2026 Edexcel IAS-Level Further Mathematics (XFM01) Practice Paper with Answers

For the equation \(2x^2 - 5x + 6 = 0\), we have \(\alpha + \beta = \frac{5}{2}\) and \(\alpha\beta = 3\). We want to find a quadratic equation with roots \(u = \alpha^2 + 1\) and \(v = \beta^2 + 1\). First, find the sum of the new roots: \(u + v = (\alpha^2 + 1) + (\beta^2 + 1) = \alpha^2 + \beta^2 + 2\). Using the identity \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\), we get \(u + v = (\alpha + \beta)^2 - 2\alpha\beta + 2 = \left(\frac{5}{2}\right)^2 - 2(3) + 2 = \frac{25}{4} - 6 + 2 = \frac{9}{4}\). Next, find the product of the new roots: \(uv = (\alpha^2 + 1)(\beta^2 + 1) = \alpha^2\beta^2 + \alpha^2 + \beta^2 + 1 = (\alpha\beta)^2 + (\alpha+\beta)^2 - 2\alpha\beta + 1 = (3)^2 + \left(\frac{5}{2}\right)^2 - 2(3) + 1 = 9 + \frac{25}{4} - 6 + 1 = \frac{41}{4}\). The new quadratic equation is given by \(x^2 - (u + v)x + uv = 0\), which is \(x^2 - \frac{9}{4}x + \frac{41}{4} = 0\). Multiplying by 4 to obtain integer coefficients yields \(4x^2 - 9x + 41 = 0\).

B1: States \(\alpha + \beta = \frac{5}{2}\) and \(\alpha\beta = 3\) (both must be correct). (1 mark) M1.5: Attempts to find the sum of the new roots in terms of \(\alpha + \beta\) and \(\alpha\beta\) and substitutes values. (1.5 marks) M1.5: Attempts to find the product of the new roots in terms of \(\alpha + \beta\) and \(\alpha\beta\) and substitutes values. (1.5 marks) A1: Obtains the correct sum \(\frac{9}{4}\) and product \(\frac{41}{4}\). (1 mark) A0.5: Forms a correct quadratic equation with integer coefficients, e.g., \(4x^2 - 9x + 41 = 0\). (0.5 marks)

(a) Evaluate \(f(x)\) at the interval endpoints: \(f(2) = 2^3 - 3(2) - 4 = -2\) and \(f(3) = 3^3 - 3(3) - 4 = 14\). Using linear interpolation: \(\frac{\alpha - 2}{3 - \alpha} \approx \frac{0 - f(2)}{f(3) - 0} = \frac{2}{14} = \frac{1}{7}\). Solving for \(\alpha\) gives \(7(\alpha - 2) = 3 - \alpha \implies 8\alpha = 17 \implies \alpha = 2.125\). (b) Find the derivative of \(f(x)\): \(f'(x) = 3x^2 - 3\). With \(x_1 = 2.2\), we calculate: \(f(2.2) = 2.2^3 - 3(2.2) - 4 = 0.048\) and \(f'(2.2) = 3(2.2)^2 - 3 = 11.52\). Applying the Newton-Raphson formula: \(x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 2.2 - \frac{0.048}{11.52} = 2.2 - \frac{1}{240} \approx 2.195833...\). To 3 decimal places, \(x_2 \approx 2.196\).

(a) M1.5: Correct linear interpolation setup using \(f(2) = -2\) and \(f(3) = 14\). (1.5 marks) A1: For obtaining \(\alpha = 2.125\). (1 mark) (b) M1.5: Correct differentiation to find \(f'(x) = 3x^2 - 3\) and evaluation of \(f(2.2)\) and \(f'(2.2)\). (1.5 marks) M1: Correct application of the Newton-Raphson formula. (1 mark) A0.5: Obtains \(2.196\) (rounded to 3 decimal places). (0.5 marks)

(a) Substituting \(t = 2\) into the coordinates of \(P\) gives \(P(3(2)^2, 6(2)) = P(12, 12)\). For the parabola \(y^2 = 12x\), we have \(4a = 12 \implies a = 3\). The equation of the directrix is \(x = -a \implies x = -3\). (b) Differentiating \(y^2 = 12x\) with respect to \(x\) gives \(2y \frac{dy}{dx} = 12 \implies \frac{dy}{dx} = \frac{6}{y}\). At \(P(12, 12)\), the gradient of the tangent is \(m = \frac{6}{12} = \frac{1}{2}\). The equation of the tangent is \(y - 12 = \frac{1}{2}(x - 12) \implies y = \frac{1}{2}x + 6\) (or \(x - 2y + 12 = 0\)). (c) The directrix is \(x = -3\). Substituting \(x = -3\) into the tangent equation gives \(y = \frac{1}{2}(-3) + 6 = 4.5\). Thus, the coordinates of \(Q\) are \((-3, 4.5)\).

(a) B0.5: Correct coordinates for \(P(12, 12)\). (0.5 marks) B1: Correct directrix equation \(x = -3\). (1 mark) (b) M1: Method to find the gradient of the tangent at \(P\) and find the equation. (1 mark) A1: Correct equation of the tangent, e.g. \(y = \frac{1}{2}x + 6\) or equivalent. (1 mark) (c) M1: Substituting \(x = -3\) into their tangent equation to find the y-coordinate. (1 mark) A1: Correct coordinates for \(Q(-3, 4.5)\). (1 mark)

(a) The area scale factor for the transformation is given by \(|\det(\mathbf{M})|\). Since \(\text{Area of image} = |\det(\mathbf{M})| \times \text{Area of object}\), we have \(72 = |\det(\mathbf{M})| \times 3 \implies |\det(\mathbf{M})| = 24\). Thus, the two possible values of the determinant are \(24\) and \(-24\). (b) The determinant of \(\mathbf{M}\) is \(\det(\mathbf{M}) = k^2 - 25\). Case 1: \(k^2 - 25 = 24 \implies k^2 = 49 \implies k = \pm 7\). Case 2: \(k^2 - 25 = -24 \implies k^2 = 1 \implies k = \pm 1\). Therefore, the possible values of \(k\) are \(-7, -1, 1, 7\).

(a) M1: For stating or using that the scale factor of area is \(|\det(\mathbf{M})|\). (1 mark) A0.5: Concludes \(\det(\mathbf{M}) = \pm 24\). (0.5 marks) (b) M1: Finds correct expression for \(\det(\mathbf{M}) = k^2 - 25\). (1 mark) M1: Sets up the two equations \(k^2 - 25 = 24\) and \(k^2 - 25 = -24\). (1 mark) A1: Solves to get \(k = \pm 7\). (1 mark) A1: Solves to get \(k = \pm 1\). (1 mark)

(a)

Differentiating \( y^2 = 8x \) implicitly with respect to \( x \):
\[ 2y \frac{dy}{dx} = 8 \implies \frac{dy}{dx} = \frac{4}{y} \]

At \( P(2t^2, 4t) \), the gradient of the tangent is:
\[ m = \frac{4}{4t} = \frac{1}{t} \]

Using the equation of a straight line:
\[ y - 4t = \frac{1}{t}(x - 2t^2) \]

Multiply through by \( t \):
\[ ty - 4t^2 = x - 2t^2 \implies ty = x + 2t^2 \] (as required)

(b)

The parabola \( y^2 = 8x \) is of the form \( y^2 = 4ax \), so \( 4a = 8 \implies a = 2 \).

The focus \( S \) has coordinates \( (a, 0) = (2, 0) \).

Since line \( L \) is perpendicular to the tangent, its gradient is:
\[ m' = -\frac{1}{m} = -t \]

The equation of line \( L \) is:
\[ y - 0 = -t(x - 2) \implies y = -tx + 2t \]

(c)

The equation of the directrix of \( C \) is \( x = -a \implies x = -2 \).

To find the intersection point \( R \), substitute \( x = -2 \) into the equation of \( L \):
\[ y = -t(-2) + 2t = 2t + 2t = 4t \]

Thus, the coordinates of \( R \) are \( (-2, 4t) \).

(a)
- M1: Attempts to find the gradient of the tangent, either by implicit differentiation of \( y^2 = 8x \) or by using \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4}{4t} \).
- A1: Correct gradient of \( \frac{1}{t} \).
- A1*: Completes the proof to show \( ty = x + 2t^2 \) with no errors or omissions.

(b)
- B1: Correctly identifies the focus \( S \) as \( (2, 0) \).
- M1: Uses the perpendicular gradient property to find the gradient \( -t \) and attempts to write down the equation of the line through \( S \).
- A0.5: Correct equation \( y = -tx + 2t \) or any equivalent form.

(c)
- B1: States or uses the equation of the directrix as \( x = -2 \).
- M1: Substitutes \( x = -2 \) into their equation of \( L \) to solve for \( y \).
- A1: Correct coordinates of \( R(-2, 4t) \).

From the quadratic equation \( 2x^2 - 5x + 4 = 0 \), we have:
\[ \alpha + \beta = \frac{5}{2} \]
\[ \alpha\beta = \frac{4}{2} = 2 \]

(a)
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
\[ \alpha^2 + \beta^2 = \left(\frac{5}{2}\right)^2 - 2(2) = \frac{25}{4} - 4 = \frac{9}{4} \]

(b)
\[ \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) \]
\[ \alpha^3 + \beta^3 = \left(\frac{5}{2}\right)^3 - 3(2)\left(\frac{5}{2}\right) = \frac{125}{8} - 15 = \frac{125 - 120}{8} = \frac{5}{8} \]

(c)
For the new equation with roots \( p = \frac{\alpha}{\beta^2} \) and \( q = \frac{\beta}{\alpha^2} \):

Sum of roots:
\[ p + q = \frac{\alpha}{\beta^2} + \frac{\beta}{\alpha^2} = \frac{\alpha^3 + \beta^3}{\alpha^2\beta^2} = \frac{\alpha^3 + \beta^3}{(\alpha\beta)^2} \]
\[ p + q = \frac{5/8}{2^2} = \frac{5/8}{4} = \frac{5}{32} \]

Product of roots:
\[ pq = \frac{\alpha}{\beta^2} \times \frac{\beta}{\alpha^2} = \frac{\alpha\beta}{(\alpha\beta)^2} = \frac{1}{\alpha\beta} = \frac{1}{2} \]

The quadratic equation is given by:
\[ x^2 - (p + q)x + pq = 0 \]
\[ x^2 - \frac{5}{32}x + \frac{1}{2} = 0 \]

Multiplying by 32 to obtain integer coefficients:
\[ 32x^2 - 5x + 16 = 0 \]

(a)
- M1: Identifies \( \alpha + \beta = 5/2 \) and \( \alpha\beta = 2 \).
- M1: Uses the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \).
- A0.5: Correct value of \( 9/4 \).

(b)
- M1: Uses a correct identity for \( \alpha^3 + \beta^3 \), e.g., \( (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) \) or \( (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \).
- M1: Substitutes their values of \( \alpha + \beta \), \( \alpha\beta \), and/or \( \alpha^2 + \beta^2 \) into the identity.
- A1: Correct value of \( 5/8 \).

(c)
- M1: Expresses the sum of the new roots in terms of \( \alpha^3 + \beta^3 \) and \( \alpha\beta \), and evaluates it.
- M1: Expresses the product of the new roots as \( \frac{1}{\alpha\beta} \) and evaluates it.
- A1: Obtains \( 32x^2 - 5x + 16 = 0 \) (must include "= 0" and have integer coefficients).

(a)

The transformation \( T \) is singular when \( \det(\mathbf{M}) = 0 \).
\[ \det(\mathbf{M}) = (k-1)(k+4) - (3)(-2) \]
\[ \det(\mathbf{M}) = k^2 + 3k - 4 + 6 = k^2 + 3k + 2 \]

Set \( \det(\mathbf{M}) = 0 \):
\[ k^2 + 3k + 2 = 0 \implies (k+1)(k+2) = 0 \]

Thus, \( k = -1 \) or \( k = -2 \).

(b)

When \( k = 1 \), \( \det(\mathbf{M}) = 1^2 + 3(1) + 2 = 6 \).

The determinant of the matrix represents the area scale factor of the transformation.
\[ \text{Area of image } B = \text{Area of } A \times |\det(\mathbf{M})| \]
\[ \text{Area of image } B = 6 \times 6 = 36\text{ cm}^2 \]

(c)

With \( k = 1 \), \( \mathbf{M} = \begin{pmatrix} 0 & 3 \\ -2 & 5 \end{pmatrix} \) and \( \det(\mathbf{M}) = 6 \).

The inverse matrix is:
\[ \mathbf{M}^{-1} = \frac{1}{6} \begin{pmatrix} 5 & -3 \\ 2 & 0 \end{pmatrix} = \begin{pmatrix} \frac{5}{6} & -\frac{1}{2} \\ \frac{1}{3} & 0 \end{pmatrix} \]

(a)
- M1: Attempts to calculate the determinant \( (k-1)(k+4) - (3)(-2) \).
- A1: Obtains the simplified quadratic \( k^2 + 3k + 2 \).
- M1: Sets their determinant to 0 and attempts to solve the quadratic equation.
- A0.5: Finds both correct values of \( k \): \( k = -1 \) and \( k = -2 \).

(b)
- M1: Calculates \( \det(\mathbf{M}) \) for \( k=1 \) (value is 6) and relates it to the area scale factor.
- A1: Correct area of \( 36\text{ cm}^2 \).

(c)
- M1: For attempting to find the adjugate matrix by swapping diagonal entries and changing the signs of the off-diagonal entries.
- M1: Dividing by their determinant value (6).
- A1: Correct inverse matrix \( \begin{pmatrix} \frac{5}{6} & -\frac{1}{2} \\ \frac{1}{3} & 0 \end{pmatrix} \) or \( \frac{1}{6} \begin{pmatrix} 5 & -3 \\ 2 & 0 \end{pmatrix} \).

(a)

Multiply both sides by \( z + \mathrm{i} \):
\[ z - 3 = (2 - \mathrm{i})(z + \mathrm{i}) \]
\[ z - 3 = (2 - \mathrm{i})z + \mathrm{i}(2 - \mathrm{i}) \]
\[ z - 3 = (2 - \mathrm{i})z + 2\mathrm{i} + 1 \]

Rearrange terms to group \( z \):
\[ z - (2 - \mathrm{i})z = 3 + 1 + 2\mathrm{i} \]
\[ z(1 - 2 + \mathrm{i}) = 4 + 2\mathrm{i} \]
\[ z(-1 + \mathrm{i}) = 4 + 2\mathrm{i} \]

Solve for \( z \):
\[ z = \frac{4 + 2\mathrm{i}}{-1 + \mathrm{i}} \]

Multiply numerator and denominator by the complex conjugate of the denominator, \( -1 - \mathrm{i} \):
\[ z = \frac{(4 + 2\mathrm{i})(-1 - \mathrm{i})}{(-1)^2 + 1^2} \]
\[ z = \frac{-4 - 4\mathrm{i} - 2\mathrm{i} + 2}{2} \]
\[ z = \frac{-2 - 6\mathrm{i}}{2} = -1 - 3\mathrm{i} \]

(b)

The conjugate of \( z \) is \( z^* = -1 + 3\mathrm{i} \).
\[ w = \frac{10}{-1 + 3\mathrm{i}} \]

Multiply numerator and denominator by the conjugate \( -1 - 3\mathrm{i} \):
\[ w = \frac{10(-1 - 3\mathrm{i})}{(-1)^2 + 3^2} \]
\[ w = \frac{10(-1 - 3\mathrm{i})}{10} = -1 - 3\mathrm{i} \]

(c)

An Argand diagram with:
- Horizontal real axis (Re) and vertical imaginary axis (Im).
- The point \( z = -1 - 3\mathrm{i} \) plotted in the third quadrant.
- The point \( z^* = -1 + 3\mathrm{i} \) plotted in the second quadrant, demonstrating reflectional symmetry across the real axis.

(a)
- M1: Multiplying through by \( z + \mathrm{i} \) and expanding the right-hand side.
- A1: Correctly simplifying the equation to a form with \( z \) on one side, e.g., \( z(-1 + \mathrm{i}) = 4 + 2\mathrm{i} \).
- M1: Attempting to divide by \( -1 + \mathrm{i} \) and multiplying the numerator and denominator by the conjugate \( -1 - \mathrm{i} \).
- A1: Reaching the final correct simplified answer \( z = -1 - 3\mathrm{i} \).

(b)
- B1: Stating the correct conjugate \( z^* = -1 + 3\mathrm{i} \).
- M1: Attempting to evaluate \( \frac{10}{z^*} \) by multiplying numerator and denominator by \( -1 - 3\mathrm{i} \).
- A0.5: Correct answer \( w = -1 - 3\mathrm{i} \).

(c)
- B1: Draws both axes (labeled Re and Im, or \( x \) and \( y \)) with origin labeled.
- B1: Correctly plots \( z = -1 - 3\mathrm{i} \) and \( z^* = -1 + 3\mathrm{i} \) showing correct relative positions and symmetry.

(a) Differentiating the equation of the parabola: \(2y \frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{2a}{y}\). At \(P(ap^2, 2ap)\), the gradient of the tangent is \(m = \frac{2a}{2ap} = \frac{1}{p}\). The equation of the tangent is \(y - 2ap = \frac{1}{p}(x - ap^2) \implies py - 2ap^2 = x - ap^2 \implies py = x + ap^2\). (b) Similarly, the tangent at \(Q\) is \(qy = x + aq^2\). Subtracting the two tangent equations gives \((p-q)y = a(p^2 - q^2) = a(p-q)(p+q)\). Since \(p \neq q\), we divide by \(p-q\) to obtain \(y = a(p+q)\). Substituting this into the equation of the tangent at \(P\) gives \(p(a(p+q)) = x + ap^2 \implies ap^2 + apq = x + ap^2 \implies x = apq\). Thus, the coordinates of \(T\) are \((apq, a(p+q))\). (c)(i) The gradient of the chord \(PQ\) is \(\frac{2aq - 2ap}{aq^2 - ap^2} = \frac{2a(q-p)}{a(q-p)(q+p)} = \frac{2}{p+q}\). The equation of the line \(PQ\) is \(y - 2ap = \frac{2}{p+q}(x - ap^2)\). Since \(PQ\) passes through \(S(a, 0)\), we substitute \(x = a\) and \(y = 0\): \(0 - 2ap = \frac{2}{p+q}(a - ap^2) \implies -ap(p+q) = a(1 - p^2) \implies -p^2 - pq = 1 - p^2 \implies pq = -1\). (c)(ii) Since \(x = apq\) and \(pq = -1\), we substitute to get \(x = a(-1) = -a\). This is the equation of the directrix of the parabola.

Mark Scheme: (a) M1: Differentiates \(y^2 = 4ax\) to find \(\frac{dy}{dx}\) and substitutes \(y = 2ap\) to find the gradient of the tangent. A1: Correctly derives \(py = x + ap^2\) with clear working. (b) M1: Writes down the tangent at \(Q\) and attempts to solve simultaneously with the tangent at \(P\). M1: Eliminates \(x\) or \(y\) to find one coordinate, factoring \(p^2-q^2\) correctly. A1: Fully correct derivation of both coordinates of \(T\). (c)(i) M1: Finds the gradient of \(PQ\) and sets up the equation of the chord. M1: Substitutes \(S(a, 0)\) into the line equation and attempts to simplify. A0.5: Achieves \(pq = -1\) with clear algebraic steps. (c)(ii) M1: Substitutes \(pq = -1\) into the \(x\)-coordinate of \(T\). A1: States the equation \(x = -a\) and identifies the locus as the 'directrix'.

(a) Differentiating \(xy = c^2 \implies y = c^2 x^{-1}\) gives \(\frac{dy}{dx} = -\frac{c^2}{x^2}\). At \(P\), \(x = cp\), so the gradient of the tangent is \(-\frac{c^2}{c^2 p^2} = -\frac{1}{p^2}\). Since the normal is perpendicular to the tangent, the gradient of the normal is \(p^2\). The equation of the normal is \(y - \frac{c}{p} = p^2(x - cp) \implies py - c = p^3 x - cp^4 \implies p^3 x - py = c(p^4 - 1)\). (b) At \(A\), \(y = 0 \implies p^3 x = c(p^4 - 1) \implies x = \frac{c(p^4-1)}{p^3}\). So \(A\) is \(\left(\frac{c(p^4-1)}{p^3}, 0\right)\). At \(B\), \(x = 0 \implies -py = c(p^4 - 1) \implies y = \frac{c(1-p^4)}{p}\). So \(B\) is \(\left(0, \frac{c(1-p^4)}{p}\right)\). (c) The vector \(\vec{AP} = \mathbf{x}_P - \mathbf{x}_A = \left(cp - \frac{c(p^4-1)}{p^3}, \frac{c}{p} - 0\right) = \left(\frac{c}{p^3}, \frac{c}{p}\right)\). The vector \(\vec{PB} = \mathbf{x}_B - \mathbf{x}_P = \left(0 - cp, \frac{c(1-p^4)}{p} - \frac{c}{p}\right) = \left(-cp, -cp^3\right)\). We observe that \(\vec{PB} = -p^4 \left(\frac{c}{p^3}, \frac{c}{p}\right) = -p^4 \vec{AP}\). Since the vectors are collinear, the ratio of their lengths is \(\frac{PA}{PB} = \frac{|\vec{AP}|}{|\vec{PB}|} = \frac{1}{p^4}\).

Mark Scheme: (a) M1: Differentiates \(y = c^2/x\) to find \(\frac{dy}{dx}\). A1: Finds the gradient of the tangent at \(P\) is \(-\frac{1}{p^2}\). M1: Finds the gradient of the normal is \(p^2\). M1: Sets up the equation of the normal line. A0.5: Correctly rearranges to the given form. (b) A1: Correct coordinates for \(A\). A1: Correct coordinates for \(B\). (c) M1: Attempts to find either the components of \(\vec{AP}\) and \(\vec{PB}\) or the lengths \(PA^2\) and \(PB^2\). A1: Obtains correct simplified expressions for components or lengths. M1: Sets up the ratio \(\frac{PA}{PB}\) and simplifies. A1: Correct proof shown clearly with no errors.