2024 CCEA AS-Level Further Mathematics 2330 Practice Paper with Answers

(a) \( A^2 = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} (2)(2)+(1)(0) & (2)(1)+(1)(3) \\ (0)(2)+(3)(0) & (0)(1)+(3)(3) \end{pmatrix} = \begin{pmatrix} 4 & 5 \\ 0 & 9 \end{pmatrix} \). (b) \( A^2-5A+6I = \begin{pmatrix} 4 & 5 \\ 0 & 9 \end{pmatrix} - 5\begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} + 6\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 4-10+6 & 5-5+0 \\ 0-0+0 & 9-15+6 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \), confirming the equation holds (this is consistent with the Cayley-Hamilton theorem, since \( A \) has trace \( 2+3=5 \) and determinant \( 2(3)-1(0)=6 \), giving characteristic equation \( \lambda^2-5\lambda+6=0 \)). (c) From part (b), \( A^2=5A-6I \). Multiplying both sides by \( A \): \( A^3=A(5A-6I)=5A^2-6A \); substituting \( A^2=5A-6I \) again, \( A^3=5(5A-6I)-6A=25A-30I-6A=19A-30I \), so \( p=19 \), \( q=-30 \). Then \( A^3=19\begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}-30\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 38 & 19 \\ 0 & 57 \end{pmatrix}-\begin{pmatrix} 30 & 0 \\ 0 & 30 \end{pmatrix} = \begin{pmatrix} 8 & 19 \\ 0 & 27 \end{pmatrix} \). Checking by direct multiplication, \( A^3=A^2 \times A = \begin{pmatrix} 4 & 5 \\ 0 & 9 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 8 & 4+15 \\ 0 & 27 \end{pmatrix} = \begin{pmatrix} 8 & 19 \\ 0 & 27 \end{pmatrix} \), which matches exactly, confirming the answer. Final answer: (a) \( \begin{pmatrix}4&5\\0&9\end{pmatrix} \); (b) confirmed \( =0 \); (c) \( A^3=19A-30I=\begin{pmatrix}8&19\\0&27\end{pmatrix} \).

(a) M1 for correct matrix multiplication method; W2 for the fully correct matrix (W1 if one entry is incorrect). Max 3. (b) M1 for correct substitution of \( A^2 \), \( A \) and \( I \); W2 for correctly showing the result is the zero matrix. Max 3. (c) M1 for correctly rearranging and substituting to find \( A^3=19A-30I \); W1 for the correct final matrix, ideally checked by direct multiplication. Max 2. Own figure rule applies throughout.

(a) For \( 3x^2-kx+12=0 \), comparing with \( ax^2+bx+c=0 \) (so \( a=3, b=-k, c=12 \)): \( \alpha+\beta=-\dfrac{b}{a}=\dfrac{k}{3} \) and \( \alpha\beta=\dfrac{c}{a}=\dfrac{12}{3}=4 \). (b) Substituting into \( \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta \): \( 8 = \left(\dfrac{k}{3}\right)^2 - 2(4) = \dfrac{k^2}{9}-8 \), so \( \dfrac{k^2}{9}=16 \), giving \( k^2=144 \), so \( k=12 \) or \( k=-12 \). (c) Taking the positive value, \( k=12 \): the equation becomes \( 3x^2-12x+12=0 \), which has discriminant \( b^2-4ac = (-12)^2-4(3)(12) = 144-144=0 \). Since the discriminant is zero, the roots are real and equal (a repeated root); indeed, dividing through by 3 gives \( x^2-4x+4=0=(x-2)^2 \), confirming a repeated root at \( x=2 \). Final answer: (a) \( \alpha+\beta=\frac{k}{3} \), \( \alpha\beta=4 \); (b) \( k=\pm12 \); (c) real and equal, since the discriminant is 0 (repeated root \( x=2 \)).

(a) W1 for \( \alpha+\beta=k/3 \); W1 for \( \alpha\beta=4 \). Max 2. (b) M1 for correct substitution into the given identity; M1 for correct rearrangement to \( k^2=144 \) (or equivalent); W3 for both correct values \( k=12 \) and \( k=-12 \) (W1 each, plus 1 mark for recognising both signs are valid solutions of \( k^2=144 \)). Max 5. (c) M1 for calculating the discriminant \( =0 \); W1 for the correct conclusion (real and equal) with correct reasoning. Max 2. Own figure rule applies throughout.

(a) \( \det(N) = (3)(4)-(-1)(2) = 12+2 = 14 \). (b) For a 2D linear transformation, the area scale factor of the transformation is \( |\det(N)| \); the image area is therefore \( |\det(N)| \times \text{original area} = 14 \times 5 = 70 \) square units. (c) A linear transformation preserves orientation if its determinant is positive, and reverses orientation (involves a reflection) if its determinant is negative. Since \( \det(N)=14 \), which is positive, the transformation preserves orientation. Final answer: (a) 14; (b) 70 square units; (c) orientation is preserved, since \( \det(N)>0 \).

(a) M1 for correct method; W1 for \( \det(N)=14 \). Max 2. (b) M1 for recognising the area scale factor is \( |\det(N)| \); W2 for the correct final area, 70 (own figure from (a)). Max 3. (c) W1 for the correct conclusion (preserves orientation); W2 for correct reasoning referencing the positive determinant. Max 3. Own figure rule applies throughout.

(a) Since the cubic has real coefficients, non-real roots occur in conjugate pairs; as \( 1-2i \) is a root, \( 1+2i \) must also be a root. (b) The quadratic factor from the conjugate pair is \( (z-(1-2i))(z-(1+2i)) = (z-1)^2-(2i)^2 = (z-1)^2+4 = z^2-2z+1+4 = z^2-2z+5 \). Let the remaining (real) root be \( r \); then \( z^3-4z^2+9z-10 = (z^2-2z+5)(z-r) \). Expanding the right-hand side: \( (z^2-2z+5)(z-r) = z^3-rz^2-2z^2+2rz+5z-5r = z^3-(r+2)z^2+(2r+5)z-5r \). Comparing coefficients of \( z^2 \): \( -(r+2)=-4 \), so \( r=2 \); checking against the constant term, \( -5r=-10 \) also gives \( r=2 \), and the \( z \) coefficient, \( 2r+5=2(2)+5=9 \), matches exactly, confirming \( r=2 \). This can also be checked by direct substitution: \( 2^3-4(2)^2+9(2)-10 = 8-16+18-10=0 \), confirming \( z=2 \) is indeed a root. (c) The fully factorised form is \( (z-2)(z-1+2i)(z-1-2i)=0 \), corresponding to roots \( 2 \), \( 1+2i \), \( 1-2i \). (d) \( |z| = |1-2i| = \sqrt{1^2+2^2} = \sqrt{5} \). The product of the two complex roots is \( (1-2i)(1+2i) = 1^2-(2i)^2 = 1-(-4) = 5 \); this is also exactly \( |z|^2 = (\sqrt5)^2=5 \), as expected since a complex number multiplied by its own conjugate always gives \( |z|^2 \), confirming the answer. This is consistent with part (c): the quadratic factor \( z^2-2z+5 \) has constant term 5, which equals the product of its roots (the two complex roots), matching exactly. Final answer: (a) \( 1+2i \); (b) third root \( z=2 \); (c) \( (z-2)(z-1+2i)(z-1-2i) \); (d) \( |z|=\sqrt5 \), product of complex roots \( =5 \), consistent with the constant term of the quadratic factor in (c).

(a) W1 for \( 1+2i \); W1 for the correct reason. Max 2. (b) M1 for forming the quadratic factor \( z^2-2z+5 \); M2 for a valid method to find the remaining factor/root (comparing coefficients or division); W2 for correctly identifying and verifying the third root \( z=2 \); W1 for a clear, fully justified method. Max 6. (c) W3 for the fully correct factorised form. Max 3. (d) M1 for correct calculation of \( |z| \); W1 for \( |z|=\sqrt5 \); M1 for correct calculation of the product of complex roots; W1 for the correct value, 5; W1 for correctly relating this to \( |z|^2 \) and/or the constant term in part (c). Max 5. Own figure rule applies throughout.

(a) The direction vector of \( L_1 \) is \( (1,-1,2) \) and of \( L_2 \) is \( (2,1,-1) \). For the lines to be parallel, one direction vector would need to be a scalar multiple of the other; comparing corresponding components, \( \frac{1}{2} \), \( \frac{-1}{1}=-1 \), and \( \frac{2}{-1}=-2 \) are all different, so there is no single scalar \( k \) with \( (1,-1,2)=k(2,1,-1) \), confirming \( L_1 \) and \( L_2 \) are not parallel. (b) Setting the general points of \( L_1 \) and \( L_2 \) equal component-wise: \(\mathbf{i}\): \( 1+s = 3+2t \), i.e. \( s-2t=2 \); \(\mathbf{j}\): \( 2-s = 0+t \), i.e. \( s+t=2 \); \(\mathbf{k}\): \( 0+2s = 1-t \), i.e. \( 2s+t=1 \). Solving the first two equations simultaneously: from \( s+t=2 \), \( s=2-t \); substituting into \( s-2t=2 \): \( (2-t)-2t=2 \), i.e. \( 2-3t=2 \), i.e. \( t=0 \), and then \( s=2-0=2 \). Substituting \( s=2 \), \( t=0 \) into the \(\mathbf{k}\) equation to check consistency: left-hand side \( 2s=2(2)=4 \); right-hand side \( 1-t=1-0=1 \); since \( 4\neq1 \), the values of \( s \) and \( t \) that satisfy the \(\mathbf{i}\) and \(\mathbf{j}\) equations do NOT also satisfy the \(\mathbf{k}\) equation, so there is no single point common to both lines — they do not intersect. (c) Since \( L_1 \) and \( L_2 \) have been shown to be non-parallel (part (a)) and non-intersecting (part (b)), by definition they must be skew lines — lines in three dimensions that do not lie in the same plane. Final answer: (a) not parallel, direction vectors are not scalar multiples; (b) \( s=2,t=0 \) from two components, but \( \mathbf{k} \) component gives \( 4\neq1 \), so no intersection; (c) \( L_1 \) and \( L_2 \) are skew lines.

(a) M1 for comparing corresponding components of the direction vectors; W1 for the correct conclusion (not parallel) with valid justification. Max 2. (b) M1 for correctly setting up all three component equations; M2 for correctly solving the first two equations simultaneously; W1 for correct values \( s=2,t=0 \); W2 for correctly substituting into the third equation and identifying the contradiction \( 4\neq1 \). Max 6. (c) W1 for the correct conclusion (skew lines); W2 for a full, correct justification referencing both non-parallel and non-intersecting. Max 3. Own figure rule applies throughout.

(a) \( |z| = \sqrt{(-2\sqrt3)^2+2^2} = \sqrt{12+4} = \sqrt{16} = 4 \). The reference angle is \( \arctan\left(\dfrac{2}{2\sqrt3}\right) = \arctan\left(\dfrac{1}{\sqrt3}\right) = \dfrac{\pi}{6} \); since the real part of \( z \) is negative and the imaginary part is positive, \( z \) lies in the second quadrant, so \( \arg(z) = \pi-\dfrac{\pi}{6} = \dfrac{5\pi}{6} \). So \( z = 4\left(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6}\right) \). (b) By De Moivre's theorem, \( z^3 = 4^3\left(\cos\left(3\times\dfrac{5\pi}{6}\right)+i\sin\left(3\times\dfrac{5\pi}{6}\right)\right) = 64\left(\cos\dfrac{15\pi}{6}+i\sin\dfrac{15\pi}{6}\right) = 64\left(\cos\dfrac{5\pi}{2}+i\sin\dfrac{5\pi}{2}\right) \). Since \( \dfrac{5\pi}{2}=2\pi+\dfrac{\pi}{2} \), this is equivalent to angle \( \dfrac{\pi}{2} \), so \( \cos\dfrac{5\pi}{2}=0 \) and \( \sin\dfrac{5\pi}{2}=1 \), giving \( z^3 = 64(0+i(1)) = 64i \). Verifying by direct expansion, with \( a=-2\sqrt3 \), \( b=2 \), using \( (a+bi)^3=(a^3-3ab^2)+(3a^2b-b^3)i \): \( a^3=(-2\sqrt3)^3=-8(3\sqrt3)=-24\sqrt3 \); \( 3ab^2=3(-2\sqrt3)(4)=-24\sqrt3 \); real part \( =a^3-3ab^2=-24\sqrt3-(-24\sqrt3)=0 \); \( 3a^2b=3(12)(2)=72 \) (since \( a^2=(2\sqrt3)^2=12 \)); \( b^3=8 \); imaginary part \( =3a^2b-b^3=72-8=64 \); so direct expansion gives \( z^3=0+64i=64i \), which matches the De Moivre result exactly, confirming the answer. Final answer: (a) \( |z|=4 \), \( \arg(z)=\frac{5\pi}{6} \), \( z=4(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}) \); (b) \( z^3=64i \), confirmed by two independent methods.

(a) M1 for correct method for \( |z| \); W1 for \( |z|=4 \); M1 for correct reference angle; W1 for correct \( \arg(z)=\frac{5\pi}{6} \) (with correct quadrant reasoning); W1 for the correct polar form. Max 5. (b) M1 for correct application of De Moivre's theorem; W1 for correct angle \( \frac{15\pi}{6} \) or equivalent reduced angle; W1 for the correct De Moivre result \( 64i \); M2 for a fully correct direct binomial expansion; W1 for confirming the two methods agree. Max 6. Own figure rule applies throughout.

(a) Expanding along the first row: \( \det(B) = 1\big[(1)(1)-(1)(2)\big] - 0\big[(2)(1)-(1)(0)\big] + 2\big[(2)(2)-(1)(0)\big] = 1(1-2) - 0 + 2(4-0) = -1+8=7 \). (b) The matrix of cofactors is found by computing each \( 2\times2 \) minor with the appropriate sign: \( C_{11}=(1)(1)-(1)(2)=-1 \); \( C_{12}=-[(2)(1)-(1)(0)]=-2 \); \( C_{13}=(2)(2)-(1)(0)=4 \); \( C_{21}=-[(0)(1)-(2)(2)]=4 \); \( C_{22}=(1)(1)-(2)(0)=1 \); \( C_{23}=-[(1)(2)-(0)(0)]=-2 \); \( C_{31}=(0)(1)-(2)(1)=-2 \); \( C_{32}=-[(1)(1)-(2)(2)]=3 \); \( C_{33}=(1)(1)-(0)(2)=1 \). So the cofactor matrix is \( \begin{pmatrix}-1&-2&4\\4&1&-2\\-2&3&1\end{pmatrix} \); transposing to form the adjugate: \( \text{adj}(B) = \begin{pmatrix}-1&4&-2\\-2&1&3\\4&-2&1\end{pmatrix} \); then \( B^{-1}=\dfrac{1}{\det(B)}\text{adj}(B) = \dfrac{1}{7}\begin{pmatrix}-1&4&-2\\-2&1&3\\4&-2&1\end{pmatrix} \). (c) Writing the system as \( B\begin{pmatrix}x\y\z\end{pmatrix}=\begin{pmatrix}7\\7\\0\end{pmatrix} \), so \( \begin{pmatrix}x\y\z\end{pmatrix}=B^{-1}\begin{pmatrix}7\\7\\0\end{pmatrix} = \dfrac{1}{7}\begin{pmatrix}-1&4&-2\\-2&1&3\\4&-2&1\end{pmatrix}\begin{pmatrix}7\\7\\0\end{pmatrix} = \dfrac{1}{7}\begin{pmatrix}-1(7)+4(7)-2(0)\\-2(7)+1(7)+3(0)\\4(7)-2(7)+1(0)\end{pmatrix} = \dfrac{1}{7}\begin{pmatrix}-7+28\\-14+7\\28-14\end{pmatrix} = \dfrac{1}{7}\begin{pmatrix}21\\-7\\14\end{pmatrix} = \begin{pmatrix}3\\-1\\2\end{pmatrix} \). Checking by substitution into the original equations: \( x+2z=3+2(2)=3+4=7 \) ✓; \( 2x+y+z=2(3)+(-1)+2=6-1+2=7 \) ✓; \( 2y+z=2(-1)+2=-2+2=0 \) ✓; all three confirmed. Final answer: (a) \( \det(B)=7 \); (b) \( B^{-1}=\frac{1}{7}\begin{pmatrix}-1&4&-2\\-2&1&3\\4&-2&1\end{pmatrix} \); (c) \( x=3, y=-1, z=2 \).

(a) M2 for correct determinant expansion method; W2 for the correct value, 7. Max 4. (b) M3 for correctly calculating all nine cofactors (partial credit for at least 6 correct); M2 for correctly forming the cofactor matrix and transposing to the adjugate; W3 for the fully correct inverse matrix (dividing by the determinant). Max 8. (c) M1 for correctly setting up \( B^{-1}\begin{pmatrix}7\\7\\0\end{pmatrix} \); M2 for correct matrix-vector multiplication method; W2 for the fully correct solution \( x=3,y=-1,z=2 \), ideally checked against the original equations. Max 5. Own figure rule applies throughout.

(a) First find \( \mathbf{b}\times\mathbf{c} \), with \( \mathbf{b}=(1,3,0) \) and \( \mathbf{c}=(0,1,2) \): \( \mathbf{b}\times\mathbf{c} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&3&0\\0&1&2\end{vmatrix} = \mathbf{i}[(3)(2)-(0)(1)] - \mathbf{j}[(1)(2)-(0)(0)] + \mathbf{k}[(1)(1)-(3)(0)] = \mathbf{i}(6) - \mathbf{j}(2) + \mathbf{k}(1) = (6,-2,1) \). Then, with \( \mathbf{a}=(2,0,1) \): \( \mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) = (2)(6)+(0)(-2)+(1)(1) = 12+0+1=13 \). The volume of the parallelepiped is the absolute value of this scalar triple product: \( 13 \) cubic units. (b) The volume of a tetrahedron with edges \( \mathbf{a} \), \( \mathbf{b} \), \( \mathbf{c} \) from a common vertex is \( \dfrac{1}{6} \) of the volume of the parallelepiped formed by the same three vectors, since the tetrahedron is one of six congruent-volume pieces the parallelepiped can be divided into using this construction; so the tetrahedron volume is \( \dfrac{13}{6} \approx 2.17 \) cubic units (3 s.f.). (c) The angle \( \phi \) between a line with direction \( \mathbf{d} \) and a plane with normal \( \mathbf{n} \) satisfies \( \sin\phi = \dfrac{|\mathbf{n}\cdot\mathbf{d}|}{|\mathbf{n}||\mathbf{d}|} \) (this is the complement of the angle between \( \mathbf{d} \) and \( \mathbf{n} \), since the line's angle to the plane is measured from the plane itself, not from the normal). Here \( \mathbf{n}\cdot\mathbf{d} = (1)(2)+(2)(-1)+(2)(2) = 2-2+4=4 \); \( |\mathbf{n}|=\sqrt{1+4+4}=\sqrt9=3 \); \( |\mathbf{d}|=\sqrt{4+1+4}=\sqrt9=3 \); so \( \sin\phi = \dfrac{|4|}{3\times3} = \dfrac{4}{9} \). Therefore \( \phi = \arcsin\left(\dfrac{4}{9}\right) \approx 26.3878^{\circ} \), which rounds to \( 26.4^{\circ} \) (3 s.f.). Final answer: (a) volume \( =13 \); (b) tetrahedron volume \( =\frac{13}{6}\approx2.17 \); (c) angle \( \approx26.4^{\circ} \).

(a) M2 for correct method/expansion for \( \mathbf{b}\times\mathbf{c} \); W2 for the correct cross product \( (6,-2,1) \); M2 for correct dot product method with \( \mathbf{a} \); W2 for the correct final volume, 13. Max 8. (b) M1 for correctly recognising the tetrahedron volume is \( \frac{1}{6} \) of the parallelepiped volume; W1 for the correct value, \( \frac{13}{6}\approx2.17 \) (own figure from (a)). Max 2. (c) M1 for correct formula \( \sin\phi=\frac{|\mathbf{n}\cdot\mathbf{d}|}{|\mathbf{n}||\mathbf{d}|} \) (recognising the line-plane angle formula, distinct from the line-line/plane-plane angle formula); M2 for correct dot product; M2 for correct magnitudes of \( \mathbf{n} \) and \( \mathbf{d} \); W2 for the correct unrounded angle; W2 for correct rounding to \( 26.4^{\circ} \) (3 s.f.). Max 10. Own figure rule applies throughout.

(a) In equilibrium, \( T=mg=1.5\times9.8=14.7 \) N. By Hooke's Law, \( T=\dfrac{\lambda x}{l} \), so \( 14.7=\dfrac{45x}{0.6}=75x \), giving \( x=\dfrac{14.7}{75}=0.196 \) m. (b) \( \text{EPE}=\dfrac{\lambda x^2}{2l}=\dfrac{45\times(0.196)^2}{2\times0.6}=\dfrac{45\times0.038416}{1.2}=\dfrac{1.72872}{1.2}=1.4406 \) J, which rounds to 1.44 J (3 s.f.). (c) The total (stretched) length is the natural length plus the extension: \( 0.6+0.196=0.796 \) m. Final answer: (a) \( x=0.196 \) m; (b) EPE \( \approx1.44 \) J; (c) total length \( =0.796 \) m.

(a) M1 for \( T=mg \); M1 for correct rearrangement of Hooke's Law; W1 for \( x=0.196 \) m. Max 3. (b) M1 for correct EPE formula; W2 for \( 1.44 \) J (3 s.f., own figure from (a)). Max 3. (c) W2 for correctly adding natural length and extension, \( 0.796 \) m. Max 2. Own figure rule applies throughout.

(a) A stratified sample is obtained by first dividing the population into non-overlapping groups, called strata, based on a relevant characteristic (here, year group); a sample is then taken from each stratum, with the number selected from each stratum proportional to the size of that stratum within the whole population, and (typically) selected using simple random sampling within each stratum. (b) The sampling fraction is \( \dfrac{60}{450} = \dfrac{2}{15} \), applied to each year group: Year 8: \( \dfrac{180}{450}\times60 = 24 \); Year 9: \( \dfrac{150}{450}\times60 = 20 \); Year 10: \( \dfrac{120}{450}\times60 = 16 \). Checking: \( 24+20+16=60 \), matching the required total sample size, confirming the calculation. (c) Because the sample is taken proportionally from each year group, a stratified sample guarantees that every year group is represented in the sample in proportion to its actual size in the school; a simple random sample of the same total size could, purely by chance, select too many students from one year group and too few from another, making a stratified sample generally more representative of the whole school population for this kind of grouped characteristic. Final answer: (a) sampling proportionally from each of several distinct sub-groups (strata) of the population; (b) 24, 20, 16 for Years 8, 9, 10 respectively; (c) e.g. guarantees proportional representation of each year group, which a simple random sample cannot guarantee.

(a) W1 for reference to dividing the population into strata/groups; W1 for reference to sampling proportionally from each stratum. Max 2. (b) M1 for correct method (proportion x sample size); W1 each for the three correct values, 24, 20, 16. Max 4. (c) W2 for a valid, clearly explained advantage. Max 2.

(a) At constant speed, the acceleration is zero, so by Newton's second law the driving force \( F \) balances both the resistance and the component of the total weight acting down the incline: \( F = R + (m_{car}+m_{trailer})g\sin5^{\circ} \). The combined mass is \( 1200+400=1600 \) kg, so the weight component down the slope is \( 1600\times9.8\times\sin5^{\circ} \approx 1600\times9.8\times0.08716 \approx 1366.6 \) N. So \( F \approx 600+1366.6 = 1966.6 \) N, which rounds to 1970 N (3 s.f.). (b) Power \( =Fv=1966.6\times20\approx39{,}332 \) W \( \approx39.3 \) kW (3 s.f.), using the unrounded value of \( F \) from part (a). (c) On a level road, the driving force at constant speed only needs to balance the resistance to motion, \( R=600 \) N. On the incline, gravity has a component acting back down the slope on the combined mass of car and trailer, which the driving force must also overcome in addition to the resistance, in order to maintain the same constant speed; this extra force requirement (approximately 1366.6 N here) is why a larger total driving force, and correspondingly greater power, is needed on the incline than would be needed at the same speed on a level road. Final answer: (a) \( F\approx1970 \) N; (b) \( P\approx39.3 \) kW; (c) the driving force must also overcome the weight component down the slope, not just the resistance to motion, which is the sole opposing force on a level road.

(a) M1 for correct resolution of forces at constant velocity (driving force = resistance + gravity component); M1 for correct combined mass; M1 for correct gravity component calculation; W2 for the correct final answer \( \approx1970 \) N (3 s.f.) (own figure). Max 5. (b) M1 for correct use of \( P=Fv \); W2 for the correct final answer \( \approx39.3 \) kW (3 s.f.), own figure from (a). Max 3. (c) W3 for a complete, correct explanation referencing the additional weight component down the slope that must be overcome on the incline but not on a level road. Max 3. Own figure rule applies throughout.

(a) On the first draw, there are 8 balls in total (5 red, 3 blue), so \( P(\text{red}_1)=\frac{5}{8} \) and \( P(\text{blue}_1)=\frac{3}{8} \). Since the draws are without replacement, the probabilities on the second draw depend on the outcome of the first: if the first ball was red, 7 balls remain (4 red, 3 blue), so \( P(\text{red}_2\mid\text{red}_1)=\frac{4}{7} \) and \( P(\text{blue}_2\mid\text{red}_1)=\frac{3}{7} \); if the first ball was blue, 7 balls remain (5 red, 2 blue), so \( P(\text{red}_2\mid\text{blue}_1)=\frac{5}{7} \) and \( P(\text{blue}_2\mid\text{blue}_1)=\frac{2}{7} \). (b) \( P(\text{both red}) = P(\text{red}_1)\times P(\text{red}_2\mid\text{red}_1) = \frac{5}{8}\times\frac{4}{7} = \frac{20}{56} = \frac{5}{14} \). (c) The two balls are different colours if either red-then-blue or blue-then-red occurs: \( P(\text{different}) = P(\text{red}_1)P(\text{blue}_2\mid\text{red}_1) + P(\text{blue}_1)P(\text{red}_2\mid\text{blue}_1) = \frac{5}{8}\times\frac{3}{7} + \frac{3}{8}\times\frac{5}{7} = \frac{15}{56}+\frac{15}{56} = \frac{30}{56} = \frac{15}{28} \). (d) This is read directly from the tree diagram branch for 'first red, then blue': \( P(\text{second blue}\mid\text{first red}) = \frac{3}{7} \), since after removing one red ball, 3 blue balls remain out of the 7 balls left. Final answer: (a) as described above; (b) \( \frac{5}{14} \); (c) \( \frac{15}{28} \); (d) \( \frac{3}{7} \).

(a) W1 for correct first-draw probabilities; W2 for correct second-draw (conditional) probabilities for both branches. Max 3. (b) M1 for correct method (multiplying along the red-red branch); W2 for the correct final value, \( \frac{5}{14} \). Max 3. (c) M1 for correctly identifying both relevant branches (red-blue and blue-red); M1 for correct method; W1 for the correct final value, \( \frac{15}{28} \). Max 3. (d) W2 for the correct value, \( \frac{3}{7} \), read correctly from the tree diagram/conditional reasoning. Max 2. Own figure rule applies throughout.

(a) On the incline, the component of gravity along the slope is \( mg\sin20^{\circ}=2(9.8)\sin20^{\circ}\approx6.704 \) N, and the friction force opposing motion is \( \mu mg\cos20^{\circ}=0.15(2)(9.8)\cos20^{\circ}\approx2.763 \) N. By the work-energy principle, the net work done over the 3 m equals the kinetic energy gained: \( (6.704-2.763)\times3 \approx 11.823 \) J \( = \frac{1}{2}(2)v^2 = v^2 \), so \( v^2\approx11.823 \), giving \( v\approx\sqrt{11.823}\approx3.4384 \), which rounds to 3.44 ms\(^{-1}\) (3 s.f.). (b) On the horizontal surface, the particle decelerates due to friction alone, \( \mu_2mg=0.3(2)(9.8)=5.88 \) N; by the work-energy principle, all the kinetic energy at the bottom of the incline is converted into work done against this friction as the particle comes to rest: \( \frac{1}{2}mv^2 = \mu_2mgd \), so \( d=\dfrac{v^2}{2\mu_2g}=\dfrac{11.823}{2(0.3)(9.8)}=\dfrac{11.823}{5.88}\approx2.0107 \), which rounds to 2.01 m (3 s.f.), using the unrounded value of \( v^2 \) from part (a). Final answer: (a) \( v\approx3.44 \) ms\(^{-1}\); (b) distance \( \approx2.01 \) m.

(a) M1 for correct gravity component along the incline; M1 for correct friction force on the incline; M1 for correctly setting up the work-energy equation; W3 for the correct final answer \( 3.44 \) ms\(^{-1}\) (3 s.f.) (W1 for correct \( v^2 \), W2 for correct final rounded value). Max 6. (b) M1 for correct friction force on the horizontal surface; M1 for correctly setting up the work-energy equation for the horizontal stage; W3 for the correct final answer \( 2.01 \) m (3 s.f.), using the own-figure value of \( v^2 \) from part (a). Max 5. Own figure rule applies throughout.

(a) Since there is a fixed number of independent trials (8 tosses), each with the same probability of 'success' (heads, \( p=0.3 \)), and \( X \) counts the number of successes, \( X \) follows a binomial distribution, \( X\sim B(8,0.3) \). (b) Using the binomial probability formula, \( P(X=3) = \binom{8}{3}(0.3)^3(0.7)^5 = 56\times0.027\times0.16807 \approx 0.25412 \), which rounds to 0.254 (3 s.f.). (c) \( P(X\le2) = P(X=0)+P(X=1)+P(X=2) \). \( P(X=0)=\binom{8}{0}(0.3)^0(0.7)^8\approx0.05765 \); \( P(X=1)=\binom{8}{1}(0.3)^1(0.7)^7\approx0.19765 \); \( P(X=2)=\binom{8}{2}(0.3)^2(0.7)^6\approx0.29648 \). Summing: \( 0.05765+0.19765+0.29648\approx0.55177 \), which rounds to 0.552 (3 s.f.). (d) For a binomial distribution \( B(n,p) \), the mean is \( np \) and the variance is \( np(1-p) \); here, mean \( =8\times0.3=2.4 \), and variance \( =8\times0.3\times0.7=1.68 \). Final answer: (a) \( X\sim B(8,0.3) \); (b) \( P(X=3)\approx0.254 \); (c) \( P(X\le2)\approx0.552 \); (d) mean \( =2.4 \), variance \( =1.68 \).

(a) W1 for 'binomial'; W1 for correct parameters \( n=8,p=0.3 \). Max 2. (b) M1 for correct binomial coefficient \( \binom{8}{3}=56 \); M2 for correct formula and substitution; W1 for the correct final answer, 0.254 (3 s.f.). Max 4. (c) M1 for correctly identifying the need to sum \( P(0)+P(1)+P(2) \); M1 for correct calculation of the three individual probabilities; W1 for the correct final answer, 0.552 (3 s.f.). Max 3. (d) W1 for correct mean, 2.4; W1 for correct variance, 1.68. Max 2.

(a) The radius of the circular path is \( r=L\sin\theta=0.5\times\sin30^{\circ}=0.5\times0.5=0.25 \) m. (b) Resolving vertically, the vertical component of tension balances the weight (since the particle moves in a horizontal circle, with no vertical acceleration): \( T\cos30^{\circ}=mg \), so \( T=\dfrac{mg}{\cos30^{\circ}}=\dfrac{0.3\times9.8}{\cos30^{\circ}}=\dfrac{2.94}{0.8660}\approx3.3948 \), which rounds to 3.39 N (3 s.f.). Resolving horizontally, the horizontal component of tension provides the centripetal force: \( T\sin30^{\circ}=\dfrac{mv^2}{r} \), so \( v^2=\dfrac{T\sin30^{\circ}\times r}{m}=\dfrac{3.3948\times0.5\times0.25}{0.3}\approx1.4145 \), giving \( v\approx\sqrt{1.4145}\approx1.1893 \), which rounds to 1.19 ms\(^{-1}\) (3 s.f.), using the unrounded value of \( T \). Final answer: (a) \( r=0.25 \) m; (b) \( T\approx3.39 \) N, \( v\approx1.19 \) ms\(^{-1}\).

(a) M1 for correct use of \( r=L\sin\theta \); W1 for \( r=0.25 \) m. Max 2. (b) M1 for correct vertical resolution \( T\cos\theta=mg \); W1 for \( T\approx3.39 \) N; M1 for correct horizontal resolution \( T\sin\theta=\frac{mv^2}{r} \); M1 for correctly substituting to find \( v^2 \); W3 for the correct final answer \( v\approx1.19 \) ms\(^{-1}\) (3 s.f.) (W1 for correct \( v^2 \), W2 for correct rounded final value), own figure from the T value found. Max 7. Own figure rule applies throughout.

(a) \( \sum x = 2+4+6+8+10=30 \); \( \sum y = 3+5+6+8+9=31 \); \( \sum xy = (2)(3)+(4)(5)+(6)(6)+(8)(8)+(10)(9) = 6+20+36+64+90 = 216 \); \( \sum x^2 = 4+16+36+64+100=220 \); \( \sum y^2 = 9+25+36+64+81=215 \). (b) Using the formula \( r = \dfrac{n\sum xy - \sum x\sum y}{\sqrt{\left(n\sum x^2-(\sum x)^2\right)\left(n\sum y^2-(\sum y)^2\right)}} \), with \( n=5 \): numerator \( = 5(216)-(30)(31) = 1080-930=150 \); \( n\sum x^2-(\sum x)^2 = 5(220)-900=1100-900=200 \); \( n\sum y^2-(\sum y)^2 = 5(215)-(31)^2=1075-961=114 \); denominator \( = \sqrt{200\times114}=\sqrt{22{,}800}\approx150.997 \). So \( r \approx \dfrac{150}{150.997} \approx 0.99340 \), which rounds to 0.993 (3 s.f.), indicating a very strong positive linear correlation between \( x \) and \( y \). Final answer: (a) as listed above; (b) \( r\approx0.993 \).

(a) W1 for each of the five correct sums. Max 5. (b) M1 for correct formula for \( r \); M2 for correct substitution and calculation of numerator and denominator; W1 for the correct final value, 0.993 (3 s.f.). Max 4. Own figure rule applies throughout.

(a) As \( P \) descends \( h=1.5 \) m, \( Q \) moves the same distance, 1.5 m, up the incline (string inextensible). Applying conservation of energy: the loss in gravitational PE of \( P \) is converted into the kinetic energy of both particles (moving with common speed \( v \)), the gain in gravitational PE of \( Q \) (as it rises up the slope), and the work done against friction on \( Q \). Loss in PE of \( P \): \( m_Pgh=6(9.8)(1.5)=88.2 \) J. Gain in PE of \( Q \): \( m_Qgh\sin25^{\circ}=2(9.8)(1.5)\sin25^{\circ}\approx12.425 \) J. Normal reaction on \( Q \): \( N=m_Qg\cos25^{\circ}=2(9.8)\cos25^{\circ}\approx17.763 \) N; work done against friction: \( \mu N h=0.2(17.763)(1.5)\approx5.329 \) J. Kinetic energy gained: \( \frac{1}{2}(m_P+m_Q)v^2=\frac{1}{2}(8)v^2=4v^2 \). By conservation of energy: \( 88.2 = 4v^2 + 12.425 + 5.329 \), so \( 4v^2 \approx 88.2-17.754 = 70.446 \), giving \( v^2\approx17.6115 \), so \( v\approx\sqrt{17.6115}\approx4.1966 \), which rounds to 4.20 ms\(^{-1}\) (3 s.f.). (b) One key modelling assumption is that the pulley is smooth (frictionless) and light (massless); this means the pulley does not exert any resistive torque and does not absorb any energy as the string passes over it, and it also means the tension in the string is the same on both sides of the pulley. This is essential for the energy conservation calculation above, since it allows the total mechanical energy of the whole system (P and Q together) to be conserved, aside from the explicitly modelled loss to friction on \( Q \); if the pulley were rough or had mass, additional energy would be lost to the pulley itself, and this simple energy equation would no longer be valid without further adjustment. Final answer: (a) \( v\approx4.20 \) ms\(^{-1}\); (b) e.g. the pulley is smooth and light, meaning no energy is lost at the pulley and the string tension is equal on both sides, which is essential for the energy conservation equation used.

(a) M1 for correct loss in PE of P; M1 for correct gain in PE of Q; M1 for correct normal reaction and friction force on Q; M1 for correct work done against friction; M1 for correctly setting up the overall energy conservation equation; W3 for the correct final answer \( 4.20 \) ms\(^{-1}\) (3 s.f.) (W1 for correct \( v^2 \), W2 for correct rounding). Max 8. (b) W1 for stating a valid modelling assumption (e.g. smooth/light pulley, particles, no air resistance); W2 for a correct, clearly linked explanation of its significance to the calculation. Max 3. Own figure rule applies throughout.

(a) Standardising, \( Z=\dfrac{182-175}{7}=\dfrac{7}{7}=1 \), so \( P(H>182)=P(Z>1) \); from standard normal tables, \( P(Z>1)\approx0.1587 \), which rounds to 0.159 (3 s.f.). (b) Standardising both bounds: for \( H=168 \), \( Z=\dfrac{168-175}{7}=\dfrac{-7}{7}=-1 \); for \( H=182 \), \( Z=1 \) (as in part (a)). So \( P(1681) \approx 1-2(0.1587)=1-0.3173=0.6827 \), which rounds to 0.683 (3 s.f.) (this is the well-known result that approximately 68.3% of a normal distribution lies within one standard deviation of the mean). (c) \( P(H>h)=0.10 \) means \( h \) is the value such that only the top 10% of the distribution lies above it; from standard normal tables, the z-value with \( P(Z>z)=0.10 \) is \( z\approx1.2816 \). Converting back to the original scale: \( h=\mu+z\sigma=175+1.2816\times7\approx175+8.971\approx183.97 \), which rounds to 184 cm (nearest cm). Final answer: (a) \( P(H>182)\approx0.159 \); (b) \( P(168

(a) M1 for correct standardisation; W1 for correct z-value, 1; W2 for the correct final probability, 0.159 (3 s.f.). Max 4. (b) M1 for correct standardisation of both bounds; M1 for correct use of symmetry (or direct table values) to combine; W2 for the correct final probability, 0.683 (3 s.f.). Max 4. (c) M1 for correctly identifying the required z-value (\( z\approx1.28 \)) from \( P(Z>z)=0.10 \); M1 for correct use of \( h=\mu+z\sigma \); W1 for the correct final answer, 184 cm (nearest cm). Max 3. Own figure rule applies throughout.