2024 Cambridge IGCSE Mathematics - Additional (0606) Practice Paper with Answers

The amplitude \(a\) is calculated as half the difference between the maximum and minimum values:
\(a = \frac{5 - (-1)}{2} = 3\).

The vertical shift \(c\) is the average of the maximum and minimum values:
\(c = \frac{5 + (-1)}{2} = 2\).

Since \(b > 0\) and the graph completes one full period of the cosine function between \(0^\circ\) and \(120^\circ\), the period is \(120^\circ\).
Using the formula for period, \(\frac{360^\circ}{b} = 120^\circ\), we find:
\(b = \frac{360^\circ}{120^\circ} = 3\).

Thus, the positive constants are \(a = 3\), \(b = 3\), and \(c = 2\).

B1: for \(a = 3\)
B1: for \(c = 2\)
B1: for \(b = 3\)

Using logarithmic properties:
\(\log_3((2x - 1)(x + 1)) = \log_3((x + 2)^2) - \log_3(3)\)

\(\log_3(2x^2 + x - 1) = \log_3\left(\frac{(x+2)^2}{3}\right)\)

Equating the arguments:
\(2x^2 + x - 1 = \frac{x^2 + 4x + 4}{3}\)

\(3(2x^2 + x - 1) = x^2 + 4x + 4\)

\(6x^2 + 3x - 3 = x^2 + 4x + 4\)

\(5x^2 - x - 7 = 0\)

Solving using the quadratic formula:
\(x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(5)(-7)}}{2(5)}\)

\(x = \frac{1 \pm \sqrt{1 + 140}}{10} = \frac{1 \pm \sqrt{141}}{10}\)

Since \(\log_3(2x - 1)\) requires \(2x - 1 > 0 \implies x > 0.5\), the negative root is extraneous.

Therefore, the only valid solution is:
\(x = \frac{1 + \sqrt{141}}{10}\).

M1: Use power/addition rules of logarithms to obtain \(\log_3((2x-1)(x+1)) = \log_3\left(\frac{(x+2)^2}{3}\right)\) or equivalent.
M1: Equate arguments to form a quadratic equation, e.g., \(5x^2 - x - 7 = 0\).
M1: Solve their 3-term quadratic equation to find two roots.
A1: Identify the correct positive root \(x = \frac{1 + \sqrt{141}}{10}\) and reject the negative extraneous root.

First, find the value of \(x\) when \(y = 2\):
\(2 = \frac{10}{2x + 1} \implies 2x + 1 = 5 \implies x = 2\).

Next, differentiate \(y\) with respect to \(x\):
\(y = 10(2x+1)^{-1}\)
\(\frac{\text{d}y}{\text{d}x} = -10(2x+1)^{-2} \cdot 2 = -\frac{20}{(2x+1)^2}\).

Evaluate \(\frac{\text{d}y}{\text{d}x}\) at \(x = 2\):
\(\frac{\text{d}y}{\text{d}x} = -\frac{20}{(2(2)+1)^2} = -\frac{20}{25} = -0.8\).

Use the chain rule to find \(\frac{\text{d}y}{\text{d}t}\):
\(\frac{\text{d}y}{\text{d}t} = \frac{\text{d}y}{\text{d}x} \cdot \frac{\text{d}x}{\text{d}t}\)
\(\frac{\text{d}y}{\text{d}t} = -0.8 \cdot 0.3 = -0.24\).

Thus, the rate of change of \(y\) is \(-0.24\) units per second.

M1: for finding \(x = 2\) when \(y = 2\).
M1: for correctly differentiating to find \(\frac{\text{d}y}{\text{d}x} = -\frac{20}{(2x+1)^2}\) and substituting \(x=2\) to find \(\frac{\text{d}y}{\text{d}x} = -0.8\).
A1: for using the chain rule with \(\frac{\text{d}x}{\text{d}t} = 0.3\) to obtain \(-0.24\).

To find the stationary points, we first differentiate \(y = \frac{\mathrm{e}^{2x - 1}}{x^2}\) with respect to \(x\) using the quotient rule:

\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2\mathrm{e}^{2x - 1}(x^2) - \mathrm{e}^{2x - 1}(2x)}{(x^2)^2}\)

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

For a stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):

\(\frac{2\mathrm{e}^{2x - 1}(x - 1)}{x^3} = 0 \implies x - 1 = 0 \implies x = 1\)

Substituting \(x = 1\) back into the curve equation to find \(y\):

\(y = \frac{\mathrm{e}^{2(1) - 1}}{1^2} = \mathrm{e}\)

So the coordinates of the stationary point are \((1, \mathrm{e})\).

To determine the nature of the stationary point, we look at the sign of the gradient on either side of \(x=1\):

For \(x = 0.9\), \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2\mathrm{e}^{0.8}(-0.1)}{0.9^3} < 0\).

For \(x = 1.1\), \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2\mathrm{e}^{1.2}(0.1)}{1.1^3} > 0\).

Since the gradient changes from negative to positive, the stationary point \((1, \mathrm{e})\) is a local minimum.

M1: Attempt to differentiate \(y\) using the quotient or product rule.
A1: Correct derivative \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2\mathrm{e}^{2x - 1}(x - 1)}{x^3}\).
M1: Equates their derivative to 0 to solve for \(x\).
A1: Finds \(x = 1\) and \(y = \mathrm{e}\).
M1: Clear method to determine nature (first derivative test or second derivative test).
A1: Shows that \((1, \mathrm{e})\) is a minimum.

Using the trigonometric identity \(\tan^2(\theta) = \sec^2(\theta) - 1\):

\(\int_{0}^{\frac{\pi}{6}} \tan^2(2x) \, \mathrm{d}x = \int_{0}^{\frac{\pi}{6}} (\sec^2(2x) - 1) \, \mathrm{d}x\)

Now we integrate each term:

\(\int (\sec^2(2x) - 1) \, \mathrm{d}x = \left[ \frac{1}{2}\tan(2x) - x \right]_{0}^{\frac{\pi}{6}}\)

Substituting the upper limit \(x = \frac{\pi}{6}\):

\(\frac{1}{2}\tan\left(2 \cdot \frac{\pi}{6}\right) - \frac{\pi}{6} = \frac{1}{2}\tan\left(\frac{\pi}{3}\right) - \frac{\pi}{6} = \frac{\sqrt{3}}{2} - \frac{\pi}{6}\)

Substituting the lower limit \(x = 0\):

\(\frac{1}{2}\tan(0) - 0 = 0\)

Thus, the exact value is \(\frac{\sqrt{3}}{2} - \frac{\pi}{6}\).

M1: Uses identity \(\tan^2(2x) = \sec^2(2x) - 1\).
M1: Integrates \(\sec^2(2x)\) to obtain \(k\tan(2x)\) where \(k = \frac{1}{2}\).
A1: Correct integral \(\frac{1}{2}\tan(2x) - x\).
M1: Applies limits correctly to their integrated expression.
A1: Obtains the exact answer \(\frac{\sqrt{3}}{2} - \frac{\pi}{6}\).

(i) First, find \(\frac{\mathrm{d}y}{\mathrm{d}x}\):

\(\frac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 - 18x + 12\)

Set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) to find the stationary points:

\(6(x^2 - 3x + 2) = 0 \implies 6(x - 1)(x - 2) = 0 \implies x = 1 \text{ or } x = 2\).

Find the corresponding \(y\)-values:

For \(x = 1\): \(y = 2(1)^3 - 9(1)^2 + 12(1) + c = 5 + c\).

For \(x = 2\): \(y = 2(2)^3 - 9(2)^2 + 12(2) + c = 4 + c\).

So the stationary points are \((1, 5 + c)\) and \((2, 4 + c)\).

(ii) At \(x = 3\), the point on the curve is \(P(3, c + 9)\).

The gradient of the tangent at \(x = 3\) is:

\(\frac{\mathrm{d}y}{\mathrm{d}x} = 6(3)^2 - 18(3) + 12 = 12\).

Therefore, the gradient of the normal at \(x = 3\) is \(-\frac{1}{12}\).

The equation of the normal is:

\(y - (c + 9) = -\frac{1}{12}(x - 3)\)

Since the normal passes through the origin \((0, 0)\):

\(0 - (c + 9) = -\frac{1}{12}(0 - 3) \implies -(c + 9) = \frac{1}{4} \implies c = -\frac{37}{4}\).

Part (i):
M1: Differentiates curve correctly to get \(\frac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 - 18x + 12\).
A1: Solves \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) to find \(x = 1, 2\) and coordinates \((1, 5+c)\), \((2, 4+c)\).

Part (ii):
B1: Finds \(y = c + 9\) at \(x = 3\).
M1: Finds gradient of tangent and hence normal is \(-\frac{1}{12}\).
M1: Formulates normal equation and substitutes \((0,0)\).
A1: Obtains \(c = -\frac{37}{4}\) (or \(-9.25\)).

(i) The volume of a cylinder is given by \(V = \pi r^2 h = 108\pi\), which gives:

\(h = \frac{108}{r^2}\)

The total surface area of a solid cylinder is:

\(A = 2\pi r^2 + 2\pi r h\)

Substituting \(h = \frac{108}{r^2}\) into the area formula:

\(A = 2\pi r^2 + 2\pi r \left(\frac{108}{r^2}\right) = 2\pi r^2 + \frac{216\pi}{r}\) [AG]

(ii) To find the rate of change of area, we use the chain rule:

\(\frac{\mathrm{d}A}{\mathrm{d}t} = \frac{\mathrm{d}A}{\mathrm{d}r} \times \frac{\mathrm{d}r}{\mathrm{d}t}\)

First, differentiate \(A\) with respect to \(r\):

\(\frac{\mathrm{d}A}{\mathrm{d}r} = 4\pi r - \frac{216\pi}{r^2}\)

When \(r = 6\):

\(\frac{\mathrm{d}A}{\mathrm{d}r} = 4\pi(6) - \frac{216\pi}{6^2} = 24\pi - 6\pi = 18\pi\)

Given \(\frac{\mathrm{d}r}{\mathrm{d}t} = 0.5 \text{ cm/s}\):

\(\frac{\mathrm{d}A}{\mathrm{d}t} = 18\pi \times 0.5 = 9\pi \approx 28.3 \text{ cm}^2/\text{s}\).

Part (i):
M1: Uses volume formula to express \(h\) in terms of \(r\).
A1: Correct substitution and algebraic simplification to show the given formula for \(A\).

Part (ii):
M1: Differentiates \(A\) with respect to \(r\).
A1: Correctly evaluates \(\frac{\mathrm{d}A}{\mathrm{d}r} = 18\pi\) at \(r=6\).
M1: Applies chain rule \(\frac{\mathrm{d}A}{\mathrm{d}t} = \frac{\mathrm{d}A}{\mathrm{d}r} \times \frac{\mathrm{d}r}{\mathrm{d}t}\).
A1: Obtains \(9\pi\) (or awrt \(28.3\)).

First, find the points of intersection by setting the curve equation equal to the line equation:

\(8 - x^2 = 2x \implies x^2 + 2x - 8 = 0\)

\((x + 4)(x - 2) = 0 \implies x = -4 \text{ or } x = 2\).

So the limits of integration are \(x = -4\) and \(x = 2\).

Area = \(\int_{-4}^{2} \left((8 - x^2) - 2x\right) \, \mathrm{d}x\)

\(\int_{-4}^{2} (8 - 2x - x^2) \, \mathrm{d}x = \left[ 8x - x^2 - \frac{x^3}{3} \right]_{-4}^{2}\)

Substitute the upper limit \(x = 2\):

\(8(2) - (2)^2 - \frac{2^3}{3} = 16 - 4 - \frac{8}{3} = \frac{28}{3}\)

Substitute the lower limit \(x = -4\):

\(8(-4) - (-4)^2 - \frac{(-4)^3}{3} = -32 - 16 - \left(-\frac{64}{3}\right) = -\frac{80}{3}\)

Subtracting the lower limit value from the upper limit value:

Area = \(\frac{28}{3} - \left(-\frac{80}{3}\right) = \frac{108}{3} = 36\).

M1: Equates equations and attempts to solve the quadratic to find limits.
A1: Obtains \(x = -4\) and \(x = 2\).
M1: Correct formulation of integration \(\int_{-4}^{2} (8 - x^2 - 2x) \, \mathrm{d}x\).
A1: Correct integration to get \(8x - x^2 - \frac{x^3}{3}\).
M1: Substitutes their limits correctly into their integrated expression.
A1: Obtains area = 36.

To find the stationary points, we differentiate \(y = \mathrm{e}^{-2x} \cos(2x)\) using the product rule:

\(\frac{\mathrm{d}y}{\mathrm{d}x} = (-2\mathrm{e}^{-2x})\cos(2x) + \mathrm{e}^{-2x}(-2\sin(2x))\)

\(\frac{\mathrm{d}y}{\mathrm{d}x} = -2\mathrm{e}^{-2x}(\cos(2x) + \sin(2x))\)

For stationary points, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):

\(-2\mathrm{e}^{-2x}(\cos(2x) + \sin(2x)) = 0\)

Since \(\mathrm{e}^{-2x} \ne 0\):

\(\cos(2x) + \sin(2x) = 0 \implies \sin(2x) = -\cos(2x) \implies \tan(2x) = -1\)

Since \(0 \le x \le \pi\), we have \(0 \le 2x \le 2\pi\).

The solutions for \(\tan(2x) = -1\) in this range are:

\(2x = \frac{3\pi}{4}\) and \(2x = \frac{7\pi}{4}\)

Thus, the exact \(x\)-coordinates are:

\(x = \frac{3\pi}{8}\) and \(x = \frac{7\pi}{8}\).

M1: Attempts product rule to differentiate \(y\).
A1: Correct derivative \(\frac{\mathrm{d}y}{\mathrm{d}x} = -2\mathrm{e}^{-2x}(\cos(2x) + \sin(2x))\).
M1: Equates their derivative to 0 and simplifies to \(\tan(2x) = -1\).
M1: Finds valid angles for \(2x\) in the interval \([0, 2\pi]\).
A1: Obtains exact coordinates \(x = \frac{3\pi}{8}\) and \(x = \frac{7\pi}{8}\).

(i) Let \(y = \frac{x}{\sqrt{4x + 1}} = \frac{x}{(4x+1)^{1/2}}\).
Using the quotient rule with \(u = x\) and \(v = (4x + 1)^{1/2}\):

\(\frac{\mathrm{d}u}{\mathrm{d}x} = 1\)

\(\frac{\mathrm{d}v}{\mathrm{d}x} = \frac{1}{2}(4x + 1)^{-1/2}(4) = \frac{2}{\sqrt{4x + 1}}\)

\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\sqrt{4x + 1}(1) - x\left(\frac{2}{\sqrt{4x+1}}\right)}{4x+1}\)

Multiplying the numerator and denominator by \(\sqrt{4x+1}\):

\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(4x+1) - 2x}{(4x+1)^{3/2}} = \frac{2x + 1}{(4x+1)^{3/2}}\). [AG]

(ii) Notice that the integrand can be rewritten by factoring out 3:

\(\frac{6x + 3}{(4x+1)^{3/2}} = 3 \left( \frac{2x + 1}{(4x+1)^{3/2}} \right)\)

Therefore, using the result from part (i):

\(\int_{0}^{2} \frac{6x + 3}{(4x + 1)^{3/2}} \, \mathrm{d}x = 3 \left[ \frac{x}{\sqrt{4x + 1}} \right]_{0}^{2}\)

Evaluating at the limits:

At \(x = 2\): \(3 \left(\frac{2}{\sqrt{4(2) + 1}}\right) = 3 \left(\frac{2}{3}\right) = 2\)

At \(x = 0\): \(3 \left(\frac{0}{\sqrt{1}}\right) = 0\)

Thus, the exact value is \(2 - 0 = 2\).

Part (i):
M1: Attempt quotient rule (or product rule with \((4x+1)^{-1/2}\)).
M1: Correctly differentiates \((4x+1)^{1/2}\).
A1: Correct simplification to reach \(\frac{2x + 1}{(4x+1)^{3/2}}\).

Part (ii):
M1: Recognises the integral is \(3 \int \frac{2x + 1}{(4x+1)^{3/2}} \, \mathrm{d}x\).
A1: Correctly integrates to get \(3 \left( \frac{x}{\sqrt{4x + 1}} \right)\).
A1: Substitutes limits 0 and 2 correctly to obtain the final exact value of 2.

(a) Given \(y = \frac{1}{3}(2x-1)^3 - 8x\), we differentiate with respect to \(x\):

\(\frac{dy}{dx} = \frac{1}{3} \cdot 3(2x-1)^2 \cdot 2 - 8 = 2(2x-1)^2 - 8\).

To find the stationary points, set \(\frac{dy}{dx} = 0\):

\(2(2x-1)^2 - 8 = 0 \implies (2x-1)^2 = 4 \implies 2x-1 = \pm 2\).

This gives two values for \(x\):

\(2x - 1 = 2 \implies 2x = 3 \implies x = 1.5\),

\(2x - 1 = -2 \implies 2x = -1 \implies x = -0.5\).

Now, find the corresponding \(y\)-coordinates:

For \(x = 1.5\):
\(y = \frac{1}{3}(2(1.5)-1)^3 - 8(1.5) = \frac{1}{3}(2)^3 - 12 = \frac{8}{3} - 12 = -\frac{28}{3} \approx -9.33\).

For \(x = -0.5\):
\(y = \frac{1}{3}(2(-0.5)-1)^3 - 8(-0.5) = \frac{1}{3}(-2)^3 + 4 = -\frac{8}{3} + 4 = \frac{4}{3} \approx 1.33\).

So the stationary points are \((1.5, -\frac{28}{3})\) and \((-0.5, \frac{4}{3})\).

(b) To determine the nature of these points, find the second derivative:

\(\frac{d^2y}{dx^2} = \frac{d}{dx}[2(2x-1)^2 - 8] = 4(2x-1) \cdot 2 = 8(2x-1)\).

At \(x = 1.5\):
\(\frac{d^2y}{dx^2} = 8(2(1.5)-1) = 8(2) = 16 > 0\).
Since the second derivative is positive, \((1.5, -\frac{28}{3})\) is a local minimum point.

At \(x = -0.5\):
\(\frac{d^2y}{dx^2} = 8(2(-0.5)-1) = 8(-2) = -16 < 0\).
Since the second derivative is negative, \((-0.5, \frac{4}{3})\) is a local maximum point.

Part (a):
- M1: Attempt to differentiate \(y\) with respect to \(x\) (at least one term correct, applying chain rule to the bracketed term).
- A1: Correct first derivative \(\frac{dy}{dx} = 2(2x-1)^2 - 8\).
- M1: Set \(\frac{dy}{dx} = 0\) and solve for \(x\).
- A1: Both pairs of coordinates correct: \((1.5, -\frac{28}{3})\) (or \((1.5, -9.33)\)) and \((-0.5, \frac{4}{3})\) (or \((-0.5, 1.33)\)).

Part (b):
- M1: Differentiate \(\frac{dy}{dx}\) to find the second derivative.
- A1: Correct second derivative expression: \(\frac{d^2y}{dx^2} = 16x - 8\) or \(8(2x-1)\).
- A1: Correctly evaluates \(\frac{d^2y}{dx^2}\) at \(x = 1.5\) and concludes it is a minimum point (with supporting working).
- A1: Correctly evaluates \(\frac{d^2y}{dx^2}\) at \(x = -0.5\) and concludes it is a maximum point (with supporting working).

(a) For the geometric progression:
The sum to infinity is \(S_\infty = \frac{a}{1-r} = 8 \implies a = 8(1-r)\).
The third term of the geometric progression is \(u_3 = a r^2\).

For the arithmetic progression:
The sum of the first 5 terms is \(S_5 = \frac{5}{2}(2a + 4d) = 5(a + 2d) = 5 \implies a + 2d = 1\).
The third term of the arithmetic progression is \(v_3 = a + 2d = 1\).

Since the third term of the geometric progression equals the third term of the arithmetic progression:
\(u_3 = v_3 \implies a r^2 = 1\).

Substitute \(a = 8(1-r)\) into \(a r^2 = 1\):
\(8(1-r)r^2 = 1 \implies 8r^2(1-r) = 1 \implies 8r^2 - 8r^3 = 1 \implies 8r^3 - 8r^2 + 1 = 0\).

(b) We are given that \(r\) is a rational number. Let's find the rational root of the cubic equation \(8r^3 - 8r^2 + 1 = 0\).

By testing simple rational roots of the form \(\frac{p}{q}\) where \(p\) divides 1 and \(q\) divides 8 (so \(r \in \{\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{8}\}\)):
Testing \(r = \frac{1}{2}\):
\(8(\frac{1}{2})^3 - 8(\frac{1}{2})^2 + 1 = 8(\frac{1}{8}) - 8(\frac{1}{4}) + 1 = 1 - 2 + 1 = 0\).
Thus, \(r = \frac{1}{2} = 0.5\) is a root.

Using polynomial division or synthetic division, we can factorise the cubic expression:
\(8r^3 - 8r^2 + 1 = (2r-1)(4r^2 - 2r - 1) = 0\).

The quadratic factor \(4r^2 - 2r - 1 = 0\) gives non-rational roots:
\(r = \frac{2 \pm \sqrt{20}}{8} = \frac{1 \pm \sqrt{5}}{4}\).
Since \(r\) must be rational, we have:
\(r = 0.5\).

Now, substitute \(r = 0.5\) into the expression for \(a\):
\(a = 8(1-0.5) = 8(0.5) = 4\).

Finally, substitute \(a = 4\) into \(a + 2d = 1\):
\(4 + 2d = 1 \implies 2d = -3 \implies d = -1.5\).

Part (a):
- B1: States or uses \(a = 8(1-r)\) from the sum to infinity formula.
- B1: States or uses \(a + 2d = 1\) from the AP sum formula (or identifies the 3rd term of the AP as \(a+2d = 1\)).
- M1: Equates the third terms: \(ar^2 = a+2d\) and sets \(ar^2 = 1\).
- A1: Correctly substitutes \(a = 8(1-r)\) and simplifies to obtain the given cubic equation: \(8r^3 - 8r^2 + 1 = 0\).

Part (b):
- M1: Attempts to find a rational root of the cubic equation, or factorise it as \((2r-1)(4r^2-2r-1)=0\).
- A1: Identifies the rational root \(r = 0.5\) (or \(\frac{1}{2}\)).
- A1: Finds the correct value of \(a = 4\).
- A1: Finds the correct value of \(d = -1.5\).
- B1: Confirms that other roots of the cubic are irrational, ensuring \(r = 0.5\) is the unique rational solution.

The midpoint of \(PQ\) is \(\left(\frac{2+6}{2}, \frac{-1+7}{2}\right) = (4, 3)\).
The gradient of the line \(PQ\) is \(\frac{7 - (-1)}{6 - 2} = \frac{8}{4} = 2\).
The gradient of the perpendicular bisector is \(-\frac{1}{2}\).
The equation of the perpendicular bisector is:
\(y - 3 = -\frac{1}{2}(x - 4)\)
\(y = -\frac{1}{2}x + 5\)

M1: For finding the midpoint \((4, 3)\)
M1: For finding the gradient of the perpendicular line as \(-\frac{1}{2}\)
A1: For the correct equation \(y = -\frac{1}{2}x + 5\) oe

For the equation to have two distinct real roots, the discriminant must be greater than zero:
\((k-2)^2 - 4(1)(k+1) > 0\)
\(k^2 - 4k + 4 - 4k - 4 > 0\)
\(k^2 - 8k > 0\)
\(k(k-8) > 0\)
Therefore, \(k < 0\) or \(k > 8\).

M1: For using the discriminant \(b^2 - 4ac > 0\) correctly with their coefficients.
A1: For simplifying to the quadratic inequality \(k^2 - 8k > 0\).
M1: For finding the critical values \(0\) and \(8\).
A1: For the correct set of values \(k < 0\) or \(k > 8\) (allow equivalent notation).

Substitute \(y = 3x - 5\) into the second equation:
\(x^2 + (3x - 5)^2 = 5\)
\(x^2 + 9x^2 - 30x + 25 = 5\)
\(10x^2 - 30x + 20 = 0\)
Divide by 10:
\(x^2 - 3x + 2 = 0\)
\((x - 1)(x - 2) = 0\)
So, \(x = 1\) or \(x = 2\).

When \(x = 1\), \(y = 3(1) - 5 = -2\).
When \(x = 2\), \(y = 3(2) - 5 = 1\).

So the solutions are \(x = 1, y = -2\) and \(x = 2, y = 1\).

M1: For substituting the linear equation into the quadratic equation to obtain a single quadratic in one variable.
A1: For simplifying to the correct quadratic equation \(x^2 - 3x + 2 = 0\) oe.
M1: For solving their quadratic equation to find two values of \(x\) (or \(y\)).
A1: For both correct pairs of solutions: \((1, -2)\) and \((2, 1)\).

First, express the vector \(\overrightarrow{AB}\) in terms of \(p\):
\(\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (p\mathbf{i} + 2\mathbf{j}) - (3\mathbf{i} - 4\mathbf{j}) = (p-3)\mathbf{i} + 6\mathbf{j}\).

The magnitude of \(\overrightarrow{AB}\) is given by:
\(|\overrightarrow{AB}| = \sqrt{(p-3)^2 + 6^2} = 10\).

Squaring both sides:
\((p-3)^2 + 36 = 100\)
\((p-3)^2 = 64\)

Since \(p > 0\), we take the positive square root:
\(p - 3 = 8 \implies p = 11\).
(The other solution \(p - 3 = -8 \implies p = -5\) is rejected as \(p > 0\)).

M1: For finding the vector \(\overrightarrow{AB} = (p-3)\mathbf{i} + 6\mathbf{j}\) oe.
M1: For setting up the magnitude equation \((p-3)^2 + 6^2 = 10^2\) and attempting to solve for \(p\).
A1: For the correct value \(p = 11\).

Express \(\vec{OP}\) in two ways:

Using \(\vec{AP} = \lambda\vec{AN}\):
\(\vec{OP} = \vec{OA} + \vec{AP} = \mathbf{a} + \lambda\vec{AN}\).
Since \(\vec{ON} = \frac{1}{3}\mathbf{b}\), we have \(\vec{AN} = \vec{ON} - \vec{OA} = \frac{1}{3}\mathbf{b} - \mathbf{a}\).
Thus, \(\vec{OP} = \mathbf{a} + \lambda\left(\frac{1}{3}\mathbf{b} - \mathbf{a}\right) = (1 - \lambda)\mathbf{a} + \frac{1}{3}\lambda\mathbf{b}\).

Using \(\vec{BP} = \mu\vec{BM}\):
\(\vec{OP} = \vec{OB} + \vec{BP} = \mathbf{b} + \mu\vec{BM}\).
Since \(\vec{OM} = \frac{1}{2}\mathbf{a}\), we have \(\vec{BM} = \vec{OM} - \vec{OB} = \frac{1}{2}\mathbf{a} - \mathbf{b}\).
Thus, \(\vec{OP} = \mathbf{b} + \mu\left(\frac{1}{2}\mathbf{a} - \mathbf{b}\right) = \frac{1}{2}\mu\mathbf{a} + (1 - \mu)\mathbf{b}\).

Equating the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\):
\(1 - \lambda = \frac{1}{2}\mu\)
\(\frac{1}{3}\lambda = 1 - \mu\)

From the first equation, \(\mu = 2 - 2\lambda\).
Substitute into the second equation:
\(\frac{1}{3}\lambda = 1 - (2 - 2\lambda) \implies \frac{1}{3}\lambda = 2\lambda - 1 \implies \frac{5}{3}\lambda = 1 \implies \lambda = \frac{3}{5}\).
Using \(\lambda = \frac{3}{5}\), we find \(\mu = 2 - 2\left(\frac{3}{5}\right) = \frac{4}{5}\).

Substituting \(\lambda = \frac{3}{5}\) back into the expression for \(\vec{OP}\):
\(\vec{OP} = \left(1 - \frac{3}{5}\right)\mathbf{a} + \frac{1}{3}\left(\frac{3}{5}\right)\mathbf{b} = \frac{2}{5}\mathbf{a} + \frac{1}{5}\mathbf{b}\).

B1: Finds \(\vec{AN} = \frac{1}{3}\mathbf{b} - \mathbf{a}\) or \(\vec{BM} = \frac{1}{2}\mathbf{a} - \mathbf{b}\)
M1: Expresses \(\vec{OP}\) in terms of \(\mathbf{a}\), \(\mathbf{b}\) and either \(\lambda\) or \(\mu\)
M1: Sets up simultaneous equations by equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\)
A1: Solves for \(\lambda = \frac{3}{5}\)
A1: Solves for \(\mu = \frac{4}{5}\)
A1: Expresses \(\vec{OP} = \frac{2}{5}\mathbf{a} + \frac{1}{5}\mathbf{b}\)

Using the double angle identity \(\sin 2A = 2 \sin A \cos A\), we can rewrite the equation as:
\(2(2 \sin 2\theta \cos 2\theta) = \sqrt{3}\)
\(2 \sin 4\theta = \sqrt{3}\)
\(\sin 4\theta = \frac{\sqrt{3}}{2}\).

Given that \(0 \le \theta \le \pi\), the range for \(4\theta\) is \(0 \le 4\theta \le 4\pi\).
Find the principal and secondary angles for \(\sin 4\theta = \frac{\sqrt{3}}{2}\):
In the interval \([0, 2\pi]\):
\(4\theta = \frac{\pi}{3}\) (1st quadrant)
\(4\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\) (2nd quadrant)

In the interval \([2\pi, 4\pi]\):
\(4\theta = \frac{\pi}{3} + 2\pi = \frac{7\pi}{3}\)
\(4\theta = \frac{2\pi}{3} + 2\pi = \frac{8\pi}{3}\)

Dividing each angle by 4:
\(\theta = \frac{\pi}{12}\)
\(\theta = \frac{2\pi}{12} = \frac{\pi}{6}\)
\(\theta = \frac{7\pi}{12}\)
\(\theta = \frac{8\pi}{12} = \frac{2\pi}{3}\)

So the solutions are \(\theta = \frac{\pi}{12}, \frac{\pi}{6}, \frac{7\pi}{12}, \frac{2\pi}{3}\).

B1: Uses the double angle formula to rewrite the equation as \(2 \sin 4\theta = \sqrt{3}\)
M1: Identifies the correct range for \(4\theta\) as \([0, 4\pi]\)
M1: Finds the two primary solutions for \(4\theta\) in \([0, 2\pi]\): \(\frac{\pi}{3}\) and \(\frac{2\pi}{3}\)
A1: Finds the other two solutions in \([2\pi, 4\pi]\): \(\frac{7\pi}{3}\) and \(\frac{8\pi}{3}\)
A2: Dividing by 4 to obtain the final four exact solutions: \(\theta = \frac{\pi}{12}, \frac{\pi}{6}, \frac{7\pi}{12}, \frac{2\pi}{3}\) (deduct 1 mark if there are extra incorrect solutions in the range)

(a) Integrate \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\) to find \(\frac{\mathrm{d}y}{\mathrm{d}x}\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \int (12x - 14) \mathrm{d}x = 6x^2 - 14x + C\).
Since there is a stationary point at \(x = 1\), \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) when \(x = 1\):
\(6(1)^2 - 14(1) + C = 0 \implies -8 + C = 0 \implies C = 8\).
So, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 - 14x + 8\).

Integrate \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to find \(y\):
\(y = \int (6x^2 - 14x + 8) \mathrm{d}x = 2x^3 - 7x^2 + 8x + D\).
Since the curve passes through \((1, 2)\):
\(2(1)^3 - 7(1)^2 + 8(1) + D = 2 \implies 3 + D = 2 \implies D = -1\).
Thus, the equation of the curve is \(y = 2x^3 - 7x^2 + 8x - 1\).

(b) To find the other stationary point, set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\(6x^2 - 14x + 8 = 0 \implies 2(3x^2 - 7x + 4) = 0 \implies 2(3x - 4)(x - 1) = 0\).
The stationary points are at \(x = 1\) and \(x = \frac{4}{3}\).
For \(x = \frac{4}{3}\), find the \(y\)-coordinate:
\(y = 2\left(\frac{4}{3}\right)^3 - 7\left(\frac{4}{3}\right)^2 + 8\left(\frac{4}{3}\right) - 1 = \frac{128}{27} - \frac{112}{9} + \frac{32}{3} - 1 = \frac{53}{27}\).
So the other stationary point is \(\left(\frac{4}{3}, \frac{53}{27}\right)\).

Second derivative test:
At \(x = \frac{4}{3}\), \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12\left(\frac{4}{3}\right) - 14 = 16 - 14 = 2\).
Since \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0\), the stationary point \(\left(\frac{4}{3}, \frac{53}{27}\right)\) is a local minimum.

Part (a) [5 marks]:
M1: Integrates \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\) to get \(\frac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 - 14x + C\)
M1: Uses the stationary point condition at \(x = 1\) to find \(C = 8\)
M1: Integrates \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to get \(y = 2x^3 - 7x^2 + 8x + D\)
M1: Uses the point \((1, 2)\) to find \(D = -1\)
A1: Writes the correct final equation: \(y = 2x^3 - 7x^2 + 8x - 1\)

Part (b) [2 marks]:
B1: Identifies \(x = \frac{4}{3}\) and calculates \(y = \frac{53}{27}\) (or equivalent decimal \(\approx 1.96\))
B1: Uses the second derivative test \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2 > 0\) to correctly conclude it is a local minimum

From the first equation:
\(2 \log_3 x - \log_3 y = 1\)
\(\log_3(x^2) - \log_3 y = 1\)
\(\log_3\left(\frac{x^2}{y}\right) = 1\)
\(\frac{x^2}{y} = 3^1 = 3\)
\(y = \frac{x^2}{3}\).

From the second equation:
\(3^{x-y} = \frac{1}{9} = 3^{-2}\)
\(x - y = -2 \implies y = x + 2\).

Equating the two expressions for \(y\):
\(\frac{x^2}{3} = x + 2\)
\(x^2 = 3x + 6\)
\(x^2 - 3x - 6 = 0\).

Using the quadratic formula:
\(x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-6)}}{2(1)} = \frac{3 \pm \sqrt{9 + 24}}{2} = \frac{3 \pm \sqrt{33}}{2}\).

Since the term \(\log_3 x\) is only defined for \(x > 0\), we must reject the negative root:
\(x = \frac{3 + \sqrt{33}}{2}\) (since \(\sqrt{33} \approx 5.74\), this value is positive, whereas \(\frac{3 - \sqrt{33}}{2}\) is negative).

Now, find \(y\):
\(y = x + 2 = \frac{3 + \sqrt{33}}{2} + 2 = \frac{7 + \sqrt{33}}{2}\).
Since \(y > 0\), this solution is valid.

Thus, the exact solutions are:
\(x = \frac{3 + \sqrt{33}}{2}\), \(y = \frac{7 + \sqrt{33}}{2}\).

M1: Applies log laws correctly to write \(\frac{x^2}{y} = 3\)
M1: Resolves indices to get \(x - y = -2\)
M1: Eliminates one variable (e.g., \(y\)) to form a quadratic equation in terms of \(x\)
A1: Obtains the correct quadratic equation: \(x^2 - 3x - 6 = 0\)
M1: Solves the quadratic equation using the quadratic formula (or completing the square)
A1: Selects the positive root \(x = \frac{3 + \sqrt{33}}{2}\) and explicitly rejects the negative root (or explains why it is rejected)
A1: Finds the correct exact value for \(y = \frac{7 + \sqrt{33}}{2}\)

(a) Given the curve equation \( y = 3 + (2x - 1)^{-2} \).

Differentiating with respect to \( x \) using the chain rule:
\( \frac{dy}{dx} = -2(2x - 1)^{-3} \cdot 2 = -4(2x - 1)^{-3} \).

At the point where \( x = 1 \):
\( y = 3 + (2(1) - 1)^{-2} = 3 + 1 = 4 \),
\( \frac{dy}{dx} = -4(2(1) - 1)^{-3} = -4 \).

The equation of the tangent is:
\( y - 4 = -4(x - 1) \implies y = -4x + 8 \).

(b) To find where the tangent line intersects the \( x \)-axis, substitute \( y = 0 \):
\( 0 = -4x + 8 \implies 4x = 8 \implies x = 2 \).

Thus, the coordinates of the intersection point are \( (2, 0) \).

(c) The exact area of the bounded region can be found by calculating the area under the curve from \( x = 1 \) to \( x = 2 \), and subtracting the area of the right-angled triangle formed below the tangent line.

First, calculate the area under the curve:
\( \int_{1}^{2} \left( 3 + (2x - 1)^{-2} \right) dx = \left[ 3x - \frac{1}{2}(2x - 1)^{-1} \right]_{1}^{2} \)
\( = \left( 3(2) - \frac{1}{2}(2(2) - 1)^{-1} \right) - \left( 3(1) - \frac{1}{2}(2(1) - 1)^{-1} \right) \)
\( = \left( 6 - \frac{1}{6} \right) - \left( 3 - \frac{1}{2} \right) \)
\( = \frac{35}{6} - \frac{5}{2} = \frac{35}{6} - \frac{15}{6} = \frac{20}{6} = \frac{10}{3} \).

Next, the area under the tangent line from \( x = 1 \) to \( x = 2 \) forms a right-angled triangle with vertices \( (1,0) \), \( (1,4) \), and \( (2,0) \):
\( \text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2 - 1) \times 4 = 2 \).

Subtract the triangle's area from the area under the curve:
\( \text{Area of region} = \frac{10}{3} - 2 = \frac{4}{3} \).

(a)
* M1: For attempting to differentiate the curve using the chain rule.
* A1: Correct derivative \( \frac{dy}{dx} = -4(2x - 1)^{-3} \).
* A1: Correct tangent equation \( y = -4x + 8 \) or equivalent.

(b)
* B1: Correct coordinates \( (2, 0) \).

(c)
* B1: For correct integration of the curve to obtain \( 3x - \frac{1}{2}(2x - 1)^{-1} \).
* M1: For correct substitution of the limits 1 and 2 into their integrated expression.
* M1: For subtracting the area of the triangle (or the integral of the tangent line) from the area under the curve.
* A1: Correct exact area of \( \frac{4}{3} \) (or equivalent).

(a) In a parallelogram \( OABC \), we have:
\( \vec{CB} = \vec{OA} = \mathbf{a} \) and \( \vec{AB} = \vec{OC} = \mathbf{c} \).

Since \( CE : EB = 1 : 2 \):
\( \vec{CE} = \frac{1}{3}\vec{CB} = \frac{1}{3}\mathbf{a} \).

Therefore,
\( \vec{OE} = \vec{OC} + \vec{CE} = \mathbf{c} + \frac{1}{3}\mathbf{a} \).

Since \( AD : DB = 3 : 1 \):
\( \vec{AD} = \frac{3}{4}\vec{AB} = \frac{3}{4}\mathbf{c} \).

Therefore,
\( \vec{OD} = \vec{OA} + \vec{AD} = \mathbf{a} + \frac{3}{4}\mathbf{c} \).

Now, express \( \vec{CD} \):
\( \vec{CD} = \vec{OD} - \vec{OC} = \left(\mathbf{a} + \frac{3}{4}\mathbf{c}\right) - \mathbf{c} = \mathbf{a} - \frac{1}{4}\mathbf{c} \).

(b) Expressing \( \vec{OP} \) using the first relation \( \vec{OP} = \lambda \vec{OE} \):
\( \vec{OP} = \lambda \left(\mathbf{c} + \frac{1}{3}\mathbf{a}\right) = \frac{1}{3}\lambda \mathbf{a} + \lambda \mathbf{c} \).

Expressing \( \vec{OP} \) using the second relation \( \vec{CP} = \mu \vec{CD} \):
\( \vec{OP} = \vec{OC} + \vec{CP} = \mathbf{c} + \mu \vec{CD} = \mathbf{c} + \mu \left(\mathbf{a} - \frac{1}{4}\mathbf{c}\right) = \mu \mathbf{a} + \left(1 - \frac{1}{4}\mu\right)\mathbf{c} \).

(c) Equating the components of \( \mathbf{a} \) and \( \mathbf{c} \) from both vector expressions for \( \vec{OP} \):
1) \( \frac{1}{3}\lambda = \mu \implies \lambda = 3\times \mu \)
2) \( \lambda = 1 - \frac{1}{4}\mu \)

Substitute equation (1) into equation (2):
\( 3\times \mu = 1 - \frac{1}{4}\mu \implies \frac{13}{4}\mu = 1 \implies \mu = \frac{4}{13} \).

Then substitute \( \mu \) back into (1) to find \( \lambda \):
\( \lambda = 3\left(\frac{4}{13}\right) = \frac{12}{13} \).

Since \( \vec{OP} = \frac{12}{13}\vec{OE} \), the point \( P \) lies \( \frac{12}{13} \) of the way along the line segment \( OE \).
Therefore, the ratio in which \( P \) divides \( OE \) is:
\( OP : PE = 12 : 1 \).

(a)
* B1: Correct vector expression for \( \vec{OE} = \mathbf{c} + \frac{1}{3}\mathbf{a} \).
* M1: For writing a correct vector pathway to find \( \vec{CD} \) (e.g., \( \vec{CO} + \vec{OA} + \vec{AD} \)).
* A1: Correct vector expression for \( \vec{CD} = \mathbf{a} - \frac{1}{4}\mathbf{c} \).

(b)
* B1: Correct first expression \( \vec{OP} = \frac{1}{3}\lambda \mathbf{a} + \lambda \mathbf{c} \).
* B1: Correct second expression \( \vec{OP} = \mu \mathbf{a} + \left(1 - \frac{1}{4}\mu\right)\mathbf{c} \).

(c)
* M1: For equating coefficients of \( \mathbf{a} \) and \( \mathbf{c} \) to set up two simultaneous equations.
* A1: Correct value of \( \mu = \frac{4}{13} \) (or equivalent decimal \( \approx 0.308 \)).
* A1: Correct value of \( \lambda = \frac{12}{13} \) (or equivalent decimal \( \approx 0.923 \)).
* B1: Correct division ratio \( OP : PE = 12 : 1 \).