Pre-Calculus — Semester B Practice

Question 1 of 10

TEKS 5A-5C Hard

A geometric sequence has recursive form a₁ = 3, aₙ = 2·aₙ₋₁. What is the EXPLICIT form?

A aₙ = 3 + 2(n−1)

B aₙ = 3 · 2ⁿ

C aₙ = 3 · 2ⁿ⁻¹

D aₙ = 2 · 3ⁿ⁻¹

Explanation

Convert recursive to explicit: the recursion "multiply previous by 2, starting at 3" generates 3, 6, 12, 24, … Geometric with a₁ = 3 and r = 2. Explicit: aₙ = a₁ · r^(n−1) = 3 · 2^(n−1). Verify a₁ = 3·1 = 3 ✓; a₂ = 3·2 = 6 ✓.

Question 2 of 10

TEKS 5D-5I Medium

Find the MAGNITUDE of the vector v = ⟨3, 4⟩.

A |v| = 12

B |v| = 25

C |v| = 7

D |v| = 5

Explanation

Vector magnitude: |⟨a, b⟩| = √(a² + b²). For ⟨3, 4⟩: √(9 + 16) = √25 = 5. 3-4-5 right triangle again — the magnitude is the length of the hypotenuse from the origin to the endpoint.

Question 3 of 10

TEKS 4A-4G Easy Diagram

In the 3-4-5 right triangle shown, the angle θ is at the bottom-right vertex (adjacent to side b = 4 and opposite side a = 3). What is sin θ?

A sin θ = 5/3

B sin θ = 3/4 (opposite/adjacent)

C sin θ = 3/5 (opposite/hypotenuse)

D sin θ = 4/5 (adjacent/hypotenuse)

Explanation

SOH-CAH-TOA: sin θ = opposite / hypotenuse. From angle θ's perspective: opposite side = 3 (vertical leg), hypotenuse = 5. So sin θ = 3/5. (B) is cos θ; (C) is tan θ; (D) inverts the ratio.

Question 4 of 10

TEKS 4H-4K Medium

Solve sin x = 1/2 on the interval [0, 2π].

A x = π/3 and x = 2π/3

B x = π/6 and x = 5π/6

C x = π/6 only

D x = π/2

Explanation

Find both solutions in [0, 2π] where sine = 1/2. Reference angle: sin⁻¹(1/2) = π/6. Sine is positive in Quadrants I and II: Q1 solution is x = π/6; Q2 solution is x = π − π/6 = 5π/6. So x = π/6 and 5π/6.

Question 5 of 10

TEKS 4A-4G Medium Diagram

On the unit circle shown, what are the coordinates of the point corresponding to θ = π/2 (90°)?

A (1, 0)

B (0, 1)

C (0, −1)

D (−1, 0)

Explanation

Unit circle definition: at angle θ, the point on the circle is (cos θ, sin θ). At θ = π/2: cos(π/2) = 0, sin(π/2) = 1. So point = (0, 1) — the top of the circle. (B) is θ = 0; (D) is θ = π.

Question 6 of 10

TEKS 5D-5I Medium

What is the VERTEX of the parabola y = 2(x − 3)² + 5?

A (2, 3)

B (3, −5)

C (−3, 5)

D (3, 5)

Explanation

Vertex form: y = a(x − h)² + k has vertex at (h, k). Here h = 3, k = 5, so vertex = (3, 5). The leading coefficient a = 2 controls how narrow/wide the parabola is (not the vertex).

Question 7 of 10

TEKS 4A-4G Medium Diagram

The graph above shows y = sin(x). What is the PERIOD of this function?

A 4π

B π

C 2π

D π/2

Explanation

Period of sin(x): the smallest positive T such that sin(x + T) = sin(x) for all x. From the graph: one complete wave cycle spans 2π (e.g., from 0 to 2π, the function goes 0 → 1 → 0 → −1 → 0). For sin(Bx): period = 2π/|B|.

Question 8 of 10

TEKS 4H-4K Hard

Verify the identity by simplifying: sin²x · (1 + cot²x).

A Cannot be simplified

B Equals sin²x

C Equals 1 — using 1 + cot²x = csc²x and csc²x = 1/sin²x

D Equals tan²x

Explanation

Apply identity 1 + cot²x = csc²x: sin²x · csc²x = sin²x · (1/sin²x) = 1. Or step by step: sin²x · (1 + cos²x/sin²x) = sin²x + cos²x = 1 (by Pythagorean).

Question 9 of 10

TEKS 5A-5C Easy Diagram

The bars show an arithmetic sequence: 5, 8, 11, 14, 17. Find the formula for the nth term aₙ.

A aₙ = 5 + 3(n − 1) = 3n + 2

B aₙ = 5 · 3ⁿ

C aₙ = 5 + 3n

D aₙ = 5n

Explanation

Arithmetic sequence formula: aₙ = a₁ + d(n − 1), where a₁ is the first term and d is the common difference. Here a₁ = 5, d = 3. So aₙ = 5 + 3(n − 1) = 3n + 2. Verify: a₁ = 3(1)+2 = 5 ✓; a₂ = 3(2)+2 = 8 ✓; a₃ = 11 ✓.

Question 10 of 10

TEKS 4H-4K Easy

Which is the PYTHAGOREAN IDENTITY?

A sin²θ + cos²θ = 1

B tan θ = sin θ · cos θ

C sin(2θ) = 2sin θ

D sin θ + cos θ = 1

Explanation

Pythagorean identity: sin²θ + cos²θ = 1 — comes directly from the unit circle (x² + y² = 1 with x = cos θ, y = sin θ). Derived: dividing through by cos²θ gives tan²θ + 1 = sec²θ; dividing by sin²θ gives 1 + cot²θ = csc²θ.

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