2023 Cambridge IGCSE Mathematics - Additional (0606) 模擬試題連答案詳解

At \( x = -1 \), \( y = \frac{\ln(2(-1)+3)}{(-1)^2} = \frac{\ln(1)}{1} = 0 \). So the point on the curve is \( (-1, 0) \). To find the gradient of the tangent, we use the quotient rule: \( \frac{dy}{dx} = \frac{x^2 \left( \frac{2}{2x+3} \right) - 2x \ln(2x+3)}{x^4} \). Substituting \( x = -1 \): \( \frac{dy}{dx} = \frac{(-1)^2 \left( \frac{2}{1} \right) - 2(-1)\ln(1)}{(-1)^4} = \frac{2 - 0}{1} = 2 \). The gradient of the tangent at \( x = -1 \) is 2, so the gradient of the normal is \( -\frac{1}{2} \). The equation of the normal is \( y - 0 = -\frac{1}{2}(x - (-1)) \), which simplifies to \( 2y = -x - 1 \), or \( x + 2y + 1 = 0 \).

M1: For attempt at quotient rule to find \( \frac{dy}{dx} \). A1: Correct derivative expression. M1: Substitute \( x = -1 \) to find gradient of tangent. A1: Correct tangent gradient of 2 and normal gradient of \( -0.5 \). M1: Use of normal gradient and point \( (-1, 0) \) to find line equation. A1: Correct equation in integer form: \( x + 2y + 1 = 0 \).

Using the identity \( \sin^2 x = 1 - \cos^2 x \), we substitute to get: \( 6(1 - \cos^2 x) - \cos x - 4 = 0 \), which simplifies to \( 6 - 6\cos^2 x - \cos x - 4 = 0 \) and rearranges to \( 6\cos^2 x + \cos x - 2 = 0 \). Factorising the quadratic equation gives \( (3\cos x + 2)(2\cos x - 1) = 0 \). This yields two cases: 1) \( \cos x = \frac{1}{2} \), giving basic angle \( 60^\circ \), so \( x = 60^\circ \) and \( x = 360^\circ - 60^\circ = 300^\circ \). 2) \( \cos x = -\frac{2}{3} \), giving basic angle \( \cos^{-1}(\frac{2}{3}) \approx 48.19^\circ \). Since cosine is negative in the second and third quadrants, \( x = 180^\circ - 48.19^\circ = 131.8^\circ \) and \( x = 180^\circ + 48.19^\circ = 228.2^\circ \). Thus, the solutions are \( 60^\circ \), \( 131.8^\circ \), \( 228.2^\circ \), and \( 300^\circ \).

M1: Use of \( \sin^2 x = 1 - \cos^2 x \) to form a quadratic in terms of \( \cos x \). A1: Correct quadratic equation \( 6\cos^2 x + \cos x - 2 = 0 \). M1: Solve the quadratic to find \( \cos x = 1/2 \) and \( \cos x = -2/3 \). A1: For finding correct values \( x = 60^\circ \) and \( 300^\circ \). M1: For finding second quadrant angle \( 131.8^\circ \) or third quadrant angle \( 228.2^\circ \). A1: Correct final set of answers: \( 60^\circ, 131.8^\circ, 228.2^\circ, 300^\circ \).

For a geometric progression, the n-th term is \( u_n = a r^{n-1} \). We are given: \( u_3 = a r^2 = 12 \) and \( u_6 = a r^5 = 1.5 \). Dividing the two equations: \( \frac{a r^5}{a r^2} = \frac{1.5}{12} \implies r^3 = \frac{1}{8} \implies r = 0.5 \). Substituting \( r = 0.5 \) back into the first equation: \( a (0.5)^2 = 12 \implies a (0.25) = 12 \implies a = 48 \). The sum to infinity is given by \( S_\infty = \frac{a}{1 - r} = \frac{48}{1 - 0.5} = 96 \).

M1: Set up equations \( a r^2 = 12 \) and \( a r^5 = 1.5 \). M1: Solve for \( r \) by division. A1: Correct common ratio \( r = 0.5 \). A1: Correct first term \( a = 48 \). M1: Correctly apply the sum to infinity formula \( S_\infty = \frac{a}{1-r} \). A1: Correct final sum of 96.

Using the factor theorem, \( p(2) = 0 \implies 2(2)^3 + a(2)^2 + b(2) - 12 = 0 \implies 16 + 4a + 2b - 12 = 0 \implies 2a + b = -2 \) (Equation 1). Using the remainder theorem, \( p(-1) = -18 \implies 2(-1)^3 + a(-1)^2 + b(-1) - 12 = -18 \implies -2 + a - b - 12 = -18 \implies a - b = -4 \) (Equation 2). Adding Equation 1 and Equation 2 gives \( 3a = -6 \implies a = -2 \). Substituting \( a = -2 \) into Equation 2: \( -2 - b = -4 \implies b = 2 \). Thus, \( p(x) = 2x^3 - 2x^2 + 2x - 12 \). Since \( x - 2 \) is a factor, we can write \( p(x) = (x - 2)(2x^2 + kx + 6) \). Expanding and matching coefficients of \( x^2 \) gives \( -4 + k = -2 \implies k = 2 \). So, \( p(x) = (x - 2)(2x^2 + 2x + 6) = 2(x - 2)(x^2 + x + 3) \). The quadratic \( x^2 + x + 3 \) has discriminant \( 1^2 - 4(1)(3) = -11 < 0 \), so it cannot be factorised further.

M1: Attempt to apply the factor theorem with \( p(2) = 0 \). A1: Obtain correct linear equation \( 2a + b = -2 \). M1: Attempt to apply the remainder theorem with \( p(-1) = -18 \). A1: Obtain correct linear equation \( a - b = -4 \) and solve for \( a = -2 \) and \( b = 2 \). M1: Perform polynomial division or algebraic matching to factorise. A1: Obtain complete factorization \( 2(x - 2)(x^2 + x + 3) \).

(a) The gradient of the straight line is \( m = \frac{8.4 - 4.8}{5 - 2} = \frac{3.6}{3} = 1.2 \). Using the point-slope form with \( (2, 4.8) \): \( \ln y - 4.8 = 1.2(x - 2) \implies \ln y = 1.2x + 2.4 \). (b) Taking the natural logarithm of both sides of \( y = Ab^x \) gives \( \ln y = \ln A + x \ln b \). Comparing this with the linear equation \( \ln y = 1.2x + 2.4 \), we identify the gradient \( \ln b = 1.2 \implies b = e^{1.2} \approx 3.32 \). The vertical intercept is \( \ln A = 2.4 \implies A = e^{2.4} \approx 11.02 \).

M1: Find the gradient of the line using the coordinates. A1: Correct linear equation \( \ln y = 1.2x + 2.4 \). M1: Take natural logarithms of \( y = Ab^x \) to obtain \( \ln y = \ln A + x \ln b \). M1: Relate the coefficients to solve for \( A \) and \( b \). A1: Correct value of \( b = 3.32 \). A1: Correct value of \( A = 11.02 \).

(a) We find the vectors \( \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (5\mathbf{i} + 2\mathbf{j}) - (\mathbf{i} - 2\mathbf{j}) = 4\mathbf{i} + 4\mathbf{j} \). Also, \( \overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (k\mathbf{i} + 8\mathbf{j}) - (5\mathbf{i} + 2\mathbf{j}) = (k-5)\mathbf{i} + 6\mathbf{j} \). Since \( A \), \( B \), and \( C \) are collinear, \( \overrightarrow{BC} = \lambda \overrightarrow{AB} \) for some scalar \( \lambda \). Comparing components: \( 6 = 4\lambda \implies \lambda = 1.5 \). Then for the \( \mathbf{i} \) components: \( k - 5 = 1.5(4) = 6 \implies k = 11 \). (b) The vector \( \overrightarrow{AB} = 4\mathbf{i} + 4\mathbf{j} \). Its magnitude is \( |\overrightarrow{AB}| = \sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2} \). The unit vector in the direction of \( \overrightarrow{AB} \) is \( \frac{1}{4\sqrt{2}}(4\mathbf{i} + 4\mathbf{j}) = \frac{1}{\sqrt{2}}\mathbf{i} + \frac{1}{\sqrt{2}}\mathbf{j} \) (or approximately \( 0.707\mathbf{i} + 0.707\mathbf{j} \)).

M1: Find correct vector expression for \( \overrightarrow{AB} \). M1: Find correct vector expression for \( \overrightarrow{BC} \) or \( \overrightarrow{AC} \) in terms of \( k \). M1: Set up a ratio or scalar multiple equation to solve for \( k \). A1: Correct value of \( k = 11 \). M1: Find magnitude of \( \overrightarrow{AB} \). A1: Correct unit vector \( \frac{1}{\sqrt{2}}\mathbf{i} + \frac{1}{\sqrt{2}}\mathbf{j} \).

(a) To find \( f^{-1}(x) \), set \( y = e^{2x} - 3 \). Rearrange to solve for \( x \): \( e^{2x} = y + 3 \implies 2x = \ln(y + 3) \implies x = \frac{1}{2}\ln(y + 3) \). Hence, \( f^{-1}(x) = \frac{1}{2}\ln(x + 3) \). The domain of \( f^{-1}(x) \) is the range of \( f(x) \). Since \( e^{2x} > 0 \) for all \( x \), \( f(x) > -3 \), meaning the domain of \( f^{-1} \) is \( x > -3 \). (b) We find the composite function: \( fg(x) = f(\ln(x+4)) = e^{2\ln(x+4)} - 3 = (e^{\ln(x+4)})^2 - 3 = (x+4)^2 - 3 \). To solve \( fg(x) = 13 \): \( (x+4)^2 - 3 = 13 \implies (x+4)^2 = 16 \implies x+4 = \pm 4 \). Since \( x > -4 \), we choose the positive root: \( x+4 = 4 \implies x = 0 \).

M1: Swap variables or set \( y = e^{2x} - 3 \) to make \( x \) the subject. A1: Correct expression \( f^{-1}(x) = \frac{1}{2}\ln(x + 3) \). A1: State correct domain \( x > -3 \). M1: Form composite function \( fg(x) = (x+4)^2 - 3 \). M1: Solve equation \( (x+4)^2 - 3 = 13 \). A1: Correct final answer \( x = 0 \) (rejecting \( x = -8 \) due to domain restrictions).

First, write the indices in Equation 1 with a common base of 2: \( 2^{3x} \times (2^2)^y = 2^{-3} \implies 2^{3x + 2y} = 2^{-3} \). Equating exponents gives \( 3x + 2y = -3 \) (Equation 1). Next, rewrite the logarithmic Equation 2 in exponential form: \( \log_3(x + 2y) = 2 \implies x + 2y = 3^2 \implies x + 2y = 9 \) (Equation 2). Subtracting Equation 2 from Equation 1: \( (3x + 2y) - (x + 2y) = -3 - 9 \implies 2x = -12 \implies x = -6 \). Substitute \( x = -6 \) back into Equation 2: \( -6 + 2y = 9 \implies 2y = 15 \implies y = 7.5 \). Checking both solutions, they are valid.

M1: Express the first equation with a base of 2. A1: Correct linear equation \( 3x + 2y = -3 \). M1: Convert logarithmic equation to exponential form. A1: Correct linear equation \( x + 2y = 9 \). M1: Apply an appropriate elimination/substitution method to solve the simultaneous equations. A1: Correct solutions: \( x = -6 \) and \( y = 7.5 \).

(i) We use the product rule to differentiate \(y = (3x+1)(2x-1)^{1/2}\).
Let \(u = 3x+1 \implies \frac{\mathrm{d}u}{\mathrm{d}x} = 3\).
Let \(v = (2x-1)^{1/2} \implies \frac{\mathrm{d}v}{\mathrm{d}x} = \frac{1}{2}(2x-1)^{-1/2} \cdot 2 = (2x-1)^{-1/2}\).

Using the product rule:
\(\frac{\mathrm{d}y}{\mathrm{d}x} = u \frac{\mathrm{d}v}{\mathrm{d}x} + v \frac{\mathrm{d}u}{\mathrm{d}x}\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = (3x+1)(2x-1)^{-1/2} + 3(2x-1)^{1/2}\)

To combine into a single fraction with denominator \(\sqrt{2x-1}\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3x+1}{\sqrt{2x-1}} + \frac{3(2x-1)}{\sqrt{2x-1}}\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3x+1+6x-3}{\sqrt{2x-1}} = \frac{9x-2}{\sqrt{2x-1}}\).
So \(a = 9\) and \(b = 2\).

(ii) From part (i), we have:
\(\frac{\mathrm{d}}{\mathrm{d}x} \left( (3x+1)\sqrt{2x-1} \right) = \frac{9x-2}{\sqrt{2x-1}}\)

Integrating both sides with respect to \(x\):
\((3x+1)\sqrt{2x-1} = \int \frac{9x-2}{\sqrt{2x-1}} \mathrm{d}x\)
\((3x+1)\sqrt{2x-1} = \int \left( \frac{9x}{\sqrt{2x-1}} - \frac{2}{\sqrt{2x-1}} \right) \mathrm{d}x\)

Rearranging to make \(\int \frac{9x}{\sqrt{2x-1}} \mathrm{d}x\) the subject:
\(\int \frac{9x}{\sqrt{2x-1}} \mathrm{d}x = (3x+1)\sqrt{2x-1} + \int \frac{2}{\sqrt{2x-1}} \mathrm{d}x\)

Now, find the integral of the second term:
\(\int \frac{2}{\sqrt{2x-1}} \mathrm{d}x = \int 2(2x-1)^{-1/2} \mathrm{d}x = \frac{2(2x-1)^{1/2}}{(-1/2 + 1) \cdot 2} = \frac{2(2x-1)^{1/2}}{1} = 2\sqrt{2x-1}\).

Substituting this back:
\(\int \frac{9x}{\sqrt{2x-1}} \mathrm{d}x = (3x+1)\sqrt{2x-1} + 2\sqrt{2x-1} + C\)

Factoring out \(\sqrt{2x-1}\):
\(\int \frac{9x}{\sqrt{2x-1}} \mathrm{d}x = (3x + 1 + 2)\sqrt{2x-1} + C = 3(x+1)\sqrt{2x-1} + C\).

Part (i):
- M1: For using the product rule to differentiate, with one term of the form \(k(2x-1)^{-1/2}\).
- A1: For obtaining correct derivative components: \(3(2x-1)^{1/2} + (3x+1)(2x-1)^{-1/2}\).
- A1: For simplifying to the required form \(\frac{9x-2}{\sqrt{2x-1}}\).

Part (ii):
- M1: For recognizing the relationship between differentiation and integration and expressing \(\int \frac{9x}{\sqrt{2x-1}} \mathrm{d}x\) in terms of \(y\) and \(\int \frac{2}{\sqrt{2x-1}} \mathrm{d}x\).
- A1: For integrating \(\int \frac{2}{\sqrt{2x-1}} \mathrm{d}x\) correctly to get \(2\sqrt{2x-1}\).
- A1: For the final correct expression \(3(x+1)\sqrt{2x-1} + C\) (or equivalent form, must include \(+C\)).

We use the identity \(\sec^2 x = 1 + \tan^2 x\) to rewrite the equation in terms of \(\tan x\):
\(3(1 + \tan^2 x) - 2 \tan x = 4\)
\(3 + 3 \tan^2 x - 2 \tan x = 4\)
\(3 \tan^2 x - 2 \tan x - 1 = 0\)

Let \(u = \tan x\), then the quadratic equation is:
\(3u^2 - 2u - 1 = 0\)
We can factorize this:
\((3u + 1)(u - 1) = 0\)

Thus:
\(\tan x = 1\) or \(\tan x = -\frac{1}{3}\)

Case 1: \(\tan x = 1\)
For \(0 \le x \le 2\pi\):
\(x = \frac{\pi}{4}\) (in the first quadrant)
\(x = \pi + \frac{\pi}{4} = \frac{5\pi}{4}\) (in the third quadrant)

Case 2: \(\tan x = -\frac{1}{3}\)
Since \(\tan x\) is negative, \(x\) lies in the second and fourth quadrants.
The basic angle is \(\alpha = \tan^{-1}\left(\frac{1}{3}\right) \approx 0.32175\) radians.

In the second quadrant:
\(x = \pi - \alpha \approx 3.14159 - 0.32175 = 2.8198 \approx 2.82\)

In the fourth quadrant:
\(x = 2\pi - \alpha \approx 6.28318 - 0.32175 = 5.9614 \approx 5.96\)

Therefore, the solutions in the interval \(0 \le x \le 2\pi\) are:
\(x = \frac{\pi}{4}\), \(x \approx 2.82\), \(x = \frac{5\pi}{4}\), \(x \approx 5.96\).

- M1: For using the identity \(\sec^2 x = 1 + \tan^2 x\) to obtain a quadratic equation in \(\tan x\).
- A1: For obtaining the correct quadratic equation \(3 \tan^2 x - 2 \tan x - 1 = 0\).
- M1: For solving the quadratic equation to get \(\tan x = 1\) and \(\tan x = -\frac{1}{3}\).
- A1: For finding the correct exact solutions \(x = \frac{\pi}{4}\) and \(x = \frac{5\pi}{4}\).
- A1: For finding the basic angle \(\alpha \approx 0.322\) or correct quadrant values.
- A1: For both decimal solutions \(x \approx 2.82\) and \(x \approx 5.96\) correct to 2 decimal places. (Deduct 1 mark for any extra values within the range, or if solutions outside range are not rejected).

(a) To find the range of \(\mathrm{f}(x)\), we can express it in completed square form:
\(\mathrm{f}(x) = 2(x^2 - 2x) + 5 = 2\left[(x - 1)^2 - 1\right] + 5 = 2(x - 1)^2 + 3\).

Since the domain is \(x \ge 1\), the minimum value of \(\mathrm{f}(x)\) occurs at \(x = 1\), where \(\mathrm{f}(1) = 3\).
For all \(x \ge 1\), \(2(x - 1)^2 \ge 0\), so \(\mathrm{f}(x) \ge 3\).
Thus, the range of \(\mathrm{f}\) is \(\mathrm{f}(x) \ge 3\).

(b) Since \(\mathrm{f}(x)\) is a quadratic function with its vertex at \(x = 1\), restricting its domain to \(x \ge 1\) means it is a strictly increasing function. Thus, \(\mathrm{f\)} is a one-to-one function, which guarantees that the inverse function \(\mathrm{f}^{-1}(x)\) exists.

To find the inverse function, let \(y = \mathrm{f}(x)\):
\(y = 2(x - 1)^2 + 3\)
\(y - 3 = 2(x - 1)^2\)
\(\frac{y-3}{2} = (x - 1)^2\)
Since \(x \ge 1\), we take the positive square root:
\(x - 1 = \sqrt{\frac{y-3}{2}}\)
\(x = 1 + \sqrt{\frac{y-3}{2}}\)
Thus, \(\mathrm{f}^{-1}(x) = 1 + \sqrt{\frac{x-3}{2}}\) for \(x \ge 3\).

(c) To solve \(\mathrm{fg}(x) = 5\), we substitute \(\mathrm{g}(x)\) into \(\mathrm{f}\):
\(\mathrm{f}(\mathrm{g}(x)) = 5\)
Using the completed square form of \(\mathrm{f}\):
\(2(\mathrm{g}(x) - 1)^2 + 3 = 5\)
\(2(\mathrm{g}(x) - 1)^2 = 2\)
\((\mathrm{g}(x) - 1)^2 = 1\)
\(\mathrm{g}(x) - 1 = 1\) or \(\mathrm{g}(x) - 1 = -1\)
This gives:
\(\mathrm{g}(x) = 2\) or \(\mathrm{g}(x) = 0\).

However, the domain of \(\mathrm{f}(x)\) is \(x \ge 1\), so the input to \(\mathrm{f}\), which is \(\mathrm{g}(x)\), must satisfy \(\mathrm{g}(x) \ge 1\).
Thus, we reject \(\mathrm{g}(x) = 0\).
We only solve \(\mathrm{g}(x) = 2\):
\(\ln(x + 2) = 2\)
\(x + 2 = e^2\)
\(x = e^2 - 2\).

Part (a):
- B1: For completing the square or finding the vertex coordinate at \((1, 3)\).
- B1: For range \(\mathrm{f}(x) \ge 3\) (accept \(y \ge 3\), do not accept \(x \ge 3\)).

Part (b):
- B1: For stating that \(\mathrm{f\)} is a one-to-one function (or equivalent explanation showing it is increasing for \(x \ge 1\)).
- M1: For attempting to make \(x\) the subject of \(y = 2(x-1)^2+3\) or \(y = 2x^2-4x+5\).
- A1: For \(\mathrm{f}^{-1}(x) = 1 + \sqrt{\frac{x-3}{2}}\) (must be in terms of \(x\), and the positive square root must be selected).

Part (c):
- M1: For setting up the equation \(2(\mathrm{g}(x)-1)^2 + 3 = 5\) (or equivalent in expanded form) and obtaining \(\mathrm{g}(x) = 2\) or \(\mathrm{g}(x) = 0\).
- M1: For identifying that \(\mathrm{g}(x) \ge 1\) must hold due to the domain of \(\mathrm{f}\), hence rejecting \(\mathrm{g}(x) = 0\) (or rejecting the solution \(x = -1\) because it gives \(\mathrm{g}(-1) = 0 < 1\)).
- A1: For the unique correct solution \(x = e^2 - 2\) (accept \(e^2 - 2\) or approx 5.39).

(a) The terms of the arithmetic progression are given by:
\(T_1 = a\)
\(T_3 = a + 2d\)
\(T_7 = a + 6d\)

The terms of the geometric progression are given by:
\(G_1 = a\)
\(G_2 = ar\)
\(G_3 = ar^2\)

Since \(T_3 = G_2\):
\(a + 2d = ar \implies 2d = ar - a \implies 2d = a(r-1)\)

Hence, \(d = \frac{a(r-1)}{2}\). (This is shown)

(b) Since \(T_7 = G_3\):
\(a + 6d = ar^2\)

Substituting \(6d = 3(2d) = 3a(r-1)\) into this equation:
\(a + 3a(r-1) = ar^2\)

Assuming \(a \neq 0\), we divide both sides by \(a\):
\(1 + 3(r-1) = r^2\)
\(1 + 3r - 3 = r^2\)
\(r^2 - 3r + 2 = 0\)
\((r-1)(r-2) = 0\)

Hence, \(r = 1\) or \(r = 2\). (This is shown)

(c)(i) The sum of the first 5 terms of a geometric progression is:
\(S_5 = \frac{a(r^5 - 1)}{r - 1}\)

With \(r = 2\) and \(S_5 = 155\):
\(155 = \frac{a(2^5 - 1)}{2 - 1}\)
\(155 = a(31) \implies a = 5\).

Using \(d = \frac{a(r-1)}{2}\) with \(a = 5\) and \(r = 2\):
\(d = \frac{5(2-1)}{2} = 2.5\).

(c)(ii) The sum of the first 12 terms of the arithmetic progression is:
\(S_{12} = \frac{12}{2}\left[2a + (12-1)d\right]\)
\(S_{12} = 6\left[2(5) + 11(2.5)\right]\)
\(S_{12} = 6\left[10 + 27.5\right] = 6(37.5) = 225\).

Part (a):
- M1: For expressing \(T_3\) in terms of \(a\) and \(d\), and \(G_2\) in terms of \(a\) and \(r\), and equating them: \(a + 2d = ar\).
- A1: For correctly rearranging to show \(d = \frac{a(r-1)}{2}\).

Part (b):
- M1: For expressing \(T_7 = G_3\) as \(a + 6d = ar^2\) and substituting \(d = \frac{a(r-1)}{2}\).
- A1: For obtaining and solving the quadratic equation \(r^2 - 3r + 2 = 0\) to get \(r = 1\) or \(r = 2\).

Part (c)(i):
- M1: For using the sum formula of a GP with \(r=2\) to find \(a\).
- A1: For \(a = 5\).
- A1: For \(d = 2.5\).

Part (c)(ii):
- M1: For using the sum formula of an AP with \(n=12\), and their values of \(a\) and \(d\).
- A1: For \(225\).

(a) Let \(y = (3x-1)e^{-2x}\). Using the product rule \(\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}\):\
\(u = 3x - 1 \implies \frac{du}{dx} = 3\)\
\(v = e^{-2x} \implies \frac{dv}{dx} = -2e^{-2x}\\)\
\(\frac{dy}{dx} = (3x-1)(-2e^{-2x}) + (3)(e^{-2x}) = e^{-2x}(-6x + 2 + 3) = (5-6x)e^{-2x}\).\
\
(b) From part (a), we have \(\int (5-6x)e^{-2x} \\, dx = (3x-1)e^{-2x} + C'\).\
We can split the integral: \(5\int e^{-2x} \\, dx - 6\int x e^{-2x} \\, dx = (3x-1)e^{-2x} + C'\).\
Since \(\int e^{-2x} \\, dx = -\frac{1}{2}e^{-2x}\), we get:\
\(-\frac{5}{2}e^{-2x} - 6\int x e^{-2x} \\, dx = (3x-1)e^{-2x} + C'\)\
\(-6\int x e^{-2x} \\, dx = (3x - 1 + 2.5)e^{-2x} + C' = (3x + 1.5)e^{-2x} + C'\)\
Dividing by \(-6\):\
\(\int x e^{-2x} \\, dx = -\frac{1}{6}(3x + 1.5)e^{-2x} + C = -\frac{1}{4}(2x + 1)e^{-2x} + C\).

(a)\
M1: Attempt to apply product rule with correct derivative of \(e^{-2x}\) or \(3x-1\).\
A1: Obtain correct unsimplified derivative: \(3e^{-2x} - 2(3x-1)e^{-2x}\).\
A1: Simplify correctly to \((5-6x)e^{-2x}\).\
(b)\
M1: Recognise integration relation from part (a): \(\int (5-6x)e^{-2x} \\, dx = (3x-1)e^{-2x}\).\
M1: Integrate \(5e^{-2x}\) to get \(-\frac{5}{2}e^{-2x}\).\
M1: Set up equation to make \(\int x e^{-2x} \\, dx\) the subject.\
A1: Obtain intermediate equation, e.g., \(-6 \int x e^{-2x} \\, dx = (3x + 1.5)e^{-2x}\).\
A1: Fully correct simplification to show the given result including arbitrary constant.

(a) The total surface area of a solid cylinder is \(A = 2\pi r^2 + 2\pi r h\).\
Given \(A = 150\pi\):\
\(2\pi r^2 + 2\pi r h = 150\pi \implies r^2 + rh = 75 \implies h = \frac{75 - r^2}{r}\).\
The volume of the cylinder is \(V = \pi r^2 h\).\
Substituing \(h\):\
\(V = \pi r^2 \left(\frac{75 - r^2}{r}\right) = \pi r(75 - r^2) = 75\pi r - \pi r^3\).\
\
(b) Differentiating \(V\) with respect to \(r\):\
\(\frac{dV}{dr} = 75\pi - 3\pi r^2\).\
For a stationary value, \(\frac{dV}{dr} = 0\):\
\(75\pi - 3\pi r^2 = 0 \implies 3\pi r^2 = 75\pi \implies r^2 = 25\).\
Since \(r > 0\), \(r = 5\).\
\
(c) The stationary value of \(V\) is:\
\(V = 75\pi(5) - \pi(5)^3 = 375\pi - 125\pi = 250\pi\).\
To find the nature, find the second derivative:\
\(\frac{d^2V}{dr^2} = -6\pi r\).\
At \(r = 5\), \(\frac{d^2V}{dr^2} = -30\pi < 0\), which confirms that the stationary value is a maximum.

(a)\
M1: Uses the correct surface area formula to write \(2\pi r^2 + 2\pi rh = 150\pi\).\
M1: Rearranges to make \(h\) the subject: \(h = \frac{75-r^2}{r}\).\
A1: Substitutes into the volume formula and simplifies to show the given expression.\
(b)\
M1: Differentiates \(V\) correctly to obtain \(\frac{dV}{dr} = 75\pi - 3\pi r^2\).\
M1: Sets \(\frac{dV}{dr} = 0\) and solves for \(r\).\
A1: Obtains \(r = 5\) (rejecting \(r = -5\)).\
(c)\
A1: Calculates the volume as \(250\pi\).\
A1: Finds \(\frac{d^2V}{dr^2} = -6\pi r\) and uses it (or a sign test) at \(r = 5\) to show it is a maximum.

(a) Expressing the LHS with a common denominator:\
\(\frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} = \frac{\sin^2 \theta + (1 + \cos \theta)^2}{\sin \theta (1 + \cos \theta)}\\)\
\(= \frac{\sin^2 \theta + 1 + 2\cos \theta + \cos^2 \theta}{\sin \theta (1 + \cos \theta)}\\)\
Since \(\sin^2 \theta + \cos^2 \theta = 1\):\
\(= \frac{1 + 1 + 2\cos \theta}{\sin \theta (1 + \cos \theta)} = \frac{2(1 + \cos \theta)}{\sin \theta (1 + \cos \theta)} = \frac{2}{\sin \theta} = 2\csc \theta\).\
\
(b) Using the identity from part (a), the equation is equivalent to:\
\(2\csc 2x = 3\)\
\(\csc 2x = 1.5 \implies \sin 2x = \frac{2}{3}\).\
Since \(0^\circ \le x \le 180^\circ\), we have \(0^\circ \le 2x \le 360^\circ\).\
Find the reference angle: \(2x = \sin^{-1}(2/3) \approx 41.81^\circ\).\
So, \(2x = 41.81^\circ\) or \(2x = 180^\circ - 41.81^\circ = 138.19^\circ\).\
Solving for \(x\):\
\(x \approx 20.9^\circ\) or \(x \approx 69.1^\circ\) (to 1 decimal place).

(a)\
M1: Combines fractions to get \(\frac{\sin^2\theta + (1+\cos\theta)^2}{\sin\theta(1+\cos\theta)}\).\
M1: Expands and applies \(\sin^2\theta + \cos^2\theta = 1\).\
A1: Factorises numerator as \(2(1+\cos\theta)\).\
A1: Cancels terms correctly to obtain \(2\csc\theta\).\
(b)\
M1: Uses identity to write \(2\csc 2x = 3\) or \(\sin 2x = \frac{2}{3}\).\
M1: Finds one valid value of \(2x\) (e.g. \(41.8^\circ\) or \(138.2^\circ\)).\
A1: Obtains \(x \approx 20.9^\circ\) (accept 20.9).\
A1: Obtains \(x \approx 69.1^\circ\) (accept 69.1) and no other values in range.

(a) For an arithmetic progression, the \(k\)-th term is \(T_k = a + (k-1)d\).\
Given:\
\(T_3 = a + 2d = 20\)\
\(T_7 = a + 6d = 44\)\
Subtracting the first equation from the second:\
\(4d = 24 \implies d = 6\).\
Substitute \(d = 6\) back into the first equation:\
\(a + 2(6) = 20 \implies a + 12 = 20 \implies a = 8\).\
So, \(a = 8\) and \(d = 6\).\
\
(b) Let the terms of the geometric progression be \(G_1, G_2, G_3\).\
\(G_1 = T_1 = a = 8\)\
\(G_2 = T_3 = 20\)\
\(G_3 = T_n = a + (n-1)d = 8 + 6(n-1) = 6n + 2\).\
Since these form a geometric progression:\
\(\frac{G_2}{G_1} = \frac{G_3}{G_2}\\)\
\(\frac{20}{8} = \frac{6n + 2}{20}\\)\
\(2.5 = \frac{6n + 2}{20}\\)\
\(50 = 6n + 2 \implies 6n = 48 \implies n = 8\).

(a)\
M1: Sets up simultaneous equations \(a + 2d = 20\) and \(a + 6d = 44\).\
M1: Eliminates one variable to solve for \(a\) or \(d\).\
A1: Obtains both \(a = 8\) and \(d = 6\).\
(b)\
M1: Identifies \(G_1 = 8\) and \(G_2 = 20\).\
M1: Computes common ratio \(r = 2.5\) or sets up the ratio equation \(\frac{20}{8} = \frac{T_n}{20}\).\
M1: Expresses \(T_n\) as \(8 + 6(n-1)\).\
M1: Sets up linear equation in \(n\): \(50 = 8 + 6(n-1)\).\
A1: Obtains \(n = 8\).

(a) Since \(2x - 3\) is a factor, by the Factor Theorem, \(p(1.5) = 0\):\
\(2(1.5)^3 + a(1.5)^2 + b(1.5) - 15 = 0\)\
\(6.75 + 2.25a + 1.5b - 15 = 0 \implies 2.25a + 1.5b = 8.25\)\
Multiplying by 4 gives: \(9a + 6b = 33 \implies 3a + 2b = 11\) (Equation 1).\
\
By the Remainder Theorem, since dividing by \(x+2\) gives remainder \(-21\), \(p(-2) = -21\):\
\(2(-2)^3 + a(-2)^2 + b(-2) - 15 = -21\)\
\(-16 + 4a - 2b - 15 = -21 \implies 4a - 2b = 10 \implies 2a - b = 5\) (Equation 2).\
\
From Equation 2, \(b = 2a - 5\). Substituting into Equation 1:\
\(3a + 2(2a - 5) = 11 \implies 7a - 10 = 11 \implies 7a = 21 \implies a = 3\).\
Then \(b = 2(3) - 5 = 1\).\
\
(b) With \(a = 3\) and \(b = 1\), \(p(x) = 2x^3 + 3x^2 + x - 15\).\
Since \(2x-3\) is a factor, we can perform polynomial division or use inspection to write:\
\(2x^3 + 3x^2 + x - 15 = (2x - 3)(x^2 + cx + 5)\).\
Comparing coefficients of \(x^2\):\
\(3 = 2c - 3 \implies 2c = 6 \implies c = 3\).\
So, \(p(x) = (2x - 3)(x^2 + 3x + 5)\).\
To solve \(p(x) = 0\):\
\(2x - 3 = 0 \implies x = 1.5\).\
For \(x^2 + 3x + 5 = 0\), the discriminant is \(\Delta = 3^2 - 4(1)(5) = 9 - 20 = -11 < 0\), meaning there are no other real roots.\
Thus, the only real solution is \(x = 1.5\).

(a)\
M1: Applies Factor Theorem to write \(p(1.5) = 0\) (or equivalent equation).\
M1: Applies Remainder Theorem to write \(p(-2) = -21\) (or equivalent equation).\
M1: Solves the system of two simultaneous equations in \(a\) and \(b\).\
A1: Obtains both \(a = 3\) and \(b = 1\).\
(b)\
M1: Performs division or matches coefficients to find the quadratic factor.\
A1: Obtains the correct factor \(x^2 + 3x + 5\).\
M1: Evaluates discriminant of the quadratic or attempts to solve, concluding no real roots.\
A1: States \(x = 1.5\) (or \(3/2\)) as the only real solution.

(a) Let \(Y = \ln y\) and \(X = x^2\). The points are \((2, 5)\) and \((6, 13)\).\
Gradient \(m = \frac{13 - 5}{6 - 2} = \frac{8}{4} = 2\).\
Using point-slope form with \((2, 5)\):\
\(Y - 5 = 2(X - 2) \implies Y = 2X + 1\).\
Substituting back \(Y\) and \(X\):\
\(\ln y = 2x^2 + 1\).\
\
(b) Exponentiating both sides of the equation from part (a):\
\(y = e^{2x^2 + 1}\).\
Using index laws:\
\(y = e^1 \cdot e^{2x^2} = e \cdot e^{2x^2}\).\
Thus, \(A = e\) and \(k = 2\).\
\
(c) When \(x = 3\), we substitute into our expression for \(y\):\
\(y = e^{2(3)^2 + 1} = e^{2(9) + 1} = e^{19}\).

(a)\
M1: Finds the gradient \(m = 2\).\
M1: Uses straight line equation formula with \(Y\) and \(X\).\
A1: Obtains \(\ln y = 2x^2 + 1\).\
(b)\
M1: Converts from logarithmic form to exponential form: \(y = e^{2x^2+1}\).\
M1: Splits exponential term using law of indices: \(e \cdot e^{2x^2}\).\
A1: Identifies \(A = e\) (or approx 2.72) and \(k = 2\).\
(c)\
M1: Substitutes \(x = 3\) into their expression.\
A1: Obtains exact answer \(e^{19}\).

(a) Let \(u = 3^x\). Then \(3^{2x+1} = 3 \cdot (3^x)^2 = 3u^2\).\
The equation becomes:\
\(3u^2 - 10u + 3 = 0\)\
\((3u - 1)(u - 3) = 0 \implies u = \frac{1}{3}\) or \(u = 3\).\
When \(u = 3^x = 3 \implies x = 1\).\
Calculates \(u = 3^x = \frac{1}{3} \implies x = -1\).\
So, the solutions are \(x = 1\) and \(x = -1\).\
\
(b) Use the change of base formula on the second log term:\
\(2\log_4(y + 3) = 2 \cdot \frac{\log_2(y+3)}{\log_2 4} = 2 \cdot \frac{\log_2(y+3)}{2} = \log_2(y+3)\).\
Substitute this back into the equation:\
\(\log_2(y - 1) + \log_2(y + 3) = 3\)\
\(\log_2[(y-1)(y+3)] = 3\)\
\((y - 1)(y + 3) = 2^3 = 8\)\
\(y^2 + 2y - 3 = 8 \implies y^2 + 2y - 11 = 0\).\
Using the quadratic formula:\
\(y = \frac{-2 \pm \sqrt{2^2 - 4(1)(-11)}}{2} = \frac{-2 \pm \sqrt{48}}{2} = -1 \pm 2\sqrt{3}\).\
Since the logarithm is only defined for positive arguments, we must have \(y - 1 > 0 \implies y > 1\).\
Evaluating the roots: \(-1 - 2\sqrt{3} \approx -4.46\) (rejected) and \(-1 + 2\sqrt{3} \approx 2.46\) (valid).\
Thus, the only solution is \(y = -1 + 2\sqrt{3}\).

(a)\
M1: Uses substitution \(u = 3^x\) to form a quadratic equation.\
M1: Correctly factorises or solves the quadratic to get \(u = 1/3\) and \(u = 3\).\
A1: Obtains \(x = 1\).\
A1: Obtains \(x = -1\).\
(b)\
M1: Applies change of base formula correctly to get \(\log_2(y+3)\).\
M1: Uses sum law of logarithms to write \(\log_2[(y-1)(y+3)] = 3\).\
A1: Resolves logarithms to obtain quadratic \(y^2+2y-11=0\).\
A1: Solves quadratic and rejects negative root to find \(y = -1 + 2\sqrt{3}\) (or approx 2.46).

(a) \(f(x) = 2x^2 - 8x + 5\) is a quadratic function, which has a many-to-one mapping on \(\mathbb{R}\). Since it is not a one-to-one function, its inverse does not exist.\
Completing the square:\
\(f(x) = 2(x^2 - 4x) + 5 = 2(x-2)^2 - 8 + 5 = 2(x-2)^2 - 3\).\
The axis of symmetry is at \(x = 2\). For the function to be one-to-one on the restricted domain \(x \ge k\), we must have \(k \ge 2\).\
Thus, the smallest value of \(k\) is 2.\
\
(b) Let \(y = 2(x-2)^2 - 3\) for \(x \ge 2\).\
\(y + 3 = 2(x-2)^2 \implies (x-2)^2 = \frac{y+3}{2}\).\
Taking the positive square root because \(x \ge 2 \implies x-2 \ge 0\):\
\(x - 2 = \sqrt{\frac{y+3}{2}} \implies x = 2 + \sqrt{\frac{y+3}{2}}\).\
Hence, \(f^{-1}(x) = 2 + \sqrt{\frac{x+3}{2}}\).\
The domain of \(f^{-1}(x)\) is the range of \(f(x)\). For \(x \ge 2\), the minimum value of \(f(x)\) is \(-3\).\
Thus, the domain of \(f^{-1}(x)\) is \(x \ge -3\).\
\
(c) \(fg(x) = 11 \implies f(3x-1) = 11\).\
Using completed square form: \(2(3x-1 - 2)^2 - 3 = 11\)\
\(2(3x - 3)^2 = 14\)\
\((3x - 3)^2 = 7\)\
\(3x - 3 = \pm \sqrt{7}\)\
\(3x = 3 \pm \sqrt{7} \implies x = 1 \pm \frac{\sqrt{7}}{3}\).\
So \(x \approx 1.88\) or \(x \approx 0.118\).

(a)\
B1: States that \(f\) is many-to-one (or not one-to-one) on \(\mathbb{R}\).\
B1: Finds \(k = 2\).\
(b)\
M1: Attempts to complete the square or sets \(y = 2x^2-8x+5\) and uses quadratic formula to make \(x\) the subject.\
M1: Resolves the square root sign correctly using the domain restriction.\
A1: Obtains \(f^{-1}(x) = 2 + \sqrt{\frac{x+3}{2}}\) (or equivalent).\
A1: States domain is \(x \ge -3\).\
(c)\
M1: Forms equation \(2(3x-1)^2 - 8(3x-1) + 5 = 11\) and attempts to solve.\
A1: Obtains \(x = 1 \pm \frac{\sqrt{7}}{3}\) (or equivalent decimals: 1.88 and 0.118).

**(a)**
We use the product rule with \(u = 3x-2\) and \(v = (2x+1)^{1/2}\).

First, find the derivatives of both parts:
\(\frac{\mathrm{d}u}{\mathrm{d}x} = 3\)

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

Now apply the product rule \(\frac{\mathrm{d}y}{\mathrm{d}x} = u\frac{\mathrm{d}v}{\mathrm{d}x} + v\frac{\mathrm{d}u}{\mathrm{d}x}\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = (3x-2)\left(\frac{1}{\sqrt{2x+1}}\right) + 3\sqrt{2x+1}\)

Combine the terms over a common denominator of \(\sqrt{2x+1}\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(3x-2) + 3(2x+1)}{\sqrt{2x+1}}\)

\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3x-2 + 6x+3}{\sqrt{2x+1}} = \frac{9x+1}{\sqrt{2x+1}}\)

So, \(a = 9\) and \(b = 1\).

**(b)**
We want to find \(\int \frac{15x+4}{\sqrt{2x+1}} \, \mathrm{d}x\).

We write the numerator \(15x+4\) in terms of the numerator from part (a), which is \(9x+1\):
\(15x+4 = p(9x+1) + q\)

Equating coefficients of \(x\):
\(9p = 15 \implies p = \frac{5}{3}\)

Equating constant terms:
\(p + q = 4 \implies \frac{5}{3} + q = 4 \implies q = \frac{7}{3}\)

Thus, we can split the integral:
\(\int \frac{15x+4}{\sqrt{2x+1}} \, \mathrm{d}x = \int \left( \frac{5}{3} \cdot \frac{9x+1}{\sqrt{2x+1}} + \frac{7}{3} \cdot \frac{1}{\sqrt{2x+1}} \right) \mathrm{d}x\)

\(= \frac{5}{3} \int \frac{9x+1}{\sqrt{2x+1}} \, \mathrm{d}x + \frac{7}{3} \int (2x+1)^{-1/2} \, \mathrm{d}x\)

Using the result from part (a):
\(\int \frac{9x+1}{\sqrt{2x+1}} \, \mathrm{d}x = (3x-2)\sqrt{2x+1}\)

Also, integrating the second term:
\(\int (2x+1)^{-1/2} \, \mathrm{d}x = \frac{(2x+1)^{1/2}}{\frac{1}{2} \cdot 2} = \sqrt{2x+1}\)

Substituting these back:
\(\int \frac{15x+4}{\sqrt{2x+1}} \, \mathrm{d}x = \frac{5}{3}(3x-2)\sqrt{2x+1} + \frac{7}{3}\sqrt{2x+1} + C\)

Factor out \(\sqrt{2x+1}\):
\(= \sqrt{2x+1} \left[ \frac{5}{3}(3x-2) + \frac{7}{3} \right] + C\)

\(= \sqrt{2x+1} \left[ 5x - \frac{10}{3} + \frac{7}{3} \right] + C\)

\(= (5x-1)\sqrt{2x+1} + C\)

**(a)**
- **M1**: For applying the product rule to \(y = (3x-2)\sqrt{2x+1}\), with correct derivative of at least one term (chain rule correctly applied for the second term).
- **A1**: For obtaining \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3x-2}{\sqrt{2x+1}} + 3\sqrt{2x+1}\) or equivalent.
- **M1**: For putting the terms over a common denominator of \(\sqrt{2x+1}\).
- **A1**: For obtaining the correct simplified form \(\frac{9x+1}{\sqrt{2x+1}}\) (explicitly stating \(a=9\) and \(b=1\)).

**(b)**
- **M1**: For expressing \(15x+4\) in the form \(p(9x+1) + q\) (or attempting substitution \(u = \sqrt{2x+1}\)).
- **A1**: For finding the correct values \(p = \frac{5}{3}\) and \(q = \frac{7}{3}\) (or obtaining the correct integral in terms of \(u\): \(\int (\frac{15}{2}u^2 - \frac{7}{2}) \, \mathrm{d}u\)).
- **M1**: For integrating both parts of their split expression, utilizing the result of part (a) for the first part.
- **A1**: For obtaining the fully simplified final answer \((5x-1)\sqrt{2x+1} + C\) (constant of integration must be present).

**(a)**
Start with the Left-Hand Side (LHS) of the identity:
\(\text{LHS} = \frac{\cos\theta}{1 - \tan\theta} + \frac{\sin\theta}{1 - \cot\theta}\)

Express \(\tan\theta\) and \(\cot\theta\) in terms of \(\sin\theta\) and \(\cos\theta\):
\(\text{LHS} = \frac{\cos\theta}{1 - \frac{\sin\theta}{\cos\theta}} + \frac{\sin\theta}{1 - \frac{\cos\theta}{\sin\theta}}\)

Simplify the denominators of both fractions:
\(\text{LHS} = \frac{\cos\theta}{\frac{\cos\theta - \sin\theta}{\cos\theta}} + \frac{\sin\theta}{\frac{\sin\theta - \cos\theta}{\sin\theta}}\)

Multiply the numerators by the reciprocals of the denominators:
\(\text{LHS} = \frac{\cos^2\theta}{\cos\theta - \sin\theta} + \frac{\sin^2\theta}{\sin\theta - \cos\theta}\)

Since \(\sin\theta - \cos\theta = -(\cos\theta - \sin\theta)\), we can write:
\(\text{LHS} = \frac{\cos^2\theta}{\cos\theta - \sin\theta} - \frac{\sin^2\theta}{\cos\theta - \sin\theta}\)

Combine the fractions over the common denominator:
\(\text{LHS} = \frac{\cos^2\theta - \sin^2\theta}{\cos\theta - \sin\theta}\)

Factor the numerator using the difference of squares:
\(\text{LHS} = \frac{(\cos\theta - \sin\theta)(\cos\theta + \sin\theta)}{\cos\theta - \sin\theta}\)

Cancel the common factor \(\cos\theta - \sin\theta\):
\(\text{LHS} = \cos\theta + \sin\theta = \text{RHS}\)

**(b)**
Using the identity proven in part (a) with \(\theta = 2x - \frac{\pi}{6}\), the equation becomes:
\(\sin\left(2x - \frac{\pi}{6}\right) + \cos\left(2x - \frac{\pi}{6}\right) = \sqrt{2}\cos\left(2x - \frac{\pi}{6}\right)\)

Subtract \(\cos\left(2x - \frac{\pi}{6}\right)\) from both sides:
\(\sin\left(2x - \frac{\pi}{6}\right) = (\sqrt{2} - 1)\cos\left(2x - \frac{\pi}{6}\right)\)

Divide by \(\cos\left(2x - \frac{\pi}{6}\right)\):
\(\tan\left(2x - \frac{\pi}{6}\right) = \sqrt{2} - 1\)

Let \(\theta = 2x - \frac{\pi}{6}\).
Given \(0 \le x \le \pi\), the range for \(\theta\) is:
\(0 \le 2x \le 2\pi \implies -\frac{\pi}{6} \le 2x - \frac{\pi}{6} \le 2\pi - \frac{\pi}{6}\)
\(-\frac{\pi}{6} \le \theta \le \frac{11\pi}{6}\)

Find the principal value of \(\theta\):
\(\theta = \tan^{-1}(\sqrt{2} - 1) = \frac{\pi}{8}\)

Since the tangent function is positive, the other solution in the range is in the third quadrant:
\(\theta = \frac{\pi}{8} + \pi = \frac{9\pi}{8}\)

Now solve for \(x\) for each case:

1) \(2x - \frac{\pi}{6} = \frac{\pi}{8} \implies 2x = \frac{\pi}{8} + \frac{\pi}{6} = \frac{7\pi}{24} \implies x = \frac{7\pi}{48}\)

2) \(2x - \frac{\pi}{6} = \frac{9\pi}{8} \implies 2x = \frac{9\pi}{8} + \frac{\pi}{6} = \frac{31\pi}{24} \implies x = \frac{31\pi}{48}\)

Both values lie within the interval \([0, \pi]\).

**(a)**
- **M1**: For substituting \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) and \(\cot\theta = \frac{\cos\theta}{\sin\theta}\).
- **M1**: For simplifying both fractions to get common denominators or equivalent common denominator forms (e.g. \(\cos\theta - \sin\theta\)).
- **A1**: For combining into the single fraction \(\frac{\cos^2\theta - \sin^2\theta}{\cos\theta - \sin\theta}\).
- **A1**: For factoring the numerator \(\cos^2\theta - \sin^2\theta\) as \((\cos\theta-\sin\theta)(\cos\theta+\sin\theta)\) and completing the proof to obtain \(\sin\theta+\cos\theta\).

**(b)**
- **M1**: For using the identity from part (a) to rewrite the equation as \(\sin\theta + \cos\theta = \sqrt{2}\cos\theta\) where \(\theta = 2x - \frac{\pi}{6}\).
- **M1**: For reducing the equation to the form \(\tan\left(2x - \frac{\pi}{6}\right) = \sqrt{2} - 1\).
- **A1**: For obtaining one correct solution for \(x\) (either \(x = \frac{7\pi}{48}\) or \(x = \frac{31\pi}{48}\); accept decimal equivalents \(x \approx 0.458\) or \(x \approx 2.03\)).
- **A1**: For obtaining the second correct solution for \(x\) and no other solutions within the range.