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Question 1 [2021 Exam 2 Section A Q1]

The period of the function with rule \( y = \tan\left(\frac{\pi x}{2}\right) \) is

  • A. 1
  • B. 2
  • C. 4
  • D. 2π
  • E. 4π
Correct answer: B
67% of students across the state got this question right.
State distribution: A: 3% | B: 67% | C: 22% | D: 6% | E: 2% | N/A: 0%
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Question 2 [2021 Exam 2 Section A Q2]

The graph of \( y = \log_e(x) + \log_e(2x) \), where \( x > 0 \), is identical, over the same domain, to the graph of

  • A. \( y = 2\log_e\left(\frac{1}{2}x\right) \)
  • B. \( y = 2\log_e(2x) \)
  • C. \( y = \log_e(2x^2) \)
  • D. \( y = \log_e(3x) \)
  • E. \( y = \log_e(4x) \)
Correct answer: C
81% of students across the state got this question right.
State distribution: A: 3% | B: 11% | C: 81% | D: 4% | E: 1% | N/A: 0%
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Question 3 [2021 Exam 2 Section A Q3]

A box contains many coloured glass beads.
A random sample of 48 beads is selected and it is found that the proportion of blue-coloured beads in this sample is 0.125
Based on this sample, a 95% confidence interval for the proportion of blue-coloured glass beads is

  • A. (0.0314, 0.2186)
  • B. (0.0465, 0.2035)
  • C. (0.0018, 0.2482)
  • D. (0.0896, 0.1604)
  • E. (0.0264, 0.2136)
Correct answer: A
72% of students across the state got this question right.
State distribution: A: 72% | B: 7% | C: 6% | D: 8% | E: 5% | N/A: 1%
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Question 4 [2021 Exam 2 Section A Q4]

The maximum value of the function \( h: [0, 2] \to R, h(x) = (x-2)e^x \) is

  • A. \(-e\)
  • B. 0
  • C. 1
  • D. 2
  • E. \(e\)
Correct answer: B
58% of students across the state got this question right.
State distribution: A: 10% | B: 58% | C: 9% | D: 18% | E: 4% | N/A: 0%
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Question 5 [2021 Exam 2 Section A Q5]

Consider the following four functional relations.

\( f(x) = f(-x) \quad -f(x) = f(-x) \quad f(x) = -f(x) \quad (f(x))^2 = f(x^2) \)

The number of these functional relations that are satisfied by the function \( f: R \to R, f(x) = x \) is

  • A. 0
  • B. 1
  • C. 2
  • D. 3
  • E. 4
Correct answer: C
73% of students across the state got this question right.
State distribution: A: 7% | B: 12% | C: 73% | D: 7% | E: 2% | N/A: 0%
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Question 6 [2021 Exam 2 Section A Q6]

The probability of winning a game is 0.25
The probability of winning a game is independent of winning any other game.
If Ben plays 10 games, the probability that he will win exactly four times is closest to

  • A. 0.1460
  • B. 0.2241
  • C. 0.9219
  • D. 0.0781
  • E. 0.7759
Correct answer: A
88% of students across the state got this question right.
State distribution: A: 88% | B: 4% | C: 2% | D: 3% | E: 2% | N/A: 0%
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Question 7 [2021 Exam 2 Section A Q7]

The tangent to the graph of \( y = x^3 - ax^2 + 1 \) at \( x = 1 \) passes through the origin.
The value of \(a\) is

  • A. \( \frac{1}{2} \)
  • B. 1
  • C. \( \frac{3}{2} \)
  • D. 2
  • E. \( \frac{5}{2} \)
Correct answer: B
56% of students across the state got this question right.
State distribution: A: 4% | B: 56% | C: 20% | D: 16% | E: 3% | N/A: 1%
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Question 8 [2021 Exam 2 Section A Q8]

The graph of the function \(f\) is shown below.

Graph of the function f

The graph corresponding to \(f'\) is

Graphs for options A, B, C, D, and E
Correct answer: E
40% of students across the state got this question right.
State distribution: A: 19% | B: 9% | C: 3% | D: 29% | E: 40% | N/A: 0%

The gradient is decreasing and positive over the interval \( (a,\infty) \).
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Question 9 [2021 Exam 2 Section A Q9]

Let \( g(x) = x + 2 \) and \( f(x) = x^2 - 4 \).
If h is the composite function given by \( h: [-5, -1) \to R, h(x) = f(g(x)) \), then the range of \(h\) is

  • A. (-3, 5]
  • B. [-3, 5)
  • C. (-3, 5)
  • D. (-4, 5]
  • E. [-4, 5]
Correct answer: E
56% of students across the state got this question right.
State distribution: A: 24% | B: 7% | C: 4% | D: 10% | E: 56% | N/A: 0%
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Question 10 [2021 Exam 2 Section A Q10]

Consider the functions \( f(x) = \sqrt{x+2} \) and \( g(x) = \sqrt{1-2x} \), defined over their maximal domains.
The maximal domain of the function \(h = f + g\) is

  • A. \( \left[-2, \frac{1}{2}\right) \)
  • B. \( [-2, \infty) \)
  • C. \( (-\infty, -2] \cup \left[\frac{1}{2}, \infty\right) \)
  • D. \( \left[-2, \frac{1}{2}\right] \)
  • E. [-2, 1]
Correct answer: D
70% of students across the state got this question right.
State distribution: A: 16% | B: 4% | C: 7% | D: 70% | E: 3% | N/A: 0%
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Question 11 [2021 Exam 2 Section A Q11]

If \( \int_0^a f(x)dx = k \), then \( \int_0^a (3f(x) + 2)dx \) is

  • A. \(3k + 2a\)
  • B. \(3k\)
  • C. \(k + 2a\)
  • D. \(k + 2\)
  • E. \(3k + 2\)
Correct answer: A
67% of students across the state got this question right.
State distribution: A: 67% | B: 6% | C: 5% | D: 2% | E: 20% | N/A: 0%
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Question 12 [2021 Exam 2 Section A Q12]

For a certain species of bird, the proportion of birds with a crest is known to be \( \frac{3}{5} \).
Let \( \hat{P} \) be the random variable representing the proportion of birds with a crest in samples of size \(n\) for this specific bird.
The smallest sample size for which the standard deviation of \( \hat{P} \) is less than 0.08 is

  • A. 7
  • B. 27
  • C. 37
  • D. 38
  • E. 43
Correct answer: D
54% of students across the state got this question right.
State distribution: A: 6% | B: 14% | C: 23% | D: 54% | E: 2% | N/A: 1%
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Question 13 [2021 Exam 2 Section A Q13]

The value of an investment, in dollars, after n months can be modelled by the function

\( f(n) = 2500 \times (1.004)^n \)

where \( n \in \{0, 1, 2, ...\} \).
The average rate of change of the value of the investment over the first 12 months is closest to

  • A. $10.00 per month.
  • B. $10.20 per month.
  • C. $10.50 per month.
  • D. $125.00 per month.
  • E. $127.00 per month.
Correct answer: B
80% of students across the state got this question right.
State distribution: A: 4% | B: 80% | C: 9% | D: 5% | E: 2% | N/A: 0%
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Question 14 [2021 Exam 2 Section A Q14]

A value of \(k\) for which the average value of \( y = \cos\left(kx - \frac{\pi}{2}\right) \) over the interval [0, π] is equal to the average value of \( y= \sin(x)\) over the same interval is

  • A. \( \frac{1}{6} \)
  • B. \( \frac{1}{5} \)
  • C. \( \frac{1}{4} \)
  • D. \( \frac{1}{3} \)
  • E. \( \frac{1}{2} \)
Correct answer: E
63% of students across the state got this question right.
State distribution: A: 6% | B: 6% | C: 14% | D: 10% | E: 63% | N/A: 1%
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Question 15 [2021 Exam 2 Section A Q15]

Four fair coins are tossed at the same time.
The outcome for each coin is independent of the outcome for any other coin.
The probability that there is an equal number of heads and tails, given that there is at least one head, is

  • A. \( \frac{1}{2} \)
  • B. \( \frac{1}{3} \)
  • C. \( \frac{3}{4} \)
  • D. \( \frac{2}{5} \)
  • E. \( \frac{4}{7} \)
Correct answer: D
48% of students across the state got this question right.
State distribution: A: 14% | B: 11% | C: 18% | D: 48% | E: 8% | N/A: 1%

\(X\sim\mathrm{Bi}(4,\tfrac12)\), and \(\Pr(X=2\mid X\ge1)=\frac{\Pr(X=2)}{\Pr(X\ge1)}=\frac{0.375}{0.9375}=\frac25\).
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Question 16 [2021 Exam 2 Section A Q16]

Let \( \cos(x) = \frac{3}{5} \) and \( \sin^2(y) = \frac{25}{169} \), where \( x \in \left[\frac{3\pi}{2}, 2\pi\right] \) and \( y \in \left[\frac{3\pi}{2}, 2\pi\right] \).
The value of \(\sin(x) + \cos(y)\) is

  • A. \( \frac{8}{65} \)
  • B. \( -\frac{112}{65} \)
  • C. \( \frac{112}{65} \)
  • D. \( -\frac{8}{65} \)
  • E. \( \frac{64}{65} \)
Correct answer: A
31% of students across the state got this question right.
State distribution: A: 31% | B: 11% | C: 32% | D: 9% | E: 15% | N/A: 1%

\(\cos x=\frac35\), \(\sin y=-\frac5{13}\), \(\cos y=\frac{12}{13}\), and \(\sin x=-\frac45\). Hence \(\sin x+\cos y=-\frac45+\frac{12}{13}=\frac8{65}\).
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Question 17 [2021 Exam 2 Section A Q17]

A discrete random variable \(X\) has a binomial distribution with a probability of success of \(p = 0.1\) for \(n\) trials, where \(n > 2\).
If the probability of obtaining at least two successes after \(n\) trials is at least 0.5, then the smallest possible value of \(n\) is

  • A. 15
  • B. 16
  • C. 17
  • D. 18
  • E. 19
Correct answer: C
57% of students across the state got this question right.
State distribution: A: 9% | B: 18% | C: 57% | D: 10% | E: 5% | N/A: 1%
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Question 18 [2021 Exam 2 Section A Q18]

Let \( f: R \to R, f(x) = (2x-1)(2x+1)(3x-1) \) and \( g: (-\infty, 0) \to R, g(x) = x \log_e(-x) \).
The maximum number of solutions for the equation \( f(x-k) = g(x) \), where \( k \in R \), is

  • A. 0
  • B. 1
  • C. 2
  • D. 3
  • E. 4
Correct answer: D
39% of students across the state got this question right.
State distribution: A: 8% | B: 23% | C: 22% | D: 39% | E: 6% | N/A: 1%

With \(f(x)=(2x-1)(2x+1)(3x-1)\) and \(g(x)=x\log_e(-x)\), \(f(x-k)=g(x)\) can have a maximum of three solutions when the graph of \(f\) is translated to the left; the report gives \(k=-1\) as an example.
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Question 19 [2021 Exam 2 Section A Q19]

Which one of the following functions is differentiable for all real values of \(x\)?

  • A. \( f(x) = \begin{cases} x & x < 0 \\ -x & x \ge 0 \end{cases} \)

  • B. \( f(x) = \begin{cases} x & x < 0 \\ -x & x > 0 \end{cases} \)

  • C. \( f(x) = \begin{cases} 8x+4 & x < 0 \\ (2x+1)^2 & x \ge 0 \end{cases} \)

  • D. \( f(x) = \begin{cases} 2x+1 & x < 0 \\ (2x+1)^2 & x \ge 0 \end{cases} \)

  • E. \( f(x) = \begin{cases} 4x+1 & x < 0 \\ (2x+1)^2 & x \ge 0 \end{cases} \)
Correct answer: E
35% of students across the state got this question right.
State distribution: A: 12% | B: 13% | C: 19% | D: 20% | E: 35% | N/A: 1%

For option E, \(f(x)=4x+1\) for \(x<0\) and \(f(x)=(2x+1)^2\) for \(x\ge0\). The one-sided function limits at \(0\) are both 1 and the one-sided derivative limits are both 4, so the function is differentiable for all real \(x\).
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Question 20 [2021 Exam 2 Section A Q20]

Let \(A\) and \(B\) be two independent events from a sample space.
If \(\Pr(A) = p\), \(\Pr(B) = p²\) and \(\Pr(A) + \Pr(B) = 1\), then \(\Pr(A' ∪ B)\) is equal to

  • A. \(1 - p - p²\)
  • B. \(p² - p³\)
  • C. \(p - p³\)
  • D. \(1 - p + p³\)
  • E. \(1 - p - p² + p³\)
Correct answer: D
39% of students across the state got this question right.
State distribution: A: 20% | B: 17% | C: 11% | D: 39% | E: 12% | N/A: 1%

\(\Pr(A\cap B)=\Pr(A)\Pr(B)=p^3\). Hence \(\Pr(A\cap B^{\prime})=p-p^3\), so \(\Pr(A^{\prime}\cup B)=1-\Pr(A\cap B^{\prime})=1-p+p^3\).
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Question 1 [2021 Exam 2 Section B Q1]

A rectangular sheet of cardboard has a width of \(h\) centimetres. Its length is twice its width. Squares of side length \(x\) centimetres, where \(x > 0\), are cut from each of the corners, as shown in the diagram below.

Diagram of a rectangular sheet of cardboard with squares cut from the corners.

The sides of this sheet of cardboard are then folded up to make a rectangular box with an open top, as shown in the diagram below. Assume that the thickness of the cardboard is negligible and that \(V_{box} > 0\).

Diagram of an open-top rectangular box.

A box is to be made from a sheet of cardboard with \(h = 25\) cm.

a. Show that the volume, \(V_{box}\), in cubic centimetres, is given by \(V_{box}(x) = 2x(25 – 2x)(25 – x)\). 1 mark

Answer
\(V=x(h-2x)(2h-2x)=x(25-2x)(50-2x)=2x(25-2x)(25-x)\).

State performance
0 marks: 29%
1 marks: 71%
State average: 0.7 / 1
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b. State the domain of \(V_{box}\). 1 mark

Answer
\((0,12.5)\).

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1 marks: 42%
State average: 0.4 / 1
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c. Find the derivative of \(V_{box}\) with respect to \(x\). 1 mark

Answer
\(V_{box}^{\prime}(x)=12x^2-300x+1250\).

State performance
0 marks: 10%
1 marks: 90%
State average: 0.9 / 1
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d. Calculate the maximum possible volume of the box and for which value of \(x\) this occurs. 3 marks

Answer
Solve \(V^{\prime}(x)=0\): \(x=-\frac{25(\sqrt3-3)}6=-\frac{25\sqrt3}{6}+\frac{25}{2}\). Then \(V_{\max}=\frac{15625\sqrt3}{9}\).

State performance
0 marks: 17%
1 marks: 13%
2 marks: 21%
3 marks: 49%
State average: 2.0 / 3
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e. Waste minimisation is a goal when making cardboard boxes. Percentage wasted is based on the area of the sheet of cardboard that is cut out before the box is made. Find the percentage of the sheet of cardboard that is wasted when \(x = 5\). 2 marks

Answer
\(\frac{4\times 5^2}{25\times 50}\times100\%=8\%\).

State performance
0 marks: 35%
1 marks: 19%
2 marks: 46%
State average: 1.1 / 2
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Now consider a box made from a rectangular sheet of cardboard where \(h > 0\) and the box's length is still twice its width.

f.

i. Let \(V_{box}\) be the function that gives the volume of the box. State the domain of \(V_{box}\) in terms of \(h\). 1 mark

Answer
\((0,\frac h2)\).

State performance
0 marks: 67%
1 marks: 33%
State average: 0.4 / 1
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ii. Find the maximum volume for any such rectangular box, \(V_{box}\), in terms of \(h\). 3 marks

Answer
\(V=x(h-2x)(2h-2x)\), \(V^{\prime}(x)=0\), so \(x=-\frac{h(\sqrt3-3)}6\), and \(V_{\max}=\frac{\sqrt3h^3}{9}\).

State performance
0 marks: 42%
1 marks: 13%
2 marks: 12%
3 marks: 33%
State average: 1.4 / 3
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g. Now consider making a box from a square sheet of cardboard with side lengths of \(h\) centimetres. Show that the maximum volume of the box occurs when \(x = \frac{h}{6}\). 2 marks

Answer
\(V=x(h-2x)^2\). From \(V^{\prime}(x)=0\), \(x=\frac h2\) or \(x=\frac h6\). Since the domain is \((0,\frac h2)\), the maximum occurs at \(x=\frac h6\).

State performance
0 marks: 55%
1 marks: 10%
2 marks: 35%
State average: 0.8 / 2
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Question 2 [2021 Exam 2 Section B Q2]

Four rectangles of equal width are drawn and used to approximate the area under the parabola \(y = x^2\) from \(x = 0\) to \(x = 1\). The heights of the rectangles are the values of the graph of \(y = x^2\) at the right endpoint of each rectangle, as shown in the graph below.

Graph of y=x^2 from x=0 to x=1 with four upper rectangles approximating the area.

a. State the width of each of the rectangles shown above. 1 mark

Answer
\(0.25\).

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0 marks: 4%
1 marks: 96%
State average: 1.0 / 1
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b. Find the total area of the four rectangles shown above. 1 mark

Answer
\(\frac{15}{32}\).

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1 marks: 60%
State average: 0.6 / 1
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c. Find the area between the graph of \(y = x^2\), the x-axis and the line \(x = 1\). 2 marks

Answer
\(\int_0^1 x^2\,dx=\frac13\).

State performance
0 marks: 14%
1 marks: 6%
2 marks: 80%
State average: 1.7 / 2
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d. The graph of \(f\) is shown below.

Graph of a function f from x=-3 to x=3.

Approximate \(\int_{-2}^{2} f(x)dx\) using four rectangles of equal width and the right endpoint of each rectangle. 1 mark

Answer
\(-2\).

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1 marks: 16%
State average: 0.2 / 1
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e. Parts of the graphs of \(y = x^2\) and \(y = \sqrt{x}\) are shown below.

Shaded area between the graphs of y=x^2 and y=sqrt(x) from x=0 to x=1.

Find the area of the shaded region. 1 mark

Answer
\(\frac13\).

State performance
0 marks: 12%
1 marks: 88%
State average: 0.9 / 1
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f. The graph of \(y = x^2\) is transformed to the graph of \(y = ax^2\), where \(a \in (0, 2]\). Find the values of \(a\) such that the area defined by the region(s) bounded by the graphs of \(y = ax^2\) and \(y = \sqrt{x}\) and the lines \(x = 0\) and \(x = a\) is equal to \(\frac{1}{3}\). Give your answer correct to two decimal places. 4 marks

Answer
\(a=1.00\). If \(a<1\): \(\int_0^a(\sqrt{x}-ax^2)\,dx=\frac13\), giving \(a=0.77\). If \(a>1\): \(\int_0^{a^{-2/3}}(\sqrt{x}-ax^2)\,dx+\int_{a^{-2/3}}^a(ax^2-\sqrt{x})\,dx=\frac13\), giving \(a=1.13\).

State performance
0 marks: 48%
1 marks: 42%
2 marks: 3%
3 marks: 5%
4 marks: 2%
State average: 0.7 / 4
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Question 3 [2021 Exam 2 Section B Q3]

Let \(q(x) = \log_e(x^2 - 1) - \log_e(1 - x)\).

a. State the maximal domain and the range of \(q\). 2 marks

Answer
Domain \(( -\infty,-1)\); range \(\mathbb{R}\).

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1 marks: 37%
2 marks: 36%
State average: 1.1 / 2
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b.

i. Find the equation of the tangent to the graph of \(q\) when \(x = -2\). 1 mark

Answer
\(y=-x-2\).

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1 marks: 74%
State average: 0.8 / 1
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ii. Find the equation of the line that is perpendicular to the graph of \(q\) when \(x = -2\) and passes through the point \((-2, 0)\). 1 mark

Answer
\(y=x+2\).

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1 marks: 65%
State average: 0.7 / 1
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Let \(p(x) = e^{-2x} - 2e^{-x} + 1\).

c. Explain why \(p\) is not a one-to-one function. 1 mark

Answer
\(p\) fails the horizontal line test (equivalently, it is many-to-one: there are two \(x\)-values for some \(y\)-values).

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1 marks: 66%
State average: 0.7 / 1
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d. Find the gradient of the tangent to the graph of \(p\) at \(x = a\). 1 mark

Answer
\(p^{\prime}(a)=2(e^a-1)e^{-2a}\).

State performance
0 marks: 33%
1 marks: 67%
State average: 0.7 / 1
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The diagram below shows parts of the graph of \(p\) and the line \(y = x + 2\).

Graph of p(x) and the line y=x+2.

The line \(y = x + 2\) and the tangent to the graph of \(p\) at \(x = a\) intersect with an acute angle of \(\theta\) between them.

e. Find the value(s) of \(a\) for which \(\theta = 60^\circ\). Give your answer(s) correct to two decimal places. 3 marks

Answer
\(p^{\prime}(a)=-\tan(75^\circ)=\tan(105^\circ)\), giving \(a=-0.67\); and \(p^{\prime}(a)=-\tan(15^\circ)=\tan(165^\circ)\), giving \(a=-0.11\).

State performance
0 marks: 83%
1 marks: 4%
2 marks: 9%
3 marks: 4%
State average: 0.4 / 3
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f. Find the x-coordinate of the point of intersection between the line \(y = x + 2\) and the graph of \(p\), and hence find the area bounded by \(y = x + 2\), the graph of \(p\) and the x-axis, both correct to three decimal places. 3 marks

Answer
\(x=-0.750\), and \(\int_{-2}^{-0.750\ldots}(x+2)\,dx+\int_{-0.750\ldots}^{0}p(x)\,dx=1.038\).

State performance
0 marks: 41%
1 marks: 23%
2 marks: 7%
3 marks: 29%
State average: 1.3 / 3
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Question 4 [2021 Exam 2 Section B Q4]

A teacher coaches their school's table tennis team. The teacher has an adjustable ball machine that they use to help the players practise. The speed, measured in metres per second, of the balls shot by the ball machine is a normally distributed random variable \(W\). The teacher sets the ball machine with a mean speed of 10 metres per second and a standard deviation of 0.8 metres per second.

a. Determine \(\Pr(W \ge 11)\), correct to three decimal places. 1 mark

Answer
\(0.106\).

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1 marks: 78%
State average: 0.8 / 1
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b. Find the value of \(k\), in metres per second, which 80% of ball speeds are below. Give your answer in metres per second, correct to one decimal place. 1 mark

Answer
\(10.7\) metres per second.

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0 marks: 36%
1 marks: 64%
State average: 0.7 / 1
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The teacher adjusts the height setting for the ball machine. The machine now shoots balls high above the table tennis table. Unfortunately, with the new height setting, 8% of balls do not land on the table. Let \(\hat{P}\) be the random variable representing the sample proportion of balls that do not land on the table in random samples of 25 balls.

c. Find the mean and the standard deviation of \(\hat{P}\). 2 marks

Answer
\(E(\hat P)=0.08=\frac{2}{25}\), and \(\operatorname{sd}(\hat P)=\frac{\sqrt{46}}{125}\).

State performance
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1 marks: 24%
2 marks: 33%
State average: 0.9 / 2
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d. Use the binomial distribution to find \(\Pr(\hat{P} > 0.1)\), correct to three decimal places. 2 marks

Answer
\(X\sim\mathrm{Bi}(25,0.08)\). Since \(\hat P>0.1\) is equivalent to \(X>2.5\), \(\Pr(X\ge3)=0.323\).

State performance
0 marks: 42%
1 marks: 18%
2 marks: 40%
State average: 1.0 / 2
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The teacher can also adjust the spin setting on the ball machine. The spin, measured in revolutions per second, is a continuous random variable \(X\) with the probability density function \[ f(x) = \begin{cases} \frac{x}{500} & 0 \le x < 20 \\ \frac{50-x}{750} & 20 \le x \le 50 \\ 0 & \text{elsewhere} \end{cases} \]

e. Find the maximum possible spin applied by the ball machine, in revolutions per second. 1 mark

Answer
\(50\) revolutions per second.

State performance
0 marks: 79%
1 marks: 21%
State average: 0.2 / 1
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f. Find the median spin, in revolutions per second, correct to one decimal place. 2 marks

Answer
For the median \(m\), \(\int_0^m f(x)\,dx=\frac12\). Equivalently, \(\int_{20}^m\frac{50-x}{750}\,dx=0.1\). Hence \(m=22.6\) revolutions per second.

State performance
0 marks: 51%
1 marks: 11%
2 marks: 37%
State average: 0.9 / 2
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g. Find the standard deviation of the spin, in revolutions per second, correct to one decimal place. 3 marks

Answer
\(\sigma=\sqrt{\int_0^{50}x^2f(x)\,dx-\left(\int_0^{50}xf(x)\,dx\right)^2}=10.3\) revolutions per second.

State performance
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1 marks: 9%
2 marks: 6%
3 marks: 32%
State average: 1.2 / 3
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h. The teacher adjusts the spin setting so that the median spin becomes 30 revolutions per second. This will transform the original probability density function \(f\) to a new probability density function \(g\), where \(g(x) = af(\frac{x}{b})\). Find the values of \(a\) and \(b\) for which the new median spin is 30 revolutions per second, giving your answer correct to two decimal places. 2 marks

Answer
Use \(\int_0^{30}a f(\frac{x}{b})\,dx=\frac12\) and \(\int_0^{50b}a f(\frac{x}{b})\,dx=1\). This gives \(a=0.75\) and \(b=1.33\).

State performance
0 marks: 86%
1 marks: 12%
2 marks: 2%
State average: 0.2 / 2
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Question 5 [2021 Exam 2 Section B Q5]

Part of the graph of \(f: \mathbb{R} \rightarrow \mathbb{R}, f(x) = \sin(\frac{x}{2}) + \cos(2x)\) is shown below.

Graph of the function f(x) = sin(x/2) + cos(2x).

a. State the period of \(f\). 1 mark

Answer
\(4\pi\).

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1 marks: 71%
State average: 0.7 / 1
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b. State the minimum value of \(f\), correct to three decimal places. 1 mark

Answer
\(-1.722\).

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1 marks: 61%
State average: 0.6 / 1
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c. Find the smallest positive value of \(h\) for which \(f(h - x) = f(x)\). 1 mark

Answer
\(2\pi\).

State performance
0 marks: 79%
1 marks: 21%
State average: 0.2 / 1
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Consider the set of functions of the form \(g_a: \mathbb{R} \rightarrow \mathbb{R}, g_a(x) = \sin(\frac{x}{a}) + \cos(ax)\), where \(a\) is a positive integer.

d. State the value of \(a\) such that \(g_a(x) = f(x)\) for all \(x\). 1 mark

Answer
\(a=2\).

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1 marks: 68%
State average: 0.7 / 1
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e.

i. Find an antiderivative of \(g_a\) in terms of \(a\). 1 mark

Answer
\(-a\cos(\frac{x}{a})+\frac{\sin(ax)}{a}\).

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1 marks: 50%
State average: 0.5 / 1
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ii. Use a definite integral to show that the area bounded by \(g_a\) and the x-axis over the interval \([0, 2a\pi]\) is equal above and below the x-axis for all values of \(a\). 3 marks

Answer
\(\int_0^{2a\pi}g_a(x)\,dx=\frac{\sin(2a^2\pi)}{a}=0\) for all positive integers \(a\). Therefore the signed areas above and below the x-axis are equal in magnitude.

State performance
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2 marks: 16%
3 marks: 12%
State average: 0.9 / 3
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f. Explain why the maximum value of \(g_a\) cannot be greater than 2 for all values of \(a\) and why the minimum value of \(g_a\) cannot be less than –2 for all values of \(a\). 1 mark

Answer
\(\sin(kx)\) and \(\cos(kx)\) each have maximum 1 and minimum -1 for all real \(k\). Therefore their sum cannot exceed 2 or be less than -2.

State performance
0 marks: 87%
1 marks: 13%
State average: 0.2 / 1
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g. Find the greatest possible minimum value of \(g_a\). 1 mark

Answer
\(-\sqrt2\).

State performance
0 marks: 98%
1 marks: 2%
State average: 0.0 / 1
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