2024 Edexcel IAS-Level Further Mathematics (XFM01) Practice Paper with Answers

(a) Given \(\frac{z_1}{z_2} = 1 + i\), we can write \(z_1 = (1 + i)z_2\). Substituting the expressions for \(z_1\) and \(z_2\): \(p+2i = (1+i)(3-qi) = 3 - qi + 3i - qi^2 = (3+q) + (3-q)i\). Equating the imaginary parts: \(2 = 3 - q \Rightarrow q = 1\). Equating the real parts: \(p = 3 + q \Rightarrow p = 3 + 1 = 4\). (b) Since \(p=4\) and \(q=1\), we have \(z_1 = 4+2i\) and \(z_2 = 3-i\). The modulus of \(z_1\) is \(|z_1| = \sqrt{4^2+2^2} = \sqrt{20}\) and the modulus of \(z_2\) is \(|z_2| = \sqrt{3^2+(-1)^2} = \sqrt{10}\). The modulus of the product is \(|z_1 z_2| = |z_1||z_2| = \sqrt{20} \times \sqrt{10} = \sqrt{200} = 10\sqrt{2}\).

(a) M1: Attempts to multiply both sides by \(3-qi\) and expands the right-hand side. A1: Obtains the correct expanded form \((3+q) + (3-q)i\). M1: Equates imaginary parts to find \(q\). A1: Correct value of \(q = 1\). A1: Correct value of \(p = 4\). (b) M1: Attempts to find the modulus of either \(z_1\), \(z_2\) or the product \(z_1 z_2\). A1: Correct modulus \(10\sqrt{2}\) or equivalent exact form.

From the given quadratic equation, we have \(\alpha + \beta = \frac{5}{2}\) and \(\alpha\beta = 2\). (a) Using the algebraic identity: \(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\). Substituting the sum and product values: \(\alpha^3 + \beta^3 = \left(\frac{5}{2}\right)^3 - 3(2)\left(\frac{5}{2}\right) = \frac{125}{8} - 15 = \frac{5}{8}\). (b) Let the roots of the new equation be \(\gamma = \frac{\alpha}{\beta^2}\) and \(\delta = \frac{\beta}{\alpha^2}\). The sum of the roots is \(\gamma + \delta = \frac{\alpha}{\beta^2} + \frac{\beta}{\alpha^2} = \frac{\alpha^3 + \beta^3}{\alpha^2\beta^2} = \frac{\alpha^3 + \beta^3}{(\alpha\beta)^2} = \frac{5/8}{4} = \frac{5}{32}\). The product of the roots is \(\gamma\delta = \frac{\alpha}{\beta^2} \times \frac{\beta}{\alpha^2} = \frac{1}{\alpha\beta} = \frac{1}{2}\). The new quadratic equation is given by \(x^2 - (\gamma+\delta)x + \gamma\delta = 0 \Rightarrow x^2 - \frac{5}{32}x + \frac{1}{2} = 0\). Multiplying by 32 to obtain integer coefficients: \(32x^2 - 5x + 16 = 0\).

(a) B1: Correctly states \(\alpha + \beta = \frac{5}{2}\) and \(\alpha\beta = 2\). M1: Uses a valid algebraic identity for \(\alpha^3 + \beta^3\) in terms of sum and product. M1: Substitutes the values into their identity. A1: Correct value of \(\frac{5}{8}\). (b) M1: Finds an expression for the sum of the new roots and substitutes values to find its numerical value. A1: Correct sum \(\frac{5}{32}\). B1: Correct product \(\frac{1}{2}\). A1: Correct quadratic equation with integer coefficients: \(32x^2 - 5x + 16 = 0\) (including the '= 0' part).

(a) Evaluating \(f(x)\) at the boundaries: \(f(2.1) = 2(2.1)^3 - 6(2.1)^2 + 3(2.1) + 1 = 18.522 - 26.46 + 6.3 + 1 = -0.638\). \(f(2.3) = 2(2.3)^3 - 6(2.3)^2 + 3(2.3) + 1 = 24.334 - 31.74 + 6.9 + 1 = 0.494\). Since \(f(x)\) is a continuous function and there is a change of sign between \(f(2.1) < 0\) and \(f(2.3) > 0\), a root \(\alpha\) must lie in the interval \([2.1, 2.3]\). (b) First differentiate: \(f'(x) = 6x^2 - 12x + 3\). At \(x_0 = 2.2\): \(f(2.2) = 2(2.2)^3 - 6(2.2)^2 + 3(2.2) + 1 = -0.144\), and \(f'(2.2) = 6(2.2)^2 - 12(2.2) + 3 = 5.64\). Applying the Newton-Raphson formula: \(x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 2.2 - \frac{-0.144}{5.64} \approx 2.22553\). To 3 decimal places, this is \(2.226\). (c) Using linear interpolation on \([2.2, 2.3]\): \(\frac{\alpha - 2.2}{2.3 - \alpha} = \frac{|f(2.2)|}{|f(2.3)|} = \frac{0.144}{0.494}\). Solving for \(\alpha\): \(\alpha - 2.2 = \frac{0.144}{0.144+0.494}(2.3 - 2.2) = \frac{0.144}{0.638} \times 0.1 \approx 0.02257\), which gives \(\alpha \approx 2.22257\). To 3 decimal places, this is \(2.223\).

(a) M1: Attempts to evaluate both \(f(2.1)\) and \(f(2.3)\). A1: Explains that the change of sign and continuity implies a root in the interval. (b) M1: Differentiates \(f(x)\) to get \(f'(x) = 6x^2 - 12x + 3\). A1: Correctly evaluates \(f(2.2)\) and \(f'(2.2)\). M1: Applies the Newton-Raphson formula correctly. A1: Correct answer to 3 d.p. (2.226). (c) M1: Sets up a correct linear interpolation ratio on \([2.2, 2.3]\). A1: Correct answer to 3 d.p. (2.223).

(a) Differentiating \(y^2 = 12x\) with respect to \(x\) implicitly gives: \(2y \frac{dy}{dx} = 12 \Rightarrow \frac{dy}{dx} = \frac{6}{y}\). At the point \(P(3t^2, 6t)\), the gradient of the tangent is \(m_T = \frac{6}{6t} = \frac{1}{t}\). Hence, the gradient of the normal is \(m_N = -t\). The equation of the normal at \(P\) is: \(y - 6t = -t(x - 3t^2) \Rightarrow y - 6t = -tx + 3t^3 \Rightarrow y + tx = 3t^3 + 6t\). (b) The normal cuts the \(x\)-axis at \(Q\), so the \(y\)-coordinate of \(Q\) is 0. Substituting \(y = 0\) into the normal equation: \(tx = 3t^3 + 6t\). Since \(t \neq 0\), we can divide by \(t\) to get: \(x = 3t^2 + 6\). Substituting \(t = 2\) yields: \(x = 3(2)^2 + 6 = 12 + 6 = 18\). Therefore, the coordinates of \(Q\) are \((18, 0)\).

(a) M1: Differentiates \(y^2 = 12x\) implicitly or uses parametric coordinates to find \(\frac{dy}{dx}\). A1: Correct gradient of the tangent is \(\frac{1}{t}\). M1: Uses the perpendicular gradient rule to find the normal gradient \(-t\) and attempts to use the straight-line equation with \(P(3t^2, 6t)\). A1*: Fully correct derivation leading to the given equation. (b) M1: Sets \(y=0\) in the normal equation to express \(x\) in terms of \(t\). M1: Substitutes \(t = 2\) into their expression. A1: Correct coordinates of \(Q\) are \((18, 0)\).

(a) Multiplying matrices \(\mathbf{A}\) and \(\mathbf{B}\): \(\mathbf{A}\mathbf{B} = \begin{pmatrix} 2 & k \\ -1 & 3 \end{pmatrix} \begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 2(3)+k(1) & 2(1)+k(2) \\ -1(3)+3(1) & -1(1)+3(2) \end{pmatrix} = \begin{pmatrix} 6+k & 2+2k \\ 0 & 5 \end{pmatrix}\). (b) The determinant of \(\mathbf{A}\mathbf{B}\) is: \(\det(\mathbf{A}\mathbf{B}) = (6+k)(5) - (2+2k)(0) = 5(6+k) = 30 + 5k\). Given \(\det(\mathbf{A}\mathbf{B}) = 35\), we solve: \(30 + 5k = 35 \Rightarrow 5k = 5 \Rightarrow k = 1\). Alternatively, using properties of determinants: \(\det(\mathbf{A}) = 2(3) - k(-1) = 6+k\) and \(\det(\mathbf{B}) = 3(2) - 1(1) = 5\). Since \(\det(\mathbf{A}\mathbf{B}) = \det(\mathbf{A})\det(\mathbf{B})\), we have \(5(6+k) = 35 \Rightarrow 6+k = 7 \Rightarrow k = 1\).

(a) M1: Attempts matrix multiplication of \(\mathbf{A}\mathbf{B}\). A1: At least two correct elements. A1: Fully correct matrix. (b) M1: Attempts to find the determinant of \(\mathbf{A}\mathbf{B}\) from part (a) or uses \(\det(\mathbf{A})\det(\mathbf{B})\). A1: Obtains the correct expression \(30+5k\) or equivalent. M1: Equates their expression to 35 and solves for \(k\). A1: Correct value of \(k = 1\).

(a) \(\det(\mathbf{M}) = 1(4) - (-2)(3) = 4 + 6 = 10\). (b) Since the determinant represents the area scale factor of the transformation, we have: \(A' = 10A\). (c) The coordinates of the vertices of \(T\) can be found by multiplying the coordinates of the vertices of \(T'\) by the inverse matrix \(\mathbf{M}^{-1}\). First, find \(\mathbf{M}^{-1}\): \(\mathbf{M}^{-1} = \frac{1}{10} \begin{pmatrix} 4 & 2 \\ -3 & 1 \end{pmatrix}\). Now apply this transformation to the three vertices of \(T'\): For \((0, 0)\): \(\mathbf{M}^{-1} \begin{pmatrix} 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\). For \((2, 6)\): \(\frac{1}{10} \begin{pmatrix} 4 & 2 \\ -3 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ 6 \end{pmatrix} = \frac{1}{10} \begin{pmatrix} 8 + 12 \\ -6 + 6 \end{pmatrix} = \frac{1}{10} \begin{pmatrix} 20 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \end{pmatrix}\). For \((-2, 14)\): \(\frac{1}{10} \begin{pmatrix} 4 & 2 \\ -3 & 1 \end{pmatrix} \begin{pmatrix} -2 \\ 14 \end{pmatrix} = \frac{1}{10} \begin{pmatrix} -8 + 28 \\ 6 + 14 \end{pmatrix} = \frac{1}{10} \begin{pmatrix} 20 \\ 20 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \end{pmatrix}\). Thus, the vertices of the original triangle \(T\) are \((0, 0)\), \((2, 0)\), and \((2, 2)\).

(a) M1: Correct application of the 2x2 determinant formula. A1: Correct determinant value of 10. (b) B1: Correctly states \(A' = 10A\). (c) M1: Attempts to find \(\mathbf{M}^{-1}\) by swapping elements and changing signs. A1: Correct inverse matrix \(\mathbf{M}^{-1} = \frac{1}{10} \begin{pmatrix} 4 & 2 \\ -3 & 1 \end{pmatrix}\). M1: Multiplies at least one non-zero vertex of \(T'\) by \(\mathbf{M}^{-1}\). A1: Finds at least one correct non-zero vertex. A1: All three correct vertices: \((0, 0)\), \((2, 0)\), and \((2, 2)\).

(a) Expanding the summand: \(\sum_{r=1}^{n} r(r+3) = \sum_{r=1}^{n} (r^2 + 3r) = \sum_{r=1}^{n} r^2 + 3 \sum_{r=1}^{n} r\). Using standard summation formulas: \(\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)\) and \(\sum_{r=1}^{n} r = \frac{1}{2}n(n+1)\). Therefore: \(\sum_{r=1}^{n} r(r+3) = \frac{1}{6}n(n+1)(2n+1) + \frac{3}{2}n(n+1) = \frac{1}{6}n(n+1)\left[ (2n+1) + 9 \right] = \frac{1}{6}n(n+1)(2n+10) = \frac{1}{3}n(n+1)(n+5)\). (b) We can write: \(\sum_{r=10}^{30} r(r+3) = \sum_{r=1}^{30} r(r+3) - \sum_{r=1}^{9} r(r+3)\). For \(n = 30\): \(\sum_{r=1}^{30} r(r+3) = \frac{1}{3}(30)(31)(35) = 10 \times 31 \times 35 = 10850\). For \(n = 9\): \(\sum_{r=1}^{9} r(r+3) = \frac{1}{3}(9)(10)(14) = 3 \times 10 \times 14 = 420\). Thus: \(\sum_{r=10}^{30} r(r+3) = 10850 - 420 = 10430\).

(a) M1: Splits the sum into \(\sum r^2\) and \(\sum r\). M1: Substitutes correct standard formulas for both. M1: Factors out a common factor of at least \(n(n+1)\). A1*: Correct simplification leading to the given answer. (b) M1: Uses the identity \(\sum_{r=10}^{30} = \sum_{r=1}^{30} - \sum_{r=1}^{9}\). M1: Evaluates their sum for \(n = 30\) using the formula from part (a). A1: Correct values for both individual sums (10850 and 420). A1: Correct final answer of 10430.

Let \(f(n) = 5^{2n} - 1\). Step 1: Base Case. For \(n = 1\), \(f(1) = 5^2 - 1 = 25 - 1 = 24\). Since 24 is divisible by 24, the statement is true for \(n = 1\). Step 2: Inductive Step. Assume the statement is true for \(n = k\), so that \(f(k) = 5^{2k} - 1 = 24m\) for some integer \(m\). Now consider \(f(k+1)\): \(f(k+1) = 5^{2(k+1)} - 1 = 5^{2k+2} - 1 = 5^2(5^{2k}) - 1 = 25(5^{2k}) - 1\). We can rewrite this as: \(f(k+1) = 25(5^{2k} - 1) + 24 = 25(24m) + 24 = 24(25m + 1)\). Since \(25m + 1\) is an integer, \(f(k+1)\) is divisible by 24. Step 3: Conclusion. Since the statement is true for \(n = 1\), and if it is true for \(n = k\) then it is also true for \(n = k+1\), the statement is true for all positive integers \(n\) by mathematical induction.

B1: Evaluates \(f(1)\) to show it is divisible by 24. M1: States the inductive assumption for \(n=k\). M1: Sets up the expression for \(f(k+1)\). A1: Obtains \(25(5^{2k}) - 1\) or equivalent. M1: Manipulates the expression to isolate a term containing \(5^{2k} - 1\) or factors out 24. A1: Correctly shows that \(f(k+1)\) is a multiple of 24. A1: Gives a complete and correct induction conclusion.

(a) The equation of the hyperbola is \(y = 16x^{-1}\). Differentiating with respect to \(x\): \(\frac{\mathrm{d}y}{\mathrm{d}x} = -16x^{-2} = -\frac{16}{x^2}\). At the point \(P\left(4t, \frac{4}{t}\right)\): \(\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{16}{(4t)^2} = -\frac{1}{t^2}\). The gradient of the normal to \(H\) at \(P\) is \(t^2\). Using the equation of a straight line: \(y - \frac{4}{t} = t^2 (x - 4t)\). Multiplying by \(t\): \(ty - 4 = t^3(x - 4t) \implies ty - 4 = t^3 x - 4t^4\). Rearranging gives: \(t^3 x - ty = 4t^4 - 4 = 4(t^4 - 1)\), as required. (b) Substitute \(y = x\) into the normal equation: \(t^3 x - tx = 4(t^4 - 1) \implies x t(t^2 - 1) = 4(t^2 - 1)(t^2 + 1)\). Dividing by \(t^2 - 1\) (assuming \(t^2 \neq 1\)): \(x t = 4(t^2 + 1) \implies x = \frac{4(t^2 + 1)}{t} = 4t + \frac{4}{t}\). Since \(y = x\), the coordinates of \(Q\) are \(\left(4t + \frac{4}{t}, 4t + \frac{4}{t}\right)\).

(a) M1: Differentiates hyperbola equation to obtain form \(k x^{-2}\). A1: Correctly evaluates gradient of tangent as \(-\frac{1}{t^2}\). M1: Uses negative reciprocal for normal gradient and attempts straight line equation. A1*: Correctly completes algebra to show the given equation. (b) M1: Substitutes \(y = x\) into the normal equation. M1: Factorises out or cancels \(t^2 - 1\) to solve for \(x\). A1: Correct coordinates of \(Q\), fully simplified.

(a) \(\sum_{r=1}^n r(3r - 1) = 3\sum_{r=1}^n r^2 - \sum_{r=1}^n r = 3 \cdot \frac{1}{6}n(n+1)(2n+1) - \frac{1}{2}n(n+1) = \frac{1}{2}n(n+1)(2n+1) - \frac{1}{2}n(n+1)\). Factorising out \(\frac{1}{2}n(n+1)\): \(= \frac{1}{2}n(n+1)[(2n+1) - 1] = \frac{1}{2}n(n+1)(2n) = n^2(n+1)\), as required. (b) Write the sum as \(\sum_{r=1}^{2n} r(3r - 1) - \sum_{r=1}^{n-1} r(3r - 1)\). Using the result from part (a): \(= (2n)^2(2n+1) - (n-1)^2(n) = 4n^2(2n+1) - n(n^2 - 2n + 1) = 8n^3 + 4n^2 - n^3 + 2n^2 - n = 7n^3 + 6n^2 - n = n(7n^2 + 6n - 1) = n(7n-1)(n+1)\), as required.

(a) M1: Splits sum and substitutes standard formulae. A1: Correct unsimplified expression. M1: Factorises out common terms. A1*: Correctly shows the given identity with no errors. (b) M1: Splits sum into two sums from 1 to \(2n\) and 1 to \(n-1\). A1: Substitutes formula to get \(4n^2(2n+1) - n(n-1)^2\). M1: Factorises out \(n\) and expands/simplifies the remaining quadratic. A1*: Correctly factorises to obtain \(n(7n - 1)(n + 1)\).