2022 CCEA AS-Level Mathematics 2210 Practice Paper with Answers

Substitute \( y = x+1 \) into the circle's equation: \( x^2 + (x+1)^2 = 25 \). Expanding: \( x^2 + x^2 + 2x + 1 = 25 \), so \( 2x^2 + 2x - 24 = 0 \), which simplifies to \( x^2 + x - 12 = 0 \). Factorising: \( (x+4)(x-3) = 0 \), so \( x = -4 \) or \( x = 3 \). Using \( y = x+1 \): when \( x=-4 \), \( y=-3 \); when \( x=3 \), \( y=4 \). Check: \( (-4)^2+(-3)^2 = 16+9=25 \) and \( 3^2+4^2=9+16=25 \), both correct. Final answer: A = (-4, -3) and B = (3, 4).

M1: substitutes \( y=x+1 \) into \( x^2+y^2=25 \). MW1: correctly expands and simplifies to \( x^2+x-12=0 \) (or an equivalent 3-term quadratic). M1: valid method to solve the quadratic (factorising or the quadratic formula). W1: correct x-values \( x=-4 \) and \( x=3 \). W1: correct y-values found by substitution. W1: both coordinate pairs stated correctly: \( (-4,-3) \) and \( (3,4) \). Total 6 marks. Follow-through (ECF) applies to the y-values if the x-values were found by a valid method but contain an arithmetic slip.

(a) Let \( u = e^x \). The equation becomes \( u^2 - 5u + 6 = 0 \), which factorises as \( (u-2)(u-3) = 0 \), so \( u = 2 \) or \( u = 3 \). Since \( u = e^x \), this gives \( e^x = 2 \) or \( e^x = 3 \), so \( x = \ln 2 \) or \( x = \ln 3 \). (b) Note that \( e^{4x} = (e^{2x})^2 \), so letting \( v = e^{2x} \), the equation \( e^{4x} - 5e^{2x} + 6 = 0 \) becomes \( v^2 - 5v + 6 = 0 \), the same equation as in part (a) with \( u \) replaced by \( v \). So \( v = 2 \) or \( v = 3 \), giving \( e^{2x} = 2 \) or \( e^{2x} = 3 \), so \( 2x = \ln 2 \) or \( 2x = \ln 3 \), hence \( x = \tfrac{1}{2}\ln 2 \) or \( x = \tfrac{1}{2}\ln 3 \). Final answer: (a) \( x = \ln 2 \) or \( x = \ln 3 \); (b) \( x = \tfrac{1}{2}\ln 2 \) or \( x = \tfrac{1}{2}\ln 3 \).

(a) M1: sets \( u=e^x \) and forms the quadratic \( u^2-5u+6=0 \); M1: valid method to solve the quadratic (factorising or formula); W1: correct values \( u=2 \) and \( u=3 \); M1: correctly converts back using \( x=\ln u \); W1: \( x=\ln 2 \); W1: \( x=\ln 3 \). (6 marks) (b) M1: recognises \( e^{4x}=(e^{2x})^2 \) and sets \( v=e^{2x} \) (or equivalent valid method); W1: correctly obtains \( v=2 \) and \( v=3 \) (may follow through from part (a)); W1: \( x=\tfrac{1}{2}\ln 2 \); W1: \( x=\tfrac{1}{2}\ln 3 \). (4 marks) Total 10 marks.

(a) \( f(x) = 2x^2-8x+3 = 2(x^2-4x)+3 = 2[(x-2)^2-4]+3 = 2(x-2)^2-8+3 = 2(x-2)^2-5 \), so \( a=2 \), \( b=-2 \), \( c=-5 \). (b) Since \( (x-2)^2 \ge 0 \) with equality when \( x=2 \), the minimum value of \( f(x) \) is \( -5 \), occurring at \( x=2 \); the minimum point is \( (2,-5) \). (c) Because the minimum value is \( -5 \) and the parabola opens upwards (\( a=2>0 \)), the range is \( f(x) \ge -5 \). (d) The equation \( f(x)=k \) is \( 2x^2-8x+(3-k)=0 \), which has no real roots when its discriminant is negative: \( (-8)^2 - 4(2)(3-k) < 0 \), i.e. \( 64 - 24 + 8k < 0 \), i.e. \( 40+8k<0 \), i.e. \( k < -5 \). This is consistent with part (b): since the minimum value of \( f(x) \) is \( -5 \), the line \( y=k \) fails to meet the curve only when \( k \) is below the minimum, i.e. \( k<-5 \). Final answer: (a) \( f(x)=2(x-2)^2-5 \); (b) \( (2,-5) \); (c) \( f(x)\ge -5 \); (d) \( k<-5 \).

(a) M1: takes out a factor of 2 from the \( x^2 \) and \( x \) terms; M1: correct completed-square form inside the bracket; W1: correct constant term outside the bracket; W1: fully correct form \( 2(x-2)^2-5 \) stated. (4 marks) (b) MW1: correct x-coordinate \( x=2 \) (or follow through from (a)); W1: correct y-coordinate \( -5 \) and point stated as \( (2,-5) \). (2 marks) (c) M1: links the range to the minimum value found; W1: correct range \( f(x)\ge -5 \) stated with correct inequality direction. (2 marks) (d) M1: sets up \( f(x)=k \) as a quadratic equation and identifies the discriminant condition for no real roots (\( b^2-4ac<0 \)), or uses the minimum value from part (b) directly; M1: correct discriminant expression in terms of \( k \) (e.g. \( 40+8k \)) or equivalent valid method; W1: correct final answer \( k<-5 \). (3 marks) Total 11 marks.

(a) Let \( s = \sin\theta \). The equation becomes \( 2s^2+s-1=0 \), which factorises as \( (2s-1)(s+1)=0 \), so \( s=\tfrac{1}{2} \) or \( s=-1 \). For \( \sin\theta = \tfrac{1}{2} \) in \( 0^\circ \le \theta \le 360^\circ \): the principal value is \( \theta=30^\circ \), and since sine is also positive in the second quadrant, \( \theta=180^\circ-30^\circ=150^\circ \). For \( \sin\theta=-1 \): this occurs only at \( \theta=270^\circ \) in the given interval. So the solutions are \( \theta=30^\circ, 150^\circ, 270^\circ \). (b) The function \( \sin\theta \) is periodic with period \( 360^\circ \), so the pattern of 3 solutions found in part (a) repeats identically in the next \( 360^\circ \) interval, \( 360^\circ<\theta\le 720^\circ \), giving a further 3 solutions at \( \theta=390^\circ, 510^\circ, 630^\circ \). In total there are \( 3+3=6 \) solutions in \( 0^\circ \le \theta \le 720^\circ \). Final answer: (a) \( \theta=30^\circ,150^\circ,270^\circ \); (b) 6 solutions.

(a) M1: sets \( s=\sin\theta \) and forms the quadratic \( 2s^2+s-1=0 \); M1: valid method to solve the quadratic (factorising or formula); W1: correct values \( s=\tfrac12 \) and \( s=-1 \); W1: correct solutions from \( \sin\theta=\tfrac12 \) (\( \theta=30^\circ,150^\circ \)); W1: correct solution from \( \sin\theta=-1 \) (\( \theta=270^\circ \)); W1: all three solutions stated with no extras or omissions in the given range. (6 marks) (b) M1: recognises the \( 360^\circ \) periodicity of sine; W1: correctly identifies the further three solutions (\( 390^\circ,510^\circ,630^\circ \)) or equivalent reasoning; W1: correct total of 6 solutions stated. (3 marks) Total 9 marks.

(a) Using the cosine rule with \( a=BC=11 \), \( c=AB=8 \) and the included angle \( B=65^\circ \): \( AC^2 = BC^2+AB^2-2(BC)(AB)\cos(65^\circ) = 11^2+8^2-2(11)(8)\cos(65^\circ) = 121+64-176\cos(65^\circ) \). Since \( \cos(65^\circ)\approx 0.42262 \), \( AC^2 \approx 185-74.38 = 110.62 \), so \( AC \approx \sqrt{110.62} \approx 10.5 \) cm (3 s.f.). (b) Area \( = \tfrac{1}{2}(BC)(AB)\sin(65^\circ) = \tfrac12(11)(8)\sin(65^\circ) = 44\sin(65^\circ) \approx 44(0.90631) \approx 39.9 \text{ cm}^2 \) (3 s.f.). (c) Using the sine rule, \( \dfrac{BC}{\sin(BAC)} = \dfrac{AC}{\sin(B)} \), so \( \sin(BAC) = \dfrac{BC\sin(B)}{AC} = \dfrac{11\sin(65^\circ)}{10.5176} \approx \dfrac{9.9694}{10.5176} \approx 0.94789 \), giving angle \( BAC \approx 71.4^\circ \) (1 d.p.), taking the acute-angled solution since angle B is already \( 65^\circ \) and the triangle's angles must sum to \( 180^\circ \), so angle A cannot be obtuse. Check: angle \( C \approx 180^\circ-65^\circ-71.4^\circ=43.6^\circ \), and all three angles are positive, confirming consistency. Final answer: (a) \( AC\approx 10.5 \) cm; (b) area \( \approx 39.9\text{ cm}^2 \); (c) angle \( BAC\approx 71.4^\circ \).

(a) M1: correct statement of the cosine rule with the given sides and included angle; MW1: correct substitution giving \( AC^2\approx 110.6 \) (or unrounded equivalent); W1: correct final answer \( AC=10.5 \) cm (3 s.f.), follow-through (ECF) from an earlier slip accepted. (3 marks) (b) M1: correct statement of the area formula \( \tfrac12 ab\sin C \) with correct sides/angle identified; MW1: correct substitution and unrounded value; W1: correct final answer \( 39.9\text{ cm}^2 \) (3 s.f.), ECF applies. (3 marks) (c) M1: correct statement of the sine rule; M1: correct rearrangement to make \( \sin(BAC) \) the subject; W1: correct unrounded value of angle \( BAC \); W1: correct final answer \( 71.4^\circ \) (1 d.p.) with an appropriate check that angle A is acute rather than the obtuse alternative. (4 marks) Total 10 marks.

(a) Differentiating term by term: \( \dfrac{dy}{dx} = 3x^2-12x+9 \). (b) At stationary points, \( \dfrac{dy}{dx}=0 \): \( 3x^2-12x+9=0 \), i.e. \( x^2-4x+3=0 \), which factorises as \( (x-1)(x-3)=0 \), so \( x=1 \) or \( x=3 \). At \( x=1 \): \( y=1-6+9+2=6 \), giving the point \( (1,6) \). At \( x=3 \): \( y=27-54+27+2=2 \), giving the point \( (3,2) \). The second derivative is \( \dfrac{d^2y}{dx^2}=6x-12 \). At \( x=1 \): \( 6(1)-12=-6<0 \), so \( (1,6) \) is a maximum point. At \( x=3 \): \( 6(3)-12=6>0 \), so \( (3,2) \) is a minimum point. (c) At \( x=0 \): \( y=0-0+0+2=2 \), and the gradient is \( \dfrac{dy}{dx}=3(0)^2-12(0)+9=9 \). The tangent through \( (0,2) \) with gradient 9 is \( y-2=9(x-0) \), i.e. \( y=9x+2 \). (d) The normal is perpendicular to the tangent, so its gradient is \( -\dfrac{1}{9} \). Through \( (0,2) \): \( y-2=-\tfrac19(x-0) \), i.e. \( y=2-\tfrac{x}{9} \). Multiplying through by 9: \( 9y=18-x \), i.e. \( x+9y-18=0 \). Final answer: (a) \( \dfrac{dy}{dx}=3x^2-12x+9 \); (b) maximum \( (1,6) \), minimum \( (3,2) \); (c) \( y=9x+2 \); (d) \( x+9y-18=0 \).

(a) MW1: correct differentiation of each term; W1: fully correct \( \dfrac{dy}{dx}=3x^2-12x+9 \). (2 marks) (b) M1: sets \( \dfrac{dy}{dx}=0 \) and forms a 3-term quadratic; M1: valid method to solve, e.g. factorising; W1: correct x-values \( x=1,3 \); W1: correct y-values \( y=6,2 \) found by substitution; M1: finds the second derivative \( 6x-12 \) and substitutes both x-values; W1: correctly identifies \( (1,6) \) as a maximum and \( (3,2) \) as a minimum, with correct reasoning from the sign of the second derivative. (7 marks) (c) MW1: correct value \( y=2 \) at \( x=0 \); MW1: correct gradient \( 9 \) at \( x=0 \); W1: correct tangent equation \( y=9x+2 \). (3 marks) (d) M1: correct normal gradient \( -\tfrac19 \) found using the perpendicular gradient rule (ECF from candidate's tangent gradient); W1: correctly rearranges to integer form \( x+9y-18=0 \). (2 marks) Total 14 marks.

(a) Writing \( y=x^2+16x^{-1} \) and differentiating: \( \dfrac{dy}{dx}=2x-16x^{-2}=2x-\dfrac{16}{x^2} \). (b) At a stationary point, \( \dfrac{dy}{dx}=0 \): \( 2x-\dfrac{16}{x^2}=0 \), so \( 2x=\dfrac{16}{x^2} \), i.e. \( 2x^3=16 \), i.e. \( x^3=8 \), so \( x=2 \) (the only real, positive solution, consistent with the domain \( x>0 \)). At \( x=2 \): \( y=2^2+\dfrac{16}{2}=4+8=12 \), giving the point \( (2,12) \). The second derivative is \( \dfrac{d^2y}{dx^2}=2+32x^{-3}=2+\dfrac{32}{x^3} \). At \( x=2 \): \( 2+\dfrac{32}{8}=2+4=6>0 \), so \( (2,12) \) is a minimum point. (c) At \( x=4 \): \( y=4^2+\dfrac{16}{4}=16+4=20 \), and the gradient is \( \dfrac{dy}{dx}=2(4)-\dfrac{16}{16}=8-1=7 \). The tangent through \( (4,20) \) with gradient 7 is \( y-20=7(x-4) \), i.e. \( y=7x-28+20=7x-8 \). Final answer: (a) \( \dfrac{dy}{dx}=2x-\dfrac{16}{x^2} \); (b) minimum point \( (2,12) \); (c) \( y=7x-8 \).

(a) M1: writes \( \dfrac{16}{x} \) as \( 16x^{-1} \) (or equivalent) ready to differentiate; MW1: correctly differentiates \( x^2 \) to \( 2x \); W1: correctly differentiates \( 16x^{-1} \) to \( -16x^{-2} \), giving fully correct \( \dfrac{dy}{dx}=2x-\dfrac{16}{x^2} \). (3 marks) (b) M1: sets \( \dfrac{dy}{dx}=0 \) and forms an equation in \( x \); M1: valid method to solve for \( x \) (e.g. multiplying through by \( x^2 \)); W1: correct value \( x=2 \), with the positive real root correctly selected/justified; W1: correct \( y \)-value \( y=12 \); M1: finds the second derivative and substitutes \( x=2 \); W1: correctly identifies \( (2,12) \) as a minimum with correct reasoning from the positive second derivative. (7 marks) (c) MW1: correct value \( y=20 \) at \( x=4 \); MW1: correct gradient \( 7 \) at \( x=4 \); M1: correct method to form the tangent equation using the point and gradient; W1: correct final tangent equation \( y=7x-8 \). (4 marks) Total 14 marks.

(a) Integrating term by term: \( \displaystyle\int (3x^2-2x+1)\,dx = x^3-x^2+x+c \), where \( c \) is a constant of integration. (b) Using the result from (a) as \( F(x)=x^3-x^2+x \): \( \displaystyle\int_1^4 (3x^2-2x+1)\,dx = F(4)-F(1) \). \( F(4)=4^3-4^2+4=64-16+4=52 \). \( F(1)=1^3-1^2+1=1-1+1=1 \). So the definite integral equals \( 52-1=51 \). (c) \( \displaystyle\int_1^k (3x^2-2x+1)\,dx = F(k)-F(1) = (k^3-k^2+k)-1 \). Setting this equal to 104: \( k^3-k^2+k-1=104 \), i.e. \( k^3-k^2+k-105=0 \). Testing \( k=5 \): \( 125-25+5-105=0 \), so \( k=5 \) is a root, and \( (k-5) \) is a factor. Dividing, \( k^3-k^2+k-105=(k-5)(k^2+4k+21) \). The quadratic factor \( k^2+4k+21 \) has discriminant \( 4^2-4(1)(21)=16-84=-68<0 \), so it has no real roots. Therefore \( k=5 \) is the only real solution, and since \( 5>1 \), it satisfies the given condition. Final answer: (a) \( x^3-x^2+x+c \); (b) 51; (c) \( k=5 \).

(a) MW1: correctly integrates \( 3x^2 \) to \( x^3 \); MW1: correctly integrates \( -2x \) to \( -x^2 \); W1: correctly integrates \( 1 \) to \( x \) and includes \( +c \). (3 marks) (b) M1: substitutes limits into the integrated expression (ECF from part (a) if a valid antiderivative was found); MW1: correct value \( F(4)=52 \); MW1: correct value \( F(1)=1 \); W1: correct final answer \( 51 \). (4 marks) (c) M1: sets up the equation \( F(k)-F(1)=104 \) leading to a cubic in \( k \); MW1: correct cubic equation \( k^3-k^2+k-105=0 \); M1: valid method to find a root (e.g. trial of factors of 105); W1: correctly verifies \( k=5 \) is a root; M1: divides to find the quadratic factor and considers its discriminant (or an equivalent valid method) to justify there are no other real roots; W1: correct final answer \( k=5 \), with valid justification of uniqueness. (6 marks) Total 13 marks.

(a) Using the factor theorem, evaluate \( p(2) = 2(2)^3+(2)^2-13(2)+6 = 2(8)+4-26+6 = 16+4-26+6=0 \). Since \( p(2)=0 \), \( (x-2) \) is a factor of \( p(x) \). (b) Dividing \( p(x) \) by \( (x-2) \) (by algebraic long division or inspection): \( p(x) = (x-2)(2x^2+5x-3) \). The quadratic factor \( 2x^2+5x-3 \) factorises as \( (x+3)(2x-1) \), since \( (x+3)(2x-1) = 2x^2-x+6x-3=2x^2+5x-3 \). So \( p(x)=(x-2)(x+3)(2x-1) \). (c) Setting \( p(x)=0 \): \( (x-2)(x+3)(2x-1)=0 \), so \( x=2 \), \( x=-3 \), or \( x=\tfrac12 \). (d) By the remainder theorem, the remainder when \( p(x) \) is divided by \( (x+1) \) is \( p(-1) \): \( p(-1)=2(-1)^3+(-1)^2-13(-1)+6=-2+1+13+6=18 \). Final answer: (a) \( p(2)=0 \), so \( (x-2) \) is a factor; (b) \( p(x)=(x-2)(x+3)(2x-1) \); (c) \( x=2,-3,\tfrac12 \); (d) remainder \( =18 \).

(a) M1: substitutes \( x=2 \) into \( p(x) \) with correct working shown; W1: correctly obtains \( p(2)=0 \) and states the conclusion that \( (x-2) \) is therefore a factor (by the factor theorem). (2 marks) (b) M1: valid method (long division or inspection/equating coefficients) to divide \( p(x) \) by \( (x-2) \); W1: correct quadratic factor \( 2x^2+5x-3 \); M1: valid method to factorise the quadratic; W1: correct final factorisation \( (x-2)(x+3)(2x-1) \). (4 marks) (c) M1: sets each linear factor equal to zero (ECF from part (b)); W1: correct \( x=2 \); W1: correct \( x=-3 \); W1: correct \( x=\tfrac12 \). (4 marks) (d) M1: recognises the remainder theorem applies and substitutes \( x=-1 \) into \( p(x) \); MW1: correct substituted working; W1: correct final answer, remainder \( =18 \). (3 marks) Total 13 marks.

(a) Adding the components: \( \mathbf{F_1}+\mathbf{F_2} = (3-5)\mathbf{i} + (4+2)\mathbf{j} = -2\mathbf{i}+6\mathbf{j} \) N. (b) The magnitude of the resultant force is \( |\mathbf{F_1}+\mathbf{F_2}| = \sqrt{(-2)^2+6^2} = \sqrt{4+36} = \sqrt{40} \approx 6.3246 \) N. By Newton's second law, \( \mathbf{F}=m\mathbf{a} \), so the magnitude of the acceleration is \( a = \dfrac{|\mathbf{F_1}+\mathbf{F_2}|}{m} = \dfrac{\sqrt{40}}{2} \approx \dfrac{6.3246}{2} \approx 3.16 \text{ m s}^{-2} \) (3 s.f.). Final answer: (a) \( -2\mathbf{i}+6\mathbf{j} \) N; (b) acceleration \( \approx 3.16 \text{ m s}^{-2} \).

(a) MW1: correct \( \mathbf{i} \)-component \( -2 \); W1: correct \( \mathbf{j} \)-component \( 6 \), giving \( -2\mathbf{i}+6\mathbf{j} \). (2 marks) (b) M1: correct method for the magnitude of the resultant, \( \sqrt{(-2)^2+6^2} \) (ECF from part (a)); W1: correct unrounded magnitude \( \sqrt{40}\approx 6.32 \) N; M1: applies \( a=F/m \) with \( m=2 \); W1: correct final answer \( 3.16 \text{ m s}^{-2} \) (3 s.f.). (4 marks) Total 6 marks.

Let \( a \) be the magnitude of the acceleration and \( T \) the tension in the string. The 5 kg particle accelerates downwards and the 3 kg particle accelerates upwards, both with magnitude \( a \). For the 5 kg particle (taking downwards as positive): \( 5g - T = 5a \). For the 3 kg particle (taking upwards as positive): \( T - 3g = 3a \). (a) Adding the two equations eliminates \( T \): \( 5g-3g = 5a+3a \), i.e. \( 2g=8a \), so \( a=\dfrac{2g}{8}=\dfrac{2(9.8)}{8}=\dfrac{19.6}{8}=2.45 \text{ m s}^{-2} \). (b) Substituting into the 3 kg particle's equation: \( T=3g+3a=3(9.8)+3(2.45)=29.4+7.35=36.75 \) N. Check using the 5 kg equation: \( T=5g-5a=5(9.8)-5(2.45)=49-12.25=36.75 \) N, which agrees. Final answer: (a) \( a=2.45 \text{ m s}^{-2} \); (b) \( T=36.75 \) N.

M1: applies Newton's second law to the 5 kg particle, giving \( 5g-T=5a \) (or equivalent, with correct signs); M1: applies Newton's second law to the 3 kg particle, giving \( T-3g=3a \) (or equivalent, with correct signs). (a) M1: valid method to eliminate \( T \) (e.g. adding the two equations); W1: correct value \( a=2.45 \text{ m s}^{-2} \). (4 marks total for the two set-up marks plus part (a)) (b) MW1: correct substitution into either equation; W1: correct final value \( T=36.75 \) N, ideally confirmed by checking consistency with the other equation. (3 marks) Total 7 marks.

Since the particle is in equilibrium and on the point of sliding down the plane, friction acts up the plane with its maximum (limiting) value, \( F=\mu R \). (a) Resolving perpendicular to the plane, the normal reaction balances the component of weight perpendicular to the plane: \( R = mg\cos(25^\circ) = 4(9.8)\cos(25^\circ) \approx 39.2(0.90631) \approx 35.5 \) N (3 s.f.). (b) Resolving along the plane, since the particle is in equilibrium, friction balances the component of weight along the plane: \( F = mg\sin(25^\circ) = 4(9.8)\sin(25^\circ) \approx 39.2(0.42262) \approx 16.6 \) N (3 s.f.). (c) Since the particle is on the point of sliding, \( F=\mu R \), so \( \mu = \dfrac{F}{R} = \dfrac{mg\sin(25^\circ)}{mg\cos(25^\circ)} = \tan(25^\circ) \approx 0.466 \) (3 s.f.). Final answer: (a) \( R\approx 35.5 \) N; (b) \( F\approx 16.6 \) N; (c) \( \mu\approx 0.466 \).

(a) M1: correct resolution perpendicular to the plane, \( R=mg\cos(25^\circ) \); W1: correct unrounded value; W1: correct final answer \( R=35.5 \) N (3 s.f.). (3 marks) (b) M1: correct resolution along the plane using equilibrium, \( F=mg\sin(25^\circ) \); W1: correct unrounded value; W1: correct final answer \( F=16.6 \) N (3 s.f.), ECF from part (a) context not required as independently resolved. (3 marks) (c) M1: recognises limiting friction condition \( F=\mu R \); M1: correct rearrangement \( \mu=F/R \) (ECF from parts (a) and (b), or uses \( \mu=\tan25^\circ \) directly); W1: correct unrounded value; W1: correct final answer \( \mu=0.466 \) (3 s.f.). (4 marks) Total 10 marks.

(a) Using \( a=\dfrac{v-u}{t} \) with \( u=0 \), \( v=24 \), \( t=8 \): \( a=\dfrac{24-0}{8}=3 \text{ m s}^{-2} \). (b) Using \( a=\dfrac{v-u}{t} \) with \( u=24 \), \( v=0 \), \( t=6 \): \( a=\dfrac{0-24}{6}=-4 \text{ m s}^{-2} \), so the magnitude of the deceleration is \( 4 \text{ m s}^{-2} \). (c) The distance in each stage is found using \( s=\tfrac12(u+v)t \). Stage 1 (acceleration): \( s_1=\tfrac12(0+24)(8)=\tfrac12(24)(8)=96 \) m. Stage 2 (constant speed): \( s_2=24\times15=360 \) m. Stage 3 (deceleration): \( s_3=\tfrac12(24+0)(6)=\tfrac12(24)(6)=72 \) m. Total distance \( = s_1+s_2+s_3=96+360+72=528 \) m. (d) Total time \( =8+15+6=29 \) s. Average speed \( =\dfrac{\text{total distance}}{\text{total time}}=\dfrac{528}{29}\approx 18.2069\approx 18.2 \text{ m s}^{-1} \) (3 s.f.). Final answer: (a) \( 3 \text{ m s}^{-2} \); (b) \( 4 \text{ m s}^{-2} \); (c) 528 m; (d) \( 18.2 \text{ m s}^{-1} \) (3 s.f.).

(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 third stage; W1: correct magnitude \( 4 \text{ m s}^{-2} \) (accept \( -4 \) with magnitude stated). (2 marks) (c) M1: correct method (e.g. \( s=\tfrac12(u+v)t \), or area under a velocity-time graph) for stage 1; W1: correct \( s_1=96 \) m; MW1: correct \( s_2=360 \) m for the constant-speed stage; M1: correct method for stage 3; W1: correct \( s_3=72 \) m; W1: correct total distance \( 528 \) m. (6 marks) (d) M1: correct method, total distance divided by total time (ECF from part (c)); W1: correct final answer \( 18.2 \text{ m s}^{-1} \) (3 s.f.). (2 marks) Total 12 marks.

(a) All the other eight times lie close together, between 22 and 31 minutes, suggesting the runners finished at broadly similar speeds; a value of 95 minutes is more than three times the next-highest time (31 minutes) and is not a plausible finishing time for a runner completing the same race as the others, strongly suggesting a recording or data-entry error (such as a mistyped digit) rather than a genuine, unusually slow finish. (b) With the value 95 removed, the 8 remaining values in order are: 22, 24, 25, 26, 27, 28, 29, 31. Median: with 8 values, the median is the mean of the 4th and 5th values: \( \dfrac{26+27}{2}=26.5 \) minutes. Lower quartile \( Q_1 \): the median of the lower half (22, 24, 25, 26) is \( \dfrac{24+25}{2}=24.5 \). Upper quartile \( Q_3 \): the median of the upper half (27, 28, 29, 31) is \( \dfrac{28+29}{2}=28.5 \). Interquartile range \( =Q_3-Q_1=28.5-24.5=4 \) minutes. Final answer: (a) 95 is implausible compared with the other times and is very likely a data-entry error; (b) median \( =26.5 \) minutes, IQR \( =4 \) minutes.

(a) W1: gives a valid reason referring to the value being implausible/inconsistent with the rest of the data (e.g. far larger than all other times), not merely stating 'it's an outlier' without justification. (1 mark) (b) M1: correctly orders/identifies the 8 remaining values; W1: correct median \( 26.5 \); M1: valid method to find \( Q_1 \) and \( Q_3 \) (e.g. median of lower/upper half); W1: correct \( Q_1=24.5 \) and \( Q_3=28.5 \); W1: correct IQR \( =4 \). (5 marks) Total 6 marks.

(a) With \( n=6 \): \( S_{xx}=\sum x^2-\dfrac{(\sum x)^2}{n}=238-\dfrac{34^2}{6}=238-\dfrac{1156}{6}=238-192.667=45.333 \) (i.e. \( 45.3 \) to 3 s.f.). \( S_{yy}=\sum y^2-\dfrac{(\sum y)^2}{n}=6570-\dfrac{192^2}{6}=6570-\dfrac{36864}{6}=6570-6144=426 \). \( S_{xy}=\sum xy-\dfrac{(\sum x)(\sum y)}{n}=1225-\dfrac{34\times192}{6}=1225-\dfrac{6528}{6}=1225-1088=137 \). (b) The PMCC is \( r=\dfrac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\dfrac{137}{\sqrt{45.333\times426}}=\dfrac{137}{\sqrt{19312.0}}\approx\dfrac{137}{138.97}\approx 0.986 \) (3 s.f.). (c) Since \( r\approx 0.986 \) is very close to \( +1 \), this indicates a very strong positive linear correlation between the number of hours revised and the test score: students who revised for longer tended to achieve substantially higher test scores, and the relationship between the two variables is very close to a straight line (though correlation does not by itself prove that revision time causes the higher score). Final answer: (a) \( S_{xx}\approx 45.3 \), \( S_{yy}=426 \), \( S_{xy}=137 \); (b) \( r\approx 0.986 \); (c) a very strong positive linear correlation between hours revised and test score.

(a) MW1: correct \( S_{xx}\approx 45.3 \); MW1: correct \( S_{yy}=426 \); W1: correct \( S_{xy}=137 \). (3 marks) (b) M1: correct statement/use of the PMCC formula \( r=S_{xy}/\sqrt{S_{xx}S_{yy}} \) (ECF from part (a)); MW1: correct unrounded value; W1: correct final answer \( r=0.986 \) (3 s.f.). (3 marks) (c) W1: identifies the correlation as strong/very strong; W1: identifies it as positive; W1: gives a contextual interpretation in terms of hours revised and test score (e.g. more revision associated with higher scores), ideally with an appropriate caveat about correlation not implying causation. (3 marks) Total 9 marks.

(a) A binomial model requires: a fixed number of trials (here, 15 components); each trial has exactly two possible outcomes (defective or not defective); the trials are independent of one another; and the probability of a component being defective is constant (0.08) for every trial. Any two of these conditions is sufficient. (b) \( P(X=0)=\binom{15}{0}(0.08)^0(0.92)^{15}=(0.92)^{15}\approx 0.286 \) (3 s.f.). (c) \( P(X\le2)=P(X=0)+P(X=1)+P(X=2) \). \( P(X=1)=\binom{15}{1}(0.08)^1(0.92)^{14}\approx 0.373 \). \( P(X=2)=\binom{15}{2}(0.08)^2(0.92)^{13}\approx 0.227 \). So \( P(X\le2)\approx0.286+0.373+0.227=0.887 \) (3 s.f.). (d) \( P(X\ge3)=1-P(X\le2)\approx1-0.887=0.113 \) (3 s.f.). (e) For \( X\sim B(n,p) \), the mean is \( np \) and the variance is \( np(1-p) \). Mean \( =15\times0.08=1.2 \). Variance \( =15\times0.08\times0.92=1.104 \). Final answer: (a) any two valid binomial conditions; (b) \( 0.286 \); (c) \( 0.887 \); (d) \( 0.113 \); (e) mean \( =1.2 \), variance \( =1.104 \).

(a) W1, W1: one mark for each of two valid conditions (fixed number of trials; two outcomes per trial; constant probability; independent trials). (2 marks) (b) M1: correct expression \( (0.92)^{15} \); W1: correct value \( 0.286 \) (3 s.f.). (2 marks) (c) M1: correct method, summing \( P(X=0)+P(X=1)+P(X=2) \) with correct binomial coefficients; MW1: correct individual probabilities (or equivalent use of cumulative binomial tables/technology); W1: correct final answer \( 0.887 \) (3 s.f.). (3 marks) (d) M1: correct method \( 1-P(X\le2) \) (ECF from part (c)); W1: correct final answer \( 0.113 \) (3 s.f.). (2 marks) (e) MW1: correct mean \( np=1.2 \); W1: correct variance \( np(1-p)=1.104 \). (2 marks) Total 11 marks.

(a) The number studying Mathematics only is the total studying Mathematics minus those who study both: \( 55-20=35 \) students. (b) The number studying Physics only is \( 40-20=20 \). The number studying Mathematics only, Physics only, or both is \( 35+20+20=75 \). So the number studying neither is \( 100-75=25 \) students. (c) \( P(M\cap P)=\dfrac{20}{100}=0.2 \) (or \( \tfrac15 \)). (d) Two events are independent if and only if \( P(M\cap P)=P(M)\times P(P) \). Here, \( P(M)=\dfrac{55}{100}=0.55 \) and \( P(P)=\dfrac{40}{100}=0.40 \), so \( P(M)\times P(P)=0.55\times0.40=0.22 \). Since \( P(M\cap P)=0.2 \) but \( P(M)\times P(P)=0.22 \), and \( 0.2\ne0.22 \), the events are NOT independent (studying Mathematics and studying Physics are associated, in this case a student who studies one subject is very slightly less likely to also study the other than independence would predict). Final answer: (a) 35; (b) 25; (c) \( P(M\cap P)=0.2 \); (d) not independent, since \( P(M)P(P)=0.22\ne P(M\cap P)=0.2 \).

(a) W1: correct answer 35. (1 mark) (b) M1: valid method (e.g. \( 100 \) minus the sum of Mathematics-only, Physics-only and both); W1: correct answer 25. (2 marks) (c) W1: correct answer \( 0.2 \) (accept \( \tfrac15 \) or \( 20\% \)). (1 mark) (d) M1: states/uses the correct independence condition \( P(M\cap P)=P(M)\times P(P) \); M1: correctly calculates \( P(M) \) and \( P(P) \); MW1: correctly calculates \( P(M)\times P(P)=0.22 \); M1: correctly compares with \( P(M\cap P)=0.2 \) from part (c) (ECF); W1: correct conclusion stated (not independent) with valid justification referencing the inequality of the two values. (5 marks) Total 9 marks.