| Exam Board | OCR MEI |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Proofs |
| Type | Truth tables or Boolean algebra |
| Difficulty | Easy -1.8 This is a Decision Mathematics question on basic Boolean logic and truth tables, not a trigonometry proof question. The tasks involve identifying logical statements from English, finding a single counterexample line in a truth table, and simple circuit-to-logic conversions—all routine recall and mechanical application with minimal problem-solving depth. |
| Spec | 1.01b Logical connectives: congruence, if-then, if and only if |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sim c \Rightarrow e\) | B1 | Must choose correct one from the four options |
| Answer | Marks | Guidance |
|---|---|---|
| Need a row where \((c \Rightarrow \sim e)\) is true and \((\sim c \Rightarrow e)\) is false, e.g. \(c=F, e=F\): LHS: \(F \Rightarrow T = T\); RHS: \(T \Rightarrow F = F\); so biconditional is false | B1+B1+B1+B1 | B1 for identifying \(c=F, e=F\) (or equivalent); B1 for evaluating LHS correctly; B1 for evaluating RHS correctly; B1 for conclusion that the biconditional is false |
| Answer | Marks | Guidance |
|---|---|---|
| Circuit output is \((\sim x) \vee y\). This equals \(x \Rightarrow y\). Truth table verification showing all four input combinations give matching outputs for both circuit and \(x \Rightarrow y\) | M1 A1 A1 A1 A1 | M1 for attempting to trace circuit; A1 for \(\sim x\) from NOT gate; A1 for \((\sim x) \vee y\) as output; A1 for stating \(x \Rightarrow y\); A1 for complete verification |
| Answer | Marks | Guidance |
|---|---|---|
| \((\sim p \wedge \sim q) \Rightarrow r\) equivalently \(\sim(\sim p \wedge \sim q) \vee r\) | B1 B1 | B1 for correct reading of circuit as \((\sim p \wedge \sim q) \Rightarrow r\); B1 for correct statement |
| Answer | Marks | Guidance |
|---|---|---|
| \((\sim p \wedge \sim q) \Rightarrow r\) is equivalent to \(\sim r \Rightarrow \sim(\sim p \wedge \sim q)\) i.e. \(\sim r \Rightarrow (p \vee q)\). Given \(\sim r\) is true, so \(p \vee q\) is true. Given \(\sim q\) is true, so \(p\) must be true. | B1 B1 B1 B1 | B1 for using contrapositive; B1 for \(\sim r \Rightarrow (p \vee q)\); B1 for applying \(\sim r\); B1 for concluding \(p\) from \(\sim q\) |
# Question 1:
## Part (a)(i)
| $\sim c \Rightarrow e$ | B1 | Must choose correct one from the four options |
## Part (a)(ii)
| Need a row where $(c \Rightarrow \sim e)$ is true and $(\sim c \Rightarrow e)$ is false, e.g. $c=F, e=F$: LHS: $F \Rightarrow T = T$; RHS: $T \Rightarrow F = F$; so biconditional is false | B1+B1+B1+B1 | B1 for identifying $c=F, e=F$ (or equivalent); B1 for evaluating LHS correctly; B1 for evaluating RHS correctly; B1 for conclusion that the biconditional is false |
## Part (b)(i)
| Circuit output is $(\sim x) \vee y$. This equals $x \Rightarrow y$. Truth table verification showing all four input combinations give matching outputs for both circuit and $x \Rightarrow y$ | M1 A1 A1 A1 A1 | M1 for attempting to trace circuit; A1 for $\sim x$ from NOT gate; A1 for $(\sim x) \vee y$ as output; A1 for stating $x \Rightarrow y$; A1 for complete verification |
## Part (b)(ii)
| $(\sim p \wedge \sim q) \Rightarrow r$ equivalently $\sim(\sim p \wedge \sim q) \vee r$ | B1 B1 | B1 for correct reading of circuit as $(\sim p \wedge \sim q) \Rightarrow r$; B1 for correct statement |
## Part (b)(iii)
| $(\sim p \wedge \sim q) \Rightarrow r$ is equivalent to $\sim r \Rightarrow \sim(\sim p \wedge \sim q)$ i.e. $\sim r \Rightarrow (p \vee q)$. Given $\sim r$ is true, so $p \vee q$ is true. Given $\sim q$ is true, so $p$ must be true. | B1 B1 B1 B1 | B1 for using contrapositive; B1 for $\sim r \Rightarrow (p \vee q)$; B1 for applying $\sim r$; B1 for concluding $p$ from $\sim q$ |
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\begin{enumerate}[label=(\alph*)]
\item Mickey ate the last of the cheese. Minnie was put out at this. Mickey's defence was "There wasn't enough left not to eat it all".
Let "c" represent "there is enough cheese for two" and "e" represent "one person can eat all of the cheese".
\begin{enumerate}[label=(\roman*)]
\item Which of the following best captures Mickey's argument?
$$\mathrm { c } \Rightarrow \mathrm { e } \quad \mathrm { c } \Rightarrow \sim \mathrm { e } \quad \sim _ { \mathrm { c } } \Rightarrow \mathrm { e } \quad \sim _ { \mathrm { c } } \Rightarrow \sim \mathrm { e }$$
In the ensuing argument Minnie concedes that if there's a lot left then one should not eat it all, but argues that this is not an excuse for Mickey having eaten it all when there was not a lot left.
\item Prove that Minnie is right by writing down a line of a truth table which shows that
$$( c \Rightarrow \sim e ) \Leftrightarrow ( \sim c \Rightarrow e )$$
is false.\\
(You may produce the whole table if you wish, but you need to indicate a specific line of the table.)
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Show that the following combinatorial circuit is modelling an implication statement. Say what that statement is, and prove that the circuit and the statement are equivalent.\\
\includegraphics[max width=\textwidth, alt={}, center]{c3a528e4-b5fe-4bff-a77e-e3199bb225a1-2_188_533_1272_767}
\item Express the following combinatorial circuit as an equivalent implication statement.\\
\includegraphics[max width=\textwidth, alt={}, center]{c3a528e4-b5fe-4bff-a77e-e3199bb225a1-2_314_835_1599_616}
\item Explain why $( \sim \mathrm { p } \wedge \sim \mathrm { q } ) \Rightarrow \mathrm { r }$, together with $\sim \mathrm { r }$ and $\sim \mathrm { q }$, give p .
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI D2 2010 Q1 [16]}}