2025 CCEA A-Level Mathematics 2210 模擬試題連答案詳解

(a) \( S_n = \frac{n}{2}\left(2a+(n-1)d\right) = \frac{n}{2}\left(14+4(n-1)\right) = \frac{n}{2}(4n+10) = 2n^2+5n \). Setting \( S_n = 525 \) gives \( 2n^2+5n-525=0 \), as required.
(b) Using the quadratic formula, \( n = \dfrac{-5 \pm \sqrt{25+4(2)(525)}}{4} = \dfrac{-5\pm\sqrt{4225}}{4} = \dfrac{-5\pm65}{4} \). Since \( n \) must be a positive integer, \( n = \dfrac{60}{4} = 15 \).
Answer: n = 15.

(a) M1 for correct \( S_n \) formula substituted; M1 for expanding; W1 for reaching the printed result. (b) M1 for correct use of the quadratic formula (or equivalent); W1 for \( \sqrt{4225}=65 \); W1 for rejecting the negative root and stating \( n=15 \).

(a) \( x_2 = 0.5(2)+3 = 4 \); \( x_3 = 0.5(4)+3 = 5 \); \( x_4 = 0.5(5)+3 = 5.5 \).
(b) As \( n\to\infty \), \( x_{n+1}\to L \) and \( x_n \to L \), so \( L = 0.5L+3 \). Solving: \( 0.5L = 3 \), so \( L = 6 \).
Answer: L = 6.

(a) M1 for correct substitution; A1 for \( x_2,x_3 \) correct; A1 for \( x_4=5.5 \) correct. (b) M1 for setting up \( L=0.5L+3 \); A1 for correct rearrangement; A1 for \( L=6 \).

(a) \( S_\infty = \dfrac{a}{1-r} \), so \( 32 = \dfrac{8}{1-r} \), giving \( 1-r = \dfrac{8}{32} = \dfrac{1}{4} \), so \( r = \dfrac{3}{4} \), as required.
(b) 5th term \( = ar^4 = 8\left(\dfrac{3}{4}\right)^4 = 8 \times \dfrac{81}{256} = \dfrac{81}{32} \).
(c) \( S_{10} = \dfrac{a(1-r^{10})}{1-r} = \dfrac{8\left(1-(0.75)^{10}\right)}{0.25} = 32\left(1-0.0563135\ldots\right) = 30.1979\ldots \approx 30.2 \).
Answer: r = 3/4, 5th term = 81/32, S_10 ≈ 30.2 (3 s.f.).

(a) M1 for correct \( S_\infty \) formula; M1 for rearranging; W1 for reaching \( r=3/4 \). (b) M1 for \( ar^4 \) with correct values; A1 for \( 81/32 \) (accept 2.53). (c) M1 for correct \( S_n \) formula with \( n=10 \); A1 for 30.2 (accept 30.1980 unrounded).

(a) Arc length \( = r\theta = 8 \times 1.2 = 9.6 \) cm.
(b) Area \( = \frac{1}{2}r^2\theta = \frac{1}{2}(8)^2(1.2) = 38.4 \) cm\(^2\).
(c) Perimeter \( = 2r + \text{arc} = 16+9.6 = 25.6 \) cm.
(d) Area of segment \( = \text{area of sector} - \text{area of triangle } OAB = \frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta = \frac{1}{2}(8)^2(1.2-\sin1.2) = 32(1.2-0.93204) = 32(0.26796) = 8.5747\ldots \approx 8.57 \) cm\(^2\).
Answer: arc = 9.6 cm; area = 38.4 cm^2; perimeter = 25.6 cm; segment area ≈ 8.57 cm^2.

(a) M1 for \( r\theta \); A1 correct. (b) M1 for \( \frac12 r^2\theta \); A1 correct. (c) M1 for \( 2r+\text{arc} \); A1 correct. (d) M1 for triangle area \( \frac12 r^2\sin\theta \); M1 for subtracting from sector area; A1 for 8.57 (accept 8.5748).

(a) \( R\sin(\theta+\alpha) = R\sin\theta\cos\alpha+R\cos\theta\sin\alpha \). Comparing with \( 12\sin\theta+5\cos\theta \): \( R\cos\alpha=12 \), \( R\sin\alpha=5 \). Then \( R = \sqrt{5^2+12^2}=\sqrt{169}=13 \), and \( \tan\alpha = \dfrac{5}{12} \), so \( \alpha = 22.62^\circ \) (2 d.p.). So \( 5\cos\theta+12\sin\theta \equiv 13\sin(\theta+22.62^\circ) \).
(b) \( 13\sin(\theta+22.62^\circ) = 6.5 \Rightarrow \sin(\theta+22.62^\circ) = 0.5 \). Let \( \phi=\theta+22.62^\circ \), with \( \phi \in [22.62^\circ, 382.62^\circ] \). \( \sin\phi=0.5 \) gives \( \phi = 30^\circ \) or \( 150^\circ \) (the next solution \( 390^\circ \) lies outside the range). So \( \theta = 30^\circ-22.62^\circ = 7.38^\circ \) or \( \theta = 150^\circ-22.62^\circ = 127.38^\circ \).
Answer: theta = 7.38° or 127.38° (to 2 d.p.).

(a) M1 for correct expansion and comparison; A1 for \( R=13 \); M1 for \( \tan\alpha=5/12 \); A1 for \( \alpha=22.62^\circ \). (b) M1 for dividing by 13 and taking \(\arcsin\); M1 for finding both values of \( \phi \) in range; A1 each for the two correct values of \( \theta \) (2 marks); B1 for correctly rejecting/excluding any out-of-range solution.

(a) \( \cos(A+B) \equiv \cos A\cos B - \sin A \sin B \). Setting \( A=B=\theta \): \( \cos2\theta \equiv \cos^2\theta-\sin^2\theta \). Using \( \cos^2\theta \equiv 1-\sin^2\theta \): \( \cos2\theta \equiv (1-\sin^2\theta)-\sin^2\theta = 1-2\sin^2\theta \), as required.
(b) Substituting: \( (1-2\sin^2\theta)+3\sin\theta = 2 \Rightarrow -2\sin^2\theta+3\sin\theta-1=0 \Rightarrow 2\sin^2\theta-3\sin\theta+1=0 \). Factorising: \( (2\sin\theta-1)(\sin\theta-1)=0 \), so \( \sin\theta=\dfrac{1}{2} \) or \( \sin\theta=1 \). For \( \sin\theta=\dfrac12 \): \( \theta=30^\circ \) or \( 150^\circ \). For \( \sin\theta=1 \): \( \theta=90^\circ \).
Answer: theta = 30°, 90°, 150°.

(a) M1 for stating \( \cos(A+B) \) formula with \( A=B=\theta \); M1 for using \( \cos^2\theta=1-\sin^2\theta \); W2 for fully correct derivation to the printed result. (b) M1 for substitution of the identity; M1 for forming the quadratic in \( \sin\theta \); M1 for factorising/solving; A1 for \( \theta=30^\circ,150^\circ \); A1 for \( \theta=90^\circ \).

(a) By the product rule, \( \dfrac{dy}{dx} = 2x\sin3x + x^2(3\cos3x) = 2x\sin3x+3x^2\cos3x \).
(b) By the quotient rule, \( \dfrac{dy}{dx} = \dfrac{2(x^2+3)-(2x+1)(2x)}{(x^2+3)^2} = \dfrac{2x^2+6-4x^2-2x}{(x^2+3)^2} = \dfrac{-2x^2-2x+6}{(x^2+3)^2} = \dfrac{-2(x^2+x-3)}{(x^2+3)^2} \).
(c) Differentiating implicitly with respect to \( x \): \( 2x+\left(y+x\dfrac{dy}{dx}\right)+2y\dfrac{dy}{dx}=0 \). Collecting terms: \( \dfrac{dy}{dx}(x+2y) = -(2x+y) \), so \( \dfrac{dy}{dx} = \dfrac{-(2x+y)}{x+2y} \).
(d) At \( (1,2) \): \( \dfrac{dy}{dx} = \dfrac{-(2(1)+2)}{1+2(2)} = \dfrac{-4}{5} \). Tangent: \( y-2 = -\dfrac{4}{5}(x-1) \Rightarrow 5y-10=-4x+4 \Rightarrow 4x+5y=14 \).
Answer: (a) \(2x\sin3x+3x^2\cos3x\); (b) \(\dfrac{-2(x^2+x-3)}{(x^2+3)^2}\); (c) \(\dfrac{-(2x+y)}{x+2y}\); (d) gradient \(-4/5\), tangent \(4x+5y=14\).

(a) M1 for product rule structure; A1 for each correct term (2 marks). (b) M1 for quotient rule structure; M1 for correct numerator expansion; A1 for simplification; A1 for final form. (c) M1 for differentiating each term implicitly; M1 for correct product-rule term on \( xy \); M1 for collecting \( dy/dx \) terms; A1 for correct final expression. (d) M1 for substituting \( (1,2) \); A1 for gradient \(-4/5\); M1 for point-gradient line equation; M1 for rearranging; A1 for correct integer-coefficient form; W1 for confirming \((1,2)\) lies on curve (\(1+2+4=7\)).

(a) Let \( u=x, dv=e^{2x}dx \), so \( du=dx, v=\tfrac12e^{2x}\). \( \int xe^{2x}dx = \tfrac12xe^{2x}-\int\tfrac12e^{2x}dx = \tfrac12xe^{2x}-\tfrac14e^{2x}+C = \dfrac{(2x-1)e^{2x}}{4}+C \).
(b) With \( u=4-x^2 \), \( du=-2x\,dx \), so \( x\,dx=-\tfrac12du \). \( \int x\sqrt{4-x^2}\,dx = \int\sqrt{u}\left(-\tfrac12du\right) = -\tfrac12\cdot\tfrac23u^{3/2}+C = -\tfrac13(4-x^2)^{3/2}+C \).
(c) Area \( = \left[-\tfrac13(4-x^2)^{3/2}\right]_0^2 = \left(-\tfrac13(0)^{3/2}\right)-\left(-\tfrac13(4)^{3/2}\right) = 0+\tfrac13(8) = \dfrac83 \).
(d) \( V = \pi\displaystyle\int_0^2 y^2\,dx = \pi\int_0^2 x^2(4-x^2)\,dx = \pi\int_0^2(4x^2-x^4)\,dx = \pi\left[\dfrac{4x^3}{3}-\dfrac{x^5}{5}\right]_0^2 = \pi\left(\dfrac{32}{3}-\dfrac{32}{5}\right) = \pi\cdot32\left(\dfrac{5-3}{15}\right) = \dfrac{64\pi}{15} \).
Answer: (a) \(\frac{(2x-1)e^{2x}}{4}+C\); (b) \(-\frac13(4-x^2)^{3/2}+C\); (c) 8/3; (d) 64π/15.

(a) M1 for correct choice of \(u,dv\); M1 for correct \(v\); M1 for by-parts formula applied; A1 for correct integral; A1 for simplified form with \(+C\). (b) M1 for correct substitution of \(du\); M1 for expressing integral in \(u\); A1 for integrating; A1 for back-substitution; A1 for simplified form. (c) M1 for substituting limits; A1 for correct evaluation; A1 for 8/3. (d) M1 for correct volume formula \(\pi\int y^2dx\); M1 for expanding \(x^2(4-x^2)\); M1 for integrating; A1 for \(64\pi/15\).

(a) Separating variables: \( \displaystyle\int\dfrac{1}{\theta-18}\,d\theta = \displaystyle\int -k\,dt \Rightarrow \ln|\theta-18| = -kt+c \). When \( t=0,\theta=90 \): \( \ln72=c \). So \( \ln(\theta-18) = -kt+\ln72 \Rightarrow \theta-18 = 72e^{-kt} \Rightarrow \theta = 18+72e^{-kt} \), as required.
(b) At \( t=10,\theta=54 \): \( 54-18=36=72e^{-10k} \Rightarrow e^{-10k}=0.5 \Rightarrow -10k=\ln0.5 \Rightarrow k=\dfrac{\ln2}{10}=0.0693 \) (3 s.f.).
(c) At \( \theta=30 \): \( 30-18=12=72e^{-kt} \Rightarrow e^{-kt}=\dfrac16 \Rightarrow t=\dfrac{\ln6}{k}=\dfrac{1.7918}{0.069315}=25.8\ldots \approx 26 \) minutes.
(d) As \( t\to\infty \), the model predicts \( \theta\to18 \)°C but never actually reaches it, whereas in reality the liquid's temperature will settle at (or effectively equal) room temperature after a finite time; the model also assumes the room temperature stays exactly constant, which may not hold in practice.
Answer: (a) shown; (b) k ≈ 0.0693; (c) 26 minutes; (d) e.g. theta never reaches 18°C exactly under this model.

(a) M1 for separating variables; M1 for correct integration of both sides; M1 for including constant of integration; M1 for using initial condition; M1 for rearranging exponential; W1 for reaching printed result. (b) M1 for substituting given values; M1 for isolating exponential; M1 for taking logs; A1 for k=0.0693. (c) M1 for substituting theta=30; M1 for isolating exponential; M1 for solving for t; A1 for 26 minutes (accept 25 or 26 with correct working using unrounded k). (d) B1 for each valid, well-explained limitation, up to 3 marks (e.g. asymptotic behaviour never truly reaches 18°C; constant room temperature assumption; model may not hold for very large t due to other heat-loss effects).

(a) \( \dfrac{dx}{dt}=2t \), \( \dfrac{dy}{dt}=6t^2-3 \). So \( \dfrac{dy}{dx} = \dfrac{6t^2-3}{2t} = 3t-\dfrac{3}{2t} \).
(b) Horizontal tangent requires \( \dfrac{dy}{dt}=0 \) (with \( \dfrac{dx}{dt}\ne0 \)): \( 6t^2-3=0 \Rightarrow t^2=\dfrac12 \Rightarrow t=\pm\dfrac{1}{\sqrt2} \). At \( t=\dfrac{1}{\sqrt2} \): \( x=\dfrac12-1=-\dfrac12 \); \( y=2\left(\dfrac{1}{\sqrt2}\right)^3-3\left(\dfrac{1}{\sqrt2}\right) = \dfrac{1}{\sqrt2}-\dfrac{3}{\sqrt2} = -\dfrac{2}{\sqrt2}=-\sqrt2 \). At \( t=-\dfrac{1}{\sqrt2} \): \( x=-\dfrac12 \); \( y=\sqrt2 \) (by symmetry, odd function of \( t \)). So the points are \( \left(-\dfrac12,-\sqrt2\right) \) and \( \left(-\dfrac12,\sqrt2\right) \).
(c) \( \dfrac{d^2y}{dx^2} = \dfrac{1}{dx/dt}\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right) \). From (a), \( \dfrac{dy}{dx}=3t-\dfrac{3}{2}t^{-1} \), so \( \dfrac{d}{dt}\left(\dfrac{dy}{dx}\right) = 3+\dfrac{3}{2}t^{-2} = 3+\dfrac{3}{2t^2} \). Dividing by \( \dfrac{dx}{dt}=2t \): \( \dfrac{d^2y}{dx^2} = \dfrac{3+\frac{3}{2t^2}}{2t} = \dfrac{3}{2t}+\dfrac{3}{4t^3} = \dfrac{6t^2+3}{4t^3} = \dfrac{3(2t^2+1)}{4t^3} \), as required.
(d) At \( t=\dfrac{1}{\sqrt2} \) (so \( t>0 \), \( t^3>0 \)): \( \dfrac{d^2y}{dx^2} = \dfrac{3(2(1/2)+1)}{4t^3} = \dfrac{6}{4t^3}>0 \), so \( \left(-\dfrac12,-\sqrt2\right) \) is a minimum. At \( t=-\dfrac{1}{\sqrt2} \) (\( t<0 \), \( t^3<0 \)): \( \dfrac{d^2y}{dx^2}<0 \), so \( \left(-\dfrac12,\sqrt2\right) \) is a maximum.
Answer: (a) 3t - 3/(2t); (b) (-1/2,-√2) and (-1/2,√2); (c) shown; (d) (-1/2,-√2) minimum, (-1/2,√2) maximum.

(a) M1 for both \(dx/dt, dy/dt\); M1 for forming ratio; A1 for correct unsimplified form; A1 for simplified form. (b) M1 for setting \(dy/dt=0\); A1 for \(t=\pm1/\sqrt2\); M1 for substituting to find x; A1,A1 for the two correct (x,y) pairs (exact form). (c) M1 for differentiating dy/dx w.r.t. t; A1 for correct derivative; M1 for dividing by dx/dt; A1 for unsimplified form; W1 for correctly simplifying to printed result. (d) M1 for evaluating sign of d2y/dx2 at each t; A1 for correctly identifying the minimum; A1 for correctly identifying the maximum.

(a) Let \( \dfrac{7x-4}{(x-1)(x+2)} \equiv \dfrac{A}{x-1}+\dfrac{B}{x+2} \). Then \( 7x-4 = A(x+2)+B(x-1) \). At \( x=1 \): \( 3=3A\Rightarrow A=1 \). At \( x=-2 \): \( -18=-3B\Rightarrow B=6 \). So \( \dfrac{7x-4}{(x-1)(x+2)} \equiv \dfrac{1}{x-1}+\dfrac{6}{x+2} \).
(b) \( \dfrac{1}{x-1} = -\dfrac{1}{1-x} = -(1+x+x^2+\cdots) \), valid for \( |x|<1 \). \( \dfrac{6}{x+2} = \dfrac{3}{1+\frac{x}{2}} = 3\left(1-\dfrac{x}{2}+\dfrac{x^2}{4}-\cdots\right) \), valid for \( |x|<2 \). Adding: \( -(1+x+x^2) + 3\left(1-\dfrac{x}{2}+\dfrac{x^2}{4}\right) = (-1+3)+\left(-1-\dfrac32\right)x+\left(-1+\dfrac34\right)x^2 = 2-\dfrac52x-\dfrac14x^2 \). The combined expansion is valid for the more restrictive of the two intervals, i.e. \( |x|<1 \).
(c) \( |3x-2|<7 \Leftrightarrow -7<3x-2<7 \Leftrightarrow -5<3x<9 \Leftrightarrow -\dfrac53\dfrac23 \), \( y=3x-2 \), gradient \( +3 \).
Answer: (a) 1/(x-1)+6/(x+2); (b) 2 - (5/2)x - (1/4)x^2, valid |x|<1; (c) -5/3 < x < 3; (d) vertex (2/3,0), y-intercept (0,2), gradients -3 and +3.

(a) M1 for correct form; M1 for forming identity; A1,A1 for A=1, B=6 correctly found; A1 for correctly stated partial fractions. (b) M1 for rewriting each fraction in binomial-expandable form; M1 for expanding \(1/(1-x)\); A1 for correct terms; M1 for expanding \(3(1+x/2)^{-1}\); A1 for correct terms; A1 for correctly combined series; B1 for correct validity |x|<1 with justification (more restrictive interval). (c) M1 for forming compound inequality; A1 for correct bounds; A1 for fully correct inequality. (d) B1 for vertex; B1 for y-intercept; B1 for both gradients correct.

(a) \( f(2) = 8-8-2=-2<0 \); \( f(3)=27-12-2=13>0 \). Since \( f \) is continuous and changes sign on \( [2,3] \), there is a root \( \alpha \) in this interval.
(b) \( f'(x)=3x^2-4 \). \( f(2.2)=10.648-8.8-2=-0.152 \); \( f'(2.2)=3(4.84)-4=10.52 \). \( x_1 = x_0-\dfrac{f(x_0)}{f'(x_0)} = 2.2-\dfrac{-0.152}{10.52} = 2.2+0.014449\ldots = 2.2144 \) (4 d.p.).
(c) \( f(2.2144)=-0.001349\ldots \); \( f'(2.2144)=3(2.2144)^2-4=10.7147\ldots \). \( x_2 = 2.2144-\dfrac{-0.001349}{10.7147} = 2.2144+0.000126 = 2.2143 \) (4 d.p.; note the iteration is converging rapidly to \( \alpha\approx2.21432 \)).
(d) With 4 strips, \( h=\dfrac{3-2}{4}=0.25 \); \( x \)-values: 2, 2.25, 2.5, 2.75, 3. \( f(2)=-2 \), \( f(2.25)=0.390625 \), \( f(2.5)=3.625 \), \( f(2.75)=7.796875 \), \( f(3)=13 \). Trapezium rule: \( \displaystyle\int_2^3 f(x)\,dx \approx \dfrac{h}{2}\Big[(f(2)+f(3))+2\big(f(2.25)+f(2.5)+f(2.75)\big)\Big] = \dfrac{0.25}{2}\big[11+2(11.8125)\big] = 0.125(34.625) = 4.328125 \approx 4.33 \) (3 s.f.).
(e) The estimate could be made more accurate by using a larger number of (narrower) strips, since the trapezium rule's error decreases as the strip width decreases, giving a closer approximation to the curve (and hence to the true area).
Answer: (a) shown; (b) x_1 = 2.2144; (c) x_2 = 2.2143; (d) ≈ 4.33; (e) use more strips.

(a) M1 for evaluating both f(2) and f(3); A1 for correct conclusion referencing sign change/continuity. (b) M1 for correct f'(x); M1 for evaluating f(2.2) and f'(2.2); M1 for correct N-R formula; A1 for correct unrounded value; A1 for 2.2144. (c) M1 for evaluating f(x_1) and f'(x_1); M1 for correct N-R formula applied; A1 for correct unrounded value; A1 for 2.2143. (d) M1 for correct h; M1 for correct y-values; M1 for correct trapezium rule structure; A1 for 4.328125 unrounded; A1 for 4.33. (e) B1 for stating increase number of strips; B1 for valid justification.

(a) \( \mathbf{v} = \mathbf{u}+\mathbf{a}t = (3\mathbf{i}-2\mathbf{j})+4(0.5\mathbf{i}+\mathbf{j}) = (3+2)\mathbf{i}+(-2+4)\mathbf{j} = 5\mathbf{i}+2\mathbf{j} \).
(b) \( \mathbf{r} = \mathbf{u}t+\tfrac12\mathbf{a}t^2 = 4(3\mathbf{i}-2\mathbf{j})+\tfrac12(16)(0.5\mathbf{i}+\mathbf{j}) = (12\mathbf{i}-8\mathbf{j})+(4\mathbf{i}+8\mathbf{j}) = 16\mathbf{i} \).
(c) Speed \( = |\mathbf{v}| = \sqrt{5^2+2^2} = \sqrt{29} = 5.385\ldots \approx 5.39 \) m s\(^{-1}\).
Answer: (a) 5i+2j m/s; (b) 16i m (i.e. 16 m from O along the i-direction); (c) 5.39 m/s.

(a) M1 for v=u+at with correct substitution; A1 for correct vector. (b) M1 for correct SUVAT vector formula; A1 for correct expansion; A1 for correct final vector. (c) M1 for Pythagoras on the velocity components; A1 for √29 unrounded; A1 for 5.39.

(a) \( u_A = +4 \) m s\(^{-1}\); \( u_B = -2 \) m s\(^{-1}\) (opposite direction).
(b) Conservation of momentum: \( m_Au_A+m_Bu_B = m_Av_A+m_Bv_B \), with \( v_A=-1 \) (rebounds). \( 2(4)+3(-2) = 2(-1)+3v_B \Rightarrow 8-6=-2+3v_B \Rightarrow 2=-2+3v_B \Rightarrow v_B = \dfrac43 \) m s\(^{-1}\) (in \( A \)'s original direction of motion).
(c) Impulse on \( A \) \( = m_A(v_A-u_A) = 2(-1-4) = -10 \) N s. The magnitude of the impulse is 10 N s.
Answer: (a) u_A=+4, u_B=-2; (b) v_B = 4/3 m/s ≈ 1.33 m/s; (c) 10 N s.

(a) B1 for both correct with signs. (b) M1 for conservation of momentum equation with correct masses; M1 for correct substitution including v_A=-1; M1 for rearranging; A1 for v_B=4/3 (accept 1.33). (c) M1 for impulse = m(v-u) formula; A1 for correct value including sign; A1 for correct magnitude 10 N s.

(a) \( \mathbf{F} = \mathbf{F_1}+\mathbf{F_2}+\mathbf{F_3} = (6-2+1)\mathbf{i}+(-2+5-1)\mathbf{j} = 5\mathbf{i}+2\mathbf{j} \) N.
(b) By Newton's second law, \( \mathbf{a} = \mathbf{F}/m = (5\mathbf{i}+2\mathbf{j})/4 = 1.25\mathbf{i}+0.5\mathbf{j} \) m s\(^{-2}\).
(c) Since the particle starts at rest at the origin, \( \mathbf{r} = \tfrac12\mathbf{a}t^2 = \tfrac12(1.25\mathbf{i}+0.5\mathbf{j})(3^2) = 4.5(1.25\mathbf{i}+0.5\mathbf{j}) = 5.625\mathbf{i}+2.25\mathbf{j} \) m.
(d) Distance \( = |\mathbf{r}| = \sqrt{5.625^2+2.25^2} = \sqrt{31.640625+5.0625} = \sqrt{36.703125} = 6.0583\ldots \approx 6.06 \) m.
Answer: (a) 5i+2j N; (b) 1.25i+0.5j m/s^2; (c) 5.625i+2.25j m; (d) 6.06 m.

(a) M1 for summing components; A1 for correct resultant. (b) M1 for F=ma rearranged; A1 for correct acceleration. (c) M1 for r=1/2at^2 (u=0); A1 for correct substitution; A1 for correct vector. (d) M1 for Pythagoras; A1 for 6.06.

(a) \( v = \dfrac{dx}{dt} = 3t^2-12t+9 \); \( a = \dfrac{dv}{dt} = 6t-12 \).
(b) Setting \( v=0 \): \( 3t^2-12t+9=0 \Rightarrow t^2-4t+3=0 \Rightarrow (t-1)(t-3)=0 \Rightarrow t=1 \) or \( t=3 \).
(c) At \( t=1 \): \( a=6(1)-12=-6 \) m s\(^{-2}\). At \( t=3 \): \( a=6(3)-12=6 \) m s\(^{-2}\). Checking the sign of \( v \) either side of each time: for \( t \) just below 1, \( v=3(0.5)^2-12(0.5)+9=3.75>0 \); for \( t \) just above 1, \( v=3(1.5)^2-12(1.5)+9=-2.25<0 \). So \( v \) changes from positive to negative at \( t=1 \), which is consistent with the negative acceleration (deceleration through zero) found there. (At \( t=3 \), \( v \) changes from negative to positive, consistent with the positive acceleration.)
Answer: (a) v=3t^2-12t+9, a=6t-12; (b) t=1s and t=3s; (c) a(1)=-6 m/s^2, a(3)=6 m/s^2; velocity changes from positive to negative at t=1s.

(a) M1 for differentiating x; A1 for v; A1 for a. (b) M1 for setting v=0 and solving quadratic; A1 for t=1; A1 for t=3. (c) M1 for evaluating a at both times; A1 for both accelerations correct; M1 for checking sign of v either side of at least one time; A1 for correctly identifying t=1 with valid justification.

(a) The forces are: the weight of the ladder, \( 20g=196 \) N, acting vertically downward at its midpoint (3 m from \( A \) along the ladder); the weight of the person, \( 70g=686 \) N, acting vertically downward at 4 m from \( A \) along the ladder; the normal reaction of the wall, \( S \), acting horizontally at \( B \) (wall is smooth, so no friction there); the normal reaction of the ground, \( R \), acting vertically at \( A \); and the friction force, \( F \), acting horizontally at \( A \) (opposing the tendency of \( A \) to slip).
(b) Taking moments about \( A \) (clockwise positive), the moment of \( S \) (acting horizontally at height \( 6\sin65^\circ \) above \( A \)) balances the moments of the two weights (acting at horizontal distances \( 3\cos65^\circ \) and \( 4\cos65^\circ \) from \( A \)):
\( S(6\sin65^\circ) = 196(3\cos65^\circ)+686(4\cos65^\circ) \)
\( S = \dfrac{\cos65^\circ}{\sin65^\circ}\cdot\dfrac{3(196)+4(686)}{6} = \cot65^\circ\times\dfrac{588+2744}{6} = 0.46631\times555.33\ldots = 258.96\ldots \approx 259.0 \) N.
(c) Resolving vertically: \( R = 196+686 = 882 \) N (the wall reaction is horizontal, so it has no vertical component).
(d) On the point of slipping, \( F=\mu R \). Resolving horizontally: \( F=S \) (the friction balances the wall's push). So \( \mu = \dfrac{S}{R} = \dfrac{258.96}{882} = 0.2936\ldots \approx 0.294 \) (3 s.f.).
Answer: (a) as listed; (b) S ≈ 259.0 N; (c) R = 882 N; (d) mu ≈ 0.294.

(a) B1 for both weights correctly located; B1 for wall reaction correctly described; B1 for both ground reaction and friction correctly described (all 5 forces needed for full marks). (b) M1 for correct moment equation structure about A; M1 for correct moment arm for S (height); M1 for correct moment arms for both weights; A1 for correct equation; M1 for rearranging for S; A1 for 259.0 N (accept 259). (c) M1 for vertical resolution; A1 for R=882 N. (d) M1 for F=muR and horizontal resolution F=S; M1 for correct rearrangement mu=S/R; A1 for unrounded value; A1 for 0.294.

(a) \( Z=\dfrac{185-175}{7}=\dfrac{10}{7}=1.4286 \). \( P(X>185)=P(Z>1.4286)=1-\Phi(1.4286)=0.0766 \) (4 d.p.).
(b) \( Z_1=\dfrac{168-175}{7}=-1 \); \( Z_2=\dfrac{180-175}{7}=\dfrac57=0.7143 \). \( P(168h)=0.10 \Rightarrow P(Z>z)=0.10 \Rightarrow z=1.2816 \) (from tables). \( h = 175+7(1.2816) = 175+8.9708 = 183.97 \approx 184.0 \) cm (1 d.p.).
Answer: (a) 0.0766; (b) 0.6038; (c) 184.0 cm.

(a) M1 for standardising; M1 for correct tail; A1 for 0.0766. (b) M1 for both z-values; M1 for correct probability structure Phi(z2)-Phi(z1); A1 for correct component probabilities; A1 for 0.6038. (c) M1 for identifying z=1.2816 (or equivalent, from P(Z>z)=0.10); M1 for h=mean+z*sd structure; A1 for 184.0.

\( H_0: \mu=500 \); \( H_1: \mu\ne500 \) (two-tailed test). Under \( H_0 \), \( \bar{X}\sim N\left(500,\dfrac{5^2}{25}\right) \). Test statistic: \( z = \dfrac{\bar{x}-\mu_0}{\sigma/\sqrt{n}} = \dfrac{502.8-500}{5/\sqrt{25}} = \dfrac{2.8}{1} = 2.80 \). The critical value for a two-tailed test at the 5% level is \( z=\pm1.96 \). Since \( 2.80>1.96 \), the test statistic lies in the critical region, so we reject \( H_0 \). There is significant evidence, at the 5% level, that the mean weight of bags filled by the machine has changed (increased) since the service.
Answer: z = 2.80 > 1.96, so reject H0; evidence the mean weight has changed.

M1/A1 for correct hypotheses in terms of mu (2 marks); M1 for correct standard error 5/sqrt(25); M1 for correct test statistic formula; A1 for z=2.80; B1 for correct critical value ±1.96; A1 for correct conclusion in context linking comparison to decision.

(a) \( H_0:\rho=0 \) (no linear correlation in the population); \( H_1:\rho\ne0 \) (linear correlation exists), two-tailed.
(b) \( |r|=0.62 \), which is greater than the critical value 0.576. Since \( |r| \) exceeds the critical value, the result is significant at the 5% level, so we reject \( H_0 \). There is evidence, at the 5% level, of a linear correlation between revision hours and exam score.
(c) No. Correlation does not imply causation: a significant PMCC only shows a linear association between the two variables, not that one causes the other. There could be a confounding variable (for example, a student's general ability or motivation) that affects both the number of hours spent revising and the exam score.
Answer: (a) H0: rho=0, H1: rho≠0; (b) significant, reject H0, evidence of correlation; (c) No, correlation does not imply causation.

(a) B1 for H0 correct; B1 for H1 correct (in terms of rho). (b) M1 for comparing |r| with critical value; A1 for correct decision (reject H0); A1 for conclusion stated in context. (c) B1 for 'No'; B1 for stating correlation does not imply causation; B1 for a valid supporting reason/example (e.g. confounding variable).

(a) Let \( D \) be the event that a component is defective. By the law of total probability: \( P(D) = P(A)P(D|A)+P(B)P(D|B)+P(C)P(D|C) = (0.4)(0.02)+(0.35)(0.03)+(0.25)(0.05) = 0.008+0.0105+0.0125 = 0.031 \), as required.
(b) By Bayes' theorem: \( P(C|D) = \dfrac{P(C)P(D|C)}{P(D)} = \dfrac{0.0125}{0.031} = 0.40322\ldots \approx 0.403 \) (3 s.f.).
(c) For each component, \( P(D)=0.031 \), so \( P(\text{not }D)=0.969 \). \( P(\text{exactly one of two defective}) = 2\times P(D)\times P(\text{not }D) = 2(0.031)(0.969) = 0.060078 \approx 0.0601 \) (3 s.f.).
Answer: (a) 0.031 (shown); (b) 0.403; (c) 0.0601.

(a) M1 for correct structure (law of total probability); M1 for each of the three correct products (up to 2 marks, one per correct product beyond the first); A1 for correct sum; W1 for reaching printed answer 0.031 with clear working shown. (b) M1 for correct Bayes' theorem structure; A1 for correct numerator 0.0125; A1 for correct division; A1 for 0.403. (c) M1 for identifying the 'exactly one' structure with factor 2; M1 for correct probability product; A1 for 0.0601.

(a) \( H_0: p=0.3 \); \( H_1: p>0.3 \) (one-tailed, since the researcher believes the proportion has increased).
(b) Under \( H_0 \), \( X\sim B(20,0.3) \). We require the smallest \( c \) with \( P(X\ge c)<0.05 \). Computing upper-tail probabilities: \( P(X\ge9)=0.1133 \) (too large); \( P(X\ge10)=0.0480 \) (\(<0.05\)); \( P(X\ge11)=0.0171 \). So the smallest \( c \) with \( P(X\ge c)<0.05 \) is \( c=10 \); the critical region is \( X\ge10 \).
(c) The actual significance level of the test is \( P(X\ge10)=0.0480 \) (3 s.f.).
(d) Since 10 lies in the critical region \( X\ge10 \), we reject \( H_0 \). There is evidence, at the 5% significance level, that the proportion of adults locally who regularly use the fitness app has increased above 30%.
Answer: (a) H0: p=0.3, H1: p>0.3; (b) critical region X ≥ 10; (c) 0.0480; (d) reject H0 — evidence of an increase.

(a) B1 for H0 correct; B1 for H1 correct (one-tailed, in terms of p). (b) M1 for correct binomial model identified; M1 for calculating (or correctly using tables for) P(X>=9); A1 for P(X>=9)=0.1133 (too large, rejected); M1 for calculating P(X>=10); A1 for P(X>=10)=0.0480 (<0.05); A1 for correctly stating critical region X>=10. (c) B1 for identifying actual significance level = P(X>=10); B1 for 0.0480. (d) M1 for correctly locating 10 within/relative to critical region; A1 for correct conclusion in context.