2023 CCEA AS-Level Mathematics 2210 Practice Paper with Answers

Setting the two expressions for \( y \) equal: \( 2x^2-3x-1 = x+5 \). Rearranging: \( 2x^2-4x-6=0 \), which simplifies to \( x^2-2x-3=0 \). Factorising: \( (x-3)(x+1)=0 \), so \( x=3 \) or \( x=-1 \). Using \( y=x+5 \): when \( x=3 \), \( y=8 \); when \( x=-1 \), \( y=4 \). Final answer: \( (x,y)=(-1,4) \) or \( (3,8) \).

M1: sets the two expressions for \( y \) equal to each other; MW1: correctly rearranges and simplifies to a 3-term quadratic \( x^2-2x-3=0 \); M1: valid method to solve the quadratic (factorising or formula); W1: correct x-values \( x=-1,3 \); W1: correct y-values found by substitution; W1: both coordinate pairs stated correctly. Total 7 marks.

Using the law \( \log_2 A + \log_2 B = \log_2(AB) \): \( \log_2\big[(x+3)(x-1)\big] = 5 \). Converting from logarithmic to index form: \( (x+3)(x-1) = 2^5 = 32 \). Expanding: \( x^2+2x-3=32 \), so \( x^2+2x-35=0 \), which factorises as \( (x+7)(x-5)=0 \), giving \( x=-7 \) or \( x=5 \). For the original logarithms to be defined, both \( x+3>0 \) and \( x-1>0 \) are required, i.e. \( x>1 \). Since \( x=-7 \) does not satisfy \( x>1 \), it is rejected. Since \( x=5>1 \), it is valid. Check: \( \log_2(8)+\log_2(4)=3+2=5 \), which confirms the solution. Final answer: \( x=5 \).

M1: combines the two logarithms using the addition law; MW1: correctly converts to index form, \( (x+3)(x-1)=32 \); M1: correctly expands and rearranges to a 3-term quadratic \( x^2+2x-35=0 \); M1: valid method to solve the quadratic (factorising or formula); W1: correct values \( x=-7 \) and \( x=5 \); W1: correctly rejects \( x=-7 \) with valid domain reasoning (\( x>1 \) required) and states final answer \( x=5 \). Total 7 marks.

(a) By the factor theorem, evaluate \( f(1)=1^3-2(1)^2-5(1)+6=1-2-5+6=0 \). Since \( f(1)=0 \), \( (x-1) \) is a factor of \( f(x) \). (b) Dividing \( f(x) \) by \( (x-1) \): \( f(x)=(x-1)(x^2-x-6) \). The quadratic factor \( x^2-x-6 \) factorises as \( (x-3)(x+2) \), since \( (x-3)(x+2)=x^2+2x-3x-6=x^2-x-6 \). So \( f(x)=(x-1)(x-3)(x+2) \). Final answer: (a) \( f(1)=0 \), so \( (x-1) \) is a factor; (b) \( f(x)=(x-1)(x-3)(x+2) \).

(a) M1: substitutes \( x=1 \) into \( f(x) \) with working shown; W1: correctly obtains \( f(1)=0 \) and states the factor theorem conclusion. (2 marks) (b) M1: valid method (long division or inspection) to divide \( f(x) \) by \( (x-1) \); W1: correct quadratic factor \( x^2-x-6 \); M1: valid method to factorise the quadratic; W1: correct final factorisation \( (x-1)(x-3)(x+2) \); W1: fully correct answer clearly stated. (5 marks) Total 7 marks.

(a) Integrating term by term: \( \displaystyle\int(6x-x^2)\,dx = 3x^2-\dfrac{x^3}{3}+c \). (b) Setting \( y=0 \): \( 6x-x^2=0 \), i.e. \( x(6-x)=0 \), so \( x=0 \) or \( x=6 \). The other point is \( (6,0) \). (c) Using the result from (a) as \( F(x)=3x^2-\dfrac{x^3}{3} \): area \( =\displaystyle\int_0^6(6x-x^2)\,dx=F(6)-F(0) \). \( F(6)=3(36)-\dfrac{216}{3}=108-72=36 \). \( F(0)=0 \). So the area is \( 36-0=36 \) (square units); since \( y\ge0 \) throughout \( 0\le x\le6 \), this definite integral gives the enclosed area directly. (d) Setting \( 6x-x^2=5 \): rearranging gives \( x^2-6x+5=0 \), which factorises as \( (x-1)(x-5)=0 \), so \( x=1 \) or \( x=5 \). Final answer: (a) \( 3x^2-\tfrac{x^3}{3}+c \); (b) \( (6,0) \); (c) area \( =36 \); (d) \( x=1 \) or \( x=5 \).

(a) MW1: correctly integrates \( 6x \) to \( 3x^2 \); W1: correctly integrates \( -x^2 \) to \( -x^3/3 \); W1: includes \( +c \) and states the fully correct antiderivative. (3 marks) (b) M1: sets \( y=0 \) and factorises/solves; W1: correct x-values \( x=0,6 \); W1: correctly identifies and states the point \( (6,0) \). (3 marks) (c) M1: correct method, evaluating the definite integral between the correct limits (ECF from part (a)); MW1: correct \( F(6)=36 \); W1: correct \( F(0)=0 \); W1: correct area \( =36 \); W1: recognises that the curve is non-negative on \( [0,6] \) so no modulus/adjustment is needed. (5 marks) (d) M1: sets \( 6x-x^2=5 \) and rearranges to a 3-term quadratic; M1: valid method to solve; W1: correct final answers \( x=1 \) and \( x=5 \). (3 marks) Total 14 marks.

(a) The bearing from B back to A is \( 070^\circ+180^\circ=250^\circ \) (since bearings reverse by \( 180^\circ \)). The bearing from B to C is \( 152^\circ \). The angle ABC, between BA and BC, is the difference between these bearings: \( 250^\circ-152^\circ=98^\circ \). (b) Using the cosine rule in triangle ABC with \( AB=40 \), \( BC=65 \) and the included angle \( ABC=98^\circ \): \( AC^2=AB^2+BC^2-2(AB)(BC)\cos(98^\circ)=40^2+65^2-2(40)(65)\cos(98^\circ) \). Since \( \cos(98^\circ)\approx-0.13917 \), \( AC^2\approx1600+4225-5200(-0.13917)\approx5825+723.7\approx6548.7 \), so \( AC\approx\sqrt{6548.7}\approx80.9 \) km (3 s.f.). (c) Using the sine rule, \( \dfrac{BC}{\sin(BAC)}=\dfrac{AC}{\sin(ABC)} \), so \( \sin(BAC)=\dfrac{BC\sin(98^\circ)}{AC}=\dfrac{65\sin(98^\circ)}{80.9}\approx\dfrac{64.37}{80.9}\approx0.7956 \), giving angle \( BAC\approx52.7^\circ \) (taking the acute solution, since the triangle's angles must sum to \( 180^\circ \) and angle ABC is already \( 98^\circ \), so angle BAC cannot be obtuse). Since C is on the same side of the line AB as the direction of travel (the ship turns further clockwise/east at B, from bearing \( 070^\circ \) to \( 152^\circ \)), the bearing of C from A is found by adding angle BAC to the bearing of B from A: bearing of C from A \( \approx070^\circ+52.7^\circ=122.7^\circ\approx123^\circ \) (nearest degree). Final answer: (a) \( 98^\circ \); (b) \( AC\approx80.9 \) km; (c) bearing \( \approx123^\circ \).

(a) M1: correctly finds the reverse bearing from B to A (\( 250^\circ \)); M1: correct method, subtracting the bearing B to C; W1: correct answer \( 98^\circ \). (3 marks) (b) M1: correct statement of the cosine rule with the given sides and included angle; MW1: correct substitution giving \( AC^2\approx6549 \) (or unrounded equivalent); W1: correct final answer \( AC=80.9 \) km (3 s.f.), ECF from part (a). (5 marks) (c) M1: correct statement of the sine rule; M1: correct rearrangement to make \( \sin(BAC) \) the subject; W1: correct unrounded value of angle BAC (\( \approx52.7^\circ \)), with correct selection of the acute angle; M1: correct method to combine angle BAC with the initial bearing \( 070^\circ \) to find the bearing of C from A; W1: correct final answer \( 123^\circ \) (nearest degree). (6 marks) Total 14 marks.

(a) Using the binomial expansion \( (1+bx)^n=1+\binom{n}{1}bx+\binom{n}{2}(bx)^2+\binom{n}{3}(bx)^3+\ldots \) with \( n=6 \), \( b=3 \): term 1: \( 1 \); term 2: \( \binom{6}{1}(3x)=6(3x)=18x \); term 3: \( \binom{6}{2}(3x)^2=15(9x^2)=135x^2 \); term 4: \( \binom{6}{3}(3x)^3=20(27x^3)=540x^3 \). So the first four terms are \( 1+18x+135x^2+540x^3 \). (b) Using the terms from (a), \( (1+3x)^6\approx1+18x+135x^2+540x^3+\ldots \). Multiplying by \( (2-x) \) and collecting the coefficient of \( x^3 \): this comes from \( 2\times(\text{coeff of }x^3\text{ in the expansion}) + (-1)\times(\text{coeff of }x^2\text{ in the expansion}) = 2(540)+(-1)(135)=1080-135=945 \). (c) Setting \( 1+3x=1.03 \) gives \( x=0.01 \). Substituting into the expansion from (a): \( 1+18(0.01)+135(0.01)^2+540(0.01)^3 = 1+0.18+0.0135+0.00054=1.19404 \), which gives \( 1.03^6\approx1.1940 \) (4 d.p.), using the first four terms of the expansion as an approximation (the true value of \( 1.03^6 \) is \( 1.194052\ldots \), showing the approximation is accurate to the number of terms used). Final answer: (a) \( 1+18x+135x^2+540x^3 \); (b) \( 945 \); (c) \( 1.03^6\approx1.1940 \).

(a) M1: correct use of binomial coefficients \( \binom{6}{1},\binom{6}{2},\binom{6}{3} \) (or Pascal's triangle row 6); MW1: correct term \( 18x \); MW1: correct term \( 135x^2 \); W1: correct term \( 540x^3 \); W1: all four terms correctly stated together. (5 marks) (b) M1: identifies that the \( x^3 \) coefficient in the product comes from \( 2\times(x^3\text{ term}) \) and \( (-x)\times(x^2\text{ term}) \) (ECF from part (a)); MW1: correct value \( 2\times540=1080 \); MW1: correct value \( -1\times135=-135 \); W1: correct final answer \( 945 \). (4 marks) (c) M1: correctly identifies \( x=0.01 \) from \( 1+3x=1.03 \); M1: substitutes into the expansion from part (a) (ECF); MW1: correct unrounded value \( 1.19404 \); W1: correct final answer \( 1.1940 \) (4 d.p.), described as an approximation from the finite expansion. (4 marks) Total 13 marks.

(a) Let the length of the field, parallel to the wall, be \( L \) metres. Fencing is required for the two widths and the one length (not the side against the wall), so the total fencing used is \( 2x+L=80 \), giving \( L=80-2x \). The area of the rectangular field is \( A=xL=x(80-2x)=80x-2x^2 \), as required. (b) Differentiating: \( \dfrac{dA}{dx}=80-4x \). At a stationary point, \( \dfrac{dA}{dx}=0 \): \( 80-4x=0 \), so \( x=20 \). (c) Substituting \( x=20 \) into \( A=80x-2x^2 \): \( A=80(20)-2(20)^2=1600-800=800 \) m\( ^2 \). The second derivative is \( \dfrac{d^2A}{dx^2}=-4 \), which is negative for all \( x \), confirming that \( x=20 \) gives a maximum value of \( A \), namely \( 800 \) m\( ^2 \). Final answer: (a) shown; (b) \( \dfrac{dA}{dx}=80-4x \), \( x=20 \); (c) maximum area \( =800 \) m\( ^2 \), confirmed as a maximum since \( \dfrac{d^2A}{dx^2}=-4<0 \).

(a) M1: correctly identifies the total fencing equation \( 2x+L=80 \) (only three sides fenced); MW1: correctly rearranges to \( L=80-2x \); W1: correctly forms \( A=x(80-2x)=80x-2x^2 \), fully shown. (3 marks) (b) MW1: correct differentiation \( \dfrac{dA}{dx}=80-4x \); M1: sets \( \dfrac{dA}{dx}=0 \); M1: valid method to solve for \( x \); W1: correct value \( x=20 \); W1: correctly identifies this is the value that maximises area (based on context/sign reasoning at this stage, ahead of full verification in (c)). (5 marks) (c) M1: substitutes \( x=20 \) into the area expression (ECF); W1: correct maximum area \( 800 \) m\( ^2 \); M1: finds the second derivative \( -4 \); W1: correctly states the second derivative is negative; W1: correct conclusion that this confirms a maximum. (5 marks) Total 13 marks.

(a) Gradient of AB \( =\dfrac{4-2}{7-1}=\dfrac{2}{6}=\dfrac{1}{3} \). (b) Midpoint \( M=\left(\dfrac{1+7}{2},\dfrac{2+4}{2}\right)=(4,3) \). (c) A line perpendicular to AB has gradient \( -3 \) (the negative reciprocal of \( \tfrac13 \)). The perpendicular bisector passes through \( M(4,3) \) with gradient \( -3 \): \( y-3=-3(x-4) \), i.e. \( y=-3x+12+3=-3x+15 \), which rearranges to \( 3x+y-15=0 \). (d) Substituting \( C(5,-2) \) into the left-hand side of the perpendicular bisector's equation: \( 3(5)+(-2)-15=15-2-15=-2 \), which is not equal to \( 0 \), so \( C \) does NOT lie on the perpendicular bisector of AB. Since any point on the perpendicular bisector of AB is equidistant from A and B, and C is not on this line, \( CA\ne CB \); this can be confirmed directly: \( CA=\sqrt{(5-1)^2+(-2-2)^2}=\sqrt{16+16}=\sqrt{32}\approx5.66 \), and \( CB=\sqrt{(5-7)^2+(-2-4)^2}=\sqrt{4+36}=\sqrt{40}\approx6.32 \), which are indeed different. Therefore triangle ABC is NOT isosceles with \( CA=CB \). Final answer: (a) \( \tfrac13 \); (b) \( (4,3) \); (c) \( 3x+y-15=0 \); (d) C does not lie on the perpendicular bisector; \( CA\approx5.66\ne CB\approx6.32 \), so the triangle is not isosceles with \( CA=CB \).

(a) MW1: correct method; W1: correct gradient \( \tfrac13 \). (2 marks) (b) MW1: correct method; W1: correct midpoint \( (4,3) \). (2 marks) (c) M1: correctly finds the perpendicular gradient \( -3 \) (ECF from part (a)); M1: correct method to form the line through M with this gradient; MW1: correctly rearranges to the required integer form; W1: fully correct equation \( 3x+y-15=0 \). (5 marks) (d) M1: substitutes C's coordinates into the perpendicular bisector's equation (ECF from part (c)); W1: correctly obtains a non-zero value and concludes C is not on the line; M1: valid alternative/confirmatory method calculating CA and CB directly; W1: correct conclusion that \( CA\ne CB \) and the triangle is not isosceles in this way. (4 marks) Total 13 marks.

(a) Writing each surd in terms of \( \sqrt3 \): \( \sqrt{75}=\sqrt{25\times3}=5\sqrt3 \); \( \sqrt{12}=\sqrt{4\times3}=2\sqrt3 \); \( \sqrt{27}=\sqrt{9\times3}=3\sqrt3 \). So \( \sqrt{75}-\sqrt{12}+\sqrt{27}=5\sqrt3-2\sqrt3+3\sqrt3=6\sqrt3 \), giving \( k=6 \). (b) Multiplying numerator and denominator by the conjugate \( \sqrt5+\sqrt2 \): \( \dfrac{10}{\sqrt5-\sqrt2}\times\dfrac{\sqrt5+\sqrt2}{\sqrt5+\sqrt2}=\dfrac{10(\sqrt5+\sqrt2)}{(\sqrt5)^2-(\sqrt2)^2}=\dfrac{10(\sqrt5+\sqrt2)}{5-2}=\dfrac{10(\sqrt5+\sqrt2)}{3}=\dfrac{10\sqrt5+10\sqrt2}{3} \). (c) Let \( u=\sqrt{x} \) (so \( u\ge0 \) and \( x=u^2 \)). The equation becomes \( u^2-4u-5=0 \), which factorises as \( (u-5)(u+1)=0 \), so \( u=5 \) or \( u=-1 \). Since \( u=\sqrt x\ge0 \), the solution \( u=-1 \) is rejected. So \( u=5 \), giving \( x=u^2=25 \). Check: \( 25-4\sqrt{25}-5=25-20-5=0 \), confirming the solution. Final answer: (a) \( 6\sqrt3 \); (b) \( \dfrac{10\sqrt5+10\sqrt2}{3} \); (c) \( x=25 \).

(a) M1: correctly simplifies at least one surd (e.g. \( \sqrt{75}=5\sqrt3 \)); MW1: correctly simplifies a second surd; W1: correctly simplifies the third surd; W1: correct final combined answer \( 6\sqrt3 \). (4 marks) (b) M1: multiplies numerator and denominator by the conjugate \( \sqrt5+\sqrt2 \); MW1: correctly simplifies the denominator to \( 3 \) using the difference of two squares; W1: correctly expands the numerator; W1: fully correct simplified answer \( \dfrac{10\sqrt5+10\sqrt2}{3} \) (or equivalent, e.g. \( \dfrac{10(\sqrt5+\sqrt2)}{3} \)). (4 marks) (c) M1: sets \( u=\sqrt x \) and forms the quadratic \( u^2-4u-5=0 \); M1: valid method to solve the quadratic; W1: correctly rejects \( u=-1 \) with valid reasoning (\( \sqrt x\ge0 \)); W1: correct final answer \( x=25 \). (4 marks) Total 12 marks.

(a) \( |\mathbf{F}|=\sqrt{6^2+(-8)^2}=\sqrt{36+64}=\sqrt{100}=10 \) N. (b) By Newton's second law, \( a=\dfrac{|\mathbf{F}|}{m}=\dfrac{10}{3}\approx3.33 \text{ m s}^{-2} \) (3 s.f.). (c) The force has a positive \( \mathbf{i} \)-component (6) and a negative \( \mathbf{j} \)-component (\( -8 \)), so it points below the \( \mathbf{i} \)-direction. The angle between \( \mathbf{F} \) and \( \mathbf{i} \) is \( \arctan\!\left(\dfrac{8}{6}\right)\approx53.1^\circ \) (1 d.p.). Final answer: (a) \( 10 \) N; (b) \( 3.33 \text{ m s}^{-2} \); (c) \( 53.1^\circ \).

(a) M1: correct method \( \sqrt{6^2+(-8)^2} \); W1: correct answer \( 10 \) N. (2 marks) (b) W1: correct answer \( 3.33 \text{ m s}^{-2} \) (ECF from part (a)). (1 mark) (c) M1: correct method, e.g. \( \arctan(8/6) \) or equivalent; W1: correct answer \( 53.1^\circ \) (1 d.p.). (2 marks) Total 5 marks.

(a) The horizontal components of the two tensions must balance for equilibrium (the strings pull in opposite horizontal directions): \( T_1\cos40^\circ = T_2\cos65^\circ \). (b) The vertical components of the two tensions must together balance the weight of the particle: \( T_1\sin40^\circ + T_2\sin65^\circ = 20 \). (c) From (a), \( T_1=\dfrac{T_2\cos65^\circ}{\cos40^\circ} \). Substituting into (b): \( \dfrac{T_2\cos65^\circ}{\cos40^\circ}\sin40^\circ+T_2\sin65^\circ=20 \), i.e. \( T_2\left(\cos65^\circ\tan40^\circ+\sin65^\circ\right)=20 \). Evaluating the bracket: \( \cos65^\circ\approx0.42262 \), \( \tan40^\circ\approx0.83910 \), \( \sin65^\circ\approx0.90631 \), so the bracket \( \approx0.42262(0.83910)+0.90631\approx0.35462+0.90631\approx1.26093 \). So \( T_2\approx\dfrac{20}{1.26093}\approx15.9 \) N (3 s.f.). Then \( T_1=\dfrac{T_2\cos65^\circ}{\cos40^\circ}\approx\dfrac{15.9(0.42262)}{0.76604}\approx\dfrac{6.720}{0.76604}\approx8.75 \) N (3 s.f.). Final answer: \( T_1\approx8.75 \) N; \( T_2\approx15.9 \) N.

(a) M1: correctly resolves horizontally; W1: correct equation \( T_1\cos40^\circ=T_2\cos65^\circ \). (2 marks) (b) M1: correctly resolves vertically; W1: correct equation \( T_1\sin40^\circ+T_2\sin65^\circ=20 \). (2 marks) (c) M1: valid method to eliminate one unknown (e.g. substitution); MW1: correct equation in a single unknown; M1: correct method to solve for that unknown; W1: correct value \( T_2\approx15.9 \) N (3 s.f.); M1: correct substitution back to find the other unknown; W1: correct value \( T_1\approx8.75 \) N (3 s.f.). (6 marks) Total 10 marks.

(a) Using \( a=\dfrac{v-u}{t} \) with \( u=0 \), \( v=15 \), \( t=5 \): \( a=\dfrac{15-0}{5}=3 \text{ m s}^{-2} \). (b) Using \( a=\dfrac{v-u}{t} \) with \( u=15 \), \( v=0 \), \( t=3 \): \( a=\dfrac{0-15}{3}=-5 \text{ m s}^{-2} \), so the magnitude of the deceleration is \( 5 \text{ m s}^{-2} \). (c) The velocity-time graph consists of two straight-line segments forming a triangle (rising from \( (0,0) \) to \( (5,15) \), then falling to \( (8,0) \)); the total distance is the total area under the graph. Stage 1: \( s_1=\tfrac12(0+15)(5)=37.5 \) m. Stage 2: \( s_2=\tfrac12(15+0)(3)=22.5 \) m. Total distance \( =37.5+22.5=60 \) m. (d) Total time \( =5+3=8 \) s. Average speed \( =\dfrac{\text{total distance}}{\text{total time}}=\dfrac{60}{8}=7.5 \text{ m s}^{-1} \). Final answer: (a) \( 3 \text{ m s}^{-2} \); (b) \( 5 \text{ m s}^{-2} \); (c) \( 60 \) m; (d) \( 7.5 \text{ m s}^{-1} \).

(a) M1: correct use of \( a=(v-u)/t \); W1: correct answer \( 3 \text{ m s}^{-2} \). (2 marks) (b) M1: correct use of \( a=(v-u)/t \) for the second stage; W1: correct magnitude \( 5 \text{ m s}^{-2} \). (2 marks) (c) M1: correct method (area under the velocity-time graph, or \( s=\tfrac12(u+v)t \) for each stage); MW1: correct individual distances \( 37.5 \) m and \( 22.5 \) m; W1: correct total distance \( 60 \) m. (3 marks) (d) M1: correct method, total distance divided by total time (ECF from part (c)); W1: correct final answer \( 7.5 \text{ m s}^{-1} \). (2 marks) Total 9 marks.

(a) Resolving perpendicular to the plane for P (no vertical motion relative to the plane's surface), the normal reaction balances the component of P's weight perpendicular to the plane: \( R=3g\cos20^\circ\approx3(9.8)(0.93969)\approx27.6 \) N (3 s.f.). (b) Since P is moving (up the plane), friction acts at its maximum (kinetic) value, opposing P's motion, i.e. down the plane: \( F=\mu R=0.25(27.6)\approx6.91 \) N (3 s.f.). (c) For P, taking up the plane as positive: \( T-3g\sin20^\circ-F=3a \). For Q, taking downwards (the direction Q moves) as positive: \( 5g-T=5a \). Adding the two equations to eliminate \( T \): \( 5g-3g\sin20^\circ-F=8a \). Substituting \( g=9.8 \), \( 3g\sin20^\circ\approx3(9.8)(0.34202)\approx10.06 \), and \( F\approx6.91 \): \( 49-10.06-6.91\approx32.03=8a \), so \( a\approx\dfrac{32.03}{8}\approx4.00 \text{ m s}^{-2} \) (3 s.f.). (d) Substituting back into Q's equation: \( T=5g-5a\approx49-5(4.00)\approx49-20.02\approx29.0 \) N (3 s.f.) (using the unrounded value of \( a \) for accuracy). Final answer: (a) \( R\approx27.6 \) N; (b) \( F\approx6.91 \) N; (c) \( a\approx4.00 \text{ m s}^{-2} \); (d) \( T\approx29.0 \) N.

(a) M1: correct resolution perpendicular to the plane, \( R=3g\cos20^\circ \); W1: correct answer \( R\approx27.6 \) N (3 s.f.). (2 marks) (b) M1: correct use of \( F=\mu R \) (ECF from part (a)); W1: correct answer \( F\approx6.91 \) N (3 s.f.). (2 marks) (c) M1: correct equation of motion for P, with correct signs for tension, weight component and friction (opposing the direction of motion); M1: correct equation of motion for Q; M1: valid method to eliminate \( T \) (e.g. adding the equations); MW1: correct unrounded value of \( a \); W1: correct final answer \( a\approx4.00 \text{ m s}^{-2} \) (3 s.f.). (5 marks) (d) M1: correct substitution into either equation of motion to find \( T \) (ECF from part (c)); W1: correct final answer \( T\approx29.0 \) N (3 s.f.), ideally checked for consistency using the other particle's equation. (2 marks) Total 11 marks.

(a) A systematic sample is obtained by selecting members from a sampling frame (here, the alphabetical list of students) at regular intervals, starting from a randomly chosen point within the first interval; every \( k \)th person on the list is then selected, where \( k \) is the sampling interval. (b) The sampling interval is \( k=\dfrac{\text{population size}}{\text{sample size}}=\dfrac{1200}{60}=20 \); so every 20th student on the list would be selected (after a random starting point between 1 and 20). (c) A systematic sample is generally quicker and more straightforward to carry out than a simple random sample, since it does not require generating or looking up 60 separate random numbers, and it tends to spread the sample evenly across the whole list rather than risking selections being clustered together, as could happen with a simple random sample. Final answer: (a) selection at a fixed interval from a randomly chosen starting point; (b) sampling interval \( =20 \); (c) it is quicker/simpler to implement and spreads the sample evenly across the list.

(a) W1: identifies selection at a regular/fixed interval; W1: mentions a random starting point within the sampling frame. (2 marks) (b) M1: correct method, population divided by sample size; W1: correct answer \( 20 \). (2 marks) (c) W1: any valid, clearly explained advantage (e.g. simplicity/speed of selection; even spread across the sampling frame), not merely restating the definition. (1 mark) Total 5 marks.

(a) Using midpoints \( 5, 15, 25, 35, 45 \) with frequencies \( 6, 14, 18, 9, 3 \) (total \( n=50 \)): \( \sum fx = 6(5)+14(15)+18(25)+9(35)+3(45) = 30+210+450+315+135 = 1140 \). Mean \( = \dfrac{\sum fx}{n} = \dfrac{1140}{50} = 22.8 \) minutes. (b) \( \sum fx^2 = 6(5^2)+14(15^2)+18(25^2)+9(35^2)+3(45^2) = 6(25)+14(225)+18(625)+9(1225)+3(2025) = 150+3150+11250+11025+6075 = 31650 \). Variance \( = \dfrac{\sum fx^2}{n} - \text{mean}^2 = \dfrac{31650}{50} - 22.8^2 = 633 - 519.84 = 113.16 \). Standard deviation \( = \sqrt{113.16} \approx 10.6 \) minutes (3 s.f.). (c) Frequency density \( = \dfrac{\text{frequency}}{\text{class width}} \). For the class \( 20

(a) M1: uses correct midpoints; MW1: correct \( \sum fx=1140 \); W1: correct mean \( 22.8 \). (3 marks) (b) M1: correct method for \( \sum fx^2 \); MW1: correct value \( \sum fx^2=31650 \); M1: correct use of the variance formula \( \sum fx^2/n-\text{mean}^2 \) (ECF from part (a)); W1: correct final standard deviation \( 10.6 \) (3 s.f.). (4 marks) (c) M1: correct method, frequency divided by class width; W1: correct answer \( 1.8 \). (2 marks) Total 9 marks.

(a) \( P(X=3)=\binom{20}{3}(0.12)^3(0.88)^{17}\approx0.224 \) (3 s.f.). (b) \( P(X\le3)=P(X=0)+P(X=1)+P(X=2)+P(X=3) \). Computing each term (or using cumulative binomial tables/technology): \( P(X\le3)\approx0.787 \) (3 s.f.). (c) \( P(X\ge4)=1-P(X\le3)\approx1-0.787=0.213 \) (3 s.f.). (d) For \( X\sim B(n,p) \), mean \( =np \) and variance \( =np(1-p) \). Mean \( =20(0.12)=2.4 \). Variance \( =20(0.12)(0.88)=2.112 \), so standard deviation \( =\sqrt{2.112}\approx1.45 \) (3 s.f.). Final answer: (a) \( 0.224 \); (b) \( 0.787 \); (c) \( 0.213 \); (d) mean \( =2.4 \), standard deviation \( \approx1.45 \).

(a) M1: correct binomial expression with correct binomial coefficient; W1: correct final answer \( 0.224 \) (3 s.f.). (3 marks) (b) M1: correct method, summing \( P(X=0) \) to \( P(X=3) \); MW1: correct individual/cumulative probabilities (or correct use of tables/technology); W1: correct final answer \( 0.787 \) (3 s.f.). (4 marks) (c) M1: correct method \( 1-P(X\le3) \) (ECF from part (b)); W1: correct final answer \( 0.213 \) (3 s.f.). (2 marks) (d) MW1: correct mean \( np=2.4 \); W1: correct standard deviation \( \sqrt{np(1-p)}\approx1.45 \) (3 s.f.). (3 marks) Total 12 marks.

(a) \( P(B)=\dfrac{68}{120} \), \( P(C)=\dfrac{45}{120} \), \( P(B\cap C)=\dfrac{15}{120} \). Using the addition rule: \( P(B\cup C)=P(B)+P(C)-P(B\cap C)=\dfrac{68}{120}+\dfrac{45}{120}-\dfrac{15}{120}=\dfrac{98}{120} \), which simplifies (dividing by 2) to \( \dfrac{49}{60} \). (b) Two events are mutually exclusive if they cannot occur together, i.e. if \( P(B\cap C)=0 \). Here, \( P(B\cap C)=\dfrac{15}{120}=0.125\ne0 \) (indeed, we are directly told 15 students study both subjects), so the events are NOT mutually exclusive. (c) Two events are independent if and only if \( P(B\cap C)=P(B)\times P(C) \). Here, \( P(B)\times P(C)=\dfrac{68}{120}\times\dfrac{45}{120}=0.5\overline{6}\times0.375=0.2125 \). Since \( P(B\cap C)=0.125 \) but \( P(B)\times P(C)=0.2125 \), and these are not equal, the events are NOT independent. Final answer: (a) \( P(B\cup C)=\dfrac{49}{60} \); (b) not mutually exclusive, since \( P(B\cap C)=0.125\ne0 \); (c) not independent, since \( P(B)P(C)=0.2125\ne P(B\cap C)=0.125 \).

(a) M1: correct statement/use of the addition rule; MW1: correct unsimplified value \( \dfrac{98}{120} \); W1: correctly simplified to \( \dfrac{49}{60} \). (3 marks) (b) M1: states/uses the correct condition for mutual exclusivity (\( P(B\cap C)=0 \)); W1: correct conclusion (not mutually exclusive) with valid numerical justification. (2 marks) (c) M1: states/uses the correct condition for independence (\( P(B\cap C)=P(B)P(C) \)); M1: correctly calculates \( P(B) \) and \( P(C) \); MW1: correctly calculates \( P(B)\times P(C)=0.2125 \); W1: correct conclusion (not independent) with valid comparison to \( P(B\cap C)=0.125 \). (4 marks) Total 9 marks.