2025 Cambridge IAL Mathematics (9709) Practice Paper with Answers

To find where the line and curve do not intersect, we set their equations equal to each other:
\[x^2 - 2x + 5 = kx - 4\]
Rearranging into standard quadratic form \(ax^2 + bx + c = 0\):
\[x^2 - (2+k)x + 9 = 0\]
For no points of intersection, the discriminant of this quadratic equation must be strictly negative (\(\Delta < 0\)):
\[\Delta = b^2 - 4ac < 0\]
\[(-(2+k))^2 - 4(1)(9) < 0\]
\[(k+2)^2 - 36 < 0\]
\[(k+2)^2 < 36\]
Taking the square root on both sides:
\[-6 < k + 2 < 6\]
Subtracting 2 from all parts:
\[-8 < k < 4\]

M1: Equating the line and curve and rearranging into a quadratic equation in \(x\).
A1: Correct quadratic equation \(x^2 - (2+k)x + 9 = 0\) (or equivalent).
M1: Using \(b^2 - 4ac < 0\) with their coefficients.
A1: Correct quadratic inequality in \(k\), e.g., \(k^2 + 4k - 32 < 0\) or \((k+2)^2 < 36\).
M1: Finding the critical values \(k = -8\) and \(k = 4\).
A1: Correct final inequality \(-8 < k < 4\).

(i) Completing the square:
\[\text{f}(x) = 2(x^2 - 6x) + 11 = 2\left[(x-3)^2 - 9\right] + 11 = 2(x-3)^2 - 18 + 11 = 2(x-3)^2 - 7\]
Thus, the vertex of the parabola is at \((3, -7)\).
For the function \(\text{f}\) to have an inverse, it must be a one-to-one function. Since the domain is defined as \(x \le a\), the largest possible value of \(a\) is the \(x\)-coordinate of the vertex.
Therefore, the largest value of \(a\) is \(3\).

(ii) To find the inverse, let \(y = 2(x-3)^2 - 7\) with domain \(x \le 3\).
Rearranging to make \(x\) the subject:
\[y + 7 = 2(x-3)^2\]
\[\frac{y+7}{2} = (x-3)^2\]
Since \(x \le 3\), \(x - 3 \le 0\), so we choose the negative square root:
\[x - 3 = -\sqrt{\frac{y+7}{2}}\]
\[x = 3 - \sqrt{\frac{y+7}{2}}\]
Replacing \(x\) with \(\text{f}^{-1}(x)\) and \(y\) with \(x\):
\[\text{f}^{-1}(x) = 3 - \sqrt{\frac{x+7}{2}}\]
The domain of \(\text{f}^{-1}\) is the range of \(\text{f}\). Since the minimum value of \(\text{f}(x)\) is \(-7\) and \(x \le 3\), the range of \(\text{f}\) is \(y \ge -7\).
Therefore, the domain of \(\text{f}^{-1}\) is \(x \ge -7\).

(i)
M1: Completing the square to obtain \(2(x-3)^2 + r\).
A1: Correct completed square form: \(2(x-3)^2 - 7\).
B1: Correctly stating that the largest value of \(a\) is \(3\).

(ii)
M1: Setting \(y = 2(x-3)^2 - 7\) (or using their expression) and attempting to make \(x\) the subject.
M1: Correctly choosing the negative square root due to the domain restriction \(x \le 3\).
A1: Correct expression \(\text{f}^{-1}(x) = 3 - \sqrt{\frac{x+7}{2}}\).
B1: Correctly stating the domain as \(x \ge -7\) (accept \([ -7, \infty )\)).

(i) First, find the midpoint \(M\) of \(AB\):
\[M = \left(\frac{-1 + 3}{2}, \frac{5 + 11}{2}\right) = (1, 8)\]
Next, find the gradient of \(AB\):
\[m_{AB} = \frac{11 - 5}{3 - (-1)} = \frac{6}{4} = \frac{3}{2}\]
The gradient of the perpendicular bisector is the negative reciprocal:
\[m_{\perp} = -\frac{2}{3}\]
The equation of the perpendicular bisector is:
\[y - 8 = -\frac{2}{3}(x - 1) \implies 3(y - 8) = -2(x - 1) \implies 2x + 3y = 26\]
To find the coordinates of \(C\), we find the intersection of the perpendicular bisector with the line \(y = x + 2\). Substitute \(y = x + 2\) into \(2x + 3y = 26\):
\[2x + 3(x + 2) = 26 \implies 5x + 6 = 26 \implies 5x = 20 \implies x = 4\]
Substitute \(x = 4\) back into \(y = x + 2\):
\[y = 4 + 2 = 6\]
So the coordinates of \(C\) are \((4, 6)\).

(ii) The radius \(r\) of the circle is the distance from the center \(C(4, 6)\) to any of the points, say \(A(-1, 5)\):
\[r^2 = (4 - (-1))^2 + (6 - 5)^2 = 5^2 + 1^2 = 25 + 1 = 26\]
The equation of the circle is:
\[(x - 4)^2 + (y - 6)^2 = 26\]

(i)
M1: Finding the midpoint and gradient of \(AB\).
M1: Finding the equation of the perpendicular bisector of \(AB\).
A1: Correct perpendicular bisector equation: \(2x + 3y = 26\) (or equivalent).
M1: Solving simultaneously with \(y = x + 2\) to find the coordinates of \(C\).
A1: Correct coordinates of \(C(4, 6)\).

(ii)
M1: Calculating the radius squared, \(r^2\), using their \(C\) and either \(A\) or \(B\).
A1: Correct circle equation: \((x-4)^2 + (y-6)^2 = 26\).

(i) The perimeter \(P\) of a sector is given by:
\[P = 2r + r\theta\]
We are given that \(P = 16\) cm, so:
\[2r + r\theta = 16 \implies r\theta = 16 - 2r \implies \theta = \frac{16 - 2r}{r}\]
The area \(A\) of a sector is given by:
\[A = \frac{1}{2}r^2\theta\]
Substitute \(\theta = \frac{16 - 2r}{r}\) into the area formula:
\[A = \frac{1}{2}r^2\left(\frac{16 - 2r}{r}\right) = \frac{1}{2}r(16 - 2r) = 8r - r^2\]
(This completes the proof.)

(ii) Given that \(A = 12\):
\[8r - r^2 = 12 \implies r^2 - 8r + 12 = 0\]
Factorising the quadratic equation:
\[(r - 2)(r - 6) = 0\]
So the two possible values of \(r\) are \(r = 2\) and \(r = 6\).

Using \(\theta = \frac{16 - 2r}{r}\):
- If \(r = 2\):
\[\theta = \frac{16 - 2(2)}{2} = \frac{12}{2} = 6\text{ radians}\]

- If \(r = 6\):
\[\theta = \frac{16 - 2(6)}{6} = \frac{4}{6} = \frac{2}{3}\text{ radians}\]

(i)
M1: Expressing the perimeter of the sector as \(2r + r\theta = 16\) and making \(\theta\) (or \(r\theta\)) the subject.
M1: Substituting this expression into the area formula \(A = \frac{1}{2}r^2\theta\).
A1: Successfully obtaining the given expression \(A = 8r - r^2\) with clear steps shown.

(ii)
M1: Equating \(8r - r^2 = 12\) and forming a three-term quadratic equation.
A1: Correct values of \(r = 2\) and \(r = 6\).
M1: Attempting to find the corresponding values of \(\theta\) using a valid formula.
A1: Correct corresponding values of \(\theta = 6\) and \(\theta = \frac{2}{3}\) (or approximately \(0.667\)).

(i) Substitute \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) into the equation:
\[3\left(\frac{\sin\theta}{\cos\theta}\right) + 2\cos\theta = 0\]
Multiply the entire equation by \(\cos\theta\) (noting \(\cos\theta \ne 0\)):
\[3\sin\theta + 2\cos^2\theta = 0\]
Use the Pythagorean identity \[\cos^2\theta = 1 - \sin^2\theta\]:
\[3\sin\theta + 2(1 - \sin^2\theta) = 0\]
\[3\sin\theta + 2 - 2\sin^2\theta = 0\]
Multiply by \(-1\) to rearrange into the required form:
\[2\sin^2\theta - 3\sin\theta - 2 = 0\]
(This completes the proof.)

(ii) Solve the quadratic in \(\sin\theta\):
Let \(u = \sin\theta\):
\[2u^2 - 3u - 2 = 0 \implies (2u + 1)(u - 2) = 0\]
This gives \(\sin\theta = -\frac{1}{2}\) or \(\sin\theta = 2\).

Since \(\sin\theta\) must lie in the range \([-1, 1]\), the equation \(\sin\theta = 2\) has no solutions.

For \(\sin\theta = -\frac{1}{2}\) in the interval \(0 \le \theta \le 2\pi\):
The basic angle is \(\alpha = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}\).
Since sine is negative, \(\theta\) lies in the third and fourth quadrants:
\[\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}\]
\[\theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}\]

(i)
M1: Substituting \(\tan\theta = \frac{\sin\theta}{\cos\theta}\).
M1: Using \(\cos^2\theta = 1 - \sin^2\theta\) to form a quadratic in \(\sin\theta\).
A1: Showing all steps clearly to arrive at \(2\sin^2\theta - 3\sin\theta - 2 = 0\).

(ii)
M1: Factorising or using the quadratic formula to solve for \(\sin\theta\).
A1: Identifying that \(\sin\theta = -\frac{1}{2}\) is the only valid option (rejecting \(\sin\theta = 2\)).
M1: Finding at least one correct value of \(\theta\) in radians.
A1: Both correct values \(\theta = \frac{7\pi}{6}\) and \(\theta = \frac{11\pi}{6}\) (and no other values in the interval).

(i) The 1st, 3rd, and 11th terms of the arithmetic progression are:
\[T_1 = a\]
\[T_3 = a + 2d\]
\[T_{11} = a + 10d\]
Since these are consecutive terms of a geometric progression, the common ratio \(r\) is constant:
\[r = \frac{a+2d}{a} = \frac{a+10d}{a+2d}\]
Cross-multiplying gives:
\[(a+2d)^2 = a(a+10d)\]
\[a^2 + 4ad + 4d^2 = a^2 + 10ad\]
\[4d^2 = 6ad\]
Since \(d \ne 0\), we can divide both sides by \(2d\):
\[2d = 3a \implies d = 1.5a\]
Substitute \(d = 1.5a\) back into the formula for the common ratio \(r\):
\[r = \frac{a + 2(1.5a)}{a} = \frac{a + 3a}{a} = \frac{4a}{a} = 4\]
(This completes the proof.)

(ii) We are given that the 5th term of the arithmetic progression is \(21\):
\[T_5 = a + 4d = 21\]
Substitute \(d = 1.5a\) into this equation:
\[a + 4(1.5a) = 21 \implies a + 6a = 21 \implies 7a = 21 \implies a = 3\]
Using \(a = 3\), we find \(d\):
\[d = 1.5(3) = 4.5\]
The sum of the first \(n\) terms of an arithmetic progression is given by:
\[S_n = \frac{n}{2}\left[2a + (n-1)d\right]\]
For \(n = 20\):
\[S_{20} = \frac{20}{2}\left[2(3) + 19(4.5)\right] = 10\left[6 + 85.5\right] = 10(91.5) = 915\]

(i)
M1: Setting up the geometric progression ratio relation, e.g., \((a+2d)^2 = a(a+10d)\).
M1: Simplifying to find a relation between \(a\) and \(d\).
A1: Correct relation, e.g., \(2d = 3a\) or equivalent.
A1: Showing that \(r = 4\) clearly using the relation.

(ii)
M1: Setting up \(a + 4d = 21\) and substituting their relation to find \(a\) and \(d\).
A1: Correct values of \(a = 3\) and \(d = 4.5\).
M1: Using the AP sum formula for \(n = 20\).
A1: Correct final sum of \(915\).

(i) Write \(y\) in index form:
\[y = 8(x+2)^{-1} + 2x\]
Differentiate with respect to \(x\):
\[\frac{\text{d}y}{\text{d}x} = -8(x+2)^{-2} + 2 = -\frac{8}{(x+2)^2} + 2\]
Differentiate a second time:
\[\frac{\text{d}^2y}{\text{d}x^2} = 16(x+2)^{-3} = \frac{16}{(x+2)^3}\]

(ii) To find the stationary points, set \(\frac{\text{d}y}{\text{d}x} = 0\):
\[-\frac{8}{(x+2)^2} + 2 = 0 \implies \frac{8}{(x+2)^2} = 2 \implies (x+2)^2 = 4\]
Taking square roots:
\[x+2 = 2 \implies x = 0\]
\[x+2 = -2 \implies x = -4\]

Find the corresponding \(y\)-coordinates:
- For \(x = 0\):
\[y = \frac{8}{0+2} + 2(0) = 4 \implies (0, 4)\]
- For \(x = -4\):
\[y = \frac{8}{-4+2} + 2(-4) = -4 - 8 = -12 \implies (-4, -12)\]

Determine the nature using the second derivative:
- At \(x = 0\):
\[\frac{\text{d}^2y}{\text{d}x^2} = \frac{16}{(0+2)^3} = \frac{16}{8} = 2\]
Since \(\frac{\text{d}^2y}{\text{d}x^2} > 0\), the point \((0, 4)\) is a minimum point.

- At \(x = -4\):
\[\frac{\text{d}^2y}{\text{d}x^2} = \frac{16}{(-4+2)^3} = \frac{16}{-8} = -2\]
Since \(\frac{\text{d}^2y}{\text{d}x^2} < 0\), the point \((-4, -12)\) is a maximum point.

(i)
M1: Attempting to differentiate, with at least one term correct.
A1: Correct first derivative: \(-\frac{8}{(x+2)^2} + 2\).
A1: Correct second derivative: \(\frac{16}{(x+2)^3}\).

(ii)
M1: Setting their \(\frac{\text{d}y}{\text{d}x} = 0\) and attempting to solve for \(x\).
A1: Correct \(x\)-values: \(0\) and \(-4\).
M1: Finding both corresponding \(y\)-coordinates.
M1: Substituting their \(x\)-values into their second derivative to determine nature.
A1: Correctly concluding \((0, 4)\) is a minimum and \((-4, -12)\) is a maximum.

(i) Write \(y\) as \(y = (3x+1)^{1/2}\).
Differentiate using the chain rule:
\[\frac{\text{d}y}{\text{d}x} = \frac{1}{2}(3x+1)^{-1/2} \cdot 3 = \frac{3}{2\sqrt{3x+1}}\]
At the point \(P(5, 4)\), the gradient of the tangent is:
\[m_T = \frac{3}{2\sqrt{3(5)+1}} = \frac{3}{2\sqrt{16}} = \frac{3}{8}\]
The gradient of the normal, \(m_N\), is the negative reciprocal:
\[m_N = -\frac{8}{3}\]
The equation of the normal at \(P(5, 4)\) is:
\[y - 4 = -\frac{8}{3}(x - 5)\]
Multiply by 3:
\[3y - 12 = -8x + 40 \implies 8x + 3y = 52\]

(ii) The volume \(V\) of the solid of revolution is given by:
\[V = \pi \int_{1}^{5} y^2 \text{ d}x\]
Since \(y = \sqrt{3x+1}\), we have \(y^2 = 3x+1\). Thus:
\[V = \pi \int_{1}^{5} (3x + 1) \text{ d}x\]
Integrate with respect to \(x\):
\[V = \pi \left[ \frac{3x^2}{2} + x \right]_{1}^{5}\]
Evaluate the limit at \(x = 5\):
\[\frac{3(5)^2}{2} + 5 = 37.5 + 5 = 42.5\]
Evaluate the limit at \(x = 1\):
\[\frac{3(1)^2}{2} + 1 = 1.5 + 1 = 2.5\]
Subtract the limits:
\[V = \pi (42.5 - 2.5) = 40\pi\]

(i)
M1: Attempting to differentiate \((3x+1)^{1/2}\) using chain rule.
A1: Correct derivative \(\frac{3}{2\sqrt{3x+1}}\).
M1: Finding the perpendicular gradient at \(x = 5\).
A1: Correct equation of the normal in the form \(8x + 3y = 52\) (or any integer multiple thereof).

(ii)
M1: Using the volume formula \(\pi \int y^2 \text{ d}x\) with limits \(1\) and \(5\).
A1: Correct integration to obtain \(\frac{3x^2}{2} + x\).
M1: Substituting limits correctly into their integrated expression.
A1: Correct exact volume of \(40\pi\).

Let the first term of the arithmetic progression (AP) be \(a\) and the common difference be \(d\). The 11th, 3rd, and 1st terms of the AP are: \(T_{11} = a + 10d\), \(T_3 = a + 2d\), and \(T_1 = a\). Since these are the 1st, 2nd, and 3rd terms of a geometric progression (GP), we have: \((a+2d)/(a+10d) = a/(a+2d)\). Cross-multiplying gives: \((a+2d)^2 = a(a+10d)\) which expands to \(a^2 + 4ad + 4d^2 = a^2 + 10ad\). Simplifying, we get \(4d^2 - 6ad = 0\), which factors to \(2d(2d - 3a) = 0\). Since the common difference \(d\) is non-zero, we must have \(2d = 3a\), which implies \(d = 1.5a\). The common ratio, \(r\), is: \(r = a/(a+2d) = a/(a + 2(1.5a)) = a/(4a) = 1/4\). For part (ii), the sum of the first 15 terms of the AP is given by: \(S_{15} = (15/2)(2a + 14d) = 15(a + 7d) = 345\), which simplifies to \(a + 7d = 23\). Substituting \(d = 1.5a\) into this equation yields: \(a + 7(1.5a) = 23\), so \(11.5a = 23\), which gives \(a = 2\). Thus, \(d = 1.5(2) = 3\). The first term of the GP, \(G_1\), is \(G_1 = a + 10d = 2 + 10(3) = 32\). The sum to infinity, \(S_{\infty}\), of the GP is: \(S_{\infty} = G_1/(1-r) = 32/(1 - 1/4) = 32/(3/4) = 128/3\).

Part (i): M1 for setting up the GP ratio equation. A1 for simplifying to find \(2d = 3a\). A1 for completing the proof to show \(r = 1/4\). Part (ii): M1 for using the sum of AP formula with \(d = 1.5a\). A1 for obtaining \(a = 2\) and \(G_1 = 32\). M1 for applying the sum to infinity formula. A1 for obtaining 128/3.

(i) First, write the equation of the curve as: \(y = 3x + 12x^{-2}\). Differentiating with respect to \(x\) gives: \(dy/dx = 3 - 24x^{-3} = 3 - 24/x^3\). At a stationary point, \(dy/dx = 0\), which means \(3 - 24/x^3 = 0\), so \(3 = 24/x^3\), yielding \(x^3 = 8\) and thus \(x = 2\). Find the \(y\)-coordinate: \(y = 3(2) + 12/(2^2) = 6 + 3 = 9\). So the stationary point is \((2, 9)\). To determine the nature of this stationary point, find the second derivative: \(d^2y/dx^2 = 72x^{-4} = 72/x^4\). At \(x = 2\), \(d^2y/dx^2 = 72/(2^4) = 72/16 = 4.5\). Since \(d^2y/dx^2 > 0\), the stationary point \((2, 9)\) is a minimum. (ii) We are given \(dx/dt = 0.4\). We need to find \(dy/dt\) when \(x = 4\). By the chain rule: \(dy/dt = (dy/dx) * (dx/dt)\). At \(x = 4\), \(dy/dx = 3 - 24/(4^3) = 3 - 24/64 = 3 - 0.375 = 2.625\). Thus, \(dy/dt = 2.625 * 0.4 = 1.05\) units per second.

Part (i): M1 for differentiating to find \(dy/dx\). A1 for finding \(x = 2\). A1 for obtaining \(y = 9\). M1 for finding the second derivative and evaluating at \(x = 2\). A1 for concluding it is a minimum. Part (ii): M1 for evaluating \(dy/dx\) at \(x = 4\). M1 for using the chain rule. A1 for obtaining 1.05.

(i) To find the points of intersection, set the equations equal to each other: \(\sqrt{2x + 1} = (1/2)x + 1\). Squaring both sides gives: \(2x + 1 = ((1/2)x + 1)^2\), which expands to \(2x + 1 = (1/4)x^2 + x + 1\). Rearranging the terms yields: \((1/4)x^2 - x = 0\), which factors to \(x((1/4)x - 1) = 0\). So \(x = 0\) or \(x = 4\). Corresponding \(y\)-coordinates are: when \(x = 0\), \(y = 1\), and when \(x = 4\), \(y = 3\). The intersection points are \((0, 1)\) and \((4, 3)\). (ii) The area of the region enclosed is given by: \(Area = \int_{0}^{4} ( \sqrt{2x + 1} - ((1/2)x + 1) ) dx\). Integrating the curve term: \(\int_{0}^{4} (2x+1)^{1/2} dx = [ (2x+1)^{3/2} / (3/2 * 2) ]_{0}^{4} = [ (1/3)(2x+1)^{3/2} ]_{0}^{4}\). At \(x = 4\), this is \((1/3)(9)^{3/2} = 9\). At \(x = 0\), this is \((1/3)(1)^{3/2} = 1/3\). So, the curve integral is \(9 - 1/3 = 26/3\). Integrating the line term: \(\int_{0}^{4} ((1/2)x + 1) dx = [ (1/4)x^2 + x ]_{0}^{4} = (4 + 4) - 0 = 8\). The enclosed area is \(26/3 - 8 = 2/3\).

Part (i): M1 for equating the expressions and squaring. A1 for solving to obtain \(x = 0\) and \(x = 4\). A1 for giving both coordinates. Part (ii): M1 for integrating \((2x+1)^{1/2}\). A1 for obtaining \((1/3)(2x+1)^{3/2}\). M1 for applying the limits 0 and 4. M1 for integrating the line. A1 for obtaining the correct exact area of 2/3.