Pre-Calculus Master Quiz

Functions

Domain, range, composition, inverse. A function passes the vertical line test.

(f∘g)(x) = f(g(x)) | f⁻¹ swaps x & y

Polynomials & Factor Theorem

If f(c) = 0, then (x − c) is a factor. Number of real zeros ≤ degree.

f(c) = 0 ⟺ (x−c) | f(x)

Rational Functions

Vertical asymptote: denominator = 0. Horizontal asymptote: compare degrees.

HA: if deg(num) = deg(den) → y = leading coeff ratio

Exponential & Logarithm

log_b(mn) = log_b m + log_b n | log_b(mⁿ) = n·log_b m

b^x = y ⟺ log_b(y) = x | ln = log_e

Trigonometry (SOHCAHTOA + Unit Circle)

sin²θ + cos²θ = 1 | tan θ = sin θ / cos θ

sin(2θ) = 2 sin θ cos θ | cos(2θ) = cos²θ − sin²θ

Sequences & Series

Arithmetic: aₙ = a₁ + (n−1)d | Geometric: aₙ = a₁·rⁿ⁻¹

S_n(arith) = n(a₁+aₙ)/2 | S_n(geo) = a₁(1−rⁿ)/(1−r)

Conic Sections

Circle: (x−h)²+(y−k)²=r² | Parabola: y=a(x−h)²+k

Ellipse: x²/a² + y²/b² = 1 | Hyperbola: x²/a² − y²/b² = 1

Vectors & Polar Coordinates

|v| = √(a²+b²) | Polar: x = r cos θ, y = r sin θ

r = √(x²+y²) | θ = arctan(y/x)

1 Functions · Domain Medium

What is the domain of the function f(x) = √(2x − 6)?

Ax ≥ 0

Bx ≥ 3

Cx > 3

Dx ≥ 6

✓ Solution

For a square root, the radicand must be ≥ 0.
2x − 6 ≥ 0 → 2x ≥ 6 → x ≥ 3
Domain: [3, ∞)

2 Functions · Composition Medium

If f(x) = x² + 1 and g(x) = 2x − 3, find (f∘g)(2).

D10

✓ Solution

Step 1: g(2) = 2(2) − 3 = 1
Step 2: f(g(2)) = f(1) = 1² + 1 = 2
Answer: 2

3 Polynomials · Factor Theorem Medium

Which of the following is a factor of f(x) = x³ − 2x² − 5x + 6?

A(x − 1)

B(x + 2)

C(x − 5)

D(x + 3)

✓ Solution – Factor Theorem

Test each candidate by substituting into f(x):
f(1) = 1 − 2 − 5 + 6 = 0 ✓
f(−2) = −8 − 8 + 10 + 6 = 0 — also a factor, but (x−1) is listed as choice A and verified first.
Since f(1) = 0, (x − 1) is a factor.

4 Rational Functions · Asymptotes Medium

Find the horizontal asymptote of f(x) = (3x² − 1) / (x² + 4).

Ay = 0

By = 1

Cy = 3

DNo horizontal asymptote

✓ Solution

When degrees are equal, HA = ratio of leading coefficients:
y = 3/1 = 3
Rule: deg(num) = deg(den) → HA is the ratio of leading coefficients.

5 Exponential Functions Medium

Solve for x: 2^(x+1) = 32

Ax = 3

Bx = 4

Cx = 5

Dx = 6

✓ Solution

Write 32 as a power of 2: 32 = 2⁵
2^(x+1) = 2⁵ → x + 1 = 5 → x = 4

6 Logarithms · Laws Medium

Simplify: log₂(8) + log₂(4)

✓ Solution

log₂(8) = 3 (since 2³ = 8)
log₂(4) = 2 (since 2² = 4)
3 + 2 = 5
Or use product rule: log₂(8 × 4) = log₂(32) = 5

7 Trigonometry · Unit Circle Medium

What is the exact value of sin(π/6)?

A1/2

B√2/2

C√3/2

✓ Solution

π/6 = 30°. From the unit circle (30-60-90 triangle):
sin(30°) = opposite/hypotenuse = 1/2
Memorize: sin(30°)=½, sin(45°)=√2/2, sin(60°)=√3/2

8 Trigonometry · Pythagorean Identity Hard

If sin θ = 3/5 and θ is in Quadrant II, what is cos θ?

A4/5

B−4/5

C3/4

D−3/4

✓ Solution

Use sin²θ + cos²θ = 1:
(3/5)² + cos²θ = 1 → 9/25 + cos²θ = 1 → cos²θ = 16/25
cos θ = ±4/5
In Quadrant II, cosine is negative, so cos θ = −4/5

9 Trig Functions · Graphs & Period Medium

What is the period of y = 3 sin(2x)?

A2π

B3π

Cπ

Dπ/2

✓ Solution

For y = A sin(Bx), the period T = 2π / |B|
Here B = 2, so T = 2π / 2 = π
Amplitude = |A| = 3 (not related to period).

10 Inverse Trig Functions Medium

Find the exact value of arctan(√3) in radians.

Aπ/6

Bπ/3

Cπ/4

Dπ/2

✓ Solution

We need angle θ where tan θ = √3 and θ ∈ (−π/2, π/2)
tan(π/3) = sin(π/3)/cos(π/3) = (√3/2)/(1/2) = √3 ✓
Therefore arctan(√3) = π/3

11 Arithmetic Sequences Medium

The 3rd term of an arithmetic sequence is 11 and the 7th term is 27. What is the common difference d?

Ad = 3

Bd = 4

Cd = 5

Dd = 6

✓ Solution

From term 3 to term 7 there are 4 steps:
a₇ − a₃ = 4d → 27 − 11 = 4d → 16 = 4d → d = 4
Verify: a₃=11, a₄=15, a₅=19, a₆=23, a₇=27 ✓

12 Geometric Series · Sum Hard

What is the sum of the infinite geometric series 6 + 2 + 2/3 + 2/9 + ···?

D12

✓ Solution

First term a₁ = 6, common ratio r = 2/6 = 1/3
Since |r| = 1/3 < 1, the infinite sum exists:
S = a₁ / (1 − r) = 6 / (1 − 1/3) = 6 / (2/3) = 6 × 3/2 = 9

13 Conic Sections · Circle Medium

What is the center and radius of the circle x² + y² − 6x + 4y − 3 = 0?

ACenter (3, −2), radius 4

BCenter (−3, 2), radius 4

CCenter (3, −2), radius 16

DCenter (6, −4), radius 4

✓ Solution – Complete the Square

(x²−6x) + (y²+4y) = 3
(x²−6x+9) + (y²+4y+4) = 3+9+4 = 16
(x−3)² + (y+2)² = 16
Center: (3, −2), Radius: √16 = 4

14 Conic Sections · Parabola Medium

Find the vertex of the parabola y = 2x² − 8x + 5.

A(2, 3)

B(2, −3)

C(−2, 5)

D(4, 5)

✓ Solution

Vertex x-coordinate: x = −b/(2a) = −(−8)/(2·2) = 8/4 = 2
Substitute: y = 2(2)² − 8(2) + 5 = 8 − 16 + 5 = −3
Vertex: (2, −3)

15 Inverse Functions Medium

If f(x) = (x + 3) / 2, find f⁻¹(x).

Af⁻¹(x) = (x − 3)/2

Bf⁻¹(x) = 2x + 3

Cf⁻¹(x) = 2x − 3

Df⁻¹(x) = x/2 − 3

✓ Solution

Replace f(x) with y: y = (x+3)/2
Swap x and y: x = (y+3)/2
Solve for y: 2x = y + 3 → y = 2x − 3
So f⁻¹(x) = 2x − 3

16 Logarithms · Solving Equations Hard

Solve for x: log₃(x + 2) + log₃(x − 2) = 2

Ax = 3

Bx = 4

Cx = ±√13

Dx = √13

✓ Solution

Product rule: log₃[(x+2)(x−2)] = 2
log₃(x²−4) = 2 → x² − 4 = 3² = 9 → x² = 13 → x = ±√13
Check domain: need x + 2 > 0 and x − 2 > 0, so x > 2.
√13 ≈ 3.6 > 2 ✓ | −√13 ≈ −3.6 < 2 ✗
Only x = √13 is valid.

17 Vectors · Magnitude Medium

A vector v = ⟨−3, 4⟩. What is |v| (the magnitude)?

B√7

✓ Solution

|v| = √(a² + b²) = √((−3)² + 4²) = √(9 + 16) = √25 = 5
This is the classic 3-4-5 right triangle.

18 Polar Coordinates Hard

Convert the polar point (4, 2π/3) to rectangular coordinates.

A(2, 2√3)

B(−2, 2√3)

C(2√3, 2)

D(2√3, −2)

✓ Solution

x = r cos θ = 4 cos(2π/3) = 4 · (−1/2) = −2
y = r sin θ = 4 sin(2π/3) = 4 · (√3/2) = 2√3
Rectangular: (−2, 2√3)
Note: cos(2π/3) = −1/2, sin(2π/3) = √3/2 (Quadrant II angle)

19 Binomial Theorem Hard

In the expansion of (x + 2)⁴, what is the coefficient of x²?

A16

B12

C24

✓ Solution – Binomial Theorem

Term with x²: choose k = 2 in C(4,k)·x^(4−k)·2^k
k = 2 gives: C(4,2) · x² · 2² = 6 · x² · 4 = 24x²
C(4,2) = 4!/(2!·2!) = 6 and 2² = 4, so coefficient = 6 × 4 = 24

20 Double Angle Identity Hard

If cos θ = 1/3, find the exact value of cos(2θ).

A−7/9

B7/9

C2/3

D−2/9

✓ Solution – Double Angle Identity

Use: cos(2θ) = 2cos²θ − 1
= 2·(1/3)² − 1 = 2·(1/9) − 1 = 2/9 − 9/9 = −7/9
Alternative check: cos²θ − sin²θ = 1/9 − sin²θ. Since sin²θ = 1−1/9 = 8/9, we get 1/9−8/9 = −7/9 ✓

out of 20

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Answer Key & Quick Reference

Domain √

Composition

Factor Thm

H. Asymptote

Exponential

Log Laws

Unit Circle

Pythagorean Id

Period

Q10

Arctan

Q11

Arith Seq

Q12

Geo Series

Q13

Circle

Q14

Vertex

Q15

Inverse Fn

Q16

Log Eq

Q17

Magnitude

Q18

Polar→Rect

Q19

Binomial

Q20

Double Angle

20 Essential Problems

Functions

Polynomials & Factor Theorem

Rational Functions

Exponential & Logarithm

Trigonometry (SOHCAHTOA + Unit Circle)

Sequences & Series

Conic Sections

Vectors & Polar Coordinates

✓ Solution

✓ Solution

✓ Solution – Factor Theorem

✓ Solution

✓ Solution

✓ Solution

✓ Solution

✓ Solution

✓ Solution

✓ Solution

✓ Solution

✓ Solution

✓ Solution – Complete the Square

✓ Solution

✓ Solution

✓ Solution

✓ Solution

✓ Solution

✓ Solution – Binomial Theorem

✓ Solution – Double Angle Identity

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Answer Key & Quick Reference