2022 CCEA A-Level Mathematics 2210 模擬試題連答案詳解

(a) 6th term \( = ar^5 = 20(0.6)^5 = 1.5552 \approx 1.56 \) (3 s.f.).
(b) \( S_\infty = \dfrac{a}{1-r} = \dfrac{20}{0.4} = 50 \).
(c) \( S_n = \dfrac{a(1-r^n)}{1-r} = 50(1-0.6^n) \). Require \( 50(1-0.6^n)>49 \Rightarrow 1-0.6^n>0.98 \Rightarrow 0.6^n<0.02 \). Trying values: at \( n=7 \), \( S_7 = 50(1-0.6^7)=48.600 \) (not yet over 49); at \( n=8 \), \( S_8 = 50(1-0.6^8)=49.160 \) (exceeds 49). So the smallest such \( n \) is \( n=8 \).
Answer: (a) 1.56; (b) 50; (c) n = 8.

(a) M1 for ar^5 structure; A1 for 1.56. (b) M1 for correct formula; A1 for 50. (c) M1 for correct Sn formula; M1 for forming the inequality; M1 for trial evaluation at consecutive integer values of n; A1 for n=8 with supporting evidence (S7<49

Let \( s=\sin\theta \). Then \( 2s^2-s-1=0 \), which factorises as \( (2s+1)(s-1)=0 \), so \( s=-\dfrac12 \) or \( s=1 \).
For \( \sin\theta=1 \): \( \theta=\dfrac{\pi}{2} \).
For \( \sin\theta=-\dfrac12 \): the reference angle is \( \dfrac{\pi}{6} \); since sine is negative in the third and fourth quadrants, \( \theta=\pi+\dfrac{\pi}{6}=\dfrac{7\pi}{6} \) or \( \theta=2\pi-\dfrac{\pi}{6}=\dfrac{11\pi}{6} \).
Answer: theta = pi/2, 7pi/6, 11pi/6.

M1 for factorising (or using the quadratic formula); A1 for both correct values of sin(theta); M1 for identifying theta=pi/2; M1 for correct reference angle pi/6; A1 for 7pi/6; A1 for 11pi/6.

(a) \( \ln1=0 \), \( \ln1.5=0.4055 \), \( \ln2=0.6931 \), \( \ln2.5=0.9163 \), \( \ln3=1.0986 \) (all 4 d.p.).
(b) With \( h=0.5 \): \( \displaystyle\int_1^3\ln x\,dx \approx \dfrac{h}{2}\Big[(y_0+y_4)+2(y_1+y_2+y_3)\Big] = \dfrac{0.5}{2}\Big[(0+1.0986)+2(0.4055+0.6931+0.9163)\Big] = 0.25\big[1.0986+2(2.0149)\big] = 0.25(5.1284) = 1.2821 \approx 1.28 \) (3 s.f.).
Answer: (a) as above; (b) ≈ 1.28.

(a) B1 for all five values correct (allow B1 for 3-4 correct, full B1 for 5 correct — up to 3 marks across accuracy/completeness). (b) M1 for correct h; M1 for correct trapezium rule structure; A1 for correct bracket evaluation; A1 for 1.2821 unrounded; A1 for 1.28.

(a) From \( x=3t-1 \), \( t=\dfrac{x+1}{3} \). Substituting into \( y=t^2+2t \): \( y = \left(\dfrac{x+1}{3}\right)^2+2\left(\dfrac{x+1}{3}\right) = \dfrac{(x+1)^2}{9}+\dfrac{6(x+1)}{9} = \dfrac{(x+1)^2+6(x+1)}{9} = \dfrac{(x+1)\big[(x+1)+6\big]}{9} = \dfrac{(x+1)(x+7)}{9} \), as required.
(b) At \( t=1 \): \( x=3(1)-1=2 \). \( \dfrac{dx}{dt}=3 \), \( \dfrac{dy}{dt}=2t+2 \), so at \( t=1 \), \( \dfrac{dy}{dt}=4 \). \( \dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{4}{3} \).
Answer: (a) shown; (b) dy/dx = 4/3.

(a) M1 for making t the subject; M1 for substituting into y; M1 for expanding; M1 for factorising numerator; W1 for reaching the printed result. (b) M1 for both dx/dt and dy/dt; M1 for evaluating at t=1; A1 for dy/dx=4/3.

(a) \( fg(x)=f(g(x))=f(x^2+1)=3(x^2+1)-2=3x^2+1 \).
(b) \( gf(x)=g(f(x))=g(3x-2)=(3x-2)^2+1=9x^2-12x+4+1=9x^2-12x+5 \).
(c) \( 3x^2+1=9x^2-12x+5 \Rightarrow 0=6x^2-12x+4 \Rightarrow 3x^2-6x+2=0 \) (dividing by 2). Using the quadratic formula: \( x=\dfrac{6\pm\sqrt{36-24}}{6}=\dfrac{6\pm\sqrt{12}}{6}=\dfrac{6\pm2\sqrt3}{6}=1\pm\dfrac{\sqrt3}{3} \).
Answer: (a) 3x^2+1; (b) 9x^2-12x+5; (c) x = 1 ± √3/3.

(a) M1 for correct composition order; A1 for correct expression. (b) M1 for correct composition order; A1 for correct expanded expression. (c) M1 for forming the equation and simplifying to a quadratic; M1 for correct use of the quadratic formula (or equivalent); A1,A1 for the two correct exact solutions.

(a) \( \tan(A+B)\equiv\dfrac{\tan A+\tan B}{1-\tan A\tan B} \). With \( A=45^\circ,B=\theta \), and \( \tan45^\circ=1 \): \( \tan(45^\circ+\theta)=\dfrac{1+\tan\theta}{1-(1)(\tan\theta)}=\dfrac{1+\tan\theta}{1-\tan\theta} \), as required.
(b) \( \dfrac{1+\tan\theta}{1-\tan\theta}=3 \Rightarrow 1+\tan\theta=3(1-\tan\theta)=3-3\tan\theta \Rightarrow 4\tan\theta=2 \Rightarrow \tan\theta=0.5 \). So \( \theta=\arctan(0.5)=26.565\ldots^\circ\approx26.6^\circ \) (1 d.p.). (The next solution, \( \theta=206.6^\circ \), lies outside \( [0^\circ,180^\circ] \).)
(c) \( R\sin(\theta-\alpha)=R\sin\theta\cos\alpha-R\cos\theta\sin\alpha \). Comparing with \( 3\sin\theta-\sqrt3\cos\theta \): \( R\cos\alpha=3 \), \( R\sin\alpha=\sqrt3 \). \( R=\sqrt{3^2+(\sqrt3)^2}=\sqrt{9+3}=\sqrt{12}=2\sqrt3 \). \( \tan\alpha=\dfrac{\sqrt3}{3}=\dfrac{1}{\sqrt3} \Rightarrow \alpha=30^\circ \). So \( 3\sin\theta-\sqrt3\cos\theta\equiv2\sqrt3\sin(\theta-30^\circ) \).
(d) The maximum value of \( 2\sqrt3\sin(\theta-30^\circ) \) is \( 2\sqrt3 \), occurring when \( \sin(\theta-30^\circ)=1 \), i.e. \( \theta-30^\circ=90^\circ \Rightarrow \theta=120^\circ \).
Answer: (a) shown; (b) theta ≈ 26.6°; (c) R=2√3, alpha=30°; (d) max = 2√3 at theta = 120°.

(a) M1 for stating tan(A+B) formula; M1 for substituting A=45,B=theta with tan45=1; W2 for fully correct derivation. (b) M1 for setting expression equal to 3 and cross-multiplying; M1 for isolating tan(theta); A1 for tan(theta)=0.5; A1 for theta=26.6 (with rejection of out-of-range solution). (c) M1 for expansion and comparison; A1 for R=2√3; M1 for tan(alpha) and A1 for alpha=30° (A1 folded into the 3 available marks: award M1 for correct method, A1,A1 for R and alpha both exactly correct). (d) B1 for max value 2√3; M1 for setting sin(theta-30)=1; A1 for theta=120°.

(a) \( \dfrac{x}{4}=\cos\theta \), \( \dfrac{y}{3}=\sin\theta \). Since \( \cos^2\theta+\sin^2\theta=1 \): \( \left(\dfrac{x}{4}\right)^2+\left(\dfrac{y}{3}\right)^2=1 \Rightarrow \dfrac{x^2}{16}+\dfrac{y^2}{9}=1 \), as required.
(b) \( \dfrac{dx}{d\theta}=-4\sin\theta \), \( \dfrac{dy}{d\theta}=3\cos\theta \). \( \dfrac{dy}{dx}=\dfrac{3\cos\theta}{-4\sin\theta}=-\dfrac34\cot\theta \).
(c) At \( \theta=\dfrac{\pi}{3} \) (\( 60^\circ \)): \( x=4\cos60^\circ=2 \); \( y=3\sin60^\circ=\dfrac{3\sqrt3}{2} \). Gradient: \( \dfrac{dy}{dx}=-\dfrac34\cot60^\circ=-\dfrac34\times\dfrac{1}{\sqrt3}=-\dfrac{\sqrt3}{4} \) (rationalising). Tangent: \( y-\dfrac{3\sqrt3}{2}=-\dfrac{\sqrt3}{4}(x-2) \). Multiplying both sides by 4: \( 4y-6\sqrt3=-\sqrt3(x-2)=-\sqrt3x+2\sqrt3 \Rightarrow \sqrt3x+4y=8\sqrt3 \).
Answer: (a) shown; (b) dy/dx = -(3/4)cot(theta); (c) √3x+4y=8√3.

(a) M1 for expressing cos,sin in terms of x,y; M1 for using the identity; W1 for the printed result. (b) M1 for both dx/d(theta) and dy/d(theta); M1 for forming the ratio; A1 for correct unsimplified form; A1 for simplified form. (c) M1 for evaluating x,y at theta=pi/3; A1 for correct coordinates; M1 for evaluating the gradient; A1 for correct gradient; M1 for point-gradient equation; M1 for clearing fractions/surds; A1 for the fully correct exact tangent equation.

(a) \( 3x^2+x+2 \equiv A(x-1)^2+B(x+1)(x-1)+C(x+1) \). At \( x=1 \): \( 3+1+2=6=2C \Rightarrow C=3 \). At \( x=-1 \): \( 3-1+2=4=4A \Rightarrow A=1 \). Comparing coefficients of \( x^2 \): \( 3=A+B \Rightarrow B=3-1=2 \). So \( \dfrac{3x^2+x+2}{(x+1)(x-1)^2} \equiv \dfrac{1}{x+1}+\dfrac{2}{x-1}+\dfrac{3}{(x-1)^2} \).
(b) \( \displaystyle\int\left[\dfrac{1}{x+1}+\dfrac{2}{x-1}+\dfrac{3}{(x-1)^2}\right]dx = \ln|x+1|+2\ln|x-1|-\dfrac{3}{x-1}+C \) (using \( \int(x-1)^{-2}dx=-(x-1)^{-1} \)).
Answer: (a) A=1, B=2, C=3; (b) ln|x+1| + 2ln|x-1| - 3/(x-1) + C.

(a) M1 for correct identity setup; M1 for substituting x=1; A1 for C=3; M1 for substituting x=-1; A1 for A=1; M1 for comparing coefficients (or a third substitution); A1 for B=2; A1 for correctly stated partial fractions. (b) M1 for integrating 1/(x+1) term; M1 for integrating 2/(x-1) term; M1 for integrating 3/(x-1)^2 term; A1 for each correctly integrated term (up to 3 marks in total for the three terms).

(a) \( h(1)=1-1-3=-3<0 \); \( h(2)=8-2-3=3>0 \). Since \( h \) is continuous and changes sign on \( [1,2] \), there is a root \( \beta \) in this interval.
(b) \( x^3-x-3=0 \Rightarrow x^3=x+3 \Rightarrow x=\sqrt[3]{x+3} \), giving the iteration \( x_{n+1}=\sqrt[3]{x_n+3} \), as required.
(c) \( x_1=\sqrt[3]{1.5+3}=\sqrt[3]{4.5}=1.650964\ldots\approx1.6510 \). \( x_2=\sqrt[3]{1.650964+3}=\sqrt[3]{4.650964}=1.669223\ldots\approx1.6692 \). \( x_3=\sqrt[3]{1.669223+3}=\sqrt[3]{4.669223}=1.671404\ldots\approx1.6714 \).
(d) \( h(1.665)=(1.665)^3-1.665-3=-0.0492\ldots<0 \). \( h(1.675)=(1.675)^3-1.675-3=0.0244\ldots>0 \). Since \( h(x) \) changes sign between \( x=1.665 \) and \( x=1.675 \), the root \( \beta \) lies in this interval, and so \( \beta=1.67 \) correct to 3 significant figures (consistent with the iteration converging towards \( x_3=1.6714\to1.671665\ldots \) as \( n \) increases).
Answer: (a) shown; (b) shown; (c) x1=1.6510, x2=1.6692, x3=1.6714; (d) beta = 1.67 (3 s.f.), verified.

(a) M1 for evaluating h(1) and h(2); A1 for correct conclusion. (b) M1 for rearranging to x^3=x+3; M1 for taking cube roots; W1 for stating the iterative formula. (c) M1 for correct method applied; A1 for x1; A1 for x2; A1 for x3 (all to 4dp); M1 for consistent iterative use of previous value; A1 combined for full accuracy across all three (mark scheme awards up to 6 across correct x1,x2,x3 with method). (d) M1 for evaluating h(1.665); A1 for correct sign/value; M1 for evaluating h(1.675); A1 for correct sign/value and conclusion linking the sign change to beta=1.67 (3 s.f.).

(a) By the product rule, \( \dfrac{dy}{dx} = e^{-x}+x(-e^{-x}) = e^{-x}(1-x) \). Since \( e^{-x}\ne0 \) for all \( x \), \( \dfrac{dy}{dx}=0 \) only when \( 1-x=0 \), i.e. \( x=1 \); this is the single stationary point. For \( x<1 \), \( 1-x>0 \) so \( dy/dx>0 \); for \( x>1 \), \( 1-x<0 \) so \( dy/dx<0 \). The gradient changes from positive to negative, so \( x=1 \) gives a maximum, with \( y=1\cdot e^{-1}=\dfrac1e \).
(b) Let \( u=x \), \( dv=e^{-x}dx \), so \( du=dx \), \( v=-e^{-x}\). \( \displaystyle\int xe^{-x}dx = -xe^{-x}-\int(-e^{-x})dx = -xe^{-x}+\int e^{-x}dx = -xe^{-x}-e^{-x}+C = -(x+1)e^{-x}+C \), as required.
(c) Area \( = \displaystyle\int_0^2 xe^{-x}dx = \Big[-(x+1)e^{-x}\Big]_0^2 = \big(-3e^{-2}\big)-\big(-1\big) = 1-3e^{-2} \) (\( \approx0.594 \)).
(d) The solid is generated by rotating \( y=xe^{-x} \) about the \( x \)-axis, so \( V=\pi\displaystyle\int_0^2 y^2\,dx = \pi\displaystyle\int_0^2(xe^{-x})^2dx = \pi\displaystyle\int_0^2x^2e^{-2x}dx \), as required. Integrating by parts with \( u=x^2,dv=e^{-2x}dx \) (so \( du=2x\,dx,v=-\tfrac12e^{-2x} \)): \( \displaystyle\int x^2e^{-2x}dx = -\tfrac12x^2e^{-2x}+\displaystyle\int xe^{-2x}dx \). For \( \int xe^{-2x}dx \), let \( u=x,dv=e^{-2x}dx \) (\( du=dx,v=-\tfrac12e^{-2x} \)): \( \int xe^{-2x}dx = -\tfrac12xe^{-2x}+\tfrac12\int e^{-2x}dx = -\tfrac12xe^{-2x}-\tfrac14e^{-2x}+C \). Combining: \( \displaystyle\int x^2e^{-2x}dx = -\tfrac12x^2e^{-2x}-\tfrac12xe^{-2x}-\tfrac14e^{-2x}+C = -\dfrac{(2x^2+2x+1)}{4}e^{-2x}+C \). Evaluating from 0 to 2: at \( x=2 \): \( -\dfrac{(8+4+1)}{4}e^{-4}=-\dfrac{13}{4}e^{-4} \); at \( x=0 \): \( -\dfrac14 \). So \( \displaystyle\int_0^2x^2e^{-2x}dx = -\dfrac{13}{4}e^{-4}-\left(-\dfrac14\right) = \dfrac14-\dfrac{13}{4}e^{-4} = \dfrac{1-13e^{-4}}{4} \). Hence \( V=\pi\cdot\dfrac{1-13e^{-4}}{4} = \dfrac{\pi(1-13e^{-4})}{4} \) (\( \approx0.598 \)).
Answer: (a) max at x=1, y=1/e; (b) shown; (c) 1-3e^{-2} (≈0.594); (d) V = π(1-13e^{-4})/4 (≈0.598).

(a) M1 for product rule; A1 for correct derivative; M1 for setting dy/dx=0 and arguing e^{-x}≠0; A1 for x=1; M1 for sign-test of gradient either side; A1 for correctly identifying maximum; B1 for y=1/e stated. (b) M1 for correct choice of u,dv; M1 for correct v; M1 for applying by-parts formula; A1 for correct integral of e^{-x}; W1 for the printed result. (c) M1 for substituting limits into the given antiderivative; M1 for correct evaluation at x=2; A1 for correct evaluation at x=0; A1 for 1-3e^{-2} (exact). (d) B1 for stating/justifying V=pi*int y^2 dx; M1 for squaring to get x^2e^{-2x}; M1 for by-parts on x^2e^{-2x} (correct u,dv); A1 for reducing to int xe^{-2x}dx; M1 for by-parts on xe^{-2x} (correct u,dv); A1 for that integral; A1 for combined antiderivative; M1 for substituting limits; A1 for exact value of the definite integral; A1 for final exact V.

(a) Volume: \( V=\pi r^2h=500 \Rightarrow h=\dfrac{500}{\pi r^2} \). Surface area (base circle, no lid, plus curved surface): \( A=\pi r^2+2\pi rh = \pi r^2+2\pi r\left(\dfrac{500}{\pi r^2}\right) = \pi r^2+\dfrac{1000}{r} \), as required. A suitable domain is \( r>0 \), since \( r \) is a physical radius (a length) and must be strictly positive for the can (and hence \( h=500/(\pi r^2) \)) to exist.
(b) \( \dfrac{dA}{dr}=2\pi r-\dfrac{1000}{r^2} \). Setting \( \dfrac{dA}{dr}=0 \): \( 2\pi r=\dfrac{1000}{r^2} \Rightarrow r^3=\dfrac{1000}{2\pi}=\dfrac{500}{\pi} \Rightarrow r=\sqrt[3]{\dfrac{500}{\pi}} = \dfrac{5\sqrt[3]4}{\sqrt[3]\pi} \) (exact form, equivalent to \( 5\times2^{2/3}\pi^{-1/3} \)). Numerically, \( r=5.41926\ldots\approx5.42 \) cm (3 s.f.).
(c) \( A_{\min}=\pi r^2+\dfrac{1000}{r} \) at \( r=5.4193 \): \( A_{\min}=276.79\ldots\approx277\text{ cm}^2 \) (3 s.f.).
(d) \( \dfrac{d^2A}{dr^2}=2\pi+\dfrac{2000}{r^3} \). At \( r=5.4193 \): substituting \( r^3=500/\pi \), \( \dfrac{2000}{r^3}=\dfrac{2000\pi}{500}=4\pi \), so \( \dfrac{d^2A}{dr^2}=2\pi+4\pi=6\pi>0 \). Since the second derivative is positive at this value of \( r \), it confirms a minimum.
(e) By the chain rule, \( \dfrac{dA}{dt}=\dfrac{dA}{dr}\times\dfrac{dr}{dt} \). At the value of \( r \) found in part (b), \( \dfrac{dA}{dr}=0 \) (by definition of the stationary point), so \( \dfrac{dA}{dt}=0\times0.02=0 \), regardless of the value of \( dr/dt \). This is because \( A \) is momentarily stationary with respect to \( r \) at this exact radius, so to first order a small change in \( r \) produces no change in \( A \).
Answer: (a) shown, r>0; (b) r = 5(4)^{1/3}/π^{1/3} ≈ 5.42 cm; (c) ≈ 277 cm^2; (d) 6π > 0, minimum confirmed; (e) dA/dt = 0.

(a) M1 for V=pi r^2 h; M1 for h=500/(pi r^2); M1 for correct surface area expression (base + curved, no top); M1 for substituting h; W1 for reaching printed result; B1 for stating r>0; B1 for valid justification. (b) M1 for differentiating; A1 for correct dA/dr; M1 for setting dA/dr=0; M1 for rearranging to r^3=500/pi; A1 for correct exact form of r; A1 for correct 3sf value; (2 further marks for full clean method/accuracy across the derivation). (c) M1 for substituting r into A; A1 for unrounded value; A1 for 277. (d) M1 for correct second derivative; M1 for substituting r (or r^3) correctly; A1 for positive value with correct conclusion. (e) M1 for stating chain rule dA/dt = dA/dr * dr/dt; A1 for recognising dA/dr=0 at this r; A1 for dA/dt=0; B1 for clear justification/explanation.

(a) \( u=+8 \) m s\(^{-1}\); \( v=-5 \) m s\(^{-1}\).
(b) Impulse \( = m(v-u) = 0.5(-5-8) = 0.5(-13) = -6.5 \) N s. The magnitude of the impulse is 6.5 N s (directed away from the wall).
(c) Average force \( = \dfrac{\text{impulse}}{\text{time}} = \dfrac{6.5}{0.02} = 325 \) N.
Answer: (a) u=+8, v=-5; (b) 6.5 N s; (c) 325 N.

(a) B1 for both signs correct. (b) M1 for impulse=m(v-u) with correct values; A1 for correct signed value; A1 for correct magnitude. (c) M1 for force=impulse/time; A1 for 325 N.

(a) \( \mathbf{v}=\mathbf{u}+\mathbf{a}t = (\mathbf{i}+3\mathbf{j})+6(-0.5\mathbf{i}+0.5\mathbf{j}) = (1-3)\mathbf{i}+(3+3)\mathbf{j} = -2\mathbf{i}+6\mathbf{j} \).
(b) \( \mathbf{r}=\mathbf{r_0}+\mathbf{u}t+\tfrac12\mathbf{a}t^2 = (2\mathbf{i}-\mathbf{j})+6(\mathbf{i}+3\mathbf{j})+\tfrac12(36)(-0.5\mathbf{i}+0.5\mathbf{j}) = (2\mathbf{i}-\mathbf{j})+(6\mathbf{i}+18\mathbf{j})+(-9\mathbf{i}+9\mathbf{j}) = -\mathbf{i}+26\mathbf{j} \).
(c) Speed \( = |\mathbf{v}| = \sqrt{(-2)^2+6^2} = \sqrt{40} = 6.3246\ldots\approx6.32 \) m s\(^{-1}\).
Answer: (a) -2i+6j; (b) -i+26j; (c) 6.32 m/s.

(a) M1 for v=u+at; A1 for correct vector. (b) M1 for correct SUVAT structure; A1 for correct expansion; A1 for correct final vector. (c) M1 for Pythagoras; A1 for 6.32.

(a) \( a=\dfrac{dv}{dt}=6-2t \). At \( t=2 \): \( a=6-4=2 \) m s\(^{-2}\).
(b) Maximum velocity occurs when \( \dfrac{dv}{dt}=0 \): \( 6-2t=0\Rightarrow t=3 \). Since \( a=6-2t \) changes from positive to negative at \( t=3 \), this is a maximum. \( v(3)=6(3)-3^2=18-9=9 \) m s\(^{-1}\).
(c) Since \( v=6t-t^2=t(6-t)\ge0 \) for \( 0\le t\le6 \), the particle does not change direction in \( [0,4] \), so distance travelled equals displacement: \( \displaystyle\int_0^4(6t-t^2)dt = \left[3t^2-\dfrac{t^3}{3}\right]_0^4 = 3(16)-\dfrac{64}{3} = 48-\dfrac{64}{3} = \dfrac{80}{3} = 26.67\ldots\approx26.7 \) m (3 s.f.).
Answer: (a) 2 m/s^2; (b) 9 m/s at t=3s; (c) 26.7 m.

(a) M1 for differentiating v; A1 for a=2. (b) M1 for setting dv/dt=0; A1 for t=3; M1 for justifying maximum (sign change of a, or second derivative); A1 for v=9. (c) M1 for recognising v>=0 throughout so distance=displacement; M1 for correct integral set up; M1 for correct antiderivative; A1 for 80/3 unrounded; A1 for 26.7.

(a) \( \mathbf{v}(t)=\mathbf{u}+\mathbf{a}t = (8\mathbf{i}+6\mathbf{j})+t(-9.8\mathbf{j}) = 8\mathbf{i}+(6-9.8t)\mathbf{j} \).
(b) \( \mathbf{r}(t)=\mathbf{u}t+\tfrac12\mathbf{a}t^2 = (8\mathbf{i}+6\mathbf{j})t+\tfrac12(-9.8\mathbf{j})t^2 = 8t\mathbf{i}+(6t-4.9t^2)\mathbf{j} \).
(c) The particle returns to the height of \( O \) when the \( \mathbf{j} \)-component of \( \mathbf{r}(t) \) is zero (other than at \( t=0 \)): \( 6t-4.9t^2=0\Rightarrow t(6-4.9t)=0\Rightarrow t=0 \) or \( t=\dfrac{6}{4.9}=1.2245\ldots\approx1.22 \) s.
(d) Horizontal distance \( =8t=8(1.2245)=9.796\ldots\approx9.80 \) m (3 s.f.).
(e) At \( t=1.2245 \): \( \mathbf{j} \)-component of velocity \( =6-9.8(1.2245)=-6 \) (this is expected: by symmetry of projectile motion, the vertical velocity when returning to launch height is the negative of the initial vertical velocity). The \( \mathbf{i} \)-component remains \( 8 \) (unchanged, since there is no horizontal acceleration). Speed \( =\sqrt{8^2+(-6)^2}=\sqrt{64+36}=\sqrt{100}=10 \) m s\(^{-1}\).
Answer: (a) v(t)=8i+(6-9.8t)j; (b) r(t)=8ti+(6t-4.9t^2)j; (c) t ≈ 1.22 s; (d) ≈ 9.80 m; (e) 10 m/s.

(a) M1 for v=u+at structure; A1 for correct expression. (b) M1 for correct SUVAT vector structure; A1 for correct expansion; A1 for correct final expression. (c) M1 for setting j-component to zero; M1 for solving/factorising; A1 for t=1.22 (rejecting t=0). (d) M1 for substituting t into i-component; A1 for 9.80 m. (e) M1 for finding the velocity components at this t (or using the symmetry argument); A1 for speed=10 m/s.

(a) The forces are: the weight of the beam, \( 12g=117.6 \) N, acting vertically downward at its midpoint, \( 2.5 \) m from \( A \); the weight of the load, \( 8g=78.4 \) N, acting vertically downward at \( B \), \( 5 \) m from \( A \); the reaction \( R_A \), acting vertically upward at \( A \); and the reaction \( R_C \), acting vertically upward at \( C \), \( 4 \) m from \( A \).
(b) Taking moments about \( A \) (clockwise positive), the upward moment of \( R_C \) balances the downward moments of the two weights: \( R_C(4) = 117.6(2.5)+78.4(5) = 294+392 = 686 \Rightarrow R_C = \dfrac{686}{4} = 171.5 \) N.
(c) Resolving vertically: \( R_A+R_C = 117.6+78.4 = 196 \Rightarrow R_A = 196-171.5 = 24.5 \) N.
(d) Since \( R_A=24.5 \) N \( >0 \), the support at \( A \) is indeed pushing up on the beam, confirming the beam remains in contact with (does not lift off) the support at \( A \). One modelling assumption made is that the beam is rigid and the supports are smooth, so that each support exerts only a single vertical (normal) reaction force on the beam, with no horizontal friction component and no resisting moment.
Answer: (a) as listed; (b) R_C = 171.5 N; (c) R_A = 24.5 N; (d) R_A=24.5>0 confirms contact; assumption: smooth supports giving purely vertical reactions.

(a) B1 for both weights correctly located; B1 for reaction at A correctly described; B1 for reaction at C correctly described. (b) M1 for correct moment equation structure; M1 for correct moment arms for both weights; A1 for correct equation; M1 for rearranging; A1 for R_C=171.5. (c) M1 for vertical resolution; A1 for correct equation; A1 for R_A=24.5. (d) M1 for checking sign of R_A; A1 for correct conclusion; B1 for a valid stated assumption.

(a) \( H_0:\rho=0 \); \( H_1:\rho\ne0 \) (two-tailed).
(b) \( |r|=0.71 \), which exceeds the critical value \( 0.6215 \), so the result is significant at the 10% level and we reject \( H_0 \). There is evidence, at the 10% level, of a linear correlation between rainfall and crop yield.
(c) A Type I error in this context would be rejecting \( H_0 \) (i.e. concluding that there is a linear correlation between rainfall and crop yield) when, in fact, there is no linear correlation in the population.
Answer: (a) H0: rho=0, H1: rho≠0; (b) significant, reject H0; (c) Type I error: wrongly concluding correlation exists when it does not.

(a) B1 for H0; B1 for H1. (b) M1 for comparing |r| to critical value; A1 for correct decision; A1 for conclusion in context. (c) B1 for identifying it as incorrectly rejecting H0; B1 for correct contextual description.

(a) \( Z=\dfrac{335-330}{4}=1.25 \). \( P(X>335)=P(Z>1.25)=1-\Phi(1.25)=1-0.8944=0.1056 \) (4 d.p.).
(b) \( Z_1=\dfrac{325-330}{4}=-1.25 \); \( Z_2=\dfrac{328-330}{4}=-0.5 \). \( P(325

(a) M1 for standardising; M1 for correct tail; A1 for 0.1056. (b) M1 for both z-values; M1 for correct probability structure; A1 for correct component probabilities; A1 for 0.2029. (c) M1 for identifying z=-2.0537 (or equivalent from P(Z

(a) By the addition law, \( n(M\cup P)=n(M)+n(P)-n(M\cap P) = 120+90-50 = 160 \). Number studying neither \( = 200-160=40 \).
(b) \( P(M|P) = \dfrac{P(M\cap P)}{P(P)} = \dfrac{50/200}{90/200} = \dfrac{50}{90} = \dfrac{5}{9} \).
(c) \( P(M)=\dfrac{120}{200}=0.6 \); \( P(P)=\dfrac{90}{200}=0.45 \); \( P(M)\times P(P)=0.6\times0.45=0.27 \). \( P(M\cap P)=\dfrac{50}{200}=0.25 \). Since \( P(M)\times P(P)=0.27\ne0.25=P(M\cap P) \), the events \( M \) and \( P \) are NOT statistically independent.
Answer: (a) 40; (b) 5/9; (c) not independent (0.27 ≠ 0.25).

(a) M1 for addition law; M1 for correct union; A1 for n(union)=160; A1 for neither=40. (b) M1 for correct conditional probability structure; A1 for correct calculation; A1 for 5/9 in simplest form. (c) M1 for calculating P(M) and P(P); M1 for calculating P(M)P(P); A1 for 0.27; M1 for calculating P(M∩P)=0.25; A1 for correct comparison and conclusion (not independent).

(a) \( H_0: p=0.25 \); \( H_1: p<0.25 \) (one-tailed, since the inspector suspects the proportion is lower).
(b) Since we are testing whether the proportion is lower than claimed, we compare the given probability \( P(X\le1)=0.0802 \) with the significance level \( 0.05 \). Since \( 0.0802>0.05 \), the result of obtaining 1 or fewer premium bulbs in a sample of 15 is not sufficiently unlikely under \( H_0 \) to be significant at the 5% level. We therefore do not reject \( H_0 \): there is insufficient evidence, at the 5% level, that the true proportion of premium bulbs is lower than 25%.
(c) The 5% significance level is the probability of incorrectly rejecting \( H_0 \) when it is in fact true — that is, the probability of concluding that the true proportion of premium bulbs is lower than 25% when, in reality, it is still 25%.
Answer: (a) H0: p=0.25, H1: p<0.25; (b) not significant (0.0802>0.05), do not reject H0; (c) probability of a Type I error (wrongly rejecting H0).

(a) B1 for H0; B1 for H1 (one-tailed, correct direction). (b) M1 for comparing 0.0802 with 0.05; A1 for correct decision (do not reject H0); M1 for stating conclusion structure; A1 for conclusion fully in context. (c) M1 for identifying it as a probability of incorrect rejection; A1 for linking to Type I error; M1 for correctly identifying which hypothesis/direction; A1 for full contextual statement.

\( H_0:\mu=25 \); \( H_1:\mu\ne25 \) (two-tailed). Under \( H_0 \), \( \bar X\sim N\left(25,\dfrac{0.8^2}{16}\right) \). Test statistic: \( z=\dfrac{\bar x-\mu_0}{\sigma/\sqrt n}=\dfrac{24.6-25}{0.8/\sqrt{16}}=\dfrac{-0.4}{0.2}=-2.00 \). The critical value for a two-tailed test at the 1% level is \( z=\pm2.576 \). Since \( |-2.00|=2.00<2.576 \), the test statistic does not lie in the critical region, so we do not reject \( H_0 \). There is no significant evidence, at the 1% level, that the mean length of rods produced by the machine differs from 25 cm.
Answer: z = -2.00, |z| < 2.576, so do not reject H0; no evidence the mean length differs from 25 cm.

B1,B1 for correct hypotheses (2 marks); M1 for correct standard error 0.8/sqrt(16); M1 for correct test statistic formula; A1 for z=-2.00 (3 marks); B1 for correct critical value ±2.576; A1 for correct conclusion in context linking comparison to decision (2 marks, total 10).