2025 Edexcel IAS-Level Further Mathematics (XFM01) 模擬試題連答案詳解

(a) A matrix is singular when its determinant is equal to zero. This gives \(\det(\mathbf{M}) = (k+1)(k-4) - (-2)(3) = k^2 - 4k + k - 4 + 6 = k^2 - 3k + 2\). Setting the determinant to zero, we have \(k^2 - 3k + 2 = 0\), which factors as \((k-1)(k-2) = 0\). Therefore, \(k = 1\) or \(k = 2\). (b) For \(k = 3\), we have \(\mathbf{M} = \begin{pmatrix} 4 & -2 \\ 3 & -1 \end{pmatrix}\). The determinant of \(\mathbf{M}\) is \(\det(\mathbf{M}) = 4(-1) - (-2)(3) = 2\). The inverse matrix is \(\mathbf{M}^{-1} = \frac{1}{2} \begin{pmatrix} -1 & 2 \\ -3 & 4 \end{pmatrix}\). Given \(\mathbf{M}\mathbf{P} = \begin{pmatrix} 6 & -4 \\ 5 & 2 \end{pmatrix}\), left-multiplying both sides by \(\mathbf{M}^{-1}\) gives \(\mathbf{P} = \mathbf{M}^{-1} \begin{pmatrix} 6 & -4 \\ 5 & 2 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} -1 & 2 \\ -3 & 4 \end{pmatrix} \begin{pmatrix} 6 & -4 \\ 5 & 2 \end{pmatrix}\). Performing the matrix multiplication yields \(\begin{pmatrix} -1 & 2 \\ -3 & 4 \end{pmatrix} \begin{pmatrix} 6 & -4 \\ 5 & 2 \end{pmatrix} = \begin{pmatrix} (-1)(6) + (2)(5) & (-1)(-4) + (2)(2) \\ (-3)(6) + (4)(5) & (-3)(-4) + (4)(2) \end{pmatrix} = \begin{pmatrix} 4 & 8 \\ 2 & 20 \end{pmatrix}\). Dividing each entry by the determinant 2, we find \(\mathbf{P} = \begin{pmatrix} 2 & 4 \\ 1 & 10 \end{pmatrix}\).

Part (a): M1: Attempts to find the determinant of \(\mathbf{M}\) in terms of \(k\), setting \((k+1)(k-4) - (3)(-2)\) or equivalent. A1: Correctly simplifies the determinant to obtain \(k^2 - 3k + 2\). A1: Solves \(k^2 - 3k + 2 = 0\) to obtain both \(k = 1\) and \(k = 2\). Part (b): M1: Substitutes \(k = 3\) to find \(\mathbf{M}\) and attempts to find \(\mathbf{M}^{-1}\) (must show division of the adjoint matrix by their determinant). M1: Complete correct method of left-multiplying the given matrix by their \(\mathbf{M}^{-1}\). A1: Correctly finds the matrix \(\mathbf{P} = \begin{pmatrix} 2 & 4 \\ 1 & 10 \end{pmatrix}\).

(a) Evaluate \(f(x)\) at the boundaries of the interval:

\[f(1.8) = (1.8)^4 - 5(1.8) - 3 = 10.4976 - 9 - 3 = -1.5024\]

\[f(2) = 2^4 - 5(2) - 3 = 16 - 10 - 3 = 3\]

Since \(f(1.8) < 0\) and \(f(2) > 0\), there is a change of sign. Since \(f(x)\) is continuous on the interval \([1.8, 2]\), there must be a root \(\alpha\) in this interval.

(b) Using linear interpolation on \([1.8, 2]\):

\[\frac{\alpha - 1.8}{2 - \alpha} \approx \frac{|f(1.8)|}{f(2)}\]

\[\frac{\alpha - 1.8}{2 - \alpha} = \frac{1.5024}{3} = 0.5008\]

\[\alpha - 1.8 = 0.5008(2 - \alpha)\]

\[\alpha - 1.8 = 1.0016 - 0.5008\alpha\]

\[1.5008\alpha = 2.8016\]

\[\alpha = \frac{2.8016}{1.5008} \approx 1.8667377...\]

To 3 decimal places, \(\alpha \approx 1.867\).

(c) First, find the derivative of \(f(x)\):

\[f'(x) = 4x^3 - 5\]

Evaluate \(f'(1.8)\):

\[f'(1.8) = 4(1.8)^3 - 5 = 4(5.832) - 5 = 23.328 - 5 = 18.328\]

Using the Newton-Raphson formula with \(x_0 = 1.8\):

\[x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}\]

\[x_1 = 1.8 - \frac{-1.5024}{18.328}\]

\[x_1 = 1.8 + 0.0819729...\]

\[x_1 \approx 1.8819729...\]

To 3 decimal places, \(x_1 \approx 1.882\).

(a)
- M1: Attempts to evaluate \(f(1.8)\) and \(f(2)\) by substituting into the formula (at least one evaluation must be correct or show explicit substitution).
- A1: Evaluates both correctly with \(f(1.8) = -1.5024\) (or \(-1.50\)) and \(f(2) = 3\), and concludes that the change of sign (along with continuity) implies a root in the interval.

(b)
- M1: For a correct linear interpolation method. For example, setting up a correct expression for the approximation: \(1.8 + \frac{1.5024}{1.5024 + 3}(0.2)\) or equivalent.
- A1: For a correct intermediate equation or unsimplified fraction, e.g., \(1.5008x = 2.8016\) or \(x = 1.8 + \frac{0.30048}{4.5024}\).
- A1: Correct answer to 3 d.p.: \(1.867\) (awarded only if M1 is scored).

(c)
- M1: For differentiating \(f(x)\) to obtain \(f'(x) = kx^3 - 5\), where \(k \ne 0\).
- A1: Correct derivative \(f'(x) = 4x^3 - 5\).
- M1: Correct application of the Newton-Raphson process: \(x_1 = 1.8 - \frac{f(1.8)}{f'(1.8)}\) using their values.
- A1: Correct second approximation to 3 d.p.: \(1.882\) (awarded only if both M marks are scored).

(a) Comparing \(2x^2 - 5x + 4 = 0\) with the standard form \(ax^2 + bx + c = 0\), we have:
\(\alpha + \beta = -\frac{b}{a} = \frac{5}{2}\)
\(\alpha\beta = \frac{c}{a} = \frac{4}{2} = 2\)

Now, we express \(\frac{1}{\alpha} + \frac{1}{\beta}\) with a common denominator:
\(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}\)

Substituting the values of \(\alpha + \beta\) and \(\alpha\beta\):
\(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\frac{5}{2}}{2} = \frac{5}{4}\)

(b) Let the new roots be \(\gamma = \alpha - \frac{3}{\beta}\) and \(\delta = \beta - \frac{3}{\alpha}\).

First, find the sum of the new roots, \(\gamma + \delta\):
\(\gamma + \delta = \left(\alpha - \frac{3}{\beta}\right) + \left(\beta - \frac{3}{\alpha}\right)\)
\(\gamma + \delta = (\alpha + \beta) - 3\left(\frac{1}{\alpha} + \frac{1}{\beta}\right)\)

Using \(\alpha + \beta = \frac{5}{2}\) and the result from part (a):
\(\gamma + \delta = \frac{5}{2} - 3\left(\frac{5}{4}\right)\)
\(\gamma + \delta = \frac{10}{4} - \frac{15}{4} = -\frac{5}{4}\)

Next, find the product of the new roots, \(\gamma\delta\):
\(\gamma\delta = \left(\alpha - \frac{3}{\beta}\right)\left(\beta - \frac{3}{\alpha}\right)\)
\(\gamma\delta = \alpha\beta - 3 - 3 + \frac{9}{\alpha\beta}\)
\(\gamma\delta = \alpha\beta - 6 + \frac{9}{\alpha\beta}\)

Using \(\alpha\beta = 2\):
\(\gamma\delta = 2 - 6 + \frac{9}{2}\)
\(\gamma\delta = -4 + \frac{9}{2} = \frac{1}{2}\)

The new quadratic equation is given by:
\(x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0\)
\(x^2 - \left(-\frac{5}{4}\right)x + \frac{1}{2} = 0\)
\(x^2 + \frac{5}{4}x + \frac{1}{2} = 0\)

Multiply through by 4 to obtain integer coefficients:
\(4x^2 + 5x + 2 = 0\)

**Part (a)**
* **M1**: States \(\alpha + \beta = \frac{5}{2}\) and \(\alpha\beta = 2\) (or implied by subsequent working) and attempts to express \(\frac{1}{\alpha} + \frac{1}{\beta}\) as \(\frac{\alpha+\beta}{\alpha\beta}\).
* **A1**: Correct value of \(\frac{5}{4}\) (or \(1.25\)).

**Part (b)**
* **M1**: Attempts to find the sum of the new roots in terms of \(\alpha, \beta\) and/or the result from part (a).
* **A1ft**: Correct sum of \(\gamma + \delta = -\frac{5}{4}\) (follow through their part (a) value).
* **M1**: Attempts to expand and find the product of the new roots in terms of \(\alpha\beta\).
* **A1**: Correct product of \(\gamma\delta = \frac{1}{2}\).
* **M1**: Uses \(x^2 - (\text{sum})x + \text{product} = 0\) with their sum and product, and multiplies by a suitable integer to obtain integer coefficients.
* **A1**: Correct equation \(4x^2 + 5x + 2 = 0\) (or equivalent integer multiple). Must include \(= 0\).

(a) To express \(\frac{z_1}{z_2}\) in the form \(x + iy\), we multiply the numerator and the denominator by the complex conjugate of the denominator, \(3 + i\):

\[\frac{z_1}{z_2} = \frac{p + 2i}{3 - i} = \frac{(p + 2i)(3 + i)}{(3 - i)(3 + i)}\]

Expand the numerator:

\[(p + 2i)(3 + i) = 3p + pi + 6i + 2i^2\]

Since \(i^2 = -1\):

\[= (3p - 2) + (p + 6)i\]

Expand the denominator:

\[(3 - i)(3 + i) = 9 - i^2 = 9 - (-1) = 10\]

Therefore:

\[\frac{z_1}{z_2} = \frac{3p - 2}{10} + \frac{p + 6}{10}i\]

(b) We are given that \(\left|\frac{z_1}{z_2}\right| = \sqrt{2}\).
Using the modulus property \(\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}\):

\[|z_1| = \sqrt{p^2 + 2^2} = \sqrt{p^2 + 4}\]
\[|z_2| = \sqrt{3^2 + (-1)^2} = \sqrt{10}\]

So,

\[\frac{\sqrt{p^2 + 4}}{\sqrt{10}} = \sqrt{2}\]

Square both sides:

\[\frac{p^2 + 4}{10} = 2\]
\[p^2 + 4 = 20\]
\[p^2 = 16\]
\[p = \pm 4\]

Alternatively, using the result from part (a):

\[\left(\frac{3p - 2}{10}\right)^2 + \left(\frac{p + 6}{10}\right)^2 = 2\]
\[\frac{9p^2 - 12p + 4 + p^2 + 12p + 36}{100} = 2\]
\[10p^2 + 40 = 200 \implies p^2 = 16 \implies p = \pm 4\]

(c) For the positive value of \(p\), we use \(p = 4\).
Substituting \(p = 4\) into the expression for \(\frac{z_1}{z_2}\):

\[\frac{z_1}{z_2} = \frac{3(4) - 2}{10} + \frac{4 + 6}{10}i = \frac{10}{10} + \frac{10}{10}i = 1 + i\]

This complex number lies in the first quadrant of the Argand diagram.

\[\arg(1 + i) = \arctan\left(\frac{1}{1}\right) = \frac{\pi}{4}\]

Part (a):
- M1: Multiplies numerator and denominator by \(3 + i\).
- A1: Obtains a correct real part or imaginary part of the fraction.
- A1: Fully correct expression \(\frac{3p - 2}{10} + \frac{p + 6}{10}i\) or equivalent.

Part (b):
- M1: Employs a valid method to find the modulus of \(\frac{z_1}{z_2}\) and equates it to \
\sqrt{2}\.
- A1: Obtains the simplified equation \(p^2 = 16\) (or equivalent).
- A1: Deduces both \(p = 4\) and \(p = -4\).

Part (c):
- M1: Substitute \(p = 4\) into their expression from part (a) to obtain a numerical complex number and attempts to find its argument using \(\arctan\left(\frac{y}{x}\right)\).
- A1: Identifies the correct exact argument of \(\frac{\pi}{4}\) (or \(0.25\pi\)).

First, expand the summand: \((3r-1)(r-2) = 3r^2 - 7r + 2\). Now, split the sum and apply the standard formulas for \(\sum_{r=1}^n r^2\), \(\sum_{r=1}^n r\), and \(\sum_{r=1}^n 1\): \(\sum_{r=1}^n (3r-1)(r-2) = \sum_{r=1}^n (3r^2 - 7r + 2) = 3\sum_{r=1}^n r^2 - 7\sum_{r=1}^n r + \sum_{r=1}^n 2 = 3\left(\frac{1}{6}n(n+1)(2n+1)\right) - 7\left(\frac{1}{2}n(n+1)\right) + 2n = \frac{1}{2}n(n+1)(2n+1) - \frac{7}{2}n(n+1) + 2n\). Factorizing out \(\frac{1}{2}n\): \(= \frac{1}{2}n [ (n+1)(2n+1) - 7(n+1) + 4 ] = \frac{1}{2}n [ (2n^2 + 3n + 1) - (7n + 7) + 4 ] = \frac{1}{2}n [ 2n^2 - 4n - 2 ] = n(n^2 - 2n - 1)\) (as required). For the second part of the question: \(\sum_{r=10}^{40} (3r-1)(r-2) = \sum_{r=1}^{40} (3r-1)(r-2) - \sum_{r=1}^{9} (3r-1)(r-2)\). Using the derived formula with \(n = 40\): \(\sum_{r=1}^{40} (3r-1)(r-2) = 40(40^2 - 2(40) - 1) = 40(1600 - 80 - 1) = 40(1519) = 60760\). Using the derived formula with \(n = 9\): \(\sum_{r=1}^{9} (3r-1)(r-2) = 9(9^2 - 2(9) - 1) = 9(81 - 18 - 1) = 9(62) = 558\). Subtracting the two results: \(60760 - 558 = 60202\).

Part 1 (5 marks): M1: Expands bracket to obtain \(3r^2 - 7r + 2\) or equivalent. M1: Attempts to use standard summation formulas for \(\sum r^2\) and \(\sum r\). M1: Factorizes out at least \(\frac{1}{2}n\) or equivalent common factor. A1: Simplifies the terms in the bracket to a correct quadratic expression, e.g., \(2n^2 - 4n - 2\). A1: Correctly simplifies to the given result \(n(n^2 - 2n - 1)\) with no errors seen (cso). Part 2 (2 marks): M1: Statement or clear use of \(\sum_{r=10}^{40} = \sum_{r=1}^{40} - \sum_{r=1}^{9}\). A1: Correct final value of 60202.

(a) The equation of the hyperbola is \(y = \dfrac{c^2}{x} = c^2x^{-1}\).

We differentiate to find the gradient of the tangent:
\[\frac{\text{d}y}{\text{d}x} = -c^2x^{-2} = -\frac{c^2}{x^2}\]

At the point \(P\left(ct, \dfrac{c}{t}\right)\), we have \(x = ct\), so the gradient of the tangent is:
\[m_T = -\frac{c^2}{(ct)^2} = -\frac{1}{t^2}\]

Since the normal is perpendicular to the tangent, the gradient of the normal, \(m_N\), satisfies \(m_T \times m_N = -1\). Thus:
\[m_N = t^2\]

The equation of the normal at \(P\left(ct, \dfrac{c}{t}\right)\) is:
\[y - \frac{c}{t} = t^2(x - ct)\]

Multiplying the entire equation by \(t\) to clear fractions:
\[ty - c = t^3(x - ct)\]
\[ty - c = t^3x - ct^4\]

Rearranging to the required form gives:
\[t^3x - ty = c(t^4 - 1)\]

(b) Since \(Q\left(cq, \dfrac{c}{q}\right)\) lies on the normal at \(P\), its coordinates satisfy the equation of the normal:
\[t^3(cq) - t\left(\frac{c}{q}\right) = c(t^4 - 1)\]

Since \(c \neq 0\), we can divide through by \(c\):
\[t^3q - \frac{t}{q} = t^4 - 1\]

Multiply through by \(q\) to clear fractions:
\[t^3q^2 - t = (t^4 - 1)q\]
\[t^3q^2 - (t^4 - 1)q - t = 0\]

We factorise this quadratic in \(q\):
\[t^3q^2 - t^4q + q - t = 0\]
\[t^3q(q - t) + 1(q - t) = 0\]
\[(t^3q + 1)(q - t) = 0\]

Since \(Q\) is a distinct point from \(P\), we have \(q \neq t\).

Therefore:
\[t^3q + 1 = 0 \implies q = -\frac{1}{t^3}\]

(c) Given \(t = 2\), we find the parameter \(q\) for the point \(Q\):
\[q = -\frac{1}{2^3} = -\frac{1}{8}\]

The coordinates of \(Q\) are:
\[Q\left(cq, \frac{c}{q}\right) = Q\left(c\left(-\frac{1}{8}\right), \frac{c}{-\frac{1}{8}}\right) = \left(-\frac{1}{8}c, -8c\right)\]

**(a)**
* **M1**: Differentiates \(y = \frac{c^2}{x}\) to find \(\frac{\text{d}y}{\text{d}x} = -\frac{c^2}{x^2}\) and substitutes \(x = ct\) to find the gradient of the tangent.
* **A1**: Correct tangent gradient \(-\frac{1}{t^2}\) and deduces the normal gradient is \(t^2\).
* **M1**: Uses \(y - y_1 = m(x - x_1)\) with their normal gradient and point \(P\left(ct, \frac{c}{t}\right)\), and multiplies by \(t\) to eliminate fractions.
* **A1***: Correctly obtains the given equation \(t^3x - ty = c(t^4 - 1)\) with no errors shown. (This is a "show that" question, so working must be convincing).

**(b)**
* **M1**: Substitutes coordinates of \(Q\left(cq, \frac{c}{q}\right)\) into the normal equation and divides by \(c\).
* **M1**: Obtains a quadratic equation in \(q\), e.g. \(t^3q^2 - (t^4-1)q - t = 0\), and attempts to factorise it, recognising \((q-t)\) is a factor.
* **A1***: Reaches \(q = -\frac{1}{t^3}\) with complete and clear reasoning, explaining why \(q \neq t\).

**(c)**
* **M1**: Substitutes \(t = 2\) into the formula from part (b) to find \(q = -\frac{1}{8}\).
* **A1**: Deduces correct coordinates \(\left(-\frac{1}{8}c, -8c\right)\) or equivalent.

For (a), the area of triangle \(R\) with vertices \(O(0,0)\), \(P(2,1)\), and \(Q(-1,3)\) is given by \(\frac{1}{2} |2(3) - 1(-1)| = \frac{1}{2} |6 + 1| = 3.5\). For (b), using \(\text{Area}(R') = |\det \mathbf{A}| \times \text{Area}(R)\), we have \(28 = |\det \mathbf{A}| \times 3.5\), which gives \(|\det \mathbf{A}| = 8\). The determinant of \(\mathbf{A}\) is \(\det \mathbf{A} = k(k-5) - (2)(-1) = k^2 - 5k + 2\). Therefore, \(k^2 - 5k + 2 = \pm 8\). Case 1: \(k^2 - 5k + 2 = 8 \implies k^2 - 5k - 6 = 0 \implies (k-6)(k+1) = 0 \implies k = 6\) or \(k = -1\). Case 2: \(k^2 - 5k + 2 = -8 \implies k^2 - 5k + 10 = 0\). The discriminant is \((-5)^2 - 4(1)(10) = -15 < 0\), so there are no real roots. Thus, the only real values of \(k\) are \(k = 6\) and \(k = -1\). For (c), since \(k > 0\), we have \(k = 6\). The matrix is \(\mathbf{A} = \begin{pmatrix} 6 & 2 \\ -1 & 1 \end{pmatrix}\) and \(\det \mathbf{A} = 8\). The inverse is \(\mathbf{A}^{-1} = \frac{1}{8} \begin{pmatrix} 1 & -2 \\ 1 & 6 \end{pmatrix} = \begin{pmatrix} 0.125 & -0.25 \\ 0.125 & 0.75 \end{pmatrix}\). For (d), \(S = \mathbf{A}^{-1} S' = \frac{1}{8} \begin{pmatrix} 1 & -2 \\ 1 & 6 \end{pmatrix} \begin{pmatrix} 4 \\ -3 \end{pmatrix} = \frac{1}{8} \begin{pmatrix} 1(4) - 2(-3) \\ 1(4) + 6(-3) \end{pmatrix} = \frac{1}{8} \begin{pmatrix} 10 \\ -14 \end{pmatrix} = \begin{pmatrix} 1.25 \\ -1.75 \end{pmatrix}\). So the coordinates of \(S\) are \((1.25, -1.75)\).

Part (a): M1: Attempts to calculate the area of triangle \(R\) using a valid formula such as \(\frac{1}{2} |x_1 y_2 - x_2 y_1|\). A1: Correctly shows the area is 3.5. Part (b): M1: Relates the areas using \(\text{Area}(R') = |\det \mathbf{A}| \times \text{Area}(R)\) to find \(|\det \mathbf{A}| = 8\). M1: Calculates \(\det \mathbf{A} = k^2 - 5k + 2\) and sets it equal to \(\pm 8\). A1: Solves \(k^2 - 5k + 2 = 8\) to find \(k = 6\) and \(k = -1\). A1: Shows that \(k^2 - 5k + 2 = -8\) has no real solutions and concludes the only real values are \(k = 6\) and \(k = -1\). Part (c): M1: Chooses \(k = 6\) and writes the matrix \(\mathbf{A}\). M1: Applies a correct method to find the inverse of a 2x2 matrix. A1: Correctly obtains \(\mathbf{A}^{-1} = \frac{1}{8} \begin{pmatrix} 1 & -2 \\ 1 & 6 \end{pmatrix}\) or equivalent. Part (d): M1: Multiplies their \(\mathbf{A}^{-1}\) by \(\begin{pmatrix} 4 \\ -3 \end{pmatrix}\). A1: Obtains the correct coordinates \((1.25, -1.75)\) or equivalent fractions.

(a) Base case: For \( n = 1 \): LHS = \( 1^2(1+1) = 2 \), RHS = \( \frac{1}{12}(1)(2)(3)(4) = 2 \). Since LHS = RHS, the statement is true for \( n = 1 \). Inductive hypothesis: Assume the statement is true for \( n = k \), where \( k \) is a positive integer, so \( \sum_{r=1}^{k} r^2(r+1) = \frac{1}{12}k(k+1)(k+2)(3k+1) \). Inductive step: For \( n = k+1 \), \( \sum_{r=1}^{k+1} r^2(r+1) = \sum_{r=1}^{k} r^2(r+1) + (k+1)^2(k+2) = \frac{1}{12}k(k+1)(k+2)(3k+1) + (k+1)^2(k+2) = \frac{1}{12}(k+1)(k+2)[k(3k+1) + 12(k+1)] = \frac{1}{12}(k+1)(k+2)[3k^2 + 13k + 12] = \frac{1}{12}(k+1)(k+2)(k+3)(3k+4) = \frac{1}{12}(k+1)((k+1)+1)((k+1)+2)(3(k+1)+1) \). This is of the same form as the original statement with \( n = k+1 \). Conclusion: If the statement is true for \( n = k \), then it is true for \( n = k+1 \). Since it is true for \( n = 1 \), it is true for all \( n \in \mathbb{Z}^+\) by mathematical induction. (b) We can write \( \sum_{r=11}^{20} r^2(r+1) = \sum_{r=1}^{20} r^2(r+1) - \sum_{r=1}^{10} r^2(r+1) \). Using the formula from part (a): For \( n = 20 \), \( S_{20} = \frac{1}{12}(20)(21)(22)(61) = 46970 \). For \( n = 10 \), \( S_{10} = \frac{1}{12}(10)(11)(12)(31) = 3410 \). Thus, the sum is \( 46970 - 3410 = 43560 \).

Part (a) [6 Marks] - B1: Show LHS = RHS = 2 for \( n = 1 \) and state 'true for \( n = 1 \)' or equivalent. - M1: Assume the induction hypothesis is true for \( n = k \) and write \( S_{k+1} = S_k + u_{k+1} \). - M1: Correct algebraic substitution and attempt to factorize out at least \( (k+1)(k+2) \). - A1: Obtain the correct quadratic factor \( 3k^2 + 13k + 12 \). - A1: Factorize the quadratic to show the expression in the form \( \frac{1}{12}(k+1)(k+2)(k+3)(3k+4) \). - A1: Complete and convincing induction conclusion. Part (b) [4 Marks] - M1: State that the required sum is \( S_{20} - S_{10} \). - M1: Attempt to substitute \( n = 20 \) and \( n = 10 \) into the formula from part (a). - A1: Obtain \( S_{20} = 46970 \) and \( S_{10} = 3410 \). - A1: Correct final answer of \( 43560 \).

(a) Differentiating \(y^2 = 12x\) implicitly with respect to \(x\):
\[2y \frac{\text{d}y}{\text{d}x} = 12 \implies \frac{\text{d}y}{\text{d}x} = \frac{6}{y}\]
At the point \(P(3t^2, 6t)\):
\[\frac{\text{d}y}{\text{d}x} = \frac{6}{6t} = \frac{1}{t}\]
Thus, the gradient of the tangent at \(P\) is \(\frac{1}{t}\), which means the gradient of the normal is \(-t\).

The equation of the normal at \(P\) is:
\[y - 6t = -t(x - 3t^2)\]
\[y - 6t = -tx + 3t^3 \implies y + tx = 6t + 3t^3\]
This completes the proof.

(b) To find the coordinates of \(A\), we first find the equation of the tangent at \(P\):
\[y - 6t = \frac{1}{t}(x - 3t^2) \implies ty - 6t^2 = x - 3t^2 \implies ty = x + 3t^2\]
The tangent intersects the \(x\)-axis at \(A\), where \(y = 0\):
\[0 = x + 3t^2 \implies x = -3t^2\]
So, \(A\) has coordinates \((-3t^2, 0)\).

The normal intersects the \(x\)-axis at \(B\), where \(y = 0\):
\[tx = 6t + 3t^3\]
Since \(t > 0\), we can divide by \(t\):
\[x = 6 + 3t^2\]
So, \(B\) has coordinates \((6 + 3t^2, 0)\).

(c) The distance between \(A\) and \(B\) along the \(x\)-axis is:
\[AB = (6 + 3t^2) - (-3t^2) = 6 + 6t^2\]
The height of triangle \(APB\) is the \(y\)-coordinate of \(P\), which is \(6t\).

The area of triangle \(APB\) is given by:
\[\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(6 + 6t^2)(6t) = 18t(1 + t^2) = 18t^3 + 18t\]
We are given that the area is \(72\sqrt{3}\):
\[18t^3 + 18t = 72\sqrt{3} \implies t^3 + t - 4\sqrt{3} = 0\]
Let \(t = \sqrt{3}\):
\[(\sqrt{3})^3 + \sqrt{3} = 3\sqrt{3} + \sqrt{3} = 4\sqrt{3}\]
Since \(t^3 + t\) is strictly increasing for \(t > 0\), \(t = \sqrt{3}\) is the unique positive real solution.

Substituting \(t = \sqrt{3}\) into the coordinates of \(P(3t^2, 6t)\):
\[x = 3(\sqrt{3})^2 = 9\]
\[y = 6\sqrt{3}\]
Thus, the exact coordinates of \(P\) are \((9, 6\sqrt{3})\).

**Part (a)**
* **M1**: Attempts differentiation of the parabola equation (implicitly or parametrically) to find \(\frac{\text{d}y}{\text{d}x}\) in terms of \(y\) or \(t\).
* **M1**: Uses the negative reciprocal of their gradient to write an equation of the normal using \(P(3t^2, 6t)\).
* **A1* (cso)**: Fully correct derivation leading to the given equation of the normal with no errors seen.

**Part (b)**
* **M1**: Uses a correct method to find the equation of the tangent and sets \(y = 0\) to find the \(x\)-coordinate of \(A\) in terms of \(t\), or uses standard parabola properties.
* **A1**: Both \(A(-3t^2, 0)\) and \(B(6 + 3t^2, 0)\) correct (accept coordinates or clearly defined values on the \(x\)-axis).

**Part (c)**
* **M1**: Forms a correct expression for the area of triangle \(APB\) in terms of \(t\), e.g. \(\frac{1}{2} \times (x_B - x_A) \times y_P\).
* **M1**: Equates their area to \(72\sqrt{3}\) and solves the resulting cubic equation to find the positive real value \(t = \sqrt{3}\).
* **A1**: Correct exact coordinates of \(P(9, 6\sqrt{3})\).