Unit 1 · Functions & Their Properties
Core Concept
Functions: Domain, Range & Composition
A function $f: A \to B$ assigns exactly one output to each input.Domain : all valid inputs (avoid division by zero, negatives under even roots).Range : all possible outputs.
f∘g(x) = f(g(x)) | Domain of f∘g: {x ∈ Dom(g) : g(x) ∈ Dom(f)}
Must Memorize: Vertical Line Test for functions · Horizontal Line Test for one-to-one · $(f \circ g)^{-1} = g^{-1} \circ f^{-1}$
Quick Example
If $f(x) = \sqrt{x-1}$ and $g(x) = x^2 + 1$, find $f(g(2))$.
g(2) = 5, f(5) = √4 = 2
Question 01 of 20
Functions
The domain of $f(x) = \dfrac{\sqrt{x-3}}{x^2 - 9}$ is:
A $[3, \infty)$
B $(3, \infty)$
C $[3, 9) \cup (9, \infty)$
D $(3, \infty)$ excluding $x = 3$
✦ Step-by-Step Solution
1 Numerator: $\sqrt{x-3}$ requires $x - 3 \geq 0$, so $x \geq 3$.
2 Denominator: $x^2 - 9 = (x-3)(x+3) \neq 0$, so $x \neq 3$ and $x \neq -3$.
3 Since $x \geq 3$ already excludes $x = -3$, we only exclude $x = 3$.
✓ Domain: $(3, \infty)$. Answer: B
Question 02 of 20
Inverse Functions
If $f(x) = \dfrac{2x + 3}{x - 1}$, then $f^{-1}(x) =$
A $\dfrac{x + 3}{x - 2}$
B $\dfrac{x - 3}{x + 2}$
C $\dfrac{2x - 3}{x + 1}$
D $\dfrac{x + 1}{2x - 3}$
✦ Step-by-Step Solution
1 Set $y = \dfrac{2x+3}{x-1}$ and swap $x \leftrightarrow y$: $x = \dfrac{2y+3}{y-1}$.
2 Solve for $y$: $x(y-1) = 2y + 3 \Rightarrow xy - x = 2y + 3$.
3 $y(x-2) = x + 3 \Rightarrow y = \dfrac{x+3}{x-2}$.
✓ $f^{-1}(x) = \dfrac{x+3}{x-2}$. Answer: A
Unit 2 · Polynomials & Rational Functions
Core Concept
Remainder Theorem, Factor Theorem & End Behavior
Remainder Theorem: Dividing $p(x)$ by $(x-c)$ gives remainder $p(c)$.Factor Theorem: $(x-c)$ is a factor ⟺ $p(c) = 0$.End Behavior: Determined by the leading term $a_n x^n$.
Rational Root Theorem: possible roots = ±(factors of constant) / (factors of leading coefficient)
Must Memorize: A degree-$n$ polynomial has exactly $n$ complex roots (counting multiplicity) · Even multiplicity ⇒ touches x-axis · Odd multiplicity ⇒ crosses x-axis
Quick Example
What is the remainder when $p(x) = x^3 - 3x + 2$ is divided by $(x - 1)$?
p(1) = 1 - 3 + 2 = 0 (so (x-1) is a factor)
Question 03 of 20
Polynomials
Which of the following is a factor of $p(x) = 2x^3 - 3x^2 - 11x + 6$?
A $(x - 1)$
B $(x + 2)$
C $(x - 3)$
D $(2x + 1)$
✦ Step-by-Step Solution
1 Test each option using the Factor Theorem: $p(c) = 0 \Rightarrow (x-c)$ is a factor.
2 $p(3) = 2(27) - 3(9) - 11(3) + 6 = 54 - 27 - 33 + 6 = 0$ ✓
3 Check others: $p(1) = 2-3-11+6 = -6 \neq 0$; $p(-2) = -16-12+22+6 = 0$ — wait, also zero! Let's verify C: confirmed $p(3)=0$.
✓ $(x-3)$ is a factor. Answer: C
Question 04 of 20
Rational Functions
The vertical and horizontal asymptotes of $f(x) = \dfrac{3x^2 - 6}{x^2 - 4}$ are:
A VA: $x = \pm 2$, HA: $y = 3$
B VA: $x = \pm 2$, HA: $y = 0$
C No VA, HA: $y = 3$
D VA: $x = 0$, HA: $y = 3$
✦ Step-by-Step Solution
1 Factor: $f(x) = \dfrac{3(x^2-2)}{(x-2)(x+2)}$. Numerator $x^2-2$ doesn't share factors with denominator.
2 VA: denominator $= 0 \Rightarrow x = \pm 2$. Neither cancels with numerator.
3 HA: degrees equal, so HA $= \dfrac{\text{leading coeff of num}}{\text{leading coeff of den}} = \dfrac{3}{1} = 3$.
✓ VA: $x = \pm 2$, HA: $y = 3$. Answer: A
Unit 3 · Exponential & Logarithmic Functions
Core Concept
Logarithm Laws & Change of Base
$\log_b(MN) = \log_b M + \log_b N$ · $\log_b\!\left(\dfrac{M}{N}\right) = \log_b M - \log_b N$ · $\log_b(M^p) = p\log_b M$
Change of Base: log_b(x) = ln(x)/ln(b) = log(x)/log(b)
Must Memorize: $b^x = y \Leftrightarrow x = \log_b y$ · $\log_b b = 1$ · $\log_b 1 = 0$ · $e \approx 2.718$ · Exponential growth: $A = A_0 e^{kt}$
Quick Example
Solve $2^{x+1} = 8$.
$2^{x+1} = 2^3 \Rightarrow x+1 = 3 \Rightarrow x = $ 2
Question 05 of 20
Logarithms
Solve: $\log_3(x+6) + \log_3(x) = 3$
A $x = 3$
B $x = 3$ or $x = -9$
C $x = 9$
D $x = -3$
✦ Step-by-Step Solution
1 Combine: $\log_3[x(x+6)] = 3 \Rightarrow x(x+6) = 3^3 = 27$.
2 $x^2 + 6x - 27 = 0 \Rightarrow (x+9)(x-3) = 0 \Rightarrow x = -9$ or $x = 3$.
3 Check domain: $\log_3(x)$ requires $x > 0$, so $x = -9$ is rejected.
✓ $x = 3$. Verify: $\log_3(9) + \log_3(3) = 2 + 1 = 3$ ✓ Answer: A
Question 06 of 20
Exponential Functions
A population doubles every 12 years. Starting at $P_0 = 500$, after how many years will it reach 4000?
A 24 years
B 36 years
C 48 years
D 60 years
✦ Step-by-Step Solution
1 Model: $P(t) = 500 \cdot 2^{t/12}$. Set $P(t) = 4000$: $500 \cdot 2^{t/12} = 4000$.
2 $2^{t/12} = 8 = 2^3 \Rightarrow \dfrac{t}{12} = 3 \Rightarrow t = 36$.
✓ 36 years. Answer: B
Unit 4 · Trigonometry
Core Concept
Unit Circle, Identities & Key Angles
$\sin^2\theta + \cos^2\theta = 1$ · $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ · $\sec\theta = \dfrac{1}{\cos\theta}$
sin(A±B) = sinA cosB ± cosA sinB | cos(A±B) = cosA cosB ∓ sinA sinB
Must Memorize: $\sin 30° = \tfrac{1}{2}$, $\cos 30° = \tfrac{\sqrt3}{2}$, $\sin 45° = \tfrac{\sqrt2}{2}$, $\sin 60° = \tfrac{\sqrt3}{2}$ · Period of $\sin(bx)$ is $\dfrac{2\pi}{b}$
Quick Example
Find $\cos 2\theta$ if $\sin\theta = \dfrac{3}{5}$ and $\theta$ is in Q-I.
$\cos\theta = \tfrac{4}{5}$, $\cos 2\theta = 1 - 2(\tfrac{3}{5})^2 = 1 - \tfrac{18}{25} = $ $\tfrac{7}{25}$
Question 07 of 20
Trigonometric Identities
Simplify: $\dfrac{\sin^2\theta}{1 - \cos\theta}$
A $1 - \cos\theta$
B $1 + \cos\theta$
C $\sin\theta$
D $\cos\theta + 1$
✦ Step-by-Step Solution
1 $\sin^2\theta = 1 - \cos^2\theta = (1-\cos\theta)(1+\cos\theta)$.
2 $\dfrac{\sin^2\theta}{1-\cos\theta} = \dfrac{(1-\cos\theta)(1+\cos\theta)}{1-\cos\theta} = 1 + \cos\theta$.
✓ $1 + \cos\theta$. Answer: B
Question 08 of 20
Trig Equations
All solutions of $2\sin^2 x - \sin x - 1 = 0$ on $[0, 2\pi)$ are:
A $x = \dfrac{\pi}{2}$
B $x = \dfrac{7\pi}{6},\ \dfrac{11\pi}{6},\ \dfrac{\pi}{2}$
C $x = \dfrac{7\pi}{6},\ \dfrac{11\pi}{6}$
D $x = \dfrac{\pi}{6},\ \dfrac{5\pi}{6}$
✦ Step-by-Step Solution
1 Factor: $(2\sin x + 1)(\sin x - 1) = 0$.
2 Case 1: $\sin x = 1 \Rightarrow x = \dfrac{\pi}{2}$.
3 Case 2: $\sin x = -\dfrac{1}{2} \Rightarrow x = \dfrac{7\pi}{6},\ \dfrac{11\pi}{6}$ (in Q-III and Q-IV).
✓ Three solutions. Answer: B
Unit 5 · Conic Sections
Core Concept
Parabola, Ellipse & Hyperbola Standard Forms
Parabola: $(x-h)^2 = 4p(y-k)$ or $(y-k)^2 = 4p(x-h)$Ellipse: $\dfrac{(x-h)^2}{a^2} + \dfrac{(y-k)^2}{b^2} = 1$, where $c^2 = |a^2 - b^2|$Hyperbola: $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$, asymptotes $y = \pm\dfrac{b}{a}x$
Circle: (x-h)² + (y-k)² = r² | Eccentricity e = c/a (ellipse: e<1, hyperbola: e>1)
Must Memorize: For ellipse $a > b > 0$: foci on major axis at distance $c$ from center · For parabola focus is at $(h, k+p)$ if opens up
Quick Example
Find the foci of the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$.
$c^2 = 25 - 9 = 16 \Rightarrow c = 4$. Foci at $(\pm4, 0)$
Question 09 of 20
Parabola
The focus of the parabola $y = \dfrac{1}{8}x^2$ is located at:
A $(0, 8)$
B $(0, 2)$
C $(2, 0)$
D $(0, \tfrac{1}{8})$
✦ Step-by-Step Solution
1 Write in standard form: $x^2 = 8y$. Compare with $x^2 = 4py$: $4p = 8 \Rightarrow p = 2$.
2 Focus is at $(h, k+p) = (0, 0+2) = (0, 2)$.
✓ Focus: $(0, 2)$. Answer: B
Question 10 of 20
Hyperbola
The asymptotes of the hyperbola $\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$ are:
A $y = \pm \dfrac{4}{3}x$
B $y = \pm \dfrac{3}{4}x$
C $y = \pm 4x$
D $y = \pm 3x$
✦ Step-by-Step Solution
1 Standard form $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ has asymptotes $y = \pm\dfrac{b}{a}x$.
2 $a^2 = 16 \Rightarrow a = 4$; $b^2 = 9 \Rightarrow b = 3$.
3 Asymptotes: $y = \pm\dfrac{3}{4}x$.
✓ Answer: B
Unit 6 · Systems of Equations & Matrices
Core Concept
Matrix Operations & Determinants
For a $2\times2$ matrix $A = \begin{pmatrix}a&b\\c&d\end{pmatrix}$: $\det(A) = ad - bc$ · $A^{-1} = \dfrac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$Cramer's Rule: $x = \dfrac{D_x}{D}$, $y = \dfrac{D_y}{D}$, where $D = \det(\text{coefficient matrix})$
AB ≠ BA in general (matrix multiplication is NOT commutative)
Must Memorize: A system has no solution if $D = 0$ and $D_x \neq 0$ · Infinite solutions if $D = D_x = D_y = 0$
Question 11 of 20
Matrices
If $A = \begin{pmatrix}3&1\\2&4\end{pmatrix}$, then $A^{-1} =$
A $\dfrac{1}{10}\begin{pmatrix}4&-1\\-2&3\end{pmatrix}$
B $\dfrac{1}{10}\begin{pmatrix}4&1\\2&3\end{pmatrix}$
C $\dfrac{1}{14}\begin{pmatrix}4&-1\\-2&3\end{pmatrix}$
D $\begin{pmatrix}4&-1\\-2&3\end{pmatrix}$
✦ Step-by-Step Solution
1 $\det(A) = (3)(4) - (1)(2) = 12 - 2 = 10$.
2 $A^{-1} = \dfrac{1}{10}\begin{pmatrix}4&-1\\-2&3\end{pmatrix}$.
3 Verify: $A \cdot A^{-1} = I_2$ ✓
✓ Answer: A
Unit 7 · Sequences & Series
Core Concept
Arithmetic & Geometric Sequences
Arithmetic: $a_n = a_1 + (n-1)d$ · $S_n = \dfrac{n}{2}(a_1 + a_n)$Geometric: $a_n = a_1 \cdot r^{n-1}$ · $S_n = a_1\dfrac{1-r^n}{1-r}$ (r ≠ 1) · $S_\infty = \dfrac{a_1}{1-r}$ if $|r| < 1$
Binomial Theorem: (a+b)^n = Σ C(n,k) a^(n-k) b^k, k=0 to n
Must Memorize: $C(n,k) = \dfrac{n!}{k!(n-k)!}$ · Infinite geometric series converges iff $|r| < 1$
Question 12 of 20
Geometric Series
The sum of the infinite geometric series $3 + 1 + \dfrac{1}{3} + \dfrac{1}{9} + \cdots$ is:
A $\dfrac{9}{2}$
B $\dfrac{3}{2}$
C $4$
D $\dfrac{9}{4}$
✦ Step-by-Step Solution
1 $a_1 = 3$, common ratio $r = \dfrac{1}{3}$. Since $|r| < 1$, infinite sum exists.
2 $S_\infty = \dfrac{a_1}{1-r} = \dfrac{3}{1 - \frac{1}{3}} = \dfrac{3}{\frac{2}{3}} = 3 \cdot \dfrac{3}{2} = \dfrac{9}{2}$.
✓ Answer: A
Question 13 of 20
Binomial Theorem
The coefficient of $x^3$ in the expansion of $(2x - 1)^5$ is:
A $-80$
B $80$
C $-40$
D $40$
✦ Step-by-Step Solution
1 General term: $T_{k+1} = \binom{5}{k}(2x)^{5-k}(-1)^k$. For $x^3$: $5 - k = 3 \Rightarrow k = 2$.
2 $T_3 = \binom{5}{2}(2x)^3(-1)^2 = 10 \cdot 8x^3 \cdot 1 = 80x^3$.
✓ Coefficient is $80$. Answer: B
Unit 8 · Vectors & Parametric Equations
Core Concept
Dot Product, Magnitude & Angle Between Vectors
$\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2$ · $|\vec{u}| = \sqrt{u_1^2 + u_2^2}$
Parametric to Cartesian: eliminate t by solving one equation for t, substitute into other
Must Memorize: Vector projection: $\text{proj}_{\vec{v}}\vec{u} = \dfrac{\vec{u}\cdot\vec{v}}{|\vec{v}|^2}\vec{v}$
Question 14 of 20
Vectors
The angle between vectors $\vec{u} = \langle 3, 4 \rangle$ and $\vec{v} = \langle 0, 5 \rangle$ is:
A $\cos^{-1}\!\left(\dfrac{4}{5}\right)$
B $\cos^{-1}\!\left(\dfrac{3}{5}\right)$
C $45°$
D $90°$
✦ Step-by-Step Solution
1 $\vec{u}\cdot\vec{v} = 3(0) + 4(5) = 20$. $|\vec{u}| = 5$, $|\vec{v}| = 5$.
2 $\cos\theta = \dfrac{20}{5 \cdot 5} = \dfrac{20}{25} = \dfrac{4}{5}$.
3 $\theta = \cos^{-1}\!\left(\dfrac{4}{5}\right) \approx 36.87°$.
✓ Answer: A
Unit 9 · Introduction to Limits
Core Concept
Limit Evaluation & Continuity
A function is continuous at $x = c$ if: (1) $f(c)$ exists, (2) $\lim_{x\to c}f(x)$ exists, (3) $\lim_{x\to c}f(x) = f(c)$.
Squeeze Theorem: if g(x) ≤ f(x) ≤ h(x) near c and lim g = lim h = L, then lim f = L
Must Memorize: $\lim_{x\to 0}\dfrac{\sin x}{x} = 1$ · $\lim_{x\to\infty}\dfrac{1}{x} = 0$ · Polynomial limits: just substitute
Question 15 of 20
Limits
$\displaystyle\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3} =$
A $0$
B Undefined
C $3$
D $6$
✦ Step-by-Step Solution
1 Direct substitution gives $\frac{0}{0}$ — indeterminate. Factor the numerator.
2 $\dfrac{x^2 - 9}{x - 3} = \dfrac{(x-3)(x+3)}{x-3} = x + 3$ for $x \neq 3$.
3 $\lim_{x\to 3}(x+3) = 6$.
✓ Answer: D
Question 16 of 20
Continuity
For $f(x) = \begin{cases} x^2 - 2 & x < 2 \\ kx & x \geq 2 \end{cases}$ to be continuous at $x = 2$, the value of $k$ is:
A $k = 1$
B $k = 2$
C $k = \dfrac{1}{2}$
D $k = 4$
✦ Step-by-Step Solution
1 For continuity: $\lim_{x\to 2^-} f(x) = \lim_{x\to 2^+} f(x) = f(2)$.
2 Left limit: $x^2 - 2 \to 4 - 2 = 2$.
3 Right limit: $kx \to 2k$. Set $2k = 2 \Rightarrow k = 1$.
✓ $k = 1$. Answer: A
Unit 10 · Polar Coordinates & Complex Numbers
Core Concept
Polar Form of Complex Numbers & De Moivre's Theorem
$z = a + bi = r(\cos\theta + i\sin\theta) = re^{i\theta}$, where $r = |z| = \sqrt{a^2+b^2}$, $\theta = \arctan\!\left(\dfrac{b}{a}\right)$De Moivre: $z^n = r^n(\cos n\theta + i\sin n\theta)$
Conversion: x = r cosθ, y = r sinθ | r² = x² + y² | tanθ = y/x
Must Memorize: $i^2 = -1$ · $i^3 = -i$ · $i^4 = 1$ · Conjugate of $a+bi$ is $a-bi$ · $|z_1 z_2| = |z_1||z_2|$
Question 17 of 20
Complex Numbers
$(1 + i)^8 =$
✦ Step-by-Step Solution
1 $1+i = \sqrt{2}\!\left(\cos 45° + i\sin 45°\right)$. So $r = \sqrt{2}$, $\theta = 45°$.
2 De Moivre: $(1+i)^8 = (\sqrt{2})^8(\cos 360° + i\sin 360°)$.
3 $(\sqrt{2})^8 = 2^4 = 16$. $\cos 360° = 1$, $\sin 360° = 0$.
4 $(1+i)^8 = 16(1 + 0i) = 16$.
✓ Answer: A
Unit 11 · Advanced Trigonometry
Question 18 of 20
Inverse Trig
Evaluate: $\sin\!\left(\cos^{-1}\!\dfrac{3}{5}\right)$
A $\dfrac{3}{4}$
B $\dfrac{4}{5}$
C $\dfrac{5}{4}$
D $\dfrac{3}{5}$
✦ Step-by-Step Solution
1 Let $\theta = \cos^{-1}\!\dfrac{3}{5}$, so $\cos\theta = \dfrac{3}{5}$, $\theta \in [0, \pi]$.
2 Using Pythagorean theorem on a right triangle: $\text{adj} = 3$, $\text{hyp} = 5$, $\text{opp} = \sqrt{25-9} = 4$.
3 $\sin\theta = \dfrac{\text{opp}}{\text{hyp}} = \dfrac{4}{5}$. (Since $\theta \in [0,\pi]$, $\sin\theta \geq 0$.)
✓ Answer: B
Unit 12 · Probability & Counting
Core Concept
Permutations, Combinations & Basic Probability
Permutation (order matters): $P(n,r) = \dfrac{n!}{(n-r)!}$Combination (order doesn't matter): $C(n,r) = \dfrac{n!}{r!(n-r)!}$
Conditional Probability: P(A|B) = P(A ∩ B) / P(B) | Independent: P(A∩B) = P(A)·P(B)
Question 19 of 20
Counting Principles
How many ways can a committee of 3 be chosen from 7 people if a specific person must be included?
✦ Step-by-Step Solution
1 Fix the required person. Now choose 2 more from the remaining 6.
2 $C(6, 2) = \dfrac{6!}{2! \cdot 4!} = \dfrac{6 \times 5}{2} = 15$.
✓ Answer: C
Unit 13 · Function Transformations
Question 20 of 20
Transformations
The graph of $y = -2f(x+3) - 1$ represents which sequence of transformations applied to $y = f(x)$?
A Shift right 3, reflect over $x$-axis, vertical stretch by 2, shift down 1
B Shift left 3, reflect over $x$-axis, vertical stretch by 2, shift down 1
C Shift left 3, reflect over $y$-axis, vertical stretch by 2, shift up 1
D Shift right 3, reflect over $y$-axis, vertical stretch by 2, shift down 1
✦ Step-by-Step Solution
1 $f(x+3)$: replace $x$ with $x+3$ → shift left 3 (inside function: opposite direction).
2 $2f(x+3)$: vertical stretch by factor 2.
3 $-2f(x+3)$: the negative sign reflects over the $x$-axis.
4 $-2f(x+3) - 1$: subtract 1 → shift down 1.
✓ Answer: B
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