2026 Cambridge IAL Mathematics (9709) Practice Paper with Answers

Let \(u = x^{\frac{1}{3}}\). The equation becomes a quadratic in \(u\):
\[2u^2 - 5u - 12 = 0\]
Factorising the quadratic:
\[(2u + 3)(u - 4) = 0\]
This gives:
\[u = 4 \quad \text{or} \quad u = -\frac{3}{2}\]
Since \(u = x^{\frac{1}{3}}\), we cube both sides to find \(x\):
For \(u = 4\):
\[x = 4^3 = 64\]
For \(u = -\frac{3}{2}\):
\[x = \left(-\frac{3}{2}\right)^3 = -\frac{27}{8} = -3.375\]
Thus, the solutions are \(x = 64\) and \(x = -\frac{27}{8}\).

M1: Introduce a substitution such as \(u = x^{\frac{1}{3}}\) (or factorise directly) to form a 3-term quadratic equation in \(u\).

A1: Obtain correct roots \(u = 4\) and \(u = -\frac{3}{2}\) (or factors \((2u+3)(u-4)\)).

M1: Correct method of cubing at least one value of \(u\) to find \(x\).

A1: Both values correct: \(x = 64\) and \(x = -\frac{27}{8}\) (or \(-3.375\)).

Step 1: Apply the horizontal stretch with scale factor \(\frac{1}{2}\).
This replaces \(x\) with \(2x\):
\[y_1 = \text{f}(2x) = 2(2x)^2 - 4(2x) + 1 = 8x^2 - 8x + 1\]

Step 2: Apply the translation by vector \(\begin{pmatrix} -3 \\ 5 \end{pmatrix}\).
This replaces \(x\) with \(x - (-3) = x + 3\) and adds \(5\) to the function:
\[\text{g}(x) = 8(x + 3)^2 - 8(x + 3) + 1 + 5\]

Expand the expression:
\[8(x^2 + 6x + 9) - 8(x + 3) + 6 = 8x^2 + 48x + 72 - 8x - 24 + 6\]
\[\text{g}(x) = 8x^2 + 40x + 54\]

M1: Replace \(x\) by \(2x\) in \(\text{f}(x)\) to obtain \(2(2x)^2 - 4(2x) + 1\) (or \(8x^2 - 8x + 1\)).

M1: Apply the translation by substituting \(x + 3\) for \(x\) and adding \(5\) to the resulting expression.

A1: Correct unsimplified or partially simplified algebraic form: \(8(x+3)^2 - 8(x+3) + 6\) (or equivalent).

A1: Fully correct simplified polynomial \(8x^2 + 40x + 54\).

(a) Midpoint of \(AB\):
\[ M = \left( \frac{1 + 9}{2}, \frac{-2 + 4}{2} \right) = (5, 1) \]

Gradient of \(AB\):
\[ m_{AB} = \frac{4 - (-2)}{9 - 1} = \frac{6}{8} = \frac{3}{4} \]

Gradient of the perpendicular bisector:
\[ m_{\perp} = -\frac{1}{\frac{3}{4}} = -\frac{4}{3} \]

Equation of the perpendicular bisector:
\[ y - 1 = -\frac{4}{3}(x - 5) \]
\[ 3(y - 1) = -4(x - 5) \implies 3y - 3 = -4x + 20 \]
\[ 4x + 3y = 23 \]

(b) The centre of the circle lies on both the perpendicular bisector of \(AB\) and the line \(y = 2x + 1\).

Substitute \(y = 2x + 1\) into \(4x + 3y = 23\):
\[ 4x + 3(2x + 1) = 23 \]
\[ 4x + 6x + 3 = 23 \]
\[ 10x = 20 \implies x = 2 \]

Find the \(y\)-coordinate of the centre:
\[ y = 2(2) + 1 = 5 \]
So the centre is \(C(2, 5)\).

Find the radius \(r\) using the distance from \(C(2, 5)\) to \(A(1, -2)\):
\[ r^2 = (2 - 1)^2 + (5 - (-2))^2 = 1^2 + 7^2 = 1 + 49 = 50 \]

Therefore, the equation of the circle is:
\[ (x - 2)^2 + (y - 5)^2 = 50 \]

(a)
- M1: For calculating the midpoint \((5, 1)\) and gradient of \(AB\) (\(\frac{3}{4}\)).
- M1: For using \(m_{\perp} = -\frac{1}{m_{AB}}\) with their midpoint in a straight line equation formula.
- A1: For correct equation in the required form: \(4x + 3y = 23\) (or any integer multiple thereof).

(b)
- M1: For solving simultaneously the equation of the perpendicular bisector and \(y = 2x + 1\).
- A1: For obtaining correct centre \((2, 5)\).
- M1: For calculating \(r^2\) using their centre and point \(A\) or \(B\).
- A1: For the correct equation of the circle \((x - 2)^2 + (y - 5)^2 = 50\) (or \(x^2 + y^2 - 4x - 10y - 21 = 0\)).

(a) Complete the square for the equation of the circle:
\[ (x - 4)^2 - 16 + (y - 2)^2 - 4 + 11 = 0 \]
\[ (x - 4)^2 + (y - 2)^2 = 9 \]

Centre is \((4, 2)\) and radius is \(\sqrt{9} = 3\).

(b) Method 1 (Substitution and Discriminant):
Substitute \(y = mx + 5\) into the circle equation \((x - 4)^2 + (y - 2)^2 = 9\):
\[ (x - 4)^2 + (mx + 5 - 2)^2 = 9 \]
\[ (x - 4)^2 + (mx + 3)^2 = 9 \]
\[ x^2 - 8x + 16 + m^2x^2 + 6mx + 9 = 9 \]
\[ (1 + m^2)x^2 + (6m - 8)x + 16 = 0 \]

Since the line is a tangent to the circle, the quadratic equation must have exactly one real root, so the discriminant \(b^2 - 4ac = 0\):
\[ (6m - 8)^2 - 4(1 + m^2)(16) = 0 \]
\[ 36m^2 - 96m + 64 - 64(1 + m^2) = 0 \]
\[ 36m^2 - 96m + 64 - 64 - 64m^2 = 0 \]
\[ -28m^2 - 96m = 0 \]
\[ -4m(7m + 24) = 0 \]

\[ m = 0 \quad \text{or} \quad m = -\frac{24}{7} \]

Method 2 (Perpendicular Distance):
The perpendicular distance from the centre \((4, 2)\) to the line \(mx - y + 5 = 0\) equals the radius \(3\):
\[ \frac{|m(4) - 2 + 5|}{\sqrt{m^2 + (-1)^2}} = 3 \]
\[ \frac{|4m + 3|}{\sqrt{m^2 + 1}} = 3 \]
\[ |4m + 3| = 3\sqrt{m^2 + 1} \]

Squaring both sides:
\[ (4m + 3)^2 = 9(m^2 + 1) \]
\[ 16m^2 + 24m + 9 = 9m^2 + 9 \]
\[ 7m^2 + 24m = 0 \]
\[ m(7m + 24) = 0 \]
\[ m = 0 \quad \text{or} \quad m = -\frac{24}{7} \]

(a)
- B1: For centre \((4, 2)\).
- B1: For radius \(3\).

(b)
- M1: For substituting \(y = mx + 5\) into the circle equation to obtain a 3-term quadratic in \(x\) OR setting up the perpendicular distance formula equating to the radius.
- A1: For obtaining correct simplified quadratic equation \((1+m^2)x^2 + (6m-8)x + 16 = 0\) OR correct equation \((4m+3)^2 = 9(m^2+1)\).
- M1: For applying \(b^2 - 4ac = 0\) and attempting to solve the resulting quadratic equation in \(m\).
- A1: For both correct values: \(m = 0\) and \(m = -\frac{24}{7}\) (or \(-3.43\)).

(a) Using the binomial expansion formula:
\[ \left(2 - \frac{1}{2}x\right)^5 = 2^5 + \binom{5}{1}(2^4)\left(-\frac{1}{2}x\right) + \binom{5}{2}(2^3)\left(-\frac{1}{2}x\right)^2 + \dots \]
Evaluating each term:
\[ = 32 + 5(16)\left(-\frac{1}{2}x\right) + 10(8)\left(\frac{1}{4}x^2\right) + \dots \]
\[ = 32 - 40x + 20x^2 \]

(b) Consider the product:
\[ (1 + ax)\left(32 - 40x + 20x^2 + \dots\right) \]
The terms contributing to the coefficient of \(x^2\) are:
\[ 1 \times (20x^2) + (ax) \times (-40x) = (20 - 40a)x^2 \]
Setting the coefficient of \(x^2\) to 0:
\[ 20 - 40a = 0 \implies 40a = 20 \implies a = \frac{1}{2} \]

(a)
B1: For the first term \(32\).
M1: For a correct method to obtain either the \(x\) term or the \(x^2\) term (e.g. evaluating \(\binom{5}{1}2^4(-\frac{1}{2}x)\) or \(\binom{5}{2}2^3(-\frac{1}{2}x)^2\)).
A1: For fully correct simplified terms \(-40x + 20x^2\).

(b)
M1: For identifying and combining the two terms producing \(x^2\): \(1(20) + a(-40)\) (allow FT from part (a)).
M1: For equating their coefficient of \(x^2\) to 0 and solving for \(a\).
A1: For \(a = \frac{1}{2}\) (or \(0.5\)).

(a) The 1st, 5th, and 17th terms of the arithmetic progression are:
\[ T_1 = a, \quad T_5 = a + 4d, \quad T_{17} = a + 16d \]
Since these terms form a geometric progression, the common ratio satisfies:
\[ \frac{a + 4d}{a} = \frac{a + 16d}{a + 4d} \]
Cross-multiplying gives:
\[ (a + 4d)^2 = a(a + 16d) \]
\[ a^2 + 8ad + 16d^2 = a^2 + 16ad \]
\[ 16d^2 - 8ad = 0 \implies 8d(2d - a) = 0 \]
Since \(d \neq 0\), it follows that \(2d - a = 0 \implies a = 2d\).

The common ratio \(r\) is:
\[ r = \frac{a + 4d}{a} = \frac{2d + 4d}{2d} = \frac{6d}{2d} = 3 \]

(b) The formula for the sum of the first \(n\) terms of an arithmetic progression is:
\[ S_n = \frac{n}{2}\left[2a + (n - 1)d\right] \]
For \(n = 20\):
\[ S_{20} = \frac{20}{2}\left[2a + 19d\right] = 10(2a + 19d) \]
Substituting \(a = 2d\):
\[ S_{20} = 10\left(2(2d) + 19d\right) = 10(23d) = 230d \]
We are given \(S_{20} = 460\):
\[ 230d = 460 \implies d = 2 \]
Then:
\[ a = 2d = 2(2) = 4 \]

(a)
M1: For setting up a correct equation using the GP condition, e.g. \((a + 4d)^2 = a(a + 16d)\) or \(\frac{a + 4d}{a} = \frac{a + 16d}{a + 4d}\).
A1: For correct expansion and algebraic steps leading to \(a = 2d\) with no errors seen [AG].
B1: For obtaining \(r = 3\).

(b)
M1: For using the AP sum formula \(S_{20} = \frac{20}{2}[2a + 19d] = 460\) and substituting \(a = 2d\) (or \(d = \frac{1}{2}a\)) to form an equation in one variable.
A1: For \(d = 2\).
A1: For \(a = 4\).

(a)
Starting with the left-hand side:
\[ \text{LHS} = \frac{\sin\theta(1 + \cos\theta) - \sin\theta(1 - \cos\theta)}{(1 - \cos\theta)(1 + \cos\theta)} \]
Expand the numerator and simplify the denominator using \(1 - \cos^2\theta = \sin^2\theta\):
\[ \text{LHS} = \frac{\sin\theta + \sin\theta\cos\theta - \sin\theta + \sin\theta\cos\theta}{1 - \cos^2\theta} = \frac{2\sin\theta\cos\theta}{\sin^2\theta} \]
Cancelling \(\sin\theta\) and using \(\tan\theta = \frac{\sin\theta}{\cos\theta}\):
\[ \text{LHS} = \frac{2\cos\theta}{\sin\theta} = \frac{2}{\tan\theta} = \text{RHS} \]

(b)
Using the identity from part (a):
\[ \frac{2}{\tan\theta} = 3\tan\theta \]
\[ \tan^2\theta = \frac{2}{3} \]
\[ \tan\theta = \pm\sqrt{\frac{2}{3}} \]
Finding the basic angle:
\[ \alpha = \tan^{-1}\left(\sqrt{\frac{2}{3}}\right) \approx 39.232^\circ \]
For \(\tan\theta = \sqrt{\frac{2}{3}}\):
\[ \theta = 39.2^\circ, \quad \theta = 180^\circ + 39.232^\circ = 219.2^\circ \]
For \(\tan\theta = -\sqrt{\frac{2}{3}}\):
\[ \theta = 180^\circ - 39.232^\circ = 140.8^\circ, \quad \theta = 360^\circ - 39.232^\circ = 320.8^\circ \]
Hence, \(\theta = 39.2^\circ, 140.8^\circ, 219.2^\circ, 320.8^\circ\) (to 1 decimal place).

(a)
- M1: Combines terms over a common denominator \(1 - \cos^2\theta\) (or \((1-\cos\theta)(1+\cos\theta)\)) and applies \(1 - \cos^2\theta = \sin^2\theta\).
- A1: Fully correct proof with all intermediate algebraic steps shown leading to \(\frac{2}{\tan\theta}\).

(b)
- M1: Substitutes identity to form \(\tan^2\theta = \frac{2}{3}\) (or equivalent) and attempts to find square roots.
- A1: Obtains any two correct angles (e.g. \(39.2^\circ\) and \(140.8^\circ\) or \(219.2^\circ\) and \(320.8^\circ\)).
- A1: Obtains all four correct angles \(\theta = 39.2^\circ, 140.8^\circ, 219.2^\circ, 320.8^\circ\) (correct to 1 d.p.) and no extra values in the range.

(a)
Write the equation of the curve in index form:
\[ y = 16(2x + 1)^{-1} + 2x \]

Differentiate with respect to \( x \) using the chain rule:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = 16 \times (-1)(2x + 1)^{-2} \times 2 + 2 = -32(2x + 1)^{-2} + 2 = 2 - \frac{32}{(2x + 1)^2} \]

Differentiate again to find the second derivative:
\[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -32 \times (-2)(2x + 1)^{-3} \times 2 = 128(2x + 1)^{-3} = \frac{128}{(2x + 1)^3} \]

(b)
At a stationary point, \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \):
\[ 2 - \frac{32}{(2x + 1)^2} = 0 \]
\[ (2x + 1)^2 = \frac{32}{2} = 16 \]

Since \( x > -\frac{1}{2} \), \( 2x + 1 > 0 \):
\[ 2x + 1 = 4 \implies 2x = 3 \implies x = \frac{3}{2} \]

Substitute \( x = \frac{3}{2} \) into the original equation to find the \( y \)-coordinate:
\[ y = \frac{16}{2\left(\frac{3}{2}\right) + 1} + 2\left(\frac{3}{2}\right) = \frac{16}{4} + 3 = 4 + 3 = 7 \]

So the stationary point is \( \left(\frac{3}{2}, 7\right) \).

To determine its nature, evaluate the second derivative at \( x = \frac{3}{2} \):
\[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{128}{(2(1.5) + 1)^3} = \frac{128}{4^3} = \frac{128}{64} = 2 \]

Since \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0 \), the stationary point is a minimum.

(c)
The function is increasing when \( \frac{\mathrm{d}y}{\mathrm{d}x} > 0 \):
\[ 2 - \frac{32}{(2x + 1)^2} > 0 \implies (2x + 1)^2 > 16 \]
Given the domain \( x > -\frac{1}{2} \), this yields \( 2x + 1 > 4 \implies x > \frac{3}{2} \) (or \( x \geqslant \frac{3}{2} \)).

(a)
- B1: Obtain \( -32(2x + 1)^{-2} \) (or equivalent from correct application of the chain rule).
- B1: Obtain \( + 2 \) for differentiating \( 2x \).
- B1: Obtain \( 128(2x + 1)^{-3} \) (or equivalent) for the second derivative.

(b)
- M1: Set their \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \) and attempt to solve for \( x \).
- A1: Obtain coordinates \( \left(\frac{3}{2}, 7\right) \) or \( (1.5, 7) \).
- A1: Identify the nature as minimum with valid justification (e.g., \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2 > 0 \)).

(c)
- B1: State \( x > \frac{3}{2} \) or \( x \geqslant \frac{3}{2} \) (or equivalent inequality/interval notation).

(a)
Rewrite the equation of the curve as \(y = 12(x + 2)^{-2}\).
Differentiating with respect to \(x\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = -24(x + 2)^{-3} = -\frac{24}{(x + 2)^3}\]
At \(x = 0\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{24}{2^3} = -3\]
The equation of the tangent at \(P(0, 3)\) is:
\[y - 3 = -3(x - 0) \implies y = -3x + 3\]
To find \(Q\), set \(y = 0\):
\[0 = -3x + 3 \implies x = 1\]
So the coordinates of \(Q\) are \((1, 0)\).

(b)
The region is bounded above by the curve and below by the tangent from \(x = 0\) to \(x = 2\).
\[\text{Area} = \int_{0}^{2} \left( \frac{12}{(x + 2)^2} - (-3x + 3) \right) \mathrm{d}x\]
Integrating each term:
\[\int_{0}^{2} 12(x + 2)^{-2}\,\mathrm{d}x = \left[ -\frac{12}{x + 2} \right]_{0}^{2} = \left(-\frac{12}{4}\right) - \left(-\frac{12}{2}\right) = -3 - (-6) = 3\]
\[\int_{0}^{2} (-3x + 3)\,\mathrm{d}x = \left[ -\frac{3}{2}x^2 + 3x \right]_{0}^{2} = \left(-\frac{3}{2}(4) + 3(2)\right) - 0 = -6 + 6 = 0\]
Therefore,
\[\text{Area} = 3 - 0 = 3\]

(c)
The volume of revolution about the \(x\)-axis is given by:
\[V = \pi \int_{0}^{2} y^2 \,\mathrm{d}x = \pi \int_{0}^{2} \left(\frac{12}{(x + 2)^2}\right)^2 \mathrm{d}x = \pi \int_{0}^{2} 144(x + 2)^{-4}\,\mathrm{d}x\]
Integrating:
\[\int_{0}^{2} 144(x + 2)^{-4}\,\mathrm{d}x = \left[ 144 \cdot \frac{(x + 2)^{-3}}{-3} \right]_{0}^{2} = \left[ -\frac{48}{(x + 2)^3} \right]_{0}^{2}\]
Substituting limits:
\[\left( -\frac{48}{(4)^3} \right) - \left( -\frac{48}{(2)^3} \right) = -\frac{48}{64} - \left(-\frac{48}{8}\right) = -\frac{3}{4} + 6 = \frac{21}{4}\]
Thus, the exact volume is:
\[V = \frac{21}{4}\pi\]

(a)
- M1: Differentiates \(y = 12(x+2)^{-2}\) to obtain \(k(x+2)^{-3}\).
- A1: Finds gradient \(m = -3\) and states correct tangent equation \(y = -3x + 3\) (or equivalent).
- A1: Correct coordinates \(Q(1, 0)\) or \(x = 1, y = 0\).

(b)
- M1: Integrates curve \(12(x+2)^{-2}\) to obtain form \(c(x+2)^{-1}\).
- A1: Correctly evaluates curve integral: \(\left[-\frac{12}{x+2}\right]_0^2 = 3\).
- M1: Sets up subtraction of line from curve (or finds area under curve minus net area under line/triangles) with limits \(0\) to \(2\).
- A1: Obtains exact area \(3\).

(c)
- M1: Uses \(V = \pi \int y^2 \,\mathrm{d}x\) with \(y^2 = 144(x+2)^{-4}\).
- A1: Correct integral \(-48(x+2)^{-3}\) or \(-\frac{48}{(x+2)^3}\) (condone omission of \(\pi\) here).
- M1: Applies limits \(0\) and \(2\) correctly into integrated expression.
- A1: Obtains exact answer \(\frac{21}{4}\pi\) (or \(5.25\pi\)).

(a)

Expanding \((2x + p)^2 + q = 4x^2 + 4px + p^2 + q\).

Comparing coefficients with \(4x^2 - 12x + 13\):
\[ 4p = -12 \implies p = -3 \]
\[ p^2 + q = 13 \implies (-3)^2 + q = 13 \implies 9 + q = 13 \implies q = 4 \]

So, \(4x^2 - 12x + 13 = (2x - 3)^2 + 4\).

(b)

A function has an inverse if and only if it is one-one.
The vertex of the parabola \(y = (2x - 3)^2 + 4\) occurs at \(2x - 3 = 0 \implies x = \frac{3}{2}\).

For \(\text{g}\) to be one-one on \(x \leqslant k\), the greatest value of \(k\) is \(k = \frac{3}{2}\).

(c)

Let \(y = (2x - 3)^2 + 4\) for \(x \leqslant \frac{3}{2}\).

\[ (2x - 3)^2 = y - 4 \]

Since \(x \leqslant \frac{3}{2}\), we have \(2x - 3 \leqslant 0\), which requires the negative square root:
\[ 2x - 3 = -\sqrt{y - 4} \]
\[ 2x = 3 - \sqrt{y - 4} \]
\[ x = \frac{3 - \sqrt{y - 4}}{2} \]

Therefore, \(\text{g}^{-1}(x) = \frac{3 - \sqrt{x - 4}}{2}\).

The domain of \(\text{g}^{-1}\) is the range of \(\text{g}\).
Since \((2x - 3)^2 \geqslant 0\), the minimum value of \(\text{g}(x)\) is \(4\).
Hence, the domain of \(\text{g}^{-1}\) is \(x \geqslant 4\) (or \(x \in \mathbb{R}, x \geqslant 4\)).

(d)

Method 1:
Using \(\text{gh}(x) = 20 \implies \text{h}(x) = \text{g}^{-1}(20)\):
\[ \text{h}(x) = \frac{3 - \sqrt{20 - 4}}{2} = \frac{3 - 4}{2} = -\frac{1}{2} \]
\[ 2x + 1 = -\frac{1}{2} \implies 2x = -\frac{3}{2} \implies x = -\frac{3}{4} \]

Method 2:
Using \(\text{g}(\text{h}(x)) = (2(2x + 1) - 3)^2 + 4 = (4x - 1)^2 + 4 = 20\):
\[ (4x - 1)^2 = 16 \implies 4x - 1 = \pm 4 \]
Since the domain of \(\text{g}\) is \(u \leqslant \frac{3}{2}\), we require \(\text{h}(x) \leqslant \frac{3}{2} \implies 2x + 1 \leqslant \frac{3}{2} \implies 4x - 1 \leqslant 0\).
Thus, \(4x - 1 = -4 \implies 4x = -3 \implies x = -\frac{3}{4}\).
(The root \(4x - 1 = 4 \implies x = \frac{5}{4}\) gives \(\text{h}(\frac{5}{4}) = \frac{7}{2} > \frac{3}{2}\), so it is rejected).

(a)
- B1: For \(p = -3\)
- B1: For \(q = 4\)

(b)
- B1: For \(k = \frac{3}{2}\) (or \(1.5\))

(c)
- M1: For correct algebraic rearrangement to make \((2x - 3)^2\) the subject: \((2x - 3)^2 = y - 4\) or \((2y - 3)^2 = x - 4\)
- M1: For selecting the negative square root due to the domain condition \(x \leqslant \frac{3}{2}\)
- A1: For \(\text{g}^{-1}(x) = \frac{3 - \sqrt{x - 4}}{2}\) (must be an expression in \(x\))
- B1: For domain \(x \geqslant 4\) (or \([4, \infty)\))

(d)
- M1: For setting \(2x + 1 = \text{g}^{-1}(20)\) or solving \((4x - 1)^2 + 4 = 20\)
- A1: For \(x = -\frac{3}{4}\) (or \(-0.75\)) only (must discard \(x = \frac{5}{4}\) if found)