Free VCE Mathematical Methods Units 3&4 practice questions with worked solutions

25 original exam-style questions for VCE Mathematical Methods Units 3&4, tagged by area of study, each with the answer and a full worked solution. Try each one first, then open the solution. Written for the current study design; always check your school's and VCAA's own material for exam rules.

Algebra and function

Q1 · Algebra and function · Circular functions: solving equations

Solve \(2\cos(2x)=\sqrt{3}\) for \(0\le x\le \pi\).

Answer and worked solution
Answer. \(x=\dfrac{\pi}{12}\) or \(x=\dfrac{11\pi}{12}\)
  1. Divide by 2: \(\cos(2x)=\dfrac{\sqrt3}{2}\).
  2. As \(0\le x\le\pi\), the argument runs over \(0\le 2x\le 2\pi\).
  3. \(\cos\theta=\dfrac{\sqrt3}{2}\) gives \(\theta=\dfrac{\pi}{6}\) or \(\theta=2\pi-\dfrac{\pi}{6}=\dfrac{11\pi}{6}\).
  4. Halve each: \(x=\dfrac{\pi}{12}\) or \(x=\dfrac{11\pi}{12}\).
Q2 · Algebra and function · Inverse functions, logarithmic functions

Let \(f(x)=\log_e(2x-4)+1\). (a) State the maximal domain of \(f\). (b) Find the rule for \(f^{-1}\) and state its domain.

Answer and worked solution
Answer. (a) \((2,\infty)\). (b) \(f^{-1}(x)=\dfrac{e^{x-1}}{2}+2\), domain \(\mathbb{R}\).
  1. (a) The logarithm needs \(2x-4\gt 0\), so \(x\gt 2\): domain \((2,\infty)\).
  2. (b) Swap and solve: \(x=\log_e(2y-4)+1\Rightarrow e^{x-1}=2y-4\Rightarrow y=\dfrac{e^{x-1}+4}{2}\).
  3. The range of \(f\) is \(\mathbb{R}\) (the log takes every real value), so \(\operatorname{dom} f^{-1}=\mathbb{R}\).
Q3 · Algebra and function · Transformations of functions

The graph of \(y=\sqrt{x}\) is reflected in the \(x\)-axis, dilated by a factor of 2 from the \(y\)-axis, then translated 3 units up. Find the rule of the image and its range.

Answer and worked solution
Answer. \(y=3-\sqrt{\dfrac{x}{2}}\), range \((-\infty,3]\)
  1. Track a general point \((t,\sqrt t)\). Reflection in the \(x\)-axis: \((t,-\sqrt t)\).
  2. Dilation by 2 from the \(y\)-axis doubles \(x\): \((2t,-\sqrt t)\). Translation up 3: \((2t,\,3-\sqrt t)\).
  3. So with \(X=2t\): \(Y=3-\sqrt{X/2}\), i.e. \(y=3-\sqrt{x/2}\) for \(x\ge0\).
  4. \(\sqrt{x/2}\ge 0\), so \(y\le 3\): range \((-\infty,3]\), with the endpoint at \((0,3)\).
Q4 · Algebra and function · Polynomials: factor theorem

The polynomial \(P(x)=x^3+ax^2+bx-6\) has \((x-1)\) and \((x+2)\) as factors. Find \(a\) and \(b\), then factorise \(P(x)\) fully.

Answer and worked solution
Answer. \(a=4,\ b=1\); \(P(x)=(x-1)(x+2)(x+3)\)
  1. \(P(1)=0\): \(1+a+b-6=0\Rightarrow a+b=5\).
  2. \(P(-2)=0\): \(-8+4a-2b-6=0\Rightarrow 2a-b=7\).
  3. Add the two equations: \(3a=12\), so \(a=4\) and \(b=1\).
  4. Product of roots of a monic cubic is \(-(-6)=6\): \(1\times(-2)\times r=6\Rightarrow r=-3\). So \(P(x)=(x-1)(x+2)(x+3)\).
Q5 · Algebra and function · Logarithm laws and equations

Solve \(2\log_3(x)-\log_3(x-2)=2\).

Answer and worked solution
Answer. \(x=3\) or \(x=6\)
  1. Domain first: both logs need \(x\gt 2\).
  2. Combine: \(\log_3\!\left(\dfrac{x^2}{x-2}\right)=2\Rightarrow \dfrac{x^2}{x-2}=9\).
  3. \(x^2=9x-18\Rightarrow x^2-9x+18=0\Rightarrow (x-3)(x-6)=0\).
  4. Both roots exceed 2, so both are valid: \(x=3\) or \(x=6\).
Q6 · Algebra and function · Exponential equations (quadratic form)

Solve \(4^x-6\cdot 2^x+8=0\) for \(x\).

Answer and worked solution
Answer. \(x=1\) or \(x=2\)
  1. Write \(4^x=(2^x)^2\) and let \(u=2^x\gt0\): \(u^2-6u+8=0\).
  2. \((u-2)(u-4)=0\), so \(u=2\) or \(u=4\).
  3. \(2^x=2\Rightarrow x=1\); \(2^x=4\Rightarrow x=2\).
Q7 · Algebra and function · Composite functions and domains

Let \(f(x)=x^2+1\) for \(x\in\mathbb{R}\) and \(g(x)=\sqrt{x-2}\) for \(x\ge 2\). Find the rule and maximal domain of \(g\circ f\).

Answer and worked solution
Answer. \(g(f(x))=\sqrt{x^2-1}\), domain \((-\infty,-1]\cup[1,\infty)\)
  1. \(g(f(x))=\sqrt{(x^2+1)-2}=\sqrt{x^2-1}\).
  2. For \(g\circ f\) to exist, the range of \(f\) must lie in the domain of \(g\): need \(f(x)\ge 2\), i.e. \(x^2+1\ge2\).
  3. \(x^2\ge 1\Rightarrow x\le -1\) or \(x\ge 1\).
Q8 · Algebra and function · Piecewise (hybrid) functions: continuity

Let \(f(x)=\begin{cases}x^2+kx, & x\lt 2\\ 3x-1, & x\ge 2\end{cases}\). Find \(k\) so that \(f\) is continuous at \(x=2\), and decide whether \(f\) is then differentiable at \(x=2\).

Answer and worked solution
Answer. \(k=\dfrac12\); not differentiable at \(x=2\) (left gradient \(4.5\), right gradient \(3\))
  1. Continuity: left limit \(=4+2k\), value \(=3(2)-1=5\). So \(4+2k=5\Rightarrow k=\dfrac12\).
  2. Gradients at 2: left \(2x+k=4+\tfrac12=4.5\); right \(3\).
  3. The gradients differ, so there is a corner: continuous but not differentiable at \(x=2\).
Q9 · Algebra and function · Simultaneous linear equations with a parameter

Consider \(mx+2y=6\) and \(3x+(m-1)y=6\). Find all values of \(m\) for which there is not exactly one solution, and say how many solutions each case has.

Answer and worked solution
Answer. \(m=3\): infinitely many solutions; \(m=-2\): no solution
  1. No unique solution when the gradients match: \(m(m-1)-2\cdot3=0\Rightarrow m^2-m-6=0\Rightarrow(m-3)(m+2)=0\).
  2. \(m=3\): equations are \(3x+2y=6\) and \(3x+2y=6\), the same line, so infinitely many solutions.
  3. \(m=-2\): equations are \(-2x+2y=6\), i.e. \(y-x=3\), and \(3x-3y=6\), i.e. \(y-x=-2\): parallel, no solution.

Calculus

Q10 · Calculus · Differentiation: product rule, stationary points

Let \(f(x)=xe^{-2x}\). Find the coordinates of the stationary point and determine its nature.

Answer and worked solution
Answer. \(\left(\dfrac12,\dfrac{1}{2e}\right)\), a local maximum
  1. Product rule: \(f'(x)=e^{-2x}-2xe^{-2x}=(1-2x)e^{-2x}\).
  2. \(f'(x)=0\Rightarrow x=\dfrac12\) (since \(e^{-2x}\gt0\)). \(f\left(\tfrac12\right)=\tfrac12e^{-1}=\dfrac{1}{2e}\).
  3. \(f'(x)\) is positive for \(x\lt\frac12\) and negative for \(x\gt\frac12\): a local maximum. (Check: \(f''(x)=(4x-4)e^{-2x}\), \(f''(\tfrac12)=-2e^{-1}\lt 0\).)
Q11 · Calculus · Differentiation: chain rule, tangent lines

Find the equation of the tangent to \(y=\log_e(x^2+1)\) at the point where \(x=1\).

Answer and worked solution
Answer. \(y=x-1+\log_e 2\)
  1. Chain rule: \(\dfrac{dy}{dx}=\dfrac{2x}{x^2+1}\). At \(x=1\) the gradient is \(\dfrac22=1\).
  2. Point: \(y=\log_e2\) at \(x=1\).
  3. Tangent: \(y-\log_e2=1\,(x-1)\Rightarrow y=x-1+\log_e2\).
Q12 · Calculus · Integration: reverse chain rule

Evaluate \(\displaystyle\int_0^1 x\,e^{x^2}\,dx\).

Answer and worked solution
Answer. \(\dfrac{e-1}{2}\)
  1. Since \(\dfrac{d}{dx}\left(e^{x^2}\right)=2xe^{x^2}\), an antiderivative of \(xe^{x^2}\) is \(\tfrac12e^{x^2}\).
  2. \(\left[\tfrac12e^{x^2}\right]_0^1=\tfrac12e-\tfrac12=\dfrac{e-1}{2}\).
Q13 · Calculus · Area between curves

Find the area of the region enclosed by \(y=x^2\) and \(y=6-x\).

Answer and worked solution
Answer. \(\dfrac{125}{6}\) square units
  1. Intersections: \(x^2=6-x\Rightarrow x^2+x-6=0\Rightarrow (x+3)(x-2)=0\), so \(x=-3\) and \(x=2\).
  2. On \((-3,2)\) the line is above the parabola. Area \(=\displaystyle\int_{-3}^{2}(6-x-x^2)\,dx\).
  3. \(=\left[6x-\tfrac{x^2}{2}-\tfrac{x^3}{3}\right]_{-3}^{2}=\left(12-2-\tfrac83\right)-\left(-18-\tfrac92+9\right)=\tfrac{22}{3}+\tfrac{27}{2}=\dfrac{125}{6}\).
Q14 · Calculus · Average value of a function

Find the average value of \(f(x)=e^{2x}\) on the interval \([0,\log_e 3]\).

Answer and worked solution
Answer. \(\dfrac{4}{\log_e 3}\)
  1. Average value \(=\dfrac{1}{b-a}\displaystyle\int_a^b f(x)\,dx=\dfrac{1}{\log_e3}\int_0^{\log_e3}e^{2x}dx\).
  2. \(\displaystyle\int_0^{\log_e3}e^{2x}dx=\left[\tfrac12e^{2x}\right]_0^{\log_e3}=\tfrac12\left(e^{2\log_e3}-1\right)=\tfrac12(9-1)=4\).
  3. So the average value is \(\dfrac{4}{\log_e3}\approx 3.64\).
Q15 · Calculus · Optimisation (maximum and minimum problems)

An open-topped box has a square base of side \(x\) cm and volume \(32\ \text{cm}^3\). Show the surface area is \(S=x^2+\dfrac{128}{x}\) and find the dimensions that minimise it, and the minimum area.

Answer and worked solution
Answer. Base \(4\times4\) cm, height \(2\) cm; minimum \(S=48\ \text{cm}^2\)
  1. Height \(h=\dfrac{32}{x^2}\). Surface = base \(x^2\) + four sides \(4xh=\dfrac{128}{x}\), so \(S=x^2+\dfrac{128}{x}\).
  2. \(S'(x)=2x-\dfrac{128}{x^2}=0\Rightarrow x^3=64\Rightarrow x=4\).
  3. \(S''(x)=2+\dfrac{256}{x^3}\gt0\), so this is a minimum. \(h=\dfrac{32}{16}=2\); \(S(4)=16+32=48\).
Q16 · Calculus · Antidifferentiation with an initial condition

A function has \(f'(x)=4e^{2x}-\sin(x)\) and \(f(0)=5\). Find \(f(x)\).

Answer and worked solution
Answer. \(f(x)=2e^{2x}+\cos(x)+2\)
  1. Antidifferentiate: \(f(x)=2e^{2x}+\cos(x)+c\) (since \(\frac{d}{dx}\cos x=-\sin x\)).
  2. \(f(0)=2+1+c=5\Rightarrow c=2\).
  3. \(f(x)=2e^{2x}+\cos(x)+2\).
Q17 · Calculus · Differentiation: circular functions

Let \(y=x\sin(2x)\). Find the gradient of the curve at \(x=\dfrac{\pi}{3}\).

Answer and worked solution
Answer. \(\dfrac{\sqrt3}{2}-\dfrac{\pi}{3}\)
  1. Product and chain rules: \(\dfrac{dy}{dx}=\sin(2x)+2x\cos(2x)\).
  2. At \(x=\tfrac\pi3\): \(\sin\tfrac{2\pi}{3}=\tfrac{\sqrt3}{2}\) and \(\cos\tfrac{2\pi}{3}=-\tfrac12\).
  3. Gradient \(=\tfrac{\sqrt3}{2}+2\cdot\tfrac\pi3\cdot\left(-\tfrac12\right)=\dfrac{\sqrt3}{2}-\dfrac\pi3\approx-0.18\).

Probability and statistics

Q18 · Probability and statistics · Binomial distribution

\(X\) is binomial with \(n=10\) and \(p=0.3\). Find \(\Pr(X\ge2)\), correct to four decimal places.

Answer and worked solution
Answer. \(0.8507\)
  1. Use the complement: \(\Pr(X\ge2)=1-\Pr(X=0)-\Pr(X=1)\).
  2. \(\Pr(X=0)=0.7^{10}\approx0.02825\) and \(\Pr(X=1)=10(0.3)(0.7)^9\approx0.12106\).
  3. \(\Pr(X\ge2)\approx1-0.02825-0.12106=0.8507\).
Q19 · Probability and statistics · Conditional probability

For events \(A\) and \(B\): \(\Pr(B)=0.5\), \(\Pr(A\mid B)=0.4\) and \(\Pr(A\cup B)=0.75\). Find \(\Pr(B\mid A)\).

Answer and worked solution
Answer. \(\dfrac49\)
  1. \(\Pr(A\cap B)=\Pr(A\mid B)\Pr(B)=0.4\times0.5=0.2\).
  2. \(\Pr(A\cup B)=\Pr(A)+\Pr(B)-\Pr(A\cap B)\Rightarrow 0.75=\Pr(A)+0.5-0.2\Rightarrow\Pr(A)=0.45\).
  3. \(\Pr(B\mid A)=\dfrac{0.2}{0.45}=\dfrac49\). (Not equal to \(\Pr(B)\), so \(A\) and \(B\) are not independent.)
Q20 · Probability and statistics · Continuous random variables: probability density functions

A continuous random variable \(X\) has pdf \(f(x)=kx(4-x)\) for \(0\le x\le4\), and 0 elsewhere. Find \(k\) and \(\Pr(X\lt1)\).

Answer and worked solution
Answer. \(k=\dfrac{3}{32}\); \(\Pr(X\lt1)=\dfrac{5}{32}\)
  1. Total probability is 1: \(\displaystyle\int_0^4 k(4x-x^2)dx=k\left[2x^2-\tfrac{x^3}{3}\right]_0^4=k\left(32-\tfrac{64}{3}\right)=\tfrac{32k}{3}=1\), so \(k=\dfrac3{32}\).
  2. \(\Pr(X\lt1)=\dfrac{3}{32}\left[2x^2-\tfrac{x^3}{3}\right]_0^1=\dfrac{3}{32}\left(2-\tfrac13\right)=\dfrac{3}{32}\cdot\dfrac53=\dfrac{5}{32}\).
Q21 · Probability and statistics · Mean and variance of a continuous random variable

For the pdf in the previous question, \(f(x)=\dfrac{3}{32}x(4-x)\) on \([0,4]\), find \(E(X)\) and \(\operatorname{Var}(X)\).

Answer and worked solution
Answer. \(E(X)=2\); \(\operatorname{Var}(X)=\dfrac45\)
  1. \(E(X)=\displaystyle\int_0^4 x f(x)\,dx=\tfrac3{32}\int_0^4(4x^2-x^3)dx=\tfrac3{32}\left(\tfrac{256}{3}-64\right)=\tfrac3{32}\cdot\tfrac{64}{3}=2\). (The pdf is symmetric about \(x=2\), which agrees.)
  2. \(E(X^2)=\tfrac3{32}\displaystyle\int_0^4(4x^3-x^4)dx=\tfrac3{32}\left(256-\tfrac{1024}{5}\right)=\tfrac3{32}\cdot\tfrac{256}{5}=\tfrac{24}{5}\).
  3. \(\operatorname{Var}(X)=E(X^2)-[E(X)]^2=\tfrac{24}{5}-4=\dfrac45\).
Q22 · Probability and statistics · Normal distribution: finding a standard deviation

\(X\) is normally distributed with mean 50 and \(\Pr(X\gt60)=0.2\). Find the standard deviation, correct to two decimal places.

Answer and worked solution
Answer. \(\sigma\approx11.88\)
  1. \(\Pr(X\gt60)=0.2\Rightarrow\Pr(X\lt60)=0.8\), so the \(z\)-value of 60 is \(\text{invNorm}(0.8)\approx0.8416\).
  2. \(z=\dfrac{60-50}{\sigma}\Rightarrow\sigma=\dfrac{10}{0.8416}\approx11.88\).
Q23 · Probability and statistics · Normal distribution: probabilities

Scores on a test are normally distributed with mean 100 and standard deviation 15. Find the probability that a score lies between 85 and 130, correct to four decimal places.

Answer and worked solution
Answer. \(0.8186\)
  1. Standardise: \(z=\dfrac{85-100}{15}=-1\) and \(z=\dfrac{130-100}{15}=2\).
  2. \(\Pr(-1\lt Z\lt2)=\Pr(Z\lt2)-\Pr(Z\lt-1)\approx0.97725-0.15866=0.8186\).
Q24 · Probability and statistics · Confidence intervals for a population proportion

In a random sample of 200 students, 70 say they study with a timer. Use \(z=1.96\) to find an approximate 95% confidence interval for the population proportion, with endpoints to three decimal places.

Answer and worked solution
Answer. \((0.284,\ 0.416)\)
  1. \(\hat p=\dfrac{70}{200}=0.35\).
  2. Standard error \(=\sqrt{\dfrac{\hat p(1-\hat p)}{n}}=\sqrt{\dfrac{0.35\times0.65}{200}}\approx0.03373\).
  3. Interval: \(0.35\pm1.96(0.03373)=0.35\pm0.0661\), i.e. \((0.284,\,0.416)\).
Q25 · Probability and statistics · Sample proportions: distribution of \(\hat P\)

A population has proportion \(p=0.4\). Random samples of size 100 are taken. Find the mean and the standard deviation of the sample proportion \(\hat P\).

Answer and worked solution
Answer. Mean \(0.4\); standard deviation \(\sqrt{0.0024}\approx0.0490\)
  1. \(E(\hat P)=p=0.4\).
  2. \(\operatorname{sd}(\hat P)=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.4\times0.6}{100}}=\sqrt{0.0024}\approx0.0490\).
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