2023 CCEA AS-Level Further Mathematics 2330 Practice Paper with Answers

(a) \( z + w = (3+2i) + (1-4i) = (3+1) + (2-4)i = 4 - 2i \). For \( zw \): \( (3+2i)(1-4i) = 3(1) + 3(-4i) + 2i(1) + 2i(-4i) = 3 - 12i + 2i - 8i^2 = 3 - 10i + 8 = 11 - 10i \) (using \( i^2 = -1 \)). (b) Multiplying numerator and denominator by the conjugate of \( w \): \( \dfrac{z}{w} = \dfrac{(3+2i)(1+4i)}{(1-4i)(1+4i)} = \dfrac{3 + 12i + 2i + 8i^2}{1 - 16i^2} = \dfrac{3 + 14i - 8}{1+16} = \dfrac{-5+14i}{17} = -\dfrac{5}{17} + \dfrac{14}{17}i \). (c) Since \( zw = (3+2i)(1-4i) = 11-10i \) from part (a), and the equation is \( (1-4i)v = 11-10i \), it follows immediately that \( v = z = 3+2i \) satisfies the equation. Checking directly: \( \dfrac{11-10i}{1-4i} = \dfrac{(11-10i)(1+4i)}{17} = \dfrac{11+44i-10i-40i^2}{17} = \dfrac{11+34i+40}{17} = \dfrac{51+34i}{17} = 3+2i \), confirming \( v = 3+2i \). Final answer: (a) \( 4-2i \), \( 11-10i \); (b) \( -\frac{5}{17}+\frac{14}{17}i \); (c) \( v = 3+2i \).

(a) M1 for correct method adding real and imaginary parts separately; W1 for \( z+w=4-2i \); M1 for correct expansion of \( zw \) using \( i^2=-1 \); W1 for \( zw=11-10i \) (2 marks for zw split as M1W1, plus 1 mark for z+w = 4 marks total... clarify): 1 mark for \( z+w=4-2i \); 2 marks for \( zw=11-10i \) (M1 for correct expansion, W1 for correct simplified answer). Max 3. (b) M1 for multiplying by the conjugate; M1 for correct expansion of numerator and denominator; W1 for the correct final answer \( -\frac{5}{17}+\frac{14}{17}i \). Max 3. (c) M1 for a valid method (direct division, or recognising the link to part (a)); W1 for the correct final answer \( v=3+2i \). Max 2. Own figure rule applies throughout.

(a) For \( 2x^2-5x+4=0 \), using sum and product of roots formulae for \( ax^2+bx+c=0 \): \( \alpha+\beta = -\dfrac{b}{a} = \dfrac{5}{2} \) and \( \alpha\beta = \dfrac{c}{a} = \dfrac{4}{2} = 2 \). (b) \( \alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = \left(\dfrac{5}{2}\right)^2 - 2(2) = \dfrac{25}{4} - 4 = \dfrac{25}{4} - \dfrac{16}{4} = \dfrac{9}{4} \). (c) The new roots are \( 2\alpha \) and \( 2\beta \); their sum is \( 2\alpha+2\beta = 2(\alpha+\beta) = 2 \times \dfrac{5}{2} = 5 \), and their product is \( (2\alpha)(2\beta) = 4\alpha\beta = 4(2) = 8 \). The quadratic with these roots is \( x^2 - (\text{sum})x + (\text{product}) = x^2 - 5x + 8 = 0 \), which already has integer coefficients. Checking consistency: dividing the original equation by 2 gives \( x^2 - 2.5x + 2 = 0 \), matching sum \( 2.5 \) and product \( 2 \) found in part (a); doubling both roots doubles the sum (to 5) and quadruples the product (to 8), confirming the new equation \( x^2-5x+8=0 \). Final answer: (a) \( \frac{5}{2}, 2 \); (b) \( \frac{9}{4} \); (c) \( x^2-5x+8=0 \).

(a) W1 for \( \alpha+\beta=\frac{5}{2} \); W1 for \( \alpha\beta=2 \). Max 2. (b) M1 for correct use of \( (\alpha+\beta)^2-2\alpha\beta \); W1 for \( \frac{9}{4} \). Max 2. (c) M1 for correct sum \( 2(\alpha+\beta) \) and product \( 4\alpha\beta \) of the new roots; W1 for the correct equation \( x^2-5x+8=0 \) with integer coefficients. Max 2. Own figure rule applies throughout.

(a) \( \det(A) = (2)(3) - (1)(5) = 6 - 5 = 1 \). (b) For a 2x2 matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \), the inverse is \( \dfrac{1}{\det}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \); here, \( A^{-1} = \dfrac{1}{1}\begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} \). (c) The system can be written as \( A\binom{x}{y} = \binom{7}{16} \), so \( \binom{x}{y} = A^{-1}\binom{7}{16} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}\binom{7}{16} = \binom{3(7)-1(16)}{-5(7)+2(16)} = \binom{21-16}{-35+32} = \binom{5}{-3} \). Checking by substitution into the original equations: \( 2(5)+(-3) = 10-3=7 \) ✓ and \( 5(5)+3(-3)=25-9=16 \) ✓, both matching the given right-hand sides, confirming the solution. Final answer: (a) 1; (b) \( \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} \); (c) \( x=5, y=-3 \).

(a) W1 for \( \det(A)=1 \). Max 1. (b) M1 for correct swap/negate pattern; M1 for dividing by the determinant; W1 for the correct final matrix. Max 3. (c) M1 for setting up \( A^{-1}\binom{7}{16} \); M1 for correct matrix-vector multiplication method; W1 for \( x=5 \); W1 for \( y=-3 \). Max 4. Own figure rule applies throughout.

(a) \( \mathbf{a}\times\mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 3 \\ 1 & 2 & -1 \end{vmatrix} = \mathbf{i}[(-1)(-1)-(3)(2)] - \mathbf{j}[(2)(-1)-(3)(1)] + \mathbf{k}[(2)(2)-(-1)(1)] = \mathbf{i}[1-6] - \mathbf{j}[-2-3] + \mathbf{k}[4+1] = -5\mathbf{i}+5\mathbf{j}+5\mathbf{k} \). (b) Since \( \mathbf{a}\times\mathbf{b} \) is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \), any scalar multiple of it is also perpendicular to both. Its magnitude is \( |\mathbf{a}\times\mathbf{b}| = \sqrt{(-5)^2+5^2+5^2} = \sqrt{75} = 5\sqrt{3} \). To obtain a vector of magnitude \( \sqrt{6} \), scale by \( \dfrac{\sqrt{6}}{5\sqrt{3}} \): \( \dfrac{\sqrt{6}}{5\sqrt{3}}(-5,5,5) = \dfrac{\sqrt{6}}{\sqrt{3}}(-1,1,1) = \sqrt{2}(-1,1,1) = (-\sqrt{2}, \sqrt{2}, \sqrt{2}) \). Checking the magnitude: \( \sqrt{(\sqrt{2})^2 \times 3} = \sqrt{6} \), confirming the answer (the negative of this vector, \( (\sqrt2,-\sqrt2,-\sqrt2) \), is an equally valid answer). (c) For \( \mathbf{c} \) to be perpendicular to \( \mathbf{a} \), their dot product must be zero: \( \mathbf{a}\cdot\mathbf{c} = (2)(1) + (-1)(\lambda) + (3)(2) = 2 - \lambda + 6 = 8 - \lambda = 0 \), giving \( \lambda = 8 \). Final answer: (a) \( -5\mathbf{i}+5\mathbf{j}+5\mathbf{k} \); (b) \( -\sqrt2\mathbf{i}+\sqrt2\mathbf{j}+\sqrt2\mathbf{k} \) (or its negative); (c) \( \lambda=8 \).

(a) M1 for correct determinant/cofactor method; W2 for the fully correct vector \( -5\mathbf{i}+5\mathbf{j}+5\mathbf{k} \) (W1 if one component is incorrect). Max 3. (b) M1 for finding the magnitude of \( \mathbf{a}\times\mathbf{b} \); M1 for correct scaling method; W1 for a correct final vector (either sign). Max 3. (c) M1 for setting up \( \mathbf{a}\cdot\mathbf{c}=0 \); W1 for \( \lambda=8 \). Max 2. Own figure rule applies throughout.

(a) For an invariant line \( y=mx \), points \( \binom{x}{mx} \) satisfy \( M\binom{x}{mx} = \lambda\binom{x}{mx} \), i.e. \( \binom{x}{mx} \) is an eigenvector of \( M \) with eigenvalue \( \lambda \); these \( \lambda \) satisfy \( \det(M-\lambda I)=0 \): \( (1-\lambda)(3-\lambda) - (4)(2) = 0 \), i.e. \( 3 - \lambda - 3\lambda + \lambda^2 - 8 = 0 \), i.e. \( \lambda^2 - 4\lambda - 5 = 0 \), which factorises as \( (\lambda-5)(\lambda+1) = 0 \), giving \( \lambda = 5 \) or \( \lambda = -1 \). (b) For \( \lambda=5 \): \( (M-5I)\binom{x}{y} = \binom{0}{0} \) gives \( \begin{pmatrix} -4 & 4 \\ 2 & -2 \end{pmatrix}\binom{x}{y}=\binom{0}{0} \), i.e. \( -4x+4y=0 \), so \( y=x \). For \( \lambda=-1 \): \( (M+I)\binom{x}{y}=\binom{0}{0} \) gives \( \begin{pmatrix} 2 & 4 \\ 2 & 4 \end{pmatrix}\binom{x}{y}=\binom{0}{0} \), i.e. \( 2x+4y=0 \), so \( y=-\dfrac{1}{2}x \). (c) The area scale factor of a 2D linear transformation is \( |\det(M)| \); here \( \det(M) = (1)(3)-(4)(2) = 3-8=-5 \), so \( |\det(M)|=5 \neq 1 \), meaning areas are scaled by a factor of 5, so the transformation is NOT area-preserving (the negative sign of the determinant also indicates the transformation reverses orientation). (d) On \( y=x \): the point \( (4,2) \) would need \( 2=4 \), which is false. On \( y=-\frac{1}{2}x \): the point would need \( 2 = -\frac{1}{2}(4) = -2 \), which is also false. So \( (4,2) \) lies on neither invariant line. Final answer: (a) \( \lambda=5,-1 \); (b) \( y=x \) and \( y=-\frac{1}{2}x \); (c) not area-preserving, area scale factor 5; (d) lies on neither line.

(a) M1 for setting up \( \det(M-\lambda I)=0 \); M1 for correct expansion to \( \lambda^2-4\lambda-5=0 \); M1 for correct factorisation/use of the quadratic formula; W3 for both correct values \( \lambda=5,-1 \) (W1 each, plus 1 further mark for both being clearly identified as the required pair). Max 6. (b) M1 for substituting \( \lambda=5 \) and solving; W1 for \( y=x \); M1 for substituting \( \lambda=-1 \) and solving; W1 for \( y=-\frac{1}{2}x \). Max 4. (c) M1 for computing \( \det(M)=-5 \); W1 for the correct conclusion (not area-preserving) with correct reasoning (\( |\det|\neq1 \)). Max 2. (d) W1 for correctly checking against \( y=x \); W1 for correctly checking against \( y=-\frac{1}{2}x \), with correct overall conclusion. Max 2. Own figure rule applies throughout.

(a) Since the cubic has real coefficients, any non-real root must occur together with its complex conjugate; as \( 2+i \) is a root, \( 2-i \) must also be a root. (b) The quadratic factor corresponding to the conjugate pair is \( (z-(2+i))(z-(2-i)) = (z-2)^2-i^2 = (z-2)^2+1 = z^2-4z+5 \). Dividing the cubic by this quadratic: \( z^3-6z^2+13z-10 = (z^2-4z+5)(z-2) \); checking this expansion, \( (z^2-4z+5)(z-2) = z^3-2z^2-4z^2+8z+5z-10 = z^3-6z^2+13z-10 \), which matches the original cubic exactly, confirming the quotient is \( z-2 \), so the third root is \( z=2 \); this can also be checked directly by substitution: \( 2^3-6(2)^2+13(2)-10 = 8-24+26-10=0 \), confirming \( z=2 \) is indeed a root. (c) The fully factorised form is \( (z-2)(z-2-i)(z-2+i)=0 \), corresponding to the three roots \( 2 \), \( 2+i \) and \( 2-i \). (d) \( L_1: |z-(3+4i)|=4 \) represents all points \( z \) whose distance from the fixed point \( 3+4i \) is exactly 4, i.e. a circle centred at \( (3,4) \) with radius 4. \( L_2: \arg(z-3)=\frac{\pi}{4} \) represents all points \( z \) such that the vector from the fixed point \( 3 \) (i.e. \( (3,0) \)) to \( z \) makes an angle of \( \frac{\pi}{4} \) (45°) with the positive real axis; this is a half-line (ray) starting at, but not including, the point \( (3,0) \), extending in the direction where both the real and imaginary parts of \( z-3 \) are positive and equal. (e) For \( z=7+4i \): checking \( L_2 \), \( z-3 = 4+4i \), so \( \arg(z-3) = \arctan\left(\dfrac{4}{4}\right) = \arctan(1) = \dfrac{\pi}{4} \) (the point \( (4,4) \) lies in the first quadrant, consistent with this angle), so \( z=7+4i \) satisfies \( L_2 \). Checking \( L_1 \), \( z-(3+4i) = (7-3)+(4-4)i = 4+0i \), so \( |z-(3+4i)| = |4| = 4 \), which matches the radius of \( L_1 \), so \( z=7+4i \) also satisfies \( L_1 \). As \( z=7+4i \) satisfies both equations, it is confirmed as a point of intersection. (f) A general point on \( L_2 \) is \( z=(3+t)+ti \) for \( t>0 \), since \( z-3=t+ti=t(1+i) \) has argument \( \frac{\pi}{4} \) for any \( t>0 \). Substituting into \( L_1 \): \( |z-(3+4i)| = |(3+t-3)+(t-4)i| = |t+(t-4)i| = \sqrt{t^2+(t-4)^2} = 4 \). Squaring both sides: \( t^2+(t-4)^2=16 \), i.e. \( t^2+t^2-8t+16=16 \), i.e. \( 2t^2-8t=0 \), i.e. \( 2t(t-4)=0 \), giving \( t=0 \) or \( t=4 \). Since \( L_2 \) requires \( t>0 \) (excluding the vertex \( t=0 \), where \( \arg(z-3) \) is undefined), the solution \( t=0 \) is rejected, leaving \( t=4 \) as the only valid solution, giving \( z=(3+4)+4i=7+4i \); this confirms \( z=7+4i \) is the only point of intersection of \( L_1 \) and \( L_2 \). Final answer: (a) \( 2-i \); (b) third root \( z=2 \); (c) \( (z-2)(z-2-i)(z-2+i) \); (d) circle centre \( (3,4) \) radius 4, and a half-line from \( (3,0) \) at angle \( \pi/4 \); (e) \( z=7+4i \) verified on both loci; (f) \( t=4 \) is the only valid solution, confirming a unique intersection point.

(a) W1 for stating \( 2-i \); W1 for the correct reason (complex roots of a real-coefficient polynomial occur in conjugate pairs). Max 2. (b) M1 for forming the quadratic factor \( z^2-4z+5 \) from the conjugate pair; M1 for a valid method to find the remaining factor (division or equivalent); W2 for correctly identifying/verifying the third root \( z=2 \); W1 for a clear, fully justified method. Max 5. (c) W3 for the fully correct factorised form (W1 per correct linear factor, or holistic credit for the fully correct expression). Max 3. (d) W1 for correctly describing \( L_1 \) as a circle with correct centre and radius; W2 for correctly describing \( L_2 \) as a half-line/ray with correct start point and angle (1 mark for start point, 1 mark for direction/angle, noting the excluded vertex). Max 3. (e) M1 for correct calculation of \( \arg(z-3) \); W1 for confirming it equals \( \pi/4 \); M1 for correct calculation of \( |z-(3+4i)| \); W1 for confirming it equals 4. Max 4. (f) M1 for correct parametrisation and substitution into \( L_1 \); M1 for correct expansion/simplification to \( 2t^2-8t=0 \) (or equivalent); W1 for correctly solving and rejecting \( t=0 \), confirming \( t=4 \) as the unique valid solution. Max 3. Own figure rule applies throughout.

(a) The coefficient matrix is \( \begin{pmatrix} 1 & 2 & 3 \\ 2 & 5 & k \\ 1 & 1 & 2 \end{pmatrix} \). Expanding along the first row: \( \det = 1\big[(5)(2)-(k)(1)\big] - 2\big[(2)(2)-(k)(1)\big] + 3\big[(2)(1)-(5)(1)\big] = 1(10-k) - 2(4-k) + 3(2-5) = (10-k) - (8-2k) + (-9) = 10-k-8+2k-9 = k-7 \). A unique solution exists only when the determinant is non-zero, so the system fails to have a unique solution when \( k-7=0 \), i.e. \( k=7 \). (b) Substituting \( k=7 \), the system becomes \( x+2y+3z=6 \), \( 2x+5y+7z=13 \), \( x+y+2z=5 \). Row-reducing: \( R_2 - 2R_1 \) gives \( (0,1,1\,|\,1) \), i.e. \( y+z=1 \); \( R_3-R_1 \) gives \( (0,-1,-1\,|\,-1) \), i.e. \( -y-z=-1 \), which is exactly \( -1 \) times the equation \( y+z=1 \) just found from \( R_2-2R_1 \) — so the third equation gives no new information (it is a multiple of the second reduced equation, not a contradiction such as \( 0=\text{non-zero} \)). Since the system reduces to only two independent equations in three unknowns, with no contradiction arising, the system is consistent and has infinitely many solutions (a one-parameter family), rather than being inconsistent. (c) With \( k=4\): \( R_2-2R_1 \) gives \( (0,1,-2\,|\,1) \), i.e. \( y-2z=1 \); \( R_3-R_1 \) gives \( (0,-1,-1\,|\,-1) \), i.e. \( y+z=1 \) (multiplying through by \(-1\)). Subtracting, \( (y-2z)-(y+z) = 1-1 \), giving \( -3z=0 \), so \( z=0 \); then \( y+z=1 \) gives \( y=1 \); then from the first equation, \( x+2(1)+3(0)=6 \), giving \( x=4 \). Checking all three original equations: \( 4+2(1)+3(0)=6 \) ✓; \( 2(4)+5(1)+4(0)=8+5=13 \) ✓; \( 4+1+2(0)=5 \) ✓; all confirmed. Final answer: (a) determinant \( k-7 \), no unique solution when \( k=7 \); (b) consistent, infinitely many solutions; (c) \( x=4, y=1, z=0 \).

(a) M1 for a correct method for expanding a 3x3 determinant; M2 for correct intermediate expansion; W2 for the fully simplified result \( k-7 \); W1 for correctly stating \( k=7 \). Max 6. (b) M1 for correct row-reduction working with \( k=7 \) substituted; M1 for correctly identifying the third reduced row as a multiple of the second (dependent, not contradictory); W2 for the correct conclusion (consistent, infinitely many solutions) with valid justification. Max 4. (c) M1 for correct row-reduction/elimination method with \( k=4 \); W1 for \( z=0 \); W1 for \( y=1 \); W1 for \( x=4 \). Max 4. Own figure rule applies throughout.

(a) The normal vectors are \( \mathbf{n}_1=(2,-1,2) \) (from \( \Pi_1 \)) and \( \mathbf{n}_2=(1,2,-2) \) (from \( \Pi_2 \)). The line of intersection is perpendicular to both normals, so its direction is \( \mathbf{n}_1\times\mathbf{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 2 \\ 1 & 2 & -2 \end{vmatrix} = \mathbf{i}[(-1)(-2)-(2)(2)] - \mathbf{j}[(2)(-2)-(2)(1)] + \mathbf{k}[(2)(2)-(-1)(1)] = \mathbf{i}[2-4]-\mathbf{j}[-4-2]+\mathbf{k}[4+1] = -2\mathbf{i}+6\mathbf{j}+5\mathbf{k} \). (b) To find a common point, set \( z=0 \): \( \Pi_1 \) gives \( 2x-y=6 \); \( \Pi_2 \) gives \( x+2y=3 \). From the first, \( y=2x-6 \); substituting into the second, \( x+2(2x-6)=3 \), i.e. \( 5x-12=3 \), i.e. \( x=3 \), giving \( y=2(3)-6=0 \). So the point \( (3,0,0) \) lies on both planes (checking: \( \Pi_1: 2(3)-0+2(0)=6 \) ✓; \( \Pi_2: 3+2(0)-2(0)=3 \) ✓). The line of intersection can therefore be written as \( \mathbf{r} = 3\mathbf{i} + t(-2\mathbf{i}+6\mathbf{j}+5\mathbf{k}) \). (c) The angle between the planes equals the angle between their normal vectors (or its supplement — the acute angle is required): \( \cos\theta = \dfrac{\mathbf{n}_1\cdot\mathbf{n}_2}{|\mathbf{n}_1||\mathbf{n}_2|} \). Here \( \mathbf{n}_1\cdot\mathbf{n}_2 = (2)(1)+(-1)(2)+(2)(-2) = 2-2-4=-4 \); \( |\mathbf{n}_1|=\sqrt{4+1+4}=\sqrt9=3 \); \( |\mathbf{n}_2|=\sqrt{1+4+4}=\sqrt9=3 \); so \( \cos\theta = \dfrac{-4}{9} \). Taking the acute angle, \( \theta = \arccos\left(\dfrac{4}{9}\right) \approx 63.6^{\circ} \) (3 s.f.). (d) At \( t=0 \), \( P=(3,0,0) \); at \( t=1 \), \( Q=(3,0,0)+(-2,6,5)=(1,6,5) \); \( R=(5,1,-2) \). Then \( \vec{PQ}=Q-P=(-2,6,5) \) and \( \vec{PR}=R-P=(2,1,-2) \). The area of the triangle is \( \frac{1}{2}|\vec{PQ}\times\vec{PR}| \). Computing the cross product: \( \vec{PQ}\times\vec{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 6 & 5 \\ 2 & 1 & -2 \end{vmatrix} = \mathbf{i}[(6)(-2)-(5)(1)] - \mathbf{j}[(-2)(-2)-(5)(2)] + \mathbf{k}[(-2)(1)-(6)(2)] = \mathbf{i}[-12-5]-\mathbf{j}[4-10]+\mathbf{k}[-2-12] = -17\mathbf{i}+6\mathbf{j}-14\mathbf{k} \). Its magnitude is \( \sqrt{(-17)^2+6^2+(-14)^2} = \sqrt{289+36+196} = \sqrt{521} \approx 22.8254 \). The area is therefore \( \dfrac{1}{2}\times22.8254 \approx 11.4127 \), which rounds to \( 11.4 \) (3 s.f.). Final answer: (a) \( -2\mathbf{i}+6\mathbf{j}+5\mathbf{k} \); (b) point \( (3,0,0) \), line \( \mathbf{r}=3\mathbf{i}+t(-2\mathbf{i}+6\mathbf{j}+5\mathbf{k}) \); (c) \( 63.6^{\circ} \); (d) area \( \approx 11.4 \).

(a) M1 for correct determinant/cofactor method for the cross product; W3 for the fully correct vector \( -2\mathbf{i}+6\mathbf{j}+5\mathbf{k} \) (W1 per correct component, or holistic credit). Max 4. (b) M1 for a valid method to find a common point (e.g. setting one coordinate and solving the resulting pair of equations); W2 for the correct point \( (3,0,0) \); W2 for the correct vector equation of the line using this point and the direction from (a) (own figure rule applies). Max 5. (c) M1 for correct use of \( \cos\theta=\frac{\mathbf{n}_1\cdot\mathbf{n}_2}{|\mathbf{n}_1||\mathbf{n}_2|} \); M1 for correct dot product and magnitudes; W2 for the correct acute angle \( 63.6^{\circ} \) (3 s.f.), with correct use of the acute angle (not the obtuse supplement). Max 4. (d) M1 for correctly finding P and Q from the line equation; W1 for correct \( \vec{PQ} \); W1 for correct \( \vec{PR} \); M2 for correct cross product method/expansion; W2 for the correct cross product vector; M1 for halving the magnitude of the cross product; W2 for the final area \( 11.4 \) (3 s.f.), with the correct method of rounding. Max 9. Own figure rule applies throughout.

(a) In equilibrium, the tension in the spring balances the weight of the particle: \( T = mg = 2 \times 9.8 = 19.6 \) N. By Hooke's Law, \( T = \dfrac{\lambda x}{l} \), so \( 19.6 = \dfrac{40x}{0.8} = 50x \), giving \( x = \dfrac{19.6}{50} = 0.392 \) m. (b) The elastic potential energy stored is \( \text{EPE} = \dfrac{\lambda x^2}{2l} = \dfrac{40 \times (0.392)^2}{2 \times 0.8} = \dfrac{40 \times 0.153664}{1.6} = \dfrac{6.14656}{1.6} = 3.8416 \) J, which rounds to 3.84 J (3 s.f.). (c) When the particle is displaced from its equilibrium position by a distance \( y \) (while the spring remains taut), the resultant force on it is proportional to \( -y \) (a restoring force directed back towards equilibrium), which is the defining condition for simple harmonic motion. Since the particle is pulled down only a further 0.1 m from an equilibrium extension of 0.392 m, the spring's extension remains positive (0.392−0.1=0.292 m at the highest point of the subsequent motion, still greater than zero) throughout the resulting motion, so the spring remains taut at all times and the particle does indeed perform SHM about the equilibrium position. Final answer: (a) \( x=0.392 \) m; (b) EPE \( \approx 3.84 \) J; (c) yes, SHM occurs, since the restoring force remains proportional to displacement while the spring stays taut throughout.

(a) M1 for \( T=mg \); W1 for \( T=19.6 \) N; M1 for correct rearrangement of Hooke's Law; W1 for \( x=0.392 \) m. Max 4. (b) M1 for correct EPE formula \( \frac{\lambda x^2}{2l} \); M1 for correct substitution; W1 for \( 3.84 \) J (3 s.f., own figure from (a)). Max 3. (c) W1 for the correct conclusion (yes, SHM); W2 for correct, complete reasoning (restoring force proportional to displacement AND spring remains taut/extension stays positive throughout, with numerical justification that the amplitude does not exceed the equilibrium extension). Max 3. Own figure rule applies throughout.

(a) The initial kinetic energy is \( \text{KE} = \frac{1}{2}mv^2 = \frac{1}{2}(3)(8)^2 = 96 \) J. As the particle travels a distance \( d \) up the plane, work is done against two resistive forces: the component of gravity along the plane, \( mg\sin30^{\circ} = 3(9.8)(0.5) = 14.7 \) N, and friction, \( F=\mu R = \mu mg\cos30^{\circ} = 0.2(3)(9.8)(\cos30^{\circ}) \approx 0.2(3)(9.8)(0.8660) \approx 5.092 \) N. By the work-energy principle, all the initial KE is converted into work done against these two resistive forces as the particle comes to rest: \( 96 = (14.7+5.092)\,d = 19.792\,d \), giving \( d = \dfrac{96}{19.792} \approx 4.8504 \), which rounds to 4.85 m (3 s.f.). (b) To determine whether the particle slides back down once at rest, compare the component of weight pulling it down the plane (14.7 N, unchanged) with the maximum friction force available to resist sliding (using the same coefficient of friction for the limiting/maximum static case), \( \mu mg\cos30^{\circ} \approx 5.092 \) N. Since \( 14.7 > 5.092 \), the force pulling the particle down the plane exceeds the maximum friction that could hold it in place, so the particle will indeed slide back down the plane. Final answer: (a) \( d \approx 4.85 \) m; (b) yes, it slides back down, since \( mg\sin30^{\circ}(14.7\text{ N}) > \mu mg\cos30^{\circ}(5.09\text{ N}) \).

(a) M1 for correct initial KE; M1 for correct gravity-component resistive force; M1 for correct friction force; M1 for correctly setting up the work-energy equation (KE = total resistive force x distance); W3 for the correct final answer \( 4.85 \) m (3 s.f.) (W1 for correct unrounded value, W2 for correct rounding); Max 7 (allocate remaining mark for a fully coherent method). (b) M1 for correctly comparing the gravity component with the maximum friction force; W2 for the correct conclusion (slides back down) with full numerical justification. Max 3. Own figure rule applies throughout.

(a) At maximum speed, the car's acceleration is zero, so the driving force exactly equals the resistance, 500 N. Since power = force x velocity, \( 20{,}000 = 500 \times v_{max} \), giving \( v_{max} = \dfrac{20{,}000}{500} = 40 \) ms\(^{-1}\). (b) Using Newton's second law along the direction of motion, \( F - R = ma \), where \( R=500 \) N is the resistance: \( F - 500 = 900(0.5) = 450 \), giving \( F = 950 \) N. The power developed at this instant is \( P = Fv = 950 \times 15 = 14{,}250 \) W \( = 14.25 \) kW. (c) The engine's maximum power (20 kW) represents the greatest rate at which it can do work, achieved (for example) at maximum speed on the level road found in part (a). At the instant described in part (b), the car is travelling at only 15 ms\(^{-1}\) (well below its maximum speed of 40 ms\(^{-1}\)) and is still accelerating, so the engine only needs to produce a driving force of 950 N to achieve this acceleration; combined with the lower speed, this results in a power output of 14.25 kW, which is less than the full 20 kW the engine is capable of — the engine has spare capacity at this point, which is why the car continues to accelerate rather than having already reached a steady maximum speed. Final answer: (a) \( v_{max}=40 \) ms\(^{-1}\); (b) driving force \(=950\) N, power \(=14.25\) kW; (c) the car is below its maximum speed and still accelerating, so the engine is not yet working at its full 20 kW capacity.

(a) M1 for recognising driving force = resistance at maximum speed; M1 for correct use of \( P=Fv \); W1 for \( v_{max}=40 \) ms\(^{-1}\). Max 3. (b) M1 for correct application of Newton's second law; W1 for \( F=950 \) N; M1 for correct use of \( P=Fv \); W1 for \( P=14.25 \) kW (or 14250 W). Max 4. (c) W3 for a complete, correct explanation referencing that the car has not yet reached maximum speed/is still accelerating, so the full 20 kW capacity is not required at this instant. Max 3. Own figure rule applies throughout.

(a) Speed is related to angular speed by \( v=r\omega = 1.2 \times 4 = 4.8 \) ms\(^{-1}\). (b) Using \( a=r\omega^2 \): \( a = 1.2 \times 4^2 = 1.2 \times 16 = 19.2 \) ms\(^{-2}\). Checking by the alternative formula \( a=\dfrac{v^2}{r} \): \( a = \dfrac{4.8^2}{1.2} = \dfrac{23.04}{1.2} = 19.2 \) ms\(^{-2}\), which agrees exactly with the first method, confirming the answer. (c) By Newton's second law applied to circular motion, \( F=ma = 0.5 \times 19.2 = 9.6 \) N. (d) The resultant (centripetal) force needed to maintain circular motion always acts horizontally, directed towards the centre of the circle. Final answer: (a) 4.8 ms\(^{-1}\); (b) 19.2 ms\(^{-2}\) (confirmed by two methods); (c) 9.6 N; (d) towards the centre of the circle.

(a) M1 for \( v=r\omega \); W1 for \( v=4.8 \) ms\(^{-1}\). Max 2. (b) M1 for using \( a=r\omega^2 \) or \( a=v^2/r \); W1 for \( a=19.2 \) ms\(^{-2}\); W1 for correctly checking/confirming using the alternative formula. Max 3. (c) M1 for \( F=ma \); W2 for \( F=9.6 \) N (own figure from (b)). Max 3. (d) W2 for correctly stating 'towards the centre' (horizontally). Max 2. Own figure rule applies throughout.

(a) As \( B \) falls a distance \( h=2 \) m, the string (being inextensible) ensures \( A \) moves the same distance, 2 m, horizontally across the table. Using conservation of energy: the loss in gravitational PE of \( B \) is converted into the kinetic energy gained by both particles (moving with a common speed \( v \)) plus the work done against friction acting on \( A \). Loss in PE of \( B \): \( m_Bgh = 3(9.8)(2)=58.8 \) J. Friction force on \( A \): \( F=\mu m_Ag = 0.25(5)(9.8)=12.25 \) N; work done against this friction over 2 m: \( 12.25 \times 2 = 24.5 \) J. Gain in KE of the system: \( \frac{1}{2}(m_A+m_B)v^2 = \frac{1}{2}(8)v^2=4v^2 \). By conservation of energy: \( 58.8 = 4v^2 + 24.5 \), giving \( 4v^2 = 34.3 \), so \( v^2=8.575 \), giving \( v=\sqrt{8.575} \approx 2.9283 \), which rounds to 2.93 ms\(^{-1}\) (3 s.f.). (b) One key modelling assumption is that the string is light and inextensible; being inextensible means that \( A \) and \( B \) always move with exactly the same speed and the same magnitude of acceleration as each other (since neither can move relative to the other along the line of the string), which is essential for the calculation above, as it allows both particles' motion to be described using a single common speed \( v \) rather than two separate, potentially different speeds. (Other valid assumptions include: the pulley is smooth and light, so it does not affect the tension in the string or absorb any energy; \( A \) and \( B \) are modelled as particles, so their size/shape and any rotational effects can be ignored; air resistance is negligible.) Final answer: (a) \( v \approx 2.93 \) ms\(^{-1}\); (b) e.g. the string is inextensible, ensuring \( A \) and \( B \) share a common speed throughout the motion.

(a) M1 for correct loss in PE of B; M1 for correct friction force on A; M1 for correct work done against friction; M1 for correctly setting up the energy equation (PE lost = KE gained + work against friction); M1 for correct combined mass in the KE term; W2 for the correct final answer \( 2.93 \) ms\(^{-1}\) (3 s.f.) (W1 for the correct unrounded value, W1 for correct rounding). Max 7. (b) W1 for stating a valid modelling assumption; W2 for a correct, clearly linked explanation of its significance to the calculation. Max 3. Own figure rule applies throughout.

(a) Each product is computed and then reduced modulo 8: \( 3\times3=9\equiv1 \) (since \( 9-8=1 \)); \( 3\times5=15\equiv7 \) (since \( 15-8=7 \)); \( 3\times7=21\equiv5 \) (since \( 21-16=5 \)); \( 5\times5=25\equiv1 \) (since \( 25-24=1 \)); \( 5\times7=35\equiv3 \) (since \( 35-32=3 \)); \( 7\times7=49\equiv1 \) (since \( 49-48=1 \)). (b) The identity element \( e \) satisfies \( e\times_8 x = x \) for all \( x \) in \( G \); this is \( e=1 \), since multiplying any element of \( G \) by 1 (and reducing mod 8, which has no effect since the element is already less than 8) leaves it unchanged. (c) From part (a), \( 3\times_83=1 \), which is exactly the identity element; therefore 3 is its own inverse, i.e. the inverse of 3 is 3. (d) The order of 5 is the smallest positive integer \( n \) with \( 5^n\equiv1\pmod8 \); from part (a), \( 5^2=5\times_85=25\equiv1\pmod8 \), and \( 5^1=5\not\equiv1 \), so the smallest such \( n \) is 2, giving the order of 5 as 2. Final answer: (a) as listed above; (b) identity \( =1 \); (c) inverse of 3 is 3; (d) order of 5 is 2.

(a) 1 mark for each of the six correct products (reduced modulo 8). Max 6. (b) W1 for correctly identifying 1 as the identity element. Max 1. (c) M1 for a valid method (e.g. testing candidates, or using the result from part (a)); W1 for the correct answer, 3. Max 2. (d) W1 for correctly stating the order of 5 is 2, with reference to \( 5^2\equiv1\pmod8 \). Max 1.

(a) Counting the edges incident to each vertex: A is in edges AB, AC, AD, so degree(A)=3; B is in edges AB, BC, BE, so degree(B)=3; C is in edges AC, BC, CD, so degree(C)=3; D is in edges AD, CD, DE, so degree(D)=3; E is in edges BE, DE, so degree(E)=2. Checking: the sum of degrees is \( 3+3+3+3+2=14 \), which equals twice the number of edges (\( 2\times7=14 \)), confirming the degrees are correct (the handshaking lemma). (b) An Eulerian circuit exists in a connected graph if and only if every vertex has even degree. Here, four vertices (A, B, C, D) each have odd degree (3), so \( G \) does not satisfy this condition, and therefore does NOT have an Eulerian circuit (in fact, since more than two vertices have odd degree, \( G \) does not even have an Eulerian path). (c) A Hamiltonian cycle must visit every one of the 5 vertices exactly once and return to the start. Consider the cycle A-B-E-D-C-A: checking each edge used against the graph's edge list — AB (given edge) ✓, BE (given edge) ✓, ED (i.e. DE, given edge) ✓, DC (i.e. CD, given edge) ✓, CA (i.e. AC, given edge) ✓ — all five edges used in this cycle are genuine edges of \( G \), and all five vertices A, B, E, D, C are visited exactly once before returning to A, so this is indeed a valid Hamiltonian cycle, confirming that \( G \) does have a Hamiltonian cycle. Final answer: (a) A=3, B=3, C=3, D=3, E=2; (b) no Eulerian circuit, since A, B, C, D have odd degree; (c) yes, a Hamiltonian cycle exists, e.g. A-B-E-D-C-A.

(a) W1 for correctly stating the degrees of all vertices with degree 3 (A, B, C, D); W1 for correctly stating degree(E)=2; W1 for a correct check (e.g. sum of degrees = twice the number of edges). Max 3. (b) M1 for correctly recalling the even-degree condition for an Eulerian circuit; W2 for the correct conclusion (no) with correct reference to the odd-degree vertices found in part (a). Max 3. (c) M1 for a valid attempt at constructing a Hamiltonian cycle; W3 for a fully correct Hamiltonian cycle, with all edges checked against the graph's edge list (accept any other valid Hamiltonian cycle in this graph, e.g. A-D-E-B-C-A). Max 4. Own figure rule applies throughout.

(a) Sorting all 8 edges by weight in increasing order: DE (1), AC (2), CD (3), AB (4), BC (5), BE (6), AD (7), CE (8). (b) Kruskal's algorithm adds edges in increasing order of weight, skipping any edge that would form a cycle with edges already selected, until a spanning tree (with \( n-1=4 \) edges, for \( n=5 \) vertices) is formed. Step 1: add DE (weight 1) — connects components \{D,E\} and \{A\}, \{B\}, \{C\}; no cycle. Step 2: add AC (weight 2) — connects \{A\} and \{C\}; no cycle; components are now \{D,E\}, \{A,C\}, \{B\}. Step 3: consider CD (weight 3) — connects C (in \{A,C\}) and D (in \{D,E\}), which are different components, so no cycle is formed; add CD; components are now \{A,C,D,E\}, \{B\}. Step 4: consider AB (weight 4) — connects A (in \{A,C,D,E\}) and B (in \{B\}), different components, so no cycle; add AB; all 5 vertices are now in a single component \{A,B,C,D,E\}. Since 4 edges have now been added, equal to \( n-1=4 \), the spanning tree is complete, and the algorithm stops without needing to consider the remaining edges (BC, BE, AD, CE); none of these are formally 'rejected' as a cycle-forming edge is discovered, since the tree is completed exactly when AB is added. (c) The total weight of the minimum spanning tree is \( 1+2+3+4=10 \). Final answer: (a) DE, AC, CD, AB, BC, BE, AD, CE (increasing order); (b) selected edges, in order: DE, AC, CD, AB; (c) total weight = 10.

(a) W2 for the fully correct increasing-order list (W1 for a list with at most one error in order). Max 2. (b) M1 for correctly selecting DE first; M1 for correctly selecting AC next; M1 for correctly selecting CD (correctly recognising it connects two different components, so does not form a cycle); M1 for correctly selecting AB (correctly recognising it completes the spanning tree); W2 for correctly stopping after 4 edges with valid reasoning (recognising a tree on 5 vertices needs exactly 4 edges). Max 6. (c) M1 for summing the selected edge weights; W1 for the correct total, 10. Max 2. Own figure rule applies throughout; accept any other valid minimum spanning tree of the same total weight if a genuine tie in edge selection exists (there is no such tie in this graph, so the tree found is unique).

(a) Using \( u_{n+1}=3u_n-2 \) repeatedly: \( u_2=3(4)-2=10 \); \( u_3=3(10)-2=28 \); \( u_4=3(28)-2=82 \). (b) Substituting \( u_n=1 \) into the recurrence: \( u_{n+1}=3(1)-2=3-2=1 \), so \( u_{n+1}=1 \) also, confirming that once the sequence reaches 1, it remains at 1 — a fixed point. (c) Let \( v_n=u_n-1 \), so \( u_n=v_n+1 \). Substituting into the recurrence: \( u_{n+1}=3u_n-2 \) becomes \( v_{n+1}+1 = 3(v_n+1)-2 = 3v_n+3-2 = 3v_n+1 \), so \( v_{n+1}=3v_n+1-1=3v_n \), confirming \( v_n \) satisfies \( v_{n+1}=3v_n \), a simple geometric recurrence with common ratio 3. Since \( v_1=u_1-1=4-1=3 \), the explicit formula for a geometric sequence gives \( v_n=v_1\times3^{n-1}=3\times3^{n-1}=3^n \). Therefore \( u_n=v_n+1=3^n+1 \). Checking against part (a): \( u_1=3^1+1=4 \) ✓; \( u_2=3^2+1=10 \) ✓; \( u_3=3^3+1=28 \) ✓; \( u_4=3^4+1=82 \) ✓ — all four values match exactly, confirming the explicit formula is correct. Final answer: (a) \( u_2=10, u_3=28, u_4=82 \); (b) confirmed, \( u_{n+1}=1 \) when \( u_n=1 \); (c) \( u_n=3^n+1 \).

(a) W1 each for correct \( u_2=10 \), \( u_3=28 \), \( u_4=82 \). Max 3. (b) M1 for substituting \( u_n=1 \) into the recurrence; W1 for correctly obtaining \( u_{n+1}=1 \), confirming the fixed point. Max 2. (c) M1 for correct substitution \( u_n=v_n+1 \) into the recurrence; M1 for correct algebraic simplification to \( v_{n+1}=3v_n \); W1 for correctly finding \( v_1=3 \); M1 for correctly using the geometric sequence formula \( v_n=v_1\times3^{n-1} \) (or equivalent); W1 for the correct final formula \( u_n=3^n+1 \), ideally with a check against part (a). Max 5. Own figure rule applies throughout.

(a) All three terms of \( E \) share the common factor \( C \); factoring this out: \( E = C(\overline{A}\,\overline{B} + \overline{A}\,B + A\,\overline{B}) \). Within the bracket, the first two terms share the common factor \( \overline{A} \): \( \overline{A}\,\overline{B}+\overline{A}\,B = \overline{A}(\overline{B}+B) = \overline{A}(1) = \overline{A} \), using the complement law \( \overline{B}+B=1 \). So the bracket becomes \( \overline{A} + A\overline{B} \). By the absorption law \( X + \overline{X}Y = X+Y \) (applied here with \( X=\overline{A} \), so \( \overline{X}=A \)): \( \overline{A}+A\overline{B} = \overline{A}+\overline{B} \). So \( E = C(\overline{A}+\overline{B}) \), which expands (distributing \( C \)) to \( E=\overline{A}\,C+\overline{B}\,C \). (b) Constructing the truth table for the original expression \( E=\overline{A}\,\overline{B}\,C+\overline{A}\,B\,C+A\,\overline{B}\,C \) and the simplified expression \( \overline{A}\,C+\overline{B}\,C \) for all 8 combinations of A, B, C (0=false, 1=true) gives, in each case, an identical result. For example: A=0,B=0,C=1: original \( = (1)(1)(1)+(1)(0)(1)+(0)(1)(1) = 1+0+0=1 \); simplified \( = (1)(1)+(1)(1)=1+1=1 \) (using OR, so still 1) — both give 1, matching. A=1,B=1,C=1: original \( = (0)(0)(1)+(0)(1)(1)+(1)(0)(1)=0 \); simplified \( = (0)(1)+(0)(1)=0 \) — both give 0, matching. A=1,B=0,C=0: original \( = (0)(1)(0)+(0)(0)(0)+(1)(1)(0)=0 \); simplified \( = (0)(0)+(1)(0)=0 \) — both give 0, matching. Checking every one of the remaining combinations in the same way confirms the original and simplified expressions agree on all 8 rows, verifying the simplification in part (a) is correct. Final answer: (a) \( E=\overline{A}\,C+\overline{B}\,C \); (b) the truth tables for the original and simplified expressions agree exactly on all 8 rows.

(a) M1 for correctly factoring out \( C \); M1 for correctly applying the complement law to simplify \( \overline{A}\,\overline{B}+\overline{A}B \) to \( \overline{A} \); M1 for correctly applying the absorption law to simplify \( \overline{A}+A\overline{B} \) to \( \overline{A}+\overline{B} \); W3 for the correct final simplified expression \( \overline{A}C+\overline{B}C \) (or equivalent correct form), with each algebraic law correctly named/justified. Max 6. (b) M1 for a correctly constructed truth table for the original expression across all 8 rows; W3 for correctly verifying agreement with the simplified expression for at least three combinations, with correct working shown for each. Max 4. Accept any equivalent, fully correct simplification method (e.g. using a Karnaugh map) for part (a).