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IB Math AA SL study materials
IB Math AA SL study resources

IB Math AA SL practice questions

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Question 1Number and algebra

An arithmetic sequence has first term 5 and common difference 3. What is the 12th term of the sequence?

  1. 44
  2. 41
  3. 36
  4. 38 Correct answer.

Explanation

The nth term of an arithmetic sequence is u_n = u_1 + (n − 1)d. Substituting u_1 = 5, d = 3 and n = 12 gives u_12 = 5 + 11 × 3 = 38.

Question 2Number and algebra

What is the value of log_3 81?

  1. 4 Correct answer.
  2. 3
  3. 27
  4. 9

Explanation

log_3 81 asks for the exponent to which 3 must be raised to obtain 81. Since 3^4 = 81, the value is 4.

Question 3Number and algebra

A geometric sequence has second term 6 and fifth term 48. What is the common ratio?

  1. 3
  2. 8
  3. 4
  4. 2 Correct answer.

Explanation

Using u_n = u_1 r^(n−1), u_2 = u_1 r = 6 and u_5 = u_1 r^4 = 48. Dividing gives r^3 = 8, so r = 2.

Question 4Number and algebra

Find the coefficient of x^3 in the expansion of (2 + x)^5.

  1. 10
  2. 40 Correct answer.
  3. 80
  4. 20

Explanation

By the binomial theorem, the term in x^3 is C(5,3) × 2^2 × x^3. Since C(5,3) = 10 and 2^2 = 4, the coefficient is 10 × 4 = 40.

Question 5Number and algebra

Expressed in the form a + bi, the product (3 + 2i)(1 − 4i) is:

  1. 5 − 10i
  2. 11 + 10i
  3. 11 − 10i Correct answer.
  4. −5 − 10i

Explanation

Expand: 3×1 + 3×(−4i) + 2i×1 + 2i×(−4i) = 3 − 12i + 2i − 8i^2. Since i^2 = −1, this becomes 3 − 10i + 8 = 11 − 10i.

Question 6Number and algebra

Evaluate the sum from k = 1 to 4 of (3k − 1).

  1. 24
  2. 30
  3. 26 Correct answer.
  4. 22

Explanation

Substituting k = 1, 2, 3, 4 gives the terms 2, 5, 8 and 11. Their sum is 2 + 5 + 8 + 11 = 26.

Question 7Number and algebra

Solve log_2 x + log_2(x − 2) = 3.

  1. x = 4 Correct answer.
  2. x = 4 or x = −2
  3. x = −2
  4. x = 2

Explanation

Combining the logarithms gives log_2[x(x − 2)] = 3, so x(x − 2) = 2^3 = 8. This leads to x^2 − 2x − 8 = 0, or (x − 4)(x + 2) = 0. Because logarithms require x > 0 and x − 2 > 0, only x = 4 is valid.

Question 8Number and algebra

A geometric series has first term 8 and common ratio 1/4. Find its sum to infinity.

  1. 10
  2. 32/3 Correct answer.
  3. 8
  4. 24

Explanation

For a geometric series with |r| < 1, the sum to infinity is S∞ = a/(1 − r). Here a = 8 and r = 1/4, so S∞ = 8/(1 − 1/4) = 8/(3/4) = 32/3.

Question 9Number and algebra

Let z = 1 + i√3. What is the argument of z^2, given in the interval [0, 2π)?

  1. π/3
  2. 4π/3
  3. 2π/3 Correct answer.
  4. 5π/6

Explanation

For z = 1 + i√3, the modulus is 2 and the argument is π/3 because tan θ = √3. Squaring a complex number doubles its argument, so arg(z^2) = 2π/3, which lies in [0, 2π).

Question 10Number and algebra

Find the constant term in the expansion of (x + 2/x^2)^6.

  1. 15
  2. 120
  3. 60 Correct answer.
  4. 240

Explanation

The general term is C(6,r) x^(6−r) (2/x^2)^r = C(6,r) 2^r x^(6−3r). For a constant term, the exponent of x must be zero, so 6 − 3r = 0 and r = 2. The constant term is C(6,2) × 2^2 = 15 × 4 = 60.

Question 11Number and algebra

Simplify (2^5 × 2^{-3}) / 2^4.

  1. 4
  2. -2
  3. 1/4 Correct answer.
  4. 1/2

Explanation

Use the exponent laws a^m × a^n = a^{m+n} and a^m / a^n = a^{m-n}. The exponent is 5 + (-3) - 4 = -2, so the value is 2^{-2} = 1/4.

Question 12Number and algebra

For the complex number z = 5 - 12i, find |z|.

  1. 7
  2. 13 Correct answer.
  3. √119
  4. 17

Explanation

The modulus of a + bi is √(a^2 + b^2). Here |z| = √(5^2 + (-12)^2) = √(25 + 144) = √169 = 13.

Question 13Number and algebra

Find the value of log_4(1/8).

  1. -1/2
  2. 3/2
  3. -2/3
  4. -3/2 Correct answer.

Explanation

Write 1/8 as 2^{-3} and 4 as 2^2. Then log_4(1/8) = log_4(2^{-3}) = -3 log_4 2 = -3 × 1/2 = -3/2.

Question 14Number and algebra

An arithmetic sequence has first term 7 and common difference -2. Find the sum of its first 20 terms.

  1. 240
  2. -260
  3. -240 Correct answer.
  4. -220

Explanation

Use S_n = n/2 [2a + (n - 1)d]. With a = 7, d = -2 and n = 20, S_20 = 10[14 + 19(-2)] = 10(14 - 38) = -240.

Question 15Number and algebra

A geometric sequence has first term 3 and common ratio 2. Find the smallest value of n for which the nth term is greater than 500.

  1. 11
  2. 8
  3. 10
  4. 9 Correct answer.

Explanation

The nth term is 3 × 2^{n-1}. The inequality 3 × 2^{n-1} > 500 is equivalent to 2^{n-1} > 166.67. Since 2^7 = 128 and 2^8 = 256, the smallest possible exponent is 8, so n - 1 = 8 and n = 9.

Question 16Number and algebra

Solve 3^(2x - 1) = 27^(x - 2).

  1. 5 Correct answer.
  2. -5
  3. 3
  4. 4

Explanation

Write 27 as 3^3, so 27^(x - 2) = 3^(3x - 6). Equating exponents gives 2x - 1 = 3x - 6, hence x = 5.

Question 17Number and algebra

Find the coefficient of x^4 in the expansion of (3 - 2x)^6.

  1. 1458
  2. -2160
  3. 2160 Correct answer.
  4. 4860

Explanation

The term containing x^4 is C(6,4) × 3^2 × (-2x)^4. This coefficient is 15 × 9 × 16 = 2160, and the even power of -2 makes it positive.

Question 18Number and algebra

Express (2 + i)/(1 - 2i) in the form a + bi.

  1. i Correct answer.
  2. 1
  3. -i
  4. -1

Explanation

Multiply numerator and denominator by the conjugate 1 + 2i. The numerator becomes (2 + i)(1 + 2i) = 2 + 4i + i + 2i^2 = 5i, while the denominator is 1^2 + 2^2 = 5. Thus the quotient is i.

Question 19Number and algebra

Solve log_{1/2}(x + 3) > -2.

  1. (-3, ∞)
  2. (-∞, 1)
  3. (-3, 1) Correct answer.
  4. (1, ∞)

Explanation

The logarithm requires x + 3 > 0, so x > -3. Because the base 1/2 is less than 1, the inequality reverses when converting to exponents: x + 3 < (1/2)^{-2} = 4, so x < 1. Combining gives -3 < x < 1.

Question 20Number and algebra

Find the sum of all integers from 1 to 200 inclusive that are divisible by 3 but not by 5.

  1. 6633
  2. 1365
  3. 5268 Correct answer.
  4. 5000

Explanation

The multiples of 3 up to 200 are 3, 6, ..., 198, with sum 3(1 + 2 + ... + 66) = 3 × 66 × 67 / 2 = 6633. The multiples of 15 up to 200 sum to 15(1 + ... + 13) = 15 × 91 = 1365. Subtracting gives 6633 - 1365 = 5268.

Question 21Number and algebra

The sum of the first 9 terms of the geometric series 2 + 6 + 18 + ... is

  1. 39366
  2. 13122
  3. 19683
  4. 19682 Correct answer.

Explanation

The first term is 2 and the common ratio is 3, so the finite geometric sum is S_9 = 2(3^9 − 1)/(3 − 1). Since 3^9 = 19683, this gives S_9 = 19683 − 1 = 19682.

Question 22Number and algebra

The expression (8^(2/3) × 4^(−1/2)) equals

  1. 2 Correct answer.
  2. 1
  3. 8
  4. 4

Explanation

Using powers of 2, 8^(2/3) = (2^3)^(2/3) = 2^2 = 4 and 4^(−1/2) = (2^2)^(−1/2) = 2^(−1) = 1/2. Their product is 4 × 1/2 = 2.

Question 23Number and algebra

For z = 4 − 3i, what is the value of z multiplied by its complex conjugate?

  1. 25 Correct answer.
  2. 16 − 9i
  3. 5
  4. 7

Explanation

For z = a + bi, the product z z* equals a^2 + b^2. Here a = 4 and b = −3, so z z* = 4^2 + (−3)^2 = 16 + 9 = 25.

Question 24Number and algebra

Given log_a 2 = p and log_a 5 = q, log_a 20 can be expressed as

  1. p + q + 2
  2. p^2 + q
  3. 2(p + q)
  4. 2p + q Correct answer.

Explanation

Since 20 = 4 × 5 = 2^2 × 5, the logarithm laws give log_a 20 = log_a(2^2) + log_a 5 = 2 log_a 2 + log_a 5. Substituting the given values yields 2p + q.

Question 25Number and algebra

In the expansion of (1 − 3x)^8, the coefficient of x^2 is

  1. −252
  2. 252 Correct answer.
  3. 224
  4. 28

Explanation

The general term is C(8, r)(−3x)^r. For x^2, take r = 2, giving C(8, 2)(−3)^2 = 28 × 9 = 252.

Question 26Number and algebra

An arithmetic sequence has u_4 = 11 and u_9 = 26. What is u_1?

  1. 8
  2. 5
  3. 2 Correct answer.
  4. 3

Explanation

The common difference is d = (u_9 − u_4)/(9 − 4) = (26 − 11)/5 = 3. Then u_1 = u_4 − 3d = 11 − 9 = 2.

Question 27Number and algebra

A geometric series has second term 3 and sum to infinity 12. What is its common ratio?

  1. 1/2 Correct answer.
  2. 2/3
  3. 3/4
  4. 1/4

Explanation

Let the first term be a and the common ratio be r. Then ar = 3 and a/(1 − r) = 12. Substituting a = 3/r gives 3/[r(1 − r)] = 12, so r(1 − r) = 1/4, which gives (2r − 1)^2 = 0 and hence r = 1/2.

Question 28Number and algebra

The complex number z has modulus 2 and argument π/3. What is z^3 in Cartesian form?

  1. −8i
  2. 8
  3. −8 Correct answer.
  4. 8i

Explanation

By De Moivre’s theorem, z^3 has modulus 2^3 = 8 and argument 3 × π/3 = π. Thus z^3 = 8(cos π + i sin π) = 8(−1 + 0i) = −8.

Question 29Number and algebra

How many positive integers less than 1000 have all distinct digits?

  1. 729
  2. 648
  3. 900
  4. 738 Correct answer.

Explanation

Count separately by number of digits. There are 9 one-digit numbers, 9 × 9 = 81 two-digit numbers with distinct digits, and 9 × 9 × 8 = 648 three-digit numbers with distinct digits. The total is 9 + 81 + 648 = 738.

Question 30Number and algebra

The equation 4^x − 5·2^x + 4 = 0 has solution set

  1. {1, 2}
  2. {2, 4}
  3. {0, 1}
  4. {0, 2} Correct answer.

Explanation

Since 4^x = (2^x)^2, let y = 2^x. Then y^2 − 5y + 4 = 0, so (y − 1)(y − 4) = 0, giving y = 1 or y = 4. Thus 2^x = 1 gives x = 0, and 2^x = 4 gives x = 2.

Question 31Number and algebra

An amount of $5000 is invested at an annual interest rate of 4%, compounded annually. What is the value of the investment after 2 years?

  1. $5408 Correct answer.
  2. $5500
  3. $5400
  4. $5200

Explanation

The value after two years is 5000(1 + 0.04)^2 = 5000(1.04)^2 = 5408. Compound interest multiplies the balance by 1.04 each year, so the second year's interest is earned on the increased balance.

Question 32Number and algebra

The complex number z satisfies z + 2z̄ = 9 + 3i, where z̄ is the complex conjugate of z. Find z.

  1. -3 + 3i
  2. 1 - i
  3. 3 + 3i
  4. 3 - 3i Correct answer.

Explanation

Write z = a + bi, so z̄ = a - bi. Then z + 2z̄ = 3a - bi, and equating real and imaginary parts gives 3a = 9 and -b = 3, so a = 3 and b = -3. Hence z = 3 - 3i.

Question 33Functions

The function f is defined by f(x) = √(3 − x) for real values of x. What is the domain of f?

  1. x ≥ 3
  2. x > 3
  3. x ≤ 3 Correct answer.
  4. x < 3

Explanation

A square-root expression is real only when its radicand is non-negative, so require 3 − x ≥ 0. Rearranging gives x ≤ 3, so the domain is all real numbers satisfying this inequality.

Question 34Functions

The function f is defined by f(x) = 2x + 5. Which expression gives f⁻¹(x)?

  1. (x + 5)/2
  2. (x − 5)/2 Correct answer.
  3. x/2 − 5
  4. 2x − 5

Explanation

Write y = 2x + 5, swap x and y to obtain x = 2y + 5, then solve for y. This gives y = (x − 5)/2, so f⁻¹(x) = (x − 5)/2.

Question 35Functions

The function f is defined by f(x) = x² − 4 for x ∈ ℝ. What is the range of f?

  1. [0, ∞)
  2. [−4, ∞) Correct answer.
  3. (−∞, −4]
  4. (−4, ∞)

Explanation

Since x² ≥ 0 for every real x, subtracting 4 gives x² − 4 ≥ −4. The minimum value −4 occurs at x = 0, so the range is [−4, ∞).

Question 36Functions

The functions f and g are defined by f(x) = x² + 1 and g(x) = 2x − 3. What is the value of f(g(2))?

  1. 5
  2. 1
  3. 2 Correct answer.
  4. −1

Explanation

First evaluate the inner function: g(2) = 2(2) − 3 = 1. Then substitute into f: f(1) = 1² + 1 = 2.

Question 37Functions

The graph of y = f(x) is transformed to the graph of y = f(2x) − 3. Which description of the transformation is correct?

  1. A horizontal compression by factor 2 and a vertical translation 3 units down
  2. A horizontal compression by factor 1/2 and a vertical translation 3 units up
  3. A horizontal stretch by factor 1/2 and a vertical translation 3 units down
  4. A horizontal compression by factor 1/2 and a vertical translation 3 units down Correct answer.

Explanation

Replacing x by 2x inside f compresses the graph horizontally by factor 1/2, because each x-coordinate is halved. Subtracting 3 outside f then translates the graph 3 units down.

Question 38Functions

The function f is defined by f(x) = 2x² − 8x + 5. What is the minimum value of f(x)?

  1. 3
  2. −11
  3. −3 Correct answer.
  4. 5

Explanation

The vertex occurs at x = −b/(2a) = 8/4 = 2. Then f(2) = 2(4) − 16 + 5 = −3, and since the leading coefficient is positive this is the minimum value.

Question 39Functions

The function f is defined by f(x) = e^(2x − 1) for x ∈ ℝ. Which expression gives f⁻¹(x)?

  1. (ln x − 1)/2
  2. (ln x + 1)/2 Correct answer.
  3. ln((x + 1)/2)
  4. 2 ln x + 1

Explanation

Start from y = e^(2x − 1) and take natural logarithms: ln y = 2x − 1. Solving for x gives x = (ln y + 1)/2, so f⁻¹(x) = (ln x + 1)/2 with domain x > 0.

Question 40Functions

The function f is defined by f(x) = (2x − 1)/(x + 3). What are the equations of the asymptotes of the graph of y = f(x)?

  1. x = −3 and y = 2 Correct answer.
  2. x = 3 and y = 2
  3. x = 2 and y = −3
  4. x = −3 and y = −1/3

Explanation

The denominator x + 3 is zero at x = −3, producing a vertical asymptote there. For large |x| the fraction behaves like 2x/x = 2, so y = 2 is the horizontal asymptote.

Question 41Functions

The functions f and g are defined by f(x) = √(x − 1) and g(x) = 1/(x − 4). What is the domain of the composite function f(g(x))?

  1. [5, ∞)
  2. (4, 5] Correct answer.
  3. (4, 5)
  4. [4, 5]

Explanation

The inner function requires x ≠ 4, and the outer square root requires 1/(x − 4) − 1 ≥ 0. This inequality is equivalent to (5 − x)/(x − 4) ≥ 0, whose solution is 4 < x ≤ 5.

Question 42Functions

The function f is defined by f(x) = x² − 4x + 7 for x ≥ 2. Which expression gives f⁻¹(x)?

  1. 2 + √(x − 3) Correct answer.
  2. 2 − √(x − 3)
  3. −2 + √(x − 3)
  4. √(x − 3) + 4

Explanation

Complete the square: f(x) = (x − 2)² + 3. With x ≥ 2, the range is [3, ∞) and the graph is the right branch, so solving y = (x − 2)² + 3 gives x = 2 + √(y − 3). Thus f⁻¹(x) = 2 + √(x − 3).

Question 43Functions

The function f is defined by f(x) = (x + 1)/(x − 2) for x ≠ 2. What is the value of f(3)?

  1. 5
  2. 2
  3. 3
  4. 4 Correct answer.

Explanation

Substitute x = 3 into the numerator and denominator: f(3) = (3 + 1)/(3 − 2) = 4/1 = 4. The restriction x ≠ 2 does not affect the input x = 3.

Question 44Functions

The function f has domain {x ∈ ℝ : −2 ≤ x < 3}. What is the domain of the function g defined by g(x) = f(x + 4)?

  1. −2 ≤ x < 3
  2. −2 ≤ x < 7
  3. 2 ≤ x < 7
  4. −6 ≤ x < −1 Correct answer.

Explanation

For f(x + 4) to be defined, the input x + 4 must lie in the interval [−2, 3). Solving −2 ≤ x + 4 < 3 gives −6 ≤ x < −1.

Question 45Functions

The function f is defined by f(x) = |x − 3| + 2 for x ∈ ℝ. What is the range of f?

  1. y ≥ 2 Correct answer.
  2. y ≥ 3
  3. y ≥ 0
  4. y ≤ 2

Explanation

The absolute value |x − 3| is always at least 0, so f(x) is always at least 2. The value 2 occurs at x = 3, and there is no upper bound, so the range is y ≥ 2.

Question 46Functions

The functions f and g are defined by f(x) = 3x − 2 and g(x) = x² + 1. Which expression gives f(g(x))?

  1. 3x² + 3
  2. 3x² − 1
  3. 9x² − 12x + 5
  4. 3x² + 1 Correct answer.

Explanation

Compute g(x) first: g(x) = x² + 1. Then f(g(x)) = 3(x² + 1) − 2 = 3x² + 3 − 2 = 3x² + 1.

Question 47Functions

The function f is defined by f(x) = (x − 1)/(x + 2) for x ≠ −2. Which expression gives f⁻¹(x)?

  1. (x + 2)/(x − 1)
  2. (1 − x)/(2x + 1)
  3. (2x − 1)/(x + 1)
  4. (2x + 1)/(1 − x) Correct answer.

Explanation

Let y = (x − 1)/(x + 2). Rearranging gives y(x + 2) = x − 1, so x(y − 1) = −1 − 2y and x = (2y + 1)/(1 − y). Hence f⁻¹(x) = (2x + 1)/(1 − x), with x ≠ 1.

Question 48Functions

The graph of y = f(x) has a local minimum at the point (2, −5). The graph of y = f(x − 3) + 4 is a transformation of y = f(x). What are the coordinates of the corresponding local minimum?

  1. (5, −9)
  2. (5, −1) Correct answer.
  3. (−1, −1)
  4. (−1, −9)

Explanation

Replacing x by x − 3 translates the graph 3 units to the right, so the x-coordinate becomes 2 + 3 = 5. Adding 4 translates the graph 4 units up, so the y-coordinate becomes −5 + 4 = −1.

Question 49Functions

The function f is defined by f(x) = ln(5 − 2x). What is the domain of f?

  1. x ≥ 5/2
  2. x < 5/2 Correct answer.
  3. x > 5/2
  4. x ≤ 5/2

Explanation

The argument of the logarithm must be strictly positive, so 5 − 2x > 0. Solving this inequality gives 2x < 5, hence x < 5/2.

Question 50Functions

The functions f and g are defined by f(x) = 2x + 1 and g(x) = x² − 3. What are the solutions of f(g(x)) = 0?

  1. x = ±1/2
  2. x = ±√3
  3. x = ±√(7/2)
  4. x = ±√(5/2) Correct answer.

Explanation

First form f(g(x)) = 2(x² − 3) + 1 = 2x² − 5. Setting this equal to 0 gives x² = 5/2, so x = ±√(5/2).