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

First, find the \( y \)-coordinate when \( x = 1 \):
\( y = 3(1)^2 - 4\ln(1) = 3 \).
So the point of contact is \( (1, 3) \).

Now, find the derivative \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = 6x - \frac{4}{x} \).

At \( x = 1 \), the gradient of the tangent is:
\( m_t = 6(1) - \frac{4}{1} = 2 \).

The gradient of the normal, \( m_n \), is the negative reciprocal of the tangent gradient:
\( m_n = -\frac{1}{2} \).

Now use the equation of a straight line with point \( (1, 3) \) and gradient \( -\frac{1}{2} \):
\( y - 3 = -\frac{1}{2}(x - 1) \)
\( 2y - 6 = -x + 1 \)
\( x + 2y = 7 \).

M1: For finding \( \frac{dy}{dx} = 6x - \frac{4}{x} \) and substituting \( x = 1 \) to obtain the tangent gradient.
M1: For using the perpendicular gradient rule \( m_1 m_2 = -1 \) and setting up the equation of the line passing through \( (1, y(1)) \).
A1: For the correct equation in the required form: \( x + 2y = 7 \) (or any equivalent integer form like \( -x - 2y = -7 \)).

Using the subtraction law of logarithms:
\( \log_3\left(\frac{x + 4}{x - 2}\right) = 2 \)

Convert from logarithmic form to exponential form:
\( \frac{x + 4}{x - 2} = 3^2 \)
\( \frac{x + 4}{x - 2} = 9 \)

Multiply both sides by \( x - 2 \):
\( x + 4 = 9(x - 2) \)
\( x + 4 = 9x - 18 \)
\( 8x = 22 \)
\( x = \frac{22}{8} = 2.75 \) (or \( \frac{11}{4} \))

M1: For applying the division rule of logarithms to obtain \( \log_3\left(\frac{x + 4}{x - 2}\right) = 2 \).
M1: For removing the logarithm to obtain the linear equation \( \frac{x + 4}{x - 2} = 9 \).
A1: For the correct answer \( x = 2.75 \) or \( x = \frac{11}{4} \).

We need to form a 5-digit number: _ _ _ _ _

1. First digit must be odd: The odd digits available are {1, 3, 5, 7} (4 choices).
2. Last digit must be even: The even digits available are {2, 4, 6} (3 choices).
3. These two choices are independent because the sets of odd and even digits do not overlap.

After choosing the first and last digits, we have used 2 of our 7 digits. There are 5 digits remaining.
We need to choose and arrange 3 digits for the remaining 3 middle positions from the 5 remaining digits:
\( ^5P_3 = 5 \times 4 \times 3 = 60 \) ways.

Total number of combinations = \( 4 \times 60 \times 3 = 720 \).

M1: For identifying and writing down the number of choices for the first (4 choices) and last (3 choices) positions.
M1: For calculating the permutations of the remaining 3 positions from 5 digits (\( ^5P_3 = 60 \)).
A1: For multiplying these to get the correct final answer of 720.

Let \( Y = y \) and \( X = x^2 \). The relationship is represented by a straight line:
\( Y = mX + c \)

We are given two points on this line: \( (X_1, Y_1) = (2, 5) \) and \( (X_2, Y_2) = (6, 17) \).

First, find the gradient \( m \):
\( m = \frac{17 - 5}{6 - 2} = \frac{12}{4} = 3 \)

Now substitute one point to find the intercept \( c \):
\( 5 = 3(2) + c \Rightarrow c = -1 \)

So the linear relationship is:
\( y = 3x^2 - 1 \)

When \( x = 3 \), \( x^2 = 9 \). Substitute \( x^2 = 9 \) into the equation:
\( y = 3(9) - 1 = 26 \).

M1: For finding the gradient of the straight line, \( m = 3 \).
M1: For formulating the full relationship \( y = 3x^2 - 1 \) by finding the intercept \( c = -1 \).
A1: For substitute \( x = 3 \) (or \( x^2 = 9 \)) to find \( y = 26 \).

Rearrange the equation:
\( \sin(2\theta - 30^\circ) = \frac{\sqrt{3}}{2} \)

Let \( \phi = 2\theta - 30^\circ \). Since \( 0^\circ \le \theta \le 180^\circ \), the range for \( \phi \) is:
\( 2(0^\circ) - 30^\circ \le \phi \le 2(180^\circ) - 30^\circ \)
\( -30^\circ \le \phi \le 330^\circ \)

The basic angle for \( \sin(\phi) = \frac{\sqrt{3}}{2} \) is \( 60^\circ \).
Since \( \sin(\phi) \) is positive, \( \phi \) can be in the 1st and 2nd quadrants:
\( \phi = 60^\circ \) or \( \phi = 180^\circ - 60^\circ = 120^\circ \)

Both values lie within the interval \( -30^\circ \le \phi \le 330^\circ \).

Now solve for \( \theta \):
For \( 2\theta - 30^\circ = 60^\circ \):
\( 2\theta = 90^\circ \Rightarrow \theta = 45^\circ \)

For \( 2\theta - 30^\circ = 120^\circ \):
\( 2\theta = 150^\circ \Rightarrow \theta = 75^\circ \)

M1: For finding the basic angle of \( 60^\circ \) (or \( \frac{\pi}{3} \) radians).
M1: For setting up both equations \( 2\theta - 30^\circ = 60^\circ \) and \( 2\theta - 30^\circ = 120^\circ \).
A1: For obtaining both correct angles \( \theta = 45^\circ \) and \( \theta = 75^\circ \) (and no extra values in range).

For the curve to lie entirely above the \( x \)-axis, the quadratic expression \( 3x^2 + kx + 12 \) must be positive for all real values of \( x \).
Since the coefficient of \( x^2 \) is \( 3 > 0 \), this condition is met if and only if the equation \( 3x^2 + kx + 12 = 0 \) has no real roots. Therefore, the discriminant \( D \) must be less than zero.

\( D = b^2 - 4ac < 0 \)
\( k^2 - 4(3)(12) < 0 \)
\( k^2 - 144 < 0 \)

Solving the quadratic inequality:
\( (k - 12)(k + 12) < 0 \)
This gives the range:
\( -12 < k < 12 \)

M1: For stating or using the condition \( b^2 - 4ac < 0 \).
M1: For establishing the inequality \( k^2 - 144 < 0 \) and finding critical values \( \pm 12 \).
A1: For the correct set of values: \( -12 < k < 12 \).

First, express the vector \( 3\mathbf{a} + \mathbf{b} \) in terms of \( k \):
\( 3\mathbf{a} + \mathbf{b} = 3\begin{pmatrix} -1 \\ -4 \end{pmatrix} + \begin{pmatrix} k \\ 4 \end{pmatrix} = \begin{pmatrix} -3 \\ -12 \end{pmatrix} + \begin{pmatrix} k \\ 4 \end{pmatrix} = \begin{pmatrix} k - 3 \\ -8
\end{pmatrix} \)

Now find the magnitude of this vector:
\( |3\mathbf{a} + \mathbf{b}| = \sqrt{(k - 3)^2 + (-8)^2} = 10 \)

Square both sides to eliminate the square root:
\( (k - 3)^2 + 64 = 100 \)
\( (k - 3)^2 = 36 \)

Taking the square root:
\( k - 3 = 6 \) or \( k - 3 = -6 \)
\( k = 9 \) or \( k = -3 \)

Since \( k \) is defined as a positive constant, we reject \( k = -3 \).
Thus, \( k = 9 \).

M1: For finding the vector expression \( 3\mathbf{a} + \mathbf{b} = \begin{pmatrix} k - 3 \\ -8 \end{pmatrix} \).
M1: For setting up the equation \( (k-3)^2 + 64 = 100 \) using the definition of magnitude.
A1: For selecting the positive solution \( k = 9 \) only.

Let \( a \) be the first term and \( r \) be the common ratio.
The second term is given by:
\( ar = 12 \Rightarrow a = \frac{12}{r} \)

The sum to infinity is given by:
\( \frac{a}{1 - r} = 64 \)

Substitute \( a = \frac{12}{r} \) into the sum to infinity formula:
\( \frac{12}{r(1 - r)} = 64 \)
\( 12 = 64r(1 - r) \)
\( 12 = 64r - 64r^2 \)

Rearrange into a standard quadratic form:
\( 64r^2 - 64r + 12 = 0 \)

Divide the entire equation by 4:
\( 16r^2 - 16r + 3 = 0 \)

Factorise the quadratic:
\( (4r - 1)(4r - 3) = 0 \)

This gives:
\( r = \frac{1}{4} \) or \( r = \frac{3}{4} \)

Both values are between \(-1\) and \(1\), so both are valid ratios for a convergent sum to infinity.

M1: For writing down the two initial equations, \( ar = 12 \) and \( \frac{a}{1-r} = 64 \).
M1: For substituting one into the other to obtain a quadratic equation in terms of \( r \) (e.g., \( 16r^2 - 16r + 3 = 0 \)).
A1: For finding both correct values: \( r = \frac{1}{4} \) (or 0.25) and \( r = \frac{3}{4} \) (or 0.75).

First, find the derivative of \(y = (3x-1)e^{-2x}\) using the product rule. Let \(u = 3x-1\) and \(v = e^{-2x}\). Then \(u' = 3\) and \(v' = -2e^{-2x}\). This gives \(\frac{dy}{dx} = 3e^{-2x} - 2(3x-1)e^{-2x} = (5-6x)e^{-2x}\). The curve crosses the \(y\)-axis at \(x = 0\). Substituting \(x = 0\) into the derivative gives the gradient of the tangent, \(m_T = (5 - 0)e^0 = 5\). The gradient of the normal, \(m_N\), is perpendicular to the tangent, so \(m_N = -\frac{1}{m_T} = -\frac{1}{5}\).

M1: For applying the product rule to differentiate \((3x-1)e^{-2x}\) (with at most one minor slip). A1: For finding the correct gradient of the tangent at \(x = 0\) as \(5\). A1: For finding the correct gradient of the normal as \(-\frac{1}{5}\) (or \(-0.2\)).

Using the addition rule for logarithms: \(\log_5((x-1)(x+3)) = \log_5(2x+1)\). Equating the arguments gives: \((x-1)(x+3) = 2x+1\). Expanding the left side: \(x^2 + 2x - 3 = 2x+1\). Subtracting \(2x+1\) from both sides: \(x^2 - 4 = 0\), which factors to \((x-2)(x+2) = 0\). This gives \(x = 2\) or \(x = -2\). Since \(\log_5(x-1)\) requires \(x > 1\) to be defined, \(x = -2\) is rejected. The only valid solution is \(x = 2\).

M1: For applying the product law of logarithms to combine the left-hand side into \(\log_5((x-1)(x+3))\). M1: For equating the arguments, simplifying to \(x^2 - 4 = 0\), and solving to find \(x = 2\) and \(x = -2\). A1: For identifying \(x = 2\) as the only valid solution and explicitly rejecting \(x = -2\).

First, find the \(y\)-coordinate at \(x = 3\):
\(y = (2(3) - 3)\sqrt{3+1} = 3\sqrt{4} = 6\).
So the point of contact is \((3, 6)\).

Next, differentiate \(y = (2x - 3)(x+1)^{1/2}\) using the product rule:
\(\frac{dy}{dx} = 2(x+1)^{1/2} + (2x-3)\left(\frac{1}{2}(x+1)^{-1/2}\right)\).

Substitute \(x = 3\) into the derivative to find the gradient of the tangent:
\(\frac{dy}{dx} = 2\sqrt{4} + \frac{2(3)-3}{2\sqrt{4}} = 4 + \frac{3}{4} = \frac{19}{4}\).

Since the normal is perpendicular to the tangent, its gradient \(m\) is:
\(m = -\frac{1}{19/4} = -\frac{4}{19}\).

Now, use the point-slope formula for the line of the normal:
\(y - 6 = -\frac{4}{19}(x - 3)\)
\(19(y - 6) = -4(x - 3)\)
\(19y - 114 = -4x + 12\)
\(4x + 19y = 126\).

M1: Substitute \(x = 3\) to find the correct \(y\)-coordinate of \(6\).
M1: Apply the product rule to differentiate \(y = (2x - 3)\sqrt{x+1}\), obtaining at least one correct term.
A1: Correctly find \(\frac{dy}{dx} = 2\sqrt{x+1} + \frac{2x-3}{2\sqrt{x+1}}\) (or any equivalent simplified form).
M1: Substitute \(x = 3\) to find the tangent gradient and use the perpendicular relation \(m_1 m_2 = -1\) to get the normal gradient of \(-\frac{4}{19}\).
A1: Obtain the correct equation of the normal in the requested form: \(4x + 19y = 126\).

Apply the power law of logarithms to the first term:
\(\log_3(x-1)^2 - \log_3(x+5) = 1\).

Apply the quotient law to combine the terms on the left-hand side:
\(\log_3\left(\frac{(x-1)^2}{x+5}\right) = 1\).

Write the logarithmic equation in exponential form:
\(\frac{(x-1)^2}{x+5} = 3^1 = 3\).

Multiply out the denominator and expand the quadratic expression:
\((x-1)^2 = 3(x+5)\)
\(x^2 - 2x + 1 = 3x + 15\)
\(x^2 - 5x - 14 = 0\).

Factorize the quadratic equation:
\((x-7)(x+2) = 0\)
This gives \(x = 7\) or \(x = -2\).

We must check the validity of these solutions in the original equation:
For \(x = -2\), the expression \(\log_3(x-1)\) becomes \(\log_3(-3)\), which is undefined. Thus, \(x = -2\) is rejected.
For \(x = 7\), both logarithmic expressions are defined.
Therefore, the only valid solution is \(x = 7\).

M1: Use the power law of logarithms to write \(2\log_3(x-1)\) as \(\log_3(x-1)^2\).
M1: Use the quotient law to express the left-hand side as a single logarithm \(\log_3\left(\frac{(x-1)^2}{x+5}\right)\).
A1: Remove logarithms correctly to obtain the algebraic equation \(\frac{(x-1)^2}{x+5} = 3\).
M1: Solve the resulting quadratic equation to find \(x = 7\) and \(x = -2\).
A1: Correctly state that \(x = 7\) is the only valid solution, with clear rejection of \(x = -2\) due to domain restrictions.

Use the trigonometric identity \(\cos^2 \theta = 1 - \sin^2 \theta\) to write the entire equation in terms of \(\sin \theta\):
\(4(1 - \sin^2 \theta) - 5\sin \theta - 2 = 0\)
\(4 - 4\sin^2 \theta - 5\sin \theta - 2 = 0\)
\(4\sin^2 \theta + 5\sin \theta - 2 = 0\).

Let \(y =
sin \theta\). The equation is \(4y^2 + 5y - 2 = 0\).
Apply the quadratic formula to solve for \(y\):
\(y = \frac{-5 \pm \sqrt{5^2 - 4(4)(-2)}}{2(4)}\)
\(y = \frac{-5 \pm \sqrt{25 + 32}}{8} = \frac{-5 \pm \sqrt{57}}{8}\).

This gives two values for \(\sin \theta\):
1. \(\sin \theta = \frac{-5 + \sqrt{57}}{8} \approx 0.3187\)
2. \(\sin \theta = \frac{-5 - \sqrt{57}}{8} \approx -1.5687\) (no real solutions since \(\sin \theta\) must be between \(-1\) and \(1\)).

Find the angles for \(\sin \theta \approx 0.3187\) in the range \(0^\circ \le \theta \le 360^\circ\):
Basic angle \(\alpha = \sin^{-1}(0.3187) \approx 18.58^\circ\).
Since \(\sin \theta\) is positive, \(\theta\) lies in the first and second quadrants:
\(\theta = 18.6^\circ\) (to 1 d.p.)
\(\theta = 180^\circ - 18.58^\circ = 161.4^\circ\) (to 1 d.p.).

M1: Use \(\cos^2 \theta = 1 - \sin^2 \theta\) to rewrite the equation in terms of \(\sin \theta\) only.
A1: Form the correct quadratic equation: \(4\sin^2 \theta + 5\sin \theta - 2 = 0\).
M1: Solve the quadratic equation to obtain \(\sin \theta \approx 0.319\) (and identify that the other solution is outside the valid range of sine).
A1: Find one correct angle: \(\theta \approx 18.6^\circ\) (or \(18.58^\circ\)).
A1: Find the second correct angle: \(\theta \approx 161.4^\circ\) (or \(161.42^\circ\)), and no other solutions within range.

First, find the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):
\(\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (5\mathbf{i} - \mathbf{j}) - (2\mathbf{i} + 3\mathbf{j}) = 3\mathbf{i} - 4\mathbf{j}\).
\(\overrightarrow{AC} = \mathbf{c} - \mathbf{a} = (k\mathbf{i} + 11\mathbf{j}) - (2\mathbf{i} + 3\mathbf{j}) = (k - 2)\mathbf{i} + 8\mathbf{j}\).

Since points \(A\), \(B\) and \(C\) are collinear, the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) are parallel. Therefore, \(\overrightarrow{AC} = \lambda \overrightarrow{AB}\) for some scalar \(\lambda\):
\((k - 2)\mathbf{i} + 8\mathbf{j} = \lambda(3\mathbf{i} - 4\mathbf{j})\).

Equating the \(\mathbf{j}\) components:
\(8 = -4\lambda \implies \lambda = -2\).

Equating the \(\mathbf{i}\) components:
\(k - 2 = 3\lambda\)
\(k - 2 = 3(-2)\)
\(k - 2 = -6\)
\(k = -4\).

M1: Find an expression for \(\overrightarrow{AB}\) or \(\overrightarrow{BC}\) or \(\overrightarrow{AC}\) in terms of unit vectors \(\mathbf{i}\) and \(\mathbf{j}\).
A1: Obtain the correct expressions for two vectors, e.g., \(\overrightarrow{AB} = 3\mathbf{i} - 4\mathbf{j}\) and \(\overrightarrow{AC} = (k-2)\mathbf{i} + 8\mathbf{j}\).
M1: State that the vectors are parallel by setting up a ratio or linear dependence equation: \(\frac{k-2}{3} = \frac{8}{-4}\) (or equivalent).
M1: Solve their linear equation to find the value of the scalar parameter \(\lambda = -2\) or directly set up the ratio calculation.
A1: Conclude with the correct value of \(k = -4\).

(a) Since the progression is geometric, the common ratio \(r\) is constant:
\(r = \frac{x}{x+4} = \frac{x-3}{x}\).

Cross-multiply to solve for \(x\):
\(x^2 = (x+4)(x-3)\)
\(x^2 = x^2 + x - 12\)
\(0 = x - 12 \implies x = 12\).

(b) Now substitute \(x = 12\) back to find the first term \(a\) and the common ratio \(r\):
First term \(a = 12 + 4 = 16\).
Second term is \(12\), so the common ratio \(r = \frac{12}{16} = \frac{3}{4} = 0.75\).

Since \(|r| < 1\), the sum to infinity, \(S_{\infty}\), is defined:
\(S_{\infty} = \frac{a}{1-r} = \frac{16}{1 - 0.75} = \frac{16}{0.25} = 64\).

M1: Set up the ratio equality for a geometric progression: \(\frac{x}{x+4} = \frac{x-3}{x}\).
A1: Correctly expand and solve the equation to show \(x = 12\).
B1: Determine the first term \(a = 16\) and the common ratio \(r = 0.75\) (or \(\frac{3}{4}\)).
M1: Use the correct sum to infinity formula \(S_{\infty} = \frac{a}{1-r}\) with their \(a\) and \(r\) (where \(|r| < 1\)).
A1: Obtain the correct sum to infinity of \(64\).

Since both sides of the inequality are non-negative, we can square both sides:
\((2x - 5)^2 < (x + 1)^2\).

Expand the brackets:
\(4x^2 - 20x + 25 < x^2 + 2x + 1\).

Rearrange the inequality to form a quadratic inequality:
\(3x^2 - 22x + 24 < 0\).

Factorize the quadratic expression on the left:
\((3x - 4)(x - 6) < 0\).

The critical values are \(x = \frac{4}{3}\) and \(x = 6\).

Since we are looking for the region where the quadratic is less than zero, the solution lies between the two critical values:
\(\frac{4}{3} < x < 6\).

M1: Attempt to square both sides of the inequality or solve the two corresponding linear boundary equations \(2x - 5 = x + 1\) and \(2x - 5 = -(x + 1)\).
A1: Obtain the correct quadratic inequality \(3x^2 - 22x + 24 < 0\) (or obtain the two correct critical values directly).
M1: Attempt to factorize their quadratic expression or use the quadratic formula to find the boundaries.
A1: Identify the correct critical values of \(x = \frac{4}{3}\) (or \(1.33\)) and \(x = 6\).
A1: Express the final range of values correctly as \(\frac{4}{3} < x < 6\) (or \(1.33 < x < 6\)).

(a) Using the product rule on \(y = (2x-3)e^{2x}\):

\(\frac{\text{d}y}{\text{d}x} = u \frac{\text{d}v}{\text{d}x} + v \frac{\text{d}u}{\text{d}x}\)

where \(u = 2x-3 \implies \frac{\text{d}u}{\text{d}x} = 2\) and \(v = e^{2x} \implies \frac{\text{d}v}{\text{d}x} = 2e^{2x}\).

\(\frac{\text{d}y}{\text{d}x} = (2x-3)(2e^{2x}) + (e^{2x})(2)\)

\(\frac{\text{d}y}{\text{d}x} = e^{2x}(4x - 6 + 2)\)

\(\frac{\text{d}y}{\text{d}x} = 4(x-1)e^{2x}\).

(b) At the stationary point, \(\frac{\text{d}y}{\text{d}x} = 0\):

\(4(x-1)e^{2x} = 0\)

Since \(e^{2x} > 0\) for all real \(x\), we have:

\(x - 1 = 0 \implies x = 1\).

Substitute \(x = 1\) into the equation of the curve to find the \(y\)-coordinate:

\(y = (2(1) - 3)e^{2(1)} = -e^2\).

So the stationary point is \((1, -e^2)\).

To determine its nature, find the second derivative:

\(\frac{\text{d}^2y}{\text{d}x^2} = 4e^{2x} + 2 \cdot 4(x-1)e^{2x} = 4(2x-1)e^{2x}\).

At \(x = 1\):

\(\frac{\text{d}^2y}{\text{d}x^2} = 4(2(1)-1)e^{2} = 4e^2\).

Since \(4e^2 > 0\), the stationary point is a minimum.

(c) The curve intersects the \(x\)-axis when \(y = 0\):

\((2x-3)e^{2x} = 0 \implies 2x - 3 = 0 \implies x = 1.5\).

The area of the region is given by:

\(\int_{1.5}^{2} (2x-3)e^{2x} \text{d}x\).

Using integration by parts \(\int u \text{d}v = uv - \int v \text{d}u\):

Let \(u = 2x-3 \implies \text{d}u = 2\text{d}x\)

Let \(\text{d}v = e^{2x}\text{d}x \implies v = \frac{1}{2}e^{2x}\)

\(\int (2x-3)e^{2x} \text{d}x = \frac{1}{2}(2x-3)e^{2x} - \int e^{2x}\text{d}x\)

\(= \frac{1}{2}(2x-3)e^{2x} - \frac{1}{2}e^{2x}\)

\(= e^{2x}(x - 1.5 - 0.5) = e^{2x}(x-2)\).

Evaluating between the limits \(1.5\) and \(2\):

\([e^{2x}(x-2)]_{1.5}^{2} = e^{4}(2-2) - e^{3}(1.5-2) = 0 - e^{3}(-0.5) = 0.5e^3\).

(a)
M1: Attempts to differentiate using the product rule.
A1: Correct unsimplified derivative, e.g., \(2(2x-3)e^{2x} + 2e^{2x}\).
A1: Correct simplified derivative \(4(x-1)e^{2x}\).

(b)
M1: Sets their \(\frac{\text{d}y}{\text{d}x} = 0\) and solves for \(x\).
A1: Correct exact coordinates \((1, -e^2)\).
M1: Valid method to find the nature of the stationary point (e.g., substituting \(x=1\) into \(\frac{\text{d}^2y}{\text{d}x^2}\) or checking signs of \(\frac{\text{d}y}{\text{d}x}\) on either side).
A1: Correctly identifies the point as a minimum with valid working.

(c)
M1: Finds the lower limit of integration by solving \(y = 0 \implies x = 1.5\).
M1: Attempts integration by parts to integrate \((2x-3)e^{2x}\).
A1: Obtains the correct integrated expression \(e^{2x}(x-2)\) and evaluates to find the exact area \(0.5e^3\) (or equivalent exact expression).

(a) LHS:

\(\frac{\sin\theta}{1 - \cos\theta} + \frac{\sin\theta}{1 + \cos\theta}\)

\(= \frac{\sin\theta(1 + \cos\theta) + \sin\theta(1 - \cos\theta)}{(1 - \cos\theta)(1 + \cos\theta)}\)

\(= \frac{\sin\theta + \sin\theta\cos\theta + \sin\theta - \sin\theta\cos\theta}{1 - \cos^2\theta}\)

\(= \frac{2\sin\theta}{\sin^2\theta} = \frac{2}{\sin\theta} = 2\text{cosec }\theta\) which is equal to RHS.

(b) Using the identity from part (a), the equation can be simplified to:

\(2\text{cosec}(2x - 0.4) = 2.5 \implies \text{cosec}(2x - 0.4) = 1.25\)

\(\sin(2x - 0.4) = 0.8\).

Since \(0 < x < \pi\), the domain of \(2x - 0.4\) is:

\(0 < 2x < 2\pi \implies -0.4 < 2x - 0.4 < 2\pi - 0.4 \approx 5.88\).

Let \(\theta = 2x - 0.4\). Since \(\sin\theta = 0.8 > 0\), \(\theta\) lies in Quadrant 1 or 2:

\(\theta = \sin^{-1}(0.8) \approx 0.9273\) rad,

\(\theta = \pi - 0.9273 \approx 2.214\) rad.

Solving for \(x\):

1) \(2x - 0.4 = 0.9273 \implies 2x = 1.3273 \implies x \approx 0.664\)

2) \(2x - 0.4 = 2.214 \implies 2x = 2.614 \implies x \approx 1.31\).

Both values are within the interval \(0 < x < \pi\).

(c) Using the identity \(\cos^2 y = 1 - \sin^2 y\):

\(3(1 - \sin^2 y) + 5\sin y - 5 = 0\)

\(3 - 3\sin^2 y + 5\sin y - 5 = 0\)

\(3\sin^2 y - 5\sin y + 2 = 0\).

Factoring the quadratic:

\((3\sin y - 2)(\sin y - 1) = 0\)

Which gives \(\sin y = \frac{2}{3}\) or \(\sin y = 1\).

If \(\sin y = 1\), then \(y = 90^\circ\).

If \(\sin y = \frac{2}{3}\), then \(y \approx 41.8^\circ\) or \(y \approx 180^\circ - 41.8^\circ = 138.2^\circ\).

Thus, the solutions in the interval \(0^\circ \le y \le 360^\circ\) are \(y = 41.8^\circ\), \(90^\circ\), and \(138.2^\circ\).

(a)
M1: Combines the fractions over a common denominator \((1-\cos\theta)(1+\cos\theta)\).
A1: Correctly simplifies the numerator to \(2\sin\theta\).
A1: Uses \(\sin^2\theta = 1-\cos^2\theta\) to simplify to \(2\text{cosec }\theta\).

(b)
M1: Formulates the equation \(\sin(2x - 0.4) = 0.8\).
B1: Finds at least one correct value for \(2x - 0.4\) in radians (e.g. \(0.927\) or \(2.21\)).
A1: Correctly finds \(x = 0.664\).
A1: Correctly finds \(x = 1.31\).

(c)
M1: Uses \(\cos^2 y = 1 - \sin^2 y\) to form a quadratic equation in \(\sin y\).
M1: Solves their quadratic equation to find \(\sin y = 1\) and \(\sin y = \frac{2}{3}\).
A1: Obtains \(y = 41.8^\circ, 90^\circ, 138.2^\circ\) (no other values in the interval, accept \(90.0^\circ\)).

To find the stationary point, we first find the derivative of \(y = x \ln x\) using the product rule. \(\frac{dy}{dx} = \ln x \cdot \frac{d}{dx}(x) + x \cdot \frac{d}{dx}(\ln x) = \ln x + 1\). Setting \(\frac{dy}{dx} = 0\) gives \(\ln x + 1 = 0\), which simplifies to \(\ln x = -1\), so \(x = e^{-1} = \frac{1}{e}\). Substituting \(x = \frac{1}{e}\) back into the equation of the curve gives \(y = \frac{1}{e} \ln\left(\frac{1}{e}\right) = -\frac{1}{e}\). Therefore, the exact coordinates of the stationary point are \(\left(\frac{1}{e}, -\frac{1}{e}\right)\).

M1: Applying product rule to find \(\frac{dy}{dx} = \ln x + 1\). A1: Setting \(\frac{dy}{dx} = 0\) and solving to find \(x = e^{-1}\) or \(x = \frac{1}{e}\). A1: Correctly calculating the corresponding \(y\)-coordinate to get \(\left(\frac{1}{e}, -\frac{1}{e}\right)\).

Let the first term be \(a\) and the common ratio be \(r\). The third term is \(ar^2 = 18\) and the sixth term is \(ar^5 = 486\). Dividing the two equations gives \(\frac{ar^5}{ar^2} = \frac{486}{18}\), which simplifies to \(r^3 = 27\), hence \(r = 3\). Substituting \(r = 3\) into \(ar^2 = 18\) gives \(9a = 18\), so \(a = 2\). The sum of the first 5 terms is \(S_5 = \frac{a(r^5 - 1)}{r - 1} = \frac{2(3^5 - 1)}{3 - 1} = 3^5 - 1 = 243 - 1 = 242\).

M1: Setting up equations for the terms and solving for \(r\) to find \(r = 3\). A1: Finding \(a = 2\). A1: Correctly calculating \(S_5 = 242\).

First, find the midpoint \(M\) of \(AB\): \(M = \left(\frac{2+8}{2}, \frac{-3+5}{2}\right) = (5, 1)\). Next, find the gradient of \(AB\): \(m = \frac{5 - (-3)}{8 - 2} = \frac{8}{6} = \frac{4}{3}\). The gradient of the perpendicular bisector is the negative reciprocal of \(m\), which is \(-\frac{3}{4}\). Using the point-slope form with point \((5, 1)\): \(y - 1 = -\frac{3}{4}(x - 5)\). Multiplying by 4 gives \(4y - 4 = -3x + 15\). Rearranging into the required form gives \(4y + 3x = 19\).

M1: Finding the midpoint \((5, 1)\) and the gradient of \(AB\) which is \(\frac{4}{3}\). M1: Using the perpendicular gradient \(-\frac{3}{4}\) in the straight line equation. A1: Correctly simplifying to the form \(4y + 3x = 19\) or equivalent integer coefficients.

Using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\), we substitute this into the equation: \(2(1 - \sin^2 \theta) + 3\sin \theta = 3\). Expanding and rearranging gives \(2 - 2\sin^2 \theta + 3\sin \theta - 3 = 0\), which simplifies to \(2\sin^2 \theta - 3\sin \theta + 1 = 0\). Factoring the quadratic gives \((2\sin \theta - 1)(\sin \theta - 1) = 0\). This yields \(\sin \theta = \frac{1}{2}\) or \(\sin \theta = 1\). For \(\sin \theta = \frac{1}{2}\), \(\theta = 30^\circ, 150^\circ\). For \(\sin \theta = 1\), \(\theta = 90^\circ\). The complete set of solutions is \(\theta = 30^\circ, 90^\circ, 150^\circ\).

M1: Substituting the identity \(\cos^2 \theta = 1 - \sin^2 \theta\) and forming a quadratic equation in \(\sin \theta\). A1: Finding the roots \(\sin \theta = \frac{1}{2}\) and \(\sin \theta = 1\). A1: Correctly finding all three solutions: \(30^\circ\), \(90^\circ\), and \(150^\circ\). Deduct 1 mark for any extra incorrect solutions in the range.

Since the committee must contain at least 3 women, there are two mutually exclusive cases. Case 1: Exactly 3 women and 1 man are chosen. The number of ways is \(\binom{6}{3} \times \binom{5}{1} = 20 \times 5 = 100\). Case 2: Exactly 4 women and 0 men are chosen. The number of ways is \(\binom{6}{4} \times \binom{5}{0} = 15 \times 1 = 15\). The total number of different committees is \(100 + 15 = 115\).

M1: Identifying the two cases and writing a correct expression for at least one case (either \(\binom{6}{3} \times \binom{5}{1}\) or \(\binom{6}{4}\)). A1: Calculating both case values correctly (100 and 15). A1: Adding the two cases to get 115.

Using the product rule of logarithms, we combine the terms on the left side: \(\log_2((x + 3)(x - 3)) = 4\). This can be rewritten in exponential form as \((x + 3)(x - 3) = 2^4\). Expanding the left side and calculating the right side gives \(x^2 - 9 = 16\), which simplifies to \(x^2 = 25\). This gives \(x = 5\) or \(x = -5\). Since the arguments of the logarithms must be positive, we must have \(x + 3 > 0\) and \(x - 3 > 0\), which means \(x > 3\). Thus, \(x = -5\) is rejected, and the only valid solution is \(x = 5\).

M1: Combining logarithms using the product rule to get \(\log_2(x^2 - 9) = 4\) or converting to exponential form. A1: Obtaining the quadratic equation \(x^2 - 9 = 16\) and solving to get \(x = \pm 5\). A1: Correctly rejecting \(x = -5\) to give the final answer \(x = 5\).

Since \(P(x)\) is exactly divisible by \(x - 2\), the remainder theorem states \(P(2) = 0\). Substituting \(x = 2\) gives \(2(2)^3 + a(2)^2 + b(2) - 6 = 0 \implies 16 + 4a + 2b - 6 = 0 \implies 4a + 2b = -10 \implies 2a + b = -5\). Since \(P(x)\) leaves a remainder of 15 when divided by \(x + 1\), \(P(-1) = 15\). Substituting \(x = -1\) gives \(2(-1)^3 + a(-1)^2 + b(-1) - 6 = 15 \implies -2 + a - b - 6 = 15 \implies a - b = 23\). We now solve the system of simultaneous equations: 1) \(2a + b = -5\) and 2) \(a - b = 23\). Adding these equations gives \(3a = 18\), so \(a = 6\). Substituting \(a = 6\) into equation 2 gives \(6 - b = 23 \implies b = -17\).

M1: Applying the remainder theorem to set up two simultaneous equations: \(2a + b = -5\) and \(a - b = 23\) (allow arithmetic errors in setup). A1: Correctly solving for \(a = 6\). A1: Correctly solving for \(b = -17\).

First, find the vector \(3\mathbf{p} + \mathbf{q}\): \(3\mathbf{p} + \mathbf{q} = 3\begin{pmatrix} 2 \\ -5 \end{pmatrix} + \begin{pmatrix} k \\ 7 \end{pmatrix} = \begin{pmatrix} 6 \\ -15 \end{pmatrix} + \begin{pmatrix} k \\ 7 \end{pmatrix} = \begin{pmatrix} 6 + k \\ -8 \end{pmatrix}\). The magnitude of this vector is given by \(\sqrt{(6 + k)^2 + (-8)^2} = 10\). Squaring both sides yields \((6 + k)^2 + 64 = 100\), which simplifies to \((6 + k)^2 = 36\). Taking the square root gives \(6 + k = 6\) or \(6 + k = -6\). This results in the possible values \(k = 0\) or \(k = -12\).

M1: Finding the vector expression \(3\mathbf{p} + \mathbf{q} = \begin{pmatrix} 6 + k \\ -8 \end{pmatrix}\). M1: Setting up the magnitude equation \((6+k)^2 + 64 = 100\) and attempting to solve for \(k\). A1: Correctly identifying both possible values: \(k = 0\) and \(k = -12\).

To find the points of intersection, we equate the line and the curve: \(kx - 5 = x^2 - 3x + 4\). Rearranging this into a standard quadratic equation gives: \(x^2 - (3+k)x + 9 = 0\). For the line and the curve to not intersect, the discriminant of this quadratic equation must be strictly less than zero: \(B^2 - 4AC < 0\). Here, \(A = 1\), \(B = -(3+k)\), and \(C = 9\). Substituting these values in: \((-(3+k))^2 - 4(1)(9) < 0\), which simplifies to \((k+3)^2 - 36 < 0\). Factoring the difference of squares gives: \((k+3-6)(k+3+6) < 0 \implies (k-3)(k+9) < 0\). Solving this quadratic inequality yields the range of values: \(-9 < k < 3\).

M1: Equating the line and the curve and rearranging into a quadratic equation in \(x\).
A1: Correct quadratic equation \(x^2 - (k+3)x + 9 = 0\) (or equivalent).
M1: Applying the discriminant condition \(b^2 - 4ac < 0\) for their quadratic.
M1: Finding the critical values of \(k\) (which are \(3\) and \(-9\)).
A1: Correct final inequality \(-9 < k < 3\).

First, use the change of base formula on the second term: \(\log_4(x-1) = \frac{\log_2(x-1)}{\log_2 4} = \frac{1}{2}\log_2(x-1)\). Substitute this back into the original equation: \(\log_2(x+3) - \frac{1}{2}\log_2(x-1) = 2\). Multiply the entire equation by 2 to eliminate the fraction: \(2\log_2(x+3) - \log_2(x-1) = 4\). Apply the laws of logarithms to combine the terms: \(\log_2 \frac{(x+3)^2}{x-1} = 4\). Convert the logarithmic equation into exponential form: \(\frac{(x+3)^2}{x-1} = 2^4 \implies \frac{(x+3)^2}{x-1} = 16\). Expand and solve the resulting quadratic equation: \((x+3)^2 = 16(x-1) \implies x^2 + 6x + 9 = 16x - 16 \implies x^2 - 10x + 25 = 0\). This factors as \((x-5)^2 = 0\), yielding the solution \(x = 5\). Checking \(x = 5\) in the original logarithmic terms shows both arguments are positive (\(8 > 0\) and \(4 > 0\)), so the solution is valid.

M1: Applying the change of base formula correctly to write \(\log_4(x-1)\) as \(\frac{1}{2}\log_2(x-1)\).
M1: Applying laws of logarithms to combine terms into a single logarithm, e.g., \(\log_2 \frac{(x+3)^2}{x-1}\).
M1: Removing logarithms to form a polynomial equation, e.g., \(\frac{(x+3)^2}{x-1} = 16\).
A1: Simplifying to the correct quadratic equation \(x^2 - 10x + 25 = 0\).
A1: Correctly solving to obtain \(x = 5\) as the only valid solution.

(i) To find \(f^{-1}(x)\), set \(y = \ln(3x - 2)\). Exponentiating both sides gives \(e^y = 3x - 2\). Rearranging for \(x\) gives \(3x = e^y + 2 \implies x = \frac{e^y + 2}{3}\). Replacing \(y\) with \(x\) gives \(f^{-1}(x) = \frac{e^x + 2}{3}\).

(ii) To solve \(g(f(x)) = 8\), substitute \(f(x)\) into \(g(x)\): \(g(f(x)) = e^{2\ln(3x-2)} - 1 = e^{\ln((3x-2)^2)} - 1 = (3x-2)^2 - 1\). Set this expression equal to 8: \((3x-2)^2 - 1 = 8 \implies (3x-2)^2 = 9\). Taking the square root of both sides gives \(3x - 2 = 3\) or \(3x - 2 = -3\). Solving these gives \(x = \frac{5}{3}\) or \(x = -\frac{1}{3}\). Since the domain of \(f\) is \(x > \frac{2}{3}\), the value \(x = -\frac{1}{3}\) must be rejected. The only valid solution is \(x = \frac{5}{3}\).

(i) M1: For attempting to make \(x\) the subject of \(y = \ln(3x-2)\).
A1: For correct expression \(f^{-1}(x) = \frac{e^x+2}{3}\).
(ii) M1: For substituting \(f(x)\) into \(g(x)\) and simplifying to \((3x-2)^2 - 1\).
M1: For setting \((3x-2)^2 - 1 = 8\) and solving the quadratic to find at least one value of \(x\).
A1: For identifying and keeping only the valid solution \(x = \frac{5}{3}\) (or \(1.67\)) and discarding the invalid solution.

Let \(Y = \ln y\) and \(X = x^2\). The relationship between \(Y\) and \(X\) is linear, represented by \(Y = mX + c\). The gradient \(m\) of the line passing through \((2, 7)\) and \((6, 15)\) is: \(m = \frac{15 - 7}{6 - 2} = \frac{8}{4} = 2\). Using the point-slope formula with the point \((2, 7)\): \(Y - 7 = 2(X - 2) \implies Y = 2X - 4 + 7 \implies Y = 2X + 3\). Replacing \(Y\) with \(\ln y\) and \(X\) with \(x^2\) gives: \(\ln y = 2x^2 + 3\). To express \(y\) in terms of \(x\), apply the exponential function to both sides: \(y = e^{2x^2 + 3}\).

M1: Finding the gradient of the straight line, \(m = 2\).
M1: Writing down the linear equation \(\ln y = 2x^2 + c\) and attempting to find \(c\) using a given point.
A1: Correct linear equation: \(\ln y = 2x^2 + 3\) (or equivalent).
M1: Attempting to convert the logarithmic equation to exponential form to make \(y\) the subject.
A1: Correct final expression: \(y = e^{2x^2 + 3}\) (or \(y = e^3 e^{2x^2}\)).

The area \(A\) is given by the definite integral of the function between the specified vertical lines: \(A = \int_{\pi/12}^{\pi/4} 3\sin(2x) \, dx\). Integrating the function gives: \(\int 3\sin(2x) \, dx = -\frac{3}{2}\cos(2x)\). Now, evaluate this expression at the limits: \(A = \left[ -\frac{3}{2}\cos(2x) \right]_{\pi/12}^{\pi/4} = \left( -\frac{3}{2}\cos\left(2 \cdot \frac{\pi}{4}\right) \right) - \left( -\frac{3}{2}\cos\left(2 \cdot \frac{\pi}{12}\right) \right)\). Simplify the angles inside the cosine function: \(A = \left( -\frac{3}{2}\cos\left(\frac{\pi}{2}\right) \right) + \frac{3}{2}\cos\left(\frac{\pi}{6}\right)\). Using the exact trigonometric values, \(\text{cos}\left(\frac{\pi}{2}\right) = 0\) and \(\text{cos}\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\), we get: \(A = 0 + \frac{3}{2}\left(\frac{\sqrt{3}}{2}\right) = \frac{3\sqrt{3}}{4}\).

M1: Recognising that the area is represented by \(\int_{\pi/12}^{\pi/4} 3\sin(2x) \, dx\).
M1: Integrating \(3\sin(2x)\) to obtain an expression of the form \(k\cos(2x)\).
A1: Correct integration: \(-\frac{3}{2}\cos(2x)\).
M1: Substituting the limits \(\frac{\pi}{4}\) and \(\frac{\pi}{12}\) into their integrated expression.
A1: Finding the exact value \(\frac{3\sqrt{3}}{4}\) (or equivalent exact form, e.g. \(0.75\sqrt{3}\)).

(i) First, calculate the vector \(\vec{AB}\): \(\vec{AB} = \vec{OB} - \vec{OA} = (15\mathbf{i} + 10\mathbf{j}) - (3\mathbf{i} + \mathbf{j}) = 12\mathbf{i} + 9\mathbf{j}\). Since \(AC:CB = 2:1\), the point \(C\) is two-thirds of the way along \(AB\) from \(A\). Therefore, \(\vec{AC} = \frac{2}{3}\vec{AB} = \frac{2}{3}(12\mathbf{i} + 9\mathbf{j}) = 8\mathbf{i} + 6\mathbf{j}\). The position vector of \(C\) is: \(\vec{OC} = \vec{OA} + \vec{AC} = (3\mathbf{i} + \mathbf{j}) + (8\mathbf{i} + 6\mathbf{j}) = 11\mathbf{i} + 7\mathbf{j}\).

(ii) To find the unit vector in the direction of \(\vec{OC}\), first calculate its magnitude: \(|\vec{OC}| = \sqrt{11^2 + 7^2} = \sqrt{121 + 49} = \sqrt{170}\). Dividing \(\vec{OC}\) by its magnitude gives the unit vector: \(\frac{1}{\sqrt{170}}(11\mathbf{i} + 7\mathbf{j})\).

(i) M1: For finding \(\vec{AB} = 12\mathbf{i} + 9\mathbf{j}\).
M1: For using \(\vec{OC} = \vec{OA} + \frac{2}{3}\vec{AB}\) or an equivalent vector ratio method.
A1: Correct position vector \(\vec{OC} = 11\mathbf{i} + 7\mathbf{j}\).
(ii) M1: For calculating the magnitude of their vector \(\vec{OC}\) using Pythagoras' Theorem.
A1: Correct unit vector \(\frac{1}{\sqrt{170}}(11\mathbf{i} + 7\mathbf{j})\) (or equivalent split form).

Let \(a\) be the first term and \(r\) be the common ratio of the geometric progression. The second term is given by: \(ar = 12 \implies a = \frac{12}{r}\). The sum to infinity is given by: \(S_{\infty} = \frac{a}{1-r} = 64\). Substitute \(a = \frac{12}{r}\) into the sum to infinity formula: \(\frac{12/r}{1-r} = 64 \implies \frac{12}{r(1-r)} = 64\). Multiply both sides by \(r(1-r)\): \(12 = 64r(1-r) \implies 12 = 64r - 64r^2\). Rearranging this into a quadratic equation gives: \(64r^2 - 64r + 12 = 0\). Divide the entire equation by 4 to simplify: \(16r^2 - 16r + 3 = 0\). Factoring the quadratic: \((4r - 1)(4r - 3) = 0\). This gives two possible values for the common ratio: \(r = \frac{1}{4}\) or \(r = \frac{3}{4}\). If \(r = \frac{1}{4}\), then \(a = \frac{12}{1/4} = 48\). If \(r = \frac{3}{4}\), then \(a = \frac{12}{3/4} = 16\). Both ratios satisfy \(|r| < 1\), meaning both progressions are valid.

M1: Using the GP term formula to write \(ar = 12\).
M1: Using the sum to infinity formula to write \(\frac{a}{1-r} = 64\).
M1: Substituting to form a quadratic equation in either \(r\) or \(a\) (e.g., \(16r^2 - 16r + 3 = 0\)).
A1: Finding the two correct values of the common ratio, \(r = \frac{1}{4}\) and \(r = \frac{3}{4}\) (or first terms, \(a = 48\) and \(a = 16\)).
A1: Correctly pairing the corresponding values of \(a\) and \(r\).

**(a)**
To find \(\frac{\text{d}y}{\text{d}x}\), we use the chain rule on the first term:
\(\frac{\text{d}}{\text{d}x}[(2x - 3)^3] = 3(2x - 3)^2 \cdot \frac{\text{d}}{\text{d}x}(2x - 3) = 6(2x - 3)^2\)
So,
\(\frac{\text{d}y}{\text{d}x} = 6(2x - 3)^2 + 4\)

To find the points where the gradient is 28, we set \(\frac{\text{d}y}{\text{d}x} = 28\):
\(6(2x - 3)^2 + 4 = 28\)
\(6(2x - 3)^2 = 24\)
\((2x - 3)^2 = 4\)
\(2x - 3 = 2\) or \(2x - 3 = -2\)

Case 1:
\(2x - 3 = 2 \implies 2x = 5 \implies x = 2.5\)
Substitute \(x = 2.5\) back into the curve equation:
\(y = (2(2.5) - 3)^3 + 4(2.5) = 2^3 + 10 = 18\)
First point: \((2.5, 18)\)

Case 2:
\(2x - 3 = -2 \implies 2x = 1 \implies x = 0.5\)
Substitute \(x = 0.5\) back into the curve equation:
\(y = (2(0.5) - 3)^3 + 4(0.5) = (-2)^3 + 2 = -6\)
Second point: \((0.5, -6)\)

**(b)**
Using the chain rule for rates of change:
\(\frac{\text{d}y}{\text{d}t} = \frac{\text{d}y}{\text{d}x} \cdot \frac{\text{d}x}{\text{d}t}\)

We are given \(\frac{\text{d}y}{\text{d}t} = 0.15\).
We need to find \(\frac{\text{d}x}{\text{d}t}\) when \(x = 2\).

First, calculate \(\frac{\text{d}y}{\text{d}x}\) at \(x = 2\):
\(\frac{\text{d}y}{\text{d}x} = 6(2(2) - 3)^2 + 4 = 6(1)^2 + 4 = 10\)

Substitute into the rate of change formula:
\(0.15 = 10 \cdot \frac{\text{d}x}{\text{d}t}\)
\(\frac{\text{d}x}{\text{d}t} = 0.015\) units per second.

**(c)**
First, find the \(y\)-coordinate at \(x = 1\):
\(y = (2(1) - 3)^3 + 4(1) = (-1)^3 + 4 = 3\)
So the point of contact is \((1, 3)\).

Next, find the gradient of the tangent at \(x = 1\):
\(\frac{\text{d}y}{\text{d}x} = 6(2(1) - 3)^2 + 4 = 10\)

The gradient of the normal is the negative reciprocal of the tangent gradient:
\(m_{\text{normal}} = -\frac{1}{10} = -0.1\)

Using the point-slope form for the normal line:
\(y - 3 = -0.1(x - 1)\)
\(y - 3 = -0.1x + 0.1\)
\(y = -0.1x + 3.1\) (or \(x + 10y = 31\))

**(a)**
M1: Differentiates using chain rule to obtain \(k(2x-3)^2 + 4\)
A1: Correct derivative \(\frac{\text{d}y}{\text{d}x} = 6(2x-3)^2 + 4\)
M1: Sets their derivative equal to 28 and solves for \(x\)
A1: Both coordinates correct: \((2.5, 18)\) and \((0.5, -6)\) (accept fractional equivalents)

**(b)**
M1: Uses \(\frac{\text{d}y}{\text{d}t} = \frac{\text{d}y}{\text{d}x} \cdot \frac{\text{d}x}{\text{d}t}\) with their \(\frac{\text{d}y}{\text{d}x}\)
A1: Finds \(\frac{\text{d}y}{\text{d}x} = 10\) when \(x = 2\)
A1: Correct rate of change \(\frac{\text{d}x}{\text{d}t} = 0.015\) (or \(\frac{3}{200}\))

**(c)**
M1: Finds the \(y\)-coordinate as 3 and the tangent gradient as 10 at \(x = 1\)
M1: Uses the negative reciprocal of their tangent gradient to write the equation of the normal
A1: Correct equation, e.g., \(y = -0.1x + 3.1\) or \(x + 10y = 31\)

**(a)**
Let the first term of the arithmetic progression (AP) be \(a\) and the common difference be \(d\).
The terms of the AP are:
\(T_1 = a\)
\(T_3 = a + 2d\)
\(T_{11} = a + 10d\)

These correspond to the first three terms of a geometric progression (GP):
\(G_1 = a\)
\(G_2 = a + 2d\)
\(G_3 = a + 10d\)

Since these are in geometric progression, there is a common ratio \(r\):
\(r = \frac{G_2}{G_1} = \frac{G_3}{G_2}\)
\(\frac{a + 2d}{a} = \frac{a + 10d}{a + 2d}\)

Cross-multiplying:
\((a + 2d)^2 = a(a + 10d)\)
\(a^2 + 4ad + 4d^2 = a^2 + 10ad\)
\(4d^2 = 6ad\)

Since the common difference \(d\) is non-zero, we can divide both sides by \(2d\):
\(2d = 3a \implies d = 1.5a\) (as required)

Now, to find the common ratio \(r\):
\(r = \frac{a + 2d}{a} = \frac{a + 2(1.5a)}{a} = \frac{a + 3a}{a} = \frac{4a}{a} = 4\)

**(b)**
The formula for the sum of the first \(n\) terms of an AP is:
\(S_n = \frac{n}{2}[2a + (n - 1)d]\)

Given \(S_{20} = 3050\):
\(3050 = \frac{20}{2}[2a + 19d]\)
\(3050 = 10[2a + 19d]\)
\(305 = 2a + 19d\)

Substitute \(d = 1.5a\):
\(305 = 2a + 19(1.5a)\)
\(305 = 2a + 28.5a\)
\(305 = 30.5a\)

\(a = \frac{305}{30.5} = 10\)

Now, find \(d\):
\(d = 1.5a = 1.5(10) = 15\)

**(c)**
The first term of the GP is \(G_1 = a = 10\).
The common ratio of the GP is \(r = 4\).
The sum of the first 6 terms of a GP is:
\(S_6 = \frac{a(r^6 - 1)}{r - 1}\)
\(S_6 = \frac{10(4^6 - 1)}{4 - 1}\)
\(S_6 = \frac{10(4096 - 1)}{3}\)
\(S_6 = \frac{10(4095)}{3}\)
\(S_6 = 10 \times 1365 = 13650\)

**(a)**
M1: Expresses the 3rd and 11th terms of the AP in terms of \(a\) and \(d\)
M1: Sets up the GP ratio equation \((a+2d)^2 = a(a+10d)\) and expands correctly
A1: Achieves \(d = 1.5a\) (or equivalent fraction) with clear working
A1: Find \(r = 4\)

**(b)**
M1: Uses the sum formula for AP with \(n=20\) and sets equal to 3050
M1: Substitutes \(d = 1.5a\) to obtain an equation in \(a\) only
A1: Obtains \(a = 10\) and \(d = 15\) (both required for this mark)

**(c)**
M1: Uses the sum formula of GP with their \(a\) and \(r\) (where \(n=6\))
M1: Correctly evaluates \(4^6\) (or equivalent working)
A1: Obtains 13650