IB DP Mathematical Studies 3.4 Paper 2

 

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Question

Consider the following statements.

     \(p\): the land has been purchased

     \(q\): the building permit has been obtained

     \(r\): the land can be used for residential purposes

Write the following argument in symbolic form.

“If the land has been purchased and the building permit has been obtained, then the land can be used for residential purposes.”[3]

a.

In your answer booklet, copy and complete a truth table for the argument in part (a).

Begin your truth table as follows.

[2]

b.

Use your truth table to determine whether the argument in part (a) is valid.

Give a reason for your decision.[2]

c.

Write down the inverse of the argument in part (a)

(i)     in symbolic form;

(ii)     in words.[4]

d.
Answer/Explanation

Markscheme

\((p \wedge q) \Rightarrow r\)     (A1)(A1)(A1)

Notes: Award (A1) for conjunction seen, award (A1) for implication seen, award (A1) for correct simple propositions in correct order (the parentheses are required). Accept \(r \Leftarrow (p \wedge q)\).

a.

     (A1)(ft)(A1)(ft)

Notes: Award (A1)(ft) for each correct column, follow through to the final column from their \((p \wedge q)\) column. For the second (A1)(ft) to be awarded there must be an implication in part (a).

Follow through from part (a).

b.

The argument is not valid since not all entries in the final column are T.     (A1)(ft)(R1)

Notes: Do not award (A1)(ft)(R0). Follow through from part (b).

Accept “The argument is not valid since \((p \wedge q) \Rightarrow r\) is not a tautology”.

c.

(i)     \(\neg (p \wedge q) \Rightarrow \neg r\)     (A1)(ft)(A1)(ft)

OR

\((\neg p \vee \neg q) \Rightarrow \neg r\)     (A1)(ft)(A1)(ft)

Notes: Award (A1)(ft) for the negation of their antecedent and the negation of their consequent, (A1)(ft) for their fully correct answer.

Follow through from part (a). Accept \(\neg r \Leftarrow \neg (p \wedge q)\) or \(\neg r \Leftarrow (\neg p \vee \neg q)\). Follow through from part (a).

(ii)     if it is not the case that the land has been purchased and the building permit has been obtained then the land can not be used for residential purposes.     (A1)(A1)(ft)

OR

if (either) the land has not been purchased or the building permit has not been obtained then the land can not be used for residential purposes.     (A1)(A1)(ft)

Notes: Award (A1) for “if… then…” seen, (A1)(ft) for correct statements in correct order. Follow through from part (d)(i).

d.

Question

A water container is made in the shape of a cylinder with internal height \(h\) cm and internal base radius \(r\) cm.

N16/5/MATSD/SP2/ENG/TZ0/06

The water container has no top. The inner surfaces of the container are to be coated with a water-resistant material.

The volume of the water container is \(0.5{\text{ }}{{\text{m}}^3}\).

The water container is designed so that the area to be coated is minimized.

One can of water-resistant material coats a surface area of \(2000{\text{ c}}{{\text{m}}^2}\).

Write down a formula for \(A\), the surface area to be coated.[2]

a.

Express this volume in \({\text{c}}{{\text{m}}^3}\).[1]

b.

Write down, in terms of \(r\) and \(h\), an equation for the volume of this water container.[1]

c.

Show that \(A = \pi {r^2}\frac{{1\,000\,000}}{r}\).[2]

d.

Show that \(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\).[2]

d.

Find \(\frac{{{\text{d}}A}}{{{\text{d}}r}}\).[3]

e.

Using your answer to part (e), find the value of \(r\) which minimizes \(A\).[3]

f.

Find the value of this minimum area.[2]

g.

Find the least number of cans of water-resistant material that will coat the area in part (g).[3]

h.
Answer/Explanation

Markscheme

\((A = ){\text{ }}\pi {r^2} + 2\pi rh\)    (A1)(A1)

Note:     Award (A1) for either \(\pi {r^2}\) OR \(2\pi rh\) seen. Award (A1) for two correct terms added together.[2 marks]

a.

\(500\,000\)    (A1)

Notes:     Units not required.[1 mark]

b.

\(500\,000 = \pi {r^2}h\)    (A1)(ft)

Notes:     Award (A1)(ft) for \(\pi {r^2}h\) equating to their part (b).

Do not accept unless \(V = \pi {r^2}h\) is explicitly defined as their part (b).[1 mark]

c.

\(A = \pi {r^2} + 2\pi r\left( {\frac{{500\,000}}{{\pi {r^2}}}} \right)\)    (A1)(ft)(M1)

Note:     Award (A1)(ft) for their \({\frac{{500\,000}}{{\pi {r^2}}}}\) seen.

Award (M1) for correctly substituting only \({\frac{{500\,000}}{{\pi {r^2}}}}\) into a correct part (a).

Award (A1)(ft)(M1) for rearranging part (c) to \(\pi rh = \frac{{500\,000}}{r}\) and substituting for \(\pi rh\) in expression for \(A\).

\(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\)    (AG)

Notes:     The conclusion, \(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\), must be consistent with their working seen for the (A1) to be awarded.

Accept \({10^6}\) as equivalent to \({1\,000\,000}\).[2 marks]

d.

\(A = \pi {r^2} + 2\pi r\left( {\frac{{500\,000}}{{\pi {r^2}}}} \right)\)    (A1)(ft)(M1)

Note:     Award (A1)(ft) for their \({\frac{{500\,000}}{{\pi {r^2}}}}\) seen.

Award (M1) for correctly substituting only \({\frac{{500\,000}}{{\pi {r^2}}}}\) into a correct part (a).

Award (A1)(ft)(M1) for rearranging part (c) to \(\pi rh = \frac{{500\,000}}{r}\) and substituting for \(\pi rh\) in expression for \(A\).

\(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\)    (AG)

Notes:     The conclusion, \(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\), must be consistent with their working seen for the (A1) to be awarded.

Accept \({10^6}\) as equivalent to \({1\,000\,000}\).[2 marks]

d.

\(2\pi r – \frac{{{\text{1}}\,{\text{000}}\,{\text{000}}}}{{{r^2}}}\)    (A1)(A1)(A1)

Note:     Award (A1) for \(2\pi r\), (A1) for \(\frac{1}{{{r^2}}}\) or \({r^{ – 2}}\), (A1) for \( – {\text{1}}\,{\text{000}}\,{\text{000}}\).[3 marks]

e.

\(2\pi r – \frac{{1\,000\,000}}{{{r^2}}} = 0\)    (M1)

Note:     Award (M1) for equating their part (e) to zero.

\({r^3} = \frac{{1\,000\,000}}{{2\pi }}\) OR \(r = \sqrt[3]{{\frac{{1\,000\,000}}{{2\pi }}}}\)     (M1)

Note:     Award (M1) for isolating \(r\).

OR

sketch of derivative function     (M1)

with its zero indicated     (M1)

\((r = ){\text{ }}54.2{\text{ }}({\text{cm}}){\text{ }}(54.1926 \ldots )\)    (A1)(ft)(G2)[3 marks]

f.

\(\pi {(54.1926 \ldots )^2} + \frac{{1\,000\,000}}{{(54.1926 \ldots )}}\)    (M1)

Note:     Award (M1) for correct substitution of their part (f) into the given equation.

\( = 27\,700{\text{ }}({\text{c}}{{\text{m}}^2}){\text{ }}(27\,679.0 \ldots )\)    (A1)(ft)(G2)[2 marks]

g.

\(\frac{{27\,679.0 \ldots }}{{2000}}\)    (M1)

Note:     Award (M1) for dividing their part (g) by 2000.

\( = 13.8395 \ldots \)    (A1)(ft)

Notes:     Follow through from part (g).

14 (cans)     (A1)(ft)(G3)

Notes:     Final (A1) awarded for rounding up their \(13.8395 \ldots \) to the next integer.[3 marks]

h.
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