Date | November 2010 | Marks available | 4 | Reference code | 10N.3srg.hl.TZ0.2 |
Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ0 |
Command term | Determine | Question number | 2 | Adapted from | N/A |
Question
Let \(f:\mathbb{Z} \times \mathbb{R} \to \mathbb{R},{\text{ }}f(m,{\text{ }}x) = {( - 1)^m}x\). Determine whether f is
(i) surjective;
(ii) injective.
P is the set of all polynomials such that \(P = \left\{ {\sum\limits_{i = 0}^n {{a_i}{x^i}|n \in \mathbb{N}} } \right\}\).
Let \(g:P \to P,{\text{ }}g(p) = xp\). Determine whether g is
(i) surjective;
(ii) injective.
Let \(h:\mathbb{Z} \to {\mathbb{Z}^ + }\), \(h(x) = \left\{ {\begin{array}{*{20}{c}}
{2x,}&{x > 0} \\
{1 - 2x,}&{x \leqslant 0}
\end{array}} \right\}\). Determine whether h is
(i) surjective;
(ii) injective.
Markscheme
(i) let \(x \in \mathbb{R}\)
for example, \(f(0,{\text{ }}x) = x\), M1
hence f is surjective A1
(ii) for example, \(f(2,{\text{ }}3) = f(4,{\text{ }}3) = 3,{\text{ but }}(2,{\text{ }}3) \ne (4,{\text{ }}3)\) M1
hence f is not injective A1
[4 marks]
(i) there is no element of P such that \(g(p) = 7\), for example R1
hence g is not surjective A1
(ii) \(g(p) = g(q) \Rightarrow xp = xq \Rightarrow p = q\), hence g is injective M1A1
[4 marks]
(i) for \(x > 0,{\text{ }}h(x) = 2,{\text{ }}4,{\text{ }}6,{\text{ }}8 \ldots \) A1
for \(x \leqslant 0,{\text{ }}h(x) = 1,{\text{ }}3,{\text{ }}5,{\text{ }}7 \ldots \) A1
therefore h is surjective A1
(ii) for \(h(x) = h(y)\), since an odd number cannot equal an even number, there are only two possibilities: R1
\(x,{\text{ }}y > 0,{\text{ }}2x = 2y \Rightarrow x = y;\) A1
\(x,{\text{ }}y \leqslant 0,{\text{ }}1 - 2x = 1 - 2y \Rightarrow x = y\) A1
therefore h is injective A1
Note: This can be demonstrated in a variety of ways.
[7 marks]
Examiners report
This was the least successfully answered question on the paper. Candidates often could quote the definitions of surjective and injective, but often could not apply the definitions in the examples.
a) Some candidates failed to show convincingly that the function was surjective, and not injective.
This was the least successfully answered question on the paper. Candidates often could quote the definitions of surjective and injective, but often could not apply the definitions in the examples.
b) Some candidates had trouble interpreting the notation used in the question, hence could not answer the question successfully.
This was the least successfully answered question on the paper. Candidates often could quote the definitions of surjective and injective, but often could not apply the definitions in the examples.
c) Many candidates failed to appreciate that the function is discrete, and hence erroneously attempted to differentiate the function to show that it is monotonic increasing, hence injective. Others who provided a graph again showed a continuous rather than discrete function.