Free practice · VCE Units 3 & 4
These 10 VCE Mathematical Methods practice questions cover the Units 3 and 4 core: functions and graphs, differentiation and integration, and probability distributions. Every question is original, written in a VCAA-style format for tech-free practice, with a full mark-by-mark scheme so you can see exactly where each method mark is earned. They are not official VCAA materials.
Work each question on paper before you read the scheme, then compare line by line and check you earned each step, not just the final number. In VCE Methods, marks are awarded for correct working even when the final answer is wrong, so clear setting-out matters.
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Let f(x) = 2x^3 - 9x^2 + 12x. Find the coordinates of the stationary points of f and state the nature of each.
f'(x) = 6x^2 - 18x + 12 (1 mark for correct derivative). Set f'(x) = 0: 6(x^2 - 3x + 2) = 0, so (x - 1)(x - 2) = 0, giving x = 1 and x = 2 (1 mark for both x-values). f(1) = 2 - 9 + 12 = 5 and f(2) = 16 - 36 + 24 = 4, so stationary points are (1, 5) and (2, 4) (1 mark for both coordinates). Testing the sign of f'(x) either side (or noting the positive leading coefficient of a cubic), (1, 5) is a local maximum and (2, 4) is a local minimum (1 mark for correct nature of both). Total 4 marks.
Evaluate the definite integral of (3x^2 - 4x + 1) with respect to x, from x = 0 to x = 2.
Antiderivative is x^3 - 2x^2 + x (1 mark). At x = 2: 8 - 8 + 2 = 2; at x = 0: 0 (1 mark for substituting both terminals). Value = 2 - 0 = 2 (1 mark). Total 3 marks.
The function f is defined by f(x) = log_e(2x - 6). State the maximal domain of f and the equation of its vertical asymptote.
Require 2x - 6 > 0, so x > 3 (1 mark for setting the argument greater than zero). Maximal domain is (3, infinity), i.e. x > 3 (1 mark). Vertical asymptote at x = 3 (1 mark). Total 3 marks.
For the function g(x) = (x - 1)e^x, use the product rule to find g'(x), and hence find the x-coordinate of the stationary point of g.
Using the product rule with u = x - 1, v = e^x: g'(x) = (1)e^x + (x - 1)e^x (1 mark for applying the product rule correctly). Simplify: g'(x) = e^x(1 + x - 1) = x e^x (1 mark). Set g'(x) = 0: since e^x > 0 for all x, x e^x = 0 gives x = 0 (1 mark). Total 3 marks.
A continuous random variable X has probability density function f(x) = kx for 0 <= x <= 2, and f(x) = 0 elsewhere. Find the value of k, then find P(X <= 1).
Total probability is 1: the integral of kx from 0 to 2 equals k[x^2/2] from 0 to 2 = k(2) = 2k (1 mark for setting up and evaluating the integral). So 2k = 1, giving k = 1/2 (1 mark). P(X <= 1) = integral of (1/2)x from 0 to 1 = (1/2)[x^2/2] from 0 to 1 = (1/2)(1/2) = 1/4 (1 mark). Total 3 marks.
Solve the equation 2sin(x) = 1 for x in the interval [0, 2pi]. Give answers as exact values.
sin(x) = 1/2 (1 mark). The base angle is pi/6 (1 mark). Sine is positive in the first and second quadrants, so x = pi/6 and x = pi - pi/6 = 5pi/6 (1 mark for both solutions in the given interval). Total 3 marks.
Let f(x) = x^2 + 2 with restricted domain x >= 0. Find the rule for the inverse function f^(-1), explain why the restriction is needed, and give the domain of f^(-1).
Over all reals f is not one-to-one, so it has no inverse; restricting to x >= 0 makes f one-to-one so the inverse exists (1 mark for a valid reason and restriction). To find the inverse, swap x and y in y = x^2 + 2 to get x = y^2 + 2, then solve: y^2 = x - 2, so y = sqrt(x - 2), taking the positive root because the restricted domain gives y >= 0 (1 mark for correct algebra and root choice). Therefore f^(-1)(x) = sqrt(x - 2) (1 mark). The domain of f^(-1) is the range of f, which is x >= 2 (1 mark). Total 4 marks.
The number of defective items in a batch is modelled by a binomial random variable X with n = 5 trials and probability of success p = 1/3. Find P(X = 2). Leave your answer as an exact fraction.
P(X = 2) = C(5,2) (1/3)^2 (2/3)^3 (1 mark for the correct binomial form with the right power on each factor). C(5,2) = 10 (1 mark). Compute: 10 x (1/9) x (8/27) = 80/243 (1 mark). Total 3 marks.
Find the equation of the tangent to the curve y = x^2 - 3x at the point where x = 2. Give your answer in the form y = mx + c.
dy/dx = 2x - 3 (1 mark). At x = 2, gradient m = 2(2) - 3 = 1 (1 mark). The point: y = 4 - 6 = -2, so the point is (2, -2) (1 mark). Tangent: y - (-2) = 1(x - 2), giving y = x - 4 (1 mark). Total 4 marks.
The function f(x) = a sin(2x), where a > 0, has a maximum value of 3. State the value of a, the period of f, and the range of f.
The maximum of sin is 1, so the maximum of f is a; since this equals 3, a = 3 (1 mark). Period = 2pi divided by the coefficient of x, which is 2, so the period = pi (1 mark). Range is from -3 to 3 inclusive, i.e. [-3, 3] (1 mark). Total 3 marks.
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