Mathematics Paper 2 June 2010

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1. A$0$, since squaring $z$ makes $z^2$ non-negative and $y$ may be zero
2. B$-32$, taking $y = -2$ and $z = 4$, the largest value $z$ itself may take
3. C$-72$, taking $y = -2$ and $z = -6$, so that $z^2$ reaches 36
4. D$-12$, taking $y = -2$ and $z$ at its smallest magnitude other than zero

1. A$12$, from $x = 6$ and $y = -6$, using the bound that belongs to $z$
2. B$8$, from $x = 6$ and $y = -2$, since subtracting $-2$ adds 2
3. C$6$, from $x = 6$ and $y = 0$, treating $y$ as though it had to be positive
4. D$4$, from $x = 6$ and $y = 2$ read off the wrong end of the bound on $y$

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1. A$\{1; 3; 7; 9\}$, which is $A \cup B$ rather than the intersection asked for
2. B$\{3; 5\}$, which is all of $A$, forgetting to intersect with $B$ as well
3. C$\{1; 7; 9\}$, which is $C$ itself rather than the complement of $C$
4. D$\{3\}$, the one element that lies in both $A$ and $B$ and outside $C$

1. A$\{2; 8\}$, which is $(A \cup B)'$ on its own, stopping before the union with $C$
2. B$\{1; 2; 7; 8; 9\}$, the two elements outside $A \cup B$ together with those of $C$
3. C$\{1; 3; 5; 7; 9\}$, which is $A \cup B$ itself rather than its complement
4. D$\{1; 7; 9\}$, which is $C$ on its own, as though the complement were empty

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1. A$(90 - x)^\circ$, treating triangle ABC as though it were right angled at C
2. B$(180 - 2x)^\circ$, which is $A\hat{B}C$ itself rather than the angle next to it
3. C$2x^\circ$, the exterior angle of triangle ABC at B, equal to the two opposite angles
4. D$x^\circ$, assuming the two angles at B are equal because $AB = BC$

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1. A$(180 - 4x)^\circ$, the third angle of triangle BCD, whose base angles are each $2x^\circ$
2. B$(180 - 2x)^\circ$, using $x^\circ$ for each base angle of triangle BCD
3. C$3x^\circ$, which is the exterior angle $D\hat{C}E$ rather than the interior one
4. D$(90 - 2x)^\circ$, as though triangle BCD contained a right angle at D

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1. A$\begin{pmatrix} 2 & -1 \\ 1,5 & 0 \end{pmatrix}$, from $2\mathbf{M} = \mathbf{N}$
2. B$\begin{pmatrix} -4 & 2 \\ -3 & 0 \end{pmatrix}$, from $\mathbf{M} = -\mathbf{N}$
3. C$\begin{pmatrix} 4 & -2 \\ 3 & 0 \end{pmatrix}$, from $\mathbf{M} = \mathbf{N}$
4. D$\begin{pmatrix} -2 & 1 \\ -1,5 & 0 \end{pmatrix}$, from $2\mathbf{M} = -\mathbf{N}$

1. A$\begin{pmatrix} 10 & 12 \\ -8 & -6 \end{pmatrix}$, with rows and columns interchanged
2. B$\begin{pmatrix} 22 & -8 \\ 12 & -6 \end{pmatrix}$, mis-adding the first product
3. C$\begin{pmatrix} 16 & 4 \\ 9 & 0 \end{pmatrix}$, squaring each entry separately
4. D$\begin{pmatrix} 10 & -8 \\ 12 & -6 \end{pmatrix}$, multiplying rows into columns

1. AInverse variation
2. BJoint variation
3. CPartial variation
4. DDirect variation

1. A$\left(-\dfrac{c}{k}; 0\right)$, which is where the line meets the other axis
2. B$(0; c)$, the value R takes when V is zero on the vertical axis
3. C$(c; 0)$, with the two coordinates written the wrong way round
4. D$(0; k)$, using the gradient in place of the intercept

1. A$(0; c)$, which is the crossing point on the vertical axis instead
2. B$\left(\dfrac{c}{k}; 0\right)$, having forgotten the sign change when c moves across
3. C$\left(-\dfrac{c}{k}; 0\right)$, solving $kV + c = 0$ for V
4. D$\left(0; -\dfrac{c}{k}\right)$, with the two coordinates written the wrong way round

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1. A$k = 2$ and $c = 4$
2. B$k = 3$ and $c = 2$
3. C$k = 0,5$ and $c = 7$
4. D$k = 1,5$ and $c = 5$

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1. A$\dfrac{449}{1798}$, giving only the case in which both pupils are boys
2. B$\dfrac{449}{899}$, adding the two like-sex cases without replacement
3. C$\dfrac{450}{899}$, using 450 in the numerator of the second fraction
4. D$\dfrac{1}{2}$, treating the second choice as if the first had been replaced

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1. A$108^\circ$, doubling the angle at R as though it stood at the centre
2. B$54^\circ$, alternate to $O\hat{R}Q$ because OS is parallel to QR
3. C$27^\circ$, halving the angle at R as though R were the centre
4. D$126^\circ$, which is $P\hat{O}S$, the angle on the other side of the line

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1. AThe perpendicular bisector of the line segment BC
2. BThe bisector of the angle BAC drawn at the vertex A
3. CThe circle whose centre is B and whose radius is BC
4. DThe line drawn through B parallel to the side CD

1. AThe circle whose centre is D and whose radius is DA
2. BThe line drawn midway between the sides AB and DC
3. CThe bisector of the angle ADC drawn at the vertex D
4. DThe perpendicular bisector of the line segment AC

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1. AAbout 5,2 cm
2. BAbout 10,0 cm
3. CAbout 2,5 cm
4. DAbout 8,0 cm

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1. A$3\mathbf{b} - 2\mathbf{a}$, going from P back to O and then on to R
2. B$2\mathbf{a} + 3\mathbf{b}$, adding the two given vectors head to tail
3. C$2\mathbf{a} - 3\mathbf{b}$, which is $\overrightarrow{RP}$, the reverse journey
4. D$\mathbf{b} - \mathbf{a}$, cancelling the numbers as though they matched

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1. A$2\mathbf{a} + \mathbf{b}$, forgetting the part of OP undone by PM
2. B$\dfrac{4}{3}\mathbf{a} + \mathbf{b}$, adding $\overrightarrow{OP}$ to $\overrightarrow{PM}$
3. C$\dfrac{2}{3}\mathbf{a} + \mathbf{b}$, taking $\overrightarrow{PM}$ as the whole route
4. D$\dfrac{2}{3}(\mathbf{a} + \mathbf{b})$, dividing both given vectors by three

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1. A$\dfrac{1}{2}$, splitting the trapezium into two pieces assumed equal
2. B$\dfrac{1}{4}$, squaring the ratio of the two parallel sides
3. C$\dfrac{2}{3}$, which is the share belonging to the other triangle
4. D$\dfrac{1}{3}$, since the two triangles are in the ratio $1 : 2$

1. AA rotation through $90^\circ$ clockwise about the origin
2. BA reflection in the line $y = 2x + 2$
3. CA reflection in the line $y = 2x - 2$
4. DA rotation through $90^\circ$ anticlockwise about $(1; 2)$

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1. A$(-0,5; 2,5)$, halving each coordinate but keeping the original signs
2. B$(0,5; 2,5)$, changing the sign of only the first coordinate
3. C$(0,5; -2,5)$, halving each coordinate and reversing both signs
4. D$(2; -10)$, doubling instead of halving and reversing both signs

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1. A$(-3; -1)$, keeping $x$ and taking $y + 2x = 5 - 6$
2. B$(-3; 11)$, adding $2x$ as though $x$ were positive
3. C$(-6; 5)$, shearing the first coordinate instead of the second
4. D$(-3; 10)$, doubling $y$ rather than adding twice $x$

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1. A$8 < x \le 10$, the class the list happens to begin with
2. B$11 < x \le 12$, the class just before the one with most students
3. C$12 < x \le 14$, the class holding 14 students, more than any other
4. D$16 < x \le 19$, the widest of the six classes given

1. A$14,0$, reading across from the cumulative frequency of 31
2. B$12,4$, reading across from a cumulative frequency of 20
3. C$11,5$, reading across from a cumulative frequency of 10 instead of 20
4. D$13,0$, taking the middle of the class that contains the median

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1. A$-2 < x \le 7$, dividing only the right hand end by 2
2. B$2 \le x < 7$, with the two inequality signs the wrong way round
3. C$5 < x \le 14$, adding 7 throughout but not dividing by 2
4. D$2 < x \le 7$, adding 7 throughout and then halving throughout

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1. A$(3x + 8)(x - 1)$
2. B$(x - 8)(3x + 1)$
3. C$(3x - 4)(x + 2)$
4. D$(3x - 8)(x + 1)$

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1. A$1,33$ metres, from the root $x = -1$ put into $3x - 5$
2. B$8$ metres, from doubling the area and forgetting the base
3. C$3$ metres, from the root $x = \dfrac{8}{3}$ put into $3x - 5$
4. D$2,67$ metres, which is the value of $x$, that is the base BC

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1. A$125$ m$^2$
2. B$175$ m$^2$
3. C$100$ m$^2$
4. D$115$ m$^2$

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1. A$450$ m$^2$
2. B$250$ m$^2$
3. C$325$ m$^2$
4. D$390$ m$^2$

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1. A$x = -2$
2. B$x = 2$
3. C$x = 3$
4. D$x = 0$

1. A$2x^2 - 7x + 3 = 0$
2. B$2x^2 + x + 3 = 0$
3. C$2x^2 + 7x + 6 = 0$
4. D$2x^2 + 7x + 3 = 0$

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Answer this when you sit the paper.