2023 CCEA A-Level Mathematics 2210 Practice Paper with Answers

(a) \( g(3) = 3^2+1 = 10 \), so \( fg(3) = f(10) = 2(10)-5 = 15 \).
(b) Let \( y = 2x-5 \), so \( x = \frac{y+5}{2} \). Hence \( f^{-1}(x) = \frac{x+5}{2} \). Since \( f \) has domain \( x \geq 0 \), its range is \( f(x) \geq -5 \) (as \( f \) is increasing). So \( f^{-1} \) has domain \( x \geq -5 \) and range \( y \geq 0 \).
(c) \( |2x-5|<9 \Leftrightarrow -9 < 2x-5 < 9 \Leftrightarrow -4 < 2x < 14 \Leftrightarrow -2 < x < 7 \).
Final answers: \( fg(3) = 15 \); \( f^{-1}(x) = \frac{x+5}{2} \), domain \( x \geq -5 \), range \( y \geq 0 \); \( -2 < x < 7 \).

(a) M1 for g(3)=10; A1 for fg(3)=15.
(b) M1 for rearranging y=2x-5 to make x the subject; A1 for f⁻¹(x)=(x+5)/2; A1 for correct domain x≥-5 and range y≥0 (both needed, condone swap of domain/range language if values correct).
(c) M1 for forming -9<2x-5<9 (or splitting into two linear inequalities); M1 for correct algebraic isolation; A1 for -2

(a) Let \( \dfrac{3x+5}{(x+1)(x+3)} \equiv \dfrac{A}{x+1} + \dfrac{B}{x+3} \). Then \( 3x+5 = A(x+3) + B(x+1) \). Setting \( x=-1 \): \( 2 = 2A \Rightarrow A=1 \). Setting \( x=-3 \): \( -4 = -2B \Rightarrow B=2 \). So \( \dfrac{3x+5}{(x+1)(x+3)} = \dfrac{1}{x+1} + \dfrac{2}{x+3} \).
(b) \( y=f(x-4)+1 \) translates the graph 4 units right (increasing the x-coordinate by 4) and 1 unit up (increasing the y-coordinate by 1): minimum at \( (2+4, -3+1) = (6, -2) \).
(c) \( y=3f(x) \) stretches vertically by scale factor 3 (x-coordinate unchanged, y-coordinate tripled): minimum at \( (2, 3\times(-3)) = (2, -9) \).
(d) \( y=f(2x) \) stretches horizontally by scale factor \( \frac12 \) (x-coordinate halved, y-coordinate unchanged): minimum at \( (2\div2, -3) = (1, -3) \).

(a) M1 for correct partial fraction setup with two unknowns; M1 for forming 3x+5=A(x+3)+B(x+1); A1 for A=1; A1 for B=2 (final answer must be written as sum of two fractions).
(b) B2 for (6,-2), B1 for one coordinate correct only.
(c) B2 for (2,-9), B1 for recognising only y-coordinate changes.
(d) B2 for (1,-3), B1 for recognising only x-coordinate changes.

(a) Let \( c = \cos\theta \). Then \( 2c^2-c-1=0 \), which factorises as \( (2c+1)(c-1)=0 \), giving \( c = -\tfrac12 \) or \( c=1 \).
For \( \cos\theta = 1 \): \( \theta = 0^\circ, 360^\circ \).
For \( \cos\theta = -\tfrac12 \): the principal value is \( \theta = 120^\circ \), and by the symmetry of cosine, the other solution in range is \( \theta = 360^\circ - 120^\circ = 240^\circ \).
So \( \theta = 0^\circ, 120^\circ, 240^\circ, 360^\circ \).
(b) Using \( \sin^2\theta + \cos^2\theta = 1 \): \( \cos^2\theta = 1 - \tfrac{9}{25} = \tfrac{16}{25} \), so \( \cos\theta = \pm\tfrac{4}{5} \). Since \( \theta \) is obtuse, \( \cos\theta < 0 \), so \( \cos\theta = -\tfrac{4}{5} \). Then \( \tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{3/5}{-4/5} = -\tfrac{3}{4} \).

(a) M1 for treating as a quadratic in cosθ; M1 for correct factorisation (2c+1)(c-1)=0 or use of formula; A1 for c=1, c=-1/2; A1 for θ=0°,360° from c=1; A1 for θ=120° from c=-1/2; A1 for θ=240° (second solution from symmetry) — all four values required for full marks, condone omission of 360° as a repeat of 0° with A1 still awarded if 0° given.
(b) M1 for cos²θ=1-sin²θ=16/25; A1 for cosθ=-4/5 with correct sign justified by obtuse angle; M1 for tanθ=sinθ/cosθ; A1 for tanθ=-3/4.

(a) \( 3\sqrt{x} = 3x^{1/2} \) integrates to \( 3 \cdot \dfrac{x^{3/2}}{3/2} = 2x^{3/2} \). \( \dfrac{2}{x^2}=2x^{-2} \) integrates to \( 2 \cdot \dfrac{x^{-1}}{-1} = -\dfrac{2}{x} \). So \( \displaystyle\int\left(3\sqrt{x}+\frac{2}{x^2}\right)dx = 2x^{3/2} - \frac{2}{x} + c \).
(b) \( \left[2x^{3/2}-\frac{2}{x}\right]_1^4 = \left(2(4)^{3/2}-\frac{2}{4}\right) - \left(2(1)^{3/2}-\frac{2}{1}\right) = (16-0.5)-(2-2) = 15.5 - 0 = 15.5 \).

(a) M1 for \( x^{1/2}\to \frac{2}{3}x^{3/2} \); A1 for \( 2x^{3/2} \) term correct; M1 for \( x^{-2}\to -x^{-1} \); A1 for \( -\frac{2}{x} \) term correct (condone omission of +c).
(b) M1 for substituting x=4; A1 for value 15.5 at x=4; M1 for substituting x=1; A1 for value 0 at x=1; A1 for final answer 15.5 (accept exact fraction 31/2).

(a) Let \( u=x \), \( dv=e^{2x}dx \), so \( du=dx \), \( v=\tfrac12 e^{2x} \). Then
\( \displaystyle\int x e^{2x}dx = \tfrac12 x e^{2x} - \int \tfrac12 e^{2x}dx = \tfrac12 x e^{2x} - \tfrac14 e^{2x} + c = \left(\dfrac{x}{2}-\dfrac14\right)e^{2x}+c \).
(b) \( \left[\left(\frac{x}{2}-\frac14\right)e^{2x}\right]_0^1 = \left(\frac12-\frac14\right)e^2 - \left(0-\frac14\right)e^0 = \frac14 e^2 + \frac14 = \dfrac{e^2+1}{4} \approx 2.10 \) (3 s.f.).
(c) Volume \( = \pi\displaystyle\int_0^1 (xe^{2x})^2\,dx = \pi\displaystyle\int_0^1 x^2 e^{4x}\,dx \). Applying integration by parts twice (u=x², dv=e^{4x}dx, then u=x, dv=e^{4x}dx for the resulting integral):
\( \displaystyle\int x^2 e^{4x}dx = \left(\dfrac{x^2}{4}-\dfrac{x}{8}+\dfrac{1}{32}\right)e^{4x}+c \).
Evaluating from 0 to 1: at \( x=1 \): \( \left(\frac14-\frac18+\frac{1}{32}\right)e^4 = \frac{5}{32}e^4 \). At \( x=0 \): \( \frac{1}{32} \). So the integral \( = \dfrac{5e^4-1}{32} \).
Volume \( = \pi \cdot \dfrac{5e^4-1}{32} \approx 26.7 \) (3 s.f.).

(a) M1 for correct choice u=x, dv=e^{2x}dx; A1 for du, v correct; M1 for applying the by-parts formula; A1 for correct integral (1/2)xe^{2x} - 1/4 e^{2x}; A1 for fully simplified form with +c.
(b) M1 for correct substitution of limits; A1 for exact value (e²+1)/4; A1 for correct numerical evaluation to 3 s.f. (2.10); A1 for correct method converting exact to decimal (dep on exact form correct).
(c) M1 for volume formula π∫y²dx; A1 for correct integrand x²e^{4x}; M1 for first application of integration by parts; M1 for second application of integration by parts (reduction of ∫xe^{4x}dx); A1 for correct antiderivative; M1 for correct substitution of limits 0 and 1; A1 for final volume 26.7 (3 s.f.), accept exact form π(5e⁴-1)/32.

(a) Let \( \dfrac{7-x}{(x+1)(x-3)} \equiv \dfrac{A}{x+1}+\dfrac{B}{x-3} \), so \( 7-x = A(x-3)+B(x+1) \). At \( x=-1 \): \( 8 = -4A \Rightarrow A=-2 \). At \( x=3 \): \( 4 = 4B \Rightarrow B=1 \). So the expression equals \( \dfrac{-2}{x+1}+\dfrac{1}{x-3} \).
(b) \( \displaystyle\int\left(\dfrac{-2}{x+1}+\dfrac{1}{x-3}\right)dx = -2\ln|x+1| + \ln|x-3| + c \).
(c) Separating variables: \( \displaystyle\int \dfrac{1}{C}\,dC = \displaystyle\int -0.5\,dt \), giving \( \ln C = -0.5t + k \), so \( C = Ae^{-0.5t} \) where \( A=e^k \). When \( t=0 \), \( C=8 \), so \( A=8 \). Hence \( C = 8e^{-0.5t} \).
When \( C=2 \): \( 2 = 8e^{-0.5t} \Rightarrow e^{-0.5t}=0.25 \Rightarrow -0.5t = \ln(0.25) = -\ln4 \Rightarrow t = 2\ln4 = \ln16 \approx 2.77 \) minutes (3 s.f.).

(a) M1 for correct partial fraction setup; M1 for forming 7-x=A(x-3)+B(x+1); A1 for A=-2; A1 for B=1.
(b) B1 for -2ln|x+1|; B1 for +ln|x-3| (ft from (a)); A1 for combined correct expression; B1 for +c (condone omission if otherwise correct, but withhold final accuracy mark if missing in a 'find the integral' context — accept either convention, mark scheme is lenient here).
(c) M1 for separating variables correctly; M1 for integrating both sides (ln C = -0.5t + k); A1 for general solution C=Ae^{-0.5t}; A1 for using C=8 at t=0 to find A=8; M1 for substituting C=2 and rearranging for t; A1 for exact value t=2ln4 (or ln16); A1 for correct value 2.77 (3 s.f.).

(a) Differentiating \( x^2+xy+y^2=7 \) implicitly with respect to x: \( 2x + \left(y + x\dfrac{dy}{dx}\right) + 2y\dfrac{dy}{dx} = 0 \) (using the product rule on xy). Collecting terms in dy/dx: \( \dfrac{dy}{dx}(x+2y) = -(2x+y) \), so \( \dfrac{dy}{dx} = -\dfrac{2x+y}{x+2y} \).
(b) At \( (1,2) \): \( 1^2+1(2)+2^2 = 1+2+4=7 \), which satisfies the equation, so \( (1,2) \) lies on the curve. Gradient: \( \dfrac{dy}{dx} = -\dfrac{2(1)+2}{1+2(2)} = -\dfrac{4}{5} \).
(c) The normal gradient is the negative reciprocal: \( -\dfrac{1}{-4/5} = \dfrac{5}{4} \). Equation: \( y-2 = \dfrac{5}{4}(x-1) \Rightarrow 4y-8 = 5x-5 \Rightarrow 5x-4y+3=0 \).
(d) The tangent is vertical where \( \dfrac{dy}{dx} \) is undefined, i.e. where the denominator \( x+2y=0 \), so \( x=-2y \). Substituting into \( x^2+xy+y^2=7 \): \( 4y^2 - 2y^2 + y^2 = 7 \Rightarrow 3y^2=7 \Rightarrow y=\pm\sqrt{7/3} = \pm\dfrac{\sqrt{21}}{3} \). Since \( x=-2y \) and we require \( x>0 \), we need \( y<0 \), so \( y=-\dfrac{\sqrt{21}}{3} \) and \( x = \dfrac{2\sqrt{21}}{3} \). The point is \( \left(\dfrac{2\sqrt{21}}{3}, -\dfrac{\sqrt{21}}{3}\right) \).

(a) M1 for differentiating x² term; M1 for product rule on xy giving y+x(dy/dx); M1 for differentiating y² term as 2y(dy/dx); M1 for collecting all dy/dx terms on one side; A1 for correct factorised form dy/dx(x+2y)=-(2x+y); A1 for final answer matching given result (AG - all steps must be shown).
(b) B1 for verification 1+2+4=7; B1 for gradient -4/5 (ft from (a)).
(c) M1 for normal gradient = negative reciprocal of tangent gradient; A1 for normal gradient 5/4; M1 for forming line equation through (1,2); A1 for correct final form 5x-4y+3=0.
(d) M1 for setting denominator x+2y=0; A1 for x=-2y; M1 for substituting into curve equation; A1 for y=±√21/3; M1 for selecting correct sign using x>0 condition; A1 for correct exact coordinates (2√21/3, -√21/3).

(a) \( \dfrac{dx}{dt}=2t \), \( \dfrac{dy}{dt}=3t^2-3 \). So \( \dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{3t^2-3}{2t} = \dfrac{3(t^2-1)}{2t} \), for \( t \neq 0 \).
(b) A horizontal tangent occurs where \( \dfrac{dy}{dt}=0 \) (and \( \dfrac{dx}{dt}\neq0 \)): \( 3t^2-3=0 \Rightarrow t=\pm1 \). At \( t=1 \): \( x=1-1=0 \), \( y=1-3=-2 \), giving \( (0,-2) \). At \( t=-1 \): \( x=1-1=0 \), \( y=-1+3=2 \), giving \( (0,2) \). (Check \( dx/dt=\pm2\neq0 \) at both, confirming genuine horizontal tangents.)
(c) At \( t=2 \): \( x=4-1=3 \), \( y=8-6=2 \). Gradient: \( \dfrac{dy}{dx}=\dfrac{3(4-1)}{2(2)}=\dfrac{9}{4} \). Tangent: \( y-2=\dfrac{9}{4}(x-3) \Rightarrow y = \dfrac{9}{4}x - \dfrac{27}{4}+2 = \dfrac{9}{4}x-\dfrac{19}{4} \).
(d) The tangent is vertical where \( \dfrac{dx}{dt}=0 \) (and \( \dfrac{dy}{dt}\neq0 \)): \( 2t=0 \Rightarrow t=0 \). At \( t=0 \), \( \dfrac{dy}{dt}=-3\neq0 \), confirming a genuine vertical tangent.

(a) M1 for dx/dt=2t; M1 for dy/dt=3t²-3; M1 for using dy/dx=(dy/dt)/(dx/dt); A1 for correct unsimplified expression; A1 for simplified 3(t²-1)/(2t).
(b) M1 for setting dy/dt=0; A1 for t=±1; A1 for one correct coordinate pair; A1 for both correct coordinate pairs (0,-2) and (0,2).
(c) B1 for x=3,y=2 at t=2; M1 for substituting t=2 into dy/dx; A1 for gradient 9/4; M1 for forming and rearranging the tangent line; A1 for y=(9/4)x-19/4.
(d) B1 for t=0; B1 for correct justification referencing dx/dt=0 with dy/dt≠0.

(a) Using the compound angle formulae: \( \cos(A+B) = \cos A\cos B - \sin A \sin B \) and \( \cos(A-B) = \cos A \cos B + \sin A \sin B \). Adding: \( \cos(A+B)+\cos(A-B) = 2\cos A \cos B \), as required.
(b) \( R\cos(\theta+\alpha) = R\cos\theta\cos\alpha - R\sin\theta\sin\alpha \). Comparing with \( 5\cos\theta - 12\sin\theta \): \( R\cos\alpha=5 \) and \( R\sin\alpha=12 \). Then \( R^2 = 5^2+12^2=169 \Rightarrow R=13 \), and \( \tan\alpha = \dfrac{12}{5}=2.4 \Rightarrow \alpha = 67.4^\circ \) (1 d.p.). So \( 5\cos\theta-12\sin\theta \equiv 13\cos(\theta+67.4^\circ) \).
(c) \( 13\cos(\theta+67.4^\circ)=6 \Rightarrow \cos(\theta+67.4^\circ) = \dfrac{6}{13}=0.4615 \). Let \( \phi=\theta+67.4^\circ \), where \( \phi \in [67.4^\circ, 427.4^\circ] \) since \( \theta \in [0^\circ,360^\circ] \). The principal value is \( \arccos(0.4615)=62.5^\circ \), which is outside the range, so the valid solutions are \( \phi = 360^\circ-62.5^\circ=297.5^\circ \) and \( \phi=360^\circ+62.5^\circ=422.5^\circ \). Then \( \theta = \phi-67.4^\circ \), giving \( \theta = 230.1^\circ \) and \( \theta=355.1^\circ \) (1 d.p.).

(a) M1 for stating both compound angle expansions correctly; M1 for adding the two expansions; A1 for sine terms cancelling; A1 for arriving at 2cosAcosB (AG - all lines must be shown).
(b) M1 for expanding Rcos(θ+α); M1 for equating coefficients Rcosα=5, Rsinα=12; A1 for R=13; A1 for α=67.4°.
(c) M1 for substituting R-form into the equation; A1 for cos(θ+67.4°)=6/13; M1 for finding principal value arccos(6/13)=62.5°; M1 for correct strategy to find solutions in extended range (360±62.5°); A1 for φ=297.5° or 422.5° (at least one); A1 for both φ values correct; A1 for θ=230.1°; A1 for θ=355.1° (both required for final mark, ft errors in α consistently).

(a) Since \( |r|=\tfrac34<1 \), the series converges: \( S_\infty = \dfrac{a}{1-r} = \dfrac{8}{1-3/4} = \dfrac{8}{1/4}=32 \).
(b) \( S_n = \dfrac{n}{2}\left[2(5)+(n-1)(3)\right] = \dfrac{n}{2}(3n+7) \). Requiring \( S_n>2000 \): \( n(3n+7)>4000 \Rightarrow 3n^2+7n-4000>0 \). Solving \( 3n^2+7n-4000=0 \) using the quadratic formula: \( n = \dfrac{-7+\sqrt{49+48000}}{6} = \dfrac{-7+\sqrt{48049}}{6} \approx 35.37 \). Testing integers: at \( n=35 \), \( S_{35}=\tfrac{35}{2}(112)=1960<2000 \); at \( n=36 \), \( S_{36}=\tfrac{36}{2}(115)=2070>2000 \). So the least \( n \) is 36.
(c) \( (1+3x)^{-2} = 1 + (-2)(3x) + \dfrac{(-2)(-3)}{2!}(3x)^2 + \cdots = 1 - 6x + \dfrac{6}{2}(9x^2)+\cdots = 1 - 6x + 27x^2 - \cdots \). The expansion is valid provided \( |3x|<1 \), i.e. \( |x|<\dfrac13 \).

(a) M1 for using S∞=a/(1-r) with correct substitution; A1 for correct value of 1-r=1/4; A1 for S∞=32.
(b) M1 for correct formula for Sn; A1 for Sn=n/2(3n+7) simplified; M1 for forming and attempting to solve the inequality 3n²+7n-4000>0; M1 for testing consecutive integer values of n; A1 for n=36 with supporting check that n=35 fails.
(c) M1 for correct general binomial term with n=-2; A1 for constant term 1; M1 for correct first-order term; A1 for -6x; M1 for correct second-order term structure; A1 for 27x²; A1 for valid range condition |3x|<1; A1 for |x|<1/3.

(a) \( f(2) = 8-8-2=-2 <0 \); \( f(3)=27-12-2=13>0 \). Since \( f \) is continuous and changes sign between \( x=2 \) and \( x=3 \), there is a root \( \alpha \) in \( [2,3] \).
(b) \( f'(x)=3x^2-4 \). The Newton-Raphson formula is \( x_{n+1}=x_n - \dfrac{f(x_n)}{f'(x_n)} = x_n - \dfrac{x_n^3-4x_n-2}{3x_n^2-4} \), as required.
(c) With \( x_0=2.3 \): \( f(2.3)=2.3^3-4(2.3)-2=12.167-9.2-2=0.967 \); \( f'(2.3)=3(2.3)^2-4=15.87-4=11.87 \). So \( x_1 = 2.3 - \dfrac{0.967}{11.87} = 2.3-0.08146\ldots = 2.21854 \), which to 5 d.p. is \( x_1=2.21853 \). Repeating with \( x_1=2.21853 \): \( f(x_1) \approx 0.01254 \), \( f'(x_1)\approx 10.7415 \), giving \( x_2 = x_1 - \dfrac{0.01254}{10.7415} \approx 2.21433 \) (5 d.p.).
(d) Evaluating \( f \) either side of 2.21432 at the 5th decimal place: \( f(2.214315) \approx -0.0000508 <0 \) and \( f(2.214325) \approx +0.0000563>0 \). Since \( f \) changes sign between 2.214315 and 2.214325, the root lies in this interval, confirming \( \alpha=2.21432 \) is correct to 5 decimal places.

(a) M1 for evaluating f(2) and f(3) with correct values -2 and 13; A1 for conclusion referencing sign change and continuity.
(b) M1 for f'(x)=3x²-4; M1 for correctly stating Newton-Raphson formula; A1 for correct substitution (AG).
(c) M1 for f(2.3) and f'(2.3) evaluated correctly; A1 for x₁=2.21853; M1 for repeating the process using x₁; A1 for f(x₁) and f'(x₁) evaluated correctly (or equivalent working); A1 for x₂=2.21433; A1 for both values given to correct 5 d.p. precision.
(d) M1 for identifying suitable bounding values around 2.21432 (e.g. 2.214315 and 2.214325); A1 for correct evaluation of f at lower bound (negative); A1 for correct evaluation of f at upper bound (positive); A1 for valid conclusion that sign change confirms root correct to 5 d.p.

(a) \( \mathbf{v} = \dfrac{d\mathbf{r}}{dt} = (3t^2-3)\mathbf{i} + 4t\mathbf{j} \).
(b) \( \mathbf{a} = \dfrac{d\mathbf{v}}{dt} = 6t\mathbf{i} + 4\mathbf{j} \).
(c) The particle moves parallel to \( \mathbf{j} \) when the \( \mathbf{i} \)-component of velocity is zero: \( 3t^2-3=0 \Rightarrow t^2=1 \Rightarrow t=1 \) (taking \( t\geq0 \)).
(d) At \( t=2 \): \( \mathbf{v} = (3(4)-3)\mathbf{i}+4(2)\mathbf{j} = 9\mathbf{i}+8\mathbf{j} \). Speed \( = |\mathbf{v}| = \sqrt{9^2+8^2} = \sqrt{145} \approx 12.0 \) ms⁻¹ (3 s.f.).

(a) M1 for differentiating each component; A1 for (3t²-3)i; A1 for 4tj.
(b) M1 for differentiating v; A1 for 6ti+4j (ft from (a)).
(c) M1 for setting i-component of v to zero; A1 for t²=1; A1 for t=1 (rejecting t=-1 as t≥0).
(d) M1 for substituting t=2 into v; A1 for v=9i+8j; M1 for speed=√(vx²+vy²); A1 for √145; A1 for 12.0 (3 s.f.).

Initial vertical component: \( u_y = 24\sin35^\circ = 13.766 \) ms⁻¹. Initial horizontal component: \( u_x = 24\cos35^\circ = 19.660 \) ms⁻¹.
(a) At greatest height, vertical velocity is zero. Using \( v^2=u_y^2-2gh \): \( 0 = 13.766^2 - 2(9.8)h \Rightarrow h = \dfrac{13.766^2}{19.6} = 9.67 \) m (3 s.f.).
(b) By symmetry, time of flight \( T = \dfrac{2u_y}{g} = \dfrac{2(13.766)}{9.8} = 2.81 \) s (3 s.f.).
(c) Horizontal range \( = u_x \times T = 19.660 \times 2.809 = 55.2 \) m (3 s.f.).
(d) The particle is modelled as a point mass (with no size), so effects such as air resistance and spin are ignored, and \( g \) is taken as constant throughout the flight.

(a) M1 for resolving initial velocity into components (uy=24sin35°); M1 for using v²=uy²-2gh with v=0; A1 for correct unsimplified expression; A1 for h=9.67 m (3 s.f.).
(b) M1 for using symmetry / T=2uy/g; A1 for T=2.81 s (3 s.f.), ft from uy.
(c) M1 for ux=24cos35°; M1 for range=ux×T; A1 for 55.2 m (3 s.f.), ft from (b).
(d) B2 for a valid modelling assumption clearly stated (e.g. no air resistance, particle treated as a point mass, g constant); B1 for a vague or partially correct statement.

The rod's own weight (60 N) acts at its midpoint, 2 m from A. The load (40 N) acts at B, 4 m from A. Let \( R_A \) and \( R_C \) be the (upward) reactions at A and C respectively.
(a) Taking moments about A (clockwise positive): \( R_C \times 3 = 60\times2 + 40\times4 = 120+160=280 \). So \( R_C = \dfrac{280}{3} \approx 93.3 \) N.
(b) Resolving vertically: \( R_A+R_C = 60+40=100 \). So \( R_A = 100 - \dfrac{280}{3} = \dfrac{300-280}{3}=\dfrac{20}{3}\approx 6.67 \) N.
(c) For the rod to rest in equilibrium on both supports, both reaction forces must be non-negative (a support can only push, not pull, on the rod). Since \( R_A = \dfrac{20}{3} \) N \( >0 \), support A is genuinely required to push upward on the rod (the rod does not tend to lift off A), confirming the rod remains in equilibrium resting on both supports.

(a) M1 for taking moments about A; A1 for correct moment of weight (60×2); A1 for correct moment of load (40×4); M1 for equation RC×3=280; A1 for RC=280/3; A1 for RC≈93.3 N (3 s.f.).
(b) M1 for resolving vertically RA+RC=100; A1 for correct substitution; A1 for RA=20/3; A1 for RA≈6.67 N (3 s.f.).
(c) B1 for stating that a support can only exert a non-negative (pushing) reaction; B1 for identifying that RA>0 is required for equilibrium on both supports; B1 for correct conclusion referencing the computed value of RA.

Take A's initial direction as positive. Before collision: \( u_A=+6 \) ms⁻¹, \( u_B=-5 \) ms⁻¹.
(a) By conservation of momentum: \( m_Au_A+m_Bu_B = (m_A+m_B)v \). \( 2(6)+3(-5) = 12-15=-3 \). So \( v = \dfrac{-3}{5} = -0.6 \) ms⁻¹, i.e. the combined sphere moves at 0.6 ms⁻¹ in B's original direction.
(b) Impulse on B = change in momentum of B \( = m_B(v-u_B) = 3(-0.6-(-5)) = 3(4.4) = 13.2 \) Ns, in the positive direction (A's original direction). By Newton's third law, the impulse exerted by A on B has the same magnitude as the impulse exerted by B on A (in the opposite sense); the magnitude of the impulse exerted by A on B is 13.2 Ns.
(c) The principle used is conservation of linear momentum: since the surface is smooth (no friction) and the spheres move horizontally, there is no external horizontal force acting on the two-sphere system during the collision, so the total momentum of the system is unchanged by the collision.

(a) M1 for total momentum before = mAuA+mBuB; A1 for correct value -3 (kg ms⁻¹); M1 for equating to (mA+mB)v; A1 for v=-0.6; A1 for correct direction stated (0.6 ms⁻¹ in B's original direction).
(b) M1 for impulse on B = mB(v-uB); A1 for v-uB=4.4 (ft from (a)); A1 for impulse=13.2 (ft); A1 for correct units Ns; A1 for correct direction/sign interpretation.
(c) B1 for naming conservation of linear momentum; B1 for correct justification referencing no external horizontal force / smooth surface.

(a) Let \( p \) be the true proportion of defective components after recalibration. \( H_0: p=0.08 \); \( H_1: p<0.08 \) (a one-tailed test, since we are testing for a reduction).
(b) \( X\sim B(60,0.08) \). \( P(X\leq2) = P(X=0)+P(X=1)+P(X=2) \)
\( = (0.92)^{60} + 60(0.08)(0.92)^{59} + \binom{60}{2}(0.08)^2(0.92)^{58} \)
\( \approx 0.1317 \) (4 d.p.).
(c) Since \( P(X\leq2)=0.1317 > 0.05 \), the result is not significant at the 5% level. There is insufficient evidence to reject \( H_0 \). We conclude there is not enough evidence, at the 5% significance level, that the recalibration has reduced the proportion of defective components.

(a) B1 for both hypotheses correctly stated in terms of p; B1 for correct definition of p and correct direction of H1 (one-tailed, p<0.08).
(b) M1 for identifying binomial model B(60,0.08); M1 for summing P(X=0)+P(X=1)+P(X=2) (or equivalent cumulative approach); A1 for correct method/set-up; A1 for P(X≤2)=0.1317 (accept 0.132 or equivalent rounding).
(c) M1 for comparing p-value to 0.05; A1 for correct decision (do not reject H0 since 0.1317>0.05); A1 for conclusion expressed in context; A1 for correct interpretation that there is insufficient evidence of a reduction (not proof that p=0.08).

(a) Let \( \mu \) be the true mean mass of bags filled by the machine after maintenance. \( H_0: \mu=500 \); \( H_1: \mu \neq 500 \) (two-tailed, since we are testing whether the mean has changed, in either direction).
(b) The sample mean \( \bar{x}=497.8 \), with \( \sigma=5 \), \( n=25 \). Test statistic: \( z = \dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}} = \dfrac{497.8-500}{5/\sqrt{25}} = \dfrac{-2.2}{1} = -2.2 \).
(c) For a two-tailed test at the 5% level, the critical values are \( z=\pm1.96 \). Since \( |{-2.2}| = 2.2 > 1.96 \), the test statistic lies in the critical region, so we reject \( H_0 \). There is evidence, at the 5% significance level, that the mean mass of bags filled by the machine has changed (decreased) after maintenance.

(a) B1 for both hypotheses correctly stated in terms of μ; B1 for correctly identifying a two-tailed test.
(b) M1 for correct standard error σ/√n = 1; M1 for correct z-formula; A1 for correct substitution; A1 for z=-2.2.
(c) B1 for correct critical values ±1.96; M1 for valid comparison |z| vs 1.96; A1 for correct decision to reject H0; A1 for conclusion expressed correctly in context.

(a) Let \( \rho \) be the population product-moment correlation coefficient between revision hours and exam score. \( H_0: \rho=0 \); \( H_1: \rho>0 \) (one-tailed, testing specifically for positive correlation).
(b) Since \( r=0.52 \) exceeds the critical value of \( 0.4409 \), the result is significant at the 5% level and we reject \( H_0 \). There is evidence, at the 5% significance level, of positive correlation between revision hours and exam score in the population.
(c) The critical value for the correlation coefficient decreases as the sample size increases, because a larger sample provides more information and reduces the variability of the sample correlation coefficient under \( H_0 \) (the sampling distribution of r becomes narrower). This means that, with a large sample such as \( n=100 \), even a relatively small sample correlation of \( r=0.21 \) can lie beyond the (much smaller) critical value and so be judged statistically significant, whereas the same value of \( r \) might not be significant for a small sample.
(d) Correlation does not imply causation: even a statistically significant correlation could arise because both variables are influenced by a third (confounding) variable, such as a student's general ability or motivation, rather than revision time directly causing improved exam scores.

(a) B1 for both hypotheses correctly stated in terms of ρ; B1 for correctly identifying a one-tailed test in the positive direction.
(b) M1 for comparing r=0.52 with critical value 0.4409; A1 for correct decision to reject H0; A1 for conclusion expressed correctly in context.
(c) B1 for stating that critical value decreases as sample size increases; B1 for correct reasoning about reduced sampling variability / increased test power with larger n; B1 for linking this correctly to why r=0.21 can be significant despite being smaller than r=0.52.
(d) B1 for stating correlation does not imply causation; B1 for a specific, plausible confounding variable or alternative explanation given in context.

Let \( X \sim N(175, 7^2) \).
(a) \( P(X>185) = P\left(Z > \dfrac{185-175}{7}\right) = P(Z>1.4286) = 1-\Phi(1.4286) \approx 0.0766 \) (3 s.f.).
(b) \( P(168h)=0.05 \), so \( P(Z>z) = 0.05 \Rightarrow z=1.6449 \) (from tables/inverse normal). Then \( h = 175 + 7(1.6449) = 175+11.51 = 186.5 \), so \( h\approx187 \) cm (nearest cm).

(a) M1 for standardising with correct z-value 1.4286; A1 for correct z (accept 1.43); A1 for P=0.0766 (accept 0.0765-0.0767).
(b) M1 for standardising both bounds; A1 for both z-values correct (-1 and 0.7143); M1 for Φ(0.7143)-Φ(-1) or equivalent; A1 for P=0.604 (accept 0.603-0.605).
(c) M1 for identifying P(Z>z)=0.05 and z=1.6449; M1 for h=175+7z; A1 for h=187 (accept 186 or 186.5 depending on rounding convention, ft from z value used).

There are 12 counters in total: 5 red (R), 7 blue (B).
(a) \( P(R_1 \cap R_2) = \dfrac{5}{12}\times\dfrac{4}{11} = \dfrac{20}{132} = \dfrac{5}{33} \).
(b) Using the law of total probability, conditioning on the first draw: \( P(R_2) = P(R_1)P(R_2|R_1) + P(B_1)P(R_2|B_1) = \dfrac{5}{12}\times\dfrac{4}{11} + \dfrac{7}{12}\times\dfrac{5}{11} = \dfrac{20}{132}+\dfrac{35}{132} = \dfrac{55}{132} = \dfrac{5}{12} \).
(c) By the definition of conditional probability: \( P(R_1|R_2) = \dfrac{P(R_1\cap R_2)}{P(R_2)} = \dfrac{5/33}{5/12} = \dfrac{5}{33}\times\dfrac{12}{5} = \dfrac{12}{33} = \dfrac{4}{11} \).

(a) M1 for P(R1)=5/12 and P(R2|R1)=4/11; A1 for correct product; A1 for 5/33 (accept equivalent forms/decimals).
(b) M1 for identifying need to condition on both cases (R1 and B1); M1 for correct computation of both branch probabilities (20/132 and 35/132); A1 for correct sum 55/132; A1 for simplified 5/12.
(c) M1 for correct conditional probability formula P(R1|R2)=P(R1∩R2)/P(R2); A1 for correct substitution (ft from (a),(b)); A1 for 4/11.