Skill Align

Skill Align QCE Mathematical Methods Paper 2 - Free Online Pack 0

Full-length Units 3&4 external-assessment-style showcase paper

Paper: Paper 2 Technology-active Question and Response Book Showcase
Reading: 5 minutes perusal
Writing: 90 minutes
Assessment: 55 marks

QCAA formula book provided; QCAA-approved technology is permitted for Paper 2. No public PDF or formula-book download is supplied with Pack 0.

Section 1

Questions 1-10 are multiple choice. Select the best answer for each question. Technology may be used where appropriate.

Question 1

1 mark

The solution of e^(0.4t)=5 is

t=((ln5) / (0.4))
t=0.4ln5
t=ln4.6
t=((0.4) / (ln5))

Question 2

1 mark

If Xsim N(50,6²), then P(X<41) is closest to

0.0228
0.0668
0.1587
0.9332

Question 3

1 mark

((d) / (dx))ln(x²+1)=

((1) / (x²+1))
2xln(x²+1)
((2x) / (x²+1))
((x²+1) / (2x))

Question 4

1 mark

The period of H(t)=200+50cos(((pi t) / (4))) is

Graph Preview

Line-graph diagram is missing enough points.

Question 5

1 mark

If Xsim B(8,0.2), then P(Xle1)=

(0.8)⁸+8(0.2)(0.8)⁷
8(0.2)(0.8)⁷
1-(0.8)⁸
(0.2)⁸+(0.8)⁸

Question 6

1 mark

For hat p=0.55, n=200 and z=1.96, the margin of error is closest to

0.035
0.069
0.055
0.096

Question 7

1 mark

The value of int_0^1e^x,dx is

1
e
e-1
ln e

Question 8

1 mark

A value of 74 from a normal distribution with mean 68 and standard deviation 4 has z-score

Graph Preview

0.67
4
6
1.5

Question 9

1 mark

For A=120(1.06)^t, the doubling time is

((ln2) / (ln1.06))
((ln1.06) / (ln2))
2ln1.06
120ln2

Question 10

1 mark

A continuous random variable is uniformly distributed on [2,8]. Its mean is

Section 2

Questions 11-19 are short response. Show working, modelling decisions and contextual interpretation.

Question 11

6 marks

A function satisfies f'(x)=3sin x+2cos(2x) and f(0)=1.

(a) 4 marks

Determine f(x).

(b) 2 marks

Determine the equation of the tangent at x=fracpi2.

Question 12

7 marks

An open cylindrical container must have volume 500pitext( cm)^3. Its base costs 4 cents per square centimetre and its curved side costs 2 cents per square centimetre. Let the radius be r cm and the height be h cm.

(a) 2 marks

Show that the cost, in dollars, can be modelled by C(r)=0.04pi r²+((20pi) / (r)), for r>0.

(b) 3 marks

Determine the radius and height that minimise the cost. Give both dimensions to two decimal places and justify that the cost is a minimum.

(c) 2 marks

A manufacturer can use only a radius of 6 cm or 7 cm, adjusting the height to retain the required volume. Recommend the cheaper radius and support the recommendation with costs.

Question 13

4 marks

In a pilot random sample of 80 voters, 32 supported a proposal. A larger survey will use a 95% approximate confidence interval with z=1.96.

(a) 1 mark

Determine the pilot estimate hat p.

(b) 2 marks

Using the pilot estimate, determine the minimum sample size needed for a margin of error no greater than 0.04.

(c) 1 mark

State one condition needed for the confidence interval to support inference about all voters.

Question 14

6 marks

Let f(x)=ln(x+1), for xge0.

(a) 1 mark

Solve f(x)=0.8, giving x to three decimal places.

(b) 2 marks

Determine the equation of the tangent to y=f(x) at x=1.

(c) 3 marks

Determine the exact area enclosed by the curve, its tangent from part (b), and the lines x=0 and x=1.

Question 15

3 marks

Battery life X, in hours, is modelled by Xsim N(120,15²).

(a) 1 mark

Determine P(X<100), to four decimal places.

(b) 2 marks

In a shipment of 250 batteries, estimate how many are expected to last between 100 and 140 hours.

Question 16

4 marks

A circular revegetation patch has radius r(t)=1+0.3t metres, where t is measured in years. Its area is A=pi r^2.

(a) 2 marks

Derive an expression for ((dA) / (dt)) in terms of r.

(b) 1 mark

Determine the rate of change of area when the radius is 4 metres.

(c) 1 mark

Determine when the radius first reaches 4 metres.

Question 17

4 marks

A sensor has an independent probability of 0.08 of failing an inspection. A batch contains 20 sensors and is replaced if at least three sensors fail.

(a) 2 marks

Determine the probability that a batch is replaced, to four decimal places.

(b) 2 marks

Assuming batches are independent, determine the probability that at least one of the next 10 batches is replaced.

Question 18

5 marks

Two surveys estimate community support for a proposal. Survey A randomly samples 400 residents and receives 168 supportive responses. Survey B is an open online poll with 1600 voluntary responses, of which 752 are supportive. Use z=1.96.

(a) 2 marks

Determine a 95% approximate confidence interval from Survey A.

(b) 1 mark

Calculate the nominal 95% interval obtained by applying the same formula to Survey B.

(c) 2 marks

A councillor argues that Survey B is more trustworthy because its calculated interval is narrower. Evaluate this claim and recommend which survey should be used for population inference.

Question 19

6 marks

A screening test is used in a population where 3% of people have a condition. For a person with the condition, the test is positive with probability 0.92. For a person without the condition, the test is negative with probability 0.94. When a positive test is repeated, the two results may be treated as independent conditional on whether the person has the condition.

(a) 2 marks

Determine the probability that a randomly selected person receives a positive result on the first test.

(b) 1 mark

Given one positive result, determine the probability that the person has the condition.

(c) 1 mark

Determine the probability that a randomly selected person receives two positive results.

(d) 2 marks

The service refers a person for further investigation after two positive results only if the probability that the person has the condition then exceeds 0.80. Determine whether the referral rule is met.

Worked Solutions And Marking Guide

Section 1 Question 1

Answer: t=((ln5) / (0.4))

Take natural logarithms and divide by 0.4.

Section 1 Question 2

Answer: 0.0668

The z-score is -1.5, so the lower-tail probability is about 0.0668.

Section 1 Question 3

Answer: ((2x) / (x²+1))

Use the chain rule for ln u.

Section 1 Question 4

Answer: 8

The period is 2pi / (pi / 4)=8.

Section 1 Question 5

Answer: (0.8)⁸+8(0.2)(0.8)⁷

Add P(X=0) and P(X=1).

Section 1 Question 6

Answer: 0.069

Compute 1.96sqrt(0.55(0.45) / 200).

Section 1 Question 7

Answer: e-1

An antiderivative is e^x.

Section 1 Question 8

Answer: 1.5

z=(74-68) / 4=1.5.

Section 1 Question 9

Answer: ((ln2) / (ln1.06))

Set the growth factor equal to 2 and take logarithms.

Section 1 Question 10

Answer: 5

The mean of a uniform distribution is the midpoint.

Section 2 Question 11

(a) f(x)=-3cos x+sin(2x)+4.

Integrating gives f(x)=-3cos x+sin(2x)+C. Since f(0)=-3+C=1, C=4.

(b) y-4=x-fracpi2.

At x=fracpi2, f(x)=4 and f'(x)=3sin(fracpi2)+2cospi=1. Hence the tangent is y-4=1(x-fracpi2).

Detailed marking criteria

Part Part (a) (4 marks)

Award one mark for each required-evidence item, up to the stated maximum. Apply follow-through when an earlier numerical result is used consistently, unless the current mark requires an independent conclusion.

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (b) (2 marks)

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Section 2 Question 12

(a) h=((500) / (r²)), so C(r)=0.04pi r²+0.02(2pi rh)=0.04pi r²+((20pi) / (r)).

From pi r^2h=500pi, h=500 / r^2. The base costs 0.04pi r² dollars and the side costs 0.02(2pi rh)=20pi / r dollars.

(b) r=√3(250)approx6.30text( cm) and happrox12.60text( cm); this gives a minimum.

C'(r)=0.08pi r-20pi / r^2. Solving C'(r)=0 gives r³=250, so rapprox6.30. Then h=500 / r^2approx12.60. Since C''(r)=0.08pi+40pi / r³>0 for r>0, the stationary value is a minimum.

(c) Use r=6text( cm): C(6)approx15.00, compared with C(7)approx15.13.

Substitution gives C(6)=0.04pi(6)²+20pi / 6approx14.996 dollars and C(7)approx15.134 dollars, so radius 6 cm is cheaper.

Detailed marking criteria

Part Part (a) (2 marks)

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (b) (3 marks)

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (c) (2 marks)

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Section 2 Question 13

(a) hat p=0.40.

The pilot estimate is 32 / 80=0.40.

(b) n=577.

Require 1.96sqrt(0.4(0.6) / n)le0.04. Squaring and rearranging gives nge576.24, so the minimum whole-number sample size is 577.

(c) The sample should be randomly selected and observations should be independent.

A random, representative selection with independent observations is required; a larger biased sample would not repair selection bias.

Detailed marking criteria

Part Part (a) (1 mark)

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (b) (2 marks)

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (c) (1 mark)

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Section 2 Question 14

(a) x=e^(0.8)-1approx1.226.

Exponentiating ln(x+1)=0.8 gives x+1=e^(0.8), hence xapprox1.226.

(b) y-ln2=frac12(x-1).

Since f(1)=ln2 and f'(x)=1 / (x+1), the gradient at 1 is 1 / 2.

(c) frac34-ln2 square units.

The logarithmic curve is concave down, so its tangent lies above it. The area is int_0¹[ln2+(x-1) / 2-ln(x+1)],dx. Using intln(x+1),dx=(x+1)ln(x+1)-(x+1) gives frac34-ln2.

Detailed marking criteria

Part Part (a) (1 mark)

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (b) (2 marks)

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (c) (3 marks)

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Section 2 Question 15

(a) P(X<100)approx0.0912.

Standardising gives z=(100-120) / 15=-1.333ldots, so the lower-tail probability is approximately 0.0912.

(b) Approximately 204 batteries.

By symmetry, P(100<X<140)approx0.8176. The expected count is 250(0.8176)=204.4, so approximately 204 batteries.

Detailed marking criteria

Part Part (a) (1 mark)

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (b) (2 marks)

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Section 2 Question 16

(a) ((dA) / (dt))=0.6pi r.

By the chain rule, dA / dt=(dA / dr)(dr / dt)=2pi r(0.3)=0.6pi r.

(b) 2.4pitext( m)^2text( per year).

Substitute r=4 into dA / dt=0.6pi r.

(c) After 10 years.

Solve 1+0.3t=4, giving t=10.

Detailed marking criteria

Part Part (a) (2 marks)

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (b) (1 mark)

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (c) (1 mark)

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Section 2 Question 17

(a) P(text(replace))approx0.2121.

For Xsim B(20,0.08), P(Xge3)=1-[P(X=0)+P(X=1)+P(X=2)]approx0.2121.

(b) Approximately 0.9078.

If qapprox0.2120538 is the unrounded result from part (a), then P(text(at least one))=1-(1-q)¹⁰approx0.9078.

Detailed marking criteria

Part Part (a) (2 marks)

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (b) (2 marks)

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Section 2 Question 18

(a) Approximately (0.3716,0.4684).

For Survey A, hat p=168 / 400=0.42. The margin is 1.96sqrt(0.42(0.58) / 400)approx0.0484, giving approximately (0.3716,0.4684).

(b) Approximately (0.4455,0.4945).

For Survey B, hat p=752 / 1600=0.47 and the calculated margin is approximately 0.0245, giving (0.4455,0.4945).

(c) Use Survey A. Survey B's voluntary-response design can create selection bias, so its narrower calculated interval does not establish greater trustworthiness for the population.

Survey A uses random selection, which supports inference to the population. Survey B's larger sample reduces the formula's nominal random sampling error, but it does not remove self-selection bias; therefore the usual confidence-interval interpretation is not justified for Survey B.

Detailed marking criteria

Part Part (a) (2 marks)

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (b) (1 mark)

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (c) (2 marks)

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Section 2 Question 19

(a) P(+)=0.0858.

The false-positive probability is 1-0.94=0.06. Hence P(+)=0.03(0.92)+0.97(0.06)=0.0276+0.0582=0.0858.

(b) Approximately 0.3217.

By conditional probability, P(Cmid+)=0.03(0.92) / 0.0858approx0.3217.

(c) P(+,+)=0.028884.

Conditional independence gives P(+,+)=0.03(0.92)²+0.97(0.06)²=0.028884.

(d) Yes. P(Cmid+,+)=((0.03(0.92)²) / (0.028884))approx0.8791>0.80.

The probability of both positives coming from a person with the condition is 0.03(0.92)²=0.025392. Dividing by P(+,+)=0.028884 gives approximately 0.8791, which exceeds the referral threshold.

Detailed marking criteria

Part Part (a) (2 marks)

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (b) (1 mark)

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (c) (1 mark)

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Part Part (d) (2 marks)

Acceptable alternatives: Equivalent exact forms and correctly rounded decimal forms are acceptable unless the question specifies otherwise.

Do not credit by itself: A correct final value without the required supporting evidence does not earn all available marks.

Diagnostic Checklist

Topic	Questions	Marks	Marks Lost	Action
Calculus, Functions and Modelling	Q1, Q3-Q4, Q7, Q9, Q11-Q12, Q14, Q16	28	___	Review derivatives, antiderivatives, logarithmic functions, optimisation, related rates and modelling.
Probability Distributions	Q2, Q5, Q8, Q10, Q15, Q17, Q19	17	___	Review binomial, normal, uniform and conditional probability models, including repeated-test reasoning.
Statistical Inference	Q6, Q13, Q18	10	___	Review confidence intervals for proportions, sample-size planning and the effect of sampling design.

QCE Mathematical Methods Units 3&4 Free Online Pack 0 — Paper 2 Technology-active Question and Response Book Showcase

Exam-pack paper structure

Read Paper 2 Technology-active Question and Response Book Showcase online

Section 1

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Section 2

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Worked Solutions And Marking Guide

Section 1 Question 1

Section 1 Question 2

Section 1 Question 3

Section 1 Question 4

Section 1 Question 5

Section 1 Question 6

Section 1 Question 7

Section 1 Question 8

Section 1 Question 9

Section 1 Question 10

Section 2 Question 11

Detailed marking criteria

Section 2 Question 12

Detailed marking criteria

Section 2 Question 13

Detailed marking criteria

Section 2 Question 14

Detailed marking criteria

Section 2 Question 15

Detailed marking criteria

Section 2 Question 16

Detailed marking criteria

Section 2 Question 17

Detailed marking criteria

Section 2 Question 18

Detailed marking criteria

Section 2 Question 19

Detailed marking criteria

Diagnostic Checklist

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