Question
The function \(f\,{\text{: }}\mathbb{Z} \to \mathbb{Z}\) is defined by \(f\left( n \right) = n + {\left( { – 1} \right)^n}\).
a.Prove that \(f \circ f\) is the identity function.[6]
b.i.Show that \(f\) is injective.[2]
b.ii.Show that \(f\) is surjective.[1]
▶️Answer/Explanation
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]
Examiners report
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Question
The set \(S\) is defined as the set of real numbers greater than 1.
The binary operation \( * \) is defined on \(S\) by \(x * y = (x – 1)(y – 1) + 1\) for all \(x,{\text{ }}y \in S\).
Let \(a \in S\).
a.Show that \(x * y \in S\) for all \(x,{\text{ }}y \in S\).[2]
b.i.Show that the operation \( * \) on the set \(S\) is commutative.[2]
b.ii.Show that the operation \( * \) on the set \(S\) is associative.[5]
c.Show that 2 is the identity element.[2]
d.Show that each element \(a \in S\) has an inverse.[3]
▶️Answer/Explanation
Markscheme
\(x,{\text{ }}y > 1 \Rightarrow (x – 1)(y – 1) > 0\) M1
\((x – 1)(y – 1) + 1 > 1\) A1
so \(x * y \in S\) for all \(x,{\text{ }}y \in S\) AG
[2 marks]
\(x * y = (x – 1)(y – 1) + 1 = (y – 1)(x – 1) + 1 = y * x\) M1A1
so \( * \) is commutative AG
[2 marks]
\(x * (y * z) = x * \left( {(y – 1)(z – 1) + 1} \right)\) M1
\( = (x – 1)\left( {(y – 1)(z – 1) + 1 – 1} \right) + 1\) (A1)
\( = (x – 1)(y – 1)(z – 1) + 1\) A1
\((x * y) * z = \left( {(x – 1)(y – 1) + 1} \right) * z\) M1
\( = \left( {(x – 1)(y – 1) + 1 – 1} \right)(z – 1) + 1\)
\( = (x – 1)(y – 1)(z – 1) + 1\) A1
so \( * \) is associative AG
[5 marks]
\(2 * x = (2 – 1)(x – 1) + 1 = x,{\text{ }}x * 2 = (x – 1)(2 – 1) + 1 = x\) M1
\(2 * x = x * 2 = 2{\text{ }}(2 \in S)\) R1
Note: Accept reference to commutativity instead of explicit expressions.
so 2 is the identity element AG
[2 marks]
\(a * {a^{ – 1}} = 2 \Rightarrow (a – 1)({a^{ – 1}} – 1) + 1 = 2\) M1
so \({a^{ – 1}} = 1 + \frac{1}{{a – 1}}\) A1
since \(a – 1 > 0 \Rightarrow {a^{ – 1}} > 1{\text{ }}({a^{ – 1}} * a = a * {a^{ – 1}})\) R1
Note: R1 dependent on M1.
so each element, \(a \in S\), has an inverse AG
[3 marks]