta,$$

with the tangent/cotangent statement restricted to defined values.

Use sine and cosine rules in a general triangle

Pair each side with its opposite angle. Use the cosine rule for two sides and their included angle, or three sides. With the sine rule, check whether a second angle gives another valid triangle.

In triangle ABC let lower-case a,b,c label the opposite sides BC, CA, AB respectively. The source sine rule is

$$\frac{\sin A}a=\frac{\sin B}b=\frac{\sin C}c.$$

All three printed cosine-rule forms are

$$\begin{aligned} c^2&=a^2+b^2-2ab\cos C,\\ b^2&=a^2+c^2-2ac\cos B,\\ a^2&=b^2+c^2-2bc\cos A. \end{aligned}$$

The general triangle diagram labels all six angles/sides; it does not require a right angle. Example 16 retains the modified exam attribution and uses the angle at B between sides 1 and 3.

Read the three trigonometric function graphs

Choose one function at a time. Read its range and one full period, then distinguish a zero from an excluded input: tangent has gaps at its poles.

The source plots each function over $[-2\pi,2\pi]$; their domains extend beyond that displayed window.

sinx

Real domain: All real x

Range: $[-1,1]$

Period: $2\pi$

Symmetry and continuity: Odd: $\sin(-x)=-\sin x$; continuous

cosx

Real domain: All real x

Range: $[-1,1]$

Period: $2\pi$

Symmetry and continuity: Even: $\cos(-x)=\cos x$; continuous

tanx

Real domain: $x\ne\pi/2+k\pi$, integer k

Range: All real values

Period: $\pi$

Symmetry and continuity: Odd: $\tan(-x)=-\tan x$; continuous on each interval between its poles

The original tangent range row says “for all real x”, but the same table explicitly excludes $x=\pm\pi/2,\pm3\pi/2,\pm5\pi/2,\ldots$. The domain above makes that qualification consistent. The tangent graph redraw preserves separate branches and asymptotes rather than joining across the poles.

Choose the principal inverse-trigonometric value

Write the principal angle interval before evaluating. An inverse function gives one selected angle; solving a trigonometric equation may require more angles in the requested interval.

Inverse trigonometric functions recover one selected angle from a ratio. They are not reciprocals of the trigonometric functions. Although infinitely many angles can have the same ratio, the calculator gives the principal value, as the source remark emphasises.

$\sin^{-1}x$

Allowed x: $[-1,1]$

Principal angle interval: $[-\pi/2,\pi/2]$

Printed example and reason: $\sin^{-1}(1/2)=\pi/6$, since $\sin(\pi/6)=1/2$

$\cos^{-1}x$

Allowed x: $[-1,1]$

Principal angle interval: $[0,\pi]$

Printed example and reason: $\cos^{-1}(1/\sqrt2)=\pi/4$, since $\cos(\pi/4)=1/\sqrt2$

$\tan^{-1}x$

Allowed x: All real x

Principal angle interval: $(-\pi/2,\pi/2)$

Printed example and reason: $\tan^{-1}\sqrt3=\pi/3$, since $\tan(\pi/3)=\sqrt3$

The three source inverse graphs retain their endpoints and monotonic directions. Inverse tangent approaches horizontal asymptotes at ±π/2 and never reaches them.

Examples 17–18 apply these principal ranges before evaluating an outer function or choosing a triangle’s height. Continue with Tutorial Question 13.

Keep the identity denominators and square-root signs

Choose an identity that contains the expression you need. Before dividing, exclude zeros of the divisor; before taking a square root, use the angle interval to choose the sign.

The basic identity is $\sin^2\theta+\cos^2\theta=1$. Dividing by cos²θ, where cosθ≠0, gives $\tan^2\theta+1=\sec^2\theta$. Dividing by sin²θ, where sinθ≠0, gives $1+\cot^2\theta=\operatorname{cosec}^2\theta$. Example 19 retains the acute degree interval that lets $\sqrt{\sec^2\theta}=\sec\theta$; generally the square root is $|\sec\theta|$.

Retain the addition and double-angle formulae

Choose the formula that reduces the number of different trigonometric functions. To reduce an even power, use a cosine double-angle form and repeat if needed.

The source labels both formula groups “in MF27”. The paired signs in cosine and tangent switch as indicated by ∓:

$$\begin{aligned} \sin(A\pm B)&=\sin A\cos B\pm\cos A\sin B,\\ \cos(A\pm B)&=\cos A\cos B\mp\sin A\sin B,\\ \tan(A\pm B)&=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}. \end{aligned}$$

The double-angle formulae follow by taking equal angles:

$$\begin{aligned} \sin2A&=2\sin A\cos A,\\ \cos2A&=\cos^2A-\sin^2A=1-2\sin^2A=2\cos^2A-1,\\ \tan2A&=\frac{2\tan A}{1-\tan^2A}. \end{aligned}$$

Tangent expressions require all terms and denominators involved to be defined. Example 20 retains the full triple-angle proof, correcting the source’s “sine terms only” callout to cosine terms. Example 21 retains both double-angle rearrangements that reduce cos⁴θ. Continue with the distinct Tutorial Question 14, which instead reduces sin⁴(θ/2).

Derive all four R-formulae

Expand the proposed shifted function and match its sine and cosine coefficients. Find the positive amplitude first, then the phase in the correct angle unit.

For positive a,b, write $a\cos x+b\sin x=R\cos(x-\alpha)$ with $R>0$ and acute α. Expanding gives

$$\begin{aligned} a\cos x+b\sin x&=R(\cos x\cos\alpha+\sin x\sin\alpha)\\&=R\cos\alpha\cos x+R\sin\alpha\sin x. \end{aligned}$$

Comparing coefficients gives (1) $R\cos\alpha=a$ and (2) $R\sin\alpha=b$. Divide (2) by (1) to obtain $\tan\alpha=b/a$, hence $\alpha=\tan^{-1}(b/a)$. Adding their squares gives

$$R^2\cos^2\alpha+R^2\sin^2\alpha=a^2+b^2 \ \Longrightarrow\ R^2=a^2+b^2 \ \Longrightarrow\ R=\sqrt{a^2+b^2}.$$

All four printed variants use this same R and α:

$$\begin{aligned} a\cos x+b\sin x&=R\cos(x-\alpha),\\ a\sin x+b\cos x&=R\sin(x+\alpha),\\ a\sin x-b\cos x&=R\sin(x-\alpha),\\ a\cos x-b\sin x&=R\cos(x+\alpha). \end{aligned}$$

Examples 22–23 preserve both extrema and the full shifted-bound chain. Their printed 56.3° and 22.6° phases are approximations; retain exact arctangents for identities, and use the same unit for the input angle and its phase. Continue with the distinct Tutorial Question 15.

Differentiate sums, products, quotients and compositions

For real constant a, the source lists

$$\frac d{dx}(a)=0,\qquad \frac d{dx}[af(x)]=a\frac d{dx}f(x),\qquad \frac d{dx}[f(x)\pm g(x)]=\frac d{dx}f(x)\pm\frac d{dx}g(x).$$

Example 24 rewrites the reciprocal and radical as powers. For differentiable u,v the product rule is

$$\frac{dy}{dx}=\frac d{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}.$$

Example 25 retains the explicitly requested product-rule method. For a nonzero denominator v, the quotient rule is

$$\frac{dy}{dx}=\frac d{dx}\left(\frac uv\right) =\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}.$$

Example 26 keeps the numerator subtraction order and the full denominator square. For y a function of u and u a function of x, the chain rule is

$$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}.$$

Example 27 applies it to a square root, distinguishing the real function’s endpoint from the open interval where its derivative is finite.

Retain the elementary derivative table

The source’s five basic/general pairs are retained below. Trigonometric derivatives use radians. The logarithm row requires x>0 in the basic form and f(x)>0 in the general form; tangent rows exclude their poles, and each inner function must be differentiable at the input.

Example 28 retains its exponential, product-with-sine and logarithm asks. Continue with Tutorial Question 16 and Tutorial Question 17.

Integrate as reverse differentiation

If $F'(x)=f(x)$, then $\int f(x)\,dx=F(x)+C$, where C is an arbitrary constant. Antiderivatives are considered on an interval where the expression is defined; additive constants may differ between disconnected intervals. The source linearity rules are

$$\begin{aligned} \int[f(x)\pm g(x)]\,dx&=\int f(x)\,dx\pm\int g(x)\,dx,\\ \int kf(x)\,dx&=k\int f(x)\,dx\quad(k\text{ constant}). \end{aligned}$$

All basic and linear-inner-function rows are retained, with a≠0 in the right column:

Power expressions use their real domains, and trigonometric identities use radians. The n=−1 case is excluded from this printed power formula; it is not obtained by dividing by zero. Example 29 retains all four asks, including its final worked part on the following source page. Continue with Tutorial Question 18.

Attempt the complete readiness tutorial

The closing tutorial is preparation for the Readiness Assessment. Its instructions are to attempt all questions and present working clearly.

The original sequence is: Questions 1–5 under Quadratic Functions and Equations (including the opening exponential rearrangement); 6 Factor Theorem; 7–8 Surds and Indices; 9–11 Exponential and Logarithmic Functions; 12 Partial Fractions; 13–15 Trigonometry; 16–18 Differentiation and Integration.

Work through all eighteen questions in the chapter Discussion set, where their printed numbers and exam references are retained.

The source’s final page is a short answer key, not full worked solutions. It omits Questions 4 and 14 entirely, and Question 15 gives only the phase/amplitude expression, omitting its requested range proof.

Full working comes from the separate tutorial-solution copy. Those solutions also explain domain restrictions and retain exact phases before rounding. The short key’s 53.1° and 1.09 values are rounded, not replacements for exact results.

For Question 16 the radical covers the whole fraction 3/(2x+1), and the key numerator is −√3; extraction alone drops those radical bars.

Both printed exam references are shown with their questions. The standalone solution copy’s JC2-header/JC1-title discrepancy remains scoped to that copy; this lecture copy prints JC1 consistently.