2023 Edexcel IAS-Level Further Mathematics (XFM01) 模擬試題連答案詳解

From the quadratic equation \(2x^2 - 3x + 4 = 0\), we can identify the sum and product of the roots: \(\alpha + \beta = \frac{3}{2}\) and \(\alpha\beta = \frac{4}{2} = 2\). We wish to find a new quadratic equation with roots \(\gamma = \frac{\alpha}{\beta^2}\) and \(\delta = \frac{\beta}{\alpha^2}\). The sum of these new roots is: \(S = \gamma + \delta = \frac{\alpha}{\beta^2} + \frac{\beta}{\alpha^2} = \frac{\alpha^3 + \beta^3}{\alpha^2\beta^2}\). Using the identity \(\alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)\), we get: \(\alpha^3 + \beta^3 = \left(\frac{3}{2}\right)^3 - 3(2)\left(\frac{3}{2}\right) = \frac{27}{8} - 9 = -\frac{45}{8}\). Since \(\alpha^2\beta^2 = (\alpha\beta)^2 = 2^2 = 4\), the sum is: \(S = \frac{-45/8}{4} = -\frac{45}{32}\). The product of the new roots is: \(P = \gamma\delta = \frac{\alpha}{\beta^2} \times \frac{\beta}{\alpha^2} = \frac{\alpha\beta}{\alpha^2\beta^2} = \frac{1}{\alpha\beta} = \frac{1}{2}\). The new quadratic equation is given by \(x^2 - Sx + P = 0\), which is \(x^2 - \left(-\frac{45}{32}\right)x + \frac{1}{2} = 0 \implies x^2 + \frac{45}{32}x + \frac{1}{2} = 0\). Multiplying by 32 to obtain integer coefficients yields: \(32x^2 + 45x + 16 = 0\).

B1: Identifies both correct sum \(\alpha + \beta = 1.5\) and product \(\alpha\beta = 2\). M1: Applies a correct algebraic identity to express \(\alpha^3 + \beta^3\) in terms of \(\alpha + \beta\) and \(\alpha\beta\). A1: Evaluates \(\alpha^3 + \beta^3 = -45/8\) or the new sum \(S = -45/32\) correctly. M1: Sets up the new product \(P = 1/(\alpha\beta)\) and evaluates it to \(1/2\). A1: Writes down the final quadratic equation with integer coefficients: \(32x^2 + 45x + 16 = 0\) (or any integer multiple thereof).

(a) We can expand the terms inside the summation and split it: \(\sum_{r=1}^{n} r(3r-1) = \sum_{r=1}^{n} (3r^2 - r) = 3\sum_{r=1}^{n} r^2 - \sum_{r=1}^{n} r\). Substituting the standard formulae \[\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)\] and \[\sum_{r=1}^{n} r = \frac{1}{2}n(n+1)\] we get: \(3 \left( \frac{1}{6}n(n+1)(2n+1) \right) - \frac{1}{2}n(n+1) = \frac{1}{2}n(n+1)(2n+1) - \frac{1}{2}n(n+1)\). Factoring out the common terms: \(\frac{1}{2}n(n+1) [ (2n+1) - 1 ] = \frac{1}{2}n(n+1)(2n) = n^2(n+1)\) as required. (b) To find the sum from \(r=10\) to \(r=25\), we use: \(\sum_{r=10}^{25} r(3r-1) = \sum_{r=1}^{25} r(3r-1) - \sum_{r=1}^{9} r(3r-1)\). Substituting \(n=25\) and \(n=9\) into our result from part (a): \(25^2(25+1) - 9^2(9+1) = 625(26) - 81(10) = 16250 - 810 = 15440\).

M1 (part a): Expands and splits the summation using standard formulae for \(\sum r^2\) and \(\sum r\). A1 (part a): Simplifies the expression correctly to a common denominator, such as \(\frac{1}{2}n(n+1)[(2n+1) - 1]\). A1 (part a): Completes the algebra to show the required result \(n^2(n+1)\) with no errors. M1 (part b): Applies the subtraction method \(\sum_{r=10}^{25} = \sum_{r=1}^{25} - \sum_{r=1}^{9}\) using the formula. A1 (part b): Calculates the correct final value of 15440.

(a) We evaluate the function at the endpoints of the interval: \(f(0.6) = (0.6)^3 - 5(0.6) + 3 = 0.216 - 3 + 3 = 0.216\) and \(f(0.7) = (0.7)^3 - 5(0.7) + 3 = 0.343 - 3.5 + 3 = -0.157\). Since \(f(x)\) is a continuous polynomial function, and there is a change of sign between \(f(0.6) > 0\) and \(f(0.7) < 0\), a root \(\alpha\) must exist in the interval \([0.6, 0.7]\). (b) Using linear interpolation over the interval \([0.6, 0.7]\), we set up the ratio: \(\frac{\alpha - 0.6}{0.7 - \alpha} = \frac{f(0.6)}{-f(0.7)} = \frac{0.216}{0.157}\). This gives: \(0.157(\alpha - 0.6) = 0.216(0.7 - \alpha) \implies 0.157\alpha - 0.0942 = 0.1512 - 0.216\alpha\). Solving for \(\alpha\): \(0.373\alpha = 0.2454 \implies \alpha = \frac{0.2454}{0.373} \approx 0.6579088...\). To 3 decimal places, \(\alpha \approx 0.658\).

M1 (part a): Calculates at least one of \(f(0.6)\) or \(f(0.7)\) correctly. A1 (part a): Finds both \(f(0.6) = 0.216\) and \(f(0.7) = -0.157\) correctly, and concludes that a root exists because of continuity and the change of sign. M1 (part b): Uses a correct linear interpolation formula or ratio method with their values. A1 (part b): Obtains a correct unrounded value such as \(0.6579...\) or equivalent fraction. A1 (part b): Correctly rounds the final approximation to 0.658.

Given the equation \(2x^2 - 5x + 6 = 0\), the sum and product of the roots are:
\(\alpha + \beta = \frac{5}{2}\)
\(\alpha \beta = 3\)

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

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

(c) For the new quadratic equation, let the roots be \(u = \frac{\alpha}{\beta^2}\) and \(v = \frac{\beta}{\alpha^2}\).
The sum of the new roots \(S\) is:
\(S = u + v = \frac{\alpha}{\beta^2} + \frac{\beta}{\alpha^2} = \frac{\alpha^3 + \beta^3}{\alpha^2 \beta^2} = \frac{-\frac{55}{8}}{(\alpha\beta)^2} = \frac{-\frac{55}{8}}{3^2} = -\frac{55}{72}\)

The product of the new roots \(P\) is:
\(P = uv = \frac{\alpha}{\beta^2} \times \frac{\beta}{\alpha^2} = \frac{\alpha\beta}{\alpha^2\beta^2} = \frac{1}{\alpha\beta} = \frac{1}{3}\)

The new quadratic equation is given by:
\(x^2 - Sx + P = 0\)
\(x^2 - \left(-\frac{55}{72}\right)x + \frac{1}{3} = 0\)
\(x^2 + \frac{55}{72}x + \frac{1}{3} = 0\)

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

(a)
M1: Applies the identity \(\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta\) with their sum and product.
A1: Correct value of \(\frac{1}{4}\) (or \(0.25\)).

(b)
M1: Attempts to express \(\alpha^3+\beta^3\) in terms of sum and product, e.g., \((\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)\).
A1: Substitutes values and simplifies correctly to get \(-\frac{55}{8}\) (or \(-6.875\)).

(c)
M1: Attempts to express the sum of the new roots in terms of \(\alpha^3+\beta^3\) and \(\alpha\beta\).
A1: Correct sum of roots of \(-\frac{55}{72}\).
B1: Correct product of roots of \(\frac{1}{3}\).
M1: Uses \(x^2 - Sx + P = 0\) with their sum and product, multiplying by a constant to obtain integer coefficients.
A1: Correct equation \(72x^2 + 55x + 24 = 0\) (or any integer multiple thereof).

Let \(z = a + \mathrm{i}b\), where \(a, b \in \mathbb{R}\). Then \(z^* = a - \mathrm{i}b\).

The given equation can be written as:
\(z = (2 - \mathrm{i})(z^* + 3\mathrm{i})\)

Substitute \(z\) and \(z^*\) into the equation:
\(a + \mathrm{i}b = (2 - \mathrm{i})(a - \mathrm{i}b + 3\mathrm{i})\)
\(a + \mathrm{i}b = (2 - \mathrm{i})(a + \mathrm{i}(3 - b))\)

Expand the right-hand side:
\((2 - \mathrm{i})(a + \mathrm{i}(3 - b)) = 2a + 2\mathrm{i}(3 - b) - \mathrm{i}a - \mathrm{i}^2(3 - b)\)
\(= 2a + \mathrm{i}(6 - 2b) - \mathrm{i}a + (3 - b)\)
\(= (2a - b + 3) + \mathrm{i}(6 - a - 2b)\)

Equating real and imaginary parts from both sides:
Real parts:
\(a = 2a - b + 3 \implies a - b + 3 = 0 \implies a = b - 3\)

Imaginary parts:
\(b = 6 - a - 2b \implies a + 3b - 6 = 0\)

Substitute \(a = b - 3\) into the imaginary parts equation:
\((b - 3) + 3b - 6 = 0\)
\(4b - 9 = 0 \implies b = \frac{9}{4}\)

Now, find \(a\):
\(a = \frac{9}{4} - 3 = -\frac{3}{4}\)

Thus, \(z = -\frac{3}{4} + \frac{9}{4}\mathrm{i}\).

M1: Substitutes \(z = a + \mathrm{i}b\) and \(z^* = a - \mathrm{i}b\) and clears the fraction.
M1: Expands the RHS, collecting real and imaginary parts (allow one sign error).
A1: Correct expanded RHS: \((2a - b + 3) + \mathrm{i}(6 - a - 2b)\).
M1: Equates real and imaginary parts to form two linear simultaneous equations.
A1: Correct equations, e.g., \(a - b + 3 = 0\) and \(a + 3b - 6 = 0\).
M1: Solves the simultaneous equations to find values for \(a\) and \(b\).
A1: Correct values of \(a = -\frac{3}{4}\) and \(b = \frac{9}{4}\).
A1: Correct final complex number in the form \(a + \mathrm{i}b\).

(a) Expanding the summand:
\((3r-1)(r+2) = 3r^2 + 5r - 2\)

Applying sum properties:
\(\sum_{r=1}^n (3r-1)(r+2) = 3\sum_{r=1}^n r^2 + 5\sum_{r=1}^n r - \sum_{r=1}^n 2\)

Substitute standard formulae:
\(= 3 \left[ \frac{1}{6}n(n+1)(2n+1) \right] + 5 \left[ \frac{1}{2}n(n+1) \right] - 2n\)
\(= \frac{1}{2}n(n+1)(2n+1) + \frac{5}{2}n(n+1) - 2n\)

Factor out \(\frac{1}{2}n\):
\(= \frac{1}{2}n \left[ (n+1)(2n+1) + 5(n+1) - 4 \right]\)
\(= \frac{1}{2}n \left[ 2n^2 + 3n + 1 + 5n + 5 - 4 \right]\)
\(= \frac{1}{2}n \left[ 2n^2 + 8n + 2 \right]\)
\(= n(n^2 + 4n + 1)\) (as required).

(b) To evaluate \(\sum_{r=10}^{20} (3r-1)(r+2)\):
\(\sum_{r=10}^{20} (3r-1)(r+2) = \sum_{r=1}^{20} (3r-1)(r+2) - \sum_{r=1}^{9} (3r-1)(r+2)\)

Using the formula from part (a):
For \(n = 20\):
\(20(20^2 + 4(20) + 1) = 20(400 + 80 + 1) = 20(481) = 9620\)

For \(n = 9\):
\(9(9^2 + 4(9) + 1) = 9(81 + 36 + 1) = 9(118) = 1062\)

Subtracting these values:
\(\sum_{r=10}^{20} (3r-1)(r+2) = 9620 - 1062 = 8558\).

(a)
B1: Expands summand to \(3r^2 + 5r - 2\).
M1: Correctly substitutes the standard sum formulae for \(\sum r^2\) and \(\sum r\).
M1: Attempts to factorise \(n\) or \(\frac{1}{2}n\) out of their algebraic sum expression.
A1: Obtains a correct simplified quadratic inside the brackets, such as \(\frac{1}{2}n(2n^2 + 8n + 2)\).
A1*: Fully correct algebraic proof leading to the given expression without errors.

(b)
M1: Uses the identity \(\sum_{r=10}^{20} u_r = \sum_{r=1}^{20} u_r - \sum_{r=1}^{9} u_r\) with their formula.
A1: Calculates both intermediate sum values correctly (9620 and 1062).
A1: Correct final answer of 8558.

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

Differentiating with respect to \(x\):
\[\frac{dy}{dx} = -c^2 x^{-2} = -\frac{c^2}{x^2}\]

At the point \(P\left(cp, \frac{c}{p}\right)\), the gradient of the tangent is:
\[m_t = -\frac{c^2}{(cp)^2} = -\frac{1}{p^2}\]

The equation of the tangent is:
\[y - \frac{c}{p} = -\frac{1}{p^2}(x - cp)\]

Multiplying through by \(p^2\):
\[p^2 y - cp = -x + cp\]
\[x + p^2 y = 2cp\]

(b) To find \(A\), substitute \(y = 0\):
\[x + p^2(0) = 2cp \implies x = 2cp \implies A(2cp, 0)\]

To find \(B\), substitute \(x = 0\):
\[0 + p^2 y = 2cp \implies y = \frac{2c}{p} \implies B\left(0, \frac{2c}{p}\right)\]

(c) The gradient of the normal to \(H\) at \(P\) is:
\[m_n = -\frac{1}{m_t} = p^2\]

The equation of the normal is:
\[y - \frac{c}{p} = p^2(x - cp)\]

Multiplying through by \(p\):
\[py - c = p^3 x - cp^4\]
\[p^3 x - py = c(p^4 - 1)\]

The normal meets the line \(y = -x\) at \(N\). Substituting \(y = -x\) into the normal equation:
\[p^3 x - p(-x) = c(p^4 - 1)\]
\[x(p^3 + p) = c(p^2 - 1)(p^2 + 1)\]
\[xp(p^2 + 1) = c(p^2 - 1)(p^2 + 1)\]

Since \(p > 0\), \(p^2 + 1 \neq 0\). Dividing both sides by \(p^2 + 1\):
\[xp = c(p^2 - 1) \implies x = \frac{c(p^2 - 1)}{p}\]

Since \(y = -x\), the y-coordinate of \(N\) is:
\[y = -\frac{c(p^2 - 1)}{p}\]

Thus, the coordinates of \(N\) are:
\[\left(\frac{c(p^2-1)}{p}, -\frac{c(p^2-1)}{p}\right)\]

(a)
- M1: Differentiates to find \(\frac{dy}{dx} = k x^{-2}\).
- A1: Correct expression for gradient at \(P\), \(m_t = -\frac{1}{p^2}\).
- M1: Uses their gradient and coordinates of \(P\) to form the equation of the tangent.
- A1*: Completes the proof to arrive at the given equation \(x + p^2 y = 2cp\) with no errors seen.

(b)
- B1: Correct coordinates for \(A(2cp, 0)\).
- B1: Correct coordinates for \(B\left(0, \frac{2c}{p}\right)\).

(c)
- B1: States or uses the gradient of the normal is \(p^2\).
- M1: Forms the equation of the normal at \(P\).
- A1: Obtains a correct simplified equation of the normal, e.g., \(p^3 x - py = c(p^4 - 1)\).
- M1: Eliminates \(y\) using \(y = -x\) and attempts to factorise/simplify to solve for \(x\).
- A1*: Correctly shows that \(x = \frac{c(p^2-1)}{p}\) and concludes with the correct coordinates of \(N\).

(a) Substitute \(z_1 = a + 3\text{i}\) into the expression for \(w\):
\[w = \frac{a + 3\text{i} + 3\text{i}}{a + 3\text{i} - 2\text{i}} = \frac{a + 6\text{i}}{a + \text{i}}\]

Multiply the numerator and denominator by the complex conjugate of the denominator, \(a - \text{i}\):
\[w = \frac{(a + 6\text{i})(a - \text{i})}{(a + \text{i})(a - \text{i})}\]
\[w = \frac{a^2 - a\text{i} + 6a\text{i} - 6\text{i}^2}{a^2 + 1}\]

Since \(\text{i}^2 = -1\):
\[w = \frac{a^2 + 6 + 5a\text{i}}{a^2 + 1}\]
\[w = \frac{a^2 + 6}{a^2 + 1} + \frac{5a}{a^2 + 1}\text{i}\]

(b) Since \(\arg(w) = \frac{\pi}{4}\), both the real part and the imaginary part of \(w\) must be positive, and:
\[\tan\left(\frac{\pi}{4}\right) = \frac{\operatorname{Im}(w)}{\operatorname{Re}(w)} = 1 \implies \operatorname{Im}(w) = \operatorname{Re}(w)\]

Therefore:
\[\frac{5a}{a^2 + 1} = \frac{a^2 + 6}{a^2 + 1}\]

Since \(a^2 + 1 > 0\) for all real \(a\), we can multiply both sides by \(a^2 + 1\):
\[5a = a^2 + 6\]
\[a^2 - 5a + 6 = 0\]
\[(a - 2)(a - 3) = 0\]

Thus, the possible values of \(a\) are \(a = 2\) or \(a = 3\).
Both values are positive, so the real and imaginary parts remain positive, confirming both solutions are valid.

(c) The larger value of \(a\) is \(a = 3\).

Substituting \(a = 3\) into \(w\):
\[w = \frac{3^2 + 6}{3^2 + 1} + \frac{5(3)}{3^2 + 1}\text{i} = \frac{15}{10} + \frac{15}{10}\text{i} = \frac{3}{2} + \frac{3}{2}\text{i}\]

Then the modulus of \(w\) is:
\[|w| = \sqrt{\left(\frac{3}{2}\right)^2 + \left(\frac{3}{2}\right)^2} = \sqrt{\frac{9}{4} + \frac{9}{4}} = \sqrt{\frac{18}{4}} = \frac{3\sqrt{2}}{2}\]

(a)
- M1: Substitutes \(z_1 = a + 3\text{i}\) into \(w\) and simplifies numerator and denominator to linear complex terms.
- M1: Multiplies numerator and denominator by \(a - \text{i}\).
- A1: Correct expansion of the numerator, showing \(a^2 + 6 + 5a\text{i}\).
- A1*: Obtains the given expression with no errors seen.

(b)
- B1: Explains that \(\arg(w) = \frac{\pi}{4} \implies \operatorname{Re}(w) = \operatorname{Im}(w) > 0\).
- M1: Equates the real and imaginary parts to form a quadratic equation in \(a\).
- A1: Obtains the correct quadratic \(a^2 - 5a + 6 = 0\).
- A1: Finds both correct solutions \(a = 2\) and \(a = 3\).

(c)
- B1: Identifies \(a = 3\) as the larger value.
- M1: Substitutes \(a = 3\) to find \(w = \frac{3}{2} + \frac{3}{2}\text{i}\) and attempts to find the modulus using Pythagoras' theorem.
- A1: Correctly simplifies to \(\frac{3\sqrt{2}}{2}\) or equivalent exact surd (e.g., \(\frac{3}{\sqrt{2}}\)).

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

(a) We use the identity:
\[\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\]
Substituting the values:
\[\alpha^2 + \beta^2 = \left(\frac{5}{2}\right)^2 - 2(2) = \frac{25}{4} - 4 = \frac{9}{4}\]

(b) We use the identity:
\[\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\]
Substituting the values:
\[\alpha^3 + \beta^3 = \left(\frac{5}{2}\right)^3 - 3(2)\left(\frac{5}{2}\right) = \frac{125}{8} - 15 = \frac{125 - 120}{8} = \frac{5}{8}\]

(c) Let the new roots be \(\gamma = \alpha^2 + \frac{1}{\beta}\) and \(\delta = \beta^2 + \frac{1}{\alpha}\).

First, find the sum of the new roots, \(S\):
\[S = \gamma + \delta = \left(\alpha^2 + \beta^2\right) + \left(\frac{1}{\alpha} + \frac{1}{\beta}\right)\]
\[S = \left(\alpha^2 + \beta^2\right) + \frac{\alpha + \beta}{\alpha\beta}\]

Using the values from earlier:
\[S = \frac{9}{4} + \frac{5/2}{2} = \frac{9}{4} + \frac{5}{4} = \frac{14}{4} = \frac{7}{2}\]

Next, find the product of the new roots, \(P\):
\[P = \gamma\delta = \left(\alpha^2 + \frac{1}{\beta}\right)\left(\beta^2 + \frac{1}{\alpha}\right)\]
\[P = \alpha^2\beta^2 + \alpha + \beta + \frac{1}{\alpha\beta}\]
\[P = (\alpha\beta)^2 + (\alpha + \beta) + \frac{1}{\alpha\beta}\]

Using the values:
\[P = 2^2 + \frac{5}{2} + \frac{1}{2} = 4 + 3 = 7\]

The new quadratic equation is given by:
\[x^2 - Sx + P = 0\]
\[x^2 - \frac{7}{2}x + 7 = 0\]

Multiplying by 2 to obtain integer coefficients:
\[2x^2 - 7x + 14 = 0\]

(a)
- B1: Identifies \(\alpha + \beta = \frac{5}{2}\) and \(\alpha\beta = 2\).
- M1: Uses the identity \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\).
- A1: Obtains \(\frac{9}{4}\).

(b)
- M1: Uses a correct identity for \(\alpha^3 + \beta^3\), e.g., \(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\).
- M1: Substitutes their values for sum and product into the identity.
- A1: Obtains \(\frac{5}{8}\).

(c)
- M1: Expresses the sum of the new roots, \(S\), in terms of \(\alpha^2+\beta^2\), \(\alpha+\beta\), and \(\alpha\beta\).
- A1: Correctly calculates \(S = \frac{7}{2}\).
- M1: Expands the product of the new roots, \(P\), and expresses it in terms of known quantities.
- A1: Correctly calculates \(P = 7\).
- A1: Obtains the equation \(2x^2 - 7x + 14 = 0\) (or any integer multiple thereof, equal to 0).