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.