Mathematics · Module 2

25 lessons · biggest chapter

Functions — full chapter

Trig, domains, monotonicity, parity, power, exponential and log — the most-tested chapter on the paper. Learn each idea, drill it, then prove it on previous-exam questions.

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After this chapter you can

Evaluate trig functions at special angles from the unit circle.
Fix signs by quadrant and use terminal-side definitions.
Apply reduction, double-angle, half-angle, and sum formulas.
Solve trig equations with two branches and interval filtering.
Read amplitude, period, and phase from $A\sin(Bx+\varphi)$ .
Find domains of roots, fractions, logs, and composite functions.
Complete the square: vertex and range of quadratics.
Write line equations from slope and points; solve intersections.
Test monotonicity and parity (including mixed terms).
Compare powers and exponentials by base and exponent.
Solve exponential and log equations with domain checks.

How to study this chapter

Read the concept in simple words.
Write down the exact formula.
Work one easy example by hand.
Name the trap for this type out loud.
Solve Basic → Intermediate → Advanced.
Finish with mixed + previous-exam questions.

Lessons

Lesson 1 of 25

Special Angles and Unit-Circle Values

Learn the idea

Five angles run the whole exam. Memorize

0°

30°

45°

60°

and

90°

once and reuse them everywhere.

On the unit circle,

\cos \theta

is the x-value and

\sin \theta

is the y-value. Then

\tan \theta = \frac{\sin \theta}{\cos \theta}

Degrees and radians match like this:

30° = \frac{\pi}{6}

45° = \frac{\pi}{4}

60° = \frac{\pi}{3}

90° = \frac{\pi}{2}

Key rules to memorize

$\sin 30° = \frac{1}{2}$ , $\cos 30° = \frac{\sqrt{3}}{2}$ , $\tan 30° = \frac{\sqrt{3}}{3}$
$\sin 45° = \frac{\sqrt{2}}{2}$ , $\cos 45° = \frac{\sqrt{2}}{2}$ , $\tan 45° = 1$
$\sin 60° = \frac{\sqrt{3}}{2}$ , $\cos 60° = \frac{1}{2}$ , $\tan 60° = \sqrt{3}$
$\sin 0° = 0$ , $\cos 0° = 1$ , $\sin 90° = 1$ , $\cos 90° = 0$

Worked example

Compute $(\sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4})) \times \tan(\frac{\pi}{4})$ .

$\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ and $\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ , so the sum is $\sqrt{2}$ .
$\tan(\frac{\pi}{4}) = 1$ , so $\sqrt{2} \times 1 = \sqrt{2}$ .

$\sqrt{2}$

Classic trap

Flipping

\sin 30°

and

\cos 30°

: sine of

30°

\frac{1}{2}

, not

\frac{\sqrt{3}}{2}

Lesson 2 of 25

Quadrant Signs and Terminal-Side Definitions

Learn the idea

A point

P(x, y)

on the terminal side with

r = \sqrt{x^{2} + y^{2}}

defines everything.

The quadrant decides the sign. Check signs before you write any value.

Quadrants I to IV go counter-clockwise starting from positive x-axis.

Key rules to memorize

$\sin \alpha = \frac{y}{r}$ , $\cos \alpha = \frac{x}{r}$ , $\tan \alpha = \frac{y}{x}$
Quadrant I: $\sin > 0$ , $\cos > 0$ , $\tan > 0$
Quadrant II: $\sin > 0$ , $\cos < 0$ , $\tan < 0$
Quadrant III: $\sin < 0$ , $\cos < 0$ , $\tan > 0$
Quadrant IV: $\sin < 0$ , $\cos > 0$ , $\tan < 0$

Worked example

$P(1, y)$ lies on the terminal side of $\alpha$ and $\tan \alpha = 2$ . Find $y$ .

Use $\tan \alpha = \frac{y}{x}$ with $x = 1$ .
So $\frac{y}{1} = 2$ , giving $y = 2$ .

$y = 2$

Classic trap

Using

\tan \alpha = \frac{y}{r}

: tangent divides

y

x

, never by

r

Lesson 3 of 25

Reduction Formulas

Learn the idea

Reduction turns big or negative angles into one acute reference angle.

The function may stay or change, and a sign may appear. Take it in two steps.

Step one: keep or co-change. Step two: fix the sign by quadrant.

Key rules to memorize

$\sin(\pi - \alpha) = \sin \alpha$ , $\cos(\pi - \alpha) = -\cos \alpha$
$\sin(\frac{\pi}{2} - \alpha) = \cos \alpha$ , $\sin(\frac{\pi}{2} + \alpha) = \cos \alpha$
$\tan(\pi + \alpha) = \tan \alpha$ , $\cos(2\pi - \alpha) = \cos \alpha$
$\sin(-\alpha) = -\sin \alpha$ , $\cos(-\alpha) = \cos \alpha$

Worked example

Compute $\sin 120°$ .

Write $120° = 180° - 60°$ , so $\sin 120° = \sin 60°$ .
$\sin 60° = \frac{\sqrt{3}}{2}$ .

$\sin 120° = \frac{\sqrt{3}}{2}$

Classic trap

Writing

\tan(\pi + \alpha) = -\tan \alpha

: the

\pi

shift keeps tangent unchanged and positive.

Lesson 4 of 25

Finding Values From Given Info

Learn the idea

One value plus the quadrant locks all the other values.

Start from

\sin^{2} \alpha + \cos^{2} \alpha = 1

to get the missing square.

Then pick the sign by quadrant, and divide for tangent.

Key rules to memorize

$\sin^{2} \alpha + \cos^{2} \alpha = 1$ links sine and cosine
Square-root gives two signs: quadrant picks the right one
$\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$ finishes the triple
Divide $\sin^{2} + \cos^{2} = 1$ by $\cos^{2}$ : $1 + \tan^{2} = \sec^{2}$ ; by $\sin^{2}$ : $1 + \cot^{2} = \csc^{2}$
Acute angle: all three of $\sin$ , $\cos$ , $\tan$ are positive

Worked example

Given $\sin \alpha = \frac{3}{5}$ with $\alpha$ in quadrant II, find $\cos \alpha$ and $\tan \alpha$ .

$\cos^{2} \alpha = 1 - \frac{9}{25} = \frac{16}{25}$ , and quadrant II gives $\cos \alpha = -\frac{4}{5}$ .
$\tan \alpha = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4}$ .

$\cos \alpha = -\frac{4}{5}$ , $\tan \alpha = -\frac{3}{4}$

Classic trap

Taking the positive root automatically: in quadrant II,

\cos \alpha

must be negative.

Lesson 5 of 25

Double-Angle Formulas

Learn the idea

Double-angle turns

2\alpha

into

\alpha

. CSCA gives one trig value and asks for

\sin 2\alpha

\cos 2\alpha

For

\cos 2\alpha

pick the form that matches what you know. If you know

\sin\alpha

, use

1 - 2\sin^{2}\alpha

Key rules to memorize

$\sin 2\alpha = 2\sin\alpha\cos\alpha$
$\cos 2\alpha = \cos^{2}\alpha - \sin^{2}\alpha$
$\cos 2\alpha = 2\cos^{2}\alpha - 1 = 1 - 2\sin^{2}\alpha$
$\tan 2\alpha = \frac{2\tan\alpha}{1 - \tan^{2}\alpha}$
Find the missing $\sin$ or $\cos$ first with $\sin^{2} + \cos^{2} = 1$

Worked example

Given $\sin\alpha = \frac{1}{4}$ , find $\cos 2\alpha$ .

Pick $\cos 2\alpha = 1 - 2\sin^{2}\alpha$ .
$\cos 2\alpha = 1 - 2 \times \frac{1}{16} = 1 - \frac{1}{8}$ .

$\frac{7}{8}$

Classic trap

Using

\cos 2\alpha = 2\cos^{2}\alpha - 1

when you only know

\sin

. Use

1 - 2\sin^{2}\alpha

instead.

Lesson 6 of 25

Half-Angle Formulas

Learn the idea

Half-angle turns

\alpha

into

\frac{\alpha}{2}

. CSCA gives

\cos\alpha

and asks for

\sin(\frac{\alpha}{2})

\cos(\frac{\alpha}{2})

The square root gives

\pm

. The quadrant of

\frac{\alpha}{2}

picks the sign. Halve the interval first.

Key rules to memorize

$\sin^{2}(\frac{\alpha}{2}) = \frac{1 - \cos\alpha}{2}$
$\cos^{2}(\frac{\alpha}{2}) = \frac{1 + \cos\alpha}{2}$
$\tan(\frac{\alpha}{2}) = \frac{\sin\alpha}{1 + \cos\alpha}$
Halve the whole interval: $\alpha \in (0, \frac{\pi}{2}) \Rightarrow \frac{\alpha}{2}$ in QI; $(\frac{\pi}{2}, \pi) \Rightarrow$ QI; $(\pi, \frac{3\pi}{2}) \Rightarrow$ QII; $(\frac{3\pi}{2}, 2\pi) \Rightarrow$ QII
If $\alpha \in (\frac{\pi}{2}, \pi)$ , then $\frac{\alpha}{2} \in (\frac{\pi}{4}, \frac{\pi}{2})$ , so QI

Worked example

$\cos\alpha = -\frac{1}{2}$ , $\alpha \in (\frac{\pi}{2}, \pi)$ . Find $\sin(\frac{\alpha}{2})$ .

$\frac{\alpha}{2}$ is in QI, so take the positive root.
$\sin^{2} = \frac{1 + \frac{1}{2}}{2} = \frac{3}{4}$ , so $\sin = \frac{\sqrt{3}}{2}$ .

$\frac{\sqrt{3}}{2}$

Classic trap

Keeping

\pm

in the answer. Acute

\frac{\alpha}{2}

means positive only.

Lesson 7 of 25

Sum and Difference Formulas

Learn the idea

Some angles split into special ones.

75^{\circ} = 45^{\circ} + 30^{\circ}

\frac{\pi}{12} = \frac{\pi}{4} - \frac{\pi}{6}

Write the split, apply the formula, then plug in the special values.

Key rules to memorize

$\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$
$\cos(A + B) = \cos A\cos B - \sin A\sin B$
$\cos(A - B) = \cos A\cos B + \sin A\sin B$
$\cos(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$

Worked example

Find $\cos 75^{\circ}$ .

$\cos(45^{\circ} + 30^{\circ}) = \cos45^{\circ}\cos30^{\circ} - \sin45^{\circ}\sin30^{\circ}$ .
$= \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \times \frac{1}{2}$ .

$\frac{\sqrt{6} - \sqrt{2}}{4}$

Classic trap

Wrong sign in

\cos(A - B)

. Minus inside becomes plus outside.

Lesson 8 of 25

Tan From Sin-Cos Ratio

Learn the idea

CSCA hides

\tan

inside

\sin

and

\cos

. One classic is

\frac{\cos - \sin}{\cos + \sin}

Divide top and bottom by

\cos\alpha

. Every

\frac{\sin}{\cos}

becomes

\tan

Key rules to memorize

$\tan\alpha = \frac{\sin\alpha}{\cos\alpha}$
Divide each term by $\cos\alpha$ to make $\tan$ .
$\frac{1 - t}{1 + t} = k$ solves to $t = \frac{1 - k}{1 + k}$
Acute $\cos = \frac{2}{3}$ gives $\sin = \frac{\sqrt{5}}{3}$ , $\tan = \frac{\sqrt{5}}{2}$

Worked example

$\frac{\cos\alpha - \sin\alpha}{\cos\alpha + \sin\alpha} = \frac{1}{3}$ . Find $\tan\alpha$ .

Divide by $\cos\alpha$ : $\frac{1 - t}{1 + t} = \frac{1}{3}$ .
$3 - 3t = 1 + t$ , so $4t = 2$ .

$\frac{1}{2}$

Classic trap

Cross-multiplying before dividing by

\cos

. Divide first, then solve for

t

Lesson 9 of 25

Sin and Cos Graphs

Learn the idea

Sine starts at

0

, rises to

1

\frac{\pi}{2}

, returns to

0

\pi

, dips to

-1

\frac{3\pi}{2}

Cosine starts at

1

, falls to

0

\frac{\pi}{2}

, to

-1

\pi

, and back to

1

2\pi

Both repeat every

2\pi

, with domain

\mathbb{R}

and range

[-1, 1]

Key rules to memorize

$\sin x$ : zeros at $x = k\pi$ , max $1$ at $x = \frac{\pi}{2} + 2k\pi$ , min $-1$ at $x = \frac{3\pi}{2} + 2k\pi$ .
$\cos x$ : zeros at $x = \frac{\pi}{2} + k\pi$ , max $1$ at $x = 2k\pi$ , min $-1$ at $x = \pi + 2k\pi$ .
Parity: $\sin(-x) = -\sin x$ (odd, origin symmetry); $\cos(-x) = \cos x$ (even, y-axis symmetry).
Odd with domain $\mathbb{R}$ passes through origin: $\sin 0 = 0$ ; even keeps $\cos 0 = 1$ .

Worked example

For $y = \sin x$ , state the max, min, and $\sin(-\frac{\pi}{6})$ .

Max is $1$ , min is $-1$ ; range is $[-1, 1]$ .
Sine is odd: $\sin(-\frac{\pi}{6}) = -\sin(\frac{\pi}{6}) = -\frac{1}{2}$ .

Max $1$ , min $-1$ , $\sin(-\frac{\pi}{6}) = -\frac{1}{2}$ .

Classic trap

Sine is odd and cosine is even:

\sin(-x) = -\sin x

but

\cos(-x) = \cos x

; do not give both the same sign.

Lesson 10 of 25

Period, Amplitude and Phase Shifts

Learn the idea

For

y = A\sin(Bx + \varphi)

|A|

sets the height and

|B|

squeezes the wave.

The wave repeats when the inside grows by

2\pi

, so the period shrinks as

|B|

grows.

The shift comes from the inside zero

Bx + \varphi = 0

, not from

\varphi

alone.

Key rules to memorize

Amplitude $= |A|$ ; range of $A\sin(Bx+\varphi)$ is $[-|A|, |A|]$ .
Sin/cos period $T = \frac{2\pi}{|B|}$ ; tan period $T = \frac{\pi}{|B|}$ .
Phase shift $= -\frac{\varphi}{B}$ (negative means right, positive means left).
Combo: $a\sin x + b\cos x$ peaks at $R = \sqrt{a^{2}+b^{2}}$ : max $R$ , min $-R$ ; e.g. $3\sin x + 4\cos x$ peaks at $5$ .
Vertical shift $+k$ moves min/max: $y = A\sin(Bx) + k$ has max $|A| + k$ , min $-|A| + k$ .

Worked example

State the amplitude and min positive period of $y = 3\sin(2x)$ .

Amplitude $= |3| = 3$ ; range is $[-3, 3]$ .
Period $T = \frac{2\pi}{|2|} = \pi$ .

Amplitude $3$ , min positive period $\pi$ .

Classic trap

Tan period is

\pi

, not

2\pi

y = \tan(3x)

has

T = \frac{\pi}{3}

; and shift is

-\frac{\varphi}{B}

, not

\varphi

. Max of

3\sin x + 4\cos x

5

, not

3+4=7

: the two peaks never align.

Lesson 11 of 25

Tan Graph Essentials

Learn the idea

Tan passes through the origin, rises through each interval, then jumps at each break.

It has no top or bottom: it runs from

-\infty

+\infty

on every branch.

Each break is a vertical asymptote where

\cos x = 0

Key rules to memorize

$\tan x = \frac{\sin x}{\cos x}$ ; undefined where $\cos x = 0$ , i.e. $x = \frac{\pi}{2} + k\pi$ .
Asymptotes $x = \frac{\pi}{2} + k\pi$ ; domain excludes these points.
Period $\pi$ : $\tan(x + \pi) = \tan x$ ; zeros at $x = k\pi$ .
Odd function: $\tan(-x) = -\tan x$ ; range is $\mathbb{R}$ (no amplitude).

Worked example

On $(-\pi, \pi)$ , where is $y = \tan x$ undefined?

$\tan x = \frac{\sin x}{\cos x}$ , undefined where $\cos x = 0$ .
Inside $(-\pi, \pi)$ that is $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$ .

Undefined at $x = -\frac{\pi}{2}, \frac{\pi}{2}$ ; both are vertical asymptotes.

Classic trap

Tan range is all reals

\mathbb{R}

, not

[-1, 1]

; and its domain is never all of

\mathbb{R}

Lesson 12 of 25

Solving Basic Trig Equations

Learn the idea

A trig equation gives a family of answers, never one angle. Find the reference angle first, then the branches.

Sine is positive in QI and QII, so

\sin x = c > 0

has two branches per period. Cosine pairs QI with QIV.

Add

2k\pi

for sin/cos (two branches) or

k\pi

for tan (one branch), then keep only what the interval asks for.

Key rules to memorize

Reference angle first: $\sin x = \frac{1}{2} \Rightarrow \alpha = \frac{\pi}{6}$ .
$\sin x = c$ : $x = \alpha + 2k\pi$ or $x = \pi - \alpha + 2k\pi$ .
$\cos x = c$ : $x = \pm\alpha + 2k\pi$ .
$\tan x = c$ : $x = \alpha + k\pi$ .
Always filter to the required interval last.

Worked example

Solve $\sin x = \frac{1}{2}$ for $x \in [0, 2\pi)$ .

Reference angle $\frac{\pi}{6}$ ; sine positive in QI and QII.
QI gives $\frac{\pi}{6}$ , QII gives $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$ ; both lie in range.

$x = \frac{\pi}{6}$ or $\frac{5\pi}{6}$

Classic trap

Reporting only

\frac{\pi}{6}

for

\sin x = \frac{1}{2}

[0, 2\pi)

. The QII branch

\frac{5\pi}{6}

counts too.

Lesson 13 of 25

Domain and Range Basics

Learn the idea

Domain is all

x

you can put in. Range is all

y

you can get out.

CSCA loves roots, fractions, and logs. Each one blocks some

x

Write each rule on its own line. Keep only

x

that pass every rule.

Key rules to memorize

$\sqrt{g(x)}$ needs $g(x) \ge 0$
$\frac{1}{g(x)}$ needs $g(x) \ne 0$
$\log_{a}(g(x))$ needs $g(x) > 0$
Range of $(x-h)^{2}+k$ is $[k,\infty)$

Worked example

Find the domain of $f(x)=\frac{1}{x}+\sqrt{1-x}$ .

Need $x \ne 0$ and $1-x \ge 0$ .
So $x \le 1$ with $x \ne 0$ .

$(-\infty,0)\cup(0,1]$

Classic trap

Fixing the root but forgetting the fraction, like giving

[2,\infty)

for

\sqrt{x-2}+\frac{1}{x-5}

instead of

[2,5)\cup(5,\infty)

Lesson 14 of 25

Quadratics: Vertex, Range and Completing the Square

Learn the idea

Complete the square to read a quadratic:

x^{2} - 4x + 1 = (x-2)^{2} - 3

. The squared part is never negative.

Vertex form

(x-h)^{2} + k

shows the vertex

(h, k)

at once.

a > 0

the parabola opens up: minimum

k

, range

[k, \infty)

. If

a < 0

it opens down: maximum

k

Key rules to memorize

$x^{2} - 4x + 1 = (x-2)^{2} - 3$ : vertex $(2, -3)$ , range $[-3, \infty)$ .
Vertex shortcut: $x = -\frac{b}{2a}$ , then $y = f(-\frac{b}{2a})$ .
$a > 0$ : range $[y_{v}, \infty)$ ; $a < 0$ : range $(-\infty, y_{v}]$ .
Check the vertex by plugging back in.

Worked example

Find the range of $f(x) = x^{2} - 4x + 1$ .

Complete the square: $f(x) = (x-2)^{2} - 3$ .
$(x-2)^{2} \ge 0$ , so $f(x) \ge -3$ , equality at $x = 2$ .

$[-3, \infty)$

Classic trap

Keeping

[-3, \infty)

for

-x^{2} + 4x - 1

. Negative

a

flips it to

(-\infty, -3]

Lesson 15 of 25

Composite and Shifted Domains

Learn the idea

For

f(g(x))

, the inside

g(x)

must stay in the domain of

f

For

\frac{1}{\lg|x-5|}

, two rules fire at once. Check both.

Shifts like

|x-5|

move the bad points. Do not guess them.

Key rules to memorize

Need $g(x)$ in the domain of $f$
Need $|x-5| > 0$ , so $x \ne 5$
Need $\lg|x-5| \ne 0$ , so $x \ne 4,6$
Solve $-1 < x^{2}-1 < 0$ as $0 < x^{2} < 1$

Worked example

If $f$ has domain $(-1,0)$ , find the domain of $f(x^{2}-1)$ .

Need $-1 < x^{2}-1 < 0$ .
Add 1: $0 < x^{2} < 1$ .
So $x \in (-1,0)\cup(0,1)$ .

$(-1,0)\cup(0,1)$

Classic trap

Only removing

x=5

for

\frac{1}{\lg|x-5|}

and missing

x=4,6

where the log is zero.

Lesson 16 of 25

Same Function Test

Learn the idea

Two functions match only if domains match and rules match.

First compare domains. Then simplify the rules.

One bad

x

is enough to say not the same.

Key rules to memorize

Check domain first, then rule
$\sqrt{x^{2}}=|x|$ , not $x$
$(\sqrt{x})^{2}$ has domain $[0,\infty)$
$\frac{x^{4}-1}{x^{2}+1}=x^{2}-1$ on $\mathbb{R}$

Worked example

Are $\frac{x^{4}-1}{x^{2}+1}$ and $x^{2}-1$ the same function?

Denominator $x^{2}+1 \ne 0$ always, so domain is $\mathbb{R}$ .
Factor top: $\frac{(x^{2}-1)(x^{2}+1)}{x^{2}+1}=x^{2}-1$ .

Yes, same function on $\mathbb{R}$

Classic trap

Saying

x

and

\sqrt{x^{2}}

are the same. At

x=-1

, one gives

-1

and one gives

1

Lesson 17 of 25

Inverse Functions

Learn the idea

Inverse swaps

x

and

y

. Write

x=

y

, then solve for

y

Lines and odd cubes keep domain

\mathbb{R}

. Log shifts keep

x

limits.

Domain of

f

becomes range of

f^{-1}

Key rules to memorize

Steps: swap $x$ and $y$ , then solve
$y=x^{3}+3$ gives $y=\sqrt[3]{x-3}$
$y=10x+3$ gives $y=\frac{x-3}{10}$
Strictly monotonic on an interval $\Rightarrow$ invertible there, and the inverse keeps the direction
$y=2+\log_{a}(x-3)$ needs $x > 3$

Worked example

Find the inverse of $y=x^{3}+3$ .

Swap: $x=y^{3}+3$ .
Solve: $y^{3}=x-3$ , so $y=\sqrt[3]{x-3}$ .

$y=\sqrt[3]{x-3}$ , $x \in \mathbb{R}$

Classic trap

Adding

x \ge 3

y=\sqrt[3]{x-3}

. Cube roots accept every real

x

Lesson 18 of 25

Monotonicity and Parity

Learn the idea

Increasing means bigger

x

gives bigger

y

. Decreasing flips it.

Even means

f(-x)=f(x)

. Odd means

f(-x)=-f(x)

Test with

f(-x)

. Never guess from one term.

Key rules to memorize

$a > 1$ : $a^{x}$ rises; $0 < a < 1$ : it falls
$|x|$ falls for $x < 0$ , rises for $x > 0$
Even: $\cos x$ , $x\sin x$ ; odd: $\sin x$ , $-x$
Products: even $\times$ even = even; odd $\times$ odd = even; even $\times$ odd = odd (so $x \cdot \sin x$ is even)
Sums keep like parity only: odd $\pm$ odd = odd, even $\pm$ even = even
Odd and defined at $0$ $\Rightarrow f(0) = 0$ ; predict symmetric values, e.g. odd $f(2) = 7 \Rightarrow f(-2) = -7$
Parity needs a symmetric domain first: defined only for $x \ge 0$ means neither
Odd plus constant is neither, like $x^{3}+1$

Worked example

Check the parity of $f(x)=x\sin x$ .

Compute $f(-x)=(-x)\sin(-x)$ .
Use $\sin(-x)=-\sin x$ : product is $x\sin x$ .

Even, since $f(-x)=f(x)$

Classic trap

Calling

x^{3}+1

odd because of

x^{3}

. The

+1

breaks it, so it is neither.

Lesson 19 of 25

Monotonicity of Lines, Cubics and Reciprocals

Learn the idea

y = mx + b

follows its slope:

m > 0

rises everywhere,

m < 0

falls everywhere.

y = x^{3}

rises on all of

\mathbb{R}

y = x^{2} + 1

does not: it falls left of

0

and rises right of

0

y = -\frac{1}{x}

rises on

(-\infty, 0)

and rises on

(0, \infty)

— it is the mirror of

\frac{1}{x}

, which falls on each branch. Either way

x = 0

splits the domain, so neither is monotone on the whole

\mathbb{R}

Key rules to memorize

$m > 0$ : line rises on $\mathbb{R}$ ; $m < 0$ : falls on $\mathbb{R}$ (e.g. $y = -x + 5$ ).
$x^{3}$ is strictly increasing on $\mathbb{R}$ .
Even powers are never monotone on $\mathbb{R}$ : split at the vertex.
$-\frac{1}{x}$ : increasing on each branch (derivative $\frac{1}{x^{2}} > 0$ ), undefined at $0$ — never claim it for the whole $\mathbb{R}$ .
Sums and compositions: increasing + increasing = increasing; two decreasing compose to increasing; mixed composes to decreasing.

Worked example

Which decreases on $(-\infty, +\infty)$ : $y = -x+5$ , $y = x^{3}$ , $y = x^{2}+1$ , $y = -\frac{1}{x}$ ?

$y = -x+5$ has slope $-1 < 0$ : falls everywhere.
$x^{3}$ rises; $x^{2}+1$ turns at $0$ ; $-\frac{1}{x}$ rises on each branch — none of the three fall.

$y = -x + 5$

Classic trap

Calling

-\frac{1}{x}

decreasing anywhere: it rises on each branch (

\frac{1}{x}

is the one that falls). And never extend a branch-claim to the whole

(-\infty, +\infty)

: the hole at

0

breaks every global claim — name each branch.

Lesson 20 of 25

Lines as Functions: Slope, Equations, Intersections

Learn the idea

Slope measures direction:

m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}

, and also

m = \tan\theta

for inclination

\theta

Point-slope builds the equation:

y - y_{1} = m(x - x_{1})

Two lines meet where their functions agree: solve

f(x) = g(x)

, then check in both. Mirror

(x, y)

across the x-axis to

(x, -y)

Key rules to memorize

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ : e.g. $(-2,3)$ , $(3,1)$ give $m = -\frac{2}{5}$ .
$m = \tan\theta$ : $60^{\circ} \Rightarrow \sqrt{3}$ ; $45^{\circ} \Rightarrow 1$ , so through $(0,2)$ that is $y = x+2$ .
Point-slope: slope $-3$ through $(1,2)$ is $y - 2 = -3(x-1)$ , i.e. $3x + y - 5 = 0$ .
Intersection: substitute one line into the other, solve, verify in both equations.
Linear-in- $n$ bonus: $a_{n} = 2n-1$ gives $1, 3, 5, \ldots$ , constant step $2$ .

Worked example

Line with slope $-3$ through $(1,2)$ : equation?

Point-slope: $y - 2 = -3(x - 1)$ .
Expand: $y = -3x + 5$ , i.e. $3x + y - 5 = 0$ .

$3x + y - 5 = 0$

Classic trap

Solving the system but checking in only one line. A slip survives unless both equations confirm it.

Lesson 21 of 25

Power Functions and Comparisons

Learn the idea

A power function looks like

y = x^{n}

Even

n

gives a U shape. Odd

n

goes through

(0,0)

For big

x

, exponential beats power, and power beats log.

Key rules to memorize

$x^{a} \cdot x^{b} = x^{a+b}$ and $(x^{a})^{b} = x^{ab}$ .
$x^{-n} = \frac{1}{x^{n}}$ for $x \ne 0$ .
$(xy)^{n} = x^{n}y^{n}$ ; $(\frac{x}{y})^{n} = \frac{x^{n}}{y^{n}}$ $(y \ne 0)$ ; $x^{1/2} = \sqrt{x}$ $(x \ge 0)$ .
$x^{m/n} =$ n-th root of $x^{m}$ .
Same exponent $>0$ : bigger base wins.
Same base $>1$ : bigger exponent wins; same base in $(0,1)$ : order flips, e.g. $(\frac{1}{2})^{3} > (\frac{1}{2})^{5}$ .
Negative exponents flip first: $2.1^{-2} > 1.2^{-2}$ is false because $\frac{1}{2.1^{2}} < \frac{1}{1.2^{2}}$ .
Plug points in to find coefficients: $y = ax^{2}$ through $(1,2)$ gives $2 = a$ , so $a = 2$ .
Unify first: radicals to fractional exponents, e.g. $(8x^{6})^{1/3} = 2x^{2}$ .

Worked example

Compare $2.1^{2/3}$ and $1.2^{2/3}$ .

Same positive exponent $\frac{2}{3}$ .
Base $2.1 > 1.2$ , so order stays.

$2.1^{2/3} > 1.2^{2/3}$

Classic trap

Never write

\sqrt{x^{2}} = x

. It is

|x|

Lesson 22 of 25

Exponential Graphs and Monotonicity

Learn the idea

y = a^{x}

needs

a > 0

and

a \ne 1

It is always positive. It passes through

(0,1)

a > 1

it rises. If

0 < a < 1

it falls.

Key rules to memorize

Domain is $R$ . Range is $(0, +\infty)$ .
$a^{x} \cdot a^{y} = a^{x+y}$ ; $\frac{a^{x}}{a^{y}} = a^{x-y}$ ; $(a^{x})^{y} = a^{xy}$ ; $(ab)^{x} = a^{x}b^{x}$ .
$a^{0} = 1$ and $a^{-x} = \frac{1}{a^{x}}$ .
$a > 1$ : increasing. $0 < a < 1$ : decreasing.
$a^{x} > 0$ always, so $a^{x} = -3$ has no root.

Worked example

Compare $2^{4.5}$ and $2^{5}$ .

Base $2 > 1$ , so $2^{x}$ is increasing.
Since $4.5 < 5$ , the left side is smaller.

$2^{4.5} < 2^{5}$

Classic trap

Base

0.75 < 1

flips order. So

0.75^{-0.2} < 0.75^{-0.4}

. From

a > b

alone,

\frac{1}{a} < \frac{1}{b}

fails (try

a=1, b=-2

a^{2} > b^{2}

fails (try

a=1, b=-1

\sin a > \sin b

fails (sine oscillates).

Lesson 23 of 25

Exponential Equations

Learn the idea

First try to make the same base on both sides.

Then set the exponents equal.

If bases differ, take logs on both sides.

Key rules to memorize

$a^{f} = a^{g}$ gives $f = g$ .
Rewrite $16 = 2^{4}$ and $\frac{1}{9} = 3^{-2}$ .
If bases differ: $x = \log_{a} b$ solves $a^{x} = b$ .
$e \approx 2.718$ . $e^{x}$ is its own derivative.

Worked example

Solve $2^{x+1} = 16$ .

Write $16 = 2^{4}$ .
Same base gives $x+1 = 4$ .

$x = 3$

Classic trap

Do not equate exponents when bases differ.

Lesson 24 of 25

Log Laws and Change of Base

Learn the idea

\log_{a} x

asks for the exponent on

a

It needs

x > 0

a > 0

a \ne 1

Condense to one log before solving.

Key rules to memorize

$\log_{a}(xy) = \log_{a} x + \log_{a} y$ .
$\log_{a}(\frac{x}{y}) = \log_{a} x - \log_{a} y$ .
$\log_{a}(x^{n}) = n \cdot \log_{a} x$ .
$\log_{a} x = \frac{\ln x}{\ln a}$ ; flipped form $\log_{a} b = \frac{1}{\log_{b} a}$ .
$\log_{a} a = 1$ , $\log_{a} 1 = 0$ and $\ln e = 1$ .
$a > 1$ : $\log_{a} x$ rises like $a^{x}$ ; $0 < a < 1$ : it falls — same direction rule as exponentials.

Worked example

Compute $\log_{2} 32 + \log_{2}(\frac{1}{4})$ .

Condense: $\log_{2}(32 \cdot \frac{1}{4}) = \log_{2} 8$ .
Since $2^{3} = 8$ , the value is $3$ .

$3$

Classic trap

Never split

\log(x+y)

into

\log x + \log y

. And never call

\log_{a} x

increasing without checking: base

0.5

means decreasing.

Lesson 25 of 25

Log Equations and Graphs

Learn the idea

y = \log_{a} x

passes through

(1,0)

It is the inverse of

y = a^{x}

f(x) = 2 + \log_{a}(x-3)

passes through

(4,2)

Key rules to memorize

$\log_{a} x = y$ means $a^{y} = x$ .
Domain is $(0, +\infty)$ . Range is $R$ .
Condense to one log per side, then equate arguments.
Check every root in the original line.
$e^{\ln x} = x$ for $x > 0$ .

Worked example

Solve $\log_{3}(x+2) = 2$ .

Rewrite: $x+2 = 3^{2} = 9$ , so $x = 7$ .
Check: $7+2 = 9 > 0$ , valid.

$x = 7$

Classic trap

Reject any root with argument

\le 0

. It is extraneous.

Master formula sheet

Trig values

$\sin 30° = \frac{1}{2}$ , $\cos 30° = \frac{\sqrt{3}}{2}$ , $\tan 30° = \frac{\sqrt{3}}{3}$
$\sin 45° = \cos 45° = \frac{\sqrt{2}}{2}$ , $\tan 45° = 1$
$\sin 60° = \frac{\sqrt{3}}{2}$ , $\cos 60° = \frac{1}{2}$ , $\tan 60° = \sqrt{3}$
$\sin^2\alpha + \cos^2\alpha = 1$ ; $\tan\alpha = \frac{\sin\alpha}{\cos\alpha}$

Identities

$\sin 2\alpha = 2\sin\alpha\cos\alpha$
$\cos 2\alpha = 1 - 2\sin^2\alpha = 2\cos^2\alpha - 1$
$\sin^2(\frac{\alpha}{2}) = \frac{1-\cos\alpha}{2}$ , $\cos^2(\frac{\alpha}{2}) = \frac{1+\cos\alpha}{2}$
$\sin(A\pm B) = \sin A\cos B \pm \cos A\sin B$
$\cos(A-B) = \cos A\cos B + \sin A\sin B$
$1 + \tan^2\theta = \sec^2\theta$ ; $1 + \cot^2\theta = \csc^2\theta$ (divide $\sin^2+\cos^2=1$ )
$\sin x = c$ : $x = \alpha+2k\pi$ or $\pi-\alpha+2k\pi$ ; $\tan x = c$ : $x = \alpha+k\pi$

Graphs

Amplitude $|A|$ ; sin/cos period $\frac{2\pi}{|B|}$ ; tan period $\frac{\pi}{|B|}$
$\sin$ odd, $\cos$ even, $\tan$ odd with asymptotes $x = \frac{\pi}{2}+k\pi$
Range $A\sin(Bx)+k$ : $[-|A|+k, |A|+k]$

Domain & parity

$\sqrt{g} \Rightarrow g \ge 0$ ; $\frac{1}{g} \Rightarrow g \ne 0$ ; $\log g \Rightarrow g > 0$
Even: $f(-x) = f(x)$ ; odd: $f(-x) = -f(x)$ ; odd at $0 \Rightarrow f(0) = 0$
Same function needs same domain AND same rule
Inverse: swap $x,y$ , then solve

Quadratics & lines

$(x-h)^{2}+k$ : vertex $(h,k)$ ; $a > 0 \Rightarrow$ min $k$ , $a < 0 \Rightarrow$ max $k$
Vertex at $x = -\frac{b}{2a}$ ; check by plugging back
Slope $m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \tan\theta$ ; line $y-y_{1} = m(x-x_{1})$
Intersection solves $f(x) = g(x)$ ; verify in both lines

Power, exp, log

$x^a x^b = x^{a+b}$ ; $x^{-n} = \frac{1}{x^n}$ ; $\sqrt{x^2} = |x|$
$a > 1$ : $a^x$ rises; $0<a<1$ : falls; always $> 0$
$a > 1$ : $\log_a x$ rises; $0<a<1$ : falls
$\log_a(xy) = \log_a x + \log_a y$ ; $\log_a(x^n) = n\log_a x$
$\log_a x = \frac{\ln x}{\ln a}$ ; reject roots with argument $\le 0$

Traps & recognition

14 classic traps

Flipping $\sin 30°$ with $\cos 30°$ .
Using $\tan = \frac{y}{r}$ instead of $\frac{y}{x}$ .
Taking the positive root without checking the quadrant.
Keeping $\pm$ in a half-angle answer.
Wrong sign in $\cos(A-B)$ .
Giving tan period $2\pi$ or range $[-1,1]$ .
Forgetting excluded points in composite/log domains.
Calling $x^3+1$ odd, or $x$ the same as $\sqrt{x^2}$ .
Restricting cube-root domains, or equating exponents across bases.
Splitting $\log(x+y)$ , or keeping log roots with argument $\le 0$ .
Calling $\log_a x$ increasing without checking a base below 1.
Reporting one trig solution and missing the second branch.
Calling $-\frac{1}{x}$ decreasing: it rises on each branch — and no branch-claim extends to all of $\mathbb{R}$ ( $x = 0$ splits it).
Comparing exponentials without checking whether the base is above or below 1.

Fast problem recognition

Trig value with π or °
Reduce to an acute special angle, fix sign by quadrant.
sin x = value on an interval
Reference angle, two branches, add 2kπ (kπ for tan), filter to the interval.
Vertex / range of a quadratic
Complete the square; sign of a decides min vs max.
Slope / line equation / intersection
m = Δy/Δx = tanθ; point-slope; solve f = g and check both.
sin 2α / cos 2α / α/2
Pick the double/half form matching the known value.
75°, π/12, non-special angle
Split into special angles, apply sum formula.
(cos±sin)/(cos±sin)
Divide by cos → single tan equation.
Period / amplitude / min
Read |A|, compute 2π/|B| (π/|B| for tan); a·sin x + b·cos x peaks at √(a²+b²).
Domain question
List each rule (root/fraction/log), intersect, keep endpoints honest.
Same/inverse function
Domains first, then swap-and-solve.
Odd/even/monotonic
Compute f(−x); test direction on the base.
Compare powers/exponentials
Same exponent → bigger base; same base → check base vs 1 first.
aˣ= / log= equations
Unify bases or condense logs; always domain-check roots.

Practice — 1000 drill problems

100 core drills plus 900 lesson-tagged extra drills, basic to exam-pattern and new-type variants, each with the thinking shown. Master these and the previous-exam set below, and the chapter is yours. Tagged by lesson so you can drill exactly what you missed.

1000 practice problems · 0/0 correct

Practice · Q1trig-values
$\sin 30° =$
Practice · Q2trig-values
$\cos 60° =$
Practice · Q3trig-values
$\tan 45° + \sin 90° =$
Practice · Q4trig-values
$\sin^{2} 45° + \cos^{2} 60° =$
Practice · Q5trig-signs
The sign of $\sin 200°$ is
Practice · Q6trig-signs
The sign of $\cos 300°$ is
Practice · Q7trig-signs
$\tan\alpha < 0$ and $\sin\alpha > 0$ : $\alpha$ is in
Practice · Q8trig-signs
$P(-3, 4)$ , $r = 5$ : $\sin\alpha =$

Previous CSCA exam questions

Real Functions questions from the Dec 2025, Jan 2026 and Apr 2026 papers — solved with the thinking shown. Answers solved by the app; items with source errors are marked and still teach the correct math. Below the core set, 12 related questions from other chapters drill the same Functions skills.

58 exam questions · 0/0 correct

Dec 2025 · Q4domain-rangesource issue
The domain of $y = \frac{1}{x+1}$ is
No correct option in the source paper
Source issue: no option shows the full domain (−∞, −1) ∪ (−1, ∞).
Correct math
True domain is $x \ne -1$ , i.e. $(-\infty,-1)\cup(-1,\infty)$ — not listed.
Full solution — 3 steps
1. Fraction rule: denominator $x+1 \ne 0$ .
2. True domain is $x \ne -1$ , i.e. $(-\infty,-1)\cup(-1,\infty)$ .
3. No listed option shows this — source issue, but the math above is correct.
Dec 2025 · Q7trigonometric-functions
$(\sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4})) \times \tan(\frac{\pi}{4}) =$
Dec 2025 · Q8even-odd-functions
Which is an odd function?
Dec 2025 · Q9power-functions
The inverse function of $y = x^3$ is
Dec 2025 · Q10monotonicity
Which function is strictly increasing?
Dec 2025 · Q15exponential-functions
Which inequality is correct?
Dec 2025 · Q16trigonometric-functions
$P(1, y)$ on terminal side of $\alpha$ , $\tan\alpha = 2$ : then $y =$
Dec 2025 · Q19trigonometric-functions
$\sin\alpha = \frac{3}{5}$ , second quadrant: $\cos\alpha =$

Related — Functions skills tested elsewhere (12)

Inclination of $y = \sqrt{3}x + 10$ : slope is $\tan\theta$ — trig definitions + special values (lessons 1–2).
Intersection of $y = 2x+1$ with $x+y+1=0$ : solving $f(x) = g(x)$ for linear functions (lesson 16).
Q symmetric about the x-axis: symmetry reading that parity work builds on (lesson 16).
Slope through two points: sign of slope = direction of a linear function (lesson 16).
$\sin\alpha$ for a line through $P(1,3)$ : terminal-side trig definitions (lesson 2).
Inclination $45^\circ$ through $(0,2)$ : $\tan45^\circ = 1$ gives $y = x+2$ (lessons 1, 16).
Intersection of two lines: solving $f(x) = g(x)$ (lesson 16).
Slope through $A(-1,1)$ , $B(2,3)$ : sign of slope = direction of a linear function (lesson 16).
Inclination $60^\circ$ : slope $= \tan60^\circ = \sqrt{3}$ (lesson 1).
$a_n = 2n-1$ : linear-in- $n$ form, monotonic like a linear function (lesson 16).
Point-slope equation: writing a linear function from slope + point (lesson 16).
Intersection of two lines: solving $f(x) = g(x)$ (lesson 16).

12 exam questions · 0/0 correct

Dec 2025 · Q14lines
The inclination angle of $y = \sqrt{3}x + 10$ is
Dec 2025 · Q28lines
Intersection of $y = 2x+1$ and $x+y+1 = 0$ is
Jan 2026 · Q6analytic-geometry
$P(1,2)$ ; Q symmetric about the x-axis: Q =
Jan 2026 · Q13lines
Line through $A(-2,3)$ , $B(3,1)$ : slope is
Jan 2026 · Q14lines
Angle between x-axis and line through $P(1,3)$ : $\sin\alpha =$
Jan 2026 · Q17lines
Inclination 45° through $(0,2)$ : equation is
Jan 2026 · Q24lines
Intersection of $3x-y+8 = 0$ , $x+2y-9 = 0$ :
Apr 2026 · Q9lines
Slope through $A(-1,1)$ , $B(2,3)$ :

Exam checklist

0/13 checked

Trigonometry

Functions & families

One-minute memory sheet

Unit circle: $\cos$ = x, $\sin$ = y
QI all +, QII sin, QIII tan, QIV cos
$\sin 2\alpha = 2\sin\alpha\cos\alpha$ ; $\cos 2\alpha = 1-2\sin^2\alpha$
Trig equation: ref angle, two branches, $+2k\pi$ (tan $+k\pi$ )
Period: $\frac{2\pi}{|B|}$ ( $\frac{\pi}{|B|}$ for tan)
$a\sin x + b\cos x$ peaks at $\sqrt{a^{2}+b^{2}}$
Vertex $x = -\frac{b}{2a}$ ; $(x-h)^{2}+k \ge k$
Slope $= \frac{\Delta y}{\Delta x} = \tan\theta$
Domain: root $\ge 0$ , fraction $\ne 0$ , log $> 0$
Even: $f(-x)=f(x)$ ; odd: $f(-x)=-f(x)$ ; odd at 0 $\Rightarrow f(0)=0$
Base > 1 rises; base < 1 falls (exp AND log)
Condense logs, unify bases, check every root

Related guides

30-day study plan → fit this chapter into a week-by-week system
75-second pacing rule → hold 75s/question once the content clicks
Top 10 mistakes → the error patterns that cost the most points

Last reviewed Sep 2026 · GetCSCA editors

Timed practice →Back to syllabus →