Prove that $f \circ f$ is the identity function.

Show that $f$ is injective.

Show that $f$ is surjective.

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]