2026 Cambridge IAL Mathematics (9709) Practice Paper with Answers

Substitute \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) into the equation:

\(3 \left(\frac{\sin \theta}{\cos \theta}\right) \sin \theta + \cos \theta = 5\)

Multiply through by \(\cos \theta\) (where \(\cos \theta \neq 0\)):

\(3 \sin^2 \theta + \cos^2 \theta = 5 \cos \theta\)

Use the identity \(\sin^2 \theta = 1 - \cos^2 \theta\):

\(3(1 - \cos^2 \theta) + \cos^2 \theta = 5 \cos \theta\)

\(3 - 3 \cos^2 \theta + \cos^2 \theta = 5 \cos \theta\)

\(3 - 2 \cos^2 \theta = 5 \cos \theta\)

Rearrange into a standard 3-term quadratic equation in \(\cos \theta\):

\(2 \cos^2 \theta + 5 \cos \theta - 3 = 0\)

Factorise the quadratic:

\((2 \cos \theta - 1)(\cos \theta + 3) = 0\)

This gives:

\(\cos \theta = \frac{1}{2}\) or \(\cos \theta = -3\)

Since \(-1 \leqslant \cos \theta \leqslant 1\), \(\cos \theta = -3\) has no real solutions.

For \(\cos \theta = \frac{1}{2}\):

\(\theta = 60^\circ\)

\(\theta = 360^\circ - 60^\circ = 300^\circ\)

Thus, the solutions in the interval \(0^\circ \leqslant \theta \leqslant 360^\circ\) are \(\theta = 60^\circ, 300^\circ\).

M1: Use \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) and multiply through by \(\cos \theta\).
M1: Substitute \(\sin^2 \theta = 1 - \cos^2 \theta\) to obtain a 3-term quadratic in \(\cos \theta\).
A1: Obtain \(2 \cos^2 \theta + 5 \cos \theta - 3 = 0\) and solve to get \(\cos \theta = \frac{1}{2}\) (rejecting \(\cos \theta = -3\)).
A1: Both correct angles \(\theta = 60^\circ\) and \(\theta = 300^\circ\) and no other values in the range.

Equate the equations of the line and the curve to find the points of intersection:

\(x^2 - 3x + 3 = kx - 1\)

Rearrange into standard quadratic form \(ax^2 + bx + c = 0\):

\(x^2 - (k + 3)x + 4 = 0\)

For the line and curve not to intersect, the quadratic equation must have no real roots, which requires the discriminant \(b^2 - 4ac < 0\):

\(b = -(k + 3)\), \(a = 1\), \(c = 4\)

\((-(k + 3))^2 - 4(1)(4) < 0\)

\((k + 3)^2 - 16 < 0\)

\(k^2 + 6k + 9 - 16 < 0\)

\(k^2 + 6k - 7 < 0\)

Factorise the quadratic in \(k\):

\((k + 7)(k - 1) < 0\)

Therefore, the set of values of \(k\) is \(-7 < k < 1\) (or \(\{k: -7 < k < 1\}\)).

M1: Equate line and curve equations and rearrange to a 3-term quadratic in \(x\).
M1: Apply discriminant condition \(b^2 - 4ac < 0\) (or find critical values using \(= 0\)).
A1: Obtain correct critical values \(k = -7\) and \(k = 1\).
A1: Correct inequality \(-7 < k < 1\) (accept interval notation \((-7, 1)\); must be strict inequalities).

Expand \((1 + ax)^n\) using the binomial theorem:

\((1 + ax)^n = 1 + n(ax) + \frac{n(n - 1)}{2}(ax)^2 + \dots = 1 + nax + \frac{n(n - 1)a^2}{2}x^2 + \dots\)

Equating coefficients with \(1 - 12x + 60x^2\):

1) Coefficient of \(x\):
\(na = -12 \implies a = -\frac{12}{n}\)

2) Coefficient of \(x^2\):
\(\frac{n(n - 1)a^2}{2} = 60 \implies n(n - 1)a^2 = 120\)

Substitute \(a = -\frac{12}{n}\) into the second equation:

\(n(n - 1)\left(-\frac{12}{n}\right)^2 = 120\)

\(n(n - 1)\left(\frac{144}{n^2}\right) = 120\)

\(\frac{144(n - 1)}{n} = 120\)

Multiply through by \(n\):

\(144(n - 1) = 120n\)

\(144n - 144 = 120n\)

\(24n = 144 \implies n = 6\)

Now substitute \(n = 6\) back to find \(a\):

\(a = -\frac{12}{6} = -2\)

Thus, \(n = 6\) and \(a = -2\).

B1: Identify the coefficient of \(x\) as \(na = -12\).
M1: State the term in \(x^2\) as \(\frac{n(n-1)}{2}a^2 = 60\) and substitute \(a = -\frac{12}{n}\) (or eliminate \(n\)).
A1: Correct value \(n = 6\).
A1: Correct value \(a = -2\).

(a) Complete the square for \(\mathrm{f}(x)\):

\(\mathrm{f}(x) = 2(x^2 - 6x) + 13\)

\(= 2\left((x - 3)^2 - 9\right) + 13\)

\(= 2(x - 3)^2 - 18 + 13\)

\(= 2(x - 3)^2 - 5\)

So \(a = 2\), \(b = -3\), \(c = -5\).

(b) To find \(\mathrm{f}^{-1}(x)\), let \(y = 2(x - 3)^2 - 5\):

\(y + 5 = 2(x - 3)^2\)

\((x - 3)^2 = \frac{y + 5}{2}\)

Since the domain of \(\mathrm{f}\) is \(x \leqslant 2\), we have \(x - 3 \leqslant -1 < 0\), so we take the negative square root:

\(x - 3 = -\sqrt{\frac{y + 5}{2}}\)

\(x = 3 - \sqrt{\frac{y + 5}{2}}\)

Therefore, \(\mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x + 5}{2}}\).

For the domain of \(\mathrm{f}^{-1}\), find the range of \(\mathrm{f}\):
Since \(x \leqslant 2\), the minimum value on the domain occurs at \(x = 2\):
\(\mathrm{f}(2) = 2(2 - 3)^2 - 5 = 2(1) - 5 = -3\).
As \(x \to -\infty\), \(\mathrm{f}(x) \to \infty\), so the range of \(\mathrm{f}\) is \(\mathrm{f}(x) \geqslant -3\).

Hence, the domain of \(\mathrm{f}^{-1}\) is \(x \geqslant -3\).

(a)
B1: State \(a = 2\) and \(b = -3\) (i.e. \(2(x - 3)^2\)).
B1: State \(c = -5\).

(b)
M1: Correctly rearrange \(y = 2(x - 3)^2 - 5\) to express \(x\) in terms of \(y\), choosing the negative square root due to domain \(x \leqslant 2\).
A1: Obtain \(\mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x + 5}{2}}\) (must be in terms of \(x\)) and state domain \(x \geqslant -3\).

(a) The \(n\)th term of an arithmetic progression is \(u_n = a + (n - 1)d\).
\(u_3 = a + 2d = 14\)
\(u_{10} = a + 9d = 49\)
Subtracting the first equation from the second:
\(7d = 35 \implies d = 5\).
Substituting \(d = 5\) into \(a + 2d = 14\):
\(a + 2(5) = 14 \implies a = 4\).

(b)(i) The sum to infinity of a geometric progression is \(S_\infty = \frac{a}{1 - r}\).
Given \(a = 4\) and \(S_\infty = 8\):
\(\frac{4}{1 - r} = 8 \implies 1 - r = \frac{4}{8} = \frac{1}{2} \implies r = \frac{1}{2}\) (or \(0.5\)).

(b)(ii) The sum of the first \(n\) terms is \(S_n = \frac{a(1 - r^n)}{1 - r}\).
For \(n = 6\):
\(S_6 = \frac{4\left(1 - \left(\frac{1}{2}\right)^6\right)}{1 - \frac{1}{2}} = 8\left(1 - \frac{1}{64}\right) = 8 \times \frac{63}{64} = \frac{63}{8} = 7.875\).

(a) M1: Forms two simultaneous equations in \(a\) and \(d\) using \(u_n = a + (n - 1)d\)
A1: Obtains \(d = 5\)
A1: Obtains \(a = 4\)

(b)(i) M1: Uses \(S_\infty = \frac{a}{1 - r}\) with their \(a\) and equates to 8
A1: Obtains \(r = \frac{1}{2}\) (or \(0.5\))

(b)(ii) M1: Applies \(S_n = \frac{a(1 - r^n)}{1 - r}\) with \(n = 6\), their \(a\) and their \(|r| < 1\)
A1: Obtains \(7.875\) or \(\frac{63}{8}\)

(a) Using the binomial theorem:
\((2 - 3x)^5 = 2^5 + \binom{5}{1} 2^4 (-3x) + \binom{5}{2} 2^3 (-3x)^2 + \dots\)
\(= 32 + 5(16)(-3x) + 10(8)(9x^2) + \dots\)
\(= 32 - 240x + 720x^2\).

(b) Expanding \((1 + ax)(2 - 3x)^5\):
\((1 + ax)(32 - 240x + 720x^2 + \dots)\)
The term in \(x^2\) is obtained from:
\(1 \times (720x^2) + (ax) \times (-240x) = (720 - 240a)x^2\).
Given that the coefficient of \(x^2\) is 280:
\(720 - 240a = 280\)
\(240a = 720 - 280\)
\(240a = 440\)
\(a = \frac{440}{240} = \frac{11}{6}\).

(a) B1: Correct first term 32
B1: Correct second term \(-240x\)
B1: Correct third term \(720x^2\)

(b) M1: Identifies the two relevant terms that multiply to give the \(x^2\) term
A1 FT: Obtains \(720 - 240a\) (ft their coefficients from (a))
M1: Sets their coefficient of \(x^2\) equal to 280 and solves a linear equation for \(a\)
A1: Obtains \(a = \frac{11}{6}\) (or exact recurring decimal \(1.8\dot{3}\))

(a) Using the chain rule to differentiate \(y\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{3} \times 3(2x - 1)^2 \times 2 - 4 = 2(2x - 1)^2 - 4\).
At \(x = 2\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 2(2(2) - 1)^2 - 4 = 2(3)^2 - 4 = 2(9) - 4 = 14\).

(b) Using the chain rule for connected rates of change:
\(\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}x} \times \frac{\mathrm{d}x}{\mathrm{d}t}\)
Given \(\frac{\mathrm{d}x}{\mathrm{d}t} = 0.05\) and at \(x = 2\), \(\frac{\mathrm{d}y}{\mathrm{d}x} = 14\):
\(\frac{\mathrm{d}y}{\mathrm{d}t} = 14 \times 0.05 = 0.7\) units per second.

(c) For \(y\) to be an increasing function, \(\frac{\mathrm{d}y}{\mathrm{d}x} \geqslant 0\):
\(2(2x - 1)^2 - 4 \geqslant 0\)
\((2x - 1)^2 \geqslant 2\)
Taking square roots:
\(2x - 1 \geqslant \sqrt{2}\) or \(2x - 1 \leqslant -\sqrt{2}\)
\(2x \geqslant 1 + \sqrt{2} \implies x \geqslant \frac{1 + \sqrt{2}}{2}\)
\(2x \leqslant 1 - \sqrt{2} \implies x \leqslant \frac{1 - \sqrt{2}}{2}\).
(Accept strict inequalities \(x > \frac{1 + \sqrt{2}}{2}\) or \(x < \frac{1 - \sqrt{2}}{2}\)).

(a) M1: Differentiates using chain rule to obtain \(k(2x - 1)^2 - 4\)
A1: Obtains gradient \(14\)

(b) M1: Uses \(\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}x} \times \frac{\mathrm{d}x}{\mathrm{d}t}\) with their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) at \(x = 2\) and \(0.05\)
A1: Obtains \(0.7\)

(c) M1: Sets their \(\frac{\mathrm{d}y}{\mathrm{d}x} \geqslant 0\) (or \(> 0\))
M1: Solves quadratic inequality \((2x - 1)^2 \geqslant 2\) for critical values \(x = \frac{1 \pm \sqrt{2}}{2}\)
A1: Obtains correct set of values: \(x \leqslant \frac{1 - \sqrt{2}}{2}\) or \(x \geqslant \frac{1 + \sqrt{2}}{2}\) (accept strict inequalities)

(a)
\(y = 2x + 8x^{-2}\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 2 - 16x^{-3} = 2 - \frac{16}{x^3}\)
\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = (-3)(-16)x^{-4} = 48x^{-4} = \frac{48}{x^4}\)

(b)
At a stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\(2 - \frac{16}{x^3} = 0 \implies x^3 = 8 \implies x = 2\)
When \(x = 2\), \(y = 2(2) + \frac{8}{2^2} = 4 + 2 = 6\)
So the stationary point is \((2, 6)\).
To determine the nature, evaluate \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\) at \(x = 2\):
\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{48}{2^4} = \frac{48}{16} = 3 > 0\)
Since the second derivative is positive, the stationary point is a local minimum.

(c)
When \(x = 1\), \(y = 2(1) + \frac{8}{1^2} = 10\).
The gradient of the tangent at \(x = 1\) is \(m_{\text{tangent}} = 2 - \frac{16}{1^3} = -14\).
The gradient of the normal is \(m_{\text{normal}} = -\frac{1}{-14} = \frac{1}{14}\).
Equation of the normal:
\(y - 10 = \frac{1}{14}(x - 1)\)
\(14(y - 10) = x - 1\)
\(14y - 140 = x - 1\)
\(x - 14y + 139 = 0\)

(a)
B1: Correct first derivative \(2 - 16x^{-3}\) (or equivalent).
B1: Correct second derivative \(48x^{-4}\) (or equivalent).

(b)
M1: Sets their \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) and solves for \(x\).
A1: Obtains \((2, 6)\) with both coordinates correct.
A1: Correctly evaluates \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\) at \(x = 2\) (giving 3) and concludes minimum.

(c)
M1: Finds gradient of normal \(-\frac{1}{m}\) at \(x = 1\) and forms equation of line with their \((1, 10)\).
A1: Obtains \(x - 14y + 139 = 0\) (or any integer multiple equivalent).

(a)
The 1st, 5th, and 17th terms of the AP are:
\(u_1 = a\)
\(u_5 = a + 4d\)
\(u_{17} = a + 16d\)
Since these form a geometric progression:
\(\frac{a + 4d}{a} = \frac{a + 16d}{a + 4d}\)
\((a + 4d)^2 = a(a + 16d)\)
\(a^2 + 8ad + 16d^2 = a^2 + 16ad\)
\(16d^2 - 8ad = 0\)
Since \(d \ne 0\), divide by \(8d\):
\(2d - a = 0 \implies d = \frac{1}{2}a\)

The common ratio \(r\) is:
\(r = \frac{a + 4d}{a} = \frac{a + 4\left(\frac{1}{2}a\right)}{a} = \frac{3a}{a} = 3\)

(b)
The sum of the first 20 terms of the AP is:
\(S_{20} = \frac{20}{2}[2a + (20 - 1)d] = 10[2a + 19d]\)
Substitute \(d = \frac{1}{2}a\):
\(S_{20} = 10\left[2a + 19\left(\frac{1}{2}a\right)\right] = 10\left[2a + 9.5a\right] = 10(11.5a) = 115a\)
Given \(S_{20} = 920\):
\(115a = 920 \implies a = 8\)

For the geometric progression:
First term \(A = a = 8\), common ratio \(r = 3\).
The sum of the first 6 terms is:
\(S_6 = \frac{8(3^6 - 1)}{3 - 1} = \frac{8(729 - 1)}{2} = 4 \times 728 = 2912\)

(a)
B1: Expresses the 5th and 17th terms in terms of \(a\) and \(d\) correctly (\(a+4d\) and \(a+16d\)).
M1: Sets up the geometric condition \((a+4d)^2 = a(a+16d)\) or equivalent ratio equation.
A1: (AG) Correct algebraic expansion leading convincingly to \(d = \frac{1}{2}a\).
A1: Obtains \(r = 3\).

(b)
M1: Applies the AP sum formula \(S_{20} = 10(2a+19d)\) with \(d = \frac{1}{2}a\) and equates to 920.
A1: Obtains \(a = 8\).
A1: Applies the GP sum formula for \(n = 6\) with their \(a\) and \(r=3\) to obtain 2912.

(a) Differentiating \(y = 4x^{\frac{1}{2}} - x\):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = 2x^{-\frac{1}{2}} - 1 \]
At the stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\[ \frac{2}{\sqrt{x}} - 1 = 0 \implies \sqrt{x} = 2 \implies x = 4 \]
When \(x = 4\), \(y = 4\sqrt{4} - 4 = 4\).
So the stationary point is \((4, 4)\).

To determine the nature, find the second derivative:
\[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -x^{-\frac{3}{2}} \]
At \(x = 4\), \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -(4)^{-\frac{3}{2}} = -\frac{1}{8} < 0\).
Therefore, the stationary point is a maximum.

(b) At the point \(P(1, 3)\), substitute \(x = 1\) into \(\frac{\mathrm{d}y}{\mathrm{d}x}\):
\[ m = 2(1)^{-\frac{1}{2}} - 1 = 1 \]
The equation of the tangent at \(P\) is:
\[ y - 3 = 1(x - 1) \implies y = x + 2 \]
To find \(A\) (the \(x\)-intercept), set \(y = 0\):
\[ 0 = x + 2 \implies x = -2 \implies A(-2, 0) \]
To find \(B\) (the \(y\)-intercept), set \(x = 0\):
\[ y = 0 + 2 = 2 \implies B(0, 2) \]
The midpoint of \(AB\) is:
\[ \left( \frac{-2 + 0}{2}, \frac{0 + 2}{2} \right) = (-1, 1) \]

(c) Find the points of intersection of \(y = 4\sqrt{x} - x\) and \(y = x\):
\[ 4\sqrt{x} - x = x \implies 4\sqrt{x} = 2x \implies 2\sqrt{x} = x \]
Squaring both sides (or factoring \(\sqrt{x}\)):
\[ 4x = x^2 \implies x(x - 4) = 0 \implies x = 0 \text{ or } x = 4 \]
The area of the enclosed region is:
\[ \text{Area} = \int_{0}^{4} ((4\sqrt{x} - x) - x) \, \mathrm{d}x = \int_{0}^{4} (4x^{\frac{1}{2}} - 2x) \, \mathrm{d}x \]
\[ = \left[ \frac{4 x^{\frac{3}{2}}}{\frac{3}{2}} - x^2 \right]_0^4 = \left[ \frac{8}{3}x^{\frac{3}{2}} - x^2 \right]_0^4 \]
Evaluating at the limits:
\[ = \left( \frac{8}{3}(4)^{\frac{3}{2}} - (4)^2 \right) - 0 = \left( \frac{8}{3}(8) - 16 \right) = \frac{64}{3} - 16 = \frac{16}{3} \]

(a)
M1: Differentiate to obtain \(k x^{-\frac{1}{2}} - 1\).
A1: Correct derivative \(2x^{-\frac{1}{2}} - 1\) and solve \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) to get \(x = 4\), \(y = 4\).
M1: Valid method to test nature (evaluating \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\) or testing gradient on either side of \(x = 4\)).
A1: Correct conclusion (Maximum) supported by correct working.

(b)
M1: Substitute \(x = 1\) into their derivative to find tangent gradient and form line equation.
A1: Correct tangent equation \(y = x + 2\) (or equivalent).
M1: Find coordinates of \(A(-2, 0)\) and \(B(0, 2)\) and apply the midpoint formula.
A1: Correct midpoint \((-1, 1)\).

(c)
M1: Equate curve and line to find limits \(x = 0\) and \(x = 4\).
M1: Integrate \(\pm(4x^{\frac{1}{2}} - 2x)\) correctly to obtain \(\frac{8}{3}x^{\frac{3}{2}} - x^2\).
DM1: Apply limits \(0\) and \(4\) to their integrated expression.
A1: Correct exact area \(\frac{16}{3}\) (or \(5\frac{1}{3}\)).

(a) Differentiating \( y = 32x^{-2} + x \):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = -64x^{-3} + 1 = 1 - \frac{64}{x^3} \]
Setting \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \):
\[ 1 - \frac{64}{x^3} = 0 \implies x^3 = 64 \implies x = 4 \]
When \( x = 4 \):
\[ y = \frac{32}{4^2} + 4 = 2 + 4 = 6 \]
Thus, the stationary point is \( (4, 6) \).
To determine its nature, find the second derivative:
\[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 192x^{-4} = \frac{192}{x^4} \]
At \( x = 4 \):
\[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{192}{256} = \frac{3}{4} > 0 \]
Since \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0 \), the stationary point is a local minimum.

(b)(i) At \( x = 2 \):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = 1 - \frac{64}{2^3} = 1 - 8 = -7 \]
The equation of the tangent at \( P(2, 10) \) is:
\[ y - 10 = -7(x - 2) \implies y = -7x + 24 \]
When \( x = 0 \), \( y = 24 \implies A(0, 24) \).
When \( y = 0 \), \( -7x + 24 = 0 \implies x = \frac{24}{7} \implies B\left(\frac{24}{7}, 0\right) \).

(b)(ii) The centre \( M \) of the circle is the midpoint of \( AB \):
\[ M = \left(\frac{0 + \frac{24}{7}}{2}, \frac{24 + 0}{2}\right) = \left(\frac{12}{7}, 12\right) \]
The radius squared \( r^2 \) is:
\[ r^2 = \left(\frac{12}{7} - 0\right)^2 + (12 - 24)^2 = \frac{144}{49} + 144 = \frac{144 + 7056}{49} = \frac{7200}{49} \]
Thus, the equation of the circle is:
\[ \left(x - \frac{12}{7}\right)^2 + (y - 12)^2 = \frac{7200}{49} \quad \text{or} \quad x^2 + y^2 - \frac{24}{7}x - 24y = 0 \]

(c) The area under the curve is given by:
\[ \text{Area} = \int_{1}^{4} \left(32x^{-2} + x\right) \mathrm{d}x = \left[ -\frac{32}{x} + \frac{1}{2}x^2 \right]_{1}^{4} \]
Evaluating at the upper limit \( x = 4 \):
\[ -\frac{32}{4} + \frac{1}{2}(4)^2 = -8 + 8 = 0 \]
Evaluating at the lower limit \( x = 1 \):
\[ -\frac{32}{1} + \frac{1}{2}(1)^2 = -32 + \frac{1}{2} = -\frac{63}{2} \]
Hence, the exact area is:
\[ \text{Area} = 0 - \left(-\frac{63}{2}\right) = \frac{63}{2} \]

(a)
M1: Attempt to differentiate with at least one term differentiated correctly.
A1: Correct first derivative \( \frac{\mathrm{d}y}{\mathrm{d}x} = -64x^{-3} + 1 \).
A1: Correct coordinates \( (4, 6) \).
B1: Correct second derivative evaluated at \( x = 4 \) with conclusion of local minimum (or valid gradient sign change method).

(b)(i)
M1: Substitute \( x = 2 \) into derivative to find gradient and form tangent line equation.
A1: Correct equation \( y = -7x + 24 \) (or equivalent).
A1: Both coordinates correct: \( A(0, 24) \) and \( B\left(\frac{24}{7}, 0\right) \).

(b)(ii)
M1: Find centre as midpoint of \( AB \) and find radius / \( r^2 \) (or use diameter form of circle).
A1: Correct equation in any standard form, e.g. \( \left(x - \frac{12}{7}\right)^2 + (y - 12)^2 = \frac{7200}{49} \) or \( 7x^2 + 7y^2 - 24x - 168y = 0 \).

(c)
M1: Integrate \( 32x^{-2} + x \) to obtain \( -32x^{-1} + \frac{1}{2}x^2 \).
M1: Correctly apply limits \( x = 1 \) and \( x = 4 \) into their integrated expression.
A1: Exact value \( \frac{63}{2} \) or \( 31.5 \).