■ Self-Study Edition

PreCalculus
Mastery Quiz

20 carefully crafted problems across all major units. Choose your answer — instant feedback.

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Unit 1
Functions & Their Graphs
01
■ Functions

Which of the following represents a one-to-one function?
Hint: Think about the Horizontal Line Test

KEY: HLT = one-to-one  |  VLT = is a function
✨ Explanation

f(x) = x³ is one-to-one because every horizontal line crosses the graph exactly once.

f(x) = x² → fails: f(2) = f(−2) = 4 ✗ f(x) = x³ → passes: if x³ = y³, then x = y ✓ f(x) = |x| → fails: |−3| = |3| = 3 ✗ f(x) = sin(x) → fails: sin(0) = sin(π) = 0 ✗

A function is one-to-one (injective) when distinct inputs always produce distinct outputs.

02
■ Functions

If f(x) = 2x − 3, what is f⁻¹(x)?

INVERSE: swap x & y, then solve for y
✨ Explanation

To find the inverse, replace f(x) with y, swap x and y, then solve for y:

y = 2x − 3 swap: x = 2y − 3 solve: x + 3 = 2y → y = (x + 3) / 2

Trick: Option A is worded confusingly — the correct arithmetic is (x + 3) ÷ 2, which matches C. Option D uses subtraction (wrong sign).

Unit 2
Polynomials & Rational Functions
03
■ Polynomials

The polynomial p(x) = x³ − 6x² + 11x − 6 has roots at x = 1 and x = 2. What is the third root?

Σ VIETA'S: sum of roots = −(b/a) = 6 for cubic ax³+bx²+...
✨ Explanation

By Vieta's formulas for x³ − 6x² + 11x − 6 = 0:

Sum of roots = 6 (from coefficient of x²: −(−6)/1) r₁ + r₂ + r₃ = 6 1 + 2 + r₃ = 6 r₃ = 3 ✓

Verify: (x−1)(x−2)(x−3) = x³ − 6x² + 11x − 6 ✓

04
■ Rational Functions

What is the horizontal asymptote of f(x) = (3x² + 1) / (x² − 4)?

HA RULE: same degree → ratio of leading coefficients
✨ Explanation

Compare degrees of numerator and denominator:

deg(top) = 2, deg(bottom) = 2 → same degree HA = leading coeff(top) / leading coeff(bottom) = 3 / 1 = 3

Rules to memorize: top < bottom → y = 0; top = bottom → y = ratio; top > bottom → no HA (oblique instead)

Unit 3
Exponential & Logarithmic Functions
05
■ Logarithms

Solve for x:   log₂(x + 3) + log₂(x − 1) = 5

+ logA + logB = log(A·B)  |  always CHECK domain!
✨ Explanation
log₂[(x+3)(x−1)] = 5 (x+3)(x−1) = 2⁵ = 32 x² + 2x − 3 = 32 x² + 2x − 35 = 0 (x + 7)(x − 5) = 0 x = −7 or x = 5

Domain check: x + 3 > 0 and x − 1 > 0 requires x > 1. So x = −7 is rejected. Only x = 5 is valid. Option D is a trap!

06
■ Exponentials

If 4ˣ = 8, what is the exact value of x?

SAME BASE TRICK: write both as powers of 2, then equate exponents
✨ Explanation
4ˣ = 8 (2²)ˣ = 2³ 2^(2x) = 2³ 2x = 3 x = 3/2 ✓

Note: Option C is also mathematically valid (change of base), but B is the exact simplified answer. Options A and D flip the fraction — common mistake!

Unit 4
Trigonometry
07
■ Trigonometry

In which quadrant is an angle θ where sin θ < 0 and cos θ > 0?

ALL STUDENTS TAKE CALCULUS: Q1=All Q2=Sin Q3=Tan Q4=Cos
✨ Explanation
cos θ > 0 → x is positive → Q1 or Q4 sin θ < 0 → y is negative → Q3 or Q4 Intersection: Quadrant IV ✓

ASTC mnemonic: All (Q1), Sin (Q2), Tan (Q3), Cos (Q4) — tells you which trig functions are positive in each quadrant.

08
■ Trigonometry

What is the exact value of sin(5π/6)?

REFERENCE ANGLE: find angle from x-axis, keep quadrant sign
✨ Explanation
5π/6 is in Quadrant II (between π/2 and π) Reference angle = π − 5π/6 = π/6 sin(π/6) = 1/2 In Q2, sin is POSITIVE → sin(5π/6) = +1/2 ✓

Common trap: students confuse 5π/6 with 7π/6 (Q3, where sin is negative).

09
■ Trig Identities

Simplify:   (1 − sin²θ) / cos θ

PYTHAGOREAN: sin²θ + cos²θ = 1  →  1−sin²θ = cos²θ
✨ Explanation
(1 − sin²θ) / cos θ = cos²θ / cos θ [since 1−sin²θ = cos²θ] = cos θ ✓

This is one of the most tested identity simplifications. Always look for the Pythagorean identity first!

10
■ Trig Equations

Solve on [0, 2π):   2cos²x − 1 = 0

DOUBLE ANGLE: 2cos²x−1 = cos(2x)  |  recognize the pattern!
✨ Explanation
2cos²x − 1 = 0 cos²x = 1/2 cos x = ±1/√2 = ±√2/2 cos x = √2/2 → x = π/4, 7π/4 cos x = −√2/2 → x = 3π/4, 5π/4 All 4 solutions: π/4, 3π/4, 5π/4, 7π/4 ✓

Common mistake: forgetting the ± and only finding 2 solutions. The equation has four solutions on [0, 2π).

Unit 5
Analytic Geometry & Conics
11
■ Conics

The equation x² / 9 + y² / 4 = 1 describes which conic section? What are the lengths of its semi-axes?

ELLIPSE: x²/a² + y²/b² = 1  |  a² bigger → major axis is horizontal
✨ Explanation
x²/9 + y²/4 = 1 → x²/3² + y²/2² = 1 → Ellipse with a = 3 (semi-major, horizontal) and b = 2 (semi-minor, vertical) ✓

Common trap: C reads 9 and 4 directly — those are a² and b², not a and b. Always take the square root!

12
■ Conics

Which equation represents a parabola opening to the right with vertex at the origin?

PARABOLA: y² = 4px opens right/left  |  x² = 4py opens up/down
✨ Explanation
y² = 4px: opens RIGHT if p > 0, LEFT if p < 0 x² = 4py: opens UP if p > 0, DOWN if p < 0 y² = 4x → p = 1 > 0 → opens right ✓

Memory shortcut: the variable that's squared tells you the axis of symmetry. y² → axis is x-axis → opens left/right.

Unit 6
Sequences, Series & Binomial Theorem
13
■ Sequences

A geometric sequence has first term a₁ = 2 and common ratio r = 3. What is the sum of the first 4 terms?

Σ GEO SUM: S_n = a₁(1 − rⁿ) / (1 − r)  when r ≠ 1
✨ Explanation
S₄ = a₁(1 − r⁴) / (1 − r) = 2(1 − 3⁴) / (1 − 3) = 2(1 − 81) / (−2) = 2(−80) / (−2) = 80 ✓ Check by listing: 2 + 6 + 18 + 54 = 80 ✓
14
■ Binomial Theorem

What is the coefficient of x³ in the expansion of (x + 2)⁵?

BINOMIAL: C(n,k) · aⁿ⁻ᵏ · bᵏ  |  use Pascal's row or nCk formula
✨ Explanation
Term with x³: k = 2 (since x^(5−k) = x³ → k = 2) T = C(5,2) · x³ · 2² = 10 · x³ · 4 = 40x³ Wait — check: x^(5−k)= x³ → 5−k = 3 → k = 2 C(5,2) = 10, 2² = 4, so coefficient = 10 × 4 = 40

Hmm wait — let me recheck: C = 80? No, 10 × 4 = 40. Option D = 40 is correct. Common mistake: students forget to raise the constant (2) to the power k.

Unit 7
Introduction to Limits
15
■ Limits

Evaluate:   lim(x→2)   (x² − 4) / (x − 2)

0/0 FORM: factor & cancel — NEVER just plug in and get 0/0
✨ Explanation
lim(x→2) (x² − 4)/(x − 2) = lim(x→2) (x+2)(x−2) / (x−2) = lim(x→2) (x + 2) = 2 + 2 = 4 ✓

Direct substitution gives 0/0 (indeterminate form). Always factor and cancel before substituting. The function is defined everywhere except x = 2, but the limit still exists!

16
■ Limits

What is   lim(x→∞)   (5x³ − 2x) / (3x³ + x²)?

INFINITY LIMIT: divide top & bottom by highest power of x
✨ Explanation
Divide numerator and denominator by x³: = (5 − 2/x²) / (3 + 1/x) As x → ∞: 2/x² → 0, 1/x → 0 = (5 − 0) / (3 + 0) = 5/3 ✓

Same degree on top and bottom → limit = ratio of leading coefficients = 5/3. This is exactly the horizontal asymptote rule!

Unit 8
Systems of Equations & Matrices
17
■ Matrices

What is the determinant of the matrix:   [[3, 1], [2, 4]]?

det 2×2: ad − bc  |  butterfly: main diagonal minus anti-diagonal
✨ Explanation
|3 1| |2 4| det = (3)(4) − (1)(2) = 12 − 2 = 10 ✓

Common trap: A is 3×4 + 1×2 = 14 (addition instead of subtraction). Always subtract cross product!

18
■ Systems

How many solutions does this system have?
2x + 4y = 6
x + 2y = 3

DEPENDENT: same line → infinitely many solutions (not "no solution"!)
✨ Explanation
2x + 4y = 6 → divide by 2 → x + 2y = 3 x + 2y = 3 (same equation!) The two equations are IDENTICAL → the same line → Infinitely many solutions ✓

If lines are identical → dependent system (infinite solutions). If lines are parallel but distinct → inconsistent system (no solution). These are commonly confused!

Unit 9
Complex Numbers
19
■ Complex Numbers

Simplify:   i⁴⁷

i CYCLE: i¹=i, i²=−1, i³=−i, i⁴=1  |  divide exponent by 4, use remainder
✨ Explanation
47 ÷ 4 = 11 remainder 3 i⁴⁷ = (i⁴)¹¹ · i³ = 1¹¹ · i³ = i³ = −i ✓ i¹ = i i² = −1 i³ = −i ← remainder 3 i⁴ = 1 (cycle resets)

The key: always divide the exponent by 4 and use the remainder. Remainder 0 → 1, Remainder 1 → i, Remainder 2 → −1, Remainder 3 → −i.

20
■ Complex Numbers

What is the modulus (absolute value) of the complex number z = 3 − 4i?

MODULUS: |z| = √(a² + b²)  |  think: Pythagorean theorem on complex plane
✨ Explanation
z = 3 − 4i → a = 3, b = −4 |z| = √(3² + (−4)²) = √(9 + 16) = √25 = 5 ✓ (3, 4, 5) is a Pythagorean triple — memorize it!

This is the classic 3-4-5 right triangle. The complex number 3 − 4i lies at point (3, −4) in the complex plane, exactly 5 units from the origin.

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