2025 Cambridge IAL Mathematics (9709) Practice Paper with Answers

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).