## Question

Let *c *be a positive, real constant. Let *G *be the set \(\{ \left. {x \in \mathbb{R}} \right| – c < x < c\} \) . The binary operation \( * \) is defined on the set *G *by \(x * y = \frac{{x + y}}{{1 + \frac{{xy}}{{{c^2}}}}}\).

Simplify \(\frac{c}{2} * \frac{{3c}}{4}\) .

State the identity element for *G *under \( * \).

For \(x \in G\) find an expression for \({x^{ – 1}}\) (the inverse of *x *under \( * \)).

Show that the binary operation \( * \) is commutative on *G *.

Show that the binary operation \( * \) is associative on *G *.

(i) If \(x,{\text{ }}y \in G\) explain why \((c – x)(c – y) > 0\) .

(ii) Hence show that \(x + y < c + \frac{{xy}}{c}\) .

Show that *G *is closed under \( * \).

Explain why \(\{ G, * \} \) is an Abelian group.

**Answer/Explanation**

## Markscheme

\(\frac{c}{2} * \frac{{3c}}{4} = \frac{{\frac{c}{2} + \frac{{3c}}{4}}}{{1 + \frac{1}{2} \cdot \frac{3}{4}}}\) *M1*

\( = \frac{{\frac{{5c}}{4}}}{{\frac{{11}}{8}}} = \frac{{10c}}{{11}}\) *A1*

*[2 marks]*

identity is 0 *A1*

*[1 mark]*

inverse is –*x* *A1*

*[1 mark]*

\(x * y = \frac{{x + y}}{{1 + \frac{{xy}}{{{c^2}}}}},{\text{ }}y * x = \frac{{y + x}}{{1 + \frac{{yx}}{{{c^2}}}}}\) *M1*

(since ordinary addition and multiplication are commutative)

\(x * y = y * x{\text{ so }} * \) is commutative *R1*

**Note: **Accept arguments using symmetry.

* *

*[2 marks]*

\((x * y) * z = \frac{{x + y}}{{1 + \frac{{xy}}{{{c^2}}}}} * z = \frac{{\left( {\frac{{x + y}}{{1 + \frac{{xy}}{{{c^2}}}}}} \right) + z}}{{1 + \left( {\frac{{x + y}}{{1 + \frac{{xy}}{{{c^2}}}}}} \right)\frac{z}{{{c^2}}}}}\) *M1*

\( = \frac{{\frac{{\left( {x + y + z + \frac{{xyz}}{{{c^2}}}} \right)}}{{\left( {1 + \frac{{xy}}{{{c^2}}}} \right)}}}}{{\frac{{\left( {1 + \frac{{xy}}{{{c^2}}} + \frac{{xz}}{{{c^2}}} + \frac{{yz}}{{{c^2}}}} \right)}}{{\left( {1 + \frac{{xy}}{{{c^2}}}} \right)}}}} = \frac{{\left( {x + y + z + \frac{{xyz}}{{{c^2}}}} \right)}}{{\left( {1 + \left( {\frac{{xy + xz + yz}}{{{c^2}}}} \right)} \right)}}\) *A1*

\(x * (y * z) = x * \left( {\frac{{y + z}}{{1 + \frac{{yz}}{{{c^2}}}}}} \right) = \frac{{x + \left( {\frac{{y + z}}{{1 + \frac{{yz}}{{{c^2}}}}}} \right)}}{{1 + \frac{x}{{{c^2}}}\left( {\frac{{y + z}}{{1 + \frac{{yz}}{{{c^2}}}}}} \right)}}\)

\( = \frac{{\frac{{\left( {x + \frac{{xyz}}{{{c^2}}} + y + z} \right)}}{{\left( {1 + \frac{{yz}}{{{c^2}}}} \right)}}}}{{\frac{{\left( {1 + \frac{{yz}}{{{c^2}}} + \frac{{xy}}{{{c^2}}} + \frac{{xz}}{{{c^2}}}} \right)}}{{\left( {1 + \frac{{yz}}{{{c^2}}}} \right)}}}} = \frac{{\left( {x + y + z + \frac{{xyz}}{{{c^2}}}} \right)}}{{\left( {1 + \left( {\frac{{xy + xz + yz}}{{{c^2}}}} \right)} \right)}}\) **A1**

since both expressions are the same \( * \) is associative *R1*

**Note**: After the initial ** M1A1**, correct arguments using symmetry also gain full marks.

* *

*[4 marks]*

(i) \(c > x{\text{ and }}c > y \Rightarrow c – x > 0{\text{ and }}c – y > 0 \Rightarrow (c – x)(c – y) > 0\) *R1AG*

* *

(ii) \({c^2} – cx – cy + xy > 0 \Rightarrow {c^2} + xy > cx + cy \Rightarrow c + \frac{{xy}}{c} > x + y{\text{ (as }}c > 0)\)

so \(x + y < c + \frac{{xy}}{c}\) *M1AG*

*[2 marks]*

if \(x,{\text{ }}y \in G{\text{ then }} – c – \frac{{xy}}{c} < x + y < c + \frac{{xy}}{c}\)

thus \( – c\left( {1 + \frac{{xy}}{{{c^2}}}} \right) < x + y < c\left( {1 + \frac{{xy}}{{{c^2}}}} \right){\text{ and }} – c < \frac{{x + y}}{{1 + \frac{{xy}}{{{c^2}}}}} < c\) *M1*

\(({\text{as }}1 + \frac{{xy}}{{{c^2}}} > 0){\text{ so }} – c < x * y < c\) *A1*

proving that *G *is closed under \( * \) *AG*

*[2 marks]*

as \(\{ G, * \} \) is closed, is associative, has an identity and all elements have an inverse *R1*

it is a group *AG*

as \( * \) is commutative *R1*

it is an Abelian group *AG*

*[2 marks]*

## Examiners report

Most candidates were able to answer part (a) indicating preparation in such questions. Many students failed to identify the command term “state” in parts (b) and (c) and spent a lot of time – usually unsuccessfully – with algebraic methods. Most students were able to offer satisfactory solutions to part (d) and although most showed that they knew what to do in part (e), few were able to complete the proof of associativity. Surprisingly few managed to answer parts (f) and (g) although many who continued to this stage, were able to pick up at least one of the marks for part (h), regardless of what they had done before. Many candidates interpreted the question as asking to prove that the group was Abelian, rather than proving that it was an Abelian group. Few were able to fully appreciate the significance in part (i) although there were a number of reasonable solutions.

Most candidates were able to answer part (a) indicating preparation in such questions. Many students failed to identify the command term “state” in parts (b) and (c) and spent a lot of time – usually unsuccessfully – with algebraic methods. Most students were able to offer satisfactory solutions to part (d) and although most showed that they knew what to do in part (e), few were able to complete the proof of associativity. Surprisingly few managed to answer parts (f) and (g) although many who continued to this stage, were able to pick up at least one of the marks for part (h), regardless of what they had done before. Many candidates interpreted the question as asking to prove that the group was Abelian, rather than proving that it was an Abelian group. Few were able to fully appreciate the significance in part (i) although there were a number of reasonable solutions.

Most candidates were able to answer part (a) indicating preparation in such questions. Many students failed to identify the command term “state” in parts (b) and (c) and spent a lot of time – usually unsuccessfully – with algebraic methods. Most students were able to offer satisfactory solutions to part (d) and although most showed that they knew what to do in part (e), few were able to complete the proof of associativity. Surprisingly few managed to answer parts (f) and (g) although many who continued to this stage, were able to pick up at least one of the marks for part (h), regardless of what they had done before. Many candidates interpreted the question as asking to prove that the group was Abelian, rather than proving that it was an Abelian group. Few were able to fully appreciate the significance in part (i) although there were a number of reasonable solutions.

## Question

The binary operation \( * \) is defined on \(\mathbb{N}\) by \(a * b = 1 + ab\).

Determine whether or not \( * \)

is closed;

is commutative;

is associative;

has an identity element.

**Answer/Explanation**

## Markscheme

\( * \) is closed *A1*

because \(1 + ab \in \mathbb{N}\) (when \(a,b \in \mathbb{N}\)) *R1*

*[2 marks]*

consider

\(a * b = 1 + ab = 1 + ba = b * a\) *M1A1*

therefore \( * \) is commutative

*[2 marks]*

**EITHER**

\(a * (b * c) = a * (1 + bc) = 1 + a(1 + bc){\text{ }}( = 1 + a + abc)\) *A1*

\((a * b) * c = (1 + ab) * c = 1 + c(1 + ab){\text{ }}( = 1 + c + abc)\) *A1*

(these two expressions are unequal when \(a \ne c\)) so \( * \) is not associative *R1*

**OR**

proof by counter example, for example

\(1 * (2 * 3) = 1 * 7 = 8\) *A1*

\((1 * 2) * 3 = 3 * 3 = 10\) *A1*

(these two numbers are unequal) so \( * \) is not associative *R1*

*[3 marks]*

let *e* denote the identity element; so that

\(a * e = 1 + ae = a\) gives \(e = \frac{{a – 1}}{a}\) (where \(a \ne 0\)) *M1*

then any valid statement such as: \(\frac{{a – 1}}{a} \notin \mathbb{N}\) or *e* is not unique *R1*

there is therefore no identity element *A1*

**Note:** Award the final ** A1** only if the previous

**is awarded.**

*R1** *

*[3 marks]*

## Examiners report

For the commutative property some candidates began by setting \(a * b = b * a\) . For the identity element some candidates confused \(e * a\) and \(ea\) stating \(ea = a\) . Others found an expression for an inverse element but then neglected to state that it did not belong to the set of natural numbers or that it was not unique.

For the commutative property some candidates began by setting \(a * b = b * a\) . For the identity element some candidates confused \(e * a\) and \(ea\) stating \(ea = a\) . Others found an expression for an inverse element but then neglected to state that it did not belong to the set of natural numbers or that it was not unique.

For the commutative property some candidates began by setting \(a * b = b * a\) . For the identity element some candidates confused \(e * a\) and \(ea\) stating \(ea = a\) . Others found an expression for an inverse element but then neglected to state that it did not belong to the set of natural numbers or that it was not unique.

## Question

The binary operation \(\Delta\) is defined on the set \(S =\) {1, 2, 3, 4, 5} by the following Cayley table.

(a) State whether *S *is closed under the operation Δ and justify your answer.

(b) State whether Δ is commutative and justify your answer.

(c) State whether there is an identity element and justify your answer.

(d) Determine whether Δ is associative and justify your answer.

(e) Find the solutions of the equation \(a\Delta b = 4\Delta b\), for \(a \ne 4\).

**Answer/Explanation**

## Markscheme

(a) yes *A1*

because the Cayley table only contains elements of *S **R1*

*[2 marks]*

* *

(b) yes *A1*

because the Cayley table is symmetric *R1*

*[2 marks]*

* *

(c) no *A1*

because there is no row (and column) with 1, 2, 3, 4, 5 *R1*

*[2 marks]*

* *

(d) attempt to calculate \((a\Delta b)\Delta c\) and \(a\Delta (b\Delta c)\) for some \(a,{\text{ }}b,{\text{ }}c \in S\) *M1*

counterexample: for example, \((1\Delta 2)\Delta 3 = 2\)

\(1\Delta (2\Delta 3) = 1\) *A1*

Δ is not associative *A1*

**Note: **Accept a correct evaluation of \((a\Delta b)\Delta c\) and \(a\Delta (b\Delta c)\) for some \(a,{\text{ }}b,{\text{ }}c \in S\) for the ** M1**.

**[3 marks]**

** **

(e) for example, attempt to enumerate \(4\Delta b\) for *b* = 1, 2, 3, 4, 5 and obtain (3, 2, 1, 4, 1) *(M1)*

find \((a,{\text{ }}b) \in \left\{ {{\text{(2, 2), (2, 3)}}} \right\}\) for \(a \ne 4\) (or equivalent) *A1A1*

**Note:** Award ** M1A1A0 **if extra ‘solutions’ are listed.

*[3 marks]*

* *

*Total [12 marks]*

## Examiners report

## Question

The binary operation \( * \) is defined for \(x,{\text{ }}y \in S = \{ 0,{\text{ }}1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5,{\text{ }}6\} \) by

\[x * y = ({x^3}y – xy)\bmod 7.\]

Find the element \(e\) such that \(e * y = y\), for all \(y \in S\).

(i) Find the least solution of \(x * x = e\).

(ii) Deduce that \((S,{\text{ }} * )\) is not a group.

Determine whether or not \(e\) is an identity element.

**Answer/Explanation**

## Markscheme

attempt to solve \({e^3}y – ey \equiv y\bmod 7\) *(M1)*

the only solution is \(e = 5\) *A1*

*[2 marks]*

(i) attempt to solve \({x^4} – {x^2} \equiv 5\bmod 7\) *(M1)*

least solution is \(x = 2\) *A1*

(ii) suppose \((S,{\text{ }} * )\) is a group with order 7 *A1*

\(2\) has order \(2\) *A1*

since \(2\) does not divide \(7\), Lagrange’s Theorem is contradicted *R1*

hence, \((S,{\text{ }} * )\) is not a group *AG*

*[5 marks]*

(\(5\) is a left-identity), so need to test if it is a right-identity:

ie, is \(y * 5 = y\)? *M1*

\(1 * 5 = 0 \ne 1\) *A1*

so \(5\) is not an identity *A1*

*[3 marks]*

*Total [10 marks]*

## Examiners report

Many candidates were not sufficiently familiar with modular arithmetic to complete this question satisfactorily. In particular, some candidates completely ignored the requirement that solutions were required to be found modulo 7, and returned decimal answers to parts (a) and (b). Very few candidates invoked Lagrange’s theorem in part (b)(ii). Some candidates were under the misapprehension that a group had to be Abelian, so tested for commutativity in part (b)(ii). It was pleasing that many candidates realised that an identity had to be both a left and right identity.

Many candidates were not sufficiently familiar with modular arithmetic to complete this question satisfactorily. In particular, some candidates completely ignored the requirement that solutions were required to be found modulo 7, and returned decimal answers to parts (a) and (b). Very few candidates invoked Lagrange’s theorem in part (b)(ii). Some candidates were under the misapprehension that a group had to be Abelian, so tested for commutativity in part (b)(ii). It was pleasing that many candidates realised that an identity had to be both a left and right identity.

Many candidates were not sufficiently familiar with modular arithmetic to complete this question satisfactorily. In particular, some candidates completely ignored the requirement that solutions were required to be found modulo 7, and returned decimal answers to parts (a) and (b). Very few candidates invoked Lagrange’s theorem in part (b)(ii). Some candidates were under the misapprehension that a group had to be Abelian, so tested for commutativity in part (b)(ii). It was pleasing that many candidates realised that an identity had to be both a left and right identity.

## Question

The set of all permutations of the elements \(1,{\text{ }}2,{\text{ }} \ldots 10\) is denoted by \(H\) and the binary operation \( \circ \) represents the composition of permutations.

The permutation \(p = (1{\text{ }}2{\text{ }}3{\text{ }}4{\text{ }}5{\text{ }}6)(7{\text{ }}8{\text{ }}9{\text{ }}10)\) generates the subgroup \(\{ G,{\text{ }} \circ \} \) of the group \(\{ H,{\text{ }} \circ \} \).

Find the order of \(\{ G,{\text{ }} \circ \} \).

State the identity element in \(\{ G,{\text{ }} \circ \} \).

Find

(i) \(p \circ p\);

(ii) the inverse of \(p \circ p\).

(i) Find the maximum possible order of an element in \(\{ H,{\text{ }} \circ \} \).

(ii) Give an example of an element with this order.

**Answer/Explanation**

## Markscheme

the order of \((G,{\text{ }} \circ )\) is \({\text{lcm}}(6,{\text{ }}4)\) *(M1)*

\( = 12\) *A1*

*[2 marks]*

\(\left( 1 \right){\rm{ }}\left( 2 \right){\rm{ }}\left( 3 \right){\rm{ }}\left( 4 \right){\rm{ }}\left( 5 \right){\rm{ }}\left( 6 \right){\rm{ }}\left( 7 \right){\rm{ }}\left( 8 \right){\rm{ }}\left( 9 \right){\rm{ }}\left( {10} \right)\) *A1*

**Note: **Accept ( ) or a word description.

**[1 mark]**

(i) \(p \circ p = (1{\text{ }}3{\text{ }}5)(2{\text{ }}4{\text{ }}6)(7{\text{ }}9)(810)\) *(M1)A1*

(ii) its inverse \( = (1{\text{ }}5{\text{ }}3)(2{\text{ }}6{\text{ }}4)(7{\text{ }}9)(810)\) *A1A1*

**Note: **Award ** A1 **for cycles of 2,

**for cycles of 3.**

*A1***[4 marks]**

(i) considering LCM of length of cycles with length \(2\), \(3\) and \(5\) *(M1)*

\(30\) *A1*

(ii) *eg*\(\;\;\;(1{\text{ }}2)(3{\text{ }}4{\text{ }}5)(6{\text{ }}7{\text{ }}8{\text{ }}9{\text{ }}10)\) *A1*

**Note: **allow FT as long as the length of cycles adds to \(10\) and their LCM is consistent with answer to part (i).

**Note: **Accept alternative notation for each part

**[3 marks]**

**Total [10 marks]**

## 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 \(a \in S\).

Show that \(x * y \in S\) for all \(x,{\text{ }}y \in S\).

Show that the operation \( * \) on the set \(S\) is commutative.

Show that the operation \( * \) on the set \(S\) is associative.

Show that 2 is the identity element.

Show that each element \(a \in S\) has an inverse.

**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]*

## Examiners report

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## 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.

**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 binary operations \( \odot \) and \( * \) are defined on \({\mathbb{R}^ + }\) by

\[a \odot b = \sqrt {ab} {\text{ and }}a * b = {a^2}{b^2}.\]

Determine whether or not

\( \odot \) is commutative;

\( * \) is associative;

\( * \) is distributive over \( \odot \) ;

\( \odot \) has an identity element.

**Answer/Explanation**

## Markscheme

\(a \odot b = \sqrt {ab} = \sqrt {ba} = b \odot a\) *A1*

since \(a \odot b = b \odot a\) it follows that \( \odot \) is commutative *R1*

*[2 marks]*

\(a * (b * c) = a * {b^2}{c^2} = {a^2}{b^4}{c^4}\) *M1A1*

\((a * b) * c = {a^2}{b^2} * c = {a^4}{b^4}{c^2}\) *A1*

these are different, therefore \( * \) is not associative *R1*

**Note:** Accept numerical counter-example.

* *

*[4 marks]*

\(a * (b \odot c) = a * \sqrt {bc} = {a^2}bc\) *M1A1*

\((a * b) \odot (a * c) = {a^2}{b^2} \odot {a^2}{c^2} = {a^2}bc\) *A1*

these are equal so \( * \) is distributive over \( \odot \) *R1*

*[4 marks]*

the identity e would have to satisfy

\(a \odot e = a\) for all *a* *M1*

now \(a \odot e = \sqrt {ae} = a \Rightarrow e = a\) *A1*

therefore there is no identity element *A1*

*[3 marks]*

## Examiners report

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