2024 CCEA A-Level Further Mathematics 2330 Practice Paper with Answers

(a) The rectangle OABC has area \( 8\times6=48\text{ cm}^2 \), and (by symmetry) its centre of mass is at its centre, \( (4,3) \).
(b) \( A=(8,0) \), \( B=(8,6) \), and since BE is horizontal with length 5 cm and E is further from O, \( E=(13,6) \). The triangle ABE has a right angle at B, with legs \( AB=6\text{ cm} \) (vertical) and \( BE=5\text{ cm} \) (horizontal), so its area is \( \tfrac12\times6\times5=15\text{ cm}^2 \). Its centroid is the mean of the vertices: \( \left(\dfrac{8+8+13}{3},\dfrac{0+6+6}{3}\right)=\left(\dfrac{29}{3},4\right)=(9.67,4) \) (3 s.f.).
(c) Treating the composite lamina as the rectangle (area 48, centroid (4,3)) together with the triangle (area 15, centroid (29/3,4)), total area \( =48+15=63\text{ cm}^2 \). \( \bar{x}=\dfrac{48(4)+15(29/3)}{63}=\dfrac{192+145}{63}=\dfrac{337}{63}=5.35 \) (3 s.f.). \( \bar{y}=\dfrac{48(3)+15(4)}{63}=\dfrac{144+60}{63}=\dfrac{204}{63}=3.24 \) (3 s.f.). So the centre of mass is at \( (5.35,\ 3.24) \).

(a) [1] correct area 48; [1] correct centroid (4,3). (b) [1] correct coordinates of A, B, E; [1] correct area of triangle (15); [1] correct method for triangle centroid (mean of vertices); [1] correct centroid (29/3, 4). (c) [1] correct total area (63); [1] correct method (area-weighted mean, ECF); [1] correct final coordinates (5.35, 3.24), both components correct.

(a) Taking B as the origin, A is at \( (0,3) \) and C is at \( (4,0) \), so triangle ABC has a right angle at B with \( AB=3\text{ m} \), \( BC=4\text{ m} \). By Pythagoras, \( AC=\sqrt{3^2+4^2}=\sqrt{25}=5\text{ m} \). The angle AC makes with the horizontal BC is \( \tan^{-1}\left(\dfrac{3}{4}\right)=36.9^{\circ} \).
(b) At joint C, three forces act: the load \( 500\text{ N} \) downward, the force in rod AC (assume tension \( T_{AC} \), pulling C towards A, i.e. at \( 36.9^{\circ} \) above the horizontal, back towards the wall), and the force in rod BC (assume tension \( T_{BC} \), pulling C towards B, i.e. horizontally towards the wall). Resolving vertically: \( T_{AC}\sin36.9^{\circ}=500 \), so \( T_{AC}=\dfrac{500}{\sin36.9^{\circ}}=\dfrac{500}{0.6}=833\text{ N} \) (since \( \sin36.9^{\circ}=3/5 \)). Resolving horizontally: \( T_{BC}+T_{AC}\cos36.9^{\circ}=0 \) (taking towards the wall as positive for both assumed tensions, noting there is no other horizontal force), so \( T_{BC}=-T_{AC}\cos36.9^{\circ}=-833\times0.8=-667\text{ N} \). Since \( T_{AC}=833\text{ N}>0 \), AC is confirmed to be in tension, magnitude 833 N. Since \( T_{BC} \) is negative (opposite to the assumed tension direction), BC is in fact in thrust (compression), magnitude 667 N.
(c) Since only rod BC meets the wall at B, the equilibrium of the (massless) pin at B requires the wall's reaction to be equal and opposite to the force the rod exerts there. As BC is in thrust, it pushes joint B horizontally into the wall; the wall must therefore push back on the framework with a reaction of \( 667\text{ N} \), horizontal, directed away from the wall (i.e. in the direction from B towards C).
(d) Similarly, only rod AC meets the wall at A. As AC is in tension, it pulls joint A towards C (down and away from the wall, at \( 36.9^{\circ} \) below the horizontal); the wall must supply an equal and opposite reaction of \( 833\text{ N} \), directed along CA (i.e. at \( 36.9^{\circ} \) above the horizontal), pointing away from C, up and into the wall.

(a) [1] correct use of Pythagoras; [1] AC=5m; [1] correct angle 36.9°. (b) [1] correct identification of the three forces at joint C; [1] correct vertical resolution equation; [1] correct value \( T_{AC}=833\text{ N} \); [1] correctly identifies AC as tension; [1] correct horizontal resolution equation; [1] correct value \( T_{BC}=667\text{ N} \) (magnitude, ECF); [1] correctly identifies BC as thrust/compression; [1] full, clearly-presented method (resolving at a single joint, correct signs used throughout). (c) [1] correct magnitude 667 N (ECF); [1] correct direction (horizontal, away from wall). (d) [1] correct magnitude 833 N (ECF); [1] correct angle stated (36.9° to horizontal); [1] correct direction description (away from C, into the wall).

(a) Using Hooke's law, the tension in the spring is \( T=\dfrac{\lambda e}{L} \), where e is the extension. At equilibrium, \( T=mg \): \( \dfrac{24e}{0.6}=0.4\times9.8=3.92 \), so \( e=\dfrac{3.92\times0.6}{24}=0.098\text{ m} \).
(b) Let x be the displacement of P below its equilibrium position at time t (x positive downward). The extension of the spring is then \( (0.098+x) \), so the tension is \( T=\dfrac{24(0.098+x)}{0.6}=40(0.098+x) \). By Newton's second law (taking downward as positive): \( m\ddot{x}=mg-T=3.92-40(0.098+x)=3.92-3.92-40x=-40x \). So \( 0.4\ddot{x}=-40x \), giving \( \ddot{x}=-100x \). This is of the standard SHM form \( \ddot{x}=-\omega^2x \) with \( \omega^2=100 \) (so \( \omega=10\text{ rad s}^{-1} \)), confirming SHM about the equilibrium position while the spring remains taut.
(c) P is released from rest at \( x=0.05\text{ m} \), so this is the amplitude: \( A=0.05\text{ m} \). Period \( T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{10}=0.628\text{ s} \) (3 s.f.).
(d) Maximum speed occurs at the equilibrium position (x=0): \( v_{max}=\omega A=10\times0.05=0.5\text{ m s}^{-1} \).
(e) Maximum extension (at the lowest point, \( x=+A \)): \( 0.098+0.05=0.148\text{ m} \). Minimum extension (at the highest point, \( x=-A \)): \( 0.098-0.05=0.048\text{ m} \). Since the minimum extension \( (0.048\text{ m}) \) is still positive (greater than zero), the spring remains extended (taut) throughout the motion and never becomes slack.

(a) [1] correct formula \( T=\lambda e/L \); [1] correct equilibrium equation \( T=mg \); [1] \( e=0.098\text{ m} \). (b) [1] correct expression for tension at displacement x, \( 40(0.098+x) \); [1] correct equation of motion \( m\ddot{x}=mg-T \); [1] correct simplification showing the \( mg \) and equilibrium tension terms cancel; [1] correct final form \( \ddot{x}=-100x \); [1] correctly identifies SHM with \( \omega^2=100 \). (c) [1] correct amplitude 0.05 m; [1] correct formula \( T=2\pi/\omega \); [1] \( T=0.628\text{ s} \). (d) [1] correct formula \( v_{max}=\omega A \); [1] \( v_{max}=0.5\text{ m s}^{-1} \). (e) [1] correct max and min extensions (0.148 m, 0.048 m); [1] correct conclusion that the spring stays taut, with valid reasoning (min extension > 0).

(a) By Newton's second law, \( m\ddot{x} \) = (net force towards O, i.e. restoring force plus resistive force, both opposing increasing x) \( =-50x-20\dot{x} \). With \( m=2 \): \( 2\ddot{x}=-50x-20\dot{x} \), so \( \ddot{x}=-25x-10\dot{x} \), i.e. \( \ddot{x}+10\dot{x}+25x=0 \), as required.
(b) The auxiliary equation is \( m^2+10m+25=0 \), i.e. \( (m+5)^2=0 \), giving a repeated root \( m=-5 \).
(c) Since the auxiliary equation has a repeated (real) root, this represents critical damping — the particle returns to its equilibrium position as quickly as possible without oscillating.
(d) For a repeated root \( m=-5 \), the general solution is \( x=(A+Bt)e^{-5t} \). At \( t=0 \): \( x=A=0.4 \). Differentiating: \( \dot{x}=Be^{-5t}-5(A+Bt)e^{-5t}=[B-5A-5Bt]e^{-5t} \). At \( t=0 \): \( \dot{x}=B-5A=0 \), so \( B=5A=5(0.4)=2.0 \). So \( x=(0.4+2.0t)e^{-5t} \).
(e) For \( t\ge0 \), the factor \( (0.4+2.0t) \) is always strictly positive (since both 0.4 and \( 2.0t \) are non-negative, and 0.4 is never zero), and \( e^{-5t}>0 \) for all finite t. So \( x(t)>0 \) for every finite \( t\ge0 \): the particle never actually reaches \( x=0 \) for any finite time. However, as \( t\to\infty \), \( e^{-5t}\to0 \) faster than \( (0.4+2.0t)\to\infty \), so \( x\to0 \); the particle approaches O asymptotically but never crosses or reaches it — this lack of oscillation about, or overshoot of, O is characteristic of critical damping.

(a) [1] correct identification of the two forces acting; [1] correct equation \( 2\ddot{x}=-50x-20\dot{x} \); [1] correctly divided through by 2 to reach the given form. (b) [1] correct auxiliary equation; [1] correct factorisation \( (m+5)^2=0 \); [1] correct repeated root \( m=-5 \). (c) [1] correctly identifies critical damping; [1] correct justification (repeated real root). (d) [1] correct general solution form \( (A+Bt)e^{-5t} \) for a repeated root; [1] correct use of \( x(0)=0.4 \) to find A; [1] correct differentiation of x(t) (product rule); [1] correct use of \( \dot{x}(0)=0 \); [1] correct value \( B=2.0 \); [1] correct final solution \( x=(0.4+2.0t)e^{-5t} \). (e) [1] correctly argues \( (0.4+2.0t)e^{-5t}>0 \) for all finite \( t\ge0 \); [1] correctly concludes the particle never reaches O in finite time; [1] correctly notes x→0 only as \( t\to\infty \) (asymptotic approach), consistent with critical damping.

(a) \( \theta=\tan^{-1}(0.3)=16.7^{\circ} \) (1 d.p.).
(b) With no friction required, the normal reaction alone provides the centripetal force: horizontally, \( N\sin\theta=\dfrac{mv_0^2}{r} \); vertically, \( N\cos\theta=mg \). Dividing: \( \tan\theta=\dfrac{v_0^2}{rg} \), so \( v_0=\sqrt{rg\tan\theta}=\sqrt{80\times9.8\times0.3}=\sqrt{235.2}=15.3\text{ m s}^{-1} \) (3 s.f.).
(c) At the point of sliding up (outward), friction \( f=\mu N \) acts down the slope, opposing the tendency to slide up. The normal reaction N is perpendicular to the road; friction acts along the road surface. Resolving vertically (N has vertical component \( N\cos\theta \) upward; friction, acting down the slope, has vertical component \( f\sin\theta \) downward; weight mg downward): \( N\cos\theta=mg+f\sin\theta=mg+\mu N\sin\theta \), so \( N(\cos\theta-\mu\sin\theta)=mg \), giving \( N=\dfrac{mg}{\cos\theta-\mu\sin\theta} \). Resolving horizontally, toward the centre (N has horizontal component \( N\sin\theta \) inward; friction, down the slope, has horizontal component \( f\cos\theta=\mu N\cos\theta \), also inward, since both act to provide extra centripetal force at maximum speed): \( N\sin\theta+\mu N\cos\theta=\dfrac{mv_{max}^2}{r} \), i.e. \( N(\sin\theta+\mu\cos\theta)=\dfrac{mv_{max}^2}{r} \). Substituting N: \( \dfrac{mg(\sin\theta+\mu\cos\theta)}{\cos\theta-\mu\sin\theta}=\dfrac{mv_{max}^2}{r} \), so \( v_{max}^2=\dfrac{rg(\sin\theta+\mu\cos\theta)}{\cos\theta-\mu\sin\theta} \). Dividing numerator and denominator by \( \cos\theta \): \( v_{max}^2=\dfrac{rg(\tan\theta+\mu)}{1-\mu\tan\theta} \), as required. Substituting values: \( v_{max}^2=\dfrac{80\times9.8\times(0.3+0.2)}{1-0.2\times0.3}=\dfrac{784\times0.5}{0.94}=\dfrac{392}{0.94}=417 \), so \( v_{max}=\sqrt{417}=20.4\text{ m s}^{-1} \) (3 s.f.).
(d) At the point of sliding down (inward), friction acts up the slope, opposing the tendency to slide down; by the same method but with the sign of the friction terms reversed: vertically, \( N\cos\theta+\mu N\sin\theta=mg \), so \( N=\dfrac{mg}{\cos\theta+\mu\sin\theta} \); horizontally, the friction's inward contribution is now subtracted (friction acts outward, up the slope, opposing the centripetal direction): \( N\sin\theta-\mu N\cos\theta=\dfrac{mv_{min}^2}{r} \), giving \( v_{min}^2=\dfrac{rg(\tan\theta-\mu)}{1+\mu\tan\theta} \). Substituting: \( v_{min}^2=\dfrac{80\times9.8\times(0.3-0.2)}{1+0.2\times0.3}=\dfrac{784\times0.1}{1.06}=\dfrac{78.4}{1.06}=74.0 \), so \( v_{min}=\sqrt{74.0}=8.60\text{ m s}^{-1} \) (3 s.f.). (Since \( \mu<\tan\theta \), this minimum speed is positive and meaningful — friction alone is not sufficient to hold the car stationary on this slope.)

(a) [1] correct angle 16.7°. (b) [1] correct pair of equations (N components) set up; [1] correctly combined (e.g. dividing) to eliminate N; [1] \( v_0=15.3\text{ m s}^{-1} \). (c) [1] correct vertical equation with friction acting down-slope; [1] correct horizontal (centripetal) equation with friction acting down-slope (inward component); [1] correct elimination of N to reach the given formula; [1] fully correct algebraic derivation shown (clear working, not just quoting the result); [1] correct substitution of values; [1] correct evaluation of \( \tan\theta+\mu \) and \( 1-\mu\tan\theta \); [1] correct value under the square root; [1] \( v_{max}=20.4\text{ m s}^{-1} \). (d) [1] correctly identifies friction acts up the slope (opposing sliding down); [1] correct vertical equation; [1] correct horizontal (centripetal) equation with friction now reducing the centripetal contribution; [1] correct formula for \( v_{min}^2 \) derived; [1] correct substitution and evaluation; [1] \( v_{min}=8.60\text{ m s}^{-1} \), with a valid comment that \( \mu<\tan\theta \) makes this a meaningful (positive) minimum speed.

(a) \( \mathbf{R}=\mathbf{F_1}+\mathbf{F_2}+\mathbf{F_3}=(4-2+1)\mathbf{i}+(3+5-4)\mathbf{j}=(3\mathbf{i}+4\mathbf{j})\text{ N} \). \( |\mathbf{R}|=\sqrt{3^2+4^2}=\sqrt{25}=5\text{ N} \).
(b) Moment of \( \mathbf{F_1} \) about O: \( (2)(3)-(0)(4)=6 \). Moment of \( \mathbf{F_2} \): \( (0)(5)-(3)(-2)=0+6=6 \). Moment of \( \mathbf{F_3} \): \( (4)(-4)-(4)(1)=-16-4=-20 \). Total moment \( =6+6-20=-8\text{ N m} \).
(c) For a system of forces with resultant \( \mathbf{R} \) and total moment M about O, the line of action of the resultant is at perpendicular distance \( d=\dfrac{|M|}{|\mathbf{R}|} \) from O (this is the single force, acting along this line, that reproduces both the resultant force and the total moment of the original system). So \( d=\dfrac{8}{5}=1.6\text{ m} \). Since the total moment \( M=-8 \) is negative (using the anticlockwise-positive convention stated), the total moment about O is clockwise.

(a) [1] correct i-component (3); [1] correct j-component (4); [1] correct magnitude 5 N. (b) [1] correct moment of F1 (6); [1] correct moment of F2 (6); [1] correct moment of F3 (-20); [1] correct total (-8 N m). (c) [1] correct formula \( d=|M|/|R| \) (ECF); [1] correct value 1.6 m; [1] correctly identifies the sign of M as negative; [1] correctly states this corresponds to a clockwise moment.

(a) Integrating \( \mathbf{a} \) with respect to t: \( \mathbf{v}=\displaystyle\int\mathbf{a}\,dt=(3t^2+C_1)\mathbf{i}+(-4t+C_2)\mathbf{j}+(2t+C_3)\mathbf{k} \). Using \( \mathbf{v}(0)=(1,5,-3) \): \( C_1=1,\ C_2=5,\ C_3=-3 \). So \( \mathbf{v}(t)=(1+3t^2)\mathbf{i}+(5-4t)\mathbf{j}+(2t-3)\mathbf{k} \).
(b) Integrating \( \mathbf{v}(t) \): \( \mathbf{r}=\displaystyle\int\mathbf{v}\,dt=(t+t^3+D_1)\mathbf{i}+(5t-2t^2+D_2)\mathbf{j}+(t^2-3t+D_3)\mathbf{k} \). Using \( \mathbf{r}(0)=(2,-1,0) \): \( D_1=2,\ D_2=-1,\ D_3=0 \). So \( \mathbf{r}(t)=(2+t+t^3)\mathbf{i}+(-1+5t-2t^2)\mathbf{j}+(t^2-3t)\mathbf{k} \).
(c) Using the expression from (a): \( \mathbf{v}(3)=(1+3(9))\mathbf{i}+(5-12)\mathbf{j}+(6-3)\mathbf{k}=(28\mathbf{i}-7\mathbf{j}+3\mathbf{k})\text{ m s}^{-1} \). For \( t>3 \), the acceleration is constant at \( (2,1,-1) \), so \( \mathbf{v}(t)=\mathbf{v}(3)+(2,1,-1)(t-3) \) for \( t\ge3 \). At \( t=5 \), \( t-3=2 \): \( \mathbf{v}(5)=(28+2(2))\mathbf{i}+(-7+1(2))\mathbf{j}+(3-1(2))\mathbf{k}=(32\mathbf{i}-5\mathbf{j}+1\mathbf{k})\text{ m s}^{-1} \).

(a) [1] correct i-component of v(t); [1] correct j-component; [1] correct k-component; [1] all three constants of integration correctly found using v(0). (b) [1] correct i-component of r(t); [1] correct j-component; [1] correct k-component; [1] all three constants correctly found using r(0). (c) [1] correct evaluation of v(3) using (a) (ECF); [1] correctly identifies constant acceleration applies for t>3; [1] correct use of \( v=v(3)+a(t-3) \) with \( t-3=2 \); [1] correct component-wise addition; [1] correct final answer \( (32,-5,1) \).

(a) Conservation of momentum for the A-B collision: \( m_Au_A+m_Bu_B=m_Av_A+m_Bv_B \): \( 2(6)+3(0)=2v_A+3v_B \), so \( 12=2v_A+3v_B \). Newton's law of restitution: \( v_B-v_A=e(u_A-u_B)=0.5(6-0)=3 \), so \( v_B=v_A+3 \). Substituting: \( 12=2v_A+3(v_A+3)=5v_A+9 \), so \( v_A=0.6\text{ m s}^{-1} \), and \( v_B=0.6+3=3.6\text{ m s}^{-1} \). (Both are positive, so both spheres continue moving in the original direction, with B now moving faster than A, so they correctly separate after the collision.)
(b) Since B now moves at \( 3.6\text{ m s}^{-1} \) towards C, which is at rest directly ahead of it, and the surface is smooth (no resistance to slow B down), B must eventually reach and collide with C. For the B-C collision, momentum conservation: \( m_Bv_B+m_Cu_C=m_Bv_B'+m_Cv_C' \): \( 3(3.6)+5(0)=3v_B'+5v_C' \), so \( 10.8=3v_B'+5v_C' \). Restitution: \( v_C'-v_B'=0.5(3.6-0)=1.8 \), so \( v_C'=v_B'+1.8 \). Substituting: \( 10.8=3v_B'+5(v_B'+1.8)=8v_B'+9 \), so \( v_B'=0.225\text{ m s}^{-1} \), and \( v_C'=0.225+1.8=2.025\text{ m s}^{-1} \).
(c) After the A-B collision, sphere A continues at a constant \( 0.6\text{ m s}^{-1} \) (nothing acts on A again until it potentially meets B, since the surface is smooth and A is not involved in the B-C collision). After the B-C collision, B's new velocity is \( 0.225\text{ m s}^{-1} \), which is less than A's velocity of \( 0.6\text{ m s}^{-1} \). Since A (behind B, in the direction of motion) is moving faster than B is now moving, A will catch up to B, so a further collision between A and B does occur.

(a) [1] correct momentum equation; [1] correct restitution equation; [1] correct substitution/elimination method; [1] \( v_A=0.6\text{ m s}^{-1} \); [1] \( v_B=3.6\text{ m s}^{-1} \); [1] valid comment/check that the velocities are consistent with the spheres separating (v_B>v_A). (b) [1] correct reasoning that B must catch C (constant velocity, smooth surface, C stationary ahead); [1] correct momentum equation for B-C; [1] correct restitution equation for B-C; [1] correct substitution/elimination; [1] \( v_B'=0.225\text{ m s}^{-1} \); [1] \( v_C'=2.025\text{ m s}^{-1} \). (c) [1] correctly identifies A's velocity is unchanged since the A-B collision (0.6 m/s); [1] correctly identifies B's new (post-second-collision) velocity (0.225 m/s, ECF); [1] correct numerical comparison of the two velocities; [1] correct conclusion that A is faster than B; [1] correctly concludes a further A-B collision occurs; [1] additional mark for a clear, fully justified explanation referencing the relative positions and velocities of A and B.

(a) By Newton's second law, taking downward as positive: \( m\dfrac{dv}{dt}=mg-0.5v \) (weight acts downward; resistance opposes the downward motion, so acts upward, i.e. negative). With \( m=0.5 \): \( 0.5\dfrac{dv}{dt}=0.5(9.8)-0.5v=4.9-0.5v \). Dividing through by 0.5: \( \dfrac{dv}{dt}=9.8-v \), i.e. \( \dfrac{dv}{dt}+v=9.8 \), as required.
(b) This is a linear first-order differential equation of the form \( \dfrac{dv}{dt}+v=9.8 \), with integrating factor \( I=e^{\int1\,dt}=e^t \). Multiplying through by \( e^t \): \( e^t\dfrac{dv}{dt}+ve^t=9.8e^t \), i.e. \( \dfrac{d}{dt}(ve^t)=9.8e^t \). Integrating both sides: \( ve^t=9.8e^t+C \), so \( v=9.8+Ce^{-t} \). Using \( v=0 \) at \( t=0 \): \( 0=9.8+C \), so \( C=-9.8 \). Hence \( v=9.8-9.8e^{-t}=9.8(1-e^{-t}) \).
(c) As \( t\to\infty \), \( e^{-t}\to0 \), so \( v\to9.8\text{ m s}^{-1} \); this is the terminal (limiting) velocity.
(d) 90% of the terminal velocity is \( 0.9\times9.8=8.82\text{ m s}^{-1} \). Setting \( 9.8(1-e^{-t})=8.82 \): \( 1-e^{-t}=0.9 \), so \( e^{-t}=0.1 \), giving \( t=-\ln(0.1)=\ln(10)=2.30\text{ s} \) (3 s.f.).

(a) [1] correctly identifies both forces (weight down, resistance up, opposing motion); [1] correct equation of motion \( 0.5\dot{v}=4.9-0.5v \); [1] correctly divided to reach the given form. (b) [1] correct identification of the equation as linear first-order; [1] correct integrating factor \( e^t \); [1] correctly multiplies through and recognises the LHS as \( d(ve^t)/dt \); [1] correctly integrates the RHS; [1] correct general solution \( v=9.8+Ce^{-t} \); [1] correct use of the initial condition \( v(0)=0 \); [1] \( C=-9.8 \); [1] final solution \( v=9.8(1-e^{-t}) \). (c) [1] correctly considers the limit as \( t\to\infty \); [1] \( v\to9.8\text{ m s}^{-1} \). (d) [1] correct equation set up for 90% of terminal velocity; [1] correctly solved for \( e^{-t}=0.1 \); [1] \( t=2.30\text{ s} \) (ECF).

(a) Cylinder: \( V=\pi r^2h=\pi(6)^2(10)=360\pi\text{ cm}^3 \). Cone: \( V=\dfrac{1}{3}\pi r^2h=\dfrac{1}{3}\pi(6)^2(8)=96\pi\text{ cm}^3 \).
(b) The cylinder's own centre of mass is at height \( \dfrac{10}{2}=5\text{ cm} \) above the base (the origin). The cone sits on top of the cylinder, so its base is at height 10 cm; its own centre of mass is \( \dfrac{8}{4}=2\text{ cm} \) above its base, i.e. at height \( 10+2=12\text{ cm} \) above the origin. Treating the composite as the two solids together: total volume \( =360\pi+96\pi=456\pi\text{ cm}^3 \). Height of composite centre of mass: \( \bar{z}=\dfrac{360\pi(5)+96\pi(12)}{456\pi}=\dfrac{1800+1152}{456}=\dfrac{2952}{456}=6.47\text{ cm} \) (3 s.f.).
(c) The solid is on the point of toppling about the lower edge of its circular base (at horizontal distance r=6 cm from the central axis) when the vertical line through the centre of mass passes exactly through that edge. This occurs when \( \tan\theta=\dfrac{r}{\bar{z}}=\dfrac{6}{6.47}=0.927 \), so \( \theta=\tan^{-1}(0.927)=42.8^{\circ} \) (3 s.f.).
(d) The sliding angle is \( \theta=\tan^{-1}(\mu)=\tan^{-1}(0.5)=26.6^{\circ} \) (3 s.f.). Since \( 26.6^{\circ} \) is reached before \( 42.8^{\circ} \) as \( \theta \) is increased from zero, the solid reaches the point of sliding first; it will therefore slide on the plane before the angle becomes large enough for it to topple.

(a) [1] correct cylinder volume; [1] correct cone volume; [1] both left in terms of π (or correct decimal equivalents). (b) [1] correct centroid height of the cylinder alone (5 cm); [1] correct centroid height of the cone alone within the composite (12 cm); [1] correct total volume (456π); [1] correct volume-weighted-mean method set up; [1] correct numerator (2952, or equivalent in π); [1] \( \bar{z}=6.47\text{ cm} \). (c) [1] correctly identifies the toppling condition \( \tan\theta=r/\bar{z} \); [1] correct substitution (ECF from (b)); [1] correct value of \( \tan\theta \); [1] correct method to find θ; [1] \( \theta=42.8^{\circ} \). (d) [1] correct sliding angle 26.6°; [1] correct comparison of the two angles; [1] correct conclusion (slides first), with valid reasoning referencing which critical angle is reached first.

(a) Since expectation is additive: \( E(T)=3E(A)+2E(O)=3(180)+2(220)=540+440=980\text{ g} \). Since all weights are independent, variances add: \( \text{Var}(T)=3\,\text{Var}(A)+2\,\text{Var}(O)=3(15^2)+2(20^2)=675+800=1475\text{ g}^2 \).
(b) A linear combination of independent normal variables is itself normally distributed, so \( T\sim N(980,1475) \). Standard deviation \( =\sqrt{1475}=38.41\text{ g} \). \( z=\dfrac{1000-980}{38.41}=0.521 \). \( P(T>1000)=P(Z>0.521)=1-\Phi(0.521)=1-0.699=0.301 \) (3 s.f.).
(c) \( \bar{A} \) is the mean of 3 independent apple weights, so \( E(\bar{A})=180 \), \( \text{Var}(\bar{A})=\dfrac{15^2}{3}=75 \). \( \bar{O} \) is the mean of 2 independent orange weights, so \( E(\bar{O})=220 \), \( \text{Var}(\bar{O})=\dfrac{20^2}{2}=200 \). Since \( D=\bar{O}-\bar{A} \): \( E(D)=E(\bar{O})-E(\bar{A})=220-180=40\text{ g} \). Since \( \bar{A} \) and \( \bar{O} \) are independent, \( \text{Var}(D)=\text{Var}(\bar{O})+\text{Var}(\bar{A})=200+75=275\text{ g}^2 \) (variances add even for a difference, since the variables are independent). So \( D\sim N(40,275) \), with standard deviation \( \sqrt{275}=16.58\text{ g} \). \( z=\dfrac{50-40}{16.58}=0.603 \). \( P(D>50)=P(Z>0.603)=1-\Phi(0.603)=1-0.727=0.273 \) (3 s.f.).

(a) [1] correct use of \( E(T)=3E(A)+2E(O) \); [1] \( E(T)=980 \); [1] correct use of \( \text{Var}(T)=3\text{Var}(A)+2\text{Var}(O) \) (variances add for independent variables); [1] correct substitution; [1] \( \text{Var}(T)=1475 \). (b) [1] correctly states T is normally distributed; [1] correct standard deviation \( \sqrt{1475}=38.4 \); [1] correct z-value; [1] correct use of the normal tables; [1] \( P(T>1000)=0.301 \). (c) [1] correct \( E(\bar A)=180,\ \text{Var}(\bar A)=75 \); [1] correct \( E(\bar O)=220,\ \text{Var}(\bar O)=200 \); [1] correct \( E(D)=40 \); [1] correct \( \text{Var}(D)=275 \) (variances add, even though D is a difference, since the variables are independent); [1] correct z-value 0.603; [1] correct use of tables; [1] \( P(D>50)=0.273 \).

(a) \( H_0 \): gender and drink preference are independent (there is no association). \( H_1 \): gender and drink preference are not independent (there is an association).
(b) Under \( H_0 \), the expected frequency for each cell is \( E=\dfrac{\text{row total}\times\text{column total}}{\text{grand total}} \). Male-Tea: \( \dfrac{100\times90}{200}=45 \). Male-Coffee: \( \dfrac{100\times110}{200}=55 \). Female-Tea: \( \dfrac{100\times90}{200}=45 \). Female-Coffee: \( \dfrac{100\times110}{200}=55 \).
(c) For each cell, \( |O-E| \) is 10 in every case (since \( |35-45|=|65-55|=|55-45|=|45-55|=10 \)), so \( (|O-E|-0.5)^2=9.5^2=90.25 \) for every cell. \( \chi^2=\dfrac{90.25}{45}+\dfrac{90.25}{55}+\dfrac{90.25}{45}+\dfrac{90.25}{55}=2.006+1.641+2.006+1.641=7.29 \) (3 s.f.).
(d) For a \( 2\times2 \) contingency table, degrees of freedom \( =(2-1)(2-1)=1 \). Since the test statistic \( \chi^2=7.29 \) is greater than the critical value \( \chi^2_{0.05,1}=3.841 \), the result is significant at the 5% level, so we reject \( H_0 \). There is significant evidence, at the 5% level, of an association between gender and drink preference in this population.

(a) [1] correct null hypothesis (independence/no association); [1] correct alternative hypothesis. (b) [1] correct method (row total × column total ÷ grand total); [1]-[1] correct values for at least 3 of 4 cells (all four should be 45, 55, 45, 55); [1] all four expected frequencies correct. (c) [1] correct |O-E| values (all 10); [1] correct application of Yates' correction (subtracting 0.5 before squaring); [1] correct individual terms (90.25/45=2.006, 90.25/55=1.641); [1] all four terms correctly calculated; [1] correct summation method; [1] \( \chi^2=7.29 \). (d) [1] correct degrees of freedom, 1; [1] correct comparison of test statistic with critical value; [1] correctly rejects H0; [1] correct conclusion stated in the context of gender and drink preference.

(a) \( H_0 \): \( \mu_A=\mu_B \) (no difference in mean completion time). \( H_1 \): \( \mu_A\ne\mu_B \) (there is a difference).
(b) \( s_p^2=\dfrac{(10-1)(12.6)+(8-1)(15.2)}{10+8-2}=\dfrac{9(12.6)+7(15.2)}{16}=\dfrac{113.4+106.4}{16}=\dfrac{219.8}{16}=13.7 \) (3 s.f.).
(c) \( t=\dfrac{25.4-22.1}{\sqrt{13.7\left(\frac{1}{10}+\frac{1}{8}\right)}}=\dfrac{3.3}{\sqrt{13.7\times0.225}}=\dfrac{3.3}{\sqrt{3.09}}=\dfrac{3.3}{1.76}=1.88 \) (3 s.f.).
(d) Degrees of freedom \( =n_A+n_B-2=10+8-2=16 \). Since \( |t|=1.88 \) is less than the critical value \( t_{0.025,16}=2.120 \), the result is not significant at the 5% level, so we do not reject \( H_0 \). There is not enough evidence, at the 5% level, of a difference in the mean completion time between the two teaching methods.
(e) For the pooled two-sample t-test to be valid: (i) both populations (completion times under Method A and under Method B) must be normally distributed; and (ii) the two population variances must be equal (this is the assumption that justifies pooling the two sample variances into a single estimate \( s_p^2 \)). The two samples must also be independent of one another (given here, as they are independent random samples).

(a) [1] correct null hypothesis; [1] correct alternative hypothesis (two-tailed). (b) [1] correct formula quoted; [1] correct numerator (219.8); [1] correct division by 16; [1] \( s_p^2=13.7 \). (c) [1] correct numerator (3.3); [1] correct use of \( 1/n_A+1/n_B \); [1] correct value inside the square root (3.09); [1] correct square root taken; [1] \( t=1.88 \) (ECF from (b)). (d) [1] correct degrees of freedom, 16; [1] correct comparison with the critical value; [1] correctly does not reject H0; [1] correct conclusion stated in context; [1] additional mark for correctly noting the test is two-tailed (comparing |t| with the given two-tailed critical value). (e) [1] each for two valid, distinct assumptions (normality of both populations; equal population variances; independence of the samples) — max 2 assumptions, [2] marks each for a correctly and fully stated assumption.

(a) The central limit theorem states that, for a sufficiently large sample size (conventionally \( n\ge30 \)), the distribution of the sample mean \( \bar{X} \) is approximately normal, with mean \( \mu \) and standard deviation \( \dfrac{\sigma}{\sqrt{n}} \) (the standard error of the mean), regardless of the shape of the underlying population distribution. This may be applied here since the sample size \( n=40 \) is at least 30.
(b) Standard error \( =\dfrac{\sigma}{\sqrt{n}}=\dfrac{8}{\sqrt{40}}=1.265 \) (3 d.p.). A 95% confidence interval is \( \bar{x}\pm1.96\times\text{SE} \): \( 502.3\pm1.96(1.265)=502.3\pm2.48 \), giving the interval \( (499.8,\ 504.8) \) (1 d.p.).
(c) A 99% confidence interval uses the critical value \( z=2.576 \): \( 502.3\pm2.576(1.265)=502.3\pm3.26 \), giving the interval \( (499.0,\ 505.6) \) (1 d.p.). This interval is wider than the 95% interval, because a higher level of confidence requires a larger margin of error (a wider range of plausible values) to be more certain of capturing the true population mean.
(d) We require the margin of error to satisfy \( 1.96\times\dfrac{8}{\sqrt{n}}\le1.5 \). Rearranging: \( \sqrt{n}\ge\dfrac{1.96\times8}{1.5}=10.45 \), so \( n\ge10.45^2=109.3 \). Since n must be a whole number and we need the margin of error to be no more than 1.5g, we round up: \( n=110 \).

(a) [1] correct statement of the CLT (sample mean approximately normal for large n, mean μ, sd σ/√n); [1] correct condition stated (n≥30, conventionally); [1] correctly confirms n=40≥30 applies here. (b) [1] correct standard error 1.265; [1] correct critical value 1.96; [1] correct margin of error (2.48, ECF); [1] correct lower limit; [1] correct upper limit; [1] interval correctly stated as (499.8, 504.8). (c) [1] correct critical value 2.576; [1] correct margin of error (3.26, ECF); [1] correct interval (499.0, 505.6); [1] correctly identifies the interval is wider; [1] correct explanation (higher confidence requires a wider interval). (d) [1] correct inequality set up; [1] correct rearrangement for √n; [1] correct value 10.45; [1] correct squaring to get n≥109.3; [1] correctly rounds UP (not to nearest) since n must give a margin of error no greater than required; [1] \( n=110 \).

Part 1
(a) The chromatic number of G is 3. A valid colouring: A = colour 1; B = colour 2; C = colour 3; D = colour 2; E = colour 3 (checking: A is adjacent to B, C, D, E — all different from colour 1 ✓; around the cycle B-C-D-E-B, adjacent vertices alternate colours 2 and 3, and since the cycle has even length 4, this alternation is consistent and valid ✓).
(b) G contains the triangle A-B-C (since A is joined to both B and C, and B is joined to C as part of the cycle). Any graph containing a triangle (three mutually adjacent vertices) requires at least 3 colours, since all three vertices of a triangle must receive different colours from one another. So 2 colours could never be enough, and the chromatic number is exactly 3 (since 3 colours were shown to suffice in (a)).

Part 2
(c) The possible cutsets (partitioning the four nodes into a set containing S and a set containing T) are: \( \{S\}\,|\,\{P,Q,T\} \): capacity \( =8+5=13 \) (edges S→P, S→Q). \( \{S,P\}\,|\,\{Q,T\} \): capacity \( =5+3+4=12 \) (edges S→Q, P→Q, P→T). \( \{S,Q\}\,|\,\{P,T\} \): capacity \( =8+9=17 \) (edges S→P, Q→T; note P→Q crosses from the 'T-side' back to the 'S-side' and does not count towards this cut). \( \{S,P,Q\}\,|\,\{T\} \): capacity \( =4+9=13 \) (edges P→T, Q→T).
(d) The minimum cutset is \( \{S,P\}\,|\,\{Q,T\} \), with capacity 12. By the max-flow min-cut theorem, the maximum possible flow from S to T equals the capacity of the minimum cutset, so the maximum flow is 12 units per hour. This can be achieved, for example, by the flow: S→P→T carrying 4 units; S→P→Q→T carrying 3 units; S→Q→T carrying 5 units. Total flow into T \( =4+3+5=12 \); checking each edge: S→P carries \( 4+3=7\le8 \) ✓; S→Q carries \( 5\le5 \) ✓; P→Q carries \( 3\le3 \) ✓; P→T carries \( 4\le4 \) ✓; Q→T carries \( 3+5=8\le9 \) ✓ — all capacities respected, confirming a valid flow of value 12.

(a) [1] correct chromatic number, 3; [1] a valid colouring given; [1] colouring checked/justified against all edges. (b) [1] correctly identifies a triangle in G (e.g. A-B-C); [1] correctly explains a triangle needs 3 mutually different colours; [1] correct conclusion that 2 colours cannot suffice. (c) [1] each for two correctly calculated cutset capacities (up to 4, one per cutset, allowing for a reasonable subset of all cuts if not all 4 combinatorial cuts are listed, provided the minimum is among those found); a fully correct table of all 4 cuts scores full marks. (d) [1] correct minimum cutset identified (12, ECF); [1] correct application of the max-flow min-cut theorem to state max flow = 12; [1] a valid explicit flow of value 12 described and shown to respect all capacities.

Starting at P, at each step the algorithm moves to the nearest town not yet visited. From P, the nearest unvisited town is Q (distance 12); move to Q. From Q, the nearest unvisited town (excluding P) is R (distance 10); move to R. From R, the nearest unvisited town is S (distance 15, compared with T at 20); move to S. From S, the only unvisited town is T (distance 14); move to T. Finally, return from T to the start, P (distance 30). Route: P → Q → R → S → T → P. Total length \( =12+10+15+14+30=81\text{ km} \).

[1] correctly identifies Q as nearest to P; [1] correctly identifies R as nearest to Q (excluding P); [1] correctly identifies S as nearest to R (excluding P, Q); [1] correctly identifies T as the only remaining town; [1] correctly includes the return leg T→P (30 km); [1] correct route stated in full, P-Q-R-S-T-P; [1] correct total length, 81 km.

(a) A: \( t_e=\dfrac{2+16+6}{6}=4.0 \), \( \sigma^2=\left(\dfrac{4}{6}\right)^2=0.444 \). B: \( t_e=\dfrac{3+20+9}{6}=5.33 \), \( \sigma^2=\left(\dfrac{6}{6}\right)^2=1.00 \). C: \( t_e=\dfrac{1+8+3}{6}=2.0 \), \( \sigma^2=\left(\dfrac{2}{6}\right)^2=0.111 \). D: \( t_e=\dfrac{4+24+9}{6}=6.17 \), \( \sigma^2=\left(\dfrac{5}{6}\right)^2=0.694 \). E: \( t_e=\dfrac{2+16+6}{6}=4.0 \), \( \sigma^2=0.444 \). F: \( t_e=\dfrac{3+20+7}{6}=5.0 \), \( \sigma^2=\left(\dfrac{4}{6}\right)^2=0.444 \).
(b) A and B have no predecessor: \( EF_A=4.0 \), \( EF_B=5.33 \). C depends on A: \( EF_C=4.0+2.0=6.0 \). D depends on A: \( EF_D=4.0+6.17=10.17 \). E depends on B, C: \( EF_E=\max(5.33,6.0)+4.0=6.0+4.0=10.0 \). F depends on D, E: \( EF_F=\max(10.17,10.0)+5.0=10.17+5.0=15.17 \). The expected project duration is 15.17 days. Comparing the two possible routes through the network: A-D-F \( =4.0+6.17+5.0=15.17 \) days; A-C-E-F \( =4.0+2.0+4.0+5.0=15.0 \) days (and B-E-F \( =5.33+4.0+5.0=14.33 \) days). The longest (critical) path is A-D-F, at 15.17 days, matching the project duration found above.
(c) Summing the variances of the activities on the critical path A-D-F: \( \sigma_{proj}^2=\sigma_A^2+\sigma_D^2+\sigma_F^2=0.444+0.694+0.444=1.58 \). Standard deviation \( =\sqrt{1.58}=1.26 \) days. Assuming the project duration \( X\sim N(15.17,\,1.58) \): \( z=\dfrac{16-15.17}{1.26}=0.662 \). \( P(X\le16)=P(Z\le0.662)=\Phi(0.662)=0.746 \) (3 s.f.).

(a) [1] each for two correctly calculated \( t_e \) values (up to 3 total, allow ECF for arithmetic); [1] each for two correctly calculated \( \sigma^2 \) values (up to 3 total) — award full [6] for all six \( t_e \) and \( \sigma^2 \) values correct. (b) [1] correct EF for A, B; [1] correct EF for C; [1] correct EF for D; [1] correct EF for E (using max of predecessors); [1] correct EF for F (using max of predecessors), giving project duration 15.17 days; [1] correctly identifies A-D-F as the critical path (with a valid comparison against the other path(s)). (c) [1] correctly identifies which activities' variances to sum (those on the critical path only); [1] correct sum, 1.58; [1] correct standard deviation, 1.26 (ECF); [1] correct z-value, 0.662 (ECF); [1] correct use of the normal distribution tables; [1] \( P(X\le16)=0.746 \).

(a) By the binomial theorem, \( \displaystyle\sum_{r=0}^n\binom{n}{r}x^r=(1+x)^n \); this is the generating function for the sequence \( a_r=\binom{n}{r} \).
(b) Substituting \( x=1 \) into the generating function: LHS \( =\displaystyle\sum_{r=0}^n\binom{n}{r}(1)^r=\sum_{r=0}^n\binom{n}{r} \); RHS \( =(1+1)^n=2^n \). Since the generating function identity \( \sum_{r=0}^n\binom{n}{r}x^r=(1+x)^n \) holds for all x, it holds in particular at \( x=1 \), giving \( \displaystyle\sum_{r=0}^n\binom{n}{r}=2^n \), as required.
(c) Differentiating both sides of \( \displaystyle\sum_{r=0}^n\binom{n}{r}x^r=(1+x)^n \) with respect to x: on the left, term-by-term differentiation gives \( \displaystyle\sum_{r=1}^n r\binom{n}{r}x^{r-1} \) (the \( r=0 \) term is constant and differentiates to zero); on the right, \( \dfrac{d}{dx}(1+x)^n=n(1+x)^{n-1} \). So \( \displaystyle\sum_{r=1}^n r\binom{n}{r}x^{r-1}=n(1+x)^{n-1} \). Substituting \( x=1 \): LHS \( =\displaystyle\sum_{r=1}^n r\binom{n}{r}(1)^{r-1}=\sum_{r=1}^n r\binom{n}{r} \); RHS \( =n(2)^{n-1} \). So \( \displaystyle\sum_{r=1}^n r\binom{n}{r}=n\cdot2^{n-1} \), as required.

(a) [1] correctly states \( (1+x)^n \). (b) [1] correctly substitutes x=1 into both sides; [1] correctly evaluates the RHS as \( 2^n \); [1] correct concluding statement linking this to the required sum. (c) [1] correctly differentiates the LHS term-by-term, obtaining \( \sum r\binom{n}{r}x^{r-1} \); [1] correctly differentiates the RHS, obtaining \( n(1+x)^{n-1} \); [1] correctly substitutes x=1 into both sides; [1] correct final result \( n\cdot2^{n-1} \) with a clear concluding statement.

(a) The forbidden cells are \( (P_1,J_1) \), \( (P_2,J_2) \), \( (P_3,J_1) \), \( (P_3,J_3) \), marked on the \( 4\times4 \) board (rows \( P_1 \)-\( P_4 \), columns \( J_1 \)-\( J_4 \); note \( P_4 \) has no forbidden cell, and \( P_3 \) has two). Since there are 4 forbidden cells and placing 1 rook on any single one of them is always non-attacking (trivially, with only 1 rook), \( r_1=4 \).
(b) Checking all \( \binom{4}{2}=6 \) pairs of forbidden cells for whether they share a row or column: \( (P_1,J_1)\text{-}(P_2,J_2) \): different row and column — non-attacking. \( (P_1,J_1)\text{-}(P_3,J_1) \): same column \( J_1 \) — attacking (excluded). \( (P_1,J_1)\text{-}(P_3,J_3) \): different row and column — non-attacking. \( (P_2,J_2)\text{-}(P_3,J_1) \): different row and column — non-attacking. \( (P_2,J_2)\text{-}(P_3,J_3) \): different row and column — non-attacking. \( (P_3,J_1)\text{-}(P_3,J_3) \): same row \( P_3 \) — attacking (excluded). So 4 of the 6 pairs are non-attacking: \( r_2=4 \).
(c) A valid non-attacking triple cannot contain both \( (P_3,J_1) \) and \( (P_3,J_3) \) (same row), and cannot contain both \( (P_1,J_1) \) and \( (P_3,J_1) \) (same column). Checking the 4 possible triples: \( \{(P_1,J_1),(P_2,J_2),(P_3,J_1)\} \) — contains the column clash, invalid. \( \{(P_1,J_1),(P_2,J_2),(P_3,J_3)\} \) — rows \( P_1,P_2,P_3 \) all different, columns \( J_1,J_2,J_3 \) all different — valid. \( \{(P_1,J_1),(P_3,J_1),(P_3,J_3)\} \) — contains the row clash, invalid. \( \{(P_2,J_2),(P_3,J_1),(P_3,J_3)\} \) — contains the row clash, invalid. So exactly 1 valid non-attacking triple: \( r_3=1 \). For \( r_4 \), all four forbidden cells would need to be placed together, but this set contains both clashing pairs identified above ( \( (P_1,J_1) \) with \( (P_3,J_1) \), and \( (P_3,J_1) \) with \( (P_3,J_3) \) ), so it is impossible to place all 4 as non-attacking rooks: \( r_4=0 \).
(d) \( N=r_0(4)!-r_1(3)!+r_2(2)!-r_3(1)!+r_4(0)! \), with \( r_0=1 \): \( N=1(24)-4(6)+4(2)-1(1)+0(1)=24-24+8-1+0=7 \).

(a) [1] forbidden cells correctly identified/marked; [1] \( r_1=4 \). (b) [1] correct systematic check of pairs (method shown, e.g. listing all 6); [1] correctly identifies the column clash \( (P_1,J_1)\text{-}(P_3,J_1) \); [1] correctly identifies the row clash \( (P_3,J_1)\text{-}(P_3,J_3) \); [1] correctly identifies the remaining 4 pairs as non-attacking; [1] \( r_2=4 \). (c) [1] correct systematic check of the 4 possible triples; [1] correctly identifies the one valid triple; [1] \( r_3=1 \); [1] correct reasoning for why no 4-cell non-attacking placement exists; [1] \( r_4=0 \) correctly stated with justification. (d) [1] correct formula quoted; [1] correct term \( r_0\times4!=24 \); [1] correct term \( r_1\times3!=24 \); [1] correct term \( r_2\times2!=8 \); [1] correct term \( r_3\times1!=1 \); [1] correct final answer, \( N=7 \).

(a) The order of a group is the total number of elements it contains.
(b) Lagrange's theorem states that the order of any subgroup of a finite group must divide the order of the group. Since \( |G|=12 \), and the divisors of 12 are 1, 2, 3, 4, 6 and 12, these are the only possible orders of subgroups of G.
(c) H consists of exactly 3 elements (identity, \( 120^{\circ} \) rotation, \( 240^{\circ} \) rotation), so \( |H|=3 \). For the \( 120^{\circ} \) rotation r: applying it once gives \( 120^{\circ} \), twice gives \( 240^{\circ} \), and three times gives \( 360^{\circ} \) (the identity); so its period is 3. Similarly, for the \( 240^{\circ} \) rotation: applying it once gives \( 240^{\circ} \), twice gives \( 480^{\circ}\equiv120^{\circ} \), and three times gives \( 720^{\circ}\equiv0^{\circ} \) (the identity); so its period is also 3.
(d) H is cyclic because all of its elements can be generated as successive powers of a single element: taking \( r= \) the \( 120^{\circ} \) rotation, \( r^0=\text{identity} \), \( r^1=120^{\circ} \), \( r^2=240^{\circ} \), which together give every element of H. So H is generated by the single element r, making it cyclic; a generator of H is the \( 120^{\circ} \) rotation (the \( 240^{\circ} \) rotation is also a valid generator, since its powers likewise produce all of H).

(a) [1] correct definition (number of elements in the group). (b) [1] correctly states Lagrange's theorem (subgroup order divides group order); [1] correct list of divisors of 12; [1] all six correct values (1, 2, 3, 4, 6, 12) stated with none incorrect/extra. (c) [1] correct order of H, 3; [1] correct period of the 120° rotation (3), with working shown; [1] correct period of the 240° rotation (3), with working shown. (d) [1] correctly explains what it means for H to be cyclic (generated by a single element); [1] correctly shows the powers of the 120° rotation produce all of H; [1] correctly states a valid generator of H.