| Date | May 2018 | Marks available | 6 | Reference code | 18M.3srg.hl.TZ0.5 |
| Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ0 |
| Command term | Prove that | Question number | 5 | Adapted from | N/A |
Question
The function \(f\,{\text{: }}\mathbb{Z} \to \mathbb{Z}\) is defined by \(f\left( n \right) = n + {\left( { - 1} \right)^n}\).
Prove that \(f \circ f\) is the identity function.
Show that \(f\) is injective.
Show that \(f\) is surjective.
Markscheme
METHOD 1
\(\left( {f \circ f} \right)\left( n \right) = n + {\left( { - 1} \right)^n} + {\left( { - 1} \right)^{n + {{\left( { - 1} \right)}^n}}}\) M1A1
\( = n + {\left( { - 1} \right)^n} + {\left( { - 1} \right)^n} \times {\left( { - 1} \right)^{{{\left( { - 1} \right)}^n}}}\) (A1)
considering \({\left( { - 1} \right)^n}\) for even and odd \(n\) M1
if \(n\) is odd, \({\left( { - 1} \right)^n} = - 1\) and if \(n\) is even, \({\left( { - 1} \right)^n} = 1\) and so \({\left( { - 1} \right)^{ \pm 1}} = - 1\) A1
\( = n + {\left( { - 1} \right)^n} - {\left( { - 1} \right)^n}\) A1
= \(n\) and so \(f \circ f\) is the identity function AG
METHOD 2
\(\left( {f \circ f} \right)\left( n \right) = n + {\left( { - 1} \right)^n} + {\left( { - 1} \right)^{n + {{\left( { - 1} \right)}^n}}}\) M1A1
\( = n + {\left( { - 1} \right)^n} + {\left( { - 1} \right)^n} \times {\left( { - 1} \right)^{{{\left( { - 1} \right)}^n}}}\) (A1)
\( = n + {\left( { - 1} \right)^n} \times \left( {1 + {{\left( { - 1} \right)}^{{{\left( { - 1} \right)}^n}}}} \right)\) M1
\({\left( { - 1} \right)^{ \pm 1}} = - 1\) R1
\(1 + {\left( { - 1} \right)^{{{\left( { - 1} \right)}^n}}} = 0\) A1
\(\left( {f \circ f} \right)\left( n \right) = n\) and so \(f \circ f\) is the identity function AG
METHOD 3
\(\left( {f \circ f} \right)\left( n \right) = f\left( {n + {{\left( { - 1} \right)}^n}} \right)\) M1
considering even and odd \(n\) M1
if \(n\) is even, \(f\left( n \right) = n + 1\) which is odd A1
so \(\left( {f \circ f} \right)\left( n \right) = f\left( {n + 1} \right) = \left( {n + 1} \right) - 1 = n\) A1
if \(n\) is odd, \(f\left( n \right) = n - 1\) which is even A1
so \(\left( {f \circ f} \right)\left( n \right) = f\left( {n - 1} \right) = \left( {n - 1} \right) + 1 = n\) A1
\(\left( {f \circ f} \right)\left( n \right) = n\) in both cases
hence \(f \circ f\) is the identity function AG
[6 marks]
suppose \(f\left( n \right) = f\left( m \right)\) M1
applying \(f\) to both sides \( \Rightarrow n = m\) R1
hence \(f\) is injective AG
[2 marks]
\(m = f\left( n \right)\) has solution \(n = f\left( m \right)\) R1
hence surjective AG
[1 mark]
