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(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).