Number and Algebra practice questions.

In a proof by mathematical induction for all integers n ≥ 1, after verifying the base case, what is assumed in the inductive step?

• P(k+1) is true
• P(k) holds for some k ≥ 1
• P(n) is true for all n ≥ 1
• P(1) is false

Answer: P(k) holds for some k ≥ 1

The inductive hypothesis assumes truth for one arbitrary k, then shows P(k) ⇒ P(k+1). Assuming it for all n would be circular reasoning.

What is the principal argument of z = −1 + i?

• π/4
• 3π/4
• −3π/4
• 5π/4

Answer: 3π/4

The principal argument is the angle θ with −π < θ ≤ π. z lies in the second quadrant with reference angle arctan(1/1) = π/4, so arg z = π − π/4 = 3π/4; 5π/4 points the right way on the wrong range, and −3π/4 is the third-quadrant angle.

How many distinct complex solutions does the equation z⁵ = 32 have?

• 1
• 5
• 10
• 32

Answer: 5

Every non-zero complex number has exactly n distinct nth roots, equally spaced by 2π/n around a circle of radius |32|^(1/5) = 2 in the Argand diagram.

Using the laws of logarithms, log 6 is equal to

• log 2 + log 3
• log 2 × log 3
• log 3 − log 2
• 2 log 3

Answer: log 2 + log 3

6 = 2 × 3, and log(ab) = log a + log b. Note that log a × log b is NOT a log law, and 2 log 3 = log 9.

In the geometric sequence 3, 6, 12, 24, … which term is equal to 96?

• 5th
• 6th
• 7th
• 8th

Answer: 6th

uₙ = 3 × 2ⁿ⁻¹. Solve 3 × 2ⁿ⁻¹ = 96 ⇒ 2ⁿ⁻¹ = 32 = 2⁵ ⇒ n = 6. Listing terms also works: 3, 6, 12, 24, 48, 96.

The coefficient of x² in the expansion of (1 + x)⁵ is

• 5
• 10
• 20
• 32

Answer: 10

By the binomial theorem the coefficient is C(5, 2) = 10. It also appears in row 5 of Pascal's triangle: 1, 5, 10, 10, 5, 1.