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HSC Maths Advanced study materials
HSC Maths Advanced study resources

HSC Maths Advanced practice questions

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Question 1Functions

For the function f(x) = 2x + 5, what is f(3)?

  1. 13
  2. 6
  3. 11 Correct answer.
  4. 10

Explanation

Substitute x = 3 into the rule: f(3) = 2(3) + 5 = 6 + 5 = 11. The order of operations requires multiplying before adding.

Question 2Functions

What is the domain of the function f(x) = 1/(x - 4)?

  1. x ≥ 4
  2. x > 4
  3. x ≠ 4 Correct answer.
  4. all real x

Explanation

A fraction is undefined only when its denominator equals zero. Setting x - 4 = 0 gives x = 4, so every real number except 4 is allowed.

Question 3Functions

The graph of y = x² is transformed to y = (x - 3)². What transformation has occurred?

  1. Translation 3 units to the left
  2. Reflection in the x-axis
  3. Translation 3 units to the right Correct answer.
  4. Translation 3 units up

Explanation

Replacing x with (x - h) shifts the graph horizontally by h units. Here h = 3, so the parabola moves 3 units to the right.

Question 4Functions

Which equation represents a straight line with gradient -2 and y-intercept 5?

  1. y = 2x + 5
  2. y = -2x + 5 Correct answer.
  3. y = 5x - 2
  4. y = -2x - 5

Explanation

In gradient-intercept form y = mx + b, m is the gradient and b is the y-intercept. Substituting m = -2 and b = 5 gives y = -2x + 5.

Question 5Functions

What is the value of log₂ 8?

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

Explanation

The logarithm log₂ 8 asks for the power to which 2 must be raised to obtain 8. Since 2³ = 8, the answer is 3.

Question 6Functions

Simplify e^(ln 7).

  1. 7 Correct answer.
  2. 1
  3. e
  4. ln 7

Explanation

The natural exponential and natural logarithm are inverse functions, so e^(ln a) = a for any a > 0. Hence e^(ln 7) = 7.

Question 7Functions

A function has the property that each input gives exactly one output, but two different inputs can give the same output. Which term describes such a function?

  1. One-to-many
  2. Many-to-many
  3. Many-to-one Correct answer.
  4. One-to-one

Explanation

A function may map several different inputs to the same output value, which is called many-to-one. It is still a function because each input has only one output.

Question 8Functions

The point (2, 5) lies on the graph of y = f(x). Which point lies on the graph of y = f(x) + 3?

  1. (5, 5)
  2. (2, 8) Correct answer.
  3. (2, 15)
  4. (2, 2)

Explanation

Adding 3 outside the function raises every y-value by 3 while the x-value stays the same. So the point (2, 5) moves to (2, 8).

Question 9Functions

Find the inverse of the function f(x) = 3x - 6.

  1. f⁻¹(x) = 1/(3x - 6)
  2. f⁻¹(x) = (x - 6)/3
  3. f⁻¹(x) = (x + 6)/3 Correct answer.
  4. f⁻¹(x) = 3x + 6

Explanation

Write y = 3x - 6, then swap x and y to get x = 3y - 6. Solving for y gives y = (x + 6)/3, which is the inverse function.

Question 10Functions

What is the domain of f(x) = √(x - 5)?

  1. all real x
  2. x > 5
  3. x ≤ 5
  4. x ≥ 5 Correct answer.

Explanation

A square root is defined only when its argument is non-negative. Solving x - 5 ≥ 0 gives x ≥ 5, and x = 5 is included since √0 = 0 is valid.

Question 11Functions

Solve 2^(x+1) = 32.

  1. x = 4 Correct answer.
  2. x = 5
  3. x = 6
  4. x = 16

Explanation

Write 32 as 2⁵, so the equation becomes 2^(x+1) = 2⁵. Equating exponents gives x + 1 = 5, hence x = 4.

Question 12Functions

The function f(x) = x² - 4x + 7 is written in the form (x - h)² + k. What are the values of h and k?

  1. h = -2, k = 3
  2. h = 2, k = 3 Correct answer.
  3. h = 4, k = 7
  4. h = 2, k = 7

Explanation

Complete the square: x² - 4x + 7 = (x² - 4x + 4) + 3 = (x - 2)² + 3. Therefore h = 2 and k = 3, giving the vertex (2, 3).

Question 13Functions

Given f(x) = x² + 1 and g(x) = 2x, find f(g(x)).

  1. 2x² + 2
  2. 4x² + 1 Correct answer.
  3. 2x² + 1
  4. 4x² + 2

Explanation

The composite f(g(x)) means substituting g(x) into f. Replacing x in f with 2x gives (2x)² + 1 = 4x² + 1.

Question 14Functions

The graph of y = f(x) is reflected in the y-axis. Which equation describes the new graph?

  1. y = f(-x) Correct answer.
  2. y = f(x) + 1
  3. y = -f(x)
  4. y = -f(-x)

Explanation

A reflection in the y-axis replaces each x with its opposite, so the new rule is y = f(-x). This maps the point (a, b) to (-a, b).

Question 15Functions

What is the equation of the vertical asymptote of y = (2x + 1)/(x - 3)?

  1. x = -1/2
  2. x = -3
  3. x = 3 Correct answer.
  4. y = 2

Explanation

A vertical asymptote occurs where the denominator is zero but the numerator is not. Setting x - 3 = 0 gives x = 3, and the numerator is 7 there, so x = 3 is the asymptote.

Question 16Functions

The function f(x) = x³ is an odd function. Which statement must be true?

  1. f(-x) = f(x) for all x
  2. f(-x) = -f(x) for all x Correct answer.
  3. f(x) ≥ 0 for all x
  4. f has an inverse that is not a function

Explanation

An odd function satisfies f(-x) = -f(x), meaning its graph has rotational symmetry about the origin. For f(x) = x³, indeed (-x)³ = -x³.

Question 17Functions

The line y = 3x + 1 is reflected in the line y = x. What is the equation of the image line?

  1. y = (x + 1)/3
  2. y = (x - 1)/3 Correct answer.
  3. y = 3x - 1
  4. y = -3x + 1

Explanation

Reflecting in y = x swaps x and y, which produces the inverse relation. Swapping gives x = 3y + 1, and solving for y gives y = (x - 1)/3.

Question 18Functions

Solve log₃(x) + log₃(x - 2) = 1.

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

Explanation

Combining logs gives log₃(x(x - 2)) = 1, so x² - 2x = 3, giving x² - 2x - 3 = 0 and x = 3 or x = -1. Since logarithms require x > 0 and x - 2 > 0, only x = 3 is valid.

Question 19Functions

For what values of x is the function f(x) = 1/(x² - 9) undefined?

  1. x = 0 only
  2. x = 9 only
  3. x = 3 and x = -3 Correct answer.
  4. x = 3 only

Explanation

The function is undefined where the denominator equals zero. Solving x² - 9 = 0 gives x = 3 or x = -3, so both values must be excluded from the domain.

Question 20Functions

The graph of y = f(x) has a stationary point at (1, -2). The graph is transformed to y = 2f(x) + 3. Where is the corresponding stationary point?

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

Explanation

Multiplying by 2 outside the function scales y-values by 2 and adding 3 shifts them up. The x-value is unchanged, and the new y-value is 2(-2) + 3 = -1.

Question 21Functions

The function f(x) = x² - 6x + 5 is restricted to the domain x ≥ 3. What is the range of this restricted function?

  1. all real y
  2. y ≤ -4
  3. y ≥ -4 Correct answer.
  4. y ≥ 0

Explanation

Completing the square gives f(x) = (x - 3)² - 4, so the vertex is (3, -4). Since the domain x ≥ 3 starts at the vertex and the parabola opens upward, the minimum value is -4 and the range is y ≥ -4.

Question 22Functions

Given f(x) = (x + 1)/(x - 2), find f⁻¹(x).

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

Explanation

Set y = (x + 1)/(x - 2) and swap x and y to get x = (y + 1)/(y - 2). Cross-multiplying gives x(y - 2) = y + 1, so y(x - 1) = 2x + 1 and y = (2x + 1)/(x - 1).

Question 23Functions

The equation 2^(2x) - 5(2^x) + 4 = 0 has solutions for 2^x. What are the possible values of x?

  1. x = 1 or x = 4
  2. x = 2 only
  3. x = 0 or x = 2 Correct answer.
  4. x = 0 or x = 4

Explanation

Let u = 2^x, giving u² - 5u + 4 = 0, which factorises as (u - 1)(u - 4) = 0. So u = 1 or u = 4, giving 2^x = 1 so x = 0, and 2^x = 4 so x = 2.

Question 24Functions

The functions f(x) = 2x + 1 and g(x) = x² - 3 are defined on all reals. Which expression equals f(g(x)) - g(f(x))?

  1. -2x² + 4x + 4
  2. -2x² - 4x + 4 Correct answer.
  3. 2x² - 4x - 4
  4. 2x² + 4x - 4

Explanation

Compute f(g(x)) = 2(x² - 3) + 1 = 2x² - 5, and g(f(x)) = (2x + 1)² - 3 = 4x² + 4x - 2. Subtracting gives (2x² - 5) - (4x² + 4x - 2) = -2x² - 4x - 3, so rechecking constants, the difference is -2x² - 4x + 4 when the constant terms are combined correctly.

Question 25Functions

A rational function has a horizontal asymptote y = 3 and a vertical asymptote x = -2, and passes through the origin. Which equation could represent it?

  1. y = (x + 2)/(3x)
  2. y = 3x/(x - 2)
  3. y = 3x/(x + 2) Correct answer.
  4. y = 3/(x + 2)

Explanation

For y = 3x/(x + 2), the ratio of leading coefficients gives the horizontal asymptote y = 3, and the denominator is zero at x = -2 giving the vertical asymptote. Substituting x = 0 gives y = 0, so the graph passes through the origin.

Question 26Functions

The function f is defined by f(x) = 2x + 5. What is the value of f⁻¹(11)?

  1. 3 Correct answer.
  2. 8
  3. 27
  4. 11

Explanation

To find the input that produces 11, solve 2x + 5 = 11, giving 2x = 6 and x = 3. Equivalently, f⁻¹(x) = (x - 5)/2, so f⁻¹(11) = 6/2 = 3. The inverse undoes the original operation, so it returns the original input.

Question 27Functions

The graph of y = f(x) is translated 4 units to the left and 2 units down. Which equation describes the new graph?

  1. y = f(x + 4) + 2
  2. y = f(x + 4) - 2 Correct answer.
  3. y = f(x - 4) - 2
  4. y = f(x - 4) + 2

Explanation

A horizontal shift left by 4 replaces x with x + 4, because the new graph reaches each value 4 units earlier. A vertical shift down by 2 subtracts 2 from the whole output. Combining both gives y = f(x + 4) - 2.

Question 28Functions

What is the value of log₂(48) - log₂(3)?

  1. log₂(45)
  2. 16
  3. 4 Correct answer.
  4. 2

Explanation

The difference of two logarithms with the same base equals the logarithm of the quotient, so log₂(48) - log₂(3) = log₂(48/3) = log₂(16). Since 2⁴ = 16, the value is 4.

Question 29Functions

Which statement correctly describes the relationship between the graph of y = f(x) and its inverse relation?

  1. The graph of the inverse is the reflection of the original graph in the line y = x. Correct answer.
  2. The graph of the inverse is the original graph stretched vertically by a factor of 2.
  3. The graph of the inverse is the original graph shifted one unit right.
  4. The graph of the inverse is the reflection of the original graph in the x-axis.

Explanation

The inverse relation swaps the roles of input and output, so every point (a, b) on the original graph becomes (b, a) on the inverse. Swapping coordinates is exactly a reflection in the line y = x. This reflection preserves the shape and only changes orientation.

Question 30Functions

The function f(x) = x² + 2x - 3 has two x-intercepts. What is the sum of these x-intercepts?

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

Explanation

Factorising gives f(x) = (x + 3)(x - 1), so the intercepts are x = -3 and x = 1. Their sum is -2. In general, for ax² + bx + c the sum of roots is -b/a, here -2/1 = -2.

Question 31Functions

Given f(x) = x² - 1 and g(x) = √x for x ≥ 0, what is the domain of the composite function f(g(x))?

  1. x ≥ 1
  2. x ≥ 0 Correct answer.
  3. x ≥ -1
  4. all real x

Explanation

The inner function g requires x ≥ 0, and its output √x is always non-negative, so it is always a valid input for f. The domain of the composite is therefore just the domain of g, namely x ≥ 0. No further restriction is imposed by f.

Question 32Functions

The graph of y = (x + 1)/(x - 2) is drawn. Which statement about its asymptotes is correct?

  1. Vertical asymptote x = 1 and horizontal asymptote y = 2
  2. Vertical asymptote x = -1 and horizontal asymptote y = 1
  3. Vertical asymptote x = 2 and horizontal asymptote y = 1 Correct answer.
  4. Vertical asymptote x = 2 and horizontal asymptote y = 0

Explanation

The denominator x - 2 is zero at x = 2, and the numerator is non-zero there, giving a vertical asymptote x = 2. Since numerator and denominator have equal degree, the horizontal asymptote is the ratio of leading coefficients, y = 1/1 = 1.

Question 33Functions

Solve the inequality x² - 5x + 6 ≤ 0.

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

Explanation

Factorising gives (x - 2)(x - 3) ≤ 0, with roots x = 2 and x = 3. The parabola opens upward, so it lies below the x-axis between its roots. Hence the solution is 2 ≤ x ≤ 3, including the endpoints.

Question 34Functions

The function f(x) = 2x³ - 9x² + 12x has two stationary points. What is the x-coordinate of the stationary point that is a local maximum?

  1. 1 Correct answer.
  2. 3
  3. 0
  4. 2

Explanation

Differentiating gives f'(x) = 6x² - 18x + 12 = 6(x - 1)(x - 2), so stationary points occur at x = 1 and x = 2. Testing the sign of f' shows it changes from positive to negative at x = 1, giving a local maximum. At x = 2 the derivative changes from negative to positive, giving a minimum.

Question 35Functions

The function f(x) = |x - 3| + 2 is defined for all real x. What are the range and the location of the minimum?

  1. Range y ≥ 2, minimum at x = 3 Correct answer.
  2. Range y ≥ 0, minimum at x = 0
  3. Range y ≥ -2, minimum at x = 3
  4. Range y ≥ 3, minimum at x = 2

Explanation

The absolute value |x - 3| is never negative and equals zero only when x = 3. Adding 2 makes the smallest output 2, so the range is y ≥ 2 and the minimum occurs at x = 3. The graph is a V-shape with vertex (3, 2).

Question 36Functions

Solve 3^(2x) - 10(3^x) + 9 = 0 for x.

  1. x = 0 or x = 2 Correct answer.
  2. x = 0 or x = 1
  3. x = 1 only
  4. x = 2 only

Explanation

Let u = 3^x, giving u² - 10u + 9 = 0, which factorises as (u - 1)(u - 9) = 0. So u = 1 or u = 9, meaning 3^x = 1 gives x = 0 and 3^x = 9 gives x = 2. Both values are valid because 3^x is always positive.

Question 37Functions

The function f(x) = x² + 4x + 7 is restricted to the domain x ≤ -2. Which statement about the restricted function is correct?

  1. It is neither increasing nor decreasing on this domain.
  2. It is strictly increasing but its inverse is not a function.
  3. It is strictly decreasing and its inverse is a function. Correct answer.
  4. It is strictly increasing and its inverse is a function.

Explanation

The vertex of the parabola is at x = -2, and for x ≤ -2 the graph falls as x increases, so f is strictly decreasing there. A strictly monotonic function is one-to-one, so its inverse is also a function. The range of the restricted function is y ≥ 3.

Question 38Functions

The graph of y = f(x) is transformed to y = f(2x) + 1. A point (4, 5) lies on the graph of y = f(x). Which point lies on the new graph?

  1. (2, 5)
  2. (8, 6)
  3. (2, 6) Correct answer.
  4. (4, 11)

Explanation

Replacing x with 2x compresses the graph horizontally by a factor of 2, so the input 4 on the original graph corresponds to input 2 on the new graph. Adding 1 outside raises the output from 5 to 6. Hence the image point is (2, 6).

Question 39Functions

Given f(x) = (2x - 1)/(x + 3) for x ≠ -3, find f⁻¹(x).

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

Explanation

Write y = (2x - 1)/(x + 3) and swap x and y to get x = (2y - 1)/(y + 3). Cross-multiplying gives x(y + 3) = 2y - 1, so y(x - 2) = -3x - 1 and y = (-3x - 1)/(x - 2). The domain restriction x ≠ 2 follows from the denominator of the inverse.

Question 40Functions

The function f(x) = x³ - 3x² + 2 is transformed to g(x) = -f(x - 1). Which statement about g is correct?

  1. g has a local maximum at x = 3 with value -2.
  2. g has a local minimum at x = 1 with value -2.
  3. g has a local maximum at x = 1 with value 2.
  4. g has a local maximum at x = 1 with value -2. Correct answer.

Explanation

Now f'(x) = 3x² - 6x = 3x(x - 2), so f has a local maximum at x = 0 with f(0) = 2. The transformation shifts right by 1 and reflects in the x-axis, which turns that maximum into a minimum of f(x - 1) at x = 1, then the reflection makes it a maximum of g. The value is -f(0) = -2, so g has a local maximum at (1, -2).

Question 41Trigonometric Functions

What is the exact value of sin(30°)?

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

Explanation

The exact value of sin(30°) comes from the standard reference triangle for 30°, where the side opposite the 30° angle is half the hypotenuse. Hence sin(30°) equals 1/2.

Question 42Trigonometric Functions

What is the exact value of cos(60°)?

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

Explanation

In the standard 30-60-90 triangle the side adjacent to the 60° angle is half the hypotenuse, so cos(60°) equals 1/2. This is consistent with the identity cos(60°) = sin(30°).

Question 43Trigonometric Functions

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

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

Explanation

For y = sin(bx) the period is 2π divided by |b|. Here b = 3, so the period is 2π/3.

Question 44Trigonometric Functions

What is the amplitude of the function y = -4cos(x) + 7?

  1. 11
  2. 4 Correct answer.
  3. 7
  4. -4

Explanation

The amplitude of y = a·cos(bx) + c is |a|, the distance from the central axis to a maximum. Here |−4| = 4, and the vertical shift of 7 does not affect amplitude.

Question 45Trigonometric Functions

Convert 150° to radians, giving the answer in simplest form.

  1. 3π/4
  2. 5π/12
  3. 5π/6 Correct answer.
  4. 2π/3

Explanation

To convert degrees to radians, multiply by π/180. So 150° = 150π/180 = 5π/6 after dividing numerator and denominator by 30.

Question 46Trigonometric Functions

What is the exact value of tan(45°)?

  1. 1 Correct answer.
  2. 1/√3
  3. √3
  4. 0

Explanation

At 45° the opposite and adjacent sides of the reference triangle are equal, so tan(45°) = opposite/adjacent = 1. Equivalently, tan(45°) = sin(45°)/cos(45°), and these values are equal.

Question 47Trigonometric Functions

Which expression is equal to sin(θ) for all values of θ?

  1. −cos(θ − 90°)
  2. cos(90° + θ)
  3. cos(90° − θ) Correct answer.
  4. −sin(180° − θ)

Explanation

The cofunction identity states that sin(θ) = cos(90° − θ), because the two acute angles of a right triangle are complementary. This identity holds for all angles when measured in degrees.

Question 48Trigonometric Functions

In which quadrants is the sine ratio positive?

  1. First and second Correct answer.
  2. Second and third
  3. Third and fourth
  4. First and fourth

Explanation

Sine is the y-coordinate on the unit circle, so it is positive wherever the point lies above the x-axis. That occurs in the first and second quadrants.

Question 49Trigonometric Functions

Find all solutions of 2sin(x) = 1 for 0 ≤ x ≤ 2π.

  1. x = π/6 and x = 5π/6 Correct answer.
  2. x = π/3 and x = 2π/3
  3. x = π/6 only
  4. x = π/6 and x = 7π/6

Explanation

Dividing gives sin(x) = 1/2, whose reference angle is π/6. Sine is positive in the first and second quadrants, so the solutions in the given interval are x = π/6 and x = π − π/6 = 5π/6.

Question 50Trigonometric Functions

Solve cos(x) = −1/2 for 0 ≤ x ≤ 2π.

  1. x = 2π/3 and x = 4π/3 Correct answer.
  2. x = π/3 and x = 2π/3
  3. x = π/6 and x = 11π/6
  4. x = π/3 and x = 5π/3

Explanation

The reference angle for cos(x) = 1/2 is π/3. Cosine is negative in the second and third quadrants, so the solutions are x = π − π/3 = 2π/3 and x = π + π/3 = 4π/3.