2024 CCEA A-Level Mathematics 2210 模拟试题及答案详解

(a) Numerator: \( 2x^2+5x-3 = (2x-1)(x+3) \). Denominator: \( x^2-9=(x-3)(x+3) \). So \( \dfrac{(2x-1)(x+3)}{(x-3)(x+3)} = \dfrac{2x-1}{x-3} \) (cancelling the common factor (x+3)), undefined at x = 3 and x = −3.
(b) \( \dfrac{2x-1}{x-3}=3 \Rightarrow 2x-1 = 3(x-3) = 3x-9 \Rightarrow -1+9 = 3x-2x \Rightarrow x = 8 \). Since 8 ≠ 3, −3, this is valid. Final answer: x = 8.

(a) [5] M1: factorise numerator (2x−1)(x+3); M1: factorise denominator (x−3)(x+3); M1: cancel common factor (x+3); A1: \( \frac{2x-1}{x-3} \); B1: states x ≠ 3 and x ≠ −3. (b) [4] M1 ft: cross-multiply using their (a), 2x−1=3(x−3); M1: correct expansion 3x−9; A1: x=8; B1: confirms x=8 valid (not an excluded value).

(a) \( u_2 = 3(5)-4=11 \); \( u_3=3(11)-4=29 \); \( u_4=3(29)-4=83 \).
(b) \( v_n = u_n - 2 \), so \( v_1 = 5-2=3 \). \( v_{n+1} = u_{n+1}-2 = (3u_n-4)-2 = 3u_n - 6 = 3(u_n-2) = 3v_n \). Since \( v_{n+1}=3v_n \) for all n, \( v_n \) is geometric with first term 3 and common ratio 3, so \( v_n = 3 \times 3^{n-1} = 3^n \). Hence \( u_n = v_n+2 = 3^n+2 \). (Check: \( u_1=3+2=5 \) ✓, \( u_2=9+2=11 \) ✓, \( u_3=27+2=29 \) ✓, \( u_4=81+2=83 \) ✓.)
(c) \( \sum_{n=1}^{6} u_n = \sum_{n=1}^{6}(3^n+2) = \sum_{n=1}^{6}3^n + 6(2) \). \( \sum_{n=1}^{6}3^n = \dfrac{3(3^6-1)}{3-1} = \dfrac{3(729-1)}{2} = \dfrac{3 \times 728}{2} = 1092 \). So the total = \( 1092+12 = 1104 \). Final answer: 1104.

(a) [3] B1 each: u₂=11, u₃=29, u₄=83. (b) [4] M1: v₁ = 3 found; M1: correct algebraic manipulation showing \( v_{n+1}=3v_n \); A1: identifies v_n geometric, ratio 3, first term 3; A1: \( u_n = 3^n+2 \). (c) [4] M1: splits sum into Σ3ⁿ and Σ2; M1: correct use of GP sum formula for Σ3ⁿ (n=1 to 6); A1: Σ3ⁿ = 1092; A1: total = 1104.

(a) Using the product rule with \( u=x^3, v=\ln(2x) \): \( \frac{dy}{dx} = 3x^2\ln(2x) + x^3 \cdot \frac{1}{x} = 3x^2\ln(2x) + x^2 \) (since \( \frac{d}{dx}\ln(2x) = \frac{2}{2x}=\frac{1}{x} \)).
(b) Write \( y = \sin 2x \cdot e^{-3x} \). Product rule: \( \frac{dy}{dx} = (2\cos 2x)(e^{-3x}) + (\sin 2x)(-3e^{-3x}) = e^{-3x}(2\cos 2x - 3\sin 2x) \).
(c) \( \int (3x-1)^5\,dx = \dfrac{(3x-1)^6}{6 \times 3} + C = \dfrac{(3x-1)^6}{18} + C \).
(d) Since \( \frac{d}{dx}(x^2) = 2x \), \( \int x e^{x^2}\,dx = \dfrac{1}{2}\int 2x\,e^{x^2}\,dx = \dfrac{1}{2}e^{x^2} + C \).

(a) [3] M1: product rule applied; M1: \( \frac{d}{dx}\ln(2x)=\frac{1}{x} \) used correctly; A1: \( 3x^2\ln(2x)+x^2 \). (b) [4] M1: rewritten as product / quotient rule set up; M1: correct derivative of sin2x (2cos2x) and of e^{-3x} (−3e^{-3x}); A1: correct unsimplified expression; A1: simplified to \( e^{-3x}(2\cos2x-3\sin2x) \). (c) [3] M1: reverse chain rule form \( (3x-1)^6/(6\times3) \); A1: \( (3x-1)^6/18 \); B1: +C. (d) [4] M1: recognises \( \frac{d}{dx}x^2=2x \); M1: correct adjustment by factor 1/2; A1: \( \frac{1}{2}e^{x^2} \); B1: +C.

(a) \( S_n = a+ar+ar^2+\dots+ar^{n-1} \). Then \( rS_n = ar+ar^2+\dots+ar^n \). Subtracting: \( S_n - rS_n = a - ar^n \), so \( S_n(1-r) = a(1-r^n) \), giving \( S_n = \dfrac{a(1-r^n)}{1-r} \), as required.
(b) For \( |r|<1 \), as \( n\to\infty \), \( r^n \to 0 \), so \( S_n \to \dfrac{a(1-0)}{1-r} = \dfrac{a}{1-r} \), i.e. \( S_\infty = \dfrac{a}{1-r} \).
(c) \( S_\infty = \dfrac{8}{1-0.75} = \dfrac{8}{0.25} = 32 \). 99% of \( S_\infty \) = 31.68. \( S_n = \dfrac{8(1-0.75^n)}{0.25} = 32(1-0.75^n) \). Require \( 32(1-0.75^n) > 31.68 \Rightarrow 1-0.75^n > 0.99 \Rightarrow 0.75^n < 0.01 \). Taking logarithms of both sides: \( n\ln(0.75) < \ln(0.01) \). Since \( \ln(0.75) < 0 \), dividing reverses the inequality: \( n > \dfrac{\ln(0.01)}{\ln(0.75)} = 16.008 \) (3 d.p.). So the smallest integer value is n = 17. (Check: \( S_{16}=31.679<31.68 \); \( S_{17}=31.759>31.68 \) ✓.) Final answer: n = 17.

(a) [4] M1: writes out S_n and rS_n; M1: correctly subtracts to get \( a-ar^n \); M1: factorises S_n(1-r); A1: correct formula stated as the conclusion of the proof. (b) [2] M1: recognises \( r^n \to 0 \) as \( n\to\infty \) for |r|<1; A1: correct conclusion \( S_\infty=a/(1-r) \). (c) [6] B1: S∞=32; B1: 99% target = 31.68; M1: correct inequality \( 0.75^n<0.01 \) formed; M1: logs taken correctly with inequality direction reversed; A1: n>16.0 (or 16.008); A1: n=17 (smallest integer, with check).

(a) \( 1-\cos2\theta = 1-(1-2\sin^2\theta)=2\sin^2\theta \). \( \sin2\theta = 2\sin\theta\cos\theta \). So \( \dfrac{1-\cos2\theta}{\sin2\theta} = \dfrac{2\sin^2\theta}{2\sin\theta\cos\theta} = \dfrac{\sin\theta}{\cos\theta} = \tan\theta \), as required.
(b) \( R\cos(\theta+\alpha) \equiv R\cos\theta\cos\alpha - R\sin\theta\sin\alpha \). Comparing with \( 5\cos\theta-12\sin\theta \): \( R\cos\alpha=5 \), \( R\sin\alpha=12 \). So \( R = \sqrt{5^2+12^2}=\sqrt{169}=13 \), and \( \tan\alpha = \dfrac{12}{5} \Rightarrow \alpha = 67.4° \) (1 d.p.). So \( 5\cos\theta-12\sin\theta \equiv 13\cos(\theta+67.4°) \).
(c) \( 13\cos(\theta+67.4°) = 6 \Rightarrow \cos(\theta+67.4°) = \dfrac{6}{13} = 0.4615 \). Principal value: \( \theta+67.4° = \cos^{-1}(0.4615) = 62.5° \) (1 d.p.). General solutions of \( \cos X = 0.4615 \): \( X = 62.5° \) or \( X = -62.5° \) (equivalently 297.5°), plus multiples of 360°. So \( \theta = 62.5°-67.4° = -4.9° \to 355.1° \) (adding 360°, to bring into range), or \( \theta = -62.5°-67.4° = -129.9° \to 230.1° \) (adding 360°). Both lie in \( 0° \le \theta \le 360° \). Final answers: \( \theta = 230.1° \) or \( \theta = 355.1° \).
(d) The maximum value of \( R\cos(\theta+\alpha) \) is R = 13, occurring when \( \cos(\theta+\alpha)=1 \), i.e. \( \theta+\alpha = 360°k \). The smallest positive θ occurs at \( \theta+67.4°=360° \Rightarrow \theta = 292.6° \) (1 d.p.) (since \( \theta+67.4°=0° \) gives a negative θ).
(e) Since \( 5\cos\theta-12\sin\theta \equiv 13\cos(\theta+\alpha) \), and \( \cos(\theta+\alpha) \) can only take values in \( [-1,1] \), the expression can only take values in \( [-13,13] \). So the equation has no solutions when \( k<-13 \) or \( k>13 \).

(a) [4] M1: \( 1-\cos2\theta=2\sin^2\theta \); M1: \( \sin2\theta=2\sin\theta\cos\theta \); M1: correctly combines/cancels; A1: reaches tanθ (identity shown, not assumed). (b) [4] M1: expands Rcos(θ+α); M1: Rcosα=5, Rsinα=12 identified; A1: R=13; A1: α=67.4°. (c) [6] M1 ft: 13cos(θ+their α)=6; M1: cos(θ+α)=6/13 evaluated; M1: principal value 62.5° found; A1 ft: θ=355.1°; A1 ft: θ=230.1°; B1: both solutions correctly placed in range 0–360° (accept either order). (d) [3] B1 ft: max value = their R (13); M1: condition θ+α=360° set up; A1 ft: θ=292.6°. (e) [3] B1: recognises range of cos(θ+α) is [−1,1]; M1: hence range of expression is [−R,R]; A1: k<−13 or k>13 stated correctly (ft their R).

(a) Using \( (1+u)^{1/2} = 1+\frac{1}{2}u+\dfrac{\frac12(\frac12-1)}{2!}u^2+\dfrac{\frac12(\frac12-1)(\frac12-2)}{3!}u^3+\dots \) with \( u=4x \):
term 1: 1.
term 2: \( \frac12(4x)=2x \).
term 3: \( \dfrac{\frac12 \times(-\frac12)}{2}(4x)^2 = \dfrac{-\frac14}{2}(16x^2) = -\dfrac{1}{8}(16x^2)=-2x^2 \).
term 4: \( \dfrac{\frac12\times(-\frac12)\times(-\frac32)}{6}(4x)^3 = \dfrac{\frac38}{6}(64x^3) = \dfrac{1}{16}(64x^3)=4x^3 \).
So \( (1+4x)^{1/2} \approx 1+2x-2x^2+4x^3 \), valid for \( |4x|<1 \), i.e. \( |x|<\dfrac14 \).
(b) At \( x=0.005 \): \( 1+2(0.005)-2(0.005)^2+4(0.005)^3 = 1+0.01-0.00005+0.0000005 = 1.0099505 \) (to 7 d.p.), so \( \sqrt{1.02} \approx 1.009951 \) (6 d.p.) — note \( 1+4(0.005)=1.02 \), confirming this estimates \( \sqrt{1.02} \).
(c) Percentage error \( = \dfrac{|1.009951-1.009950494|}{1.009950494}\times100\% \approx \dfrac{0.0000005}{1.00995}\times 100\% \approx 6.1\times10^{-5}\,\% \) (2 s.f.).

(a) [6] M1: correct binomial coefficients used with u=4x; A1: term in x (2x); A1: term in x² (−2x²); A1: term in x³ (4x³); B1: full expansion \( 1+2x-2x^2+4x^3 \); B1: validity |x|<1/4 (ft from |4x|<1). (b) [3] M1 ft: correct substitution x=0.005 into their expansion; M1: correct arithmetic evaluation; A1: 1.009951 (6 d.p., ft). (c) [2] M1 ft: correct percentage error method \( |estimate-actual|/actual \times100 \); A1 ft: answer in standard form to 2 s.f., consistent with their (b) (e.g. 6.1×10⁻⁵ %).

Let \( u=4-x^2 \), so \( \dfrac{du}{dx}=-2x \Rightarrow x\,dx = -\dfrac{1}{2}\,du \). When \( x=0, u=4 \); when \( x=1, u=3 \). The integral becomes: \( \int_{u=4}^{u=3} \sqrt{u}\left(-\dfrac12\right)du = \dfrac12\int_3^4 \sqrt{u}\,du = \dfrac12\left[\dfrac{2}{3}u^{3/2}\right]_3^4 = \dfrac{1}{3}\left[u^{3/2}\right]_3^4 = \dfrac{1}{3}\left(4^{3/2}-3^{3/2}\right) = \dfrac{1}{3}(8-3\sqrt3) \). Final answer: \( \dfrac{8-3\sqrt3}{3} \) (≈ 0.935, 3 s.f.).

[7] M1: correct derivative du/dx=−2x; M1: correct rearrangement x dx = −½du; M1: correct change of limits (x=0→u=4, x=1→u=3); M1: integral rewritten fully in terms of u; M1: correct integration \( \frac23 u^{3/2} \); A1: correct substitution of limits; A1: exact final answer \( \frac{8-3\sqrt3}{3} \) (accept equivalent exact forms).

(a) Let \( \dfrac{7x-4}{(x-2)(2x+1)} \equiv \dfrac{A}{x-2}+\dfrac{B}{2x+1} \), so \( 7x-4 \equiv A(2x+1)+B(x-2) \). Let \( x=2 \): \( 10 = A(5) \Rightarrow A=2 \). Let \( x=-\frac12 \): \( -7.5 = B(-2.5) \Rightarrow B=3 \). So \( \dfrac{7x-4}{(x-2)(2x+1)} \equiv \dfrac{2}{x-2}+\dfrac{3}{2x+1} \).
(b) Separating variables: \( \dfrac{1}{y}\,dy = \left(\dfrac{2}{x-2}+\dfrac{3}{2x+1}\right)dx \). Integrating both sides: \( \ln|y| = 2\ln|x-2| + \dfrac{3}{2}\ln|2x+1| + C \) (the second term integrates to \( \frac{3}{2}\ln|2x+1| \) using the reverse chain rule, since \( \int \frac{3}{2x+1}dx = \frac32\ln|2x+1| \)). So \( y = A(x-2)^2(2x+1)^{3/2} \), where \( A=e^C \). Using \( y=5 \) when \( x=3 \): \( 5 = A(3-2)^2(2(3)+1)^{3/2} = A(1)(7)^{3/2} = 7\sqrt7\,A \), so \( A = \dfrac{5}{7\sqrt7} \). Final answer: \( y = \dfrac{5}{7\sqrt7}(x-2)^2(2x+1)^{3/2} \) (equivalently \( y=\dfrac{5\sqrt7}{49}(x-2)^2(2x+1)^{3/2} \)).

(a) [4] M1: correct partial fraction setup with (x−2)(2x+1) denominators; M1: correct substitution method (e.g. x=2 and x=−½, or comparing coefficients); A1: A=2; A1: B=3. (b) [8] M1: separates variables correctly; M1 ft: integrates LHS to ln|y|; M1 ft: integrates \( \frac{2}{x-2} \) term to \( 2\ln|x-2| \); M1 ft: integrates \( \frac{3}{2x+1} \) term to \( \frac32\ln|2x+1| \); A1: correct combined equation with +C; M1: correctly exponentiates to y = A(x-2)²(2x+1)^{3/2} form; M1: correctly substitutes y=5, x=3 to find A; A1: fully correct final expression for y.

(a) Differentiating 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 \( \frac{dy}{dx} \) terms: \( \dfrac{dy}{dx}(x+2y) = -(2x+y) \), so \( \dfrac{dy}{dx} = -\dfrac{2x+y}{x+2y} \).
(b) At (1,2): \( 1^2+(1)(2)+2^2 = 1+2+4=7 \) ✓, so (1,2) lies on C. Gradient of tangent at (1,2): \( \dfrac{dy}{dx} = -\dfrac{2(1)+2}{1+2(2)} = -\dfrac{4}{5} \). Gradient of normal = negative reciprocal = \( \dfrac{5}{4} \). Equation of normal: \( y-2 = \dfrac54(x-1) \Rightarrow 4(y-2)=5(x-1) \Rightarrow 4y-8=5x-5 \Rightarrow 5x-4y+3=0 \).

(a) [4] M1: correct implicit differentiation of x² (2x) and y² (2y dy/dx); M1: correct product rule on xy giving \( y+x\frac{dy}{dx} \); M1: correctly collects dy/dx terms; A1: \( \frac{dy}{dx}=-\frac{2x+y}{x+2y} \). (b) [6] B1: verifies (1,2) satisfies the equation (=7); M1 ft: substitutes (1,2) into their (a); A1 ft: tangent gradient = −4/5; M1: normal gradient = 5/4 (negative reciprocal); M1: correct line equation using point (1,2) and gradient 5/4; A1: correctly rearranged to 5x−4y+3=0 (integer coefficients).

(a) \( \dfrac{dx}{dt}=2t-4 \); \( \dfrac{dy}{dt}=3t^2-6t \).
(b) \( \dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{3t^2-6t}{2t-4} = \dfrac{3t(t-2)}{2(t-2)} = \dfrac{3t}{2} \) for \( t\ne2 \) (cancelling the common factor (t−2)).
(c) A horizontal tangent requires \( \dfrac{dy}{dx}=0 \Rightarrow \dfrac{3t}{2}=0 \Rightarrow t=0 \). (Check this is genuine: at t=0, \( \frac{dy}{dt}=0 \) and \( \frac{dx}{dt}=-4\ne0 \), so the tangent is genuinely horizontal, not indeterminate.) At \( t=0 \): \( x=0-0=0 \), \( y=0-0=0 \). The point is (0, 0).
(d) \( \dfrac{d^2y}{dx^2} = \dfrac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt} = \dfrac{\frac{d}{dt}\left(\frac{3t}{2}\right)}{2t-4} = \dfrac{3/2}{2t-4} \). At \( t=0 \): \( \dfrac{d^2y}{dx^2} = \dfrac{3/2}{-4} = -\dfrac{3}{8} \), which is negative, so the point (0,0) is a local maximum.
(e) \( y=0 \Rightarrow t^3-3t^2=0 \Rightarrow t^2(t-3)=0 \Rightarrow t=0 \) (repeated root) or \( t=3 \). At \( t=0 \): point (0,0) (as found above), gradient \( \frac{dy}{dx}=\frac{3(0)}{2}=0 \) (consistent with the horizontal tangent/stationary point found in (c) — the curve touches the x-axis here). At \( t=3 \): \( x=9-12=-3 \), \( y=27-27=0 \), point (−3, 0), gradient \( \frac{dy}{dx}=\frac{3(3)}{2}=4.5 \).

(a) [2] B1: dx/dt=2t−4; B1: dy/dt=3t²−6t. (b) [3] M1 ft: dy/dx = (dy/dt)/(dx/dt); M1: correct factorisation of numerator and denominator; A1: correctly simplified to 3t/2 with cancellation shown. (c) [4] M1 ft: sets their dy/dx = 0; A1: t=0; M1: substitutes t=0 into x and y; A1: point (0,0). (d) [5] M1: attempts d²y/dx² via d(dy/dx)/dt ÷ dx/dt (or equivalent valid method, e.g. considering sign change of dy/dx); M1: correct derivative of 3t/2 (=3/2); A1 ft: correct expression for d²y/dx²; M1: evaluates at t=0; A1: correct conclusion — local maximum, with valid justification (negative second derivative, or equivalent sign-change argument). (e) [5] M1: sets y=0, factorises t²(t−3)=0; A1: t=0 and t=3 found; A1: point (−3,0) at t=3 (with (0,0) at t=0 identified, ft from (c)); M1: gradient at t=3 using 3t/2; A1: both gradients correct (0 at t=0, 4.5 at t=3).

(a) Let \( u=x, \; dv=\sin x\,dx \), so \( du=dx, \; v=-\cos x \). \( \int x\sin x\,dx = -x\cos x - \int(-\cos x)\,dx = -x\cos x + \int \cos x\,dx = -x\cos x + \sin x + C \).
(b) Since \( x\sin x \ge 0 \) for \( 0 \le x \le \pi \), the area equals the definite integral: \( \int_0^\pi x\sin x\,dx = \left[-x\cos x+\sin x\right]_0^\pi = \left(-\pi\cos\pi+\sin\pi\right)-\left(0+0\right) = \left(-\pi(-1)+0\right) - 0 = \pi \). Area = π.
(c) Area between the curve (above) and the line (below) \( = \int_0^{\pi/2}\left(x\sin x - \dfrac{2}{\pi}x\right)dx = \int_0^{\pi/2}x\sin x\,dx - \int_0^{\pi/2}\dfrac{2}{\pi}x\,dx \).
\( \int_0^{\pi/2}x\sin x\,dx = \left[-x\cos x+\sin x\right]_0^{\pi/2} = \left(-\tfrac{\pi}{2}\cos\tfrac{\pi}{2}+\sin\tfrac{\pi}{2}\right)-(0+0) = (0+1)-0=1 \).
\( \int_0^{\pi/2}\dfrac{2}{\pi}x\,dx = \dfrac{2}{\pi}\left[\dfrac{x^2}{2}\right]_0^{\pi/2} = \dfrac{1}{\pi}\left(\dfrac{\pi}{2}\right)^2 = \dfrac{1}{\pi}\cdot\dfrac{\pi^2}{4} = \dfrac{\pi}{4} \).
Area \( = 1 - \dfrac{\pi}{4} \) (≈ 0.215, 3 s.f.).

(a) [4] M1: correct choice u=x, dv=sinx dx; M1: correct du, v found; M1: correct application of the by-parts formula; A1: −xcosx+sinx+C. (b) [4] B1 ft: recognises the region has y≥0 throughout, so area = the integral directly; M1 ft: correctly evaluates their (a) at limits π and 0; A1: correct evaluation at each limit; A1: area = π. (c) [6] M1: sets up area = ∫(curve − line) over [0,π/2]; M1 ft: evaluates ∫x sinx dx over [0,π/2] = 1; M1: correctly integrates (2/π)x to (1/π)x²; A1: evaluates this integral over [0,π/2] = π/4; A1: subtracts correctly; A1: final exact answer \( 1-\frac{\pi}{4} \).

(a) \( fg(x) = f(g(x)) = f\left(\dfrac{1}{x-1}\right) = 2\left(\dfrac{1}{x-1}\right)-3 = \dfrac{2}{x-1}-3 \). Domain: \( x \in \mathbb{R}, x \ne 1 \) (inherited from the domain of g, since f is defined for all real inputs).
(b) f(x)=2x−3 is increasing, so on \( x \ge 2 \) the minimum value is at x=2: \( f(2)=2(2)-3=1 \), and f increases without bound as x increases. Range: \( f(x) \ge 1 \).
(c) Let \( y=2x-3 \Rightarrow x = \dfrac{y+3}{2} \), so \( f^{-1}(x) = \dfrac{x+3}{2} \). Since f (with domain ℝ) has range ℝ, \( f^{-1} \) has domain ℝ and range ℝ.
(d) \( fg(x)=5 \Rightarrow \dfrac{2}{x-1}-3=5 \Rightarrow \dfrac{2}{x-1}=8 \Rightarrow x-1=\dfrac{2}{8}=0.25 \Rightarrow x=1.25 \).

(a) [3] M1: correct composition f(g(x)); A1: \( \frac{2}{x-1}-3 \); B1: domain x≠1 stated. (b) [2] M1: identifies f increasing, evaluates f(2); A1: range f(x)≥1. (c) [3] M1: correct rearrangement to find inverse; A1: \( f^{-1}(x)=\frac{x+3}{2} \); B1: domain and range both correctly stated as ℝ. (d) [3] M1: correctly sets up equation using their (a); M1: correct rearrangement to x−1=0.25; A1: x=1.25.

Taking the direction of A's motion as positive, A has velocity +5 m s⁻¹ and B has velocity −2 m s⁻¹.
(a) By conservation of linear momentum: \( m_Av_A + m_Bv_B = (m_A+m_B)v \Rightarrow 2(5)+3(-2) = (2+3)v \Rightarrow 10-6 = 5v \Rightarrow v = \dfrac{4}{5} = 0.8 \) m s⁻¹, in the original direction of A's motion (since v is positive).
(b) Impulse on B = change in momentum of B \( = m_B(v-v_B) = 3(0.8-(-2)) = 3(2.8) = 8.4 \) N s.

(a) [3] M1: correct conservation of momentum equation with signs; M1: correct substitution 2(5)+3(−2); A1: v=0.8 m s⁻¹ in A's original direction (direction statement required). (b) [3] M1 ft: impulse = m_B × (v − v_B), using their v; M1: correct substitution 3(0.8−(−2)); A1 ft: 8.4 N s.

(a) Displacement \( = \int_0^4 (3t^2-2t)\,dt = \left[t^3-t^2\right]_0^4 = (64-16)-(0-0) = 48 \) m.
(b) \( v(4) = 3(4)^2-2(4) = 48-8 = 40 \) m s⁻¹. \( a=\dfrac{dv}{dt}=6t-2 \), so \( a(4)=6(4)-2=22 \) m s⁻².
(c) For \( t>4 \), P moves with constant acceleration 22 m s⁻², starting (at t=4) with velocity 40 m s⁻¹. Over the 2-second interval from t=4 to t=6, using \( s=ut+\frac12at^2 \) with u=40, a=22, t=2: \( s = 40(2)+\frac12(22)(2)^2 = 80+44 = 124 \) m. Since the velocity remains positive (and increasing) throughout this interval, distance travelled = displacement = 124 m.

(a) [4] M1: correct integral set up \( \int(3t^2-2t)dt \); M1: correct integration \( t^3-t^2 \); M1: correct limits substitution; A1: 48 m. (b) [2] B1: v(4)=40 m s⁻¹; B1: a(4)=22 m s⁻² (from correct differentiation 6t−2). (c) [3] M1 ft: correct use of \( s=ut+\frac12at^2 \) with their u=40, a=22, over 2 seconds; M1: correct substitution/evaluation; A1 ft: 124 m.

Let R_A = normal reaction at the ground (vertical), F_A = friction at the ground (horizontal), R_B = normal reaction from the wall (horizontal, since the wall is smooth). The weight W=200 N acts at the midpoint of the ladder, 3 m from A along the ladder.
(a) Taking moments about A: the moment of R_B (acting horizontally at B, height \( 6\sin60° \) above the ground) balances the moment of W (acting at the midpoint, horizontal distance \( 3\cos60° \) from A). \( R_B \times 6\sin60° = W \times 3\cos60° \Rightarrow R_B = \dfrac{3W\cos60°}{6\sin60°} = \dfrac{W\cos60°}{2\sin60°} = \dfrac{200 \times 0.5}{2 \times 0.8660} = \dfrac{100}{1.7321} = 57.7 \) N (3 s.f.).
(b) Vertically: \( R_A = W = 200 \) N (the wall's reaction is horizontal only). Horizontally: \( F_A = R_B = 57.7 \) N (3 s.f.).
(c) At the point of slipping, \( F_A = \mu R_A \Rightarrow \mu = \dfrac{F_A}{R_A} = \dfrac{57.735}{200} = 0.289 \) (3 s.f.).

(a) [4] M1: correct moment of W about A (\( 3W\cos60° \)); M1: correct moment of R_B about A (\( 6R_B\sin60° \)); M1: correct equation formed and rearranged; A1: R_B=57.7 N (3 s.f.). (b) [3] B1: R_A=200 N (vertical resolution); M1 ft: F_A=R_B (horizontal resolution); A1 ft: F_A=57.7 N. (c) [4] M1: recognises point of slipping means F_A=μR_A; M1 ft: correct substitution of their F_A and R_A; A1 ft: μ=0.289 (3 s.f.); B1: μ correctly identified as a dimensionless ratio in (0,1) (sensible check).

Resolving the initial velocity: \( u_x = 28\cos40° = 21.4 \) m s⁻¹; \( u_y = 28\sin40° = 18.0 \) m s⁻¹ (3 s.f.).
(a) At maximum height, the vertical velocity is zero: \( 0 = u_y - gt \Rightarrow t = \dfrac{u_y}{g} = \dfrac{28\sin40°}{9.8} = 1.84 \) s (3 s.f.).
(b) Maximum height \( = \dfrac{u_y^2}{2g} = \dfrac{(28\sin40°)^2}{2(9.8)} = \dfrac{323.9}{19.6} = 16.5 \) m (3 s.f.).
(c) By symmetry, the time of flight is \( 2t = \dfrac{2u_y}{g} \), and the range \( = u_x \times \dfrac{2u_y}{g} = \dfrac{2u_xu_y}{g} = \dfrac{u^2\sin(2\times40°)}{g} = \dfrac{28^2\sin80°}{9.8} = \dfrac{784\times0.9848}{9.8} = 78.8 \) m (3 s.f.).
(d) With \( x=u_xt \Rightarrow t=\dfrac{x}{u_x} \), and \( y=u_yt-\frac12gt^2 \), substituting: \( y = u_y\dfrac{x}{u_x} - \dfrac12g\dfrac{x^2}{u_x^2} \). Since \( \dfrac{u_y}{u_x}=\tan40° \) and \( u_x=28\cos40° \), this gives \( y=x\tan40°-\dfrac{gx^2}{2(28\cos40°)^2} = x\tan40°-\dfrac{gx^2}{2(28)^2\cos^240°} \), as required. Setting \( y=0 \) (landing point): \( x\left(\tan40° - \dfrac{gx}{2(28)^2\cos^240°}\right)=0 \), so \( x=0 \) (launch point) or \( x = \dfrac{2(28)^2\cos^240°\tan40°}{g} = \dfrac{2(28)^2\sin40°\cos40°}{g} = \dfrac{(28)^2\sin80°}{g} = 78.8 \) m (3 s.f.), which matches part (c).

(a) [3] M1: resolves u_y=28sin40°; M1: uses v=u−gt with v=0; A1: t=1.84 s. (b) [2] M1 ft: uses \( u_y^2/(2g) \); A1: 16.5 m. (c) [3] M1: correct range formula \( u^2\sin(2\theta)/g \) (or equivalent via time of flight × u_x); M1: correct substitution; A1: 78.8 m. (d) [3] M1: correctly eliminates t using \( t=x/u_x \) substituted into \( y=u_yt-\frac12gt^2 \); A1: correctly simplified to the given form; A1: correctly sets y=0 and solves to recover x≈78.8 m, consistent with (c).

(a) \( \mathbf{r} = \int \mathbf{v}\,dt = (4t-t^2)\mathbf{i}+(t^3-6t)\mathbf{j}+\mathbf{C} \). Since P starts at the origin, \( \mathbf{r}(0)=\mathbf{0} \Rightarrow \mathbf{C}=\mathbf{0} \). So \( \mathbf{r} = (4t-t^2)\mathbf{i}+(t^3-6t)\mathbf{j} \).
(b) P moves due east when the north (\(\mathbf{j}\)) component of velocity is zero (and the east component is positive): \( 3t^2-6=0 \Rightarrow t^2=2 \Rightarrow t=\sqrt2 \) (taking t>0). Check: at \( t=\sqrt2 \), the i-component is \( 4-2\sqrt2 \approx1.17>0 \), confirming motion is due east (not west) at this instant.
(c) At \( t=3 \): \( \mathbf{v} = (4-6)\mathbf{i}+(27-6)\mathbf{j} = -2\mathbf{i}+21\mathbf{j} \). Speed \( = \sqrt{(-2)^2+21^2} = \sqrt{4+441} = \sqrt{445} = 21.1 \) m s⁻¹ (3 s.f.).
(d) At \( t=1 \): \( \mathbf{v} = (4-2)\mathbf{i}+(3-6)\mathbf{j} = 2\mathbf{i}-3\mathbf{j} \), i.e. 2 units east and 3 units south. This direction lies in the south-east quadrant. The bearing (measured clockwise from north) is \( 180° - \tan^{-1}\left(\dfrac{2}{3}\right) = 180°-33.7° = 146.3° \), so the bearing is 146° (nearest degree).

(a) [4] M1: integrates i-component to 4t−t²; M1: integrates j-component to t³−6t; A1: correct expression (unsimplified constant included); A1: correctly uses r(0)=0 to give C=0, final answer stated. (b) [3] M1: sets j-component of v to zero; A1: t=√2 (or 1.41); B1: checks/confirms i-component positive at this t (due east, not west). (c) [3] M1 ft: finds v at t=3; M1: correct speed formula √(vi²+vj²); A1: 21.1 m s⁻¹ (3 s.f.). (d) [3] M1 ft: finds v at t=1 (2i−3j); M1: correct bearing method (e.g. 180°−arctan(2/3), or equivalent using arctan/atan2 reasoning) with quadrant correctly identified (SE); A1: 146° (nearest degree).

(a) \( H_0: \rho=0 \) (no linear correlation in the population); \( H_1: \rho>0 \) (positive linear correlation in the population) — a one-tailed test.
(b) Since the sample value \( r=0.62 \) is greater than the critical value 0.4973, the result is significant: reject \( H_0 \). There is evidence, at the 5% significance level, of positive linear correlation between the two variables.

(a) [1] B1: both hypotheses correct (H0: ρ=0; H1: ρ>0), in terms of ρ. (b) [2] M1: correct comparison — 0.62 > 0.4973; A1: correct conclusion in context — reject H0, significant evidence of positive correlation.

(a) [Tree diagram description, as text tables/diagrams cannot be embedded: first branch — P(A)=0.6, P(B)=0.4. From A: P(Defective|A)=0.05, P(Not defective|A)=0.95. From B: P(Defective|B)=0.08, P(Not defective|B)=0.92.]
(b) \( P(D) = P(A)P(D|A)+P(B)P(D|B) = (0.6)(0.05)+(0.4)(0.08) = 0.03+0.032 = 0.062 \).
(c) By Bayes' formula: \( P(B|D) = \dfrac{P(B)P(D|B)}{P(D)} = \dfrac{0.4 \times 0.08}{0.062} = \dfrac{0.032}{0.062} = \dfrac{16}{31} \approx 0.516 \) (3 s.f.).
(d) Let \( p=P(D)=0.062 \). For two independent components, P(exactly one defective) \( = 2 \times p \times (1-p) = 2(0.062)(0.938) = 0.1163 \) (4 s.f.).

(a) [2] B1: correct first-branch probabilities (0.6, 0.4); B1: correct second-branch conditional probabilities on each branch (0.05/0.95 from A; 0.08/0.92 from B). (b) [3] M1: correct structure P(A)P(D|A)+P(B)P(D|B); M1: correct substitution; A1: 0.062. (c) [4] M1 ft: correct Bayes' formula set up P(B)P(D|B)/P(D); M1: correct numerator 0.032; A1 ft: correctly evaluated using their (b); A1: 0.516 (3 s.f.) or exact fraction 16/31. (d) [4] M1 ft: recognises P(exactly one) = 2p(1−p) form; M1 ft: correct substitution using their p; A1: correct evaluation; A1: 0.116 (3 s.f., ft).

Let X~N(175, 7²) model an individual male's height.
(a) \( P(X>185) = P\left(Z>\dfrac{185-175}{7}\right) = P(Z>1.4286) = 0.0766 \) (3 s.f.).
(b) We require h such that \( P(X>h)=0.10 \), i.e. \( P(Z>z)=0.10 \Rightarrow z=1.2816 \) (from tables). \( h = 175+1.2816(7) = 175+8.97 = 184 \) cm (3 s.f.).
(c) Let \( p=P(X>185)=0.0766 \) (from (a)). The number taller than 185 cm out of 3 independent males follows \( \mathrm{B}(3,p) \). \( P(\text{exactly 2}) = \binom{3}{2}p^2(1-p) = 3(0.0766)^2(0.9234) = 0.0162 \) (3 s.f.).
(d) The sample mean of n=4 independent heights, \( \bar X \sim N\left(175, \dfrac{7^2}{4}\right) \), i.e. standard error \( = \dfrac{7}{\sqrt4}=3.5 \). \( P(\bar X>178) = P\left(Z>\dfrac{178-175}{3.5}\right) = P(Z>0.8571) = 0.196 \) (3 s.f.).

(a) [3] M1: standardises correctly, z=(185−175)/7; M1: correct z-value (1.43); A1: 0.0766 (3 s.f.). (b) [3] M1: uses P(Z>z)=0.10 to find z=1.2816 (or 1.28); M1: correct rearrangement h=175+z(7); A1: 184 cm (3 s.f.). (c) [3] M1 ft: correct binomial set-up B(3, their p); M1: correct substitution into \( 3p^2(1-p) \); A1 ft: 0.0162 (3 s.f., ft their p). (d) [3] M1: correct standard error 7/√4=3.5; M1: correct standardisation (178−175)/3.5; A1: 0.196 (3 s.f.).

Let X = number of defective components in a random sample of 25, and let p = the true proportion of defective components produced.
(a) \( H_0: p=0.1 \); \( H_1: p>0.1 \) (one-tailed test, testing the inspector's suspicion that the true proportion is higher).
(b) Under \( H_0 \), \( X \sim B(25, 0.1) \). We require the smallest c such that \( P(X\ge c)<0.05 \). Calculating cumulative probabilities: \( P(X\ge5)=0.098 \) (too large, >0.05); \( P(X\ge6)=0.0334 \) (< 0.05). So the critical region is \( X \ge 6 \) (i.e. \( \{6,7,\ldots,25\} \)), with actual significance level 0.0334 (3 s.f.).
(c) The observed value, x=6, lies within the critical region \( X\ge6 \), so the result is significant at the 5% level: reject \( H_0 \). There is evidence, at the 5% significance level, that the true proportion of defective components exceeds 10%, supporting the inspector's suspicion (contradicting the manufacturer's claim).
(d) The probability that any individual component is defective must be constant (the same, 0.10 under H₀) for every component in the sample, and whether or not one component is defective must be independent of whether any other component is defective (e.g. components are produced independently, not in a way where one defect causes/predicts another).

(a) [2] B1: p defined as the (population) proportion of defective components; B1: H0: p=0.1, H1: p>0.1, both correct. (b) [6] M1: recognises X~B(25,0.1) under H0; M1: calculates or uses P(X≥5); A1: P(X≥5)≈0.098 (>0.05, so 5 rejected); M1: calculates or uses P(X≥6); A1: P(X≥6)≈0.0334 (<0.05); A1: critical region correctly stated as X≥6. (c) [3] M1 ft: compares observed x=6 to their critical region; A1 ft: correctly concludes significant/reject H0; A1: correct conclusion stated in context (evidence proportion exceeds 10%). (d) [2] B1: constant probability of a component being defective; B1: independence between components.

Let μ be the true mean mass (in g) of bags currently produced by the machine.
(a) \( H_0: \mu=500 \); \( H_1: \mu<500 \) (one-tailed test, testing whether the mean mass has decreased).
(b) Under \( H_0 \), the sample mean \( \bar X \sim N\left(500, \dfrac{8^2}{16}\right) \), so the standard error \( = \dfrac{8}{\sqrt{16}} = 2 \). Test statistic: \( z = \dfrac{\bar x-\mu_0}{\sigma/\sqrt n} = \dfrac{496-500}{2} = \dfrac{-4}{2} = -2.00 \).
(c) Since \( z=-2.00 \) is less than the critical value \( -1.6449 \) (i.e. it lies further into the rejection region, in the left/lower tail), the result is significant at the 5% level: reject \( H_0 \). There is evidence, at the 5% significance level, that the mean mass of bags produced by the machine has decreased from 500 g.

(a) [2] B1: μ defined as the population mean mass; B1: H0: μ=500, H1: μ<500, both correct. (b) [3] M1: correct standard error 8/√16=2; M1: correct standardisation formula used; A1: z=−2.00. (c) [4] M1 ft: correctly compares their z to −1.6449; A1 ft: −2.00 < −1.6449, so result is significant; A1: correct decision — reject H0; A1: correct conclusion stated in context (evidence mean mass has decreased).