2023 Cambridge IAS-Level Mathematics (9709) 模擬試題連答案詳解

The first term is \(a = 12\) and the second term is \(ar = 9x\). This gives the common ratio \(r = \frac{9x}{12} = \frac{3x}{4}\). For a geometric progression to have a sum to infinity, the common ratio must satisfy \(|r| < 1\). Therefore, we have \(\left| \frac{3x}{4} \right| < 1\), which simplifies to \(-1 < \frac{3x}{4} < 1\). Multiplying all parts of the inequality by \(4\) and dividing by \(3\) gives \(-\frac{4}{3} < x < \frac{4}{3}\).

M1: Finding the common ratio \(r\) in terms of \(x\) (e.g. \(r = \frac{9x}{12}\)). M1: Setting up the correct condition for a sum to infinity, \(|r| < 1\) (or \(-1 < r < 1\)). A1: Correct final interval \(-\frac{4}{3} < x < \frac{4}{3}\) (or equivalent).

First, we differentiate the equation of the curve with respect to \(x\) using the chain rule: \(\frac{dy}{dx} = 4(2x - 3)^3 \times 2 - 8 = 8(2x - 3)^3 - 8\). The gradient of the curve is 0 when \(\frac{dy}{dx} = 0\). Setting the derivative to 0: \(8(2x - 3)^3 - 8 = 0\), which gives \((2x - 3)^3 = 1\). Taking the cube root of both sides gives \(2x - 3 = 1\). Solving for \(x\) gives \(2x = 4\), so \(x = 2\).

M1: For differentiating the expression to find \(\frac{dy}{dx}\) (expecting \(8(2x - 3)^3 - 8\)). M1: Setting their derivative equal to 0 and attempting to solve for \(x\). A1: Correct answer \(x = 2\).

To find the equation of the curve, we integrate \(\frac{dy}{dx}\) with respect to \(x\): \(y = \int 6(3x + 4)^{-\frac{1}{2}} dx\). Using the reverse chain rule: \(y = \frac{6(3x + 4)^{\frac{1}{2}}}{\frac{1}{2} \times 3} + c = 4(3x + 4)^{\frac{1}{2}} + c\). We are given that the curve passes through the point \((4, 15)\). Substituting these coordinates into the equation: \(15 = 4(3(4) + 4)^{\frac{1}{2}} + c \implies 15 = 4\sqrt{16} + c \implies 15 = 16 + c\). This gives \(c = -1\). Therefore, the equation of the curve is \(y = 4\sqrt{3x + 4} - 1\).

M1: Attempt to integrate \(6(3x+4)^{-1/2}\) resulting in a form \(k(3x+4)^{1/2}\) where \(k\) is a constant. M1: Substitute \((4, 15)\) into their integrated equation including the constant \(c\) to find \(c\). A1: Correct equation \(y = 4\sqrt{3x + 4} - 1\) (or equivalent).

To find the equation of the curve, we integrate the derivative:

$$y = \int \left(3x^{1/2} - kx^{-2}\right) \text{d}x$$

$$y = 2x^{3/2} + \frac{k}{x} + c$$

Substitute the first point \((1, 4)\):

$$4 = 2(1)^{3/2} + \frac{k}{1} + c \implies 4 = 2 + k + c \implies k + c = 2$$

Substitute the second point \((4, 15)\):

$$15 = 2(4)^{3/2} + \frac{k}{4} + c$$

Since \(4^{3/2} = 8\):

$$15 = 16 + \frac{k}{4} + c \implies \frac{k}{4} + c = -1$$

We now solve the two simultaneous equations:
1) \(k + c = 2 \implies c = 2 - k\)
2) \(\frac{k}{4} + c = -1\)

Substitute (1) into (2):

$$\frac{k}{4} + 2 - k = -1 \implies -\frac{3}{4}k = -3 \implies k = 4$$

Substitute \(k = 4\) back to find \(c\):

$$c = 2 - 4 = -2$$

Thus, the equation of the curve is:

$$y = 2x^{3/2} + \frac{4}{x} - 2$$

M1: Attempt integration of the derivative, increasing at least one power by 1.
A1: Correct integration with constant of integration, \(2x^{3/2} + \frac{k}{x} + c\).
M1: Substitute \((1, 4)\) to form a linear equation in \(k\) and \(c\).
M1: Substitute \((4, 15)\) to form a second linear equation in \(k\) and \(c\).
A1: Solve simultaneously to obtain \(k = 4\) and \(c = -2\).
A1: State the correct final equation of the curve.

(a) Let \(r\) be the common ratio of the geometric progression. The first three terms of the geometric progression are \(a\), \(ar\), and \(ar^2\).

These are equal to the 1st, 3rd, and 9th terms of the arithmetic progression respectively:

$$a = a$$

$$a + 2d = ar \implies 2d = a(r - 1)$$

$$a + 8d = ar^2 \implies 8d = a(r^2 - 1)$$

Dividing the two equations:

$$\frac{8d}{2d} = \frac{a(r^2 - 1)}{a(r - 1)}$$

$$4 = r + 1 \implies r = 3$$

(b) Since \(r = 3\), we have:

$$2d = a(3 - 1) \implies 2d = 2a \implies d = a$$

The sum of the first 10 terms of the arithmetic progression is given by:

$$S_{10} = \frac{10}{2}[2a + 9d] = 110$$

$$5(2a + 9d) = 110 \implies 2a + 9d = 22$$

Substitute \(d = a\):

$$2a + 9a = 22 \implies 11a = 22 \implies a = 2$$

(a)
M1: Formulate equations for the terms of the GP in terms of \(a\) and \(d\), i.e., \(a + 2d = ar\) and \(a + 8d = ar^2\).
M1: Attempt to eliminate \(d\) by division or substitution.
A1: Correctly show that \(r = 3\) with clear algebraic steps shown.

(b)
M1: Use \(r = 3\) to find the relationship between \(a\) and \(d\) (\(d = a\)).
M1: Substitute into the AP sum formula for \(S_{10}\) and set equal to 110.
A1: Solve to find \(a = 2\).

(a) We complete the square for \(\text{f}(x) = 2x^2 - 12x + 13\):

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

$$\text{f}(x) = 2[(x-3)^2 - 9] + 13$$

$$\text{f}(x) = 2(x-3)^2 - 18 + 13$$

$$\text{f}(x) = 2(x-3)^2 - 5$$

So \(p = 3\) and \(q = -5\).

(b) To find the inverse, let \(y = 2(x-3)^2 - 5\):

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

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

Since the domain of \(\text{f}\) is \(x \le 3\), \(x-3 \le 0\). Therefore, we must choose the negative square root:

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

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

Replacing \(y\) with \(x\):

$$\text{f}^{-1}(x) = 3 - \sqrt{\frac{x+5}{2}}$$

To find the domain of \(\text{f}^{-1}\), we find the range of \(\text{f}\). For \(x \le 3\), the minimum value of \(\text{f}(x)\) occurs at the vertex where \(\text{f}(3) = -5\). Since the parabola opens upwards, the range of \(\text{f}\) is \(\text{f}(x) \ge -5\).

Hence, the domain of \(\text{f}^{-1}\) is \(x \ge -5\).

(a)
M1: Attempt to complete the square by factoring out 2 and completing the square inside.
A1: Correct expression \(2(x-3)^2 - 5\).

(b)
M1: Set \(y = 2(x-3)^2 - 5\) and attempt to rearrange to make \(x\) the subject.
A1: Correctly choose the negative square root because of the domain constraint \(x \le 3\).
A1: Obtain \(\text{f}^{-1}(x) = 3 - \sqrt{\frac{x+5}{2}}\).
B1: Correctly state that the domain is \(x \ge -5\).

(a) The center \(M\) of the circle is the midpoint of the diameter \(AB\):

$$M = \left(\frac{1+5}{2}, \frac{2+10}{2}\right) = (3, 6)$$

The radius \(r\) is the distance from the center \(M(3,6)\) to \(B(5,10)\):

$$r^2 = (5-3)^2 + (10-6)^2 = 2^2 + 4^2 = 4 + 16 = 20$$

Thus, the equation of the circle \(C\) is:

$$(x-3)^2 + (y-6)^2 = 20$$

(b) The gradient of the radius \(MB\) is:

$$m_{\text{radius}} = \frac{10-6}{5-3} = \frac{4}{2} = 2$$

Since the tangent is perpendicular to the radius at the point of contact, the gradient of the tangent is:

$$m_{\text{tangent}} = -\frac{1}{2}$$

Using the point \(B(5, 10)\), the equation of the tangent is:

$$y - 10 = -\frac{1}{2}(x - 5)$$

$$2y - 20 = -x + 5$$

$$x + 2y = 25$$

(a)
M1: Find the midpoint of \(AB\) to determine the center of the circle.
M1: Calculate the radius squared \(r^2\) using distance formula.
A1: Correct equation of the circle: \((x-3)^2 + (y-6)^2 = 20\).

(b)
M1: Find the gradient of \(MB\) and state that the gradient of the tangent is its negative reciprocal.
M1: Attempt to find the equation of a line through \(B(5, 10)\) with their perpendicular gradient.
A1: Correct final equation in the required integer form: \(x + 2y = 25\) (or any integer multiple).

We rewrite the given equation using the identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\):

$$3 \left(\frac{\sin \theta}{\cos \theta}\right) \sin \theta = 8$$

$$3 \sin^2 \theta = 8 \cos \theta$$

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

$$3(1 - \cos^2 \theta) = 8 \cos \theta$$

$$3 - 3 \cos^2 \theta = 8 \cos \theta$$

$$3 \cos^2 \theta + 8 \cos \theta - 3 = 0$$

Let \(u = \cos \theta\). The quadratic equation becomes:

$$3u^2 + 8u - 3 = 0$$

$$(3u - 1)(u + 3) = 0$$

This gives two solutions:

$$\cos \theta = \frac{1}{3} \quad \text{or} \quad \cos \theta = -3$$

Since \(-1 \le \cos \theta \le 1\), the equation \(\cos \theta = -3\) has no solutions.

For \(\cos \theta = \frac{1}{3}\):

$$\theta = \cos^{-1}\left(\frac{1}{3}\right) \approx 70.5^\circ$$

In the interval \(0^\circ \le \theta \le 360^\circ\), the other solution is:

$$\theta = 360^\circ - 70.5^\circ = 289.5^\circ$$

M1: Substitute \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) to obtain an equation in terms of \(\sin\) and \(\cos\).
M1: Substitute \(\sin^2 \theta = 1 - \cos^2 \theta\) to obtain an equation only in terms of \(\cos \theta\).
A1: Correct quadratic equation \(3 \cos^2 \theta + 8 \cos \theta - 3 = 0\).
M1: Attempt to solve the quadratic equation to find \(\cos \theta\).
A1: Correctly identify \(\cos \theta = \frac{1}{3}\) as the only valid solution (explicitly or implicitly rejecting \(\cos\theta = -3\)).
A1: Both correct angles \(70.5^\circ\) and \(289.5^\circ\) (rounded to 1 d.p., accept 70.5 and 290).

\(\textbf{(a)}\)
To find the equation of the curve, we integrate the gradient function:
\(y = \int \frac{6}{\sqrt{3x + 4}} \, \mathrm{d}x = \int 6(3x + 4)^{-\frac{1}{2}} \, \mathrm{d}x\)
Using the reverse chain rule:
\(y = 6 \cdot \frac{(3x + 4)^{\frac{1}{2}}}{\frac{1}{2} \cdot 3} + C\)
\(y = 4\sqrt{3x + 4} + C\)

We are given that the curve passes through \((4, 10)\). Substituting these coordinates:
\(10 = 4\sqrt{3(4) + 4} + C\)
\(10 = 4\sqrt{16} + C\)
\(10 = 16 + C \implies C = -6\)

Thus, the equation of the curve is:
\(y = 4\sqrt{3x + 4} - 6\)

\(\textbf{(b)}\)
To find the area of the region bounded by the curve, the \(x\)-axis, and the lines \(x = 0\) and \(x = 4\), we integrate \(y\) from \(0\) to \(4\):
\(\text{Area} = \int_{0}^{4} \left(4(3x + 4)^{\frac{1}{2}} - 6\right) \mathrm{d}x\)

Integrating term-by-term:
\(\int \left(4(3x + 4)^{\frac{1}{2}} - 6\right) \mathrm{d}x = \left[ 4 \cdot \frac{(3x + 4)^{\frac{3}{2}}}{\frac{3}{2} \cdot 3} - 6x \right] = \left[ \frac{8}{9}(3x + 4)^{\frac{3}{2}} - 6x \right]\)

Now, we evaluate this expression at the limits \(4\) and \(0\):
At \(x = 4\):
\(\frac{8}{9}(3(4) + 4)^{\frac{3}{2}} - 6(4) = \frac{8}{9}(16)^{\frac{3}{2}} - 24 = \frac{8}{9}(64) - 24 = \frac{512}{9} - \frac{216}{9} = \frac{296}{9}\)

At \(x = 0\):
\(\frac{8}{9}(3(0) + 4)^{\frac{3}{2}} - 6(0) = \frac{8}{9}(4)^{\frac{3}{2}} = \frac{8}{9}(8) = \frac{64}{9}\)

Subtracting the lower limit value from the upper limit value:
\(\text{Area} = \frac{296}{9} - \frac{64}{9} = \frac{232}{9}\) (or \(25\frac{7}{9}\))

\(\textbf{(a)}\)
* **M1**: Integrate \(6(3x + 4)^{-\frac{1}{2}}\right.\) to obtain \(k(3x+4)^{\frac{1}{2}}\right.\) where \(k\) is a constant.
* **A1**: Obtain \(y = 4(3x+4)^{\frac{1}{2}} + C\) (accept without \(+ C\)).
* **M1**: Substitute \((4, 10)\) into their integrated expression to find \(C\).
* **A1**: Correctly obtain \(y = 4\sqrt{3x + 4} - 6\).

\(\textbf{(b)}\)
* **M1**: Integrate their curve equation, showing the form \(m(3x+4)^{\frac{3}{2}} - 6x\) with a non-zero constant \(m\).
* **A1**: Obtain the correct integral \(\frac{8}{9}(3x + 4)^{\frac{3}{2}} - 6x\).
* **M1**: Substitute the limits \(4\) and \(0\) into their integrated expression.
* **A1**: Correctly obtain the final exact area as \(\frac{232}{9}\) (or \(25\frac{7}{9}\); accept \(25.8\) if correct working is shown).

\(\textbf{(a)}\)
We complete the square for the expression of \(f(x)\):
\(f(x) = 2(x^2 - 6x) + 13 = 2\left((x - 3)^2 - 9\right) + 13 = 2(x - 3)^2 - 5\)
The vertex of the parabola \(y = f(x)\) is at \(x = 3\).
For \(f\) to have an inverse, it must be a one-to-one function. Since the domain is defined as \(x \le k\), the largest value of \(k\) that ensures the function is strictly decreasing (and thus one-to-one) is the x-coordinate of the vertex.
Therefore, the largest value of \(k\) is \(3\).

\(\textbf{(b)}\)
Using \(k = 3\), we have the function \(f(x) = 2(x - 3)^2 - 5\) defined for \(x \le 3\).
To find \(f^{-1}(x)\), we set \(y = f(x)\) and make \(x\) the subject:
\(y = 2(x - 3)^2 - 5\)
\(y + 5 = 2(x - 3)^2\)
\((x - 3)^2 = \frac{y + 5}{2}\)

Since \(x \le 3\), it follows that \(x - 3 \le 0\). Therefore, we must take the negative square root:
\(x - 3 = -\sqrt{\frac{y+5}{2}}\)
\(x = 3 - \sqrt{\frac{y+5}{2}}\)

Replacing \(y\) with \(x\), we get:
\(f^{-1}(x) = 3 - \sqrt{\frac{x+5}{2}}\)

The domain of \(f^{-1}\) is the range of \(f\).
Since \(x \le 3\), the minimum value of \(f(x)\) occurs at the vertex where \(y = -5\), and \(f(x) \ge -5\) for all \(x \le 3\).
Thus, the domain of \(f^{-1}(x)\) is \(x \ge -5\).

\(\textbf{(c)}\)
We wish to solve \(fg(x) = 13\), which is \(f(g(x)) = 13\).
First, let \(y = g(x)\). We solve \(f(y) = 13\) subject to the domain restriction \(y \le 3\):
\(2(y - 3)^2 - 5 = 13\)
\(2(y - 3)^2 = 18\)
\((y - 3)^2 = 9\)

Since \(y \le 3\), we must have:
\(y - 3 = -3 \implies y = 0\)

Now we set \(g(x) = y = 0\):
\(3x - 1 = 0 \implies x = \frac{1}{3}\)

(Note: If we took \(y - 3 = 3 \implies y = 6\), then \(g(x) = 6 \implies 3x - 1 = 6 \implies x = \frac{7}{3}\). However, this is rejected because \(g(x) = 6\) is outside the domain of \(f\), which requires \(x \le 3\).)

\(\textbf{(a)}\)
* **M1**: Complete the square to find the vertex coordinates or use differentiation to find the stationary point \(x = 3\).
* **A1**: State \(k = 3\).

\(\textbf{(b)}\)
* **M1**: Set \(y = 2(x - 3)^2 - 5\) (or their expression) and make a valid attempt to solve for \(x\).
* **A1**: Obtain \((x - 3)^2 = \frac{y + 5}{2}\) or equivalent.
* **M1**: Choose the negative square root correctly based on \(x \le 3\).
* **A1**: Express \(f^{-1}(x) = 3 - \sqrt{\frac{x+5}{2}}\) (must be in terms of \(x\)).
* **B1 (0.33)**: State correct domain \(x \ge -5\) (accept \([-5, \infty)\)).

\(\textbf{(c)}\)
* **M1**: Form the equation \(f(3x-1) = 13\) or solve \(f(y) = 13\) using completed square form.
* **M1**: Identify and apply the domain restriction to select only \(g(x) = 0\) (or reject \(g(x) = 6\)).
* **A1**: Obtain the unique correct solution \(x = \frac{1}{3}\).

\(\textbf{(a)}\)
The first, second, and fifth terms of the arithmetic progression (AP) are:
\(u_1 = a\)
\(u_2 = a + d\)
\(u_5 = a + 4d\)

These terms form the first three terms of a geometric progression (GP), so:
\(g_1 = a\)
\(g_2 = a + d\)
\(g_3 = a + 4d\)

Since they are in geometric progression, the ratio between successive terms is constant:
\(\frac{g_2}{g_1} = \frac{g_3}{g_2} \implies \frac{a + d}{a} = \frac{a + 4d}{a + d}\)

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

Subtracting \(a^2 + 2ad\) from both sides:
\(d^2 = 2ad\)

Since \(d \ne 0\), we can divide both sides by \(d\):
\(d = 2a\) (as required).

\(\textbf{(b)}\)
The common ratio \(r\) of the geometric progression is given by:
\(r = \frac{g_2}{g_1} = \frac{a + d}{a}\)

Substituting \(d = 2a\):
\(r = \frac{a + 2a}{a} = \frac{3a}{a} = 3\)

\(\textbf{(c)}\)
We are given \(a = 2\). Since \(d = 2a\), we have \(d = 4\).

For the arithmetic progression:
We use the sum formula \(S_n = \frac{n}{2} [2a + (n-1)d]\) with \(n = 8\):
\(S_8 = \frac{8}{2} [2(2) + (8-1)(4)]\)
\(S_8 = 4 [4 + 7(4)] = 4 [4 + 28] = 4 [32] = 128\)

For the geometric progression:
The first term is \(g_1 = a = 2\) and the common ratio is \(r = 3\).
We use the sum formula \(S'_n = \frac{g_1(r^n - 1)}{r - 1}\) with \(n = 8\):
\(S'_8 = \frac{2(3^8 - 1)}{3 - 1}\)
\(S'_8 = \frac{2(6561 - 1)}{2} = 6560\)

\(\textbf{(a)}\)
* **B1**: State the terms of the AP in terms of \(a\) and \(d\) as \(a\), \(a+d\), and \(a+4d\).
* **M1**: Formulate the geometric progression equation \(\frac{a+d}{a} = \frac{a+4d}{a+d}\) (or equivalent \((a+d)^2 = a(a+4d)\)).
* **A1**: Correctly expand to obtain \(a^2 + 2ad + d^2 = a^2 + 4ad\).
* **A1**: Simplify and divide by \(d\) (justified by \(d \ne 0\)) to obtain \(d = 2a\).

\(\textbf{(b)}\)
* **M1**: Substitute \(d = 2a\) into the expression for \(r = \frac{a+d}{a}\).
* **A1 (0.33)**: Correctly evaluate \(r = 3\).

\(\textbf{(c)}\)
* **M1**: Calculate \(d = 4\) and substitute \(a = 2, d = 4\) into the AP sum formula.
* **A1**: Obtain \(S_8 = 128\) for the AP.
* **M1**: Substitute \(g_1 = 2, r = 3\) into the GP sum formula.
* **A1**: Obtain \(S_8 = 6560\) for the GP.

Take the natural logarithm of both sides:
\(\ln(3^{2x+1}) = \ln(7^{x-1})\)

Apply the power law of logarithms:
\((2x+1)\ln 3 = (x-1)\ln 7\)

Expand both sides:
\(2x\ln 3 + \ln 3 = x\ln 7 - \ln 7\)

Rearrange to collect terms in \(x\) on one side:
\(2x\ln 3 - x\ln 7 = -\ln 7 - \ln 3\)
\(x(2\ln 3 - \ln 7) = -(\ln 7 + \ln 3)\)

Express \(x\) explicitly:
\(x = \frac{-(\ln 7 + \ln 3)}{2\ln 3 - \ln 7}\)

Calculate the numerical value:
\(x = \frac{-\ln 21}{\ln 9 - \ln 7} \approx \frac{-3.04452}{0.25131} \approx -12.114\)

To 3 significant figures, \(x = -12.1\).

M1: Take logs of both sides and use power law to obtain a linear equation in \(x\).
M1: Rearrange the equation to isolate \(x\) correctly.
A1: Obtain \(x = -12.1\) (or any value rounding to \(-12.1\)).

Use the trigonometric identity \(\sec^2 \theta = 1 + \tan^2 \theta\):
\(1 + \tan^2 \theta + 2\tan \theta = 4\)

Rearrange to form a quadratic equation in \(\tan \theta\):
\(\tan^2 \theta + 2\tan \theta - 3 = 0\)

Factorise the quadratic expression:
\((\tan \theta + 3)(\tan \theta - 1) = 0\)

This gives two possible values:
\(\tan \theta = 1\) or \(\tan \theta = -3\)

Solve for \(\theta\) within the interval \(0^\circ \le \theta \le 180^\circ\):
- For \(\tan \theta = 1\):
\(\theta = 45^\circ\)
- For \(\tan \theta = -3\):
\(\theta = 180^\circ - 71.57^\circ = 108.4^\circ\) (to 1 decimal place)

Thus, the solutions are \(\theta = 45^\circ\) and \(\theta = 108.4^\circ\).

M1: Substitute \(\sec^2 \theta = 1 + \tan^2 \theta\) and form a three-term quadratic equation in \(\tan \theta\).
A1: Correctly solve the quadratic to find \(\tan \theta = 1\) and \(\tan \theta = -3\).
A1: Obtain both \(\theta = 45^\circ\) and \(\theta = 108.4^\circ\), and no other solutions in the range.

First, use the power law of logarithms to rewrite the first term:

\(2\ln(x+2) = \ln((x+2)^2)\)

The equation becomes:

\(\ln((x+2)^2) - \ln(x-1) = \ln(2x+10)\)

Next, use the subtraction law of logarithms on the left-hand side:

\(\ln\left(\frac{(x+2)^2}{x-1}\right) = \ln(2x+10)\)

Since \(\ln(A) = \ln(B) \implies A = B\), we can remove the logarithms:

\(\frac{(x+2)^2}{x-1} = 2x+10\)

Multiply both sides by \(x-1\):

\((x+2)^2 = (2x+10)(x-1)\)

Expand both sides:

\(x^2 + 4x + 4 = 2x^2 + 8x - 10\)

Rearrange to form a quadratic equation:

\(x^2 + 4x - 14 = 0\)

Apply the quadratic formula to solve for \(x\):

\(x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-14)}}{2}\)

\(x = \frac{-4 \pm \sqrt{16 + 56}}{2} = \frac{-4 \pm \sqrt{72}}{2}\)

Simplify the radical:

\(\sqrt{72} = 6\sqrt{2}\), so \(x = \frac{-4 \pm 6\sqrt{2}}{2} = -2 \pm 3\sqrt{2}\)

This can also be written in the form \(-2 \pm \sqrt{18}\).

Since the domain of \(\ln(x-1)\) requires \(x > 1\), we must check the validity of both solutions:
- For \(x = -2 - \sqrt{18} \approx -6.24\), this is less than 1, so this solution is rejected.
- For \(x = -2 + \sqrt{18} \approx 2.24\), this is greater than 1, so this is a valid solution.

Thus, the solution is \(x = -2 + \sqrt{18}\).

M1: For applying the power law of logarithms to \(2\ln(x+2)\) to get \(\ln((x+2)^2)\).
M1: For combining the terms on the left-hand side to get \(\ln\left(\frac{(x+2)^2}{x-1}\right)\).
M1: For removing logarithms to obtain a quadratic equation in \(x\).
A1: For obtaining the correct simplified quadratic equation \(x^2 + 4x - 14 = 0\) (or equivalent).
M1: For solving their quadratic equation to find roots.
A1: For selecting the correct positive root \(x = -2 + \sqrt{18}\) (or \(-2 + 3\sqrt{2}\)) and explicitly rejecting the negative root based on the domain \(x > 1\).

(i) Let \(f(x) = e^{-0.5x} + 2x - 3\).

Evaluate \(f(x)\) at the boundaries:

\(f(1.0) = e^{-0.5} + 2(1.0) - 3 = 0.60653 + 2 - 3 = -0.39347\) (to 5 d.p.)

\(f(1.5) = e^{-0.75} + 2(1.5) - 3 = 0.47237 + 3 - 3 = 0.47237\) (to 5 d.p.)

Since there is a change of sign in the continuous function \(f(x)\) between \(x = 1.0\) and \(x = 1.5\), a root of the equation \(f(x) = 0\) lies in this interval.

(ii) Apply the iterative formula \(x_{n+1} = 1.5 - 0.5 e^{-0.5 x_n}\) starting with \(x_1 = 1.2\):

\(x_2 = 1.5 - 0.5 e^{-0.5(1.2)} = 1.5 - 0.5 e^{-0.6} = 1.22559\)

\(x_3 = 1.5 - 0.5 e^{-0.5(1.22559)} = 1.5 - 0.5 e^{-0.61280} = 1.22909\)

\(x_4 = 1.5 - 0.5 e^{-0.5(1.22909)} = 1.5 - 0.5 e^{-0.61455} = 1.22956\)

\(x_5 = 1.5 - 0.5 e^{-0.5(1.22956)} = 1.5 - 0.5 e^{-0.61478} = 1.22962\)

\(x_6 = 1.5 - 0.5 e^{-0.5(1.22962)} = 1.5 - 0.5 e^{-0.61481} = 1.22963\)

Comparing \(x_5\) and \(x_6\), both values round to \(1.230\) to 3 decimal places. Thus, the root is \(1.230\) correct to 3 decimal places.

Part (i):
M1: For evaluating \(f(1.0)\) and \(f(1.5)\) with at least one correct calculation showing a sign change.
A1: For both values calculated correctly with a concluding statement mentioning the sign change and continuity.

Part (ii):
M1: For attempting to calculate \(x_2\) using the formula.
A1: For obtaining \(x_2 = 1.22559\) and \(x_3 = 1.22909\) correct to 5 d.p.
A1: For continuing the iterations to obtain \(x_4 = 1.22956\), \(x_5 = 1.22962\), and showing convergence.
A1: For stating the final root as \(1.230\) correct to 3 decimal places.

To integrate \(6 \cos^2(x)\), we use the double-angle trigonometric identity:

\(\cos^2(x) = \frac{1 + \cos(2x)}{2}\)

Substitute this into the integrand:

\(6 \cos^2(x) - \sin(2x) = 6 \left(\frac{1 + \cos(2x)}{2}\right) - \sin(2x) = 3 + 3\cos(2x) - \sin(2x)\)

Now, integrate each term with respect to \(x\):

\(\int (3 + 3\cos(2x) - \sin(2x)) \, dx = 3x + \frac{3}{2}\sin(2x) - \left(-\frac{1}{2}\cos(2x)\right)\)

\(= 3x + \frac{3}{2}\sin(2x) + \frac{1}{2}\cos(2x)\)

Evaluate this antiderivative between the limits \(0\) and \(\frac{\pi}{4}\):

At the upper limit \(x = \frac{\pi}{4}\):

\(3\left(\frac{\pi}{4}\right) + \frac{3}{2}\sin\left(2 \cdot \frac{\pi}{4}\right) + \frac{1}{2}\cos\left(2 \cdot \frac{\pi}{4}\right) = \frac{3\pi}{4} + \frac{3}{2}\sin\left(\frac{\pi}{2}\right) + \frac{1}{2}\cos\left(\frac{\pi}{2}\right)\)

Since \(\sin\left(\frac{\pi}{2}\right) = 1\) and \(\cos\left(\frac{\pi}{2}\right) = 0\):

\(\frac{3\pi}{4} + \frac{3}{2}(1) + \frac{1}{2}(0) = \frac{3\pi}{4} + \frac{3}{2}\)

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

\(3(0) + \frac{3}{2}\sin(0) + \frac{1}{2}\cos(0) = 0 + 0 + \frac{1}{2}(1) = \frac{1}{2}\)

Subtract the lower limit value from the upper limit value:

\(\left(\frac{3\pi}{4} + \frac{3}{2}\right) - \frac{1}{2} = \frac{3\pi}{4} + 1\)

Thus, the exact value of the integral is \(\frac{3\pi}{4} + 1\).

M1: For using the identity \(\cos^2(x) = \frac{1 + \cos(2x)}{2}\) to rewrite the integrand.
A1: For obtaining the simplified integrand \(3 + 3\cos(2x) - \sin(2x)\).
M1: For integrating to obtain an expression of the form \(3x + k\sin(2x) + m\cos(2x)\).
A1: For the correct integrated terms \(\frac{3}{2}\sin(2x) + \frac{1}{2}\cos(2x)\).
M1: For substituting limits \(0\) and \(\frac{\pi}{4}\) correctly into their integrated expression.
A1: For obtaining the correct exact value \(\frac{3\pi}{4} + 1\) (or equivalent exact form, e.g., \(\frac{3\pi+4}{4}\)).

(i) Since \((x - 2)\) is a factor of \(p(x)\), by the Factor Theorem we have:
\[p(2) = 0\]
\[2(2)^3 + a(2)^2 + b(2) - 6 = 0\]
\[16 + 4a + 2b - 6 = 0 \implies 4a + 2b = -10 \implies 2a + b = -5 \quad \text{--- (Equation 1)}\]

When divided by \((x + 1)\), the remainder is \(-15\). By the Remainder Theorem:
\[p(-1) = -15\]
\[2(-1)^3 + a(-1)^2 + b(-1) - 6 = -15\]
\[-2 + a - b - 6 = -15 \implies a - b = -7 \quad \text{--- (Equation 2)}\]

Adding Equation 1 and Equation 2:
\[(2a + b) + (a - b) = -5 + (-7)\]
\[3a = -12 \implies a = -4\]

Substituting \(a = -4\) into Equation 2:
\[-4 - b = -7 \implies b = 3\]

Thus, \(a = -4\) and \(b = 3\).

(ii) Substituting the values of \(a\) and \(b\) back into \(p(x)\):
\[p(x) = 2x^3 - 4x^2 + 3x - 6\]

Since \((x - 2)\) is a factor, we can factorise by grouping or division:
\[p(x) = 2x^2(x - 2) + 3(x - 2)\]
\[p(x) = (x - 2)(2x^2 + 3)\]

Since \(2x^2 + 3 = 0\) has no real roots, the complete factorization over the real numbers is \((x - 2)(2x^2 + 3)\).

(iii) We are asked to solve the equation:
\[2(e^y)^3 - 4(e^y)^2 + 3e^y - 6 = 0\]

Letting \(x = e^y\), the equation becomes:
\[2x^3 - 4x^2 + 3x - 6 = 0\]

From part (ii), the factored form is:
\[(x - 2)(2x^2 + 3) = 0\]

Since \(2x^2 + 3 = 0\) has no real solutions (as \(x^2 = -1.5\) is impossible for real \(x\)), the only real solution is:
\[x = 2\]

Substituting back \(x = e^y\):
\[e^y = 2 \implies y = \ln 2\]

(i)
- M1: Attempt to apply the Factor Theorem with \(p(2) = 0\) to get a linear equation in \(a\) and \(b\).
- A1: Correctly obtain \(2a + b = -5\) (or equivalent).
- M1: Attempt to apply the Remainder Theorem with \(p(-1) = -15\) to get a second linear equation in \(a\) and \(b\).
- A1: Solve simultaneously to obtain \(a = -4\) and \(b = 3\).

(ii)
- M1: Attempt division or factorisation by grouping for \(2x^3 - 4x^2 + 3x - 6\).
- A1: Correctly factor out \((x - 2)\) to obtain \((x - 2)(2x^2 + 3)\).
- A1: Conclude that \(2x^2 + 3\) cannot be factorised further over real numbers.

(iii)
- M1: Recognize the substitution \(x = e^y\) and relate to the equation in part (ii).
- M1: State that \(e^y = 2\) is the only real solution (justifying that \(2e^{2y} + 3 = 0\) has no real solution).
- A1: Obtain the final exact answer \(y = \ln 2\).

(i) Let \(f(x) = 3\cos(x) - x\).
We evaluate the function at \(x = 1.1\) and \(x = 1.2\) (using radian mode):
\[f(1.1) = 3\cos(1.1) - 1.1 \approx 3(0.45360) - 1.1 = 1.36080 - 1.1 = 0.26080 > 0\]
\[f(1.2) = 3\cos(1.2) - 1.2 \approx 3(0.36236) - 1.2 = 1.08708 - 1.2 = -0.11292 < 0\]

Since there is a change of sign and the function \(f(x)\) is continuous, the equation \(3\cos(x) - x = 0\) has a root between 1.1 and 1.2, meaning the curve intersects the \(x\)-axis in this interval.

(ii) Rearranging the equation:
\[3\cos(x) - x = 0\]
\[3\cos(x) = x\]
\[\cos(x) = \frac{x}{3}\]
Taking the inverse cosine of both sides:
\[x = \arccos\left(\frac{x}{3}\right)\]

(iii) We apply the iteration formula \(x_{n+1} = \arccos\left(\frac{x_n}{3}\right)\) with \(x_1 = 1.15\):
- \(x_1 = 1.15\)
- \(x_2 = \arccos\left(\frac{1.15}{3}\right) \approx 1.17724\)
- \(x_3 = \arccos\left(\frac{1.17724}{3}\right) \approx 1.16738\)
- \(x_4 = \arccos\left(\frac{1.16738}{3}\right) \approx 1.17096\)
- \(x_5 = \arccos\left(\frac{1.17096}{3}\right) \approx 1.16966\)
- \(x_6 = \arccos\left(\frac{1.16966}{3}\right) \approx 1.17013\)
- \(x_7 = \arccos\left(\frac{1.17013}{3}\right) \approx 1.16996\)
- \(x_8 = \arccos\left(\frac{1.16996}{3}\right) \approx 1.17002\)

Since the successive approximations round to \(1.170\), the root is \(1.170\) correct to 3 decimal places.

(iv) The gradient of the line \(y = 2x + 5\) is \(m_1 = 2\).
Since the tangent is perpendicular to this line, the gradient of the tangent is:
\[m_2 = -\frac{1}{m_1} = -\frac{1}{2}\]

The gradient of the curve is found by differentiating \(y = 3\cos(x) - x\):
\[\frac{dy}{dx} = -3\sin(x) - 1\]

Setting the gradient equal to \(-\frac{1}{2}\):
\[-3\sin(x) - 1 = -\frac{1}{2}\]
\[-3\sin(x) = \frac{1}{2}\]
\[\sin(x) = -\frac{1}{6}\]

Hence, \(k = -\frac{1}{6}\).

To find the two smallest positive values of \(x\):
Since \(\sin(x) = -\frac{1}{6} < 0\), the solutions lie in the third and fourth quadrants.

The basic angle is:
\[\alpha = \arcsin\left(\frac{1}{6}\right) \approx 0.16745 \text{ radians}\]

- The first positive value (3rd quadrant) is:
\[x = \pi + \alpha \approx 3.14159 + 0.16745 = 3.3090 \approx 3.31\]

- The second positive value (4th quadrant) is:
\[x = 2\pi - \alpha \approx 6.28319 - 0.16745 = 6.1157 \approx 6.12\]

So the two smallest positive values are \(x = 3.31\) and \(x = 6.12\) (to 3 sig figs).

(i)
- M1: Substitute \(1.1\) and \(1.2\) into \(3\cos(x) - x\) (must use radian mode).
- A1: Obtain correct values of approximately \(0.261\) and \(-0.113\), and make a conclusion pointing out the change of sign.

(ii)
- B1: Complete algebraic steps to show the rearrangement to the given form.

(iii)
- M1: Calculate at least two iterations correctly.
- A1: Produce sequence of values \(1.17724, 1.16738, 1.17096, 1.16966, ...\) (allow small rounding variances in the 5th decimal place).
- A1: State the final root as \(1.170\) and show sufficient iterations to justify this accuracy.

(iv)
- M1: State that the perpendicular gradient is \(-1/2\) and differentiate the curve to get \(-3\sin(x) - 1\).
- A1: Show that \(\sin(x) = -1/6\), identifying \(k = -1/6\).
- M1: Use basic angle \(\arcsin(1/6)\) to find at least one correct quadrant solution.
- A1: Obtain both \(3.31\) and \(6.12\) correct to 3 significant figures.