2023 CCEA GCSE Further Mathematics 2330 模擬試題連答案詳解

Factorise the numerator and denominator: \( 3x^2-12 = 3(x^2-4) = 3(x-2)(x+2) \); \( x^2+x-6 = (x+3)(x-2) \). So \( \dfrac{3x^2-12}{x^2+x-6} = \dfrac{3(x-2)(x+2)}{(x+3)(x-2)} \). Cancelling the common factor \( (x-2) \) (valid for \( x\ne2 \)) gives \( \dfrac{3(x+2)}{x+3} \).

[1] numerator correctly factorised, \( 3(x-2)(x+2) \); [1] denominator correctly factorised, \( (x+3)(x-2) \); [1] common factor \( (x-2) \) correctly identified; [1] correctly cancelled; [1] final simplified answer \( \dfrac{3(x+2)}{x+3} \). Accept the equivalent unsimplified factorised form for [4] if the final cancellation is not completed.

\( 2x^2-8x+5 = 2(x^2-4x)+5 = 2\left[(x-2)^2-4\right]+5 = 2(x-2)^2-8+5 = 2(x-2)^2-3 \). So \( a=2,\ b=-2,\ c=-3 \). Since \( (x-2)^2 \ge 0 \) for all x, the expression has a minimum value of \( c=-3 \), occurring when \( (x-2)^2=0 \), i.e. when \( x=2 \).

[1] correctly factors out 2 from the x-terms; [1] correctly completes the square inside the bracket, \( (x-2)^2-4 \); [1] correct final form \( 2(x-2)^2-3 \) with a, b, c correctly identified; [1] correct minimum value -3 stated; [1] correct value \( x=2 \) stated. ECF applied to the minimum value and x if the completed-square form is correct in structure but has a numerical slip.

First solve the corresponding equation \( x^2-2x-15=0 \): factorising, \( (x-5)(x+3)=0 \), so \( x=5 \) or \( x=-3 \). Since the coefficient of \( x^2 \) is positive, the graph of \( y=x^2-2x-15 \) is a upward-opening parabola, so \( y\le0 \) between the roots. Hence the solution is \( -3 \le x \le 5 \).

[1] correct factorisation \( (x-5)(x+3) \); [1] correct roots \( x=5, x=-3 \); [1] correctly identifies that the region between the roots satisfies the inequality (upward parabola, \( \le0 \)); [1] correct final answer in set notation \( -3\le x\le5 \). Reject \( x\le-3 \) or \( x\ge5 \) (the region outside the roots) as this corresponds to \( \ge0 \), not \( \le0 \).

Rearranging, \( \sin\theta = \dfrac{1}{2} \). The principal solution is \( \theta = \sin^{-1}(0.5) = 30^{\circ} \). Since sine is also positive in the second quadrant, a second solution in the given range is \( \theta = 180^{\circ}-30^{\circ} = 150^{\circ} \). Both solutions lie within \( 0^{\circ}\le\theta\le360^{\circ} \), so the solutions are \( \theta=30^{\circ} \) and \( \theta=150^{\circ} \).

[1] correct rearrangement \( \sin\theta=0.5 \); [1] correct principal solution \( \theta=30^{\circ} \); [1] correct use of the second-quadrant identity \( 180^{\circ}-\theta \); [1] both correct solutions given, \( 30^{\circ} \) and \( 150^{\circ} \), and no extra incorrect solutions included.

Substituting \( y=x+1 \) into \( x^2+y^2=13 \): \( x^2+(x+1)^2=13 \), so \( x^2+x^2+2x+1=13 \), giving \( 2x^2+2x-12=0 \), i.e. \( x^2+x-6=0 \). Factorising: \( (x+3)(x-2)=0 \), so \( x=-3 \) or \( x=2 \). Using \( y=x+1 \): when \( x=-3 \), \( y=-2 \); when \( x=2 \), \( y=3 \). Check: \( (-3)^2+(-2)^2=9+4=13 \) ✓; \( 2^2+3^2=4+9=13 \) ✓.

[1] correct substitution of \( y=x+1 \) into the second equation; [1] correctly expanded to \( 2x^2+2x-12=0 \) or equivalent; [1] correctly simplified to \( x^2+x-6=0 \); [1] correctly factorised/solved, \( x=-3 \) or \( x=2 \); [1] both corresponding y-values correctly found; [1] both solution pairs stated correctly and clearly as (x,y) pairs.

(a) \( AB = \begin{pmatrix}2&1\\3&-2\end{pmatrix}\begin{pmatrix}1&0\\-1&2\end{pmatrix} = \begin{pmatrix}(2)(1)+(1)(-1) & (2)(0)+(1)(2) \\ (3)(1)+(-2)(-1) & (3)(0)+(-2)(2)\end{pmatrix} = \begin{pmatrix}1&2\\5&-4\end{pmatrix} \).
(b) \( \det(A) = (2)(-2)-(1)(3) = -4-3 = -7 \).

(a) [1] correct top-row entries (1, 2); [1] correct bottom-row entries (5, -4); [1] matrix correctly assembled/presented. (b) [1] correct formula \( ad-bc \) applied; [1] correct substitution; [1] correct value \( -7 \).

\( \det(M) = (3)(2)-(1)(5) = 6-5 = 1 \). So \( M^{-1} = \dfrac{1}{1}\begin{pmatrix}2&-1\\-5&3\end{pmatrix} = \begin{pmatrix}2&-1\\-5&3\end{pmatrix} \). Writing the equations as \( M\binom{x}{y}=\binom{11}{19} \), we get \( \binom{x}{y} = M^{-1}\binom{11}{19} = \begin{pmatrix}2&-1\\-5&3\end{pmatrix}\binom{11}{19} = \binom{2(11)-1(19)}{-5(11)+3(19)} = \binom{22-19}{-55+57} = \binom{3}{2} \). So \( x=3,\ y=2 \). Check: \( 3(3)+2=11 \) ✓; \( 5(3)+2(2)=15+4=19 \) ✓.

[1] correct determinant \( \det(M)=1 \); [1] correct inverse matrix \( \begin{pmatrix}2&-1\\-5&3\end{pmatrix} \); [1] correct method \( \binom{x}{y}=M^{-1}\binom{11}{19} \) set up; [1] correct matrix multiplication carried out; [1] \( x=3 \) and \( y=2 \) both correctly stated; [1] both solutions verified/checked in the original equations. ECF from an incorrect inverse carried through consistently.

Using the law of logarithms \( \log_2 A - \log_2 B = \log_2\left(\dfrac{A}{B}\right) \): \( \log_2\left(\dfrac{x+3}{x-1}\right)=2 \). Converting from logarithmic to index form: \( \dfrac{x+3}{x-1}=2^2=4 \). So \( x+3=4(x-1)=4x-4 \), giving \( 3+4=4x-x \), i.e. \( 7=3x \), so \( x=\dfrac{7}{3} \). Since \( \log_2(x+3) \) requires \( x>-3 \) and \( \log_2(x-1) \) requires \( x>1 \), the overall restriction is \( x>1 \); \( x=\dfrac{7}{3}\approx2.33 \) satisfies this, so it is a valid solution. Check: \( \dfrac{7/3+3}{7/3-1}=\dfrac{16/3}{4/3}=4 \), and \( \log_2 4=2 \) ✓.

[1] correct combination of logs into a single log using the subtraction law; [1] correct conversion to index form, \( (x+3)/(x-1)=4 \); [1] correct expansion \( x+3=4x-4 \); [1] correctly solved, \( x=7/3 \); [1] correct restriction on x stated (\( x>1 \)); [1] correctly confirms \( x=7/3 \) satisfies the restriction; [1] check/verification carried out correctly in the original equation.

\( 75 = 25\times3 = 5^2\times3 \). So \( \log_a75 = \log_a(5^2\times3) = \log_a(5^2)+\log_a3 \) (multiplication law) \( = 2\log_a5+\log_a3 \) (power law) \( = 2p+q \).

[1] correctly expresses 75 as \( 5^2\times3 \); [1] correctly applies the multiplication law of logs, \( \log_a(5^2)+\log_a3 \); [1] correctly applies the power law, \( 2\log_a5 \); [1] correct substitution of p and q; [1] final answer \( 2p+q \) correctly stated; [1] additional mark for showing each law applied explicitly and clearly (full working shown, not just the answer).

\( \dfrac{dy}{dx} = 6x^2-10x+4 \). At \( x=2 \): gradient \( = 6(2)^2-10(2)+4 = 24-20+4 = 8 \). The y-coordinate at \( x=2 \): \( y = 2(2)^3-5(2)^2+4(2)-1 = 16-20+8-1 = 3 \). So the point of contact is \( (2,3) \). Using \( y-y_1=m(x-x_1) \): \( y-3=8(x-2) \), so \( y=8x-16+3=8x-13 \).

[1] correct differentiation, \( dy/dx=6x^2-10x+4 \); [1] correct substitution of x=2 into the derivative; [1] gradient \( =8 \); [1] correct y-coordinate at x=2, \( y=3 \); [1] correct use of \( y-y_1=m(x-x_1) \) with their gradient and point; [1] correctly expanded; [1] final tangent equation \( y=8x-13 \).

The indefinite integral is \( \displaystyle\int(3x^2-4x+2)\,dx = x^3-2x^2+2x\ (+C) \). Evaluating between the limits: at \( x=3 \): \( 3^3-2(3)^2+2(3) = 27-18+6 = 15 \). At \( x=1 \): \( 1^3-2(1)^2+2(1) = 1-2+2 = 1 \). So \( \displaystyle\int_1^3(3x^2-4x+2)\,dx = 15-1 = 14 \).

[1] correct integration of \( 3x^2 \) to \( x^3 \); [1] correct integration of \( -4x \) to \( -2x^2 \); [1] correct integration of the constant 2 to \( 2x \); [1] correct evaluation at the upper limit x=3, giving 15; [1] correct evaluation at the lower limit x=1, giving 1; [1] correct final subtraction giving 14.

(a) \( \dfrac{dy}{dx} = 3x^2-12x+9 \).
(b) At a stationary point, \( \dfrac{dy}{dx}=0 \): \( 3x^2-12x+9=0 \), so \( x^2-4x+3=0 \), giving \( (x-1)(x-3)=0 \), so \( x=1 \) or \( x=3 \). At \( x=1 \): \( y=1-6+9+2=6 \). At \( x=3 \): \( y=27-54+27+2=2 \). So the stationary points are \( (1,6) \) and \( (3,2) \).
(c) \( \dfrac{d^2y}{dx^2}=6x-12 \). At \( x=1 \): \( 6(1)-12=-6<0 \), so \( (1,6) \) is a local maximum. At \( x=3 \): \( 6(3)-12=6>0 \), so \( (3,2) \) is a local minimum.
(d) When \( x=0 \), \( y=2 \), so the curve crosses the y-axis at \( (0,2) \). Since this is a positive cubic (positive \( x^3 \) coefficient), as \( x\to+\infty \), \( y\to+\infty \), and as \( x\to-\infty \), \( y\to-\infty \); the curve rises from the bottom left, reaches a local maximum at (1,6), falls to a local minimum at (3,2), then rises again to the top right.

(a) [1] correct differentiation of each term; [1] fully correct \( dy/dx=3x^2-12x+9 \). (b) [1] correctly sets \( dy/dx=0 \); [1] correctly factorises/solves for \( x=1, x=3 \); [1] correct y-value at x=1 (y=6); [1] correct y-value at x=3 (y=2). (c) [1] correct second derivative \( 6x-12 \); [1] correct evaluation at x=1 (-6) with 'maximum' correctly stated; [1] correct evaluation at x=3 (6) with 'minimum' correctly stated; [1] both natures correctly and clearly linked to the correct point. (d) [1] correct y-intercept (0,2); [1] correct description of end behaviour (falls to -∞ on the left, rises to +∞ on the right) consistent with a positive cubic.

(a) Taking logarithms to base 10 of both sides of \( N=N_0b^t \): \( \log_{10}N = \log_{10}(N_0b^t) = \log_{10}N_0+\log_{10}(b^t) = \log_{10}N_0+t\log_{10}b \), using the multiplication and power laws of logarithms. Rearranged as \( \log_{10}N = (\log_{10}b)t+\log_{10}N_0 \), this is of the linear form \( Y=mt+c \) with \( Y=\log_{10}N \), gradient \( m=\log_{10}b \), and intercept \( c=\log_{10}N_0 \); since \( N_0 \) and b are constants, the gradient and intercept are both constant, so the graph is a straight line.
(b) From (a), intercept \( =\log_{10}N_0=2.000 \), so \( N_0=10^{2.000}=100 \). Gradient \( =\log_{10}b=0.0170 \), so \( b=10^{0.0170}=1.040 \) (4 s.f.).
(c) The population doubles when \( N=2N_0 \), i.e. \( b^t=2 \). Taking logs: \( t\log_{10}b=\log_{10}2 \), so \( t=\dfrac{\log_{10}2}{\log_{10}b}=\dfrac{\log_{10}2}{0.0170}=17.7 \) hours (3 s.f.).

(a) [1] correctly takes logs of both sides; [1] correctly applies the multiplication law; [1] correctly applies the power law to get \( t\log_{10}b \); [1] correctly identifies the linear form and states gradient \( =\log_{10}b \), intercept \( =\log_{10}N_0 \). (b) [1] correct relation \( N_0=10^{intercept} \) used; [1] \( N_0=100 \); [1] correct relation \( b=10^{gradient} \) used; [1] \( b=1.040 \) (accept 1.039-1.040). (c) [1] correct condition \( b^t=2 \) (doubling); [1] correct use of logs to solve for t; [1] correct substitution of their b (or gradient); [1] \( t=17.7 \) hours (accept 17.6-17.8, ECF from (b)).

(a) After removing squares of side x from each corner and folding up the sides, the base of the box has length \( (32-2x) \) and width \( (20-2x) \), and the box has height x. So \( V = x(32-2x)(20-2x) \). Expanding: \( (32-2x)(20-2x) = 640-64x-40x+4x^2 = 640-104x+4x^2 \). So \( V = x(640-104x+4x^2) = 640x-104x^2+4x^3 = 4x^3-104x^2+640x \), as required.
(b) \( \dfrac{dV}{dx} = 12x^2-208x+640 \).
(c) At a stationary point, \( 12x^2-208x+640=0 \); dividing by 4, \( 3x^2-52x+160=0 \). Using the quadratic formula: \( x=\dfrac{52\pm\sqrt{52^2-4(3)(160)}}{2(3)} = \dfrac{52\pm\sqrt{2704-1920}}{6} = \dfrac{52\pm\sqrt{784}}{6} = \dfrac{52\pm28}{6} \). So \( x=\dfrac{80}{6}=\dfrac{40}{3} \) or \( x=\dfrac{24}{6}=4 \). Since the width of the base is \( 20-2x \), which must be positive, \( x<10 \); \( x=\dfrac{40}{3}\approx13.3 \) is rejected as it would make the width negative, so \( x=4 \) is the only valid stationary point. Checking the second derivative, \( \dfrac{d^2V}{dx^2}=24x-208 \); at \( x=4 \), this is \( 24(4)-208=96-208=-112<0 \), confirming \( x=4 \) gives a maximum.
(d) \( V_{max} = 4(4)^3-104(4)^2+640(4) = 4(64)-104(16)+2560 = 256-1664+2560 = 1152\text{ cm}^3 \).

(a) [1] correct base dimensions \( (32-2x) \) and \( (20-2x) \) identified; [1] correct expansion of the product; [1] correctly shown to equal \( 4x^3-104x^2+640x \). (b) [1] correct differentiation of each term; [1] fully correct \( dV/dx=12x^2-208x+640 \). (c) [1] correctly sets \( dV/dx=0 \) and simplifies (e.g. divides by 4); [1] correct use of the quadratic formula (or equivalent) with correct substitution; [1] both roots correctly found, \( x=4 \) and \( x=40/3 \); [1] correctly rejects \( x=40/3 \) with a valid physical reason (negative side length); [1] correct use of the second derivative to confirm x=4 is a maximum. (d) [1] correct substitution of x=4 into V; [1] correct arithmetic; [1] \( V_{max}=1152\text{ cm}^3 \) (ECF from (c)).

(a) \( \mathbf{F_1}+\mathbf{F_2} = (3+(-5))\mathbf{i}+(4+2)\mathbf{j} = (-2\mathbf{i}+6\mathbf{j})\text{ N} \).
(b) Magnitude \( = \sqrt{(-2)^2+6^2} = \sqrt{4+36} = \sqrt{40} = 6.32\text{ N} \) (3 s.f.).

(a) [1] correct i-component (-2); [1] correct j-component (6). (b) [1] correct use of Pythagoras' theorem; [1] correct squares (4 and 36) summed to give 40; [1] correct square root taken; [1] final answer 6.32 N (3 s.f.), ECF from (a).

(a) At maximum height the vertical velocity is zero. Using \( v^2=u^2-2gs \) with \( v=0,\ u=18 \): \( 0=18^2-2(10)s \), so \( s=\dfrac{324}{20}=16.2\text{ m} \).
(b) By symmetry, the time to reach maximum height equals the time to fall back down, so the total time is \( t=\dfrac{2u}{g}=\dfrac{2\times18}{10}=3.6\text{ s} \) (this can also be found from \( s=ut-\tfrac12gt^2=0 \) at return, giving \( t=2u/g \) directly).

(a) [1] correct equation of motion selected (e.g. \( v^2=u^2-2gs \)); [1] correct substitution with v=0; [1] \( s=16.2\text{ m} \). (b) [1] correct method (e.g. \( t=2u/g \), or setting \( s=0 \) in \( s=ut-\tfrac12gt^2 \)); [1] correct substitution; [1] \( t=3.6\text{ s} \).

(a) Resolving along the incline (up the slope positive), the component of P up the slope is \( P\cos25^{\circ} \), and the component of the weight down the slope is \( 40\sin25^{\circ} \). Since the plane is smooth (no friction) and the particle is in equilibrium: \( P\cos25^{\circ}=40\sin25^{\circ} \), so \( P = 40\tan25^{\circ} = 18.7\text{ N} \) (3 s.f.).
(b) Resolving perpendicular to the incline: \( R = 40\cos25^{\circ}+P\sin25^{\circ} = 40\cos25^{\circ}+18.65\sin25^{\circ} = 36.25+7.88 = 44.1\text{ N} \) (3 s.f.). As a check, resolving vertically and horizontally instead: vertically, \( R\cos25^{\circ}=40 \Rightarrow R=40/\cos25^{\circ}=44.1\text{ N} \), which agrees exactly with the value found by resolving perpendicular to the incline, confirming the result.

(a) [1] correct component of P along the incline, \( P\cos25^{\circ} \); [1] correct component of weight along the incline, \( 40\sin25^{\circ} \); [1] correct equilibrium equation formed; [1] \( P=18.7\text{ N} \). (b) [1] correct component of weight perpendicular to incline, \( 40\cos25^{\circ} \); [1] correct component of P perpendicular to incline, \( P\sin25^{\circ} \) (ECF); [1] correct equilibrium equation formed; [1] \( R=44.1\text{ N} \) (accept the equivalent method resolving vertically/horizontally with \( R=40/\cos25^{\circ} \), and accept 44.0-44.2 N).

(a) For B (moving downwards, taking downward as positive for B): \( m_Bg-T = m_Ba \), i.e. \( 30-T=3a \). For A (on a smooth table, the string pulls it horizontally): \( T=m_Aa=5a \). Adding the two equations to eliminate T: \( 30-5a=3a \Rightarrow 30=8a \Rightarrow a=3.75\text{ m s}^{-2} \).
(b) Substituting into \( T=5a \): \( T=5\times3.75=18.75\text{ N} \). As a check, using B's equation: \( T=m_Bg-m_Ba=30-3(3.75)=30-11.25=18.75\text{ N} \), which agrees.

(a) [1] correct equation of motion for A, \( T=5a \); [1] correct equation of motion for B, \( 30-T=3a \); [1] correct elimination of T; [1] correctly solved for a; [1] \( a=3.75\text{ m s}^{-2} \). (b) [1] correct substitution into either equation (ECF); [1] correct working; [1] \( T=18.75\text{ N} \), with a valid check shown using the other particle's equation.

(a) Since C is \( 1.0\text{ m} \) from B, and the beam is \( 5.0\text{ m} \) long, C is \( 5.0-1.0=4.0\text{ m} \) from A. Since the beam is uniform, its weight acts at its midpoint, \( 2.5\text{ m} \) from A.
(b) Taking moments about A (clockwise positive, with downward forces at C's support treated as the unknown upward reaction \( R_C \), and the beam's weight and the load both acting downwards, causing clockwise moments, balanced by the anticlockwise moment of \( R_C \)): \( R_C\times4.0 = 80\times2.5+60\times5.0 = 200+300 = 500 \). So \( R_C = \dfrac{500}{4.0} = 125\text{ N} \).
(c) Resolving vertically for equilibrium: \( R_A+R_C = 80+60 = 140 \), so \( R_A = 140-125 = 15\text{ N} \).

(a) [1] correct distance of C from A (4.0 m); [1] correct distance of the beam's weight from A (2.5 m, midpoint). (b) [1] correct moment of the beam's weight about A (80×2.5=200); [1] correct moment of the load about A (60×5.0=300); [1] correct moments equation \( R_C\times4.0=500 \); [1] \( R_C=125\text{ N} \). (c) [1] correct vertical equilibrium equation \( R_A+R_C=140 \); [1] \( R_A=15\text{ N} \) (ECF from (b)).

(a) Using \( v=u+at \) for B: \( 30=0+2.5t_1 \), so \( t_1=\dfrac{30}{2.5}=12\text{ s} \).
(b) Using \( s=\tfrac12at_1^2 \) (or \( v^2=2as \)): \( s_B=\tfrac12(2.5)(12)^2=\tfrac12(2.5)(144)=180\text{ m} \). (Check: \( v^2/2a=30^2/(2\times2.5)=900/5=180\text{ m} \), consistent.)
(c) A travels at a constant \( 25\text{ m s}^{-1} \), so in \( t_1=12\text{ s} \): \( s_A=25\times12=300\text{ m} \). Since \( 300\text{ m}>180\text{ m} \), car A is ahead of car B at this instant, by \( 300-180=120\text{ m} \).
(d) After \( t_1 \), B travels at a constant \( 30\text{ m s}^{-1} \), which is \( 30-25=5\text{ m s}^{-1} \) faster than A, so B closes the \( 120\text{ m} \) gap at a rate of \( 5\text{ m s}^{-1} \). Further time needed: \( t_2=\dfrac{120}{5}=24\text{ s} \). Total time since \( t=0 \): \( t_1+t_2=12+24=36\text{ s} \).
(e) Total distance travelled by A in 36 s: \( 25\times36=900\text{ m} \). Total distance travelled by B: \( 180\text{ m} \) (in the first 12 s) plus \( 30\times24=720\text{ m} \) (in the next 24 s), giving \( 180+720=900\text{ m} \). Both distances are equal (900 m), confirming that B has indeed caught up with A after a total time of 36 s.

(a) [1] correct use of \( v=u+at \); [1] \( t_1=12\text{ s} \). (b) [1] correct method (e.g. \( \tfrac12at_1^2 \) or \( v^2/2a \)); [1] correct substitution; [1] \( s_B=180\text{ m} \). (c) [1] correct method \( s_A=25\times12 \); [1] \( s_A=300\text{ m} \); [1] correctly identifies A is ahead by 120 m (ECF). (d) [1] correctly identifies the closing speed as \( 30-25=5\text{ m s}^{-1} \); [1] correct method \( t_2=\text{gap}/\text{closing speed} \); [1] \( t_2=24\text{ s} \) (ECF); [1] correct total time \( 36\text{ s} \). (e) [1] correct total distance for A (900 m, ECF); [1] correct total distance for B calculated in two stages and shown to equal 900 m (ECF), with a valid concluding statement that the distances match.

(a) \( S_{xx}=139-\dfrac{27^2}{6}=139-\dfrac{729}{6}=139-121.5=17.5 \). \( S_{yy}=20100-\dfrac{340^2}{6}=20100-\dfrac{115600}{6}=20100-19266.67=833.33 \) (2 d.p.).
(b) \( S_{xy}=1650-\dfrac{27\times340}{6}=1650-\dfrac{9180}{6}=1650-1530=120 \).
(c) \( r=\dfrac{120}{\sqrt{17.5\times833.33}}=\dfrac{120}{\sqrt{14583.3}}=\dfrac{120}{120.76}=0.994 \) (3 s.f.). Since r is very close to 1, this indicates a very strong, positive correlation between hours revised and test score.
(d) \( b=\dfrac{S_{xy}}{S_{xx}}=\dfrac{120}{17.5}=6.857\text{ (}=6.86\text{ to 3 s.f.)} \). \( \bar{x}=\dfrac{27}{6}=4.5 \), \( \bar{y}=\dfrac{340}{6}=56.67 \). Since the regression line passes through \( (\bar{x},\bar{y}) \): \( a=\bar{y}-b\bar{x}=56.67-6.857\times4.5=56.67-30.86=25.8 \) (3 s.f.). So the regression line is \( y=25.8+6.86x \).
(e) At \( x=8 \): \( y=25.8+6.86(8)=25.8+54.9=80.7 \), so the estimated score is about 81 (to the nearest whole mark). However, \( x=8 \) hours lies outside the range of the original data (2 to 7 hours), so this is an extrapolation; the estimate may be less reliable, since the linear relationship might not continue to hold, and in this case predicts a score close to (or exceeding, for larger x) the maximum possible mark of 100.

(a) [1] correct \( S_{xx}=17.5 \); [1] correct method for \( S_{yy} \); [1] \( S_{yy}=833.33 \) (accept 833.3). (b) [1] correct method; [1] \( S_{xy}=120 \). (c) [1] correct substitution into the r formula (ECF); [1] \( r=0.994 \) (accept 0.99); [1] correct comment (strong/very strong, positive). (d) [1] correct gradient \( b=6.86 \) (ECF); [1] correct use of \( \bar x,\bar y \) to find \( a=25.8 \) (ECF); [1] correctly stated regression equation. (e) [1] correct substitution x=8 into their regression line, giving ≈80.7 (ECF); [1] valid comment identifying this as extrapolation (x=8 outside the data range 2-7) and correctly noting reduced reliability.

(a) For a linear coding \( y=ax+b \), the mean transforms in exactly the same way as the data: \( \bar{y}=a\bar{x}+b \).
(b) Here \( a=3,\ b=-10 \): \( \bar{y}=3(24)-10=72-10=62 \).
(c) A constant shift (the \( +b \)) does not affect the spread of the data, but multiplying by a produces a proportional change in spread, so the standard deviation transforms as \( \sigma_y=|a|\sigma_x \) (the constant b has no effect on standard deviation). Here \( \sigma_y=|3|\times5=15 \).

(a) [1] correct formula \( \bar{y}=a\bar{x}+b \) stated. (b) [1] correct substitution a=3, b=-10; [1] \( \bar{y}=62 \). (c) [1] correctly states that b does not affect the standard deviation; [1] correct formula \( \sigma_y=|a|\sigma_x \) applied; [1] \( \sigma_y=15 \).

(a) \( \Sigma x = 32+35+29+31+38+33+30 = 228 \). Mean \( = \dfrac{228}{7} = 32.6\text{ minutes} \) (3 s.f.).
(b) \( \Sigma x^2 = 32^2+35^2+29^2+31^2+38^2+33^2+30^2 = 1024+1225+841+961+1444+1089+900 = 7484 \). Using \( \sigma=\sqrt{\dfrac{\Sigma x^2}{n}-\bar{x}^2} \): \( \sigma = \sqrt{\dfrac{7484}{7}-(32.571\ldots)^2} = \sqrt{1069.14-1060.90} = \sqrt{8.245} = 2.87\text{ minutes} \) (3 s.f.). (As a check: computing the deviations from the mean directly, \( \Sigma(x-\bar{x})^2/7 \), gives the same value of 8.245, confirming the result.)

(a) [1] correct \( \Sigma x=228 \); [1] mean \( =32.6 \) (accept 32.57, 32.6). (b) [1] correct \( \Sigma x^2=7484 \); [1] correct substitution into the given formula; [1] correct value under the square root (8.24-8.25); [1] \( \sigma=2.87\text{ minutes} \) (accept 2.86-2.87, ECF from (a)).

(a) On the first draw, \( P(\text{red})=\dfrac{5}{8} \) and \( P(\text{blue})=\dfrac{3}{8} \). Since the balls are drawn without replacement, there are 7 balls left for the second draw, and the probabilities on the second draw depend on the outcome of the first: if the first ball is red (leaving 4 red, 3 blue), \( P(\text{red}\mid\text{red})=\dfrac{4}{7} \), \( P(\text{blue}\mid\text{red})=\dfrac{3}{7} \); if the first ball is blue (leaving 5 red, 2 blue), \( P(\text{red}\mid\text{blue})=\dfrac{5}{7} \), \( P(\text{blue}\mid\text{blue})=\dfrac{2}{7} \).
(b) \( P(\text{RR})=\dfrac{5}{8}\times\dfrac{4}{7}=\dfrac{20}{56}=\dfrac{5}{14} \).
(c) The two balls are different colours if the outcome is Red-then-Blue or Blue-then-Red: \( P(\text{RB})=\dfrac{5}{8}\times\dfrac{3}{7}=\dfrac{15}{56} \); \( P(\text{BR})=\dfrac{3}{8}\times\dfrac{5}{7}=\dfrac{15}{56} \). So \( P(\text{different colours})=\dfrac{15}{56}+\dfrac{15}{56}=\dfrac{30}{56}=\dfrac{15}{28} \).

(a) [1] correct first-draw probabilities (5/8, 3/8); [1] correct second-draw (conditional) probabilities correctly dependent on the first draw. (b) [1] correct identification of the RR branch; [1] correct multiplication \( \tfrac58\times\tfrac47 \); [1] \( \tfrac{5}{14} \) correctly simplified. (c) [1] correctly identifies both relevant branches (RB and BR); [1] both branch probabilities correctly calculated (15/56 each); [1] correctly summed and simplified to \( \tfrac{15}{28} \).

(a) Each roll is an independent trial with a fixed probability \( \dfrac{1}{6} \) of 'success' (a 6), repeated a fixed number of times (8), so \( X\sim B\left(8,\dfrac{1}{6}\right) \).
(b) \( P(X=2)=\binom{8}{2}\left(\dfrac16\right)^2\left(\dfrac56\right)^6 = 28\times\dfrac{1}{36}\times0.3349 = 0.260 \) (3 s.f.).
(c) \( P(X\ge1)=1-P(X=0)=1-\left(\dfrac56\right)^8=1-0.2326=0.767 \) (3 s.f.).

(a) [1] correctly identifies a binomial distribution; [1] correct parameters \( n=8,\ p=1/6 \). (b) [1] correct use of \( \binom{8}{2} \) (=28); [1] correct substitution into the binomial formula; [1] \( P(X=2)=0.260 \) (accept 0.259-0.260). (c) [1] correct use of the complement, \( 1-P(X=0) \); [1] correct calculation of \( P(X=0)=(5/6)^8 \); [1] \( P(X\ge1)=0.767 \) (accept 0.766-0.767).

(a) \( z=\dfrac{172-165}{6}=1.167 \) (3 d.p.). From tables, \( P(Z>1.167) = 1-\Phi(1.167) = 1-0.8783 = 0.122 \) (3 s.f.).
(b) \( z_1=\dfrac{158-165}{6}=-1.167 \); \( z_2=\dfrac{170-165}{6}=0.833 \). \( P(158z)=0.10 \), so \( z \) is the value such that \( \Phi(z)=0.90 \); from tables, \( z=1.282 \) (3 d.p.). Then \( h=\mu+z\sigma = 165+1.282\times6 = 165+7.69 = 172.7\text{ cm} \) (1 d.p.).

(a) [1] correct standardisation, \( z=1.167 \); [1] correct use of \( 1-\Phi(z) \); [1] \( P=0.122 \) (accept 0.121-0.122). (b) [1] both z-values correctly calculated (-1.167 and 0.833); [1] correct use of \( \Phi(z_2)-\Phi(z_1) \) (or equivalent, e.g. subtracting tail areas); [1] \( P=0.676 \) (accept 0.675-0.677). (c) [1] correctly identifies \( \Phi(z)=0.90 \) is needed; [1] correct z-value, \( z=1.282 \) (accept 1.28); [1] \( h=172.7\text{ cm} \) (accept 172.6-172.7).

(a) Since the letters must be different and their order matters, this is a permutation: \( ^{26}P_4 = 26\times25\times24\times23 = 358\,800 \).
(b) Similarly for the digits: \( ^{10}P_2 = 10\times9 = 90 \).
(c) By the multiplication principle, the total number of passwords is the product of the number of ways of choosing each part: \( 358\,800\times90 = 32\,292\,000 \).

(a) [1] correct method \( ^{26}P_4 \) (or \( 26\times25\times24\times23 \)); [1] \( 358\,800 \). (b) [1] correct method \( ^{10}P_2 \); [1] \( 90 \). (c) [1] correct use of the multiplication principle (multiplying (a) and (b)); [1] \( 32\,292\,000 \) (ECF from (a) and (b)).

The statement \( (p\land q)\Rightarrow\lnot r \) is false only when its antecedent \( (p\land q) \) is true and its consequent \( \lnot r \) is false, i.e. only when \( p=T,\ q=T \) and \( r=T \) (since \( \lnot r \) is false exactly when \( r=T \)). The full truth table (8 rows) is:

p q r p∧q ¬r (p∧q)⇒¬r
T T T T F F
T T F T T T
T F T F T T
T F F F T T
F T T F T T
F T F F T T
F F T F T T
F F F F T T

Since the final column is true for 7 of the 8 rows and false for exactly one row (p=T, q=T, r=T), the statement is not always true, so it is not a tautology; and it is not always false, so it is not a contradiction. It is therefore neither a tautology nor a contradiction.

[1] correct column for \( p\land q \); [1] correct column for \( \lnot r \); [1] correct final column for \( (p\land q)\Rightarrow\lnot r \), correctly false only in the row p=T, q=T, r=T; [1] all 8 rows of truth-value combinations present and correctly labelled; [1] correctly concludes the statement is neither a tautology nor a contradiction; [1] correct reason given, identifying the single row where the statement is false (showing it is not a tautology) while noting it is true elsewhere (showing it is not a contradiction).

(a) With no restrictions, this is choosing 5 people from 10: \( \binom{10}{5}=252 \).
(b) Choosing 3 men from 6 and 2 women from 4, independently, and multiplying (since both must happen): \( \binom{6}{3}\times\binom{4}{2}=20\times6=120 \).
(c) 'At least 4 men' means either exactly 4 men (and 1 woman) or exactly 5 men (and 0 women), and these two cases are mutually exclusive, so their counts are added. Exactly 4 men, 1 woman: \( \binom{6}{4}\times\binom{4}{1}=15\times4=60 \). Exactly 5 men, 0 women: \( \binom{6}{5}\times\binom{4}{0}=6\times1=6 \). Total: \( 60+6=66 \).

(a) [1] correct method \( \binom{10}{5} \); [1] \( 252 \). (b) [1] correct method \( \binom{6}{3}\times\binom{4}{2} \); [1] \( 120 \). (c) [1] correctly identifies the two mutually exclusive cases (4 men+1 woman, and 5 men+0 women); [1] both cases correctly calculated (60 and 6); [1] correctly summed to give \( 66 \).

(a) Carpentry time: \( 4x+2y\le60 \). Finishing time: \( 2x+3y\le42 \). Since the number of tables and chairs cannot be negative: \( x\ge0 \) and \( y\ge0 \).
(b) Profit: \( P=30x+20y \).
(c) The carpentry constraint simplifies to \( 2x+y=30 \) (boundary). Setting \( y=0 \): \( x=15 \), giving vertex \( (15,0) \) (check this satisfies finishing: \( 2(15)+3(0)=30\le42 \) ✓, so it is on the feasible boundary). Setting \( x=0 \) in the finishing constraint \( 2x+3y=42 \): \( y=14 \), giving vertex \( (0,14) \) (check carpentry: \( 4(0)+2(14)=28\le60 \) ✓). Solving the two boundary equations simultaneously: \( 2x+y=30 \) and \( 2x+3y=42 \); subtracting, \( 2y=12 \), so \( y=6 \), and \( 2x+6=30\Rightarrow x=12 \), giving vertex \( (12,6) \). Together with the origin \( (0,0) \), the vertices of the feasible region are \( (0,0),\ (15,0),\ (12,6),\ (0,14) \).
(d) \( P(0,0)=0 \). \( P(15,0)=30(15)+20(0)=450 \). \( P(12,6)=30(12)+20(6)=360+120=480 \). \( P(0,14)=30(0)+20(14)=280 \). The maximum occurs at \( (12,6) \), so the company should make 12 tables and 6 chairs each week, giving a maximum weekly profit of £480. (Check: at (12,6), carpentry used \( =4(12)+2(6)=60 \) hours exactly, and finishing used \( =2(12)+3(6)=42 \) hours exactly, confirming both resources are fully and validly used at the optimum.)

(a) [1] correct carpentry inequality; [1] correct finishing inequality; [1] \( x\ge0 \); [1] \( y\ge0 \). (b) [1] correct profit expression. (c) [1] correct vertex (0,0) identified; [1] correct vertex (15,0) found and justified; [1] correct vertex (0,14) found and justified; [1] correct simultaneous solution method for the two boundary lines; [1] correct vertex (12,6). (d) [1] profit correctly evaluated at all four vertices (ECF); [1] correctly identifies (12,6) as the maximum; [1] correct final answer (12 tables, 6 chairs, £480 maximum profit), with a valid check that both resource constraints are satisfied.

(a) The 4-point moving averages, each the mean of 4 consecutive quarters: \( \text{MA}_1=\dfrac{20+35+50+25}{4}=32.5 \); \( \text{MA}_2=\dfrac{35+50+25+24}{4}=33.5 \); \( \text{MA}_3=\dfrac{50+25+24+39}{4}=34.5 \); \( \text{MA}_4=\dfrac{25+24+39+54}{4}=35.5 \); \( \text{MA}_5=\dfrac{24+39+54+29}{4}=36.5 \).
(b) Since the moving averages are based on an even number of quarters (4), each \( \text{MA} \) value falls between two actual quarters, so pairs of consecutive moving averages must be averaged (centred) to align with an actual quarter. The centred moving average for Year 1 Q4 is the average of \( \text{MA}_2 \) and \( \text{MA}_3 \): \( \dfrac{33.5+34.5}{2}=34.0 \). The centred moving average for Year 2 Q1 is the average of \( \text{MA}_3 \) and \( \text{MA}_4 \): \( \dfrac{34.5+35.5}{2}=35.0 \).
(c) The trend value for Year 2 Q1 is \( 35.0 \) (from (b)) and the actual sales value is \( 24 \). Using the additive model, seasonal variation \( = \text{actual}-\text{trend} = 24-35.0 = -11.0 \). (This negative value indicates that Q1 sales are typically about £11,000 below the underlying trend, consistent with Q1 being the seasonal low point in the data given.)

(a) [1] each for all five correctly calculated moving averages (32.5, 33.5, 34.5, 35.5, 36.5) — award [1] per correct value up to [5], or [3] for correct method with 1-2 arithmetic slips. (b) [1] correct centred value for Year 1 Q4 (34.0); [1] correct centred value for Year 2 Q1 (35.0), both using the correct pair of moving averages (ECF from (a)). (c) [1] correct use of the additive model formula (actual − trend); [1] correct value \( -11.0 \) (ECF from (b)), with correct sign.

(a) A has no predecessor: \( ES_A=0,\ EF_A=0+4=4 \). B has no predecessor: \( ES_B=0,\ EF_B=6 \). C depends on A: \( ES_C=EF_A=4,\ EF_C=4+3=7 \). D depends on A: \( ES_D=4,\ EF_D=4+5=9 \). E depends on B and C: \( ES_E=\max(EF_B,EF_C)=\max(6,7)=7,\ EF_E=7+4=11 \). F depends on D: \( ES_F=EF_D=9,\ EF_F=9+7=16 \). G depends on E and F: \( ES_G=\max(EF_E,EF_F)=\max(11,16)=16,\ EF_G=16+2=18 \).
(b) Working backwards from the project finish time of 18 days (the largest EF, at G, which has no successor so \( LF_G=18 \)): \( LS_G=18-2=16 \). F's only successor is G: \( LF_F=LS_G=16,\ LS_F=16-7=9 \). E's only successor is G: \( LF_E=LS_G=16,\ LS_E=16-4=12 \). D's only successor is F: \( LF_D=LS_F=9,\ LS_D=9-5=4 \). C's only successor is E: \( LF_C=LS_E=12,\ LS_C=12-3=9 \). B's only successor is E: \( LF_B=LS_E=12,\ LS_B=12-6=6 \). A's successors are C and D: \( LF_A=\min(LS_C,LS_D)=\min(9,4)=4,\ LS_A=4-4=0 \). The minimum project completion time is the earliest finish time of the final activity, G: 18 days.
(c) The critical path consists of the activities where \( ES=LS \) (zero float): A (\( ES=LS=0 \)), D (\( ES=LS=4 \)), F (\( ES=LS=9 \)), G (\( ES=LS=16 \)); their durations sum to \( 4+5+7+2=18 \), matching the project duration, confirming the critical path is A→D→F→G. Total float for C \( = LS_C-ES_C = 9-4 = 5\text{ days} \) (equivalently \( LF_C-EF_C=12-7=5 \) days).

(a) [1] correct ES/EF for A, B (0,4 and 0,6); [1] correct ES/EF for C, D (4,7 and 4,9); [1] correct ES/EF for E, F, G, correctly taking the maximum of predecessors' EF at each merge point (7,11; 9,16; 16,18). (b) [1] correct LF/LS for G and F (18,16 and 16,9); [1] correct LF/LS for E and D (16,12 and 9,4); [1] correct LF/LS for C, B, A, correctly taking the minimum of successors' LS at each split point (12,9; 12,6; 4,0), and correct project completion time of 18 days stated. (c) [1] correctly identifies all activities with zero float (A, D, F, G); [1] correctly states critical path A-D-F-G with durations summing to 18; [1] correct total float for C, 5 days, via a valid method (LS-ES or LF-EF).