2023 Cambridge IGCSE Mathematics - Additional (0606) Practice Paper with Answers

(a) Integrate the gradient function:
\(\text{f}(x) = \int \left(3x^2 - 8x^{-3}\right) \text{d}x = x^3 + 4x^{-2} + C = x^3 + \frac{4}{x^2} + C\)

Substitute the point \((2, 5)\):
\(5 = 2^3 + \frac{4}{2^2} + C\)
\(5 = 8 + 1 + C\)
\(C = -4\)

So, the equation of the curve is \(y = x^3 + \frac{4}{x^2} - 4\).

(b) At \(x = 1\):
The gradient is \(\text{f}'(1) = 3(1)^2 - \frac{8}{1^3} = 3 - 8 = -5\).
The \(y\)-coordinate is \(y = 1^3 + \frac{4}{1^2} - 4 = 1\).

The equation of the tangent is:
\(y - 1 = -5(x - 1) \implies y = -5x + 6\).

M1 for integrating \(3x^2 - \frac{8}{x^3}\) to obtain \(x^3 + \frac{k}{x^2}\)
A1 for \(y = x^3 + \frac{4}{x^2} + C\)
M1 for substituting \((2, 5)\) to find \(C\)
A1 for \(y = x^3 + \frac{4}{x^2} - 4\)
M1 for finding gradient at \(x = 1\) and finding \(y\)-coordinate at \(x = 1\)
M1 for substituting into tangent formula \(y - y_1 = m(x - x_1)\)
A1 for \(y = -5x + 6\) (or equivalent)

Rewrite \(\cos^2 \theta\) using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\):
\(4(1 - \sin^2 \theta) + 5 \sin \theta - 5 = 0\)
\(4 - 4 \sin^2 \theta + 5 \sin \theta - 5 = 0\)
\(4 \sin^2 \theta - 5 \sin \theta + 1 = 0\)

Factorise the quadratic in \(\sin \theta\):
\((4 \sin \theta - 1)(\sin \theta - 1) = 0\)

This gives:
\(\sin \theta = 0.25\) or \(\sin \theta = 1\)

For \(\sin \theta = 0.25\):
\(\theta = \sin^{-1}(0.25) \approx 14.5^\circ\)
\(\theta = 180^\circ - 14.5^\circ = 165.5^\circ\)

For \(\sin \theta = 1\):
\(\theta = 90^\circ\)

Thus, the solutions are \(\theta = 14.5^\circ, 90^\circ, 165.5^\circ\).

M1 for using \(\cos^2 \theta = 1 - \sin^2 \theta\)
A1 for obtaining \(4 \sin^2 \theta - 5 \sin \theta + 1 = 0\)
M1 for solving the quadratic to get \(\sin \theta = 0.25\) and \(\sin \theta = 1\)
A1 for \(\theta = 90^\circ\)
M1 for finding one value of \(\theta\) from \(\sin \theta = 0.25\) (approx 14.5)
A1 for the second value from \(\sin \theta = 0.25\) (approx 165.5)
A1 for no other values within the range

(a) Using the formula for the \(n\)-th term of an arithmetic progression, \(u_n = a + (n - 1)d\):
\(u_3 = a + 2d = 9\)
\(u_{11} = a + 10d = 41\)

Subtracting the first equation from the second:
\(8d = 32 \implies d = 4\)

Substituting \(d = 4\) back into the first equation:
\(a + 2(4) = 9 \implies a = 1\)

(b) Using the sum formula \(S_n = \frac{n}{2}[2a + (n - 1)d]\):
\(S_{25} = \frac{25}{2}[2(1) + (25 - 1)(4)]\)
\(S_{25} = \frac{25}{2}[2 + 24(4)] = \frac{25}{2}[2 + 96] = \frac{25}{2}(98) = 25 \times 49 = 1225\).

M1 for setting up both equations \(a + 2d = 9\) and \(a + 10d = 41\)
M1 for solving to find \(d\)
A1 for \(d = 4\)
A1 for \(a = 1\)
M1 for using the AP sum formula with \(n = 25\)
A1 for substituting correct values into the sum formula
A1 for 1225

(a) Since \(x - 2\) is a factor of \(\text{p}(x)\), by the Factor Theorem:
\(\text{p}(2) = 0\)
\(2(2)^3 - k(2)^2 + 5(2) + 6 = 0\)
\(16 - 4k + 10 + 6 = 0\)
\(32 - 4k = 0 \implies k = 8\)

(b) Using \(k = 8\), we have \(\text{p}(x) = 2x^3 - 8x^2 + 5x + 6\).
Divide \(\text{p}(x)\) by \(x - 2\) using algebraic long division:
\(\text{p}(x) = (x - 2)(2x^2 - 4x - 3)\)

Since the quadratic term \(2x^2 - 4x - 3\) cannot be factorised further with integer coefficients, the completely factorised form is \((x - 2)(2x^2 - 4x - 3)\).

M1 for using the factor theorem: \(\text{p}(2) = 0\)
A1 for completing the algebra to show \(k = 8\)
M1 for attempting to divide \(\text{p}(x)\) by \(x - 2\)
A1 for obtaining a quadratic quotient of the form \(2x^2 + px + q\)
A1 for the correct quadratic factor \(2x^2 - 4x - 3\)
A1 for writing the final factorised form \((x - 2)(2x^2 - 4x - 3)\)

(a) Using the vector equation of path \(\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t\):
\(\mathbf{r} = (3\mathbf{i} + 5\mathbf{j}) + (4\mathbf{i} - 3\mathbf{j})t = (3 + 4t)\mathbf{i} + (5 - 3t)\mathbf{j}\)

(b) When \(P\) is due east of \(O\), its position vector has no \(\mathbf{j}\) component.
Therefore, set the \(\mathbf{j}\) component to zero:
\(5 - 3t = 0 \implies t = \frac{5}{3}\) seconds (or \(1.67\) seconds).

(c) When \(t = 5\), the position vector of \(P\) is:
\(\mathbf{r} = (3 + 4(5))\mathbf{i} + (5 - 3(5))\mathbf{j} = 23\mathbf{i} - 10\mathbf{j}\)
The distance of \(P\) from \(O\) is the magnitude of this vector:
\(\text{Distance} = \sqrt{23^2 + (-10)^2} = \sqrt{529 + 100} = \sqrt{629} \approx 25.1\) m.

M1 for using \(\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t\)
A1 for \((3 + 4t)\mathbf{i} + (5 - 3t)\mathbf{j}\) (or equivalent)
M1 for setting the \(\mathbf{j}\) component to 0
A1 for \(t = 1.67\) s (or \(\frac{5}{3}\) s)
M1 for substituting \(t = 5\) into their position vector
M1 for calculating the magnitude \(\sqrt{x^2 + y^2}\)
A1 for \(\sqrt{629}\) or \(25.1\)

Using logarithmic properties, rewrite the equation:
\(2 \ln(x + 1) - \ln(5x - 3) = \ln 2\)
\(\ln((x + 1)^2) - \ln(5x - 3) = \ln 2\)
\(\ln\left(\frac{(x + 1)^2}{5x - 3}\right) = \ln 2\)

Take the exponential of both sides:
\[\frac{(x + 1)^2}{5x - 3} = 2\]
\((x + 1)^2 = 2(5x - 3)\)
\(x^2 + 2x + 1 = 10x - 6\)
\(x^2 - 8x + 7 = 0\)

Factorise the quadratic:
\((x - 1)(x - 7) = 0\)
\(x = 1\) or \(x = 7\)

Checking the domain constraints:
For the term \(\ln(5x - 3)\), we require \(5x - 3 > 0 \implies x > 0.6\).
Both \(x = 1\) and \(x = 7\) satisfy \(x > 0.6\).
Thus, both solutions are valid.

M1 for applying power law of logarithms to \(2 \ln(x + 1)\) to get \(\ln((x + 1)^2)\)
M1 for applying subtraction law of logarithms to get \(\ln\left(\frac{(x + 1)^2}{5x - 3}\right)\)
M1 for equating \(\frac{(x + 1)^2}{5x - 3} = 2\)
A1 for simplifying to \(x^2 - 8x + 7 = 0\)
M1 for solving the quadratic equation to find two roots
A1 for \(x = 1\)
A1 for \(x = 7\)

(a) To find \(\text{fg}(x)\), we substitute \(\text{g}(x)\) into \(\text{f}(x)\):
\(\text{fg}(x) = \text{f}(2x^2 - 5) = (2x^2 - 5) + 3 = 2x^2 - 2\)

(b) First find \(\text{gf}(x)\) by substituting \(\text{f}(x)\) into \(\text{g}(x)\):
\(\text{gf}(x) = \text{g}(x + 3) = 2(x + 3)^2 - 5\)
\(\text{gf}(x) = 2(x^2 + 6x + 9) - 5 = 2x^2 + 12x + 18 - 5 = 2x^2 + 12x + 13\)

Now set \(\text{fg}(x) = \text{gf}(x)\):
\(2x^2 - 2 = 2x^2 + 12x + 13\)

Subtract \(2x^2\) from both sides:
\(-2 = 12x + 13\)
\(12x = -15\)
\(x = -1.25\).

M1 for substituting \(\text{g}(x)\) into \(\text{f}(x)\)
A1 for \(2x^2 - 2\)
M1 for attempting to find \(\text{gf}(x)\)
A1 for \(2x^2 + 12x + 13\)
M1 for equating their \(\text{fg}(x)\) and \(\text{gf}(x)\) and solving for \(x\)
A1 for \(x = -1.25\) (or \(-\frac{5}{4}\))

(a) The perimeter of the sector is given by:
\(P = 2r + r\theta = 36\)
From this, we can express \(\theta\) in terms of \(r\):
\(r\theta = 36 - 2r \implies \theta = \frac{36 - 2r}{r}\)

The area of the sector is given by:
\(A = \frac{1}{2}r^2\theta = 80\)

Substitute \(\theta = \frac{36 - 2r}{r}\) into the area formula:
\(80 = \frac{1}{2}r^2\left(\frac{36 - 2r}{r}\right)\)
\(80 = \frac{1}{2}r(36 - 2r)\)
\(80 = 18r - r^2\)
\(r^2 - 18r + 80 = 0\) (as required).

(b) Factorise the quadratic equation:
\((r - 8)(r - 10) = 0\)
So, the two possible values of \(r\) are \(r = 8\) cm and \(r = 10\) cm.

For each value of \(r\), find the corresponding value of \(\theta\):
If \(r = 8\):
\(\theta = \frac{36 - 2(8)}{8} = \frac{20}{8} = 2.5\) radians.

If \(r = 10\):
\(\theta = \frac{36 - 2(10)}{10} = \frac{16}{10} = 1.6\) radians.

M1 for using perimeter formula \(2r + r\theta = 36\)
M1 for using area formula \(\frac{1}{2}r^2\theta = 80\)
M1 for substituting to eliminate \(\theta\)
A1 for showing the given quadratic \(r^2 - 18r + 80 = 0\) clearly
M1 for solving the quadratic equation to get \(r = 8\) and \(r = 10\)
A1 for \(r = 8\), \(\theta = 2.5\)
A1 for \(r = 10\), \(\theta = 1.6\)

From the first equation, we have:
$$\log_2\left(\frac{x}{y}\right) = 3$$
$$\frac{x}{y} = 2^3 = 8 \implies x = 8y$$

From the second equation, we have:
$$3^x \cdot (3^2)^y = (3^3)^5$$
$$3^{x + 2y} = 3^{15} \implies x + 2y = 15$$

Substituting $x = 8y$ into the linear equation:
$$8y + 2y = 15$$
$$10y = 15 \implies y = 1.5$$

Substituting $y = 1.5$ back into $x = 8y$:
$$x = 8(1.5) = 12$$

Thus, the solution is $x = 12$ and $y = 1.5$.

M1: For using the subtraction law of logarithms to write $\log_2(x/y) = 3$
A1: For obtaining $x = 8y$ (or equivalent)
M1: For writing the exponential equation with a common base of 3, leading to $x + 2y = 15$
A1: For obtaining the correct linear relation $x + 2y = 15$
M1: For solving the linear simultaneous equations to find $x$ and $y$
A1: For both correct answers: $x = 12$ and $y = 1.5$ (or $y = 3/2$)

(a) Using the quotient rule with $u = (4x + 1)^{1/2}$ and $v = x - 2$:
$$u' = \frac{1}{2}(4x + 1)^{-1/2} \cdot 4 = \frac{2}{\sqrt{4x + 1}}$$
$$v' = 1$$

$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{u'v - uv'}{v^2} = \frac{\frac{2}{\sqrt{4x + 1}}(x - 2) - \sqrt{4x + 1}}{(x - 2)^2}$$

Multiplying numerator and denominator by $\sqrt{4x + 1}$:
$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2(x - 2) - (4x + 1)}{(x - 2)^2 \sqrt{4x + 1}} = \frac{2x - 4 - 4x - 1}{(x - 2)^2 \sqrt{4x + 1}} = \frac{-(2x + 5)}{(x - 2)^2 \sqrt{4x + 1}}$$

(b) When $x = 6$:
$$y = \frac{\sqrt{4(6) + 1}}{6 - 2} = \frac{\sqrt{25}}{4} = \frac{5}{4}$$

Substituting $x = 6$ into the derivative:
$$m_t = \frac{-(2(6) + 5)}{(6 - 2)^2 \sqrt{4(6) + 1}} = \frac{-17}{16 \cdot 5} = -\frac{17}{80}$$

The gradient of the normal is:
$$m_n = -\frac{1}{m_t} = \frac{80}{17}$$

The equation of the normal is:
$$y - \frac{5}{4} = \frac{80}{17}(x - 6) \implies 68y - 85 = 320(x - 6)$$
$$320x - 68y - 1835 = 0$$

(a)
M1: For attempting to differentiate $u = \sqrt{4x+1}$ using chain rule (at least one term correct, e.g., $k(4x+1)^{-1/2}$)
A1: For correct derivative $u' = 2(4x+1)^{-1/2}$
M1: For applying the quotient rule correctly: $\frac{u'v - uv'}{v^2}$
A1: For obtaining an unsimplified expression of the form $\frac{\frac{2(x-2)}{\sqrt{4x+1}} - \sqrt{4x+1}}{(x-2)^2}$
A1: For fully correct algebraic simplification leading to the shown result.

(b)
M1: For substituting $x=6$ to find $y = 5/4$ and the gradient of normal $m_n = 80/17$
A1: For the correct equation of the normal: $320x - 68y - 1835 = 0$ (or any equivalent form)

Using the trigonometric identity $\sec^2 x = 1 + \tan^2 x$:
$$2(1 + \tan^2 x) + \tan x = 5$$
$$2 \tan^2 x + \tan x - 3 = 0$$

Factorizing the quadratic equation in terms of $\tan x$:
$$(2 \tan x + 3)(\tan x - 1) = 0$$

This gives:
$$\tan x = 1 \quad \text{or} \quad \tan x = -1.5$$

For $\tan x = 1$ in the range $0 \le x \le 2\pi$:
$$x = \frac{\pi}{4} \approx 0.785 \quad \text{and} \quad x = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \approx 3.93$$

For $\tan x = -1.5$:
The basic angle is $\alpha = \tan^{-1}(1.5) \approx 0.9828$ radians.
Since $\tan x$ is negative, $x$ lies in the second or fourth quadrant:
$$x = \pi - 0.9828 \approx 2.16$$
$$x = 2\pi - 0.9828 \approx 5.30$$

Thus, the solutions are $x = \frac{\pi}{4}$ (or $0.785$), $2.16$, $\frac{5\pi}{4}$ (or $3.93$), and $5.30$.

B1: For using $\sec^2 x = 1 + \tan^2 x$ correctly
M1: For forming and simplifying to a 3-term quadratic equation in $\tan x$: $2 \tan^2 x + \tan x - 3 = 0$
M1: For solving the quadratic equation to get $\tan x = 1$ and $\tan x = -1.5$
A1: For obtaining both correct values of $\tan x$
B1: For correct solutions from $\tan x = 1$: $x = \pi/4$ (or $0.785$) and $x = 5\pi/4$ (or $3.93$)
M1: For attempting to find quadrant 2 and 4 solutions for $\tan x = -1.5$ using a correct basic angle
A1: For correct solutions: $x = 2.16$ and $x = 5.30$ (allow awrt 2.16 and 5.30)

(a) The $n$-th term of an arithmetic progression is given by:
$$T_n = a + (n - 1)d$$
The 3rd term is:
$$T_3 = a + 2d = 9 \quad \text{--- (Equation 1)}$$

The sum of the first $n$ terms is given by:
$$S_n = \frac{n}{2}[2a + (n - 1)d]$$
The sum of the first 10 terms is:
$$S_{10} = 5[2a + 9d] = 165 \implies 2a + 9d = 33 \quad \text{--- (Equation 2)}$$

Multiplying Equation 1 by 2:
$$2a + 4d = 18$$

Subtracting this from Equation 2:
$$(2a + 9d) - (2a + 4d) = 33 - 18$$
$$5d = 15 \implies d = 3$$

Substituting $d = 3$ back into Equation 1:
$$a + 2(3) = 9 \implies a = 3$$

(b) The $k$-th term is given by:
$$T_k = a + (k - 1)d = 3 + (k - 1)3 = 3k$$
We are given that $T_k = 105$:
$$3k = 105 \implies k = 35$$

(a)
M1: For using the formula $T_n = a + (n-1)d$ to write $a + 2d = 9$
M1: For using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$ to write $5(2a + 9d) = 165$ (or equivalent)
A1: For simplifying to $2a + 9d = 33$
M1: For solving the simultaneous equations to find $a$ or $d$
A1: For both $a = 3$ and $d = 3$

(b)
M1: For setting up the equation $3 + (k-1)3 = 105$ using their $a$ and $d$
A1: For $k = 35$

(a) Let \(u = x^2 \ln(3x)\).
Using the product rule:
\[\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2 \ln(3x)\right) = 2x \ln(3x) + x^2 \cdot \frac{3}{3x} = 2x \ln(3x) + x\]
Since the derivative of \(x^2 \ln(3x)\) is exactly \(\mathrm{f}'(x)\), it follows by integration that \(\mathrm{f}(x) = x^2 \ln(3x) + C\).

(b) From part (a), we have:
\[\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2 \ln(3x)\right) = 2x \ln(3x) + x\]
Integrating both sides with respect to \(x\) from \(1\) to \(2\):
\[\left[ x^2 \ln(3x) \right]_{1}^{2} = \int_{1}^{2} 2x \ln(3x) \, \mathrm{d}x + \int_{1}^{2} x \, \mathrm{d}x\]
\[\left[ x^2 \ln(3x) \right]_{1}^{2} = 2\int_{1}^{2} x \ln(3x) \, \mathrm{d}x + \left[ \frac{1}{2}x^2 \right]_{1}^{2}\]
Evaluate the left-hand side:
\[2^2 \ln(6) - 1^2 \ln(3) = 4\ln(6) - \ln(3)\]
Evaluate \(\left[ \frac{1}{2}x^2 \right]_{1}^{2}\):
\[\frac{1}{2}(4) - \frac{1}{2}(1) = \frac{3}{2}\]
Substitute these back into the equation:
\[4\ln(6) - \ln(3) = 2\int_{1}^{2} x \ln(3x) \, \mathrm{d}x + \frac{3}{2}\]
\[2\int_{1}^{2} x \ln(3x) \, \mathrm{d}x = 4\ln(6) - \ln(3) - \frac{3}{2}\]
Divide by 2:
\[\int_{1}^{2} x \ln(3x) \, \mathrm{d}x = 2\ln(6) - \frac{1}{2}\ln(3) - \frac{3}{4}\]

(a)
- **M1**: For attempting product rule on \(x^2 \ln(3x)\).
- **A1**: For finding the correct derivative: \(2x \ln(3x) + x\).
- **A1**: For concluding that \(\mathrm{f}(x) = x^2 \ln(3x) + C\).

(b)
- **M1**: For expressing the integral of \(2x\ln(3x)\) or \(x\ln(3x)\) in terms of \([x^2\ln(3x)]\) and the integral of \(x\).
- **M1**: For evaluating \([x^2 \ln(3x)]_1^2\) correctly to obtain \(4\ln(6) - \ln(3)\).
- **B1**: For integrating \(x\) to get \(\frac{1}{2}x^2\) and evaluating it to \(\frac{3}{2}\).
- **M1**: For rearranging to make \(\int_1^2 x \ln(3x) \, \mathrm{d}x\) the subject.
- **A1**: For the correct exact answer: \(2\ln(6) - \frac{1}{2}\ln(3) - \frac{3}{4}\) (or equivalent).

(a) LHS \(= \frac{\sin^2 \theta + (1 - \cos \theta)^2}{\sin \theta (1 - \cos \theta)}\)
\[= \frac{\sin^2 \theta + 1 - 2\cos \theta + \cos^2 \theta}{\sin \theta (1 - \cos \theta)}\]
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\[= \frac{1 + 1 - 2\cos \theta}{\sin \theta (1 - \cos \theta)}\]
\[= \frac{2(1 - \cos \theta)}{\sin \theta (1 - \cos \theta)}\]
\[= \frac{2}{\sin \theta} = 2\csc \theta = \text{RHS}\]

(b) Let \(\theta = 2x\). From part (a), the equation can be written as:
\[2\csc 2x = 4\]
\[\csc 2x = 2 \implies \sin 2x = \frac{1}{2}\]
Since \(0^\circ \le x \le 180^\circ\), we have \(0^\circ \le 2x \le 360^\circ\).
\[2x = 30^\circ \text{ or } 2x = 150^\circ\]
\[x = 15^\circ \text{ or } x = 75^\circ\]

(a)
- **M1**: For finding a common denominator and expanding the numerator correctly.
- **M1**: For using \(\sin^2 \theta + \cos^2 \theta = 1\) to simplify the numerator to \(2 - 2\cos \theta\).
- **A1**: For factorising and cancelling \((1 - \cos \theta)\) to show \(2\csc \theta\).

(b)
- **M1**: For identifying the connection to part (a) to write \(2\csc 2x = 4\).
- **B1**: For obtaining \(\sin 2x = \frac{1}{2}\).
- **M1**: For solving \(2x = 30^\circ, 150^\circ\).
- **A1**: For \(x = 15^\circ\).
- **A1**: For \(x = 75^\circ\) (and no other solutions in range).

(a) We can rewrite the equation as:
\[3 \cdot (3^x)^2 - 10(3^x) + 3 = 0\]
Let \(u = 3^x\). The equation becomes:
\[3u^2 - 10u + 3 = 0\]
Factorising the quadratic:
\[(3u - 1)(u - 3) = 0\]
So \(u = \frac{1}{3}\) or \(u = 3\).
Substitute back \(u = 3^x\):
If \(3^x = \frac{1}{3} \implies x = -1\).
If \(3^x = 3 \implies x = 1\).

(b) Using the laws of logarithms:
\[\log_2((y + 3)(y - 1)) = 5\]
Converting to exponential form:
\[(y + 3)(y - 1) = 2^5\]
\[y^2 + 2y - 3 = 32\]
\[y^2 + 2y - 35 = 0\]
Factorising the quadratic:
\[(y + 7)(y - 5) = 0\]
So \(y = -7\) or \(y = 5\).
We must check the validity of these solutions in the original equation:
For \(\log_2(y - 1)\) to be defined, we require \(y - 1 > 0 \implies y > 1\).
Therefore, \(y = -7\) is invalid.
The only solution is \(y = 5\).

(a)
- **M1**: For recognizing it as a quadratic in \(3^x\) and using a suitable substitution \(u = 3^x\).
- **A1**: For correct factorisation \((3u-1)(u-3) = 0\) or solving to get \(u = \frac{1}{3}, 3\).
- **A1**: For \(x = -1\).
- **A1**: For \(x = 1\).

(b)
- **M1**: For applying the product rule of logarithms: \(\log_2((y+3)(y-1)) = 5\).
- **M1**: For converting to index form: \((y+3)(y-1) = 32\).
- **A1**: For solving the quadratic equation to find \(y = 5\) and \(y = -7\).
- **A1**: For rejecting \(y = -7\) and giving \(y = 5\) as the unique solution.

(a) For an arithmetic progression:
\[u_n = a + (n-1)d\]
\[S_n = \frac{n}{2}[2a + (n-1)d]\]
Given \(u_3 = 11 \implies a + 2d = 11\) (Equation 1)
Given \(S_8 = 112 \implies \frac{8}{2}[2a + 7d] = 112 \implies 4(2a + 7d) = 112 \implies 2a + 7d = 28\) (Equation 2)
From Equation 1, we have \(a = 11 - 2d\). Substituting this into Equation 2:
\[2(11 - 2d) + 7d = 28\]
\[22 - 4d + 7d = 28\]
\[3d = 6 \implies d = 2\]
Substituting \(d = 2\) back into Equation 1:
\[a = 11 - 2(2) = 7\]
So the first term is \(7\) and the common difference is \(2\).

(b) For a geometric progression:
Given \(a = 12\) and \(S_\infty = 16\).
\[S_\infty = \frac{a}{1-r} = 16\]
\[\frac{12}{1-r} = 16 \implies 1-r = \frac{12}{16} = 0.75 \implies r = 0.25 = \frac{1}{4}\]
Now find the sum of the first 4 terms:
\[S_4 = \frac{a(1-r^4)}{1-r} = 16(1 - (0.25)^4) = 16(1 - \frac{1}{256}) = 16\left(\frac{255}{256}\right) = \frac{255}{16} = 15.9375\]

(a)
- **B1**: For writing \(a + 2d = 11\).
- **B1**: For writing \(4(2a + 7d) = 112\) or \(2a + 7d = 28\).
- **M1**: For an attempt to solve the simultaneous equations to find \(a\) or \(d\).
- **A1**: For \(a = 7\) and \(d = 2\).

(b)
- **M1**: For using \(S_\infty = \frac{a}{1-r}\) with \(a = 12\) and \(S_\infty = 16\).
- **A1**: For finding \(r = 0.25\) (or \(\frac{1}{4}\)).
- **M1**: For an attempt to find \(S_4\) using the geometric series sum formula.
- **A1**: For finding \(S_4 = 15.9375\) (or \(\frac{255}{16}\)).

(a) First, find the midpoint \(M\) of \(AB\):
\[M = \left(\frac{-2+6}{2}, \frac{3+9}{2}\right) = (2, 6)\]
Next, find the gradient \(m\) of \(AB\):
\[m = \frac{9 - 3}{6 - (-2)} = \frac{6}{8} = \frac{3}{4}\]
The gradient of the perpendicular bisector, \(m_\perp\), is:
\[m_\perp = -\frac{1}{m} = -\frac{4}{3}\]
The equation of the perpendicular bisector is:
\[y - 6 = -\frac{4}{3}(x - 2)\]
\[3(y - 6) = -4(x - 2)\]
\[3y - 18 = -4x + 8\]
\[4x + 3y = 26\]

(b) Since \(C\) lies on the perpendicular bisector and has \(x\)-coordinate \(1\):
\[4(1) + 3y = 26 \implies 3y = 22 \implies y = \frac{22}{3}\]
So the coordinates of \(C\) are \((1, \frac{22}{3})\).
Now, find the area of triangle \(ABC\).
The length of the base \(AB\) is:
\[AB = \sqrt{(6 - (-2))^2 + (9 - 3)^2} = \sqrt{8^2 + 6^2} = 10\]
The altitude of the triangle is the distance from the midpoint \(M(2, 6)\) to the vertex \(C(1, \frac{22}{3})\):
\[MC = \sqrt{(1 - 2)^2 + \left(\frac{22}{3} - 6\right)^2} = \sqrt{(-1)^2 + \left(\frac{4}{3}\right)^2} = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}\]
The area of triangle \(ABC\) is:
\[\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times \frac{5}{3} = \frac{25}{3}\]

(a)
- **B1**: For finding the midpoint \(M(2, 6)\).
- **B1**: For finding the gradient of \(AB\) as \(\frac{3}{4}\).
- **M1**: For finding the perpendicular gradient as \(-\frac{4}{3}\) and attempting the equation of the line.
- **A1**: For \(4x + 3y = 26\) (or equivalent line equation).

(b)
- **B1**: For substituting \(x = 1\) to find the coordinates of \(C\left(1, \frac{22}{3}\right)\).
- **B1**: For finding the length of base \(AB = 10\).
- **M1**: For finding the height \(MC = \frac{5}{3}\) or using the coordinate area formula.
- **A1**: For correctly showing the area of the triangle is \(\frac{25}{3}\).

(a) First, find the vector \(\vec{PQ}\):
\[\vec{PQ} = \mathbf{q} - \mathbf{p} = (12\mathbf{i} - 5\mathbf{j}) - (3\mathbf{i} + 4\mathbf{j}) = 9\mathbf{i} - 9\mathbf{j}\]
Now find the magnitude of \(\vec{PQ}\):
\[|\vec{PQ}| = \sqrt{9^2 + (-9)^2} = \sqrt{81 + 81} = \sqrt{162} = 9\sqrt{2}\]
The unit vector in the direction of \(\vec{PQ}\) is:
\[\frac{\vec{PQ}}{|\vec{PQ}|} = \frac{9\mathbf{i} - 9\mathbf{j}}{9\sqrt{2}} = \frac{1}{\sqrt{2}}\mathbf{i} - \frac{1}{\sqrt{2}}\mathbf{j} = \frac{\sqrt{2}}{2}\mathbf{i} - \frac{\sqrt{2}}{2}\mathbf{j}\]

(b) The vector \(\vec{OP} = 3\mathbf{i} + 4\mathbf{j}\).
For \(\mathbf{v} = k\mathbf{i} + 6\mathbf{j}\) to be perpendicular to \(\vec{OP}\), their dot product must be zero:
\[\vec{OP} \cdot \mathbf{v} = 0\]
\[(3\mathbf{i} + 4\mathbf{j}) \cdot (k\mathbf{i} + 6\mathbf{j}) = 0\]
\[3(k) + 4(6) = 0\]
\[3k + 24 = 0\]
\[3k = -24 \implies k = -8\]

(a)
- **M1**: For finding \(\vec{PQ} = 9\mathbf{i} - 9\mathbf{j}\).
- **M1**: For calculating the magnitude of their \(\vec{PQ}\).
- **A1**: For \(|\vec{PQ}| = 9\sqrt{2}\).
- **A1**: For the correct unit vector \(\frac{1}{\sqrt{2}}\mathbf{i} - \frac{1}{\sqrt{2}}\mathbf{j}\) (or equivalent).

(b)
- **B1**: For identifying \(\vec{OP} = 3\mathbf{i} + 4\mathbf{j}\).
- **M1**: For setting up the dot product equation: \(3(k) + 4(6) = 0\).
- **M1**: For attempting to solve for \(k\).
- **A1**: For \(k = -8\).

(a) Since \(x - 2\) is a factor, by the Factor Theorem:
\[p(2) = 0\]
\[2(2)^3 + a(2)^2 + b(2) - 6 = 0\]
\[16 + 4a + 2b - 6 = 0 \implies 4a + 2b = -10 \implies 2a + b = -5\) (Equation 1) By the Remainder Theorem, when \(p(x)\) is divided by \(x+1\), the remainder is \(p(-1) = -9\): \[2(-1)^3 + a(-1)^2 + b(-1) - 6 = -9\]
\[-2 + a - b - 6 = -9 \implies a - b = -1\) (Equation 2) Add Equation 1 and Equation 2: \[(2a + b) + (a - b) = -5 + (-1)\]
\[3a = -6 \implies a = -2\]
Substitute \(a = -2\) into Equation 2:
\[-2 - b = -1 \implies b = -1\]
So \(a = -2\) and \(b = -1\).

(b) With \(a = -2\) and \(b = -1\), the polynomial is:
\[p(x) = 2x^3 - 2x^2 - x - 6\]
Since \(x - 2\) is a factor, we can perform algebraic long division or compare coefficients:
\[2x^3 - 2x^2 - x - 6 = (x - 2)(2x^2 + cx + 3)\]
Comparing \(x^2\) coefficients:
\[-4 + c = -2 \implies c = 2\]
So the quadratic factor is \(2x^2 + 2x + 3\).
Let's check the discriminant of \(2x^2 + 2x + 3\):
\[\Delta = 2^2 - 4(2)(3) = 4 - 24 = -20 < 0\]
Since the discriminant is negative, the quadratic factor has no real roots and cannot be factorised further over real numbers.
The completely factorised form is:
\[p(x) = (x - 2)(2x^2 + 2x + 3)\]

(a)
- **M1**: For applying the Factor Theorem: \(p(2) = 16 + 4a + 2b - 6 = 0\) or equivalent.
- **M1**: For applying the Remainder Theorem: \(p(-1) = -2 + a - b - 6 = -9\) or equivalent.
- **A1**: For obtaining two correct equations, e.g., \(2a + b = -5\) and \(a - b = -1\).
- **M1**: For attempting to solve the simultaneous equations for \(a\) and \(b\).
- **A1**: For \(a = -2\) and \(b = -1\).

(b)
- **M1**: For attempting to divide \(p(x)\) by \((x-2)\) to find a quadratic factor.
- **A1**: For finding the correct quadratic factor \(2x^2 + 2x + 3\).
- **A1**: For expressing the final answer as \((x-2)(2x^2 + 2x + 3)\) and noting that the quadratic cannot be factorised further (or just writing the correct expression).

(a) Let \(y = \frac{2x+1}{x-3}\).
\[y(x - 3) = 2x + 1\]
\[xy - 3y = 2x + 1\]
\[xy - 2x = 3y + 1\]
\[x(y - 2) = 3y + 1\]
\[x = \frac{3y + 1}{y - 2}\]
Thus,
\[\mathrm{f}^{-1}(x) = \frac{3x + 1}{x - 2} \quad \text{for } x \neq 2\]

(b) First find an expression for \(\mathrm{f}^2(x)\):
\[\mathrm{f}^2(x) = \mathrm{f}(\mathrm{f}(x)) = \frac{2\left(\frac{2x+1}{x-3}\right) + 1}{\frac{2x+1}{x-3} - 3}\]
Multiply the numerator and denominator by \((x-3)\):
\[\mathrm{f}^2(x) = \frac{2(2x+1) + (x-3)}{2x+1 - 3(x-3)} = \frac{4x+2+x-3}{2x+1-3x+9} = \frac{5x-1}{10-x}\]
Now set \(\mathrm{f}^2(x) = x\):
\[\frac{5x-1}{10-x} = x\]
\[5x - 1 = x(10 - x)\]
\[5x - 1 = 10x - x^2\]
\[x^2 - 5x - 1 = 0\]
Using the quadratic formula:
\[x = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(-1)}}{2} = \frac{5 \pm \sqrt{25 + 4}}{2} = \frac{5 \pm \sqrt{29}}{2}\]

(a)
- **M1**: For setting \(y = \frac{2x+1}{x-3}\) and multiplying by \((x-3)\).
- **M1**: For rearranging terms to group \(x\) on one side.
- **M1**: For factorising \(x\) and making it the subject.
- **A1**: For \(\mathrm{f}^{-1}(x) = \frac{3x+1}{x-2}\) (or equivalent).

(b)
- **M1**: For an attempt to find \(\mathrm{f}(\ \mathrm{f}(x))\) by substituting \(\mathrm{f}(x)\) into \(\mathrm{f}(x)\).
- **A1**: For simplifying \(\mathrm{f}^2(x)\) to \(\frac{5x-1}{10-x}\).
- **M1**: For equating to \(x\) and forming the quadratic equation \(x^2 - 5x - 1 = 0\).
- **A1**: For \(x = \frac{5 \pm \sqrt{29}}{2}\).

**(a)**
Rewrite the left-hand side with a base of 2:
\(\frac{2^{2x+3}}{8^{x-1}} = \frac{2^{2x+3}}{(2^3)^{x-1}} = \frac{2^{2x+3}}{2^{3x-3}}\).
Using the division law for exponents:
\(2^{(2x+3) - (3x-3)} = 2^{6-x}\).
So, the equation becomes:
\(2^{6-x} = 5^x\).
Take natural logarithms on both sides:
\(\ln(2^{6-x}) = \ln(5^x)\)
\((6-x)\ln 2 = x\ln 5\)
\(6\ln 2 - x\ln 2 = x\ln 5\)
\(6\ln 2 = x(\ln 5 + \ln 2)\)
Using log laws:
\(\ln(2^6) = x\ln(5 \times 2)\)
\(\ln 64 = x\ln 10\)
\(x = \frac{\ln 64}{\ln 10}\).

**(b)**
Use the change of base formula on \(\log_4(y-1)\):
\(\log_4(y-1) = \frac{\log_2(y-1)}{\log_2 4} = \frac{1}{2}\log_2(y-1)\).
Substitute this back into the equation:
\(\log_2(y+3) - \frac{1}{2}\log_2(y-1) = 2\).
Multiply the entire equation by 2:
\(2\log_2(y+3) - \log_2(y-1) = 4\)
\(\log_2(y+3)^2 - \log_2(y-1) = 4\)
\(\log_2\left(\frac{(y+3)^2}{y-1}\right) = 4\).
Convert to exponential form:
\(\frac{(y+3)^2}{y-1} = 2^4 = 16\)
\(y^2 + 6y + 9 = 16(y-1)\)
\(y^2 + 6y + 9 = 16y - 16\)
\(y^2 - 10y + 25 = 0\)
\((y-5)^2 = 0\)
\(y = 5\).

**(a)**
M1: For writing \(8^{x-1}\) as \(2^{3x-3}\) and simplifying LHS to \(2^{6-x}\).
M1: For taking logs of both sides to get \((6-x)\ln 2 = x\ln 5\) oe.
M1: For grouping \(x\) terms to obtain \(x(\ln 2 + \ln 5) = 6\ln 2\) or \(x\ln 10 = \ln 64\).
A1: For \(x = \frac{\ln 64}{\ln 10}\).

**(b)**
M1: For using change of base formula to write \(\log_4(y-1) = \frac{1}{2}\log_2(y-1)\) or equivalent.
M1: For applying laws of logarithms to obtain a single logarithmic expression, e.g., \(\log_2\left(\frac{(y+3)^2}{y-1}\right) = 4\).
M1: For resolving the logarithm to obtain a quadratic equation, e.g., \(y^2 - 10y + 25 = 0\).
A1: For \(y = 5\) only (must check validity, i.e., \(y-1 > 0\) and \(y+3 > 0\)).

**(a)**
Using the product rule with \(u = x^2\) and \(v = \ln(3x)\):
\(\frac{du}{dx} = 2x\)
\(\frac{dv}{dx} = \frac{1}{3x} \cdot 3 = \frac{1}{x}\)
Applying the product rule formula:
\(\frac{d}{dx} (x^2 \ln(3x)) = u \frac{dv}{dx} + v \frac{du}{dx}\)
\(= x^2 \left(\frac{1}{x}\right) + (2x)\ln(3x)\)
\(= x + 2x\ln(3x)\).

**(b)**
From part (a), we have:
\(\frac{d}{dx} [x^2 \ln(3x)] = 2x\ln(3x) + x\).
Integrating both sides with respect to \(x\):
\(x^2 \ln(3x) = \int (2x\ln(3x) + x) \, dx\)
\(x^2 \ln(3x) = 2\int x\ln(3x) \, dx + \int x \, dx\)
\(x^2 \ln(3x) = 2\int x\ln(3x) \, dx + \frac{1}{2}x^2\)
Rearranging to make the required integral the subject:
\(2\int x\ln(3x) \, dx = x^2 \ln(3x) - \frac{1}{2}x^2\)
\(\int x\ln(3x) \, dx = \frac{1}{2}x^2 \ln(3x) - \frac{1}{4}x^2\).
Now, evaluate the definite integral from 1 to 2:
\(\int_1^2 x \ln(3x) \, dx = \left[ \frac{1}{2}x^2 \ln(3x) - \frac{1}{4}x^2 \right]_1^2\)
\(= \left( \frac{1}{2}(2)^2 \ln(3(2)) - \frac{1}{4}(2)^2 \right) - \left( \frac{1}{2}(1)^2 \ln(3(1)) - \frac{1}{4}(1)^2 \right)\)
\(= (2\ln 6 - 1) - \left(\frac{1}{2}\ln 3 - \frac{1}{4}\right)\)
\(= 2\ln 6 - 1 - \frac{1}{2}\ln 3 + \frac{1}{4}\)
\(= 2\ln 6 - \frac{1}{2}\ln 3 - \frac{3}{4}\).
Thus, \(A = 2\), \(B = \frac{1}{2}\), and \(C = \frac{3}{4}\).

**(a)**
M1: For attempting to apply the product rule to differentiate \(x^2 \ln(3x)\).
A1: For correct derivative of \(\ln(3x)\) as \(\frac{1}{x}\) and \(x^2\) as \(2x\).
A1: For correct simplified expression \(2x\ln(3x) + x\).

**(b)**
M1: For integrating both sides of the derivative from part (a) or using integration by parts.
A1: For correct integral \(\int x \ln(3x) \, dx = \frac{1}{2}x^2 \ln(3x) - \frac{1}{4}x^2\).
M1: For substituting limits 2 and 1 into their integrated expression.
A1: For obtaining \(2\ln 6 - 1\) and \(\frac{1}{2}\ln 3 - \frac{1}{4}\) oe.
A1: For the final correct exact answer: \(2\ln 6 - \frac{1}{2}\ln 3 - \frac{3}{4}\).