2025 Cambridge IAL Mathematics (9709) 模擬試題連答案詳解

Let the first term of the geometric progression be \(a\) and its common ratio be \(r\).

The second term is given by:
\(ar = 24 \implies r = \frac{24}{a}\)

The sum to infinity is given by:
\(S_{\infty} = \frac{a}{1-r} = 100\)

Substitute \(r = \frac{24}{a}\) into the sum to infinity formula:
\(\frac{a}{1 - \frac{24}{a}} = 100\)

\(\frac{a^2}{a - 24} = 100\)

\(a^2 = 100(a - 24)\)

\(a^2 - 100a + 2400 = 0\)

Factorising the quadratic equation:
\((a - 40)(a - 60) = 0\)

Thus, the two possible values of the first term are \(a = 40\) and \(a = 60\).

(We can verify both values give a valid ratio: for \(a = 40\), \(r = 0.6\); for \(a = 60\), \(r = 0.4\). Both satisfy the condition \(|r| < 1\).)

M1: For formulating the equations \(ar = 24\) and \(\frac{a}{1-r} = 100\), and attempting to substitute for \(r\).
M1: For simplifying to obtain a correct three-term quadratic equation in terms of \(a\) (e.g., \(a^2 - 100a + 2400 = 0\)).
A1: For attempting to solve their quadratic equation by factorisation, completing the square, or formula.
A1: For obtaining both correct values: \(a = 40\) and \(a = 60\).

Rewrite the curve's equation as:
\(y = x + 9(x-2)^{-1}\)

Differentiate with respect to \(x\):
\(\frac{dy}{dx} = 1 - 9(x-2)^{-2} = 1 - \frac{9}{(x-2)^2}\)

At stationary points, \(\frac{dy}{dx} = 0\):
\(1 - \frac{9}{(x-2)^2} = 0\)

\((x-2)^2 = 9\)

Taking the square root on both sides:
\(x - 2 = \pm 3\)

This gives two cases:
1) \(x - 2 = 3 \implies x = 5\)
Corresponding \(y\)-coordinate: \(y = 5 + \frac{9}{5-2} = 5 + 3 = 8\)

2) \(x - 2 = -3 \implies x = -1\)
Corresponding \(y\)-coordinate: \(y = -1 + \frac{9}{-1-2} = -1 - 3 = -4\)

Hence, the coordinates of the stationary points are \((5, 8)\) and \((-1, -4)\).

M1: For attempting to differentiate the term \(\frac{9}{x-2}\) (applying chain rule or quotient rule).
A1: For obtaining the correct derivative \(\frac{dy}{dx} = 1 - \frac{9}{(x-2)^2}\).
M1: For setting their \(\frac{dy}{dx} = 0\) and attempting to solve for \(x\).
A1: For obtaining both correct points: \((5, 8)\) and \((-1, -4)\).

First, find the midpoint \(M\) of the line segment \(AB\):
\(M = \left( \frac{1+5}{2}, \frac{5+(-3)}{2} \right) = (3, 1)\)

Next, find the gradient \(m\) of the line segment \(AB\):
\(m = \frac{-3 - 5}{5 - 1} = \frac{-8}{4} = -2\)

The gradient \(m'\) of the perpendicular bisector is:
\(m' = -\frac{1}{m} = \frac{1}{2}\)

Find the equation of the perpendicular bisector passing through \(M(3, 1)\) with gradient \(\frac{1}{2}\):
\(y - 1 = \frac{1}{2}(x - 3)\)
\(y = \frac{1}{2}x - \frac{1}{2}\)

The point \(P\) is where the perpendicular bisector intersects the \(y\)-axis (set \(x = 0\)):
\(y = \frac{1}{2}(0) - \frac{1}{2} = -0.5\)

Therefore, the coordinates of \(P\) are \((0, -0.5)\).

M1: For finding the correct midpoint \((3, 1)\) and calculating the gradient of \(AB\) as \(-2\).
M1: For using \(m_1 m_2 = -1\) to find the gradient of the perpendicular bisector as \(\frac{1}{2}\).
A1: For obtaining a correct linear equation of the perpendicular bisector, e.g., \(y - 1 = 0.5(x - 3)\).
A1: For setting \(x = 0\) in their equation to obtain \(P(0, -0.5)\) or \(P\left(0, -\frac{1}{2}\right)\).

The perimeter of the sector is given by:
\(P = 2r + r\theta = 20 \implies r\theta = 20 - 2r\)

The area of the sector is given by:
\(A = \frac{1}{2}r^2\theta = 24\)

We can rewrite the area formula to substitute \(r\theta\):
\(A = \frac{1}{2}r(r\theta) = 24\)

Substitute \(r\theta = 20 - 2r\) into this equation:
\(\frac{1}{2}r(20 - 2r) = 24\)

\(r(10 - r) = 24\)

\(10r - r^2 = 24\)

Rearrange into standard quadratic form:
\(r^2 - 10r + 24 = 0\)

Factorising the quadratic equation yields:
\((r - 4)(r - 6) = 0\)

Thus, the two possible values of the radius are \(r = 4\) and \(r = 6\).

M1: For using the correct circular measure formulas for perimeter and area to write two simultaneous equations.
M1: For attempting to eliminate \(\theta\) to form a quadratic equation in terms of \(r\) only.
A1: For obtaining a correct quadratic equation, e.g., \(r^2 - 10r + 24 = 0\) (or equivalent).
A1: For solving to find both correct possible values of \(r\) (\(r = 4\) and \(r = 6\)).

Since the terms are in geometric progression, the common ratio \(r\) must be constant:
\(\frac{k}{k+2} = \frac{k-1.5}{k}\)
Cross-multiplying:
\(k^2 = (k+2)(k-1.5)\)
\(k^2 = k^2 + 0.5k - 3\)
\(0.5k = 3 \implies k = 6\).

Substituting \(k=6\):
First term \(a = 6 + 2 = 8\).
Second term \(ar = 6\).
Common ratio \(r = \frac{6}{8} = 0.75\).

Since \(|r| < 1\), the sum to infinity exists:
\(S_{\infty} = \frac{a}{1-r} = \frac{8}{1-0.75} = \frac{8}{0.25} = 32\).

- M1: For setting up the ratio equation \(\frac{k}{k+2} = \frac{k-1.5}{k}\)
- A1: For correctly solving to find \(k = 6\)
- B1: For finding the first term \(a = 8\)
- B1: For finding the common ratio \(r = 0.75\)
- M1: For using the sum to infinity formula with their \(a\) and \(r\)
- A1: For obtaining 32

(i) Expressing \(f(x)\) in completed square form:
\(f(x) = 2(x^2 - 4x) + 5 = 2\left[(x-2)^2 - 4\right] + 5 = 2(x-2)^2 - 3\).
The vertex of the quadratic curve is at \((2, -3)\).
For \(f\) to be a one-to-one function, the domain \(x \ge c\) must only cover the right-hand branch of the vertex.
Therefore, the smallest value of \(c\) is \(2\).

(ii) To find the inverse, let \(y = 2(x-2)^2 - 3\).
Rearranging to make \(x\) the subject:
\(y+3 = 2(x-2)^2 \implies \frac{y+3}{2} = (x-2)^2\)
Since \(x \ge 2\), we take the positive square root:
\(x-2 = \sqrt{\frac{y+3}{2}} \implies x = 2 + \sqrt{\frac{y+3}{2}}\)
Thus, \(f^{-1}(x) = 2 + \sqrt{\frac{x+3}{2}}\).
The domain of \(f^{-1}\) is the range of \(f\). Since the minimum value of \(f(x)\) is \(-3\) for \(x \ge 2\), the range is \(f(x) \ge -3\).
Therefore, the domain of \(f^{-1}\) is \(x \ge -3\).

- B1: Complete the square to get \(2(x-2)^2 - 3\) (or find vertex by differentiation)
- B1: State \(c = 2\)
- M1: Set \(y = f(x)\) and attempt to rearrange for \(x\)
- A1: Obtain \((x-2)^2 = \frac{y+3}{2}\) or equivalent
- A1: Write correct inverse function \(f^{-1}(x) = 2 + \sqrt{\frac{x+3}{2}}\)
- B1: State domain as \(x \ge -3\)

We rewrite the equation as \(y = 4(x-2)^{-1} + x\).
Differentiating with respect to \(x\):
\(\frac{dy}{dx} = -4(x-2)^{-2} + 1 = 1 - \frac{4}{(x-2)^2}\).
At stationary points, \(\frac{dy}{dx} = 0\):
\(1 - \frac{4}{(x-2)^2} = 0 \implies (x-2)^2 = 4\).
Taking square roots:
\(x-2 = \pm 2 \implies x = 4\) or \(x = 0\).

When \(x = 4\), \(y = \frac{4}{4-2} + 4 = 6\). Point is \((4, 6)\).
When \(x = 0\), \(y = \frac{4}{0-2} + 0 = -2\). Point is \((0, -2)\).

To determine the nature, find the second derivative:
\(\frac{d^2y}{dx^2} = 8(x-2)^{-3} = \frac{8}{(x-2)^3}\).

At \(x = 4\):
\(\frac{d^2y}{dx^2} = \frac{8}{2^3} = 1 > 0\) (Local Minimum).
At \(x = 0\):
\(\frac{d^2y}{dx^2} = \frac{8}{(-2)^3} = -1 < 0\) (Local Maximum).

- M1: Attempt to differentiate \(y\) with at least one correct term
- A1: Correctly obtain \(\frac{dy}{dx} = 1 - \frac{4}{(x-2)^2}\)
- M1: Set their \(\frac{dy}{dx} = 0\) and solve for \(x\)
- A1: Correctly find both coordinates: \((0, -2)\) and \((4, 6)\)
- M1: Correct second derivative expression and substitution of at least one of their values of \(x\)
- A1: Correctly identify \((0, -2)\) as a maximum and \((4, 6)\) as a minimum

(i) To find the points of intersection, set the curve equal to the line:
\(x^2 - 4x + 6 = 2x - 2\)
\(x^2 - 6x + 8 = 0\)
\((x-2)(x-4) = 0\)
So \(x = 2\) or \(x = 4\).
Substituting back into \(y = 2x - 2\):
When \(x = 2\), \(y = 2\). Point is \((2, 2)\).
When \(x = 4\), \(y = 6\). Point is \((4, 6)\).

(ii) The area of the enclosed region is:
\(\text{Area} = \int_{2}^{4} \left( (2x - 2) - (x^2 - 4x + 6) \right) \, dx\)
\(\text{Area} = \int_{2}^{4} (-x^2 + 6x - 8) \, dx\)
Integrating:
\(= \left[ -\frac{x^3}{3} + 3x^2 - 8x \right]_{2}^{4}\)
Evaluating at upper limit \(x = 4\):
\(-\frac{64}{3} + 3(16) - 8(4) = -\frac{16}{3}\)
Evaluating at lower limit \(x = 2\):
\(-\frac{8}{3} + 3(4) - 8(2) = -\frac{20}{3}\)
Subtracting the limits:
\(\text{Area} = -\frac{16}{3} - \left(-\frac{20}{3}\right) = \frac{4}{3}\).

- M1: Equate line and curve and attempt to solve the resulting quadratic
- A1: Obtain correct intersection points \((2, 2)\) and \((4, 6)\)
- M1: Set up the correct definite integral of \(\text{line} - \text{curve}\) with limits from their \(x\)-values
- A1: Correct integration to get \(-\frac{x^3}{3} + 3x^2 - 8x\)
- M1: Correct substitution of limits into their integrated expression
- A1: Correctly obtain final area of \(\frac{4}{3}\)

Using the trigonometric identity \(\sin^2 \theta = 1 - \cos^2 \theta\):
\(3(1 - \cos^2 \theta) + 5 \cos \theta - 1 = 0\)
\(3 - 3 \cos^2 \theta + 5 \cos \theta - 1 = 0\)
\(3 \cos^2 \theta - 5 \cos \theta - 2 = 0\)

Factorising the quadratic in \(\cos \theta\):
\((3 \cos \theta + 1)(\cos \theta - 2) = 0\)

This gives two equations:
\(\cos \theta = -\frac{1}{3}\) or \(\cos \theta = 2\) (which has no real solutions as \(|\cos \theta| \le 1\)).

For \(\cos \theta = -\frac{1}{3}\):
\(\theta = \cos^{-1}\left(-\frac{1}{3}\right) \approx 1.91\) radians.
The other solution in the range \(0 \le \theta \le 2\pi\) is:
\(\theta = 2\pi - 1.9106 \approx 4.37\) radians.

- M1: Use identity \(\sin^2 \theta = 1 - \cos^2 \theta\) to form a quadratic equation in \(\cos \theta\)
- A1: Obtain the correct quadratic equation \(3 \cos^2 \theta - 5 \cos \theta - 2 = 0\)
- M1: Attempt to solve their quadratic equation in \(\cos \theta\)
- A1: Identify \(\cos \theta = -1/3\) (and reject/ignore \(\cos \theta = 2\))
- A1: Obtain first radian solution \(\theta = 1.91\) (allow \(1.9\) to \(1.92\))
- A1: Obtain second radian solution \(\theta = 4.37\) (allow \(4.36\) to \(4.38\))

(a) Let the line of symmetry from \(O\) through \(C\) intersect the arc at \(D\). The distance \(OD = r\). The radius of the inscribed circle is \(a\), so \(CD = a\) and the distance \(OC = r - a\). Let the circle be tangent to \(OA\) at a point \(T\). In the right-angled triangle \(OTC\), the angle \(TOC = \theta/2\). Using trigonometry: \(\sin(\theta/2) = \frac{CT}{OC} = \frac{a}{r-a}\). Rearranging this: \((r-a)\sin(\theta/2) = a \implies r\sin(\theta/2) - a\sin(\theta/2) = a \implies r\sin(\theta/2) = a(1 + \sin(\theta/2))\), which gives \(a = \frac{r\sin(\theta/2)}{1 + \sin(\theta/2)}\). (b) When \(\theta = \frac{2\pi}{3}\), the half-angle is \(\theta/2 = \frac{\pi}{3}\). Since \(\sin(\pi/3) = \frac{\sqrt{3}}{2}\), we have \(a = \frac{r(\sqrt{3}/2)}{1 + \sqrt{3}/2} = \frac{r\sqrt{3}}{2+\sqrt{3}}\). Rationalising the denominator: \(a = r\sqrt{3}(2-\sqrt{3}) = r(2\sqrt{3}-3)\). The area of the inscribed circle is \(A_c = \pi a^2 = \pi r^2(2\sqrt{3}-3)^2 = \pi r^2(12 - 12\sqrt{3} + 9) = \pi r^2(21 - 12\sqrt{3})\). The area of the sector is \(A_s = \frac{1}{2} r^2 \theta = \frac{1}{2} r^2 \left(\frac{2\pi}{3}\right) = \frac{\pi r^2}{3}\). The ratio of the areas is \(\frac{A_c}{A_s} = \frac{\pi r^2(21 - 12\sqrt{3})}{\frac{1}{3}\pi r^2} = 3(21 - 12\sqrt{3}) = 63 - 36\sqrt{3}\). Thus, \(p=63\) and \(q=36\). (c) The perimeter of the inscribed circle is \(P_c = 2\pi a = 2\pi r(2\sqrt{3}-3)\). The perimeter of the sector is \(P_s = 2r + r\theta = r\left(2 + \frac{2\pi}{3}\right) = \frac{2r(3+\pi)}{3}\). The ratio of the perimeters is \(\frac{P_c}{P_s} = \frac{2\pi r(2\sqrt{3}-3)}{\frac{2r(3+\pi)}{3}} = \frac{3\pi(2\sqrt{3}-3)}{3+\pi}\).

(a) M1: For identifying \(OC = r - a\) and establishing a right-angled triangle. M1: For stating \(\sin(\theta/2) = \frac{a}{r-a}\). A1: For convincing algebraic steps leading to the given expression. (b) M1: For substituting \(\theta = 2\pi/3\) and evaluating \(\sin(\pi/3) = \sqrt{3}/2\). M1: For rationalising the denominator to get \(a = r(2\sqrt{3}-3)\). M1: For setting up the ratio of area of circle to area of sector. A1: For obtaining \(63 - 36\sqrt{3}\). (c) M1: For obtaining correct expressions for both perimeters. A1: For simplifying the ratio to \(\frac{3\pi(2\sqrt{3}-3)}{3+\pi}\) or equivalent single fraction.

(a) The terms of the arithmetic progression are given by \(u_n = a + (n-1)d\). So \(u_1 = a\), \(u_3 = a + 2d\), and \(u_{11} = a + 10d\). The terms of the geometric progression are given by \(v_n = a r^{n-1}\). So \(v_1 = a\), \(v_2 = ar\), and \(v_3 = ar^2\). Since \(u_1 = v_1 = a\), we equate the other terms: \(a + 2d = ar\) and \(a + 10d = ar^2\). From the first equation, we get \(2d = ar - a = a(r-1) \implies d = \frac{a(r-1)}{2}\). Substituting this expression for \(d\) into the second equation: \(a + 10\left(\frac{a(r-1)}{2}\right) = ar^2 \implies a + 5a(r-1) = ar^2\). Since \(d \neq 0\), we must have \(a \neq 0\). Dividing both sides by \(a\): \(1 + 5(r-1) = r^2 \implies 1 + 5r - 5 = r^2 \implies r^2 - 5r + 4 = 0\). Factoring the quadratic equation: \((r-1)(r-4) = 0\). This gives \(r = 1\) or \(r = 4\). If \(r = 1\), then \(d = 0\), which contradicts the given condition that \(d \neq 0\). Therefore, \(r = 4\). (b) Since \(a = 6\) and \(r = 4\), we substitute these into \(d = \frac{a(r-1)}{2}\) to get \(d = \frac{6(4-1)}{2} = 9\). The sum of the first 20 terms of the arithmetic progression is given by \(S_{20} = \frac{20}{2}[2a + 19d] = 10[2(6) + 19(9)] = 10[12 + 171] = 10[183] = 1830\). (c) The sum of the first 6 terms of the geometric progression is given by \(S_6 = \frac{a(r^6 - 1)}{r-1} = \frac{6(4^6 - 1)}{4-1} = \frac{6(4096 - 1)}{3} = 2(4095) = 8190\).

(a) M1: For writing \(a + 2d = ar\) and \(a + 10d = ar^2\). M1: For eliminating \(d\) to obtain an equation in \(a\) and \(r\). M1: For dividing by \(a\) (acknowledging \(a \neq 0\)) to obtain \(r^2 - 5r + 4 = 0\). A1: For solving to find \(r = 4\) and explicitly rejecting \(r = 1\) with reference to \(d \neq 0\). (b) B1: For finding \(d = 9\). M1: For using the correct AP sum formula \(S_n = \frac{n}{2}[2a + (n-1)d]\) with \(n=20\). A1: For obtaining \(1830\). (c) M1: For using the correct GP sum formula with \(n=6\), \(a=6\), and \(r=4\). A1: For obtaining \(8190\).

(a) We write the curve as \(y = 12(x^2 + 3)^{-1}\). Applying the chain rule: \(\frac{\mathrm{d}y}{\mathrm{d}x} = -12(x^2 + 3)^{-2} \cdot (2x) = -\frac{24x}{(x^2 + 3)^2}\). (b) The area \(A\) of the rectangle is given by the product of its width \(x\) and height \(y\). Thus, \(A = x y = x \left(\frac{12}{x^2 + 3}\right) = \frac{12x}{x^2 + 3}\). To find \(\frac{\mathrm{d}A}{\mathrm{d}x}\), we use the quotient rule with \(u = 12x\) and \(v = x^2 + 3\): \(u' = 12\) and \(v' = 2x\). \(\frac{\mathrm{d}A}{\mathrm{d}x} = \frac{u'v - uv'}{v^2} = \frac{12(x^2 + 3) - 12x(2x)}{(x^2 + 3)^2} = \frac{12x^2 + 36 - 24x^2}{(x^2 + 3)^2} = \frac{36 - 12x^2}{(x^2 + 3)^2}\). (c) For stationary values of \(A\), we set \(\frac{\mathrm{d}A}{\mathrm{d}x} = 0\): \(\frac{36 - 12x^2}{(x^2 + 3)^2} = 0 \implies 36 - 12x^2 = 0 \implies x^2 = 3\). Since \(x > 0\) on the positive \(x\)-axis, we have \(x = \sqrt{3}\). The maximum area is \(A = \frac{12\sqrt{3}}{(\sqrt{3})^2 + 3} = \frac{12\sqrt{3}}{6} = 2\sqrt{3}\). To justify that it is a maximum, we can examine the sign of \(\frac{\mathrm{d}A}{\mathrm{d}x}\) around \(x = \sqrt{3} \approx 1.732\): For \(x = 1\), \(\frac{\mathrm{d}A}{\mathrm{d}x} = \frac{36 - 12(1)^2}{(1+3)^2} = \frac{24}{16} > 0\). For \(x = 2\), \(\frac{\mathrm{d}A}{\mathrm{d}x} = \frac{36 - 12(2)^2}{(4+3)^2} = \frac{-12}{49} < 0\). Since the gradient changes from positive to negative as \(x\) increases through \(x = \sqrt{3}\), the area \(A\) is indeed a maximum.

(a) M1: For attempting chain rule or quotient rule on \(y\). A1: For correct derivative \(-\frac{24x}{(x^2 + 3)^2}\). (b) B1: For stating \(A = \frac{12x}{x^2 + 3}\). M1: For attempting the quotient rule (or product rule) to find \(\frac{\mathrm{d}A}{\mathrm{d}x}\). A1: For obtaining \(\frac{36 - 12x^2}{(x^2 + 3)^2}\). (c) M1: For setting \(\frac{\mathrm{d}A}{\mathrm{d}x} = 0\) and solving to find \(x = \sqrt{3}\). A1: For finding the exact area \(2\sqrt{3}\). M1: For a valid method of justification (first or second derivative test). A1: For correct justification showing it is a maximum.

The given line equation is \(3x - 4y = 12\), which can be rewritten as \(4y = 3x - 12\), or \(y = \frac{3}{4}x - 3\). This line has a gradient of \(m_1 = \frac{3}{4}\). Since the line \(L\) is perpendicular to this line, its gradient \(m_2\) satisfies \(m_1 m_2 = -1\), giving \(m_2 = -\frac{4}{3}\). The equation of \(L\) passing through \((2, -5)\) is: \(y - (-5) = -\frac{4}{3}(x - 2)\). Multiplying both sides by 3: \(3(y + 5) = -4(x - 2)\), which simplifies to \(3y + 15 = -4x + 8\). Rearranging into the required form gives \(4x + 3y + 7 = 0\).

M1: For finding the perpendicular gradient, \(m = -\frac{4}{3}\), from the gradient of the given line. A1: For the correct equation of the line in the form \(ax + by + c = 0\) with integer coefficients, e.g. \(4x + 3y + 7 = 0\) (or any integer multiple thereof).