2024 CCEA GCSE Further Mathematics 2330 模擬試題連答案詳解

\( f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3) \). Setting \( f'(x) = 0 \) gives \( x = 1 \) or \( x = 3 \). \( f(1) = 1 - 6 + 9 + 2 = 6 \), and \( f(3) = 27 - 54 + 27 + 2 = 2 \), so the turning points are \( (1, 6) \) and \( (3, 2) \). To classify: \( f''(x) = 6x - 12 \). At \( x = 1 \), \( f''(1) = 6 - 12 = -6 < 0 \), so \( (1, 6) \) is a maximum. At \( x = 3 \), \( f''(3) = 18 - 12 = 6 > 0 \), so \( (3, 2) \) is a minimum.

M1: \( f'(x) = 3x^2-12x+9 \); M1: factorises/solves to give \( x=1, x=3 \); W1: \( y \)-coordinates \( 6 \) and \( 2 \) found (so points \( (1,6) \) and \( (3,2) \)); M1: \( f''(x)=6x-12 \); W1: \( f''(1)=-6 \), correctly concludes maximum; W1: \( f''(3)=6 \), correctly concludes minimum; W1: both points fully and correctly stated with nature. [8]

(a) Using the product rule with \( u = x^2 \), \( v = (x-2)^3 \): \( \frac{du}{dx} = 2x \), \( \frac{dv}{dx} = 3(x-2)^2 \). So \( \frac{dy}{dx} = 2x(x-2)^3 + x^2 \cdot 3(x-2)^2 = (x-2)^2\left[2x(x-2) + 3x^2\right] = (x-2)^2(2x^2-4x+3x^2) = (x-2)^2(5x^2-4x) \). (b) At \( x=3 \): \( \frac{dy}{dx} = (3-2)^2(5(9)-4(3)) = 1 \times (45-12) = 33 \). (c) At \( x=3 \), \( y = 3^2(3-2)^3 = 9 \times 1 = 9 \), so the curve passes through \( (3,9) \) with gradient \( 33 \). The tangent is \( y - 9 = 33(x-3) \), i.e. \( y = 33x - 99 + 9 = 33x - 90 \).

M1: correct product rule statement with \( u,v,\frac{du}{dx},\frac{dv}{dx} \) identified; M1: correct unsimplified derivative; W1: factorised to \( (x-2)^2(5x^2-4x) \); M1: substitutes \( x=3 \) into \( \frac{dy}{dx} \); W1: gradient \( =33 \); M1: finds \( y(3)=9 \); M1: uses \( y-y_1=m(x-x_1) \) with correct point/gradient; W1: simplifies to \( y=33x-90 \). [9]

(a) After cutting corners of side \( x \) and folding up, the base of the box is a square of side \( (12-2x) \) and the height is \( x \), so \( V = x(12-2x)^2 \). (b) Expanding, \( V = x(144-48x+4x^2) = 144x - 48x^2 + 4x^3 \), so \( \frac{dV}{dx} = 144 - 96x + 12x^2 = 12(x^2-8x+12) = 12(x-2)(x-6) \). (c) Setting \( \frac{dV}{dx}=0 \) gives \( x=2 \) or \( x=6 \); since \( x=6 \) gives a base side of \( 12-12=0 \) (zero volume), the valid critical value is \( x=2 \). \( \frac{d^2V}{dx^2} = 24x-96 \), and at \( x=2 \), \( \frac{d^2V}{dx^2} = 48-96 = -48 < 0 \), confirming a maximum. Maximum volume: \( V(2) = 2(12-4)^2 = 2(8)^2 = 128 \, \text{cm}^3 \).

M1: correct expression \( V=x(12-2x)^2 \) with valid reasoning shown (base side, height); M1: expands \( V \) correctly; W1: \( \frac{dV}{dx}=144-96x+12x^2 \); W1: factorises to \( 12(x-2)(x-6) \); M1: sets \( \frac{dV}{dx}=0 \) and solves to give \( x=2,6 \); W1: rejects \( x=6 \) with valid reason; M1: uses second derivative test (or equivalent) to confirm maximum at \( x=2 \); W1: correct maximum volume \( 128 \, \text{cm}^3 \). [9]

(a) At the points of intersection, \( 6x-x^2 = 2x \), so \( 4x-x^2=0 \), i.e. \( x(4-x)=0 \), giving \( x=0 \) or \( x=4 \). (b) The curve lies above the line between \( x=0 \) and \( x=4 \), so the area is \( \int_0^4 \left[(6x-x^2)-2x\right]dx = \int_0^4 (4x-x^2)\,dx = \left[2x^2-\frac{x^3}{3}\right]_0^4 = \left(32-\frac{64}{3}\right) - 0 = \frac{96}{3}-\frac{64}{3} = \frac{32}{3} \).

M1: sets \( 6x-x^2=2x \); W1: correct intersection points \( x=0, x=4 \); M1: sets up integral of the difference \( (6x-x^2)-2x \) between the correct limits; M1: correct integration \( 2x^2-\frac{x^3}{3} \); M1: correctly substitutes limits; W1: area \( =\frac{32}{3} \) (or exact equivalent, e.g. \( 10\frac{2}{3} \)); W1: fully correct, clearly presented working throughout. [8]

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

M1: factors out \( 2 \) from the \( x^2 \) and \( x \) terms; M1: correctly completes the square inside the bracket, \( (x-2)^2-4 \); W1: correctly multiplies back out by \( 2 \); W1: correct final form \( 2(x-2)^2-5 \); W1: correct values of \( a,b,c \) stated; W1: minimum value \( -5 \) correctly stated; W1: correct value \( x=2 \) stated. [7]

Factorising, \( 2x^2-5x-3 = (2x+1)(x-3) \). Setting \( (2x+1)(x-3)=0 \) gives critical values \( x=-\frac{1}{2} \) and \( x=3 \). Since the coefficient of \( x^2 \) is positive, the graph of \( y=2x^2-5x-3 \) is a positive (upward-opening) parabola, so \( y \geq 0 \) outside the roots. Hence the solution is \( x \leq -\frac{1}{2} \) or \( x \geq 3 \), i.e. \( \{x: x\leq -\tfrac12\}\cup\{x: x\geq3\} \).

M1: attempts to factorise/use the quadratic formula; W1: correct factorisation \( (2x+1)(x-3) \); W1: correct critical values \( x=-\tfrac12 \) and \( x=3 \); M1: recognises the parabola opens upwards; W1: correctly identifies solution as outside the roots (not between); W1: correct final inequality/set notation for each branch (2 marks, 1 per branch). [7]

Substituting \( y=x+1 \) into \( x^2+y^2=25 \): \( x^2 + (x+1)^2 = 25 \), so \( x^2+x^2+2x+1=25 \), giving \( 2x^2+2x-24=0 \), i.e. \( x^2+x-12=0 \). Factorising, \( (x+4)(x-3)=0 \), so \( x=-4 \) or \( x=3 \). Using \( y=x+1 \): when \( x=-4 \), \( y=-3 \); when \( x=3 \), \( y=4 \). Check: \( (-4)^2+(-3)^2=16+9=25 \) ✓, and \( 3^2+4^2=9+16=25 \) ✓.

M1: correctly substitutes \( y=x+1 \) into the second equation; M1: expands \( (x+1)^2 \) correctly; W1: forms the correct quadratic \( x^2+x-12=0 \); M1: factorises/solves to give \( x=-4, 3 \); W1: both correct \( y \)-values found using \( y=x+1 \); W1: both solution pairs correctly and clearly stated. [7]

Factorising: \( x^2-9=(x-3)(x+3) \), \( x^2+x-6=(x+3)(x-2) \), and \( 2x-4=2(x-2) \). So the expression becomes \( \dfrac{(x-3)(x+3)}{(x+3)(x-2)} \div \dfrac{x+3}{2(x-2)} = \dfrac{(x-3)(x+3)}{(x+3)(x-2)} \times \dfrac{2(x-2)}{x+3} = \dfrac{2(x-3)(x+3)(x-2)}{(x+3)^2(x-2)} \). Cancelling the common factor \( (x-2) \), and one factor of \( (x+3) \), gives \( \dfrac{2(x-3)}{x+3} \).

M1: correctly factorises \( x^2-9 \); M1: correctly factorises \( x^2+x-6 \); W1: correctly factorises \( 2x-4 \); M1: correctly rewrites division as multiplication by the reciprocal; W1: correct combined fraction before cancelling; W1: cancels \( (x-2) \) correctly; W1: fully simplified final answer \( \dfrac{2(x-3)}{x+3} \). [7]

Combining the logarithms: \( \log_2\left[(x+3)(x-3)\right] = 4 \), so \( (x+3)(x-3) = 2^4 = 16 \), giving \( x^2-9=16 \), i.e. \( x^2=25 \), so \( x=5 \) or \( x=-5 \). For the original logarithms to be defined, we need \( x+3>0 \) and \( x-3>0 \), i.e. \( x>3 \). Since \( x=-5 \) does not satisfy this, it is rejected; only \( x=5 \) is valid. Check: \( \log_2(8)+\log_2(2)=3+1=4 \) ✓.

M1: combines the two logarithms using \( \log a+\log b=\log(ab) \); W1: correct combined equation \( \log_2[(x+3)(x-3)]=4 \); M1: converts to \( (x+3)(x-3)=16 \); W1: forms and solves \( x^2=25 \) to give \( x=\pm5 \); M1: states the domain requirement \( x>3 \); W1: correctly rejects \( x=-5 \) and states final answer \( x=5 \). [7]

(a) \( 45 = 9\times5 = 3^2\times5 \), so \( \log_a45 = \log_a(3^2)+\log_a5 = 2\log_a3+\log_a5 = 2q+p \). (b) \( \dfrac{25}{3}=\dfrac{5^2}{3} \), so \( \log_a\left(\dfrac{25}{3}\right) = \log_a(5^2)-\log_a3 = 2\log_a5-\log_a3 = 2p-q \).

M1: recognises \( 45=3^2\times5 \); M1: applies the addition and power laws of logarithms correctly; W1: correct answer \( 2q+p \); M1: recognises \( \tfrac{25}{3}=\tfrac{5^2}{3} \); M1: applies the subtraction and power laws correctly; W1: correct answer \( 2p-q \); W1: both final answers correctly and clearly stated in terms of \( p \) and \( q \) only. [7]

Taking \( \log_{10} \) of both sides of \( y=Ax^n \): \( \log_{10}y = \log_{10}A + n\log_{10}x \), which is linear in \( \log_{10}x \) with gradient \( n \) and \( y \)-intercept \( \log_{10}A \). From the two given points, the gradient is \( n = \dfrac{3.7-1.3}{2-0} = \dfrac{2.4}{2} = 1.2 \). The intercept (at \( \log_{10}x=0 \)) is \( \log_{10}A = 1.3 \), so \( A = 10^{1.3} \approx 19.95 \).

M1: takes logs of \( y=Ax^n \) to give \( \log y=\log A+n\log x \); M1: identifies gradient \( =n \) and intercept \( =\log A \); M1: correctly calculates gradient using the two points; W1: \( n=1.2 \) (or \( 1.20 \)); M1: identifies intercept \( \log_{10}A=1.3 \); M1: correctly computes \( A=10^{1.3} \); W1: \( A\approx19.95 \) (2 d.p.). [7]

(a) \( AB = \begin{pmatrix} 2(1)+1(2) & 2(0)+1(3) \\ 3(1)+(-1)(2) & 3(0)+(-1)(3) \end{pmatrix} = \begin{pmatrix} 4 & 3 \\ 1 & -3 \end{pmatrix} \). (b) \( \det(A) = (2)(-1)-(1)(3) = -2-3 = -5 \). (c) For \( A=\begin{pmatrix}a&b\c&d\end{pmatrix} \), \( A^{-1} = \dfrac{1}{\det A}\begin{pmatrix}d&-b\\-c&a\end{pmatrix} \). So \( A^{-1} = \dfrac{1}{-5}\begin{pmatrix}-1&-1\\-3&2\end{pmatrix} = \begin{pmatrix}\tfrac{1}{5}&\tfrac{1}{5}\\ \tfrac{3}{5}&-\tfrac{2}{5}\end{pmatrix} \). Check: \( A A^{-1} = \begin{pmatrix}1&0\\0&1\end{pmatrix} \).

M1: correct method for matrix multiplication; W1: correct \( AB=\begin{pmatrix}4&3\\1&-3\end{pmatrix} \); M1: correct determinant method \( ad-bc \); W1: \( \det(A)=-5 \); M1: correct inverse formula/method applied; W1: correct \( A^{-1}=\begin{pmatrix}\tfrac15&\tfrac15\\ \tfrac35&-\tfrac25\end{pmatrix} \). [7]

Let \( s=\sin\theta \). Then \( 2s^2+s-1=0 \), which factorises as \( (2s-1)(s+1)=0 \), giving \( s=\tfrac12 \) or \( s=-1 \). For \( \sin\theta = \tfrac12 \) in the range \( 0^\circ \leq \theta \leq 360^\circ \): \( \theta=30^\circ \) (first quadrant) and \( \theta=180^\circ-30^\circ=150^\circ \) (second quadrant). For \( \sin\theta=-1 \): \( \theta=270^\circ \). So the solutions are \( \theta=30^\circ, 150^\circ, 270^\circ \).

M1: substitutes \( s=\sin\theta \) to obtain a quadratic in \( s \); M1: correctly factorises \( (2s-1)(s+1) \); W1: correct values \( s=\tfrac12 \) and \( s=-1 \); M1: uses \( \sin\theta=\tfrac12 \) to find \( \theta=30^\circ \); W1: correctly finds the second solution \( \theta=150^\circ \) using the sine symmetry \( 180^\circ-30^\circ \); M1: uses \( \sin\theta=-1 \) to find \( \theta=270^\circ \); W1: recognises no further solutions from this branch; W1: all three correct solutions clearly stated with no extraneous values; W1: solutions given within the correct specified range. [10]

(a) At \( t=0 \): \( v = 3(0)^2-12(0)+9 = 9 \, \text{m/s} \). (b) The particle is at rest when \( v=0 \): \( 3t^2-12t+9=0 \), i.e. \( t^2-4t+3=0 \), which factorises as \( (t-1)(t-3)=0 \), giving \( t=1 \) or \( t=3 \) seconds.

W1: correct initial velocity \( 9 \, \text{m/s} \); M1: sets \( v=0 \) and simplifies to \( t^2-4t+3=0 \); M1: correctly factorises/solves the quadratic; W1: both correct values \( t=1 \) and \( t=3 \) stated. [4]

(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 \( = \sqrt{(-2)^2+6^2} = \sqrt{4+36} = \sqrt{40} \approx 6.32 \, \text{N} \). (c) Since the resultant has a negative \( \mathbf{i} \)-component and a positive \( \mathbf{j} \)-component, it lies in the second quadrant. The reference angle from the negative \( \mathbf{i} \)-direction is \( \arctan\left(\dfrac{6}{2}\right) = \arctan(3) \approx 71.6^\circ \), so the angle measured from the positive \( \mathbf{i} \)-direction is \( 180^\circ - 71.6^\circ = 108.4^\circ \).

M1: adds the \( \mathbf{i} \)-components and \( \mathbf{j} \)-components separately; W1: correct resultant \( -2\mathbf{i}+6\mathbf{j} \); M1: correct method \( \sqrt{(-2)^2+6^2} \); W1: magnitude \( \approx6.32 \, \text{N} \); M1: recognises the resultant lies in the second quadrant; M1: correct method for the angle, e.g. \( \arctan(6/2) \) or \( \arctan(6/-2) \) adjusted for quadrant; W1: correct reference angle \( \approx71.6^\circ \); W1: correctly adjusts for the second quadrant; W1: final angle \( \approx108.4^\circ \) clearly stated. [10]

Taking east as \( \mathbf{i} \) and north as \( \mathbf{j} \): \( \mathbf{F_1}+\mathbf{F_2} = 6\mathbf{i}+8\mathbf{j} \). For equilibrium, \( \mathbf{F_3} = -(\mathbf{F_1}+\mathbf{F_2}) = -6\mathbf{i}-8\mathbf{j} \), so \( |\mathbf{F_3}| = \sqrt{6^2+8^2} = \sqrt{100} = 10 \, \text{N} \). The bearing of \( \mathbf{F_1}+\mathbf{F_2} \) from north is \( \arctan\left(\dfrac{6}{8}\right) \approx 36.9^\circ \), so \( \mathbf{F_3} \), being in exactly the opposite direction, has bearing \( 36.9^\circ+180^\circ = 216.9^\circ \).

M1: correct expression for \( \mathbf{F_1}+\mathbf{F_2} \) as \( 6\mathbf{i}+8\mathbf{j} \); M1: uses equilibrium to find \( \mathbf{F_3}=-(\mathbf{F_1}+\mathbf{F_2}) \); W1: correct magnitude \( 10 \, \text{N} \); M1: finds the bearing of \( \mathbf{F_1}+\mathbf{F_2} \) correctly; W1: correct value \( \approx36.9^\circ \); W1: correctly adds \( 180^\circ \) to give the bearing of \( \mathbf{F_3} \) as \( 216.9^\circ \). [6]

Weight \( W = mg = 5\times10=50 \, \text{N} \). (a) Component of weight down the plane \( = W\sin30^\circ = 50\times0.5 = 25 \, \text{N} \). Normal reaction \( R = W\cos30^\circ = 50\times\dfrac{\sqrt3}{2} \approx 43.30 \, \text{N} \). (b) Maximum available friction \( F_{max} = \mu R = 0.3\times43.30 \approx 12.99 \, \text{N} \). For the block to remain at rest, friction would need to balance the \( 25 \, \text{N} \) component down the plane, but the maximum friction available (\( \approx12.99 \, \text{N} \)) is less than this. Since \( 25 > 12.99 \), the required friction exceeds what is available, so the block does NOT remain at rest; it slides down the plane.

W1: correct weight \( W=50 \, \text{N} \); M1: correct method \( W\sin30^\circ \) for the component down the plane; W1: \( 25 \, \text{N} \); M1: correct method \( W\cos30^\circ \) for the normal reaction; W1: \( R\approx43.30 \, \text{N} \); M1: correct method \( \mu R \) for maximum friction; W1: \( F_{max}\approx12.99 \, \text{N} \); M1: compares the required force (\( 25 \, \text{N} \)) with maximum friction; W1: correct conclusion (block slides) with valid justification (\( 25>12.99 \)). [10]

Taking moments about \( A \): the weight (\( 20 \, \text{N} \), acting \( 2.5 \, \text{m} \) from \( A \)) is balanced by the reaction \( R_C \) at \( C \) (\( 3 \, \text{m} \) from \( A \)): \( R_C\times3 = 20\times2.5 = 50 \), so \( R_C = \dfrac{50}{3} \approx16.67 \, \text{N} \). Resolving vertically, \( R_A+R_C=20 \), so \( R_A = 20-\dfrac{50}{3} = \dfrac{60-50}{3} = \dfrac{10}{3} \approx3.33 \, \text{N} \).

M1: takes moments about \( A \) (or \( C \)) with correct terms; W1: correct moment equation \( R_C\times3=20\times2.5 \); M1: solves for \( R_C \); W1: \( R_C\approx16.67 \, \text{N} \); M1: uses vertical equilibrium \( R_A+R_C=20 \); W1: solves for \( R_A \); W1: \( R_A\approx3.33 \, \text{N} \) correctly stated. [8]

(a) Treating the car and trailer as a single system of total mass \( 900+300=1200 \, \text{kg} \): total driving force \( 2400 \, \text{N} \), total resistance \( 150+50=200 \, \text{N} \). By Newton's second law, \( 2400-200 = 1200a \), so \( a = \dfrac{2200}{1200} \approx1.83 \, \text{m/s}^2 \). (b) Considering the trailer alone: the only horizontal forces are the tension \( T \) (forward) and resistance \( 50 \, \text{N} \) (backward). By Newton's second law: \( T-50 = 300a = 300\times\dfrac{2200}{1200} = 550 \), so \( T = 550+50 = 600 \, \text{N} \). (Check using the car alone: \( 2400-150-T=900a \Rightarrow T=2400-150-900(1.8\overline{3})=600 \, \text{N} \), which agrees.)

M1: treats car and trailer as a single system with combined mass \( 1200 \, \text{kg} \); M1: correct equation \( 2400-200=1200a \); W1: correct total resistance \( 200 \, \text{N} \) used; W1: \( a\approx1.83 \, \text{m/s}^2 \); M1: considers the trailer (or car) in isolation with correct forces identified; M1: correct equation of motion for the trailer, \( T-50=300a \); W1: correctly substitutes \( a \); W1: \( T=600 \, \text{N} \); M1: checks consistency using the car's equation of motion (or equivalent valid method); W1: full, clearly structured working with correct final answers for both (a) and (b). [12]

Using midpoints \( 2.5, 7.5, 12.5, 17.5 \) with frequencies \( 6,14,12,8 \) (\( \Sigma f=40 \)): \( \Sigma fx = 6(2.5)+14(7.5)+12(12.5)+8(17.5) = 15+105+150+140 = 410 \). Mean \( = \dfrac{410}{40} = 10.25 \) minutes. For the standard deviation: \( \Sigma fx^2 = 6(2.5)^2+14(7.5)^2+12(12.5)^2+8(17.5)^2 = 37.5+787.5+1875+2450 = 5150 \). Variance \( = \dfrac{\Sigma fx^2}{\Sigma f} - \bar{x}^2 = \dfrac{5150}{40} - 10.25^2 = 128.75-105.0625 = 23.6875 \). Standard deviation \( = \sqrt{23.6875} \approx 4.87 \) minutes.

M1: uses correct midpoints \( 2.5,7.5,12.5,17.5 \); M1: correct method \( \Sigma fx \); W1: \( \Sigma fx=410 \), mean \( =10.25 \); M1: correct method \( \Sigma fx^2 \); W1: \( \Sigma fx^2=5150 \); M1: correct variance formula \( \frac{\Sigma fx^2}{\Sigma f}-\bar x^2 \) applied; W1: standard deviation \( \approx4.87 \) minutes. [7]

(a) Since \( x \) is already in ascending order, its ranks are \( 1,2,3,4,5,6 \). Ranking \( y \) (\( 3,5,4,8,7,9 \)) from smallest to largest gives ranks \( 1,3,2,5,4,6 \) respectively. (b) Differences \( d = \) rank(x) \( - \) rank(y): \( 0,-1,1,-1,1,0 \), so \( \Sigma d^2 = 0+1+1+1+1+0=4 \). With \( n=6 \): \( r_s = 1-\dfrac{6\Sigma d^2}{n(n^2-1)} = 1-\dfrac{6(4)}{6(35)} = 1-\dfrac{24}{210} \approx 0.89 \). (c) Since \( r_s \) is close to \( 1 \), this indicates a strong positive correlation between the ranks of \( x \) and \( y \) — as \( x \) increases, \( y \) tends to increase. (d) \( \Sigma x=21 \), \( \Sigma y=36 \), \( \Sigma xy = 1(3)+2(5)+3(4)+4(8)+5(7)+6(9) = 3+10+12+32+35+54=146 \), \( \Sigma x^2 = 1+4+9+16+25+36=91 \). \( S_{xy} = \Sigma xy - \dfrac{\Sigma x\Sigma y}{n} = 146-\dfrac{21(36)}{6} = 146-126=20 \). \( S_{xx} = \Sigma x^2-\dfrac{(\Sigma x)^2}{n} = 91-\dfrac{441}{6} = 91-73.5=17.5 \). (e) \( b = \dfrac{S_{xy}}{S_{xx}} = \dfrac{20}{17.5} \approx1.14 \). \( a = \bar{y}-b\bar{x} = 6-1.142857(3.5) = 6-4=2.00 \). So the regression line is \( y \approx 2.00+1.14x \).

W1: correct ranks for \( x \); W1: correct ranks for \( y \); M1: correct differences and \( \Sigma d^2=4 \); M1: correct Spearman's formula applied; W1: \( r_s\approx0.89 \); W1: valid interpretation (strong positive correlation between ranks); M1: correct method for \( \Sigma xy \) and \( \Sigma x^2 \); W1: \( S_{xy}=20 \); W1: \( S_{xx}=17.5 \); M1: correct method \( b=S_{xy}/S_{xx} \); W1: \( b\approx1.14 \); M1: correct method \( a=\bar y-b\bar x \); W1: \( a\approx2.00 \), regression equation correctly stated. [13]

(a) There are 8 balls in total (5 red, 3 blue). First draw: \( P(\text{Red}) = \dfrac{5}{8} \), \( P(\text{Blue}) = \dfrac{3}{8} \). Second draw (without replacement) depends on the first: if the first was red, 7 balls remain (4 red, 3 blue), so \( P(\text{Red}\mid\text{Red}) = \dfrac{4}{7} \), \( P(\text{Blue}\mid\text{Red}) = \dfrac{3}{7} \); if the first was blue, 7 remain (5 red, 2 blue), so \( P(\text{Red}\mid\text{Blue}) = \dfrac{5}{7} \), \( P(\text{Blue}\mid\text{Blue}) = \dfrac{2}{7} \). (b) \( P(\text{both red}) = \dfrac{5}{8}\times\dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}\approx0.36 \). (c) \( P(\text{at least one blue}) = 1-P(\text{both red}) = 1-\dfrac{5}{14} = \dfrac{9}{14}\approx0.64 \). (d) Given the first ball is red, this is simply the branch probability \( P(\text{Red}\mid\text{Red}) = \dfrac{4}{7}\approx0.57 \).

M1: correct first-draw probabilities \( \tfrac58,\tfrac38 \); W1: correct second-draw (conditional) probabilities on all branches; M1: correct method \( \tfrac58\times\tfrac47 \) for both red; W1: \( P(\text{both red})=\tfrac{5}{14}\approx0.36 \); M1: correct method \( 1-P(\text{both red}) \); W1: \( P(\text{at least one blue})=\tfrac{9}{14}\approx0.64 \); W1: correctly identifies \( P(\text{Red}\mid\text{Red})=\tfrac47\approx0.57 \) directly from the tree. [8]

Let \( X\sim B(5,0.6) \), so \( P(X=x) = \binom{5}{x}(0.6)^x(0.4)^{5-x} \). (a) The binomial coefficient \( \binom{5}{3}=10 \) (from Pascal's triangle, row 5). \( P(X=3) = 10(0.6)^3(0.4)^2 = 10(0.216)(0.16) = 0.3456 \approx0.346 \). (b) \( P(X=4) = \binom{5}{4}(0.6)^4(0.4)^1 = 5(0.1296)(0.4) = 0.2592 \). \( P(X=5) = \binom{5}{5}(0.6)^5(0.4)^0 = 1(0.07776) = 0.07776 \). So \( P(X\geq4) = P(X=4)+P(X=5) = 0.2592+0.07776 = 0.33696 \approx0.337 \).

M1: correct binomial coefficient \( \binom53=10 \); M1: correct expression \( \binom53(0.6)^3(0.4)^2 \); W1: \( P(X=3)\approx0.346 \); M1: correct expression for \( P(X=4) \); W1: \( P(X=4)=0.2592 \); M1: correct expression for \( P(X=5) \); W1: \( P(X=5)=0.07776 \); W1: correctly sums to give \( P(X\geq4)\approx0.337 \). [9]

Standardising: \( Z = \dfrac{187-175}{8} = \dfrac{12}{8} = 1.5 \). From the standard normal distribution table, \( \Phi(1.5) = 0.9332 \), so \( P(Z>1.5) = 1-0.9332 = 0.0668 \). Hence \( P(X>187) \approx0.0668 \).

M1: correct standardisation formula \( Z=\dfrac{X-\mu}{\sigma} \); W1: \( Z=1.5 \); M1: correctly reads \( \Phi(1.5)=0.9332 \) from the table; M1: correct method \( 1-\Phi(1.5) \); W1: \( P(X>187)\approx0.0668 \). [5]

(a) Number studying only French \( = 18-8=10 \); only Spanish \( = 15-8=7 \); both \( =8 \). (b) Total in \( F\cup S = 10+7+8=25 \), so the number studying neither is \( 30-25=5 \). (c) \( P(F\cup S) = \dfrac{25}{30} = \dfrac{5}{6}\approx0.83 \). (d) \( P(S\mid F) = \dfrac{P(F\cap S)}{P(F)} = \dfrac{8/30}{18/30} = \dfrac{8}{18} = \dfrac{4}{9}\approx0.44 \).

M1: correctly calculates only-French \( =10 \) and only-Spanish \( =7 \); W1: correctly completed Venn diagram (10, 8, 7 and outside region); M1: correct method \( 30-(10+7+8) \); W1: \( 5 \) students studying neither; M1: correct method for \( P(F\cup S) \); W1: \( P(F\cup S)=\tfrac56\approx0.83 \); M1: correct conditional probability method \( P(F\cap S)/P(F) \); W1: \( P(S\mid F)=\tfrac49\approx0.44 \). [8]

By the product rule for counting, since each stage's choice is independent of the others, the total number of routes is \( 3\times2\times4 = 24 \).

M1: correctly applies the product rule (multiplies the three route counts); W1: correct answer \( 24 \). [3]

(a) Carpentry: \( 4x+2y\leq60 \). Finishing: \( x+2y\leq24 \). At least 2 tables: \( x\geq2 \). Non-negativity of chairs: \( y\geq0 \). (b) The boundary lines are \( 4x+2y=60 \) (i.e. \( 2x+y=30 \)), \( x+2y=24 \), and \( x=2 \). Intersecting \( x=2 \) with \( y=0 \) gives \( (2,0) \); intersecting \( 2x+y=30 \) with \( y=0 \) gives \( (15,0) \); intersecting \( 2x+y=30 \) with \( x+2y=24 \): substituting \( y=30-2x \) into \( x+2(30-2x)=24 \) gives \( x+60-4x=24 \), so \( -3x=-36 \), \( x=12 \), \( y=6 \), giving \( (12,6) \); intersecting \( x=2 \) with \( x+2y=24 \) gives \( 2+2y=24 \), \( y=11 \), giving \( (2,11) \) (this point satisfies \( 2x+y=4+11=15\leq30 \), so it is feasible). The vertices of \( R \) are \( (2,0) \), \( (15,0) \), \( (12,6) \) and \( (2,11) \). (c) Evaluating the profit \( P=30x+20y \) at each vertex: \( (2,0)\to £60 \); \( (15,0)\to £450 \); \( (12,6)\to 30(12)+20(6)=360+120=£480 \); \( (2,11)\to 30(2)+20(11)=60+220=£280 \). The maximum profit is \( £480 \), achieved by making 12 tables and 6 chairs per week.

W1: correct carpentry inequality \( 4x+2y\leq60 \); W1: correct finishing inequality \( x+2y\leq24 \); W1: correct inequality \( x\geq2 \); W1: correct inequality \( y\geq0 \); M1: correctly finds intersection of \( x=2 \) and \( y=0 \); M1: correctly finds intersection of \( 2x+y=30 \) and \( y=0 \); M1: correctly solves the simultaneous equations \( 2x+y=30 \) and \( x+2y=24 \); W1: correct vertex \( (12,6) \); M1: correctly finds intersection of \( x=2 \) and \( x+2y=24 \); W1: all four vertices correctly stated; M1: evaluates \( P=30x+20y \) at each vertex; W1: correctly identifies maximum profit \( £480 \) at \( (12,6) \). [13]

(a) The first 4-point moving average covers Q1-Q4 of Year 1: \( \dfrac{40+55+70+45}{4} = \dfrac{210}{4} = 52.5 \). The second covers Q2 Year 1 to Q1 Year 2: \( \dfrac{55+70+45+48}{4} = \dfrac{218}{4} = 54.5 \). (b) Since the moving average has an even number of terms (4), it must be centred: the centred moving average for Q3 Year 1 is the average of these two moving averages: \( \dfrac{52.5+54.5}{2} = 53.5 \). (c) The actual value for Q3 Year 1 is \( 70 \), so the seasonal variation \( = 70-53.5 = +16.5 \).

M1: correct method for the first 4-point moving average; W1: \( 52.5 \); M1: correct method for the second 4-point moving average; W1: \( 54.5 \); M1: correctly centres by averaging the two moving averages; W1: centred trend value \( 53.5 \); M1: correct method \( \text{actual}-\text{trend} \); W1: seasonal variation \( +16.5 \). [9]

(a) \( p=T,q=T \): \( \lnot p=F \), \( p\land q=T \), so \( \lnot p\lor(p\land q)=T \). \( p=T,q=F \): \( \lnot p=F \), \( p\land q=F \), result \( =F \). \( p=F,q=T \): \( \lnot p=T \), \( p\land q=F \), result \( =T \). \( p=F,q=F \): \( \lnot p=T \), \( p\land q=F \), result \( =T \). So the column is \( T,F,T,T \). (b) For \( \lnot p\lor q \): \( TT\to F\lor T=T \); \( TF\to F\lor F=F \); \( FT\to T\lor T=T \); \( FF\to T\lor F=T \). This gives the same column \( T,F,T,T \) as in part (a), so \( \lnot p\lor(p\land q) \) IS logically equivalent to \( \lnot p\lor q \) (they have identical truth tables for all combinations of \( p \) and \( q \)). (c) By the definition of material implication, \( p\rightarrow q \) is true except when \( p \) is true and \( q \) is false, giving the column \( T,F,T,T \) — identical to \( \lnot p\lor q \). So \( p\rightarrow q \) is also logically equivalent to \( \lnot p\lor q \) (indeed, \( p\rightarrow q \) is defined to be equivalent to \( \lnot p\lor q \)).

M1: correct \( \lnot p \) column; M1: correct \( p\land q \) column; W1: correct final column for \( \lnot p\lor(p\land q) \) — \( T,F,T,T \); M1: correctly constructs the truth table for \( \lnot p\lor q \); W1: correct column \( T,F,T,T \); W1: correctly concludes the two statements are equivalent, with valid reasoning (identical truth tables); M1: correctly constructs the truth table for \( p\rightarrow q \); W1: correct column \( T,F,T,T \); W1: correctly concludes \( p\rightarrow q \) is also equivalent to \( \lnot p\lor q \). [10]

(a) Since the order in which the committee members are chosen does not matter, this is a combination: \( \binom{7}{4} = \dfrac{7!}{4!\,3!} = 35 \). (b) If a specific named person must be included, the remaining 3 committee members must be chosen from the other 6 people: \( \binom{6}{3} = \dfrac{6!}{3!\,3!} = 20 \).

M1: recognises this as a combination (order does not matter); M1: correct method \( \binom74 \); W1: \( 35 \); M1: correctly reduces to choosing 3 from the remaining 6, \( \binom63 \); W1: \( 20 \). [5]

(a) Forward pass: \( A \): earliest start \( 0 \), earliest finish \( 0+4=4 \). \( B \): earliest start \( 0 \), earliest finish \( 0+3=3 \). \( C \) (needs \( A \)): earliest start \( =4 \), earliest finish \( =4+5=9 \). \( D \) (needs \( A \)): earliest start \( =4 \), earliest finish \( =4+2=6 \). \( E \) (needs \( B \) and \( D \)): earliest start \( =\max(3,6)=6 \), earliest finish \( =6+4=10 \). \( F \) (needs \( C \) and \( E \)): earliest start \( =\max(9,10)=10 \), earliest finish \( =10+3=13 \). So the minimum project completion time is \( 13 \) days. (b) Backward pass (starting from \( F \), latest finish \( =13 \)): \( F \): latest finish \( 13 \), latest start \( 13-3=10 \). \( E \) (only successor \( F \)): latest finish \( =10 \), latest start \( =10-4=6 \). \( C \) (only successor \( F \)): latest finish \( =10 \), latest start \( =10-5=5 \). \( D \) (only successor \( E \)): latest finish \( =6 \), latest start \( =6-2=4 \). \( B \) (only successor \( E \)): latest finish \( =6 \), latest start \( =6-3=3 \). \( A \) (successors \( C,D \)): latest finish \( =\min(5,4)=4 \), latest start \( =4-4=0 \). Activities with zero float (earliest start = latest start) are \( A \) (\( 0=0 \)), \( D \) (\( 4=4 \)), \( E \) (\( 6=6 \)) and \( F \) (\( 10=10 \)), so the critical path is \( A\to D\to E\to F \), with total duration \( 4+2+4+3=13 \) days, confirming the project completion time. (c) Total float for \( C \) \( = \) latest start \( - \) earliest start \( = 5-4=1 \) day.

M1: correct forward pass method; W1: correct earliest finish times for \( A,B,C,D \); W1: correct earliest finish times for \( E \) and \( F \) (using the maximum of preceding finishes); M1: correct backward pass method starting from \( F \); W1: correct latest start/finish times for \( F,E,C \); W1: correct latest start/finish times for \( D,B,A \) (using the minimum where an activity has multiple successors); M1: correctly identifies all zero-float activities; W1: critical path \( A\to D\to E\to F \) and project time \( 13 \) days correctly stated; M1: correct method for float, latest start \( - \) earliest start (or latest finish \( - \) earliest finish); W1: float for \( C \) \( =1 \) day. [10]