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

(a) \( \overrightarrow{AB}=\mathbf{b}-\mathbf{a}=(5-2)\mathbf{i}+(3-(-1))\mathbf{j}=3\mathbf{i}+4\mathbf{j} \). (b) \( |\overrightarrow{AB}|=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5 \). (c) \( M=\dfrac{\mathbf{a}+\mathbf{b}}{2}=\dfrac{(2+5)\mathbf{i}+(-1+3)\mathbf{j}}{2}=\dfrac{7\mathbf{i}+2\mathbf{j}}{2}=3.5\mathbf{i}+\mathbf{j} \). (d) A unit vector in the direction of \( \overrightarrow{AB} \) is \( \dfrac{\overrightarrow{AB}}{|\overrightarrow{AB}|}=\dfrac{3\mathbf{i}+4\mathbf{j}}{5}=0.6\mathbf{i}+0.8\mathbf{j} \). Final answer: (a) \( 3\mathbf{i}+4\mathbf{j} \); (b) \( 5 \); (c) \( 3.5\mathbf{i}+\mathbf{j} \); (d) \( 0.6\mathbf{i}+0.8\mathbf{j} \).

(a) MW1: correct \( \mathbf{i} \)-component; W1: correct \( \mathbf{j} \)-component. (2 marks) (b) M1: correct method (ECF from (a)); W1: correct magnitude \( 5 \). (2 marks) (c) M1: correct method \( (\mathbf{a}+\mathbf{b})/2 \); W1: correct \( M=3.5\mathbf{i}+\mathbf{j} \). (2 marks) (d) M1: correct method, dividing \( \overrightarrow{AB} \) by its magnitude (ECF); W1: correct unit vector \( 0.6\mathbf{i}+0.8\mathbf{j} \). (2 marks) Total 8 marks.

(a) Rearranging \( 2x+y-7=0 \) gives \( y=-2x+7 \), so the gradient of \( l_1 \) is \( -2 \). (b) A line perpendicular to \( l_1 \) has gradient \( \tfrac12 \) (the negative reciprocal of \( -2 \)). Through \( A(3,4) \): \( y-4=\tfrac12(x-3) \), i.e. \( y=\tfrac12x-1.5+4=\tfrac12x+2.5 \). (c) Setting the two equations equal: \( -2x+7=\tfrac12x+2.5 \). Multiplying through by 2: \( -4x+14=x+5 \), so \( 9=5x \), giving \( x=1.8 \). Substituting: \( y=\tfrac12(1.8)+2.5=0.9+2.5=3.4 \). Final answer: (a) gradient \( =-2 \); (b) \( y=\tfrac12x+2.5 \); (c) \( (1.8,3.4) \).

(a) W1: correct gradient \( -2 \). (1 mark) (b) M1: correct perpendicular gradient \( \tfrac12 \) (ECF from (a)); M1: correct method to form the equation through A; W1: fully correct equation \( y=\tfrac12x+2.5 \). (3 marks) (c) M1: sets the two equations equal; M1: valid method to solve for \( x \); W1: correct \( x=1.8 \); W1: correct \( y=3.4 \). (4 marks) Total 8 marks.

(a) Completing the square: \( x^2-6x+y^2+4y=12 \), i.e. \( (x-3)^2-9+(y+2)^2-4=12 \), i.e. \( (x-3)^2+(y+2)^2=25 \). Comparing with \( (x-a)^2+(y-b)^2=r^2 \), the centre is \( (3,-2) \) and the radius is \( \sqrt{25}=5 \). (b) The distance from the centre \( (3,-2) \) to the point \( (7,1) \) is \( \sqrt{(7-3)^2+(1-(-2))^2}=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5 \). Since this distance equals the radius \( (5) \), the point \( (7,1) \) lies exactly ON the circle. Final answer: (a) centre \( (3,-2) \), radius \( 5 \); (b) \( (7,1) \) lies on the circle.

(a) M1: correctly completes the square in \( x \); M1: correctly completes the square in \( y \); W1: correct centre \( (3,-2) \); W1: correct radius \( 5 \). (4 marks) (b) M1: correct method, distance from centre to the given point (ECF from (a)); MW1: correct distance \( 5 \); W1: correct conclusion, comparing the distance to the radius and stating the point lies on the circle. (3 marks) Total 7 marks.

(a) By the factor theorem, \( f(-2)=2(-2)^3-3(-2)^2-11(-2)+6=2(-8)-3(4)+22+6=-16-12+22+6=0 \). Since \( f(-2)=0 \), \( (x+2) \) is a factor of \( f(x) \). (b) Dividing \( f(x) \) by \( (x+2) \): \( f(x)=(x+2)(2x^2-7x+3) \). The quadratic factor \( 2x^2-7x+3 \) factorises as \( (x-3)(2x-1) \), since \( (x-3)(2x-1)=2x^2-x-6x+3=2x^2-7x+3 \). So \( f(x)=(x+2)(x-3)(2x-1) \). (c) The three roots of \( f(x)=0 \) are \( x=-2 \), \( x=\tfrac12 \) and \( x=3 \), in increasing order \( -2<\tfrac12<3 \). For any \( x \) with \( -2-2 \); the factor \( (x-3) \) is negative, since \( x<\tfrac12<3 \); and the factor \( (2x-1) \) is negative, since \( x<\tfrac12 \) means \( 2x<1 \), i.e. \( 2x-1<0 \). Therefore \( f(x)=(x+2)(x-3)(2x-1) = (\text{positive})\times(\text{negative})\times(\text{negative}) \), and the product of a positive number and two negative numbers is positive. Hence \( f(x)>0 \) for all \( x \) in \( -2

(a) M1: substitutes \( x=-2 \) into \( f(x) \) with working shown; W1: correctly obtains \( f(-2)=0 \) and states the factor theorem conclusion. (2 marks) (b) M1: valid method to divide \( f(x) \) by \( (x+2) \); W1: correct quadratic factor \( 2x^2-7x+3 \); M1: valid method to factorise the quadratic; W1: correct final factorisation \( (x+2)(x-3)(2x-1) \); W1: fully correct answer clearly stated. (5 marks) (c) M1: identifies the three roots and their order (ECF from (b)); M1: correctly determines the sign of each of the three factors within the given interval; W1: correctly combines the three signs to a positive product; W1: clear, complete concluding statement that \( f(x)>0 \) throughout the interval, with valid reasoning (not simply asserted). (4 marks) Total 11 marks.

(a) Completing the square: \( x^2+6x+11=(x+3)^2-9+11=(x+3)^2+2 \). Since \( (x+3)^2\ge0 \) for all real \( x \), it follows that \( (x+3)^2+2\ge2 \), and so \( x^2+6x+11\ge2>0 \) for all real \( x \), as required. (b) The inequality \( x^2+6x+11\ge2x+15 \) rearranges to \( x^2+4x-4\ge0 \). Using the quadratic formula, the roots of \( x^2+4x-4=0 \) are \( x=\dfrac{-4\pm\sqrt{16+16}}{2}=\dfrac{-4\pm\sqrt{32}}{2}=-2\pm2\sqrt2 \). Since the coefficient of \( x^2 \) is positive, the quadratic \( x^2+4x-4 \) is a upward-opening parabola, so it is \( \ge0 \) outside the roots: \( x\le-2-2\sqrt2 \) or \( x\ge-2+2\sqrt2 \). (c) The complementary region, where \( x^2+4x-4<0 \) (i.e. where the original inequality is reversed), lies between the two roots: \( -2-2\sqrt2

(a) M1: correct completed-square method; W1: correct form \( (x+3)^2+2 \); W1: correctly explains \( (x+3)^2\ge0 \) so the expression is \( \ge2 \); W1: complete, valid concluding statement that the expression is therefore always positive. (4 marks) (b) M1: correctly rearranges to \( x^2+4x-4\ge0 \); M1: valid method (quadratic formula) to find the roots; W1: correct roots \( -2\pm2\sqrt2 \); W1: correct final inequality with correct direction (outside the roots). (4 marks) (c) M1: recognises the complementary region lies between the roots (ECF from (b)); W1: correctly states the region is where the expression is negative; W1: correct final answer \( -2-2\sqrt2

(a) Using \( \cos^2\theta=1-\sin^2\theta \): \( 3(1-\sin^2\theta)-7\sin\theta+3=0 \), i.e. \( 3-3\sin^2\theta-7\sin\theta+3=0 \), i.e. \( -3\sin^2\theta-7\sin\theta+6=0 \). Multiplying by \( -1 \): \( 3\sin^2\theta+7\sin\theta-6=0 \). Let \( s=\sin\theta \); using the quadratic formula: \( s=\dfrac{-7\pm\sqrt{49+72}}{6}=\dfrac{-7\pm\sqrt{121}}{6}=\dfrac{-7\pm11}{6} \), giving \( s=\dfrac{4}{6}=\dfrac23 \) or \( s=\dfrac{-18}{6}=-3 \). Since \( \sin\theta \) must lie in \( [-1,1] \), the solution \( s=-3 \) is rejected. So \( \sin\theta=\dfrac23 \): the principal value is \( \theta=\arcsin\left(\tfrac23\right)\approx41.8^\circ \), and since sine is also positive in the second quadrant, \( \theta=180^\circ-41.8^\circ=138.2^\circ \). So the solutions are \( \theta\approx41.8^\circ \) or \( 138.2^\circ \). (b) The general solutions are \( \theta=41.8^\circ+360^\circ n \) or \( \theta=138.2^\circ+360^\circ n \), for integer \( n \). Checking values of \( n \) that place \( \theta \) within \( -360^\circ\le\theta\le360^\circ \): for \( n=0 \): \( 41.8^\circ \) and \( 138.2^\circ \) (both in range); for \( n=-1 \): \( 41.8^\circ-360^\circ=-318.2^\circ \) and \( 138.2^\circ-360^\circ=-221.8^\circ \) (both in range); for \( n=1 \) or \( n=-2 \), the resulting values fall outside \( [-360^\circ,360^\circ] \). So there are \( 4 \) solutions in total: \( \theta\approx-318.2^\circ,-221.8^\circ,41.8^\circ,138.2^\circ \). Final answer: (a) \( \theta\approx41.8^\circ \) or \( 138.2^\circ \); (b) 4 solutions.

(a) M1: correctly uses \( \cos^2\theta=1-\sin^2\theta \) to form an equation in \( \sin\theta \) only; MW1: correctly simplifies to \( 3\sin^2\theta+7\sin\theta-6=0 \) (or equivalent); M1: valid method to solve the quadratic; W1: correct values \( s=\tfrac23 \) with \( s=-3 \) rejected (with reasoning); W1: correct principal value \( \theta\approx41.8^\circ \); W1: correct second solution \( \theta\approx138.2^\circ \); (up to 7 marks, allocate across the above steps) (7 marks) (b) M1: recognises the \( 360^\circ \) periodicity and considers values of \( n \) beyond the given range; M1: correctly identifies the two further solutions from \( n=-1 \); W1: correctly checks no further values fall in range for other \( n \); W1: correct total of 4 solutions stated. (4 marks) Total 11 marks.

(a) Taking natural logarithms of both sides: \( \ln\left(5^{x+1}\right)=\ln\left(3^{2x}\right) \), so \( (x+1)\ln5=2x\ln3 \). Expanding: \( x\ln5+\ln5=2x\ln3 \), so \( \ln5=2x\ln3-x\ln5=x(2\ln3-\ln5) \), giving \( x=\dfrac{\ln5}{2\ln3-\ln5} \). Evaluating: \( \ln5\approx1.6094 \), \( \ln3\approx1.0986 \), so \( 2\ln3-\ln5\approx2.1972-1.6094\approx0.5878 \), giving \( x\approx\dfrac{1.6094}{0.5878}\approx2.74 \) (3 s.f.). (b) Using the laws of logarithms, \( 2\ln(x)=\ln(x^2) \), so the equation becomes \( \ln(x^2)-\ln(x+3)=\ln4 \), i.e. \( \ln\!\left(\dfrac{x^2}{x+3}\right)=\ln4 \), so \( \dfrac{x^2}{x+3}=4 \). Multiplying both sides by \( (x+3) \): \( x^2=4(x+3)=4x+12 \), i.e. \( x^2-4x-12=0 \), which factorises as \( (x-6)(x+2)=0 \), giving \( x=6 \) or \( x=-2 \). Since \( \ln(x) \) requires \( x>0 \), the solution \( x=-2 \) is not valid and must be rejected. The solution \( x=6 \) satisfies \( x>0 \) and is therefore valid. Check: \( 2\ln6-\ln9=\ln36-\ln9=\ln(36/9)=\ln4 \), confirming the solution. Final answer: (a) \( x\approx2.74 \) (3 s.f.); (b) \( x=6 \).

(a) M1: takes logarithms of both sides; MW1: correctly expands to \( x\ln5+\ln5=2x\ln3 \) (or equivalent); M1: correctly collects terms in \( x \) and factorises/rearranges; W1: correct unrounded expression/value for \( x \); W1: correct final answer \( x\approx2.74 \) (3 s.f.). (5 marks) (b) M1: correctly combines the logarithms on the left-hand side using the power and subtraction laws; MW1: correctly converts to \( \dfrac{x^2}{x+3}=4 \); M1: correctly rearranges to the 3-term quadratic \( x^2-4x-12=0 \); M1: valid method to solve the quadratic; W1: correctly rejects \( x=-2 \) with valid domain reasoning (\( x>0 \) required); W1: correct final answer \( x=6 \). (6 marks) Total 11 marks.

(a) After cutting a square of side \( x \) from each corner and folding up the sides, the base of the box is a square of side \( 20-2x \), and the height of the box is \( x \). So \( V=x(20-2x)^2 \). Expanding: \( (20-2x)^2=400-80x+4x^2 \), so \( V=x(400-80x+4x^2)=400x-80x^2+4x^3=4x^3-80x^2+400x \), as required. (b) Differentiating: \( \dfrac{dV}{dx}=12x^2-160x+400 \). (c) At a stationary point, \( \dfrac{dV}{dx}=0 \): \( 12x^2-160x+400=0 \), which simplifies (dividing by 4) to \( 3x^2-40x+100=0 \). Using the quadratic formula: \( x=\dfrac{40\pm\sqrt{1600-1200}}{6}=\dfrac{40\pm\sqrt{400}}{6}=\dfrac{40\pm20}{6} \), giving \( x=10 \) or \( x=\dfrac{10}{3} \). Since \( 00 \), confirming this is a minimum, consistent with \( V=0 \) there.) So \( x=\tfrac{10}{3} \) gives the maximum volume. (d) Substituting \( x=\tfrac{10}{3} \) into \( V=4x^3-80x^2+400x \): \( V=\dfrac{16000}{27}\approx592.59 \), so the maximum volume is approximately \( 593 \) cm\( ^3 \) (3 s.f.). Final answer: (a) shown; (b) \( \dfrac{dV}{dx}=12x^2-160x+400 \); (c) \( x=\tfrac{10}{3} \) gives the maximum; (d) maximum volume \( \approx593 \) cm\( ^3 \).

(a) M1: correctly identifies the base side as \( 20-2x \) and height as \( x \); MW1: correctly forms \( V=x(20-2x)^2 \); M1: correctly expands \( (20-2x)^2 \); W1: fully correct expanded form \( 4x^3-80x^2+400x \), clearly shown. (4 marks) (b) MW1: correct differentiation of each term; W1: fully correct \( \dfrac{dV}{dx}=12x^2-160x+400 \). (3 marks) (c) M1: sets \( \dfrac{dV}{dx}=0 \) and simplifies to a 3-term quadratic; M1: valid method to solve (quadratic formula or factorising); W1: correct values \( x=10 \) and \( x=\tfrac{10}{3} \); M1: correctly identifies \( x=\tfrac{10}{3} \) as the relevant value within the domain \( 0

(a) Setting the two expressions for \( y \) equal: \( x^2=8-x^2 \), so \( 2x^2=8 \), i.e. \( x^2=4 \), giving \( x=2 \) or \( x=-2 \). At \( x=2 \): \( y=2^2=4 \). At \( x=-2 \): \( y=(-2)^2=4 \). So the points of intersection are \( (-2,4) \) and \( (2,4) \). (b) \( (8-x^2)-x^2=8-2x^2 \). Integrating: \( \displaystyle\int(8-2x^2)\,dx=8x-\dfrac{2x^3}{3}+c \). (c) Since \( C_2 \) lies above \( C_1 \) between the two intersection points (e.g. at \( x=0 \), \( C_1 \) gives \( y=0 \) and \( C_2 \) gives \( y=8 \), so \( C_2 \) is above), the enclosed area is given by \( \displaystyle\int_{-2}^{2}\big[(8-x^2)-x^2\big]\,dx=\Big[8x-\dfrac{2x^3}{3}\Big]_{-2}^{2} \). At \( x=2 \): \( 8(2)-\dfrac{2(8)}{3}=16-\dfrac{16}{3}=\dfrac{48-16}{3}=\dfrac{32}{3} \). At \( x=-2 \): \( 8(-2)-\dfrac{2(-8)}{3}=-16+\dfrac{16}{3}=\dfrac{-48+16}{3}=-\dfrac{32}{3} \). Area \( =\dfrac{32}{3}-\left(-\dfrac{32}{3}\right)=\dfrac{64}{3} \) (square units). Final answer: (a) \( (-2,4) \) and \( (2,4) \); (b) \( 8x-\dfrac{2x^3}{3}+c \); (c) area \( =\dfrac{64}{3} \).

(a) M1: sets the two expressions for \( y \) equal; M1: correctly solves for \( x \); W1: correct x-values \( x=\pm2 \); W1: both correct coordinate pairs \( (-2,4) \) and \( (2,4) \). (4 marks) (b) MW1: correctly simplifies the integrand to \( 8-2x^2 \); MW1: correctly integrates \( 8 \) to \( 8x \); W1: correctly integrates \( -2x^2 \) to \( -\tfrac{2x^3}{3} \), with \( +c \). (4 marks) (c) M1: identifies which curve lies above the other in the interval (e.g. by testing a value); M1: sets up the definite integral with correct limits (ECF from (a)); MW1: correct evaluation at the upper limit; MW1: correct evaluation at the lower limit; M1: correct method to subtract; W1: correct final area \( \dfrac{64}{3} \); W1: fully correct working shown throughout with no sign errors. (8 marks) Total 16 marks.

(a) Adding the components: \( \mathbf{i} \)-components: \( 4-2-1=1 \); \( \mathbf{j} \)-components: \( -3+5-4=-2 \). So the resultant force is \( \mathbf{i}-2\mathbf{j} \) N. (b) The magnitude of the resultant is \( |\mathbf{i}-2\mathbf{j}|=\sqrt{1^2+(-2)^2}=\sqrt{5}\approx2.236 \) N. By Newton's second law, \( a=\dfrac{|\mathbf{F}|}{m}=\dfrac{\sqrt5}{0.5}\approx\dfrac{2.236}{0.5}\approx4.47 \text{ m s}^{-2} \) (3 s.f.). Final answer: (a) \( \mathbf{i}-2\mathbf{j} \) N; (b) \( 4.47 \text{ m s}^{-2} \).

(a) MW1: correct \( \mathbf{i} \)-component \( 1 \); W1: correct \( \mathbf{j} \)-component \( -2 \). (2 marks) (b) M1: correct method for the magnitude of the resultant (ECF from (a)); W1: correct unrounded magnitude \( \sqrt5\approx2.24 \) N; M1: applies \( a=F/m \) with \( m=0.5 \); W1: correct final answer \( 4.47 \text{ m s}^{-2} \) (3 s.f.). (4 marks) Total 6 marks.

(a) Using \( a=\dfrac{v-u}{t} \) with \( u=0 \), \( v=8 \), \( t=20 \): \( a=\dfrac{8-0}{20}=0.4 \text{ m s}^{-2} \). (b) The velocity-time graph consists of three straight-line stages; the total distance travelled is the total area under the graph. Stage 1 (acceleration): \( s_1=\tfrac12(0+8)(20)=80 \) m. Stage 2 (constant speed): \( s_2=8\times30=240 \) m. Stage 3 (deceleration): \( s_3=\tfrac12(8+0)(10)=40 \) m. Total distance \( =80+240+40=360 \) m. Final answer: (a) \( 0.4 \text{ m s}^{-2} \); (b) \( 360 \) m.

(a) M1: correct use of \( a=(v-u)/t \); W1: correct answer \( 0.4 \text{ m s}^{-2} \). (2 marks) (b) M1: correct method for stage 1 (e.g. \( s=\tfrac12(u+v)t \), or area under the graph); W1: correct \( s_1=80 \) m; MW1: correct \( s_2=240 \) m for the constant-speed stage; M1: correct method for stage 3; W1: correct \( s_3=40 \) m and correct total \( 360 \) m. (5 marks) Total 7 marks.

(a) Resolving perpendicular to the plane: the weight's component into the plane is \( 6g\cos32^\circ \), and the horizontal force \( P \) also has a component into the plane of \( P\sin32^\circ \) (since \( P \) is horizontal and the plane is inclined at \( 32^\circ \)). These are balanced by the normal reaction: \( R=6g\cos32^\circ+P\sin32^\circ \), as required. (b) Resolving along the plane: the weight's component down the plane is \( 6g\sin32^\circ \), the horizontal force's component up the plane is \( P\cos32^\circ \), and since the block is on the point of sliding down, friction acts up the plane at its limiting value \( F=\mu R \). For equilibrium along the plane: \( 6g\sin32^\circ = P\cos32^\circ + F \), i.e. \( 6g\sin32^\circ-P\cos32^\circ=\mu R=0.3\left(6g\cos32^\circ+P\sin32^\circ\right) \). (c) Substituting \( g=9.8 \): \( 6g=58.8 \). So \( 58.8\sin32^\circ-P\cos32^\circ=0.3\left(58.8\cos32^\circ+P\sin32^\circ\right) \). Evaluating \( \sin32^\circ\approx0.52992 \), \( \cos32^\circ\approx0.84805 \): \( 58.8(0.52992)-P(0.84805)=0.3\left[58.8(0.84805)+P(0.52992)\right] \), i.e. \( 31.16-0.84805P=0.3(49.865+0.52992P) \), i.e. \( 31.16-0.84805P=14.960+0.15898P \). Rearranging: \( 31.16-14.960=0.84805P+0.15898P \), i.e. \( 16.20=1.00703P \), so \( P\approx16.1 \) N (3 s.f.). Final answer: (a) shown; (b) \( 6g\sin32^\circ-P\cos32^\circ=0.3(6g\cos32^\circ+P\sin32^\circ) \); (c) \( P\approx16.1 \) N.

(a) M1: correctly identifies the weight's perpendicular component \( 6g\cos32^\circ \); W1: correctly identifies \( P \)'s perpendicular component \( P\sin32^\circ \); W1: correctly combines to show \( R=6g\cos32^\circ+P\sin32^\circ \). (3 marks) (b) M1: correctly resolves along the plane with correct terms for weight and \( P \); M1: correctly applies \( F=\mu R \) at limiting equilibrium; W1: fully correct equation formed. (3 marks) (c) M1: correct substitution of numerical values (ECF from (a) and (b)); M1: valid algebraic method to collect terms in \( P \) and solve; W1: correct final answer \( P\approx16.1 \) N (3 s.f.). (3 marks) Total 9 marks.

Let \( a \) be the magnitude of the acceleration and \( T \) the tension. The 7 kg particle accelerates downwards and the 4 kg particle accelerates upwards, both with magnitude \( a \). For the 7 kg particle (downwards positive): \( 7g-T=7a \). For the 4 kg particle (upwards positive): \( T-4g=4a \). (a) Adding the two equations to eliminate \( T \): \( 7g-4g=7a+4a \), i.e. \( 3g=11a \), so \( a=\dfrac{3(9.8)}{11}=\dfrac{29.4}{11}\approx2.67 \text{ m s}^{-2} \) (3 s.f.). (b) Substituting into the 4 kg particle's equation: \( T=4g+4a\approx4(9.8)+4(2.673)\approx39.2+10.69\approx49.9 \) N (3 s.f.) (using the unrounded value of \( a \)). (c) Using \( v^2=u^2+2as \) with \( u=0 \), \( a\approx2.673 \), \( s=2 \): \( v^2=0+2(2.673)(2)\approx10.69 \), so \( v\approx\sqrt{10.69}\approx3.27 \text{ m s}^{-1} \) (3 s.f.). (d) Since the string passes over a smooth pulley with both sides vertical, the tension \( T \) acts downwards on the pulley from each side, so the total force exerted by the string on the pulley is \( 2T\approx2(49.89)\approx99.8 \) N (3 s.f.). Final answer: (a) \( a\approx2.67 \text{ m s}^{-2} \); (b) \( T\approx49.9 \) N; (c) \( v\approx3.27 \text{ m s}^{-1} \); (d) force on pulley \( \approx99.8 \) N.

M1: correct equation of motion for the 7 kg particle; M1: correct equation of motion for the 4 kg particle. (a) M1: valid method to eliminate \( T \) (e.g. adding); W1: correct final answer \( a\approx2.67 \text{ m s}^{-2} \) (3 s.f.). (4 marks total for the two set-up marks plus part (a)) (b) MW1: correct substitution into an equation of motion; W1: correct final answer \( T\approx49.9 \) N (3 s.f.), ideally checked using the other particle's equation. (3 marks) (c) M1: correct use of \( v^2=u^2+2as \) (ECF from (a)); W1: correct final answer \( v\approx3.27 \text{ m s}^{-1} \) (3 s.f.). (3 marks) (d) M1: recognises the total force on the pulley is \( 2T \) (ECF from (b)); W1: correct final answer \( \approx99.8 \) N (3 s.f.). (3 marks) Total 13 marks.

The class frequencies are: \( 0

(a) M1: identifies the median lies in the class \( 10

(a) With \( n=5 \): \( S_{xx}=\sum x^2-\dfrac{(\sum x)^2}{n}=1375-\dfrac{75^2}{5}=1375-1125=250 \). \( S_{yy}=\sum y^2-\dfrac{(\sum y)^2}{n}=4317-\dfrac{137^2}{5}=4317-3753.8=563.2 \). \( S_{xy}=\sum xy-\dfrac{(\sum x)(\sum y)}{n}=1680-\dfrac{75\times137}{5}=1680-2055=-375 \). So \( r=\dfrac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\dfrac{-375}{\sqrt{250\times563.2}}=\dfrac{-375}{\sqrt{140800}}\approx\dfrac{-375}{375.23}\approx-0.999 \) (3 s.f.). (b) Substituting \( x=40 \) into the regression line: \( y=49.9-1.5(40)=49.9-60=-10.1 \). This prediction is unreliable for two connected reasons: first, \( x=40^\circ \)C is well outside the range of temperatures used to calculate the regression line (\( 5^\circ \)C to \( 25^\circ \)C), so using the line here involves extrapolation, and there is no guarantee the same linear relationship continues to hold at such a high temperature; second, the resulting prediction of \( -10.1 \) drinks is physically impossible (a negative number of drinks cannot be sold), which itself demonstrates that the model has broken down outside the range of the original data. Final answer: (a) \( r\approx-0.999 \); (b) predicted value \( -10.1 \), which is unreliable/impossible, since it is obtained by extrapolating far beyond the range of the original data.

(a) MW1: correct \( S_{xx}=250 \); MW1: correct \( S_{yy}=563.2 \); W1: correct \( S_{xy}=-375 \); M1: correct statement/use of the PMCC formula (ECF); W1: correct final answer \( r\approx-0.999 \) (3 s.f.). (5 marks) (b) M1: correctly substitutes \( x=40 \) into the regression equation; W1: correct value \( -10.1 \); W1: identifies that \( x=40 \) is outside the data range (extrapolation); W1: identifies the negative/impossible predicted value as evidence the prediction is unreliable. (4 marks) Total 9 marks.

(a) Selecting every 5th person who passes is an example of systematic sampling (selecting at a regular interval from the people encountered). However, the overall sample is unlikely to be representative of all adults in the town, because it only includes adults who happen to be shopping at that particular supermarket on a Tuesday morning; this excludes, for example, adults who are at work at that time, adults who shop elsewhere, or adults who do not shop at supermarkets at all, so the sample is likely to be biased towards a particular subgroup of the town's adult population (e.g. those free on weekday mornings). (b) Using the addition rule, \( P(S\cup B)=P(S)+P(B)-P(S\cap B)=\dfrac{50}{80}+\dfrac{35}{80}-\dfrac{25}{80}=\dfrac{60}{80}=0.75 \). The probability of neither event is the complement: \( P(\text{neither})=1-P(S\cup B)=1-0.75=0.25 \). Final answer: (a) systematic sampling; not representative because it only samples adults present at that supermarket on a Tuesday morning; (b) \( P(\text{neither})=0.25 \).

(a) W1: correctly identifies the method as systematic sampling; W1, W1: valid reason why the sample is not representative, referring to the specific location/time restricting who could be selected (up to 2 marks for a well-explained reason). (3 marks) (b) M1: correct use of the addition rule to find \( P(S\cup B) \); MW1: correct value \( P(S\cup B)=0.75 \); M1: correct method, \( 1-P(S\cup B) \); W1: correct final answer \( 0.25 \). (4 marks) Total 7 marks.

(a) For \( X\sim B(10,p) \), \( P(X=0)=(1-p)^{10} \). Setting this equal to the given value: \( (1-p)^{10}=0.0563 \). Taking the tenth root of both sides: \( 1-p=0.0563^{1/10}\approx0.7500 \), so \( p\approx1-0.7500=0.2500\approx0.25 \), as required. (Check: with \( p=0.25 \) exactly, \( (1-0.25)^{10}=0.75^{10}\approx0.05631 \), which rounds to \( 0.0563 \) to 3 significant figures, confirming the result.) (b) With \( p=0.25 \): \( P(X=2)=\binom{10}{2}(0.25)^2(0.75)^8 \). \( \binom{10}{2}=45 \), \( (0.25)^2=0.0625 \), \( (0.75)^8\approx0.100113 \). So \( P(X=2)\approx45\times0.0625\times0.100113\approx0.282 \) (3 s.f.). Final answer: (a) shown, \( p\approx0.25 \); (b) \( P(X=2)\approx0.282 \).

(a) M1: correctly states \( P(X=0)=(1-p)^{10} \) and sets it equal to \( 0.0563 \); M1: correct method, taking the tenth root of both sides; W1: correct value \( 1-p\approx0.75 \); W1: correct conclusion \( p\approx0.25 \), ideally with a confirming check. (4 marks) (b) M1: correct binomial expression with correct binomial coefficient \( \binom{10}{2}=45 \); MW1: correct substitution of \( p=0.25 \); W1: correct unrounded value; W1: correct final answer \( 0.282 \) (3 s.f.). (4 marks) Total 8 marks.