(i)     Sketch the graph of \(y = f(x)\) and hence justify whether or not \(f\) is a bijection.

(ii)     Show that \(A\) is a group under the binary operation of multiplication.

(iii)     Give a reason why \(B\) is not a group under the binary operation of multiplication.

(iv)     Find an example to show that \(f(a \times b) = f(a) \times f(b)\) is not satisfied for all \(a,{\text{ }}b \in A\).

(i)     Sketch the graph of \(y = g(x)\) and hence justify whether or not \(g\) is a bijection.

(ii)     Show that \(g(a + b) = g(a) \times g(b)\) for all \(a,{\text{ }}b \in \mathbb{R}\).

(iii)     Given that \(\{ \mathbb{R},{\text{ }} + \} \) and \(\{ D,{\text{ }} \times \} \) are both groups, explain whether or not they are isomorphic.

(i)          A1

Notes: Award A1 for general shape, labelled asymptotes, and showing that \(x \ne 0\).

graph shows that it is injective since it is increasing or by the horizontal line test     R1

graph shows that it is surjective by the horizontal line test     R1

Note: Allow any convincing reasoning.

so \(f\) is a bijection     A1

(ii)     closed since non-zero real times non-zero real equals non-zero real     A1R1

we know multiplication is associative     R1

identity is 1     A1

inverse of \(x\) is \(\frac{1}{x}(x \ne 0)\)     A1

hence it is a group     AG

(iii)     \(B\) does not have an identity     A2

hence it is not a group     AG

(iv)     \(f(1 \times 1) = f(1) = \frac{1}{2}\) whereas \(f(1) \times f(1) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\) is one counterexample     A2

hence statement is not satisfied     AG

[13 marks]

award A1 for general shape going through (0, 1) and with domain \(\mathbb{R}\)     A1

graph shows that it is injective since it is increasing or by the horizontal line test and graph shows that it is surjective by the horizontal line test     R1

Note: Allow any convincing reasoning.

so \(g\)  is a bijection     A1

(ii)     \(g(a + b) = {{\text{e}}^{a + b}}\) and \(g(a) \times g(b) = {{\text{e}}^a} \times {{\text{e}}^b} = {{\text{e}}^{a + b}}\)     M1A1

hence \(g(a + b) = g(a) \times g(b)\)     AG

(iii)     since \(g\) is a bijection and the homomorphism rule is obeyed     R1R1

the two groups are isomorphic     A1

[8 marks]