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Find the first differences: \(12 - 5 = 7\), \(23 - 12 = 11\), \(38 - 23 = 15\), \(57 - 38 = 19\). Find the second differences: \(11 - 7 = 4\), \(15 - 11 = 4\), \(19 - 15 = 4\). Since the second difference is constant and equal to 4, the coefficient of \(n^2\) is \(4 / 2 = 2\). Subtract \(2n^2\) from each term: For \(n=1\): \(5 - 2(1)^2 = 3\). For \(n=2\): \(12 - 2(2)^2 = 4\). For \(n=3\): \(23 - 2(3)^2 = 5\). This linear sequence \(3, 4, 5, \dots\) has \(n\)-th term \(n + 2\). Therefore, the overall \(n\)-th term is \(2n^2 + n + 2\).

M1 for finding second difference of 4 or attempting to fit quadratic form. M1 for identifying the linear part as \(n + 2\). A0.5 for \(2n^2 + n + 2\).

Using the laws of logarithms, \(\log_3((x - 2)(x + 4)) = 3\). Convert to exponential form: \((x - 2)(x + 4) = 3^3\), which expands to \(x^2 + 2x - 8 = 27\). Rearranging gives \(x^2 + 2x - 35 = 0\). Factoring the quadratic: \((x + 7)(x - 5) = 0\), which yields \(x = -7\) or \(x = 5\). Since the argument of a logarithm must be positive, \(x - 2 > 0 \implies x > 2\). Thus, we reject \(x = -7\), leaving \(x = 5\).

M1 for applying the log addition rule correctly. M1 for setting up the quadratic equation \(x^2 + 2x - 35 = 0\). A0.5 for \(x = 5\) and rejecting \(x = -7\).

First, find the gradient \(m\): \(m = \frac{-5 - 11}{5 - (-3)} = \frac{-16}{8} = -2\). Next, use the point-slope formula with \((5, -5)\): \(y - (-5) = -2(x - 5)\) which simplifies to \(y + 5 = -2x + 10\). Rearranging into the required form gives \(y = -2x + 5\).

M1 for finding the gradient \(m = -2\). M1 for substituting a point into \(y = mx + c\) to find \(c\). A0.5 for \(y = -2x + 5\).

Multiply both sides by \((5 - 2x)\): \(y(5 - 2x) = 3x + 2\). Expand the expression: \(5y - 2xy = 3x + 2\). Group all terms containing \(x\) on one side: \(5y - 2 = 3x + 2xy\). Factor out \(x\): \(5y - 2 = x(3 + 2y)\). Divide by \((3 + 2y)\) to isolate \(x\): \(x = \frac{5y - 2}{2y + 3}\).

M1 for clearing the fraction to obtain \(y(5 - 2x) = 3x + 2\). M1 for grouping \(x\) terms and factoring out \(x\). A0.5 for \(x = \frac{5y - 2}{2y + 3}\) or equivalent.

Rearranging the equation to solve for \(\mathbf{c}\): \(3\mathbf{a} - 5\mathbf{b} = 2\mathbf{c} \implies \mathbf{c} = \frac{1}{2}(3\mathbf{a} - 5\mathbf{b})\). Calculate the components: \(3\mathbf{a} = 3\begin{pmatrix} 3 \\ -4 \end{pmatrix} = \begin{pmatrix} 9 \\ -12 \end{pmatrix}\) and \(5\mathbf{b} = 5\begin{pmatrix} -1 \\ 2 \end{pmatrix} = \begin{pmatrix} -5 \\ 10 \end{pmatrix}\). Substitute these into the expression: \(3\mathbf{a} - 5\mathbf{b} = \begin{pmatrix} 9 - (-5) \\ -12 - 10 \end{pmatrix} = \begin{pmatrix} 14 \\ -22 \end{pmatrix}\). Therefore, \(\mathbf{c} = \frac{1}{2}\begin{pmatrix} 14 \\ -22 \end{pmatrix} = \begin{pmatrix} 7 \\ -11 \end{pmatrix}\).

M1 for correctly expressing \(2\mathbf{c}\) or \(\mathbf{c}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\). M1 for calculating \(3\mathbf{a} - 5\mathbf{b} = \begin{pmatrix} 14 \\ -22 \end{pmatrix}\\. A0.5 for \)\\begin{pmatrix} 7 \\ -11 \\end{pmatrix}\).

First, find the composite function \(g(f(x))\): \(g(f(x)) = g(2x - 3) = \frac{12}{(2x - 3) + 1} = \frac{12}{2x - 2} = \frac{6}{x - 1}\). Now, set this expression equal to 3: \(\frac{6}{x - 1} = 3 \implies 6 = 3(x - 1) \implies 6 = 3x - 3 \implies 3x = 9 \implies x = 3\).

M1 for obtaining the composite function \(\frac{12}{2x - 2}\) or equivalent. M1 for setting their composite function equal to 3 and solving for \(x\). A0.5 for \(3\).

First, isolate the cosine term: \(\cos(2x) = -\frac{\sqrt{3}}{2}\). Let \(\theta = 2x\). Since \(0^\circ \le x \le 180^\circ\), we have \(0^\circ \le \theta \le 360^\circ\). The reference angle for which \(\cos \theta_0 = \frac{\sqrt{3}}{2}\) is \(30^\circ\). Since cosine is negative, \(\theta\) must lie in the second or third quadrant: \(\theta = 180^\circ - 30^\circ = 150^\circ\) or \(\theta = 180^\circ + 30^\circ = 210^\circ\). Therefore, \(2x = 150^\circ \implies x = 75^\circ\) or \(2x = 210^\circ \implies x = 105^\circ\).

M1 for isolating cosine: \(\cos(2x) = -\frac{\sqrt{3}}{2}\). M1 for identifying the angles \(2x = 150^\circ\) or \(2x = 210^\circ\). A0.5 for \(75^\circ, 105^\circ\).

Let \(P\) be the original price of the bicycle. A reduction of \(15\\%\) means the sale price is \(85\\%\) of the original price: \(0.85P = 306\). Solving for \(P\): \(P = \frac{306}{0.85} = \frac{306 \times 100}{85} = \frac{306 \times 20}{17}\). Since \(306 \div 17 = 18\), we find \(P = 18 \times 20 = 360\). The original price of the bicycle was \(\\\$360\).

M1 for translating the percentage reduction into the equation \(0.85P = 306\) or equivalent. M1 for demonstrating a correct non-calculator division technique to solve for \(P\). A0.5 for \(360\).

Given the general term \(u_n = an^2 + bn - 3\):
1. For \(n = 1\):
\(u_1 = a(1)^2 + b(1) - 3 = 4\)
\(a + b - 3 = 4 \implies a + b = 7\) (Equation 1)

2. For \(n = 2\):
\(u_2 = a(2)^2 + b(2) - 3 = 15\)
\(4a + 2b - 3 = 15 \implies 4a + 2b = 18 \implies 2a + b = 9\) (Equation 2)

Subtracting Equation 1 from Equation 2:
\((2a + b) - (a + b) = 9 - 7\)
\(a = 2\)

Substitute \(a = 2\) back into Equation 1:
\(2 + b = 7 \implies b = 5\)

Thus, \(a = 2\) and \(b = 5\).

M1 for substituting \(n=1\) and \(n=2\) to write two equations in terms of \(a\) and \(b\).
M1 for a valid method to solve the simultaneous equations (elimination or substitution).
A0.5 for \(a = 2\).
A0.5 for \(b = 5\).

Using the laws of logarithms:
\(\log_3((x - 2)(x + 4)) = 3\)

Convert the logarithmic equation to its exponential form:
\((x - 2)(x + 4) = 3^3\)
\(x^2 + 2x - 8 = 27\)
\(x^2 + 2x - 35 = 0\)

Factorise the quadratic equation:
\((x + 7)(x - 5) = 0\)

This gives potential solutions:
\(x = -7\) or \(x = 5\)

We must check the domain of the logarithmic terms:
- For \(x = -7\), \(x - 2 = -9\), which is undefined since we cannot take the logarithm of a negative number.
- For \(x = 5\), both \(x - 2 = 3 > 0\) and \(x + 4 = 9 > 0\) are valid.

Thus, the only valid solution is \(x = 5\).

M1 for applying the addition rule of logarithms to obtain \(\log_3((x - 2)(x + 4)) = 3\).
M1 for converting to quadratic equation form \(x^2 + 2x - 35 = 0\).
A0.5 for solving and correctly identifying \(x = 5\) as the only valid solution (rejecting \(x = -7\)).

First, find the gradient of the line \(L\):
\(m = \frac{-5 - 7}{4 - (-2)} = \frac{-12}{6} = -2\)

Since the required line is parallel to \(L\), it has the same gradient, \(m = -2\).

Using the point-slope form with the point \((1, 3)\):
\(y - 3 = -2(x - 1)\)
\(y - 3 = -2x + 2\)
\(y = -2x + 5\)

M1 for finding the gradient of \(L\) as \(-2\).
M1 for using \(y - y_1 = m(x - x_1)\) or \(y = mx + c\) with their gradient and the point \((1, 3)\).
A0.5 for the correct final equation \(y = -2x + 5\).

First, factor out the common term, which is \(2\):
\(12x^2 - 14xy - 6y^2 = 2(6x^2 - 7xy - 3y^2)\)

Next, factorise the quadratic expression inside the parentheses, \(6x^2 - 7xy - 3y^2\):
We look for two numbers that multiply to \(6 \times (-3) = -18\) and add up to \(-7\). These numbers are \(-9\) and \(2\).

Rewrite the middle term:
\(6x^2 - 9xy + 2xy - 3y^2\)

Factor by grouping:
\(3x(2x - 3y) + y(2x - 3y) = (3x + y)(2x - 3y)\)

Putting it all together, the fully factorised expression is:
\(2(3x + y)(2x - 3y)\)

M1 for factoring out the common factor of \(2\) to get \(2(6x^2 - 7xy - 3y^2)\).
M1 for a correct method of factorising the quadratic trinomial (e.g., splitting the middle term into \(-9xy + 2xy\)).
A0.5 for the fully factorised correct expression.

Rearrange the vector equation to solve for \(\mathbf{c}\):
\(3\mathbf{a} - 4\mathbf{b} = 2\mathbf{c}\)
\(2\mathbf{c} = 3\begin{pmatrix} 4 \\ -2 \end{pmatrix} - 4\begin{pmatrix} -1 \\ 5 \end{pmatrix}\)

Calculate each term:
\(3\mathbf{a} = \begin{pmatrix} 12 \\ -6 \end{pmatrix}\)
\(4\mathbf{b} = \begin{pmatrix} -4 \\ 20 \end{pmatrix}\)

Subtract the vectors:
\(2\mathbf{c} = \begin{pmatrix} 12 - (-4) \\ -6 - 20 \end{pmatrix} = \begin{pmatrix} 16 \\ -26 \end{pmatrix}\)

Divide by 2 to find \(\mathbf{c}\):
\(\mathbf{c} = \begin{pmatrix} 8 \\ -13 \end{pmatrix}\)

M1 for rearranging the equation to express \(2\mathbf{c}\) (or \(\mathbf{c}\)) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
M1 for correctly substituting and evaluating \(3\mathbf{a} - 4\mathbf{b}\) to find \(\begin{pmatrix} 16 \\ -26 \end{pmatrix}\).
A0.5 for the correct final column vector \(\begin{pmatrix} 8 \\ -13 \end{pmatrix}\).

First, evaluate \(g(4)\):
\(g(4) = 5 - 2(4) = 5 - 8 = -3\)

Now we need to find \(f^{-1}(-3)\). Let \(f^{-1}(-3) = x\), which means \(f(x) = -3\).

Set up the equation:
\(\frac{3x - 1}{2} = -3\)
\(3x - 1 = -6\)
\(3x = -5\)
\(x = -\frac{5}{3}\)

Therefore, \(f^{-1}(g(4)) = -\frac{5}{3}\).

M1 for finding \(g(4) = -3\).
M1 for setting up the equation \(f(x) = -3\) or for finding the inverse function formula \(f^{-1}(x) = \frac{2x + 1}{3}\).
A0.5 for obtaining the final correct value of \(-\frac{5}{3}\) (or equivalent fraction/decimal).

Rearrange the equation to find \(\sin\theta\):
\(2\sin\theta = \sqrt{3} \implies \sin\theta = \frac{\sqrt{3}}{2}\)

Since \(\sin\theta\) is positive, \(\theta\) lies in the first and second quadrants.

The principal value (first-quadrant angle) where \(\sin\theta = \frac{\sqrt{3}}{2}\) is:
\(\theta = 60^\circ\)

The second-quadrant angle is:
\(\theta = 180^\circ - 60^\circ = 120^\circ\)

Both values lie in the given interval \(0^\circ \le \theta \le 360^\circ\).

M1 for rearranging to \(\sin\theta = \frac{\sqrt{3}}{2}\).
M1 for finding the basic angle of \(60^\circ\) or the second quadrant angle \(120^\circ\).
A0.5 for giving both correct solutions: \(60^\circ\) and \(120^\circ\) (and no extra incorrect solutions within the range).

Let the original price of the bicycle be \(P\).
A reduction of \(15\%\) means the sale price is \(100\% - 15\% = 85\%\) of the original price.

So, \(0.85 \times P = 306\)

Solve for \(P\):
\(P = \frac{306}{0.85} = \frac{30600}{85}\)

Simplify the fraction by dividing numerator and denominator by 5:
\(P = \frac{6120}{17}\)

Perform the division:
\(6120 \div 17 = 360\)

Thus, the original price was \(\$360\).

M1 for setting up the equation \(0.85P = 306\) (or equivalent statement like \(85\% = 306\)).
M1 for performing a correct division method, e.g., \(\frac{306}{0.85}\).
A0.5 for the correct final answer of \(360\).

(a) Using differences: Terms: 5, 12, 23, 38. First differences: 7, 11, 15. Second differences: 4, 4. Since the second differences are constant, the sequence is quadratic, and \(2a = 4 \implies a = 2\). Using \(u_1 = 5 \implies 2(1)^2 + b(1) + c = 5 \implies b + c = 3\). Using \(u_2 = 12 \implies 2(2)^2 + b(2) + c = 12 \implies 8 + 2b + c = 12 \implies 2b + c = 4\). Subtracting the first equation from the second gives \(b = 1\), which yields \(c = 2\). Therefore, \(a = 2\), \(b = 1\), and \(c = 2\). (b) The nth term is \(u_n = 2n^2 + n + 2\). For \(n = 20\): \(u_{20} = 2(20)^2 + 20 + 2 = 800 + 20 + 2 = 822\). (c) Set \(2n^2 + n + 2 = 1277 \implies 2n^2 + n - 1275 = 0\). Using the quadratic formula: \(n = \frac{-1 \pm \sqrt{1^2 - 4(2)(-1275)}}{2(2)} = \frac{-1 \pm \sqrt{10201}}{4} = \frac{-1 \pm 101}{4}\). Since \(n > 0\), we have \(n = \frac{100}{4} = 25\).

(a) [M1] for finding a second difference of 4 or setting up simultaneous equations. [A1] for \(a = 2\). [M1] for setting up a correct system for \(b\) and \(c\). [A1] for \(b = 1\). [A1] for \(c = 2\). (b) [M1] for substituting \(n=20\) into their formula. [A1] for 822. (c) [M1] for setting their quadratic expression equal to 1277. [M1] for a correct method to solve the quadratic equation. [A1.9] for \(n = 25\).

(a) Taking the base-10 logarithm on both sides of \(y = A x^b\) gives \(\log_{10} y = \log_{10}(A \cdot x^b) = \log_{10} A + \log_{10}(x^b) = b \log_{10} x + \log_{10} A\). (b) By letting \(Y = \log_{10} y\) and \(X = \log_{10} x\), the equation can be written as \(Y = b X + \log_{10} A\). This matches the slope-intercept form \(Y = mX + C\), where the gradient \(m = b\) and the vertical intercept \(C = \log_{10} A\). (c) The line passes through (1, 1.5) and (3, 5.5). The gradient \(b = \frac{5.5 - 1.5}{3 - 1} = \frac{4}{2} = 2\). The line's equation is \(Y = 2X + C\). Substituting (1, 1.5): \(1.5 = 2(1) + C \implies C = -0.5\). Since \(C = \log_{10} A\), we have \(\log_{10} A = -0.5 \implies A = 10^{-0.5} = \frac{1}{\sqrt{10}} = \frac{\sqrt{10}}{10}\).

(a) [M1] for taking logs of both sides. [A2] for \(\log_{10} y = b \log_{10} x + \log_{10} A\) (award [1] for \(\log_{10} A + \log_{10} x^b\)). (b) [M1] for relating to \(Y=mX+c\) format. [A1] for identifying gradient and intercept terms. (c) [M2] for gradient calculation \(b = 2\). [M1] for finding the intercept \(C = -0.5\). [M1] for solving \(A = 10^{-0.5}\). [A1.9] for the exact value \(A = \frac{1}{\sqrt{10}}\) or equivalent simplified surd.

(a) Gradient of \(L_1 = \frac{5 - (-3)}{6 - 2} = \frac{8}{4} = 2\). (b) Using \(y - y_1 = m(x - x_1)\) with point \(Q(6, 5)\): \(y - 5 = 2(x - 6) \implies y = 2x - 12 + 5 \implies y = 2x - 7\). (c) Since \(L_2\) is parallel to \(L_1\), its gradient is also 2. Using \(R(-4, 7)\): \(y - 7 = 2(x + 4) \implies y = 2x + 8 + 7 \implies y = 2x + 15\). To find the y-axis intersection, substitute \(x = 0\), giving \(y = 15\). The coordinates are \((0, 15)\). (d) The midpoint of \(PQ\) is \(M = \left(\frac{2+6}{2}, \frac{-3+5}{2}\right) = (4, 1)\). Since \(L_3\) is perpendicular to \(L_1\), its gradient is \(-\frac{1}{2} = -0.5\). Using \(M(4, 1)\): \(y - 1 = -0.5(x - 4) \implies y = -0.5x + 2 + 1 \implies y = -0.5x + 3\).

(a) [M1] for \(\frac{5 - (-3)}{6 - 2}\). [A1] for 2. (b) [M1] for a correct equation method using their gradient. [A1] for \(y = 2x - 7\). (c) [M1] for identifying gradient of \(L_2\) as 2. [M1] for substituting \((-4,7)\) to find the intercept. [A1] for \((0, 15)\). (d) [M1] for finding the midpoint \((4, 1)\). [M1] for using perpendicular gradient \(m = -0.5\). [A1.9] for \(y = -0.5x + 3\).

(a) From the second equation, \(y = x + 1\). Substitute this into the first equation: \(2x^2 - (x+1)^2 = 14 \implies 2x^2 - (x^2 + 2x + 1) = 14 \implies x^2 - 2x - 1 = 14 \implies x^2 - 2x - 15 = 0\). Factorising: \((x - 5)(x + 3) = 0\), which gives \(x = 5\) or \(x = -3\). Substitute these back into \(y = x + 1\): if \(x = 5\), then \(y = 6\); if \(x = -3\), then \(y = -2\). The solutions are \(x = 5, y = 6\) and \(x = -3, y = -2\). (b) Determine the highest common factor of the terms: \(\text{HCF}(12a^2b, 18ab^2) = 6ab\). Factorising gives: \(12a^2b - 18ab^2 = 6ab(2a - 3b)\). (c) Multiply both sides by \((5 - t)\): \(S(5 - t) = 3t + 2 \implies 5S - St = 3t + 2\). Group all terms involving \(t\) on one side: \(5S - 2 = 3t + St\). Factorise \(t\) out: \(5S - 2 = t(3 + S)\). Divide by \((3 + S)\) to isolate \(t\): \(t = \frac{5S - 2}{S + 3}\).

(a) [M1] for rearranging the linear equation to express one variable in terms of the other. [M1] for substituting into the quadratic equation to obtain a standard quadratic in one variable. [M1] for solving the quadratic. [A1] for one correct pair of \((x, y)\). [A1] for the second correct pair. (b) [M1] for factorising out any common factors. [A1] for \(6ab(2a - 3b)\). (c) [M1] for multiplying both sides by \((5-t)\). [M1] for collecting all terms with \(t\) on one side. [A1.9] for \(t = \frac{5S - 2}{S + 3}\) or equivalent.

(a)(i) \(\vec{AB} = \vec{AO} + \vec{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}\). (ii) \(\vec{OP} = \vec{OA} + \vec{AP} = \mathbf{a} + \frac{2}{5}\vec{AB} = \mathbf{a} + \frac{2}{5}(\mathbf{b} - \mathbf{a}) = \frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\). (iii) Since \(Q\) is the midpoint of \(OB\), \(\vec{OQ} = \frac{1}{2}\mathbf{b}\). Thus, \(\vec{AQ} = \vec{AO} + \vec{OQ} = -\mathbf{a} + \frac{1}{2}\mathbf{b}\). (b) \(\vec{PQ} = \vec{PO} + \vec{OQ} = -\vec{OP} + \vec{OQ} = -\left(\frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\right) + \frac{1}{2}\mathbf{b} = -\frac{3}{5}\mathbf{a} - \frac{2}{5}\mathbf{b} + \frac{1}{2}\mathbf{b} = -\frac{3}{5}\mathbf{a} + \frac{1}{10}\mathbf{b}\). (c) Substitute \(\mathbf{a}\) and \(\mathbf{b}\): \(\vec{PQ} = -\frac{3}{5}\begin{pmatrix} 3 \\ 4 \end{pmatrix} + \frac{1}{10}\begin{pmatrix} -5 \\ 10 \end{pmatrix} = \begin{pmatrix} -1.8 \\ -2.4 \end{pmatrix} + \begin{pmatrix} -0.5 \\ 1 \end{pmatrix} = \begin{pmatrix} -2.3 \\ -1.4 \end{pmatrix}\). The magnitude is \(|\vec{PQ}| = \sqrt{(-2.3)^2 + (-1.4)^2} = \sqrt{5.29 + 1.96} = \sqrt{7.25} \approx 2.69\).

(a)(i) [A1] for \(\mathbf{b} - \mathbf{a}\). (a)(ii) [M1] for \(\mathbf{a} + \frac{2}{5}\vec{AB}\). [A1] for \(\frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\). (a)(iii) [A1] for \(-\mathbf{a} + \frac{1}{2}\mathbf{b}\). (b) [M1] for a correct vector addition path. [M1] for combining like terms correctly to obtain \(-\frac{3}{5}\mathbf{a} + \frac{1}{10}\mathbf{b}\). (c) [M1] for finding the components of \(\vec{PQ}\) as \(\begin{pmatrix} -2.3 \\ -1.4 \end{pmatrix}\). [M1] for utilizing the Pythagoras theorem. [A1.9] for \(2.69\) (or \(\sqrt{7.25}\)).

(a) \(f(4) = \frac{2(4) + 1}{4 - 3} = \frac{9}{1} = 9\). (b) Let \(y = \frac{2x + 1}{x - 3}\). Multiply by \((x - 3)\): \(y(x - 3) = 2x + 1 \implies xy - 3y = 2x + 1\). Rearrange to collect \(x\) terms: \(xy - 2x = 3y + 1 \implies x(y - 2) = 3y + 1\). This yields \(x = \frac{3y + 1}{y - 2}\). Thus, \(f^{-1}(x) = \frac{3x + 1}{x - 2}\). (c) \(g(f(x)) = 3\left(\frac{2x + 1}{x - 3}\right) - 5 = \frac{6x + 3}{x - 3} - \frac{5(x - 3)}{x - 3} = \frac{6x + 3 - (5x - 15)}{x - 3} = \frac{x + 18}{x - 3}\). (d) First, calculate \(g(2) = 3(2) - 5 = 1\). Now solve \(f(x) = 1 \implies \frac{2x + 1}{x - 3} = 1 \implies 2x + 1 = x - 3 \implies x = -4\).

(a) [A1] for 9. (b) [M1] for setting \(y = \frac{2x+1}{x-3}\) and multiplying. [M1] for isolating \(x\) terms. [A1] for \(f^{-1}(x) = \frac{3x + 1}{x - 2}\). (c) [M1] for substituting \(f(x)\) into \(g(x)\). [M1] for writing over a common denominator. [A1] for \(\frac{x + 18}{x - 3}\). (d) [M1] for evaluating \(g(2) = 1\). [M1] for equating \(f(x) = 1\). [A1.9] for \(x = -4\).

(a)(i) The amplitude \(a\) is half the difference between the maximum and minimum values: \(a = \frac{7 - (-1)}{2} = 4\). (ii) The vertical shift \(c\) is the midline value: \(c = \frac{7 + (-1)}{2} = 3\). (iii) The period of \(y = \sin(bx)\) is \(\frac{360^\circ}{b}\). We are given the period is \(120^\circ\), so \(\frac{360^\circ}{b} = 120^\circ \implies b = 3\). (b) We solve \(4 \sin(3x) + 3 = 2\) for \(0^\circ \le x \le 120^\circ\). This simplifies to \(4 \sin(3x) = -1 \implies \sin(3x) = -0.25\). Let \(\theta = 3x\), where \(0^\circ \le \theta \le 360^\circ\). \(\sin \theta = -0.25\). The acute reference angle is \(\sin^{-1}(0.25) \approx 14.48^\circ\). Since sine is negative, \(\theta\) is in Quadrant III or IV: Quadrant III: \(\theta_1 = 180^\circ + 14.48^\circ = 194.48^\circ\), Quadrant IV: \(\theta_2 = 360^\circ - 14.48^\circ = 345.52^\circ\). Dividing these angles by 3 to find \(x\): \(x_1 = \frac{194.48^\circ}{3} \approx 64.8^\circ\), and \(x_2 = \frac{345.52^\circ}{3} \approx 115^\circ\) (to 3 significant figures).

(a)(i) [A1] for \(a = 4\). (ii) [A1] for \(c = 3\). (iii) [M1] for period formula \(360/b = 120\). [A1] for \(b = 3\). (b) [M1] for simplifying equation to \(\sin(3x) = -0.25\). [M1] for finding reference angle \(14.5^\circ\). [M1] for identifying quadrant angles \(194.5^\circ\) or \(345.5^\circ\). [A1] for \(x = 64.8^\circ\) (accept 64.82 to 64.83). [A1.9] for \(x = 115^\circ\) (accept 115.1 to 115.2).

(a) The probabilities for the first choice are Girl: 7/12, Boy: 5/12. After a girl is chosen, the remaining group has 11 students (6 girls, 5 boys), so the probabilities are Girl: 6/11, Boy: 5/11. After a boy is chosen, the remaining group has 11 students (7 girls, 4 boys), so the probabilities are Girl: 7/11, Boy: 4/11. (b)(i) \(\text{P(Both Girls)} = \frac{7}{12} \times \frac{6}{11} = \frac{42}{132} = \frac{7}{22} \approx 0.318\). (ii) \(\text{P(At least one boy)} = 1 - \text{P(GG)} = 1 - \frac{7}{22} = \frac{15}{22} \approx 0.682\). (c) We require \(\text{P(First Boy | Second Girl)} = \frac{\text{P(Boy and then Girl)}}{\text{P(Second Girl)}}\). The numerator \(\text{P(BG)} = \frac{5}{12} \times \frac{7}{11} = \frac{35}{132}\). The total probability of selecting a girl second is \(\text{P(Second Girl)} = \text{P(GG)} + \text{P(BG)} = \frac{42}{132} + \frac{35}{132} = \frac{77}{132}\). Thus, the conditional probability is \(\frac{35/132}{77/132} = \frac{35}{77} = \frac{5}{11} \approx 0.455\).

(a) [M1] for a correct tree structure showing 4 outcomes. [A1] for correct first choice branches. [A1] for correct second choice branches. (b)(i) [M1] for \(\frac{7}{12} \times \frac{6}{11}\). [A1] for \(\frac{7}{22}\) (or 0.318). (b)(ii) [M1] for \(1 - \text{their (b)(i)}\) or correct summing of branches. [A1] for \(\frac{15}{22}\) (or 0.682). (c) [M1] for setting up the conditional probability relation. [M1] for finding the denominator \(\text{P(Second Girl)} = \frac{77}{132}\). [A1.9] for \(\frac{5}{11}\) (or 0.455).

(a) Start with the logarithmic equation:
\(\log_2(x + 3) + \log_2(x - 3) = 4\)

Combine the terms using the product rule of logarithms:
\(\log_2((x + 3)(x - 3)) = 4\)

Convert to exponential form:
\((x + 3)(x - 3) = 2^4\)
\(x^2 - 9 = 16\)
\(x^2 = 25\)
\(x = 5\) or \(x = -5\)

Since the arguments of logarithms must be positive, we require \(x + 3 > 0\) and \(x - 3 > 0\), which means \(x > 3\).
Therefore, the only valid solution is \(x = 5\).

(b) From the first equation:
\(\log_y(x) = 2 \implies x = y^2\)

From the second equation:
\(\log_y(3x - 2) = 3 \implies 3x - 2 = y^3\)

Substitute \(x = y^2\) into the second equation:
\(3y^2 - 2 = y^3\)
\(y^3 - 3y^2 + 2 = 0\)

By inspection, \(y = 1\) is a root. However, the base of a logarithm cannot be 1. Thus, \(y \neq 1\).

Factorize \(y^3 - 3y^2 + 2\) by dividing by \(y - 1\):
\((y - 1)(y^2 - 2y - 2) = 0\)

Since \(y \neq 1\), solve \(y^2 - 2y - 2 = 0\) using the quadratic formula:
\(y = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2} = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}\)

Since the base of a logarithm must be positive, we reject the negative root. Thus:
\(y = 1 + \sqrt{3}\) (as \(1 - \sqrt{3} < 0\))

Now substitute \(y\) back to find \(x\):
\(x = y^2 = (1 + \sqrt{3})^2 = 1 + 2\sqrt{3} + 3 = 4 + 2\sqrt{3}\).

(a)
M1 for using log product rule: \(\log_2((x+3)(x-3)) = 4\)
M1 for converting to exponential form: \(x^2 - 9 = 16\)
A1 for finding \(x = \pm 5\)
A1 for rejecting \(-5\) and stating final answer \(x = 5\)

(b)
M1 for writing \(x = y^2\)
M1 for writing \(3x - 2 = y^3\)
M1 for substituting to obtain \(y^3 - 3y^2 + 2 = 0\)
M1 for factoring out \((y-1)\) to get \((y-1)(y^2-2y-2) = 0\)
M1 for finding roots of the quadratic part: \(y = 1 \pm \sqrt{3}\)
A1 for selecting \(y = 1 + \sqrt{3}\) with justification
A1 for determining \(x = 4 + 2\sqrt{3}\)

(a) The \(n\)-th term of arithmetic sequence A with first term \(a=5\) is:
\(u_n = 5 + (n-1)d\)
So, \(u_3 = 5 + 2d\) and \(u_7 = 5 + 6d\).

The \(n\)-th term of geometric sequence B with first term \(b=5\) is:
\(v_n = 5r^{n-1}\)
So, \(v_2 = 5r\) and \(v_3 = 5r^2\).

Since \(u_3 = v_2\):
\(5 + 2d = 5r \implies 2d = 5r - 5\) (Equation 1)

Since \(u_7 = v_3\):
\(5 + 6d = 5r^2\) (Equation 2)

Multiply Equation 1 by 3:
\(6d = 15r - 15\)

Substitute \(6d\) into Equation 2:
\(5 + (15r - 15) = 5r^2\)
\(15r - 10 = 5r^2\)

Divide by 5:
\(3r - 2 = r^2 \implies r^2 - 3r + 2 = 0\) (Showed)

(b) Solve the quadratic equation:
\((r-1)(r-2) = 0\)
Since \(r > 1\), we have \(r = 2\).

Substitute \(r = 2\) into Equation 1:
\(2d = 5(2) - 5 \implies 2d = 5 \implies d = 2.5\).

(c) The 10th term of sequence B is:
\(v_{10} = 5 \times r^9 = 5 \times 2^9 = 5 \times 512 = 2560\).

(d) The sum of the first 10 terms of sequence A is:
\(S_{10} = \frac{10}{2} (2a + 9d) = 5 (2(5) + 9(2.5)) = 5 (10 + 22.5) = 5 \times 32.5 = 162.5\).

(a)
M1 for finding correct expressions for terms: \(u_3 = 5+2d\), \(u_7 = 5+6d\), \(v_2 = 5r\), \(v_3 = 5r^2\)
M1 for substituting \(d\) or \(6d\) to form a quadratic equation in \(r\)
A1 for convincing algebraic steps showing \(r^2 - 3r + 2 = 0\)

(b)
B1 for \(r = 2\) (rejecting \(r = 1\))
B1 for \(d = 2.5\)

(c)
M1 for \(5 \times 2^9\)
A1 for 2560

(d)
M1 for using correct arithmetic sum formula \(S_n = \frac{n}{2}(2a + (n-1)d)\)
M1 for substituting \(n=10\), \(a=5\), and \(d=2.5\)
A2 for 162.5 (A1 if there is a single arithmetic slip)

(a) Let \(O\) be the center of the sector and \(C\) be the center of the inscribed circle.
By symmetry, the line from \(O\) to the midpoint of the outer arc passes through \(C\) and bisects the sector angle into two \(60^\circ\) angles.

Let \(T\) be the point of contact where the inscribed circle is tangent to one of the radii of the sector.
Thus, \(CT\) is perpendicular to the radius \(OT\), forming a right-angled triangle \(OTC\) with \(\angle COT = 60^\circ\).

Since \(CT = r\), we have:
\(\sin(60^\circ) = \frac{r}{OC} \implies OC = \frac{r}{\sin(60^\circ)} = \frac{r}{\frac{\sqrt{3}}{2}} = \frac{2r}{\sqrt{3}}\)

The line segment from \(O\) to the outer boundary along the line of symmetry is equal to the radius of the sector, which is 12 cm. This distance is also \(OC + r\).

Therefore:
\(OC + r = 12\)
\(\frac{2r}{\sqrt{3}} + r = 12\)
\(r \left(\frac{2}{\sqrt{3}} + 1\right) = 12\)
\(r \left(\frac{2 + \sqrt{3}}{\sqrt{3}}\right) = 12\)
\(r = \frac{12\sqrt{3}}{2 + \sqrt{3}}\)

To rationalize the denominator, multiply the numerator and denominator by \(2 - \sqrt{3}\):
\(r = \frac{12\sqrt{3}(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{24\sqrt{3} - 36}{4 - 3} = 24\sqrt{3} - 36\text{ cm}\). (Showed)

(b) Area of the sector:
\(A_{\text{sector}} = \frac{120}{360} \times \pi \times 12^2 = 48\pi \approx 150.796\text{ cm}^2\).
To 3 significant figures, this is \(151\text{ cm}^2\).

(c) The exact value of \(r\) is:
\(r = 24\sqrt{3} - 36 \approx 5.5692\text{ cm}\).

The area of the inscribed circle is:
\(A_{\text{circle}} = \pi r^2 \approx \pi \times (5.5692)^2 \approx 97.438\text{ cm}^2\).

The percentage of the sector's area occupied is:
\(\text{Percentage} = \frac{A_{\text{circle}}}{A_{\text{sector}}} \times 100 = \frac{97.438}{150.796} \times 100 \approx 64.617\%\).
To 3 significant figures, this is \(64.6\%\).

(a)
M1 for identifying the angle of the bisected sector is \(60^\circ\)
M1 for writing \(OC = \frac{r}{\sin(60^\circ)}\)
M1 for establishing the relation \(OC + r = 12\)
M1 for obtaining the expression \(r = \frac{12\sqrt{3}}{2 + \sqrt{3}}\)
A1 for rationalizing correctly to show \(r = 24\sqrt{3} - 36\)

(b)
M1 for \(\frac{120}{360} \times \pi \times 12^2\)
A1 for 151 (or exact \(48\pi\))

(c)
M1 for finding \(r \approx 5.57\) or \(r^2 \approx 31.0\)
M1 for calculating circle area \(\approx 97.4\)
M1 for dividing circle area by sector area and multiplying by 100
A1 for 64.6% (accept 64.6)

1. \(C_5 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 = 1 + 8 + 27 + 64 + 125 = 225\). Alternatively, since \(C_n = (T_n)^2\), \(C_5 = (T_5)^2 = 15^2 = 225\). 2. To find \(C_{12}\), first find \(T_{12} = \frac{12 \times 13}{2} = 78\). Then, \(C_{12} = (T_{12})^2 = 78^2 = 6084\). 3. Since \(C_n = (T_n)^2\) and \(T_n = \frac{n(n+1)}{2}\), we square the expression: \(C_n = \left(\frac{n(n+1)}{2}\right)^2 = \frac{n^2(n+1)^2}{4}\).

B1 for C_5 = 225. M1 for calculating T_{12} = 78. A1 for C_{12} = 6084. M1 for attempting to square the algebraic expression for T_n. A2 for the correct simplified expression \frac{n^2(n+1)^2}{4}.

1. The differences between terms are 4, 7, 10, 13, so the next differences are 16, 19, 22. This gives P_6 = 51, P_7 = 70, P_8 = 92. 2. Since second difference is 3, 2a = 3 so a = 1.5. Using P_1 = 1, we get 1.5 + b = 1, so b = -0.5. Thus P_n = \frac{3n^2 - n}{2}. 3. P_{2k} = \frac{3(2k)^2 - (2k)}{2} = 6k^2 - k. P_k = 1.5k^2 - 0.5k. P_{2k} - P_k = 4.5k^2 - 0.5k = \frac{k(9k-1)}{2}.

B1 for P_8 = 92. M1 for using second differences to set up equations. A1 for a = 1.5 and b = -0.5. M1 for substituting 2k and k into the formula. M1 for attempting subtraction. A1 for the final simplified factorised expression \frac{k(9k-1)}{2}.

1. Working term by term: u_3 = a+b, u_4 = a+2b, u_5 = 2a+3b, u_6 = 3a+5b. 2. Sum S_5 = u_1 + u_2 + u_3 + u_4 + u_5 = a + b + (a+b) + (a+2b) + (2a+3b) = 5a + 7b. 3. Compare S_5 with u_7 = u_6 + u_5 = 5a+8b. We see S_5 = u_7 - b. Since u_2 = b, the general relation is S_n = u_{n+2} - u_2.

B1 for u_5 = 2a+3b and u_6 = 3a+5b. M1 for summing the 5 terms. A1 for S_5 = 5a + 7b. M1 for calculating u_7 = 5a+8b or comparing S_n and u_{n+2}. A2 for the formula S_n = u_{n+2} - u_2.

1. The sum of Group n is n^3, so Group 6 sum is 6^3 = 216. 2. The first elements are 1, 3, 7, 13, 21. Second differences are constant at 2, so the expression is quadratic: n^2 - n + 1. 3. The elements in Group n form an arithmetic progression with n terms, first term a = n^2 - n + 1 and last term l = n^2 + n - 1. Using S_n = \frac{n}{2}(a+l), we get S = \frac{n}{2}(n^2 - n + 1 + n^2 + n - 1) = \frac{n}{2}(2n^2) = n^3.

B1 for 216. M1 for second differences of the first elements. A1 for n^2 - n + 1. M1 for using the AP sum formula S_n = \frac{n}{2}(a+l). A2 for substitute-and-simplify to show S_n = n^3 clearly.

1. S_1 = 2, S_2 = 8, S_3 = 20, S_4 = 40, S_5 = 70, S_6 = 70 + 6(7) = 112. 2. S_1 = (1*2*3)/3, S_2 = (2*3*4)/3, S_3 = (3*4*5)/3. Thus, S_n = \frac{n(n+1)(n+2)}{3}. 3. S_n - S_{n-1} = \frac{n(n+1)(n+2)}{3} - \frac{(n-1)n(n+1)}{3} = \frac{n(n+1)}{3} [(n+2) - (n-1)] = \frac{n(n+1)}{3} [3] = n(n+1).

B1 for S_6 = 112. M1 for identifying the form of the formula. A1 for S_n = \frac{n(n+1)(n+2)}{3}. M1 for setting up the difference S_n - S_{n-1}. M1 for factoring. A1 for simplifying to n(n+1).

(a) At \(t = 0\), \(P_A(0) = 150 e^0 = 150\) and \(P_B(0) = 300 e^0 = 300\). (b) Set \(P_A(t) = P_B(t)\), so \(150 e^{0.18t} = 300 e^{0.08t}\). Dividing both sides by \(150 e^{0.08t}\) gives \(e^{0.10t} = 2\). Taking the natural logarithm gives \(0.10t = \ln(2)\), which yields \(t = 10 \ln(2) \approx 6.931...\) hours. To 3 significant figures, \(t = 6.93\) hours. (c) Since Strain A has a larger growth rate parameter (0.18 compared to 0.08), once its population overtakes Strain B at \(t \approx 6.93\), the population of Strain A will remain greater than the population of Strain B for all \(t > 6.93\).

(a) 1 mark: both 150 and 300 correct. (b) 3 marks: 1 mark for equating the two population formulas, 1 mark for simplifying to exponential or logarithmic form, 1 mark for correct value 6.93. (c) 2 marks: 1 mark for stating Strain A is larger, 1 mark for relating this to its faster growth rate.

(a) At \(t = 0\), \(H_1(0) = 5.2 + 1.5 \ln(1) = 5.2\) and \(H_2(0) = 4.0 + 2.1 \ln(1) = 4.0\). (b) Set \(5.2 + 1.5 \ln(t + 1) = 4.0 + 2.1 \ln(t + 1)\). This simplifies to \(1.2 = 0.6 \ln(t + 1)\), so \(2 = \ln(t + 1)\). Converting to exponential form, \(t + 1 = e^2\), so \(t = e^2 - 1 \approx 6.39\) minutes. (c) Reaction 2 increases faster because its growth factor coefficient is larger (2.1 > 1.5).

(a) 2 marks: 1 mark for each initial pH. (b) 3 marks: 1 mark for setting equations equal, 1 mark for finding \(\ln(t+1) = 2\) or \(t = e^2 - 1\), 1 mark for correct decimal value 6.39. (c) 1 mark for stating Reaction 2 with a valid explanation based on the coefficient.

(a) Since minimum cosine is -1, the minimum temperature occurs when cosine is 1 due to the negative coefficient. \(T_A = 15 - 12(1) = 3\) and \(T_B = 18 - 15(1) = 3\). (b) Setting them equal: \(15 - 12 \cos(\frac{\pi t}{6}) = 18 - 15 \cos(\frac{\pi t}{6}) \Rightarrow 3 \cos(\frac{\pi t}{6}) = 3 \Rightarrow \cos(\frac{\pi t}{6}) = 1\). Thus \(\frac{\pi t}{6} = 0\) or \(2\pi\), giving \(t = 0\) and \(t = 12\). (c) The period is given by \(\frac{2\pi}{B}\) where \(B = \frac{\pi}{6}\), so \(\text{Period} = \frac{2\pi}{\pi / 6} = 12\) months.

(a) 2 marks: 1 mark for each minimum temperature. (b) 3 marks: 1 mark for setting equations equal, 1 mark for simplifying to cosine term equals 1, 1 mark for finding both values \(t=0\) and \(t=12\). (c) 1 mark for stating 12 months.

(a) For Type X, the optimum temperature is at the vertex \(T = -\frac{b}{2a} = -\frac{4}{2(-0.1)} = 20\). For Type Y, the optimum temperature is \(T = -\frac{3.2}{2(-0.08)} = 20\). Both reach their maximum at \(20^\circ\text{C}\). (b) Set \(-0.1 T^2 + 4T - 20 = -0.08 T^2 + 3.2T - 12\). This simplifies to \(0.02 T^2 - 0.8T + 8 = 0\). Multiplying by 50 gives \(T^2 - 40T + 400 = 0\), which factors to \((T - 20)^2 = 0\). Thus \(T = 20\).

(a) 3 marks: 1 mark for finding 20 for Type X, 1 mark for finding 20 for Type Y, 1 mark for showing mathematical method. (b) 3 marks: 1 mark for equating rates, 1 mark for forming a correct quadratic, 1 mark for solving to find \(T = 20\).

(a) Enzyme A has a larger gradient (0.6) than Enzyme B (0.2), meaning its activity increases faster with temperature. However, Enzyme B has a higher initial value at the starting temperature of the range. (b) Set \(0.6T - 5 = 0.2T + 5\), leading to \(0.4T = 10\), which gives \(T = 25\). (c) For Enzyme A to be strictly more active, we solve \(0.6T - 5 > 0.2T + 5\), which gives \(0.4T > 10 \Rightarrow T > 25\). Restricting this to the domain gives \(25 < T \le 40\).

(a) 2 marks: 1 mark for noting Enzyme A has a larger gradient/increases faster, 1 mark for noting Enzyme B has higher activity at lower temperatures. (b) 2 marks: 1 mark for setting equations equal, 1 mark for finding \(T = 25\). (c) 2 marks: 1 mark for inequality setup, 1 mark for correct final range \(25 < T \le 40\) (or \(T > 25\)).