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

(a) Since the coefficients of the cubic equation are real, the complex roots must occur in conjugate pairs. Therefore, the other complex root is \(z_2 = 3 + 2i\).

(b) Let the three roots be \(z_1 = 3 - 2i\), \(z_2 = 3 + 2i\), and \(z_3\).

Using the relation for the sum of the roots of a cubic equation:
\(z_1 + z_2 + z_3 = -\frac{\text{coefficient of } z^2}{\text{coefficient of } z^3} = -\frac{-2}{1} = 2\)

Substituting the values of \(z_1\) and \(z_2\):
\((3 - 2i) + (3 + 2i) + z_3 = 2 \implies 6 + z_3 = 2 \implies z_3 = -4\)

Using the product of the roots:
\(z_1 z_2 z_3 = -\frac{\text{constant term}}{\text{coefficient of } z^3} = -c\)
\((3 - 2i)(3 + 2i)(-4) = -c \implies (9 + 4)(-4) = -c \implies 13 \times (-4) = -c \implies -52 = -c \implies c = 52\)

Using the sum of pairwise products of the roots:
\(z_1 z_2 + z_2 z_3 + z_3 z_1 = b\)
\((3 - 2i)(3 + 2i) + (3 + 2i)(-4) + (-4)(3 - 2i) = b\)
\(13 - 12 - 8i - 12 + 8i = b \implies 13 - 24 = b \implies b = -11\)

M1: Identifies the complex conjugate root z_2 = 3 + 2i.
M1: Uses the sum of roots formula to form an equation for z_3.
A1: Obtains the correct third root z_3 = -4.
M1: Uses the product of roots formula to form an equation for c.
A1: Obtains the correct value c = 52.
M1: Uses the sum of pairwise products formula to form an equation for b.
A1: Obtains the correct value b = -11.

(a) Evaluating \(f(x)\) at the endpoints of the interval:
\(f(2) = 2^3 - 5(2) - 3 = 8 - 10 - 3 = -5\)
\(f(3) = 3^3 - 5(3) - 3 = 27 - 15 - 3 = 9\)

Since \(f(x)\) is continuous on the interval \([2, 3]\) and there is a change of sign between \(f(2) < 0\) and \(f(3) > 0\), there must be a root \(\alpha\) in the interval \([2, 3]\).

(b) Using the linear interpolation formula:
\(\alpha \approx 2 + \frac{0 - f(2)}{f(3) - f(2)} (3 - 2)\)
\(\alpha \approx 2 + \frac{5}{9 - (-5)} = 2 + \frac{5}{14} \approx 2.35714...\)

So the linear interpolation approximation to \(\alpha\) is 2.357 (to 3 d.p.).

(c) Differentiating \(f(x)\):
\(f'(x) = 3x^2 - 5\)

For \(x_0 = 2.4\):
\(f(2.4) = (2.4)^3 - 5(2.4) - 3 = 13.824 - 12 - 3 = -1.176\)
\(f'(2.4) = 3(2.4)^2 - 5 = 17.28 - 5 = 12.28\)

Applying the Newton-Raphson formula:
\(x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 2.4 - \frac{-1.176}{12.28} = 2.4 + 0.095765... \approx 2.496\) (to 3 d.p.).

M1: Evaluates both f(2) and f(3) with at least one value correct.
A1: States that the function is continuous, notes the change of sign, and concludes a root exists in the interval.
M1: Applies the linear interpolation formula correctly.
A1: Obtains 2.357.
M1: Differentiates f(x) and substitutes x_0 = 2.4 into both f(x) and f'(x).
A1: Applies the Newton-Raphson formula correctly and obtains 2.496.

(a) We find \(\mathbf{A}^2\) by matrix multiplication:
\(\mathbf{A}^2 = \begin{pmatrix} 2 & k \\ -1 & 3 \end{pmatrix} \begin{pmatrix} 2 & k \\ -1 & 3 \end{pmatrix} = \begin{pmatrix} 2(2) + k(-1) & 2(k) + k(3) \\ -1(2) + 3(-1) & -1(k) + 3(3) \end{pmatrix} = \begin{pmatrix} 4 - k & 5k \\ -5 & 9 - k \end{pmatrix}\)

(b) Substituting \(\mathbf{A}^2\), \(\mathbf{A}\), and \(\mathbf{I}\) into the equation \(\mathbf{A}^2 - 5\mathbf{A} + 7\mathbf{I} = \mathbf{0}\):
\(\begin{pmatrix} 4 - k & 5k \\ -5 & 9 - k \end{pmatrix} - 5\begin{pmatrix} 2 & k \\ -1 & 3 \end{pmatrix} + 7\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\)

\(\begin{pmatrix} 4 - k & 5k \\ -5 & 9 - k \end{pmatrix} - \begin{pmatrix} 10 & 5k \\ -5 & 15 \end{pmatrix} + \begin{pmatrix} 7 & 0 \\ 0 & 7 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\)

\(\begin{pmatrix} 4 - k - 10 + 7 & 5k - 5k + 0 \\ -5 - (-5) + 0 & 9 - k - 15 + 7 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\)

\(\begin{pmatrix} 1 - k & 0 \\ 0 & 1 - k \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\)

Equating corresponding elements yields:
\(1 - k = 0 \implies k = 1\).

M1: Attempts to multiply A by itself.
A1: At least two correct entries in A^2.
A1: Fully correct expression for A^2.
M1: Substitutes A^2, A, and I into the matrix equation.
M1: Equates the diagonal entries to zero.
A1: Obtains the correct value k = 1.

(a) Differentiating the equation of the parabola \(y^2 = 12x\) with respect to \(x\):
\(2y \frac{dy}{dx} = 12 \implies \frac{dy}{dx} = \frac{6}{y}\)

At the point \(P(3t^2, 6t)\), the gradient of the tangent is:
\(m = \frac{6}{6t} = \frac{1}{t}\)

Using the equation of a straight line, the equation of the tangent at \(P\) is:
\(y - 6t = \frac{1}{t}(x - 3t^2)\)

Multiplying both sides by \(t\):
\(ty - 6t^2 = x - 3t^2 \implies ty = x + 3t^2\)

(b) Since the tangent passes through the point \((-9, 6)\), we substitute \(x = -9\) and \(y = 6\) into the equation of the tangent:
\(6t = -9 + 3t^2 \implies 3t^2 - 6t - 9 = 0\)

Dividing the equation by 3:
\(t^2 - 2t - 3 = 0 \implies (t - 3)(t + 1) = 0\)

So the possible values of \(t\) are \(t = 3\) and \(t = -1\).

For \(t = 3\), the coordinates of \(P\) are:
\(P(3(3)^2, 6(3)) = P(27, 18)\)

For \(t = -1\), the coordinates of \(P\) are:
\(P(3(-1)^2, 6(-1)) = P(3, -6)\)

Therefore, the possible coordinates of the point of contact \(P\) are \((27, 18)\) and \((3, -6)\).

M1: Differentiates the parabola equation to find an expression for dy/dx.
M1: Evaluates the gradient at the point P.
M1: Uses their gradient and point P to form a linear equation.
A1: Obtains the correct given equation ty = x + 3t^2.
M1: Substitutes (-9, 6) into the tangent equation to form a quadratic in t.
M1: Solves the quadratic to find the values of t.
A1: Correctly identifies both points of contact: (27, 18) and (3, -6).

(a) The matrix representing a reflection in the line \(y = -x\) is:
\(\mathbf{R} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\)

The matrix representing a stretch parallel to the \(y\)-axis with scale factor 3 is:
\(\mathbf{S} = \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}\)

Since the reflection is followed by the stretch, the combined matrix \(\mathbf{M}\) is:
\(\mathbf{M} = \mathbf{S}\mathbf{R} = \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix} \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ -3 & 0 \end{pmatrix}\)

(b) To find the coordinates of \(P'\), \(Q'\), and \(R'\), we multiply the matrix \(\mathbf{M}\) by the coordinates of the original vertices:
\(P' = \begin{pmatrix} 0 & -1 \\ -3 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} -2 \\ -3 \end{pmatrix}\)
\(Q' = \begin{pmatrix} 0 & -1 \\ -3 & 0 \end{pmatrix} \begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix} -2 \\ -9 \end{pmatrix}\)
\(R' = \begin{pmatrix} 0 & -1 \\ -3 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 5 \end{pmatrix} = \begin{pmatrix} -5 \\ -3 \end{pmatrix}\)

So the vertices of \(A'\) are \(P'(-2, -3)\), \(Q'(-2, -9)\), and \(R'(-5, -3)\).

(c) The area of the original triangle \(A\) with vertices \(P(1, 2)\), \(Q(3, 2)\), and \(R(1, 5)\) is:
\(\text{Area}(A) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (3 - 1) \times (5 - 2) = 3\)

The determinant of \(\mathbf{M}\) is:
\(\det(\mathbf{M}) = (0)(0) - (-1)(-3) = -3\)

The area of the transformed triangle is:
\(\text{Area}(A') = |\det(\mathbf{M})| \times \text{Area}(A) = |-3| \times 3 = 9\)

M1: Identifies the correct transformation matrices for reflection and stretch.
M1: Multiplies the matrices in the correct order (SR).
A1: Obtains the correct combined matrix M.
M1: Attempts to multiply their matrix M by the vectors of P, Q, and R.
A1: Obtains the correct coordinates for P', Q', and R'.
M1: Finds the area of original triangle A and uses det(M) (or directly calculates the area of A').
A1: Obtains the correct area 9.

(a) From the relations between the roots and coefficients of a quadratic equation:
\(\alpha + \beta = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2} = -\frac{-5}{2} = \frac{5}{2}\)
\(\alpha\beta = \frac{\text{constant term}}{\text{coefficient of } x^2} = \frac{4}{2} = 2\)

(b) Using the algebraic identity for the sum of squares:
\(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\)
\(\alpha^2 + \beta^2 = \left(\frac{5}{2}\right)^2 - 2(2) = \frac{25}{4} - 4 = \frac{9}{4}\)

(c) Let the new roots be \(\alpha' = \frac{\alpha}{\beta}\) and \(\beta' = \frac{\beta}{\alpha}\).

The sum of the new roots is:
\(\alpha' + \beta' = \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{9/4}{2} = \frac{9}{8}\)

The product of the new roots is:
\(\alpha'\beta' = \frac{\alpha}{\beta} \times \frac{\beta}{\alpha} = 1\)

The new quadratic equation is given by:
\(x^2 - (\alpha' + \beta')x + \alpha'\beta' = 0\)
\(x^2 - \frac{9}{8}x + 1 = 0\)

Multiplying by 8 to express with integer coefficients:
\(8x^2 - 9x + 8 = 0\)

B1: Writes down correct sum of roots alpha + beta = 5/2.
B1: Writes down correct product of roots alpha * beta = 2.
M1: Uses the correct algebraic identity for the sum of squares.
A1: Obtains the correct value 9/4.
M1: Sets up the sum of the new roots and expresses it in terms of alpha and beta.
M1: Finds the product of the new roots as 1.
A1: Formulates the final quadratic equation with integer coefficients as 8x^2 - 9x + 8 = 0.

(a) Expanding the general term in the summation:
\(\sum_{r=1}^{n} r(2r-1) = \sum_{r=1}^{n} (2r^2 - r) = 2\sum_{r=1}^{n} r^2 - \sum_{r=1}^{n} r\)

Using the standard summation formulae:
\(\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)\)
\(\sum_{r=1}^{n} r = \frac{1}{2}n(n+1)\)

Substituting these into the expression:
\(= 2 \left( \frac{1}{6}n(n+1)(2n+1) \right) - \frac{1}{2}n(n+1)\)

\(= \frac{1}{3}n(n+1)(2n+1) - \frac{1}{2}n(n+1)\)

Factorising out \(\frac{1}{6}n(n+1)\):
\(= \frac{1}{6}n(n+1) [ 2(2n+1) - 3 ]\)

\(= \frac{1}{6}n(n+1)(4n + 2 - 3)\)

\(= \frac{1}{6}n(n+1)(4n-1)\) (as required).

(b) We evaluate the sum by writing it as the difference between two sums:
\(\sum_{r=11}^{20} r(2r-1) = \sum_{r=1}^{20} r(2r-1) - \sum_{r=1}^{10} r(2r-1)\)

Using the formula derived in part (a):

For \(n = 20\):
\(\sum_{r=1}^{20} r(2r-1) = \frac{1}{6}(20)(21)(4(20)-1) = \frac{1}{6}(20)(21)(79) = 10 \times 7 \times 79 = 5530\)

For \(n = 10\):
\(\sum_{r=1}^{10} r(2r-1) = \frac{1}{6}(10)(11)(4(10)-1) = \frac{1}{6}(10)(11)(39) = 5 \times 11 \times 13 = 715\)

Subtracting the two values:
\(\sum_{r=11}^{20} r(2r-1) = 5530 - 715 = 4815\)

M1: Expands the algebraic term and splits the sum.
M1: Substitutes the standard formulas for sum of r and sum of r^2.
M1: Factorises out the common terms including n(n+1).
A1: Achieves the given result cleanly.
M1: Identifies the correct split S_20 - S_10.
M1: Substitutes n=20 and n=10 into the formula.
A1: Correctly calculates 4815.

Let \(P(n)\) be the statement that \(f(n) = 3^{2n} + 7\) is divisible by 8.

**Base Case:**
For \(n = 1\):
\(f(1) = 3^{2(1)} + 7 = 9 + 7 = 16\)
Since 16 is divisible by 8 (\(16 = 8 \times 2\)), the statement is true for \(n = 1\).

**Inductive Step:**
Assume that the statement is true for \(n = k\), where \(k\) is a positive integer.
That is, \(f(k) = 3^{2k} + 7 = 8m\) for some integer \(m\).
We want to show that the statement is true for \(n = k + 1\), i.e., \(f(k+1) = 3^{2(k+1)} + 7\) is divisible by 8.

\(f(k+1) = 3^{2(k+1)} + 7 = 3^{2k+2} + 7\)
\(f(k+1) = 9(3^{2k}) + 7\)

From the inductive hypothesis, \(3^{2k} = 8m - 7\).
Substituting this into the expression for \(f(k+1)\):
\(f(k+1) = 9(8m - 7) + 7\)
\(f(k+1) = 72m - 63 + 7\)
\(f(k+1) = 72m - 56\)
\(f(k+1) = 8(9m - 7)\)

Since \(m\) is an integer, \(9m - 7\) is also an integer, which shows that \(f(k+1)\) is divisible by 8.
Therefore, if \(P(k)\) is true, then \(P(k+1)\) is also true.

**Conclusion:**
Since the statement is true for \(n = 1\), and if true for \(n = k\) it is also true for \(n = k + 1\), by mathematical induction the statement is true for all positive integers \(n\).

B1: Verifies the base case n=1 correctly showing f(1)=16 is divisible by 8.
M1: States the inductive assumption for n=k.
M1: Expresses f(k+1) as 3^(2k+2) + 7.
M1: Correctly substitutes the inductive hypothesis to write f(k+1) in terms of m.
A1: Obtains a fully correct expression showing 8 as a factor.
A1: Concludes with a complete and correct mathematical induction statement.

**(a)**
From the given quadratic equation \(2x^2 + 5x + 6 = 0\), we have:
\(\alpha + \beta = -\frac{5}{2}\) and \(\alpha\beta = \frac{6}{2} = 3\).

Using the identity:
\(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\)
\(\alpha^2 + \beta^2 = \left(-\frac{5}{2}\right)^2 - 2(3)\)
\(\alpha^2 + \beta^2 = \frac{25}{4} - 6 = \frac{1}{4}\)

**(b)**
Using the identity:
\(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\)
\(\alpha^3 + \beta^3 = \left(-\frac{5}{2}\right)^3 - 3(3)\left(-\frac{5}{2}\right)\)
\(\alpha^3 + \beta^3 = -\frac{125}{8} + \frac{45}{2}\)
\(\alpha^3 + \beta^3 = -\frac{125}{8} + \frac{180}{8} = \frac{55}{8}\)

Alternatively:
\(\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2)\)
\(\alpha^3 + \beta^3 = \left(-\frac{5}{2}\right)\left(\frac{1}{4} - 3\right)\)
\(\alpha^3 + \beta^3 = \left(-\frac{5}{2}\right)\left(-\frac{11}{4}\right) = \frac{55}{8}\)

**(c)**
Let the new roots be \(u = \frac{\alpha}{\beta^2}\) and \(v = \frac{\beta}{\alpha^2}\).

The sum of the new roots is:
\(u + v = \frac{\alpha}{\beta^2} + \frac{\beta}{\alpha^2} = \frac{\alpha^3 + \beta^3}{\alpha^2\beta^2} = \frac{\alpha^3 + \beta^3}{(\alpha\beta)^2}\)

Substituting the known values:
\(u + v = \frac{\frac{55}{8}}{3^2} = \frac{55}{72}\)

The product of the new roots is:
\(uv = \left(\frac{\alpha}{\beta^2}\right)\left(\frac{\beta}{\alpha^2}\right) = \frac{\alpha\beta}{\alpha^2\beta^2} = \frac{1}{\alpha\beta}\)

Substituting the known value:
\(uv = \frac{1}{3}\)

The new quadratic equation is given by:
\(x^2 - (u + v)x + uv = 0\)
\(x^2 - \frac{55}{72}x + \frac{1}{3} = 0\)

Multiplying by 72 to obtain integer coefficients:
\(72x^2 - 55x + 24 = 0\)

**(a)**
* **M1**: Identifies \(\alpha + \beta = -\frac{5}{2}\) and \(\alpha\beta = 3\), and attempts to use the identity \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\).
* **A1**: Correct value of \(\frac{1}{4}\) (or 0.25).

**(b)**
* **M1**: Attempts to use a valid algebraic identity for \(\alpha^3 + \beta^3\) in terms of \(\alpha + \beta\) and \(\alpha\beta\).
* **M1**: Substitutes their values for the sum and product into their expression.
* **A1**: Correct value of \(\frac{55}{8}\) (or 6.875).

**(c)**
* **M1**: Expresses the sum of the new roots as \(\frac{\alpha^3 + \beta^3}{(\alpha\beta)^2}\) and attempts to evaluate using their part (b) value.
* **M1**: Expresses the product of the new roots as \(\frac{1}{\alpha\beta}\) and evaluates to find \(\frac{1}{3}\).
* **A1**: Correct quadratic equation with integer coefficients, e.g., \(72x^2 - 55x + 24 = 0\) (must include "= 0", allow any integer multiple).

Let \(f(n) = 4^{n+1} + 5^{2n-1}\).

**Step 1: Base Case**
For \(n = 1\):
\[f(1) = 4^{1+1} + 5^{2(1)-1} = 4^2 + 5^1 = 16 + 5 = 21\]
Since 21 is divisible by 21, the statement is true for \(n = 1\).

**Step 2: Inductive Hypothesis**
Assume the statement is true for \(n = k\), where \(k\) is a positive integer.
That is, assume \(f(k) = 4^{k+1} + 5^{2k-1}\) is divisible by 21, so we can write:
\[f(k) = 21M\]
for some integer \(M\).

**Step 3: Inductive Step**
We need to show that the statement is true for \(n = k+1\).
Consider \(f(k+1)\):
\[f(k+1) = 4^{(k+1)+1} + 5^{2(k+1)-1}\]
\[f(k+1) = 4^{k+2} + 5^{2k+1}\]
\[f(k+1) = 4 \cdot 4^{k+1} + 5^2 \cdot 5^{2k-1}\]
\[f(k+1) = 4 \cdot 4^{k+1} + 25 \cdot 5^{2k-1}\]

We can rewrite this in terms of \(f(k)\):
\[f(k+1) = 4(4^{k+1} + 5^{2k-1}) + 21 \cdot 5^{2k-1}\]
\[f(k+1) = 4f(k) + 21 \cdot 5^{2k-1}\]

Using our induction hypothesis \(f(k) = 21M\):
\[f(k+1) = 4(21M) + 21 \cdot 5^{2k-1} = 21(4M + 5^{2k-1})\]

Since \(M\) and \(5^{2k-1}\) are integers (for \(k \ge 1\)), \(4M + 5^{2k-1}\) is an integer, so \(f(k+1)\) is divisible by 21.

**Step 4: Conclusion**
Since the statement is true for \(n=1\), and if true for \(n=k\) it is also true for \(n=k+1\), the statement is true for all positive integers \(n\) by mathematical induction.

**B1**: Evaluates \(f(1) = 21\) and states that it is divisible by 21.
**M1**: Writes down the expression for \(f(k+1) = 4^{k+2} + 5^{2k+1}\) (accept equivalent forms).
**M1**: Attempts to separate powers to form an expression involving \(f(k)\). For example, writing \(f(k+1) = 4 \cdot 4^{k+1} + 25 \cdot 5^{2k-1}\).
**A1**: Correctly expresses \(f(k+1)\) in a form showing divisibility by 21, such as \(f(k+1) = 4f(k) + 21 \cdot 5^{2k-1}\) or \(f(k+1) = 25f(k) - 21 \cdot 4^{k+1}\) or \(21(4M + 5^{2k-1})\).
**A1**: Explains clearly why the resulting expression is divisible by 21 (e.g., pointing out that both \(4f(k)\) and \(21 \cdot 5^{2k-1}\) are divisible by 21, or that 21 is a factor of the entire expression).
**A1**: Complete and correct proof with a concluding statement containing all essential elements: "true for \(n=1\)", "if true for \(n=k\) then true for \(n=k+1\)", and "therefore true for all positive integers \(n\)" (or equivalent terms).