Three functions mapping $\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ are defined by

$${f_1}(m,{\text{ }}n) = m - n + 4;\,\,\,{f_2}(m,{\text{ }}n) = \left| m \right|;\,\,\,{f_3}(m,{\text{ }}n) = {m^2} - {n^2}.$$

Two functions mapping $\mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}$ are defined by

$${g_1}(k) = (2k,{\text{ }}k);\,\,\,{g_2}(k) = \left( {k,{\text{ }}\left| k \right|} \right).$$

(a)     Find the range of

(i)     ${f_1} \circ {g_1}$ ;

(ii)     ${f_3} \circ {g_2}$ .

(b)     Find all the solutions of ${f_1} \circ {g_2}(k) = {f_2} \circ {g_1}(k)$ .

(c)     Find all the solutions of ${f_3}(m,{\text{ }}n) = p$ in each of the cases p =1 and p = 2 .

(a)     (i)     ${f_1} \circ {g_1}(k) = k + 4$     M1

Range $({f_1} \circ {g_1}) = \mathbb{Z}$     A1

(ii)     ${f_3} \circ {g_2}(k) = 0$     M1

Range $({f_3} \circ {g_2}) = \{ 0\}$     A1

[4 marks]

(b)     the equation to solve is

$k - \left| k \right| + 4 = \left| {2k} \right|$     M1A1

the positive solution is k = 2     A1

the negative solution is k = –1     A1

[4 marks]

(c)     the equation factorizes: $(m + n)(m - n) = p$     (M1)

for p = 1 , the possible factors over $\mathbb{Z}$ are $m + n =  \pm 1,{\text{ }}m - n =  \pm 1$     (M1)(A1)

with solutions (1, 0) and (–1, 0)     A1

for p = 2 , the possible factors over $\mathbb{Z}$ are $m + n = \pm 1,{\text{ }} \pm 2;{\text{  }}m - n = \pm 2,{\text{ }} \pm 1$     M1A1

there are no solutions over $\mathbb{Z} \times \mathbb{Z}$     A1

[7 marks]

Total [15 marks]

The majority of candidates were able to compute the composite functions involved in parts (a) and (b). Part(c) was satisfactorily tackled by a minority of candidates. There were more GDC solutions than the more obvious approach of factorizing a difference of squares. Some candidates seemed to forget that m and n belonged to the set of integers.