MetadataPastPaper.sampleTitle

(a) For the equation to have no real roots, the discriminant must be less than zero: \(\Delta = b^2 - 4ac < 0\). Here, \(a = 1\), \(b = k-3\), and \(c = k+5\). Substituting these into the inequality: \((k-3)^2 - 4(1)(k+5) < 0\), which expands to \(k^2 - 6k + 9 - 4k - 20 < 0\) and simplifies to \(k^2 - 10k - 11 < 0\). Factorizing the quadratic gives \((k - 11)(k + 1) < 0\). The critical values are \(k = -1\) and \(k = 11\). Since we require the expression to be less than zero, the solution is the region between the critical values: \(-1 < k < 11\). (b) When \(k = 4\), the quadratic expression is \(x^2 + x + 9\). Completing the square: \(x^2 + x + 9 = (x + 0.5)^2 - 0.25 + 9 = (x + 0.5)^2 + 8.75\). Hence, \(a = 0.5\) and \(b = 8.75\).

(a) M1: Attempts discriminant with correct substitution of coefficients in terms of \(k\). A1: Obtains correct simplified quadratic expression \(k^2 - 10k - 11\). M1: Solves their quadratic to find critical values. A1: Identifies critical values as \(k = -1\) and \(11\). A1: Correct inequality range \(-1 < k < 11\). (b) M1: Attempts to complete the square on \(x^2 + x + 9\) with at least \((x + 0.5)^2\) correct. A1: Fully correct expression \((x + 0.5)^2 + 8.75\) (or equivalent values like \(a = 1/2\), \(b = 35/4\)).

(a) The gradient of \(l_1\) is \(2\). Since \(l_2\) is perpendicular to \(l_1\), the gradient of \(l_2\) is \(-\frac{1}{2}\). Using the point-slope form with \(P(4, 3)\): \(y - 3 = -\frac{1}{2}(x - 4)\). Multiplying by 2: \(2y - 6 = -x + 4\), which simplifies to \(x + 2y - 10 = 0\). (b) To find the intersection \(Q\), solve the simultaneous equations: \(y = 2x\) and \(x + 2y - 10 = 0\). Substituting \(y\) into the second equation: \(x + 2(2x) - 10 = 0 \implies 5x = 10 \implies x = 2\). Then \(y = 2(2) = 4\). Thus, the coordinates of \(Q\) are \((2, 4)\). (c) The distance between \(P(4, 3)\) and \(Q(2, 4)\) is given by \(PQ = \sqrt{(4-2)^2 + (3-4)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{5}\).

(a) M1: Finds gradient of perpendicular line \(l_2\) as \(-1/2\). M1: Uses their gradient and point \(P(4, 3)\) to set up equation of the line. A1: Correct equation in requested integer form \(x + 2y - 10 = 0\). (b) M1: Substitutes \(y = 2x\) into their line equation for \(l_2\). A1: Solves to find \(x = 2\). A1: Finds corresponding \(y = 4\) to give \(Q(2, 4)\). (c) M1: Uses distance formula with their coordinates for \(P\) and \(Q\). A1: Correct exact distance of \(\sqrt{5}\).

(a) Using the trigonometric identity \(\cos^2 x = 1 - \sin^2 x\), substitute this into the equation: \(3(1 - \sin^2 x) - 5\sin x - 1 = 0 \implies 3 - 3\sin^2 x - 5\sin x - 1 = 0\), which simplifies to \(3\sin^2 x + 5\sin x - 2 = 0\). Factorizing the quadratic: \((3\sin x - 1)(\sin x + 2) = 0\). This yields \(\sin x = \frac{1}{3}\) or \(\sin x = -2\) (which has no real solutions). For \(\sin x = \frac{1}{3}\) within \(0 \le x < 360^\circ\): \(x = \arcsin\left(\frac{1}{3}\right) \approx 19.5^\circ\). The second solution is \(x = 180^\circ - 19.471^\circ \approx 160.5^\circ\). (b) Substituting \(x = 2\theta\), the interval for \(\theta\) is \(0 \le \theta < 360^\circ\), which means \(2\theta\) lies in the interval \(0 \le 2\theta < 720^\circ\). Since the original equation has 2 solutions in the interval \([0, 360^\circ)\), the equation in \(2\theta\) will have twice as many solutions in the double-sized interval, giving exactly 4 solutions.

(a) M1: Uses identity \(\cos^2 x = 1 - \sin^2 x\) to set up a quadratic equation in terms of \(\sin x\). A1: Obtains correct quadratic equation \(3\sin^2 x + 5\sin x - 2 = 0\). M1: Solves their quadratic to obtain \(\sin x = 1/3\) (rejecting \(\sin x = -2\)). A1: One correct angle of \(19.5^\circ\) (accept 19.5 or 19.47). A1: Second correct angle of \(160.5^\circ\) (accept 160.5 or 160.53). (b) M1: Explains that \(2\theta\) doubles the range of values to \(720^\circ\) resulting in two full cycles. A1: Concludes with 4 solutions.

(a) Write the term involving \(x\) with a negative fractional power: \(y = 2x^2 - 8x^{-1/2} + 3\). Differentiating with respect to \(x\): \(\frac{\text{d}y}{\text{d}x} = 4x - 8\left(-\frac{1}{2}\right)x^{-3/2} = 4x + 4x^{-3/2}\). (b) At the point where \(x = 4\), calculate the \(y\)-coordinate: \(y = 2(4)^2 - \frac{8}{\sqrt{4}} + 3 = 32 - 4 + 3 = 31\). Find the gradient of the tangent by substituting \(x = 4\) into \(\frac{\text{d}y}{\text{d}x}\): \(m = 4(4) + 4(4)^{-3/2} = 16 + \frac{4}{8} = 16.5\). The equation of the tangent line is: \(y - 31 = 16.5(x - 4) \implies y - 31 = 16.5x - 66 \implies y = 16.5x - 35\).

(a) M1: Differentiates at least one term correctly (e.g., \(2x^2 \to 4x\) or \(-8x^{-1/2} \to kx^{-3/2}\)). A1: One term fully correct. A1: Correct derivative \(4x + 4x^{-3/2}\) (or equivalent form). (b) B1: Finds \(y = 31\) when \(x = 4\). M1: Substitutes \(x = 4\) into their derivative. A1: Obtains gradient of \(16.5\) (or \(33/2\)). M1: Uses their point and gradient in the line equation formula. A1: Correct final equation in the form \(y = mx + c\) (e.g. \(y = 16.5x - 35\) or \(y = \frac{33}{2}x - 35\)).

(a) To find intersections with the \(x\)-axis, set \(y = 0\): \(9x - x^3 = 0 \implies x(9 - x^2) = 0 \implies x(3 - x)(3 + x) = 0\). The roots are \(x = -3, 0, 3\). Thus, the coordinates are \((-3, 0)\), \((0, 0)\), and \((3, 0)\). (b) The region bounded by the curve and the positive \(x\)-axis lies between \(x = 0\) and \(x = 3\). The area is given by the definite integral: \(\int_{0}^{3} (9x - x^3) \, \text{d}x = \left[ \frac{9}{2}x^2 - \frac{1}{4}x^4 \right]_{0}^{3}\). Substituting limits: \(\left(\frac{9}{2}(3)^2 - \frac{1}{4}(3)^4\right) - 0 = \frac{81}{2} - \frac{81}{4} = \frac{81}{4} = 20.25\).

(a) M1: Sets \(y = 0\) and attempts to solve the cubic equation. A1: Correct coordinates \((-3, 0)\), \((0, 0)\), and \((3, 0)\) (accept listed values of \(x\)). (b) M1: Formulates the integral with limits \(0\) and \(3\) (or their positive root from part a). M1: Integrates with at least one term increased in power. A1: Correct integration to give \(\frac{9}{2}x^2 - \frac{1}{4}x^4\). M1: Evaluates their integral using limits \(0\) and \(3\). A1: Correct exact area of \(20.25\) (or \(81/4\)).

(a) Express \(y\) from the first equation: \(y = 2x + 5\). Substitute this into the second equation: \(x^2 + x(2x + 5) + (2x + 5) = 3 \implies x^2 + 2x^2 + 5x + 2x + 5 = 3 \implies 3x^2 + 7x + 2 = 0\). Factorizing the quadratic equation: \((3x + 1)(x + 2) = 0\). This gives \(x = -2\) and \(x = -\frac{1}{3}\). For \(x = -2\): \(y = 2(-2) + 5 = 1\). For \(x = -\frac{1}{3}\): \(y = 2\left(-\frac{1}{3}\right) + 5 = \frac{13}{3}\). The intersection points are \((-2, 1)\) and \(\left(-\frac{1}{3}, \frac{13}{3}\right)\). (b) The distance between these two points is: \(d = \sqrt{\left(-\frac{1}{3} - (-2)\right)^2 + \left(\frac{13}{3} - 1\right)^2} = \sqrt{\left(\frac{5}{3}\right)^2 + \left(\frac{10}{3}\right)^2} = \sqrt{\frac{25}{9} + \frac{100}{9}} = \sqrt{\frac{125}{9}} = \frac{5\sqrt{5}}{3}\).

(a) M1: Makes \(y\) (or \(x\)) the subject of the linear equation. M1: Substitutes into the quadratic equation. A1: Reaches a correct simplified quadratic equation such as \(3x^2 + 7x + 2 = 0\). M1: Solves their quadratic to find both values of \(x\). A1: Correct \(x\) values: \(x = -2\) and \(x = -1/3\). A1: Correct corresponding \(y\) values: \(y = 1\) and \(y = 13/3\). (b) M1: Uses the distance formula with their points. A1: Obtains the correct simplified surd \(\frac{5\sqrt{5}}{3}\) (or equivalent simplified form).

(a) First find the midpoint \(M\) of \(AB\): \(M = \left(\frac{-2 + 4}{2}, \frac{5 + 1}{2}\right) = (1, 3)\). Next, find the gradient of \(AB\): \(m_{AB} = \frac{1 - 5}{4 - (-2)} = \frac{-4}{6} = -\frac{2}{3}\). The gradient of the perpendicular bisector \(l\) is the negative reciprocal of \(m_{AB}\): \(m_l = \frac{3}{2} = 1.5\). The equation of the line \(l\) is \(y - 3 = 1.5(x - 1) \implies y - 3 = 1.5x - 1.5 \implies y = 1.5x + 1.5\). (b) To find where the line crosses the \(x\)-axis, set \(y = 0\): \(0 = 1.5x + 1.5 \implies 1.5x = -1.5 \implies x = -1\). The coordinates of \(P\) are \((-1, 0)\).

(a) M1: Finds the midpoint of \(AB\). A1: Correct midpoint \((1, 3)\). M1: Finds gradient of \(AB\). M1: Determines perpendicular gradient (negative reciprocal of their gradient of \(AB\)). A1: Correct equation in \(y = mx + c\) form: \(y = 1.5x + 1.5\) (or equivalent). (b) M1: Sets \(y = 0\) and solves for \(x\). A1: Correct coordinates of \(P\) are \((-1, 0)\).

(a) The volume of a cylinder is given by \(V = \pi r^2 h = 250\pi\). Solving for \(h\): \(h = \frac{250}{r^2}\). Substituting this expression for \(h\) into the surface area equation: \(A = 2\pi r^2 + 2\pi r\left(\frac{250}{r^2}\right) = 2\pi r^2 + \frac{500\pi}{r}\). (b) Differentiating \(A\) with respect to \(r\): \(\frac{\text{d}A}{\text{d}r} = 4\pi r - \frac{500\pi}{r^2}\). To find stationary points, set \(\frac{\text{d}A}{\text{d}r} = 0 \implies 4\pi r = \frac{500\pi}{r^2} \implies r^3 = 125 \implies r = 5\). Substitute \(r = 5\) back into the surface area formula to obtain the minimum area: \(A = 2\pi(5)^2 + \frac{500\pi}{5} = 50\pi + 100\pi = 150\pi\). (c) Find the second derivative: \(\frac{\text{d}^2A}{\text{d}r^2} = 4\pi + \frac{1000\pi}{r^3}\). At \(r = 5\): \(\frac{\text{d}^2A}{\text{d}r^2} = 4\pi + \frac{1000\pi}{125} = 4\pi + 8\pi = 12\pi\). Since \(12\pi > 0\), the value is indeed a minimum.

(a) M1: Uses the volume formula \(V = \pi r^2 h = 250\pi\) to express \(h\) in terms of \(r\). A1: Correct expression \(h = 250/r^2\). A1*: Substitutes \(h\) into the surface area formula to obtain the given expression (with no errors shown). (b) M1: Attempts to differentiate \(A\) with respect to \(r\). A1: Correct derivative \(4\pi r - \frac{500\pi}{r^2}\). M1: Sets their derivative to 0 and solves for \(r\). A1: Obtains the correct minimum value \(150\pi\). (c) B1: Finds second derivative and correctly evaluates it at \(r=5\) to show it is positive (hence a minimum).

(a) Equating the line and the curve:
\( 2x^2 - 5x + 3 = kx - 5 \)
\( 2x^2 - (5 + k)x + 8 = 0 \)

Since the line is a tangent to the curve, the resulting quadratic equation has equal roots, meaning the discriminant must be zero:
\( b^2 - 4ac = 0 \)
\( (-(5+k))^2 - 4(2)(8) = 0 \)
\( (5+k)^2 - 64 = 0 \)
\( 5+k = \pm 8 \)

This gives:
\( 5+k = 8 \implies k = 3 \)
\( 5+k = -8 \implies k = -13 \)

So the values of \( k \) are \( 3 \) and \( -13 \).

(b) The larger value of \( k \) is \( 3 \).
Substituting \( k = 3 \) back into the quadratic equation:
\( 2x^2 - (5 + 3)x + 8 = 0 \)
\( 2x^2 - 8x + 8 = 0 \)
\( x^2 - 4x + 4 = 0 \)
\( (x-2)^2 = 0 \implies x = 2 \)

Substituting \( x = 2 \) into the equation of \( L_1 \):
\( y = 3(2) - 5 = 1 \)

Thus, the coordinates of the point of contact are \( (2, 1) \).

(a)
M1: Equates the equations of the line and the curve, and rearranges to form a 3-term quadratic in \( x \).
A1: Correct quadratic equation, e.g., \( 2x^2 - (5+k)x + 8 = 0 \).
dM1: Applies the discriminant condition \( b^2 - 4ac = 0 \). Dependent on the previous M mark.
A1: Correctly set-up equation in terms of \( k \), e.g., \( (5+k)^2 - 64 = 0 \) or \( k^2 + 10k - 39 = 0 \).
A1: Both correct values of \( k \): \( k = 3 \) and \( k = -13 \).

(b)
M1: Substitutes \( k = 3 \) into the quadratic equation to find a value for \( x \).
A1: Obtains \( x = 2 \).
A1: Obtains \( y = 1 \) and writes the final coordinates as \( (2, 1) \) or equivalent.

(a) Let \( f(x) = 4x^2 \). The derivative from first principles is defined as:
\( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)

Substitute \( f(x) = 4x^2 \):
\( f'(x) = \lim_{h \to 0} \frac{4(x+h)^2 - 4x^2}{h} \)
\( f'(x) = \lim_{h \to 0} \frac{4(x^2 + 2xh + h^2) - 4x^2}{h} \)
\( f'(x) = \lim_{h \to 0} \frac{4x^2 + 8xh + 4h^2 - 4x^2}{h} \)
\( f'(x) = \lim_{h \to 0} \frac{8xh + 4h^2}{h} \)
\( f'(x) = \lim_{h \to 0} (8x + 4h) \)

As \( h \to 0 \), \( 4h \to 0 \), so:
\( f'(x) = 8x \).

(b) Find the y-coordinate of \( P \) by substituting \( x = 4 \) into the equation of \( C \):
\( y = 4(4)^2 - 3(4) - \frac{16}{\sqrt{4}} = 64 - 12 - 8 = 44 \).
So \( P \) is \( (4, 44) \).

Write the equation of \( C \) in index form:
\( y = 4x^2 - 3x - 16x^{-1/2} \).

Differentiate with respect to \( x \):
\( \frac{\text{d}y}{\text{d}x} = 8x - 3 - 16\left(-\frac{1}{2}\right)x^{-3/2} \)
\( \frac{\text{d}y}{\text{d}x} = 8x - 3 + 8x^{-3/2} \).

At \( x = 4 \):
\( \frac{\text{d}y}{\text{d}x} = 8(4) - 3 + 8(4)^{-3/2} = 32 - 3 + \frac{8}{8} = 30 \).

The equation of the tangent at \( (4, 44) \) with gradient 30 is:
\( y - 44 = 30(x - 4) \)
\( y - 44 = 30x - 120 \)
\( 30x - y - 76 = 0 \).

(a)
M1: Applies the first principles formula with \( f(x) = 4x^2 \).
M1: Expands, simplifies, and divides through by \( h \) to get \( 8x + 4h \).
A1: Completes the proof rigorously, including limit notation as \( h \to 0 \).

(b)
B1: Finds the correct y-coordinate \( y = 44 \).
M1: Differentiates the curve's equation, reducing at least one power by 1.
A1: Correct derivative \( \frac{\text{d}y}{\text{d}x} = 8x - 3 + 8x^{-3/2} \).
M1: Substitutes \( x = 4 \) into their derivative to find the tangent's gradient and uses it to find the equation of the tangent.
A1: Correct final equation: \( 30x - y - 76 = 0 \) (or equivalent integer form, e.g., \( -30x + y + 76 = 0 \)).

(a) The radius \(r\) of the circle is the distance between \(A(5, -2)\) and \(B(9, 1)\):
\(r = \sqrt{(9-5)^2 + (1 - (-2))^2} = \sqrt{4^2 + 3^2} = \sqrt{25} = 5\).
Thus, the equation of \(C\) is:
\((x-5)^2 + (y+2)^2 = 25\).

(b) The gradient of the radius \(AB\) is:
\(m_{\text{radius}} = \frac{1 - (-2)}{9 - 5} = \frac{3}{4}\).
Since the tangent is perpendicular to the radius at the point of contact, its gradient is:
\(m_{\text{tangent}} = -\frac{1}{m_{\text{radius}}} = -\frac{4}{3}\).
Using the point-slope formula with \(B(9, 1)\):
\(y - 1 = -\frac{4}{3}(x - 9)\)
\(3(y - 1) = -4(x - 9)\)
\(3y - 3 = -4x + 36\)
\(4x + 3y - 39 = 0\).

(c) \(A\), \(B\), and \(P\) form a right-angled triangle \(ABP\) with the right angle at \(B\) (since the tangent is perpendicular to the radius at \(B\)).
By Pythagoras' theorem:
\(AP^2 = AB^2 + BP^2\)
\(10^2 = 5^2 + BP^2\)
\(100 = 25 + BP^2\)
\(BP^2 = 75\)
\(BP = \sqrt{75} = 5\sqrt{3}\).

(a)
M1: Correct use of distance formula or Pythagoras to find \(r^2\) or \(r\).
A1: \((x-5)^2 + (y+2)^2 = 25\) (or equivalent).

(b)
M1: Finding gradient of \(AB\) using \(\frac{y_2 - y_1}{x_2 - x_1}\).
M1: Using the perpendicular gradient rule \(m_1 m_2 = -1\).
M1: Finding the equation of the line through \(B\) using their perpendicular gradient.
A1: \(4x + 3y - 39 = 0\) (or any integer multiple thereof).

(c)
M1: Realising that triangle \(ABP\) is right-angled at \(B\) and using Pythagoras' theorem.
A1: \(5\sqrt{3}\) or exact equivalent.

(a) The sum of the first two terms is:
\(a + ar = 15 \implies a(1+r) = 15\) [Equation 1]

The sum to infinity is:
\(S_\infty = \frac{a}{1-r} = 27 \implies a = 27(1-r)\) [Equation 2]

Substitute [Equation 2] into [Equation 1]:
\(27(1-r)(1+r) = 15\)
\(27(1-r^2) = 15\)
\(1-r^2 = \frac{15}{27} = \frac{5}{9}\)
\(r^2 = 1 - \frac{5}{9} = \frac{4}{9}\)

Since \(r > 0\), we take the positive root:
\(r = \frac{2}{3}\).

Substituting \(r = \frac{2}{3}\) back into [Equation 2]:
\(a = 27\left(1 - \frac{2}{3}\right) = 27\left(\frac{1}{3}\right) = 9\).

(b) The sum of the first 5 terms is given by:
\(S_5 = \frac{a(1-r^5)}{1-r}\)
\(S_5 = \frac{9\left(1 - \left(\frac{2}{3}\right)^5\right)}{1 - \frac{2}{3}} = \frac{9\left(1 - \frac{32}{243}\right)}{\frac{1}{3}} = 27\left(\frac{211}{243}\right) = \frac{211}{9}\).

(a)
M1: Writes down equations \(a(1+r) = 15\) and \(\frac{a}{1-r} = 27\).
M1: Eliminates \(a\) to obtain a quadratic equation in \(r^2\) or \(r\).
A1: Shows that \(r = \frac{2}{3}\) clearly (must reject negative root with reason or show positive root used).
A1: \(a = 9\).

(b)
M1: Applies the sum formula \(S_n = \frac{a(1-r^n)}{1-r}\) with \(n=5\) and their \(a\) and \(r\).
M1: Correctly substitutes values into the formula.
A1: \(\frac{211}{9}\) (or \(23 \frac{4}{9}\)).

(a) Using the laws of logarithms:
\(\log_3\left(\frac{x+5}{x-1}\right) = 2\)

Converting to exponential form:
\(\frac{x+5}{x-1} = 3^2\)
\(\frac{x+5}{x-1} = 9\)

Solving for \(x\):
\(x + 5 = 9(x - 1)\)
\(x + 5 = 9x - 9\)
\(8x = 14 \implies x = \frac{14}{8} = \frac{7}{4}\).

(b) Let \(u = 3^y\). The equation becomes:
\(u^2 - 10u + 9 = 0\)

Factorising the quadratic:
\((u - 9)(u - 1) = 0\)
So \(u = 9\) or \(u = 1\).

Case 1: \(3^y = 9 \implies 3^y = 3^2 \implies y = 2\).

Case 2: \(3^y = 1 \implies 3^y = 3^0 \implies y = 0\).

Thus, the solutions are \(y = 0\) and \(y = 2\).

(a)
M1: Combines the logarithms using the division rule \(\log_a A - \log_a B = \log_a \frac{A}{B}\).
M1: Eliminates the logarithm correctly by setting \(\frac{x+5}{x-1} = 3^2\).
M1: Solves the linear equation for \(x\).
A1: \(x = \frac{7}{4}\) (or \(1.75\)).

(b)
M1: Introduces a substitution (e.g., \(u = 3^y\)) to form a quadratic equation.
M1: Factorises or uses formula to find values for \(u\) (or \(3^y\)).
A1: Correctly identifies \(3^y = 9\) and \(3^y = 1\).
A1: Finds both solutions \(y = 0\) and \(y = 2\) (no extra solutions).

(a) Using the identity \(\sin^2 \theta = 1 - \cos^2 \theta\):
\[5(1 - \cos^2 \theta) + 7\cos \theta - 7 = 0\]
\[5 - 5\cos^2 \theta + 7\cos \theta - 7 = 0\]
\[-5\cos^2 \theta + 7\cos \theta - 2 = 0\]
Multiplying the entire equation by \(-1\):
\[5\cos^2 \theta - 7\cos \theta + 2 = 0\] (as required)

(b) Let \(c = \cos \theta\). Then:
\[5c^2 - 7c + 2 = 0\]
Factorising this quadratic equation:
\[(5c - 2)(c - 1) = 0\]

This gives:
\[\cos \theta = \frac{2}{5} = 0.4 \quad \text{or} \quad \cos \theta = 1\]

Case 1: \(\cos \theta = 1\)
For \(0 \le \theta < 360^\circ\), the only solution is:
\[\theta = 0^\circ\]

Case 2: \(\cos \theta = 0.4\)
\[\theta = \arccos(0.4) \approx 66.4^\circ\]
The other solution in the range \(0 \le \theta < 360^\circ\) is:
\[\theta = 360^\circ - 66.4^\circ = 293.6^\circ\]

Thus, the solutions are \(\theta = 0^\circ, 66.4^\circ, 293.6^\circ\).

(a)
M1: Uses the trigonometric identity \(\sin^2 \theta = 1 - \cos^2 \theta\).
A1: Simplifies correctly to show the given equation (must show intermediate steps).

(b)
M1: Factorises or solves the quadratic equation in \(\cos \theta\).
A1: Obtains \(\cos \theta = 0.4\) and \(\cos \theta = 1\).
B1: Identifies \(\theta = 0^\circ\).
M1: Finds one principal angle for \(\cos \theta = 0.4\) (approx. \(66.4^\circ\)).
M1: Finds the second angle by calculating \(360^\circ - \text{their principal angle}\).
A1: Fully correct set of values: \(0^\circ, 66.4^\circ, 293.6^\circ\). Deduct 1 mark if extra solutions are given within the range.

(a) The volume \(V\) of the box is given by:
\(V = x^2 h = 36 \implies h = \frac{36}{x^2}\)

The surface area \(S\) of an open-topped box with a square base of side \(x\) and height \(h\) is:
\(S = x^2 + 4xh\)

Substituting \(h = \frac{36}{x^2}\) into the surface area equation:
\(S = x^2 + 4x\left(\frac{36}{x^2}\right) = x^2 + \frac{144}{x}\) (as required).

(b) To find the minimum surface area, we differentiate \(S\) with respect to \(x\):
\(\frac{\mathrm{d}S}{\mathrm{d}x} = 2x - \frac{144}{x^2}\)

Setting \(\frac{\mathrm{d}S}{\mathrm{d}x} = 0\) for stationary points:
\(2x - \frac{144}{x^2} = 0 \implies 2x^3 = 144 \implies x^3 = 72 \implies x = \sqrt[3]{72} \approx 4.16\text{ cm}\).

Substitute \(x = 72^{1/3}\) back into \(S\):
\(S = (72^{1/3})^2 + \frac{144}{72^{1/3}} = 72^{2/3} + 2 \times 72^{2/3} = 3 \times 72^{2/3} \approx 51.9\text{ cm}^2\).

To justify that this is a minimum, we find the second derivative:
\(\frac{\mathrm{d}^2S}{\mathrm{d}x^2} = 2 + \frac{288}{x^3}\)

At \(x = \sqrt[3]{72}\):
\(\frac{\mathrm{d}^2S}{\mathrm{d}x^2} = 2 + \frac{288}{72} = 2 + 4 = 6 > 0\).

Since the second derivative is positive, the value of \(S\) is indeed a minimum.

(a)
M1: Expresses volume \(V = x^2 h = 36\) and makes \(h\) the subject.
M1: Expresses surface area as \(S = x^2 + 4xh\).
A1: Substitutes \(h\) into the surface area expression to obtain \(S = x^2 + \frac{144}{x}\) cleanly.

(b)
M1: Differentiates \(S\) to obtain \(\frac{\mathrm{d}S}{\mathrm{d}x} = 2x - 144x^{-2}\).
M1: Equates the derivative to \(0\) and solves for \(x^3\) or \(x\).
A1: Finds \(x = \sqrt[3]{72}\) (or 3 s.f. value \(4.16\)).
M1: Evaluates the second derivative \(\frac{\mathrm{d}^2S}{\mathrm{d}x^2}\) at their value of \(x\).
A1: Shows \(\frac{\mathrm{d}^2S}{\mathrm{d}x^2} > 0\) and concludes it is a minimum.
A1: Calculates minimum area \(S \approx 51.9\text{ cm}^2\) (or exact \(3 \times 72^{2/3}\)).

(a) To find the points of intersection, set the equations equal to each other:
\(4x - x^2 = 3\)
\(x^2 - 4x + 3 = 0\)
\((x-1)(x-3) = 0\)

This gives \(x = 1\) and \(x = 3\).
Since both points lie on \(y = 3\), the coordinates of the points of intersection are \((1, 3)\) and \((3, 3)\).

(b) The area \(A\) of the bounded region is given by the integral of the difference between the curve and the line between \(x = 1\) and \(x = 3\):
\(A = \int_{1}^{3} (y_{\text{curve}} - y_{\text{line}}) \,\mathrm{d}x\)
\(A = \int_{1}^{3} (4x - x^2 - 3) \,\mathrm{d}x\)

Integrating each term:
\(A = \left[ 2x^2 - \frac{x^3}{3} - 3x \right]_{1}^{3}\)

Evaluating at the limits:
At \(x = 3\):
\(2(3)^2 - \frac{3^3}{3} - 3(3) = 18 - 9 - 9 = 0\)

At \(x = 1\):
\(2(1)^2 - \frac{1^3}{3} - 3(1) = 2 - \frac{1}{3} - 3 = -\frac{4}{3}\)

Subtracting the lower limit value from the upper limit value:
\(A = 0 - \left(-\frac{4}{3}\right) = \frac{4}{3}\).

(a)
M1: Equates curve and line equations to form a quadratic in \(x\).
A1: Solves the quadratic to find both coordinates: \((1, 3)\) and \((3, 3)\).

(b)
M1: Formulates the integral for the difference \(\int (4x - x^2 - 3) \,\mathrm{d}x\) with correct limits \(1\) and \(3\).
M1: Integrates the terms correctly, raising powers by 1.
A1: Correct integration: \(2x^2 - \frac{x^3}{3} - 3x\).
M1: Substitutes the limits \(3\) and \(1\) into their integrated expression and subtracts.
A1: Exact area of \(\frac{4}{3}\) (or equivalent exact fraction).

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

By the Remainder Theorem, since dividing by \((x+1)\) gives remainder \(-9\):
\(\mathrm{f}(-1) = -9\)
\(2(-1)^3 + a(-1)^2 + b(-1) - 6 = -9\)
\(-2 + a - b - 6 = -9 \implies a - b = -1\) [Equation 2]

Adding [Equation 1] and [Equation 2]:
\(3a = -6 \implies a = -2\)

Substituting \(a = -2\) into [Equation 2]:
\(-2 - b = -1 \implies b = -1\).

Thus, \(a = -2\) and \(b = -1\).

(b) Using the values of \(a\) and \(b\), the polynomial is:
\(\mathrm{f}(x) = 2x^3 - 2x^2 - x - 6\)

Since \((x-2)\) is a factor, we can perform algebraic division or factorise:
\(2x^3 - 2x^2 - x - 6 = (x - 2)(2x^2 + kx + 3)\)

Matching coefficients of \(x^2\):
\(-4 + k = -2 \implies k = 2\).

So, \(\mathrm{f}(x) = (x - 2)(2x^2 + 2x + 3)\).

For the quadratic factor \(2x^2 + 2x + 3 = 0\), we calculate the discriminant \(\Delta\):
\(\Delta = b^2 - 4ac = 2^2 - 4(2)(3) = 4 - 24 = -20\).

Since \(\Delta < 0\), the quadratic expression \(2x^2 + 2x + 3\) has no real roots.
Therefore, the only real root of \(\mathrm{f}(x) = 0\) is \(x = 2\).

(a)
M1: Applies the factor theorem \(\mathrm{f}(2) = 0\) to set up a linear equation in \(a\) and \(b\).
M1: Applies the remainder theorem \(\mathrm{f}(-1) = -9\) to set up a second linear equation in \(a\) and \(b\).
A1: Obtains two correct equations, e.g., \(2a + b = -5\) and \(a - b = -1\).
M1: Solves the simultaneous equations to find \(a\) and \(b\).
A1: Correct values \(a = -2\) and \(b = -1\).

(b)
M1: Factorises the cubic expression to obtain a quadratic factor of the form \(2x^2 + kx + c\).
A1: Correct quadratic factor \(2x^2 + 2x + 3\).
M1: Finds the discriminant \(\Delta\) of the quadratic factor.
A1: Shows \(\Delta = -20 < 0\) and correctly concludes that there are no other real roots, hence \(x=2\) is the only real root.

(a) Let \(n = 10\).
Then \(n^2 + n + 11 = 10^2 + 10 + 11 = 100 + 10 + 11 = 121\).
Since \(121 = 11 \times 11\), \(121\) is not a prime number.
Thus, the statement is false.
(Alternatively, choosing \(n = 11\) gives \(11^2 + 11 + 11 = 11(11 + 1 + 1) = 11 \times 13 = 143\), which is not prime).

(b) Consider the expression \((x - 3y)^2\).
Since the square of any real number is always non-negative:
\[(x - 3y)^2 \ge 0\]

Expanding the bracket:
\[x^2 - 6xy + 9y^2 \ge 0\]

Rearranging the terms gives:
\[x^2 + 9y^2 \ge 6xy\] (as required).

(c) For equality to hold:
\[x^2 + 9y^2 = 6xy \implies x^2 - 6xy + 9y^2 = 0\]
\[(x - 3y)^2 = 0\]
This is true if and only if:
\[x - 3y = 0 \implies x = 3y\]

(a)
M1: Substitutes a suitable value of \(n\) (usually \(10\) or \(11\)) into the formula.
A1: Evaluates the formula correctly (e.g. \(121\) or \(143\)) and states why it is not prime.

(b)
M1: Starts with a correct squared inequality, e.g. \((x-3y)^2 \ge 0\).
M1: Expands \((x-3y)^2\) correctly.
A1: Concludes the proof with a clear final step showing \(x^2 + 9y^2 \ge 6xy\).

(c)
M1: Identifies that equality holds when the squared term is zero: \((x-3y)^2 = 0\).
A1: States the final relationship \(x = 3y\).

**(a)**
Using the power law of logarithms on the first term:
\[ 2\log_3(x-1) = \log_3((x-1)^2) \]

Substitute this back into the equation:
\[ \log_3((x-1)^2) - \log_3(x+5) = 1 \]

Apply the subtraction/division law of logarithms:
\[ \log_3\left(\frac{(x-1)^2}{x+5}\right) = 1 \]

Using the definition of a logarithm to base 3:
\[ \frac{(x-1)^2}{x+5} = 3^1 \]
\[ \frac{x^2 - 2x + 1}{x+5} = 3 \]

Multiply both sides by \( (x+5) \):
\[ x^2 - 2x + 1 = 3(x+5) \]
\[ x^2 - 2x + 1 = 3x + 15 \]

Rearrange to form a quadratic equation set to zero:
\[ x^2 - 5x - 14 = 0 \]
(This is the required form).

**(b)**
Solve the quadratic equation from part (a):
\[ x^2 - 5x - 14 = 0 \]
\[ (x-7)(x+2) = 0 \]

This gives two potential solutions:
\[ x = 7 \quad \text{or} \quad x = -2 \]

We must check the validity of these solutions against the original logarithmic equation:
- For \( x = 7 \), the terms \( x-1 = 6 > 0 \) and \( x+5 = 12 > 0 \) are both positive, so \( x = 7 \) is a valid solution.
- For \( x = -2 \), the term \( x-1 = -3 \le 0 \). Since the logarithm of a non-positive number is undefined, \( x = -2 \) is not a valid solution.

Thus, the only valid solution is:
\[ x = 7 \]

**(a)**
* **M1**: Correctly applies the power law of logarithms to write \( 2\log_3(x-1) \) as \( \log_3((x-1)^2) \).
* **M1**: Correctly applies the subtraction/division law of logarithms to combine the terms on the left-hand side into a single logarithm.
* **M1**: Correctly removes logarithms by raising the base 3 to the power of 1, obtaining \( \frac{(x-1)^2}{x+5} = 3 \).
* **M1**: Multiplies through by \( (x+5) \) and expands the numerator \( (x-1)^2 \) to obtain a linear expression in \( x \) on the right and quadratic on the left.
* **A1**: Fully correct algebraic progression leading to the given quadratic equation \( x^2 - 5x - 14 = 0 \) with no errors shown.

**(b)**
* **M1**: Attempts to solve the quadratic equation \( x^2 - 5x - 14 = 0 \) by factorisation, completing the square, or formula, yielding two solutions (can be implied by finding \( x = 7 \) and \( x = -2 \)).
* **M1**: Explains that \( x = -2 \) is invalid because it leads to taking the logarithm of a negative number (e.g., \( \log_3(-3) \) is undefined, or stating that \( x > 1 \) is required).
* **A1**: Concludes with the single correct solution \( x = 7 \) only.

**(a)**
Start by substituting \( \tan\theta = \frac{\sin\theta}{\cos\theta} \) into the given equation:
\[ 2\sin\theta\left(\frac{\sin\theta}{\cos\theta}\right) + \frac{1}{\cos\theta} = 5 \]

Combine the terms over a common denominator:
\[ \frac{2\sin^2\theta + 1}{\cos\theta} = 5 \]

Multiply both sides by \( \cos\theta \) (since \( \cos\theta \neq 0 \) for the given domain):
\[ 2\sin^2\theta + 1 = 5\cos\theta \]

Use the trigonometric identity \( \sin^2\theta = 1 - \cos^2\theta \):
\[ 2(1 - \cos^2\theta) + 1 = 5\cos\theta \]
\[ 2 - 2\cos^2\theta + 1 = 5\cos\theta \]
\[ 3 - 2\cos^2\theta = 5\cos\theta \]

Rearrange the terms to one side to form a quadratic in \( \cos\theta \):
\[ 2\cos^2\theta + 5\cos\theta - 3 = 0 \]
(This is the required form).

**(b)**
Solve the quadratic equation:
\[ 2\cos^2\theta + 5\cos\theta - 3 = 0 \]

Factorise the quadratic:
\[ (2\cos\theta - 1)(\cos\theta + 3) = 0 \]

This gives:
\[ \cos\theta = \frac{1}{2} \quad \text{or} \quad \cos\theta = -3 \]

Since \( -1 \le \cos\theta \le 1 \) for all real values of \( \theta \), the equation \( \cos\theta = -3 \) has no solutions.

Now solve \( \cos\theta = \frac{1}{2} \) for the interval \( 0 \le \theta < 360^\circ \):
- The principal value is \( \theta = \arccos\left(\frac{1}{2}\right) = 60^\circ \).
- The second solution in the interval is \( \theta = 360^\circ - 60^\circ = 300^\circ \).

Thus, the solutions in the given range are:
\[ \theta = 60^\circ, 300^\circ \]

**(a)**
* **M1**: Substitutes \( \tan\theta = \frac{\sin\theta}{\cos\theta} \) and multiplies through by \( \cos\theta \) to obtain a linear combination of \( \sin^2\theta \) and \( \cos\theta \).
* **M1**: Uses the identity \( \sin^2\theta = 1 - \cos^2\theta \) to obtain an equation containing only cosine terms.
* **A1**: Correctly expands and rearranges the equation to the given form \( 2\cos^2\theta + 5\cos\theta - 3 = 0 \) with no errors.

**(b)**
* **M1**: Attempts to solve the quadratic equation in \( \cos\theta \) to find values for \( \cos\theta \) (by factorisation, formula, or calculator).
* **A1**: Identifies \( \cos\theta = \frac{1}{2} \) as the only source of solutions, explicitly rejecting \( \cos\theta = -3 \) (or equivalent statement).
* **A1**: Finds one correct solution, \( \theta = 60^\circ \).
* **A1**: Finds the second correct solution, \( \theta = 300^\circ \), and no other solutions in the range.

(a)
\(f(3.1) = (3.1)^3 - 3(3.1)^2 - 2(3.1) + 5 = -0.239\)
\(f(3.2) = (3.2)^3 - 3(3.2)^2 - 2(3.2) + 5 = 0.648\)
Since there is a change of sign and \(f(x)\) is continuous, there is at least one root \(\alpha\) in the interval \([3.1, 3.2]\).

(b)
\(x^3 - 3x^2 - 2x + 5 = 0\)
\(x^3 = 3x^2 + 2x - 5\)
Divide by \(x\) (since \(x \neq 0\)):
\(x^2 = 3x + 2 - \frac{5}{x}\)
Taking the positive square root:
\(x = \sqrt{3x + 2 - \frac{5}{x}}\)

(c)
Using \(x_{n+1} = \sqrt{3x_n + 2 - \frac{5}{x_n}}\) with \(x_0 = 3.1\):
\(x_1 = \sqrt{3(3.1) + 2 - \frac{5}{3.1}} = \sqrt{9.687097...} \approx 3.1124\)
\(x_2 = \sqrt{3(3.11241) + 2 - \frac{5}{3.11241}} = \sqrt{9.730761...} \approx 3.1194\)
\(x_3 = \sqrt{3(3.11942) + 2 - \frac{5}{3.11942}} = \sqrt{9.755400...} \approx 3.1234\)

(d)
Using the interval \([3.1275, 3.1285]\):
\(f(3.1275) = (3.1275)^3 - 3(3.1275)^2 - 2(3.1275) + 5 \approx -0.0044\)
\(f(3.1285) = (3.1285)^3 - 3(3.1285)^2 - 2(3.1285) + 5 \approx 0.0042\)
Since there is a sign change in the interval \([3.1275, 3.1285]\), the root lies within this interval. Hence, \(\alpha = 3.128\) correct to 3 decimal places.

(a)
- M1: Attempts to evaluate both \(f(3.1)\) and \(f(3.2)\).
- A1: Obtains correct values of \(f(3.1) \approx -0.239\) and \(f(3.2) \approx 0.648\), and states that the sign change indicates a root.

(b)
- M1: Reorders the equation to isolate the cubic term or divides by \(x\).
- A1: Full proof showing clear steps leading to the given expression.

(c)
- M1: Correctly evaluates \(x_1\) to at least 3 d.p.
- A1: \(x_1 = 3.1124\) and \(x_2 = 3.1194\).
- A1: \(x_3 = 3.1234\).

(d)
- M1: Selects the bounds \(3.1275\) and \(3.1285\) and attempts to evaluate \(f(x)\) at these points.
- A1: Correctly calculates \(f(3.1275) \approx -0.0044\) and \(f(3.1285) \approx 0.0042\) with a conclusion that the sign change shows \(\alpha = 3.128\) to 3 d.p.

(a)
Using the product rule:
\(\frac{dy}{dx} = -2e^{-2x}\ln(3x) + e^{-2x} \cdot \frac{1}{x} = e^{-2x}\left( \frac{1}{x} - 2\ln(3x) \right)\)
At a stationary point, \(\frac{dy}{dx} = 0\):
Since \(e^{-2x} \neq 0\), we have:
\(\frac{1}{x} - 2\ln(3x) = 0 \implies \frac{1}{x} = 2\ln(3x) \implies x = \frac{1}{2\ln(3x)}\)

(b)
Let \(f(x) = 2x\ln(3x) - 1\)
\(f(0.68) = 2(0.68)\ln(2.04) - 1 \approx -0.0304\)
\(f(0.69) = 2(0.69)\ln(2.07) - 1 \approx 0.0040\)
Since there is a change of sign in the continuous interval \([0.68, 0.69]\), a root \(\beta\) lies in this interval.

(c)
Using \(x_{n+1} = \frac{1}{2\ln(3x_n)}\) with \(x_0 = 0.68\):
\(x_1 = \frac{1}{2\ln(3(0.68))} = \frac{1}{2\ln(2.04)} \approx 0.7013\)
\(x_2 = \frac{1}{2\ln(3(0.70131))} = \frac{1}{2\ln(2.10393)} \approx 0.6722\)

(d)
The formula is not suitable because the values \(x_0 = 0.68\), \(x_1 = 0.7013\), and \(x_2 = 0.6722\) are oscillating and moving further away from the root \(\beta \approx 0.689\). This indicates that the sequence is diverging.

(a)
- M1: Correctly applies product rule on \(y\).
- A1: Obtains \(\frac{dy}{dx} = e^{-2x}\left( \frac{1}{x} - 2\ln(3x) \right)\).
- M1: Sets \(\frac{dy}{dx} = 0\) and removes exponential factor.
- A1: Rearranges correctly to obtain the given expression.

(b)
- M1: Defines a continuous function \(f(x)\) and evaluates it at \(0.68\) and \(0.69\).
- A1: Obtains \(f(0.68) \approx -0.0304\) and \(f(0.69) \approx 0.0040\) and concludes there is a sign change and hence a root.

(c)
- A1: \(x_1 = 0.7013\).
- A1: \(x_2 = 0.6722\).

(d)
- B1: States 'not suitable' or 'divergent' with a clear mathematical reason (e.g. values of \(x_n\) diverge or \(|g'(x)| > 1\)).

(a)
LHS \(= \frac{1 - \cos 2\theta}{\sin 2\theta}\)
Using the double-angle identities \(\cos 2\theta = 1 - 2\sin^2\theta\) and \(\sin 2\theta = 2\sin\theta\cos\theta\):
LHS \(= \frac{1 - (1 - 2\sin^2\theta)}{2\sin\theta\cos\theta} = \frac{2\sin^2\theta}{2\sin\theta\cos\theta} = \frac{\sin\theta}{\cos\theta} = \tan\theta = \text{RHS}\)

(b)
Using the identity from part (a), the equation is equivalent to:
\(\tan\theta = 2 - \theta^2 \implies \tan\theta + \theta^2 - 2 = 0\)
Let \(f(\theta) = \tan\theta + \theta^2 - 2\)
Using radians:
\(f(0.88) = \tan(0.88) + 0.88^2 - 2 \approx 1.2097 + 0.7744 - 2 = -0.0159\)
\(f(0.89) = \tan(0.89) + 0.89^2 - 2 \approx 1.2346 + 0.7921 - 2 = 0.0267\)
Since there is a sign change in the continuous interval \([0.88, 0.89]\), there is a root \(\alpha\) in the interval.

(c)
Using \(\theta_{n+1} = \arctan(2 - \theta_n^2)\) with \(\theta_0 = 0.88\):
\(\theta_1 = \arctan(2 - 0.88^2) = \arctan(1.2256) \approx 0.8865\)
\(\theta_2 = \arctan(2 - 0.88647^2) = \arctan(1.21417) \approx 0.8818\)
\(\theta_3 = \arctan(2 - 0.88177^2) = \arctan(1.22248) \approx 0.8852\)

(a)
- M1: Applies double angle identity for either \(\cos 2\theta\) or \(\sin 2\theta\).
- M1: Applies double angle identities for both and simplifies the numerator.
- A1: Clear algebraic progression to conclude \(\tan \theta\).

(b)
- M1: Re-expresses the equation as \(f(\theta) = 0\).
- A1: Evaluates both \(f(0.88)\) and \(f(0.89)\) to at least 2 significant figures, notes sign change, and draws conclusion.

(c)
- M1: Correct attempt to evaluate \(\theta_1\) in radians.
- A1: \(\theta_1 \approx 0.8865\).
- A1: \(\theta_2 \approx 0.8818\).
- A1: \(\theta_3 \approx 0.8852\).

(a)
Let \(f(x) = 3e^{x-2} - 8 + 2x\)
\(f(2.18) = 3e^{0.18} - 8 + 2(2.18) = 3(1.1972) - 8 + 4.36 = 3.5916 - 3.64 = -0.0484\)
\(f(2.19) = 3e^{0.19} - 8 + 2(2.19) = 3(1.2092) - 8 + 4.38 = 3.6277 - 3.62 = 0.0077\)
Since \(f(x)\) is continuous and there is a sign change in \([2.18, 2.19]\), there is a root \(\alpha\) in this interval.

(b)
\(3e^{x-2} - 8 + 2x = 0\)
\(3e^{x-2} = 8 - 2x\)
\(e^{x-2} = \frac{8 - 2x}{3}\)
Taking the natural logarithm of both sides:
\(x - 2 = \ln\left(\frac{8 - 2x}{3}\right)\)
\(x = 2 + \ln\left(\frac{8 - 2x}{3}\right)\)

(c)
Using \(x_{n+1} = 2 + \ln\left(\frac{8 - 2x_n}{3}\right)\) with \(x_0 = 2.18\):
\(x_1 = 2 + \ln\left(\frac{8 - 2(2.18)}{3}\right) = 2 + \ln\left(\frac{3.64}{3}\right) \approx 2.1934\)
\(x_2 = 2 + \ln\left(\frac{8 - 2(2.19337)}{3}\right) = 2 + \ln\left(\frac{3.61326}{3}\right) \approx 2.1860\)
\(x_3 = 2 + \ln\left(\frac{8 - 2(2.18601)}{3}\right) = 2 + \ln\left(\frac{3.62798}{3}\right) \approx 2.1901\)

(a)
- M1: Evaluates \(f(2.18)\) and \(f(2.19)\).
- A1: Obtains \(f(2.18) \approx -0.0484\) and \(f(2.19) \approx 0.0077\), and states that the sign change indicates the root.

(b)
- M1: Rearranges the equation to isolate the exponential term.
- M1: Takes natural logarithms of both sides to remove the exponential.
- A1: Correctly derives the given form.

(c)
- M1: Correctly substitutes \(x_0 = 2.18\) into the iterative formula.
- A1: \(x_1 \approx 2.1934\).
- A1: \(x_2 \approx 2.1860\).
- A1: \(x_3 \approx 2.1901\).

(a)
Let \(f(x) = x^3 - 6x + 3\)
\(f(0.5) = 0.125 > 0\)
\(f(0.6) = -0.384 < 0\)
Since there is a sign change and \(f(x)\) is continuous, \(\gamma\) lies in the interval \([0.5, 0.6]\).

(b)
For Scheme I: \(g_1(x) = \frac{x^3 + 3}{6} \implies g_1'(x) = \frac{x^2}{2}\)
Since \(\gamma \approx 0.52\), \(|g_1'(\gamma)| \approx \frac{0.52^2}{2} \approx 0.135 < 1\). Therefore, Scheme I converges.

For Scheme II: \(g_2(x) = (6x - 3)^{1/3} \implies g_2'(x) = \frac{1}{3}(6x - 3)^{-2/3} \times 6 = 2(6x - 3)^{-2/3}\)
Since \(\gamma \approx 0.52\), \(6\gamma - 3 \approx 0.12\), so:
\(|g_2'(\gamma)| \approx 2(0.12)^{-2/3} \approx 8.2 > 1\). Therefore, Scheme II diverges.

(c)
Using \(x_{n+1} = \frac{x_n^3 + 3}{6}\) with \(x_0 = 0.5\):
\(x_1 = \frac{0.5^3 + 3}{6} = 0.5208\) (to 4 d.p.)
\(x_2 = \frac{0.520833^3 + 3}{6} \approx 0.5235\) (to 4 d.p.)
\(x_3 = \frac{0.523538^3 + 3}{6} \approx 0.5239\) (to 4 d.p.)
Rounding \(x_3\) (or evaluating subsequent terms), we find \(\gamma = 0.524\) to 3 decimal places.

(a)
- M1: Evaluates \(f(0.5)\) and \(f(0.6)\).
- A1: Correctly calculates \(f(0.5) = 0.125\) and \(f(0.6) = -0.384\), with a valid conclusion.

(b)
- M1: Finds the derivative \(g_1'(x)\) for Scheme I.
- A1: Demonstrates that \(|g_1'(\gamma)| < 1\) (using a value in \([0.5, 0.6]\)) and states it converges.
- M1: Finds the derivative \(g_2'(x)\) for Scheme II.
- A1: Demonstrates that \(|g_2'(\gamma)| > 1\) (using a value in \([0.5, 0.6]\)) and states it diverges.

(c)
- M1: Performs the first iteration correctly.
- A1: Obtains \(x_1 = 0.5208\), \(x_2 = 0.5235\), \(x_3 = 0.5239\).
- A1: Writes down \(\gamma = 0.524\) correct to 3 decimal places.

(a)
Using the product rule:
\(\frac{dy}{dx} = 2x \cos x + x^2 (-\sin x) = 2x \cos x - x^2 \sin x\)
For a stationary point, \(\frac{dy}{dx} = 0\):
\(x(2 \cos x - x \sin x) = 0\)
Since \(0 < x < \frac{\pi}{2}\), \(x \neq 0\), we must have:
\(2 \cos x - x \sin x = 0\)
\(2 \cos x = x \sin x \implies \tan x = \frac{2}{x}\)
Taking arctangent:
\(x = \arctan\left(\frac{2}{x}\right)\)

(b)
Let \(f(x) = x - \arctan\left(\frac{2}{x}\right)\)
Using radian mode:
\(f(1.0) = 1.0 - \arctan(2) \approx 1.0 - 1.1071 = -0.1071\)
\(f(1.1) = 1.1 - \arctan\left(\frac{2}{1.1}\right) \approx 1.1 - 1.0681 = 0.0319\)
Since there is a change of sign in the continuous function, there is a root \(\alpha\) in \([1.0, 1.1]\).

(c)
Using \(x_{n+1} = \arctan\left(\frac{2}{x_n}\right)\) with \(x_0 = 1.0\):
\(x_1 = \arctan(2) \approx 1.1071\)
\(x_2 = \arctan\left(\frac{2}{1.107148}\right) \approx 1.0652\)
\(x_3 = \arctan\left(\frac{2}{1.06517}\right) \approx 1.0811\)

(a)
- M1: Differentiates \(y\) using the product rule.
- A1: Correct expression for \(\frac{dy}{dx}\).
- M1: Sets \(\frac{dy}{dx} = 0\) and divides by \(x\cos x\) to introduce \(\tan x\).
- A1: Shows complete steps leading to the given arctan formula.

(b)
- M1: Formulates the equation in the form \(f(x) = 0\) and evaluates it at both boundary points in radians.
- A1: Obtains \(f(1.0) \approx -0.1071\) and \(f(1.1) \approx 0.0319\), and makes a valid conclusion about the root.

(c)
- M1: Computes \(x_1\) correctly.
- A1: \(x_1 \approx 1.1071\) and \(x_2 \approx 1.0652\).
- A1: \(x_3 \approx 1.0811\).

(a)
LHS \(= \sec^2 \theta + \csc^2 \theta = \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta}\)
Combining fractions over a common denominator:
LHS \(= \frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} = \frac{1}{\sin^2 \theta \cos^2 \theta}\)
LHS \(= \left(\frac{1}{\cos^2 \theta}\right)\left(\frac{1}{\sin^2 \theta}\right) = \sec^2 \theta \csc^2 \theta = \text{RHS}\)

(b)
Let \(f(\theta) = \sec^2 \theta + \csc^2 \theta - 5\)
In radians:
\(f(0.5) = \frac{1}{\cos^2(0.5)} + \frac{1}{\sin^2(0.5)} - 5 \approx 1.2987 + 4.3510 - 5 = 0.6497 > 0\)
\(f(0.6) = \frac{1}{\cos^2(0.6)} + \frac{1}{\sin^2(0.6)} - 5 \approx 1.4682 + 3.1365 - 5 = -0.3953 < 0\)
Since \(f(\theta)\) is continuous and there is a sign change in the interval \([0.5, 0.6]\), there is a root in this interval.

(c)
\(\sec^2 \theta \csc^2 \theta = 5 \implies \frac{1}{\sin^2\theta \cos^2\theta} = 5 \implies \sin^2\theta \cos^2\theta = \frac{1}{5}\)
For \(0 < \theta < \frac{\pi}{2}\), \(\sin\theta\) and \(\cos\theta\) are positive:
\(\sin\theta \cos\theta = \frac{1}{\sqrt{5}} \implies \sin\theta = \frac{1}{\sqrt{5}\cos\theta}\)
\(\theta = \arcsin\left(\frac{1}{\sqrt{5}\cos\theta}\right)\)

Using the iteration with \(\theta_0 = 0.5\):
\(\theta_1 = \arcsin\left(\frac{1}{\sqrt{5}\cos(0.5)}\right) = \arcsin\left(\frac{1}{2.23607 \times 0.87758}\right) \approx 0.5348\)
\(\theta_2 = \arcsin\left(\frac{1}{\sqrt{5}\cos(0.53482)}\right) = \arcsin\left(\frac{1}{2.23607 \times 0.86037}\right) \approx 0.5466\)

(a)
- M1: Converts cosec and sec to sin and cos, then finds a common denominator.
- M1: Applies the identity \(\sin^2\theta + \cos^2\theta = 1\).
- A1: Shows final equivalence step clearly.

(b)
- M1: Translates the equation into \(f(\theta) = 0\) and attempts to evaluate at \(0.5\) and \(0.6\) in radians.
- A1: Correctly evaluates \(f(0.5) \approx 0.650\) (or any value to 2 s.f.).
- A1: Correctly evaluates \(f(0.6) \approx -0.395\) (or any value to 2 s.f.), noting the sign change and drawing a conclusion.

(c)
- M1: Derives the iterative form by taking the positive square root and applying arcsin.
- A1: \(\theta_1 \approx 0.5348\).
- A1: \(\theta_2 \approx 0.5466\).

(a)
\(f(x) = g(x) \implies \frac{2x+3}{x-1} = e^x\)
Since \(x > 1\), we can multiply by \(x-1\):
\(2x + 3 = (x-1)e^x \implies (x-1)e^x - 2x - 3 = 0\)
Let \(h(x) = (x-1)e^x - 2x - 3\)
\(h(1.9) = (0.9)e^{1.9} - 2(1.9) - 3 \approx 6.0173 - 6.8 = -0.7827\)
\(h(2.0) = (1.0)e^{2.0} - 2(2.0) - 3 \approx 7.3891 - 7.0 = 0.3891\)
Since there is a change of sign in the continuous interval \([1.9, 2.0]\), a root \(\beta\) lies in this interval.

(b)
\(\frac{2x+3}{x-1} = e^x\)
\(2x + 3 = (x-1)e^x\)
Multiply both sides by \(e^{-x}\):
\((2x + 3)e^{-x} = x - 1\)
Rearrange to isolate \(x\):
\(x = 1 + (2x + 3)e^{-x}\)

(c)
Using \(x_{n+1} = 1 + (2x_n + 3)e^{-x_n}\) with \(x_0 = 1.9\):
\(x_1 = 1 + (2(1.9) + 3)e^{-1.9} = 1 + 6.8e^{-1.9} \approx 2.0171\)
\(x_2 = 1 + (2(2.01707) + 3)e^{-2.01707} = 1 + 7.03413e^{-2.01707} \approx 1.9359\)
\(x_3 = 1 + (2(1.93586) + 3)e^{-1.93586} = 1 + 6.87171e^{-1.93586} \approx 1.9916\)

(a)
- M1: Sets \(f(x) = g(x)\) and establishes a function \(h(x) = 0\).
- M1: Evaluates \(h(1.9)\) and \(h(2.0)\).
- A1: Obtains \(h(1.9) \approx -0.783\) and \(h(2.0) \approx 0.389\) (or equivalent accurate values), and notes sign change to conclude root exists.

(b)
- M1: Clears fractions to get \(2x + 3 = (x-1)e^x\).
- M1: Multiplies by \(e^{-x}\) to isolate \(x-1\).
- A1: Correctly shows the given iterative form with no errors.

(c)
- M1: Correctly performs the first step of the iteration.
- A1: \(x_1 \approx 2.0171\).
- A1: \(x_2 \approx 1.9359\) and \(x_3 \approx 1.9916\).

(a) Using integration by parts: Let \(u = \ln(3x)\), then \(\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{x}\). Let \(\frac{\mathrm{d}v}{\mathrm{d}x} = x^2\), then \(v = \frac{1}{3}x^3\). Using the integration by parts formula: \(\int u \frac{\mathrm{d}v}{\mathrm{d}x} \, \mathrm{d}x = uv - \int v \frac{\mathrm{d}u}{\mathrm{d}x} \, \mathrm{d}x\), we get \(\int x^2 \ln(3x) \, \mathrm{d}x = \frac{1}{3}x^3 \ln(3x) - \int \frac{1}{3}x^2 \, \mathrm{d}x = \frac{1}{3}x^3 \ln(3x) - \frac{1}{9}x^3 + C\). (b) The curve intersects the \(x\)-axis where \(y = 0 \Rightarrow x^2 \ln(3x) = 0\). Since \(x > 0\), we have \(\ln(3x) = 0 \Rightarrow 3x = 1 \Rightarrow x = \frac{1}{3}\). The required area is \(\int_{1/3}^{1} x^2 \ln(3x) \, \mathrm{d}x = \left[ \frac{1}{3}x^3 \ln(3x) - \frac{1}{9}x^3 \right]_{1/3}^{1} = \left( \frac{1}{3}(1)^3 \ln(3) - \frac{1}{9}(1)^3 \right) - \left( \frac{1}{3}\left(\frac{1}{27}\right) \ln(1) - \frac{1}{9}\left(\frac{1}{27}\right) \right) = \left( \frac{1}{3}\ln(3) - \frac{1}{9} \right) - \left( 0 - \frac{1}{243} \right) = \frac{1}{3}\ln(3) - \frac{26}{243}\).

(a) M1: Applies integration by parts with correct choices \(u = \ln(3x)\) and \(v' = x^2\). A1: Obtains \(v = \frac{1}{3}x^3\) and \(u' = \frac{1}{x}\). M1: Applies parts formula to get \(\frac{1}{3}x^3 \ln(3x) - \int \frac{1}{3}x^2 \, \mathrm{d}x\). A1: Correct integration of the remaining term. A1: Correct expression with constant of integration \(+ C\). (b) M1: Recognises the lower limit of integration is \(x = 1/3\) (where \(y = 0\)). M1: Substitutes the limits \(1\) and \(1/3\) into their part (a) result. A1: Obtains the correct exact value \(\frac{1}{3}\ln(3) - \frac{26}{243}\).

(a) For the lines to intersect, there must exist values of \(\lambda\) and \(\mu\) such that: \(2 + \lambda = -1 + 2\mu\) (1), \(-1 + 3\lambda = a - \mu\) (2), \(5 - 2\lambda = 11 + 4\mu\) (3). Rearranging (1): \(\lambda - 2\mu = -3\). Rearranging (3): \(2\lambda + 4\mu = -6 \Rightarrow \lambda + 2\mu = -3\). Adding these two equations gives \(2\lambda = -6 \Rightarrow \lambda = -3\). Substituting \(\lambda = -3\) back into \(\lambda - 2\mu = -3\) gives \(-3 - 2\mu = -3 \Rightarrow \mu = 0\). Substituting \(\lambda = -3\) and \(\mu = 0\) into (2): \(-1 + 3(-3) = a - 0 \Rightarrow a = -10\). To find \(P\), substitute \(\lambda = -3\) into the equation of \(l_1\): \(\mathbf{r} = \begin{pmatrix} 2 - 3 \\ -1 - 9 \\ 5 + 6 \end{pmatrix} = \begin{pmatrix} -1 \\ -10 \\ 11 \end{pmatrix}\). So, \(P\) has coordinates \((-1, -10, 11)\). (b) The direction vectors of \(l_1\) and \(l_2\) are \(\mathbf{u} = \begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix}\) and \(\mathbf{v} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}\). \(\mathbf{u} \cdot \mathbf{v} = 1(2) + 3(-1) + (-2)(4) = 2 - 3 - 8 = -9\). \(|\mathbf{u}| = \sqrt{1^2 + 3^2 + (-2)^2} = \sqrt{14}\), \(|\mathbf{v}| = \sqrt{2^2 + (-1)^2 + 4^2} = \sqrt{21}\). \(\cos\theta = \frac{|\mathbf{u} \cdot \mathbf{v}|}{|\mathbf{u}||\mathbf{v}|} = \frac{9}{\sqrt{14}\sqrt{21}} = \frac{9}{\sqrt{294}}\). \(\theta = \arccos\left(\frac{9}{\sqrt{294}}\right) \approx 58.337^\circ \approx 58.3^\circ\).

(a) M1: Sets up simultaneous equations from the \(x\), \(y\), and \(z\) components. M1: Solves the equations to find \(\lambda\) and \(\mu\). A1: Correctly identifies \(\lambda = -3\) and \(\mu = 0\). A1: Finds \(a = -10\). A1: Correctly writes the coordinates of \(P\) as \((-1, -10, 11)\). (b) M1: Uses the scalar product formula \(\cos\theta = \frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|}\) with the direction vectors. A1: Correct evaluation of the scalar product and magnitudes: \(|\mathbf{u}\cdot\mathbf{v}| = 9\), \(|\mathbf{u}| = \sqrt{14}\), \(|\mathbf{v}| = \sqrt{21}\). A1: Correctly calculates \(\theta \approx 58.3^\circ\).

(a) Rearranging the equation for \(x\): \(t+1 = \frac{2}{x} \Rightarrow t = \frac{2}{x} - 1\). Substitute this into the equation for \(y\): \(y = \left(\frac{2}{x} - 1\right)^2 - 3 = \frac{4}{x^2} - \frac{4}{x} + 1 - 3 = \frac{4}{x^2} - \frac{4}{x} - 2\). (b) At \(t = 1\), the coordinates of the point on the curve are: \(x = \frac{2}{1+1} = 1\), \(y = 1^2 - 3 = -2\). The gradient of the tangent is \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}\). \(\frac{\mathrm{d}x}{\mathrm{d}t} = -\frac{2}{(t+1)^2}\) and \(\frac{\mathrm{d}y}{\mathrm{d}t} = 2t\). Thus, \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2t}{-\frac{2}{(t+1)^2}} = -t(t+1)^2\). At \(t = 1\), \(\frac{\mathrm{d}y}{\mathrm{d}x} = -1(1+1)^2 = -4\). The equation of the tangent is: \(y - (-2) = -4(x - 1) \Rightarrow y + 2 = -4x + 4 \Rightarrow 4x + y - 2 = 0\).

(a) M1: Expresses \(t\) in terms of \(x\). A1: Correct substitution of their \(t\) into the formula for \(y\). A1: Correctly simplified Cartesian equation \(y = \frac{4}{x^2} - \frac{4}{x} - 2\). (b) B1: Correct coordinates at \(t=1\): \((1, -2)\). M1: Calculates both \(\frac{\mathrm{d}x}{\mathrm{d}t}\) and \(\frac{\mathrm{d}y}{\mathrm{d}t}\). M1: Uses the chain rule to find \(\frac{\mathrm{d}y}{\mathrm{d}x}\) and evaluates this at \(t = 1\) to get gradient \(-4\). A1: Correctly forms and simplifies the line equation into the required integer form \(4x + y - 2 = 0\).

(a) We rewrite \(\mathrm{f}(x)\) as: \(\mathrm{f}(x) = 3(4 - 2x)^{-1/2} = 3 \cdot 4^{-1/2} \left(1 - \frac{1}{2}x\right)^{-1/2} = \frac{3}{2}\left(1 - \frac{1}{2}x\right)^{-1/2}\). Using the binomial expansion formula \((1 + y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3 + \dots\) with \(n = -1/2\) and \(y = -\frac{1}{2}x\): \(\left(1 - \frac{1}{2}x\right)^{-1/2} = 1 + \left(-\frac{1}{2}\right)\left(-\frac{1}{2}x\right) + \frac{(-1/2)(-3/2)}{2}\left(-\frac{1}{2}x\right)^2 + \frac{(-1/2)(-3/2)(-5/2)}{6}\left(-\frac{1}{2}x\right)^3 + \dots\) \(= 1 + \frac{1}{4}x + \frac{3}{32}x^2 + \frac{5}{128}x^3 + \dots\) Multiplying by \(\frac{3}{2}\): \(\mathrm{f}(x) \approx \frac{3}{2} + \frac{3}{8}x + \frac{9}{64}x^2 + \frac{15}{256}x^3\). (b) Substituting \(x = \frac{1}{8}\) into \(\mathrm{f}(x)\): \(\mathrm{f}\left(\frac{1}{8}\right) = \frac{3}{\sqrt{4 - 1/4}} = \frac{3}{\sqrt{15/4}} = \frac{6}{\sqrt{15}}\). Thus, \(\sqrt{15} = \frac{6}{\mathrm{f}(1/8)}\). Substituting \(x = \frac{1}{8}\) into our expansion: \(\mathrm{f}\left(\frac{1}{8}\right) \approx \frac{3}{2} + \frac{3}{8}\left(\frac{1}{8}\right) + \frac{9}{64}\left(\frac{1}{64}\right) + \frac{15}{256}\left(\frac{1}{512}\right) = 1.5 + 0.046875 + 0.0021972656 + 0.0001144409 = 1.5491867\). Therefore, \(\sqrt{15} \approx \frac{6}{1.5491867} \approx 3.873007 \approx 3.8730\) (to 4 decimal places).

(a) M1: Takes out factor of \(4^{-1/2}\) to obtain the leading constant \(\frac{3}{2}\). M1: Applies the binomial expansion formula with \(n = -1/2\) and term \(\pm \frac{1}{2}x\). A1: Correct unsimplified expansion with at least 3 terms. A1: Correct terms up to \(x^2\): \(\frac{3}{2} + \frac{3}{8}x + \frac{9}{64}x^2\). A1: Correct final term \(\frac{15}{256}x^3\). (b) M1: Shows that substituting \(x = 1/8\) leads to an expression involving \(\sqrt{15}\), specifically \(\sqrt{15} = \frac{6}{\mathrm{f}(1/8)}\). M1: Evaluates the expansion at \(x = 1/8\) to obtain a decimal approximation. A1: Obtains \(3.8730\) (must be accurate to 4 d.p.).

(a) Differentiating the equation with respect to \(x\) implicitly: \(\frac{\mathrm{d}}{\mathrm{d}x}(y^3) + \frac{\mathrm{d}}{\mathrm{d}x}(3xy) - \frac{\mathrm{d}}{\mathrm{d}x}(x^2) = \frac{\mathrm{d}}{\mathrm{d}x}(13)\) \(3y^2 \frac{\mathrm{d}y}{\mathrm{d}x} + \left(3y + 3x\frac{\mathrm{d}y}{\mathrm{d}x}\right) - 2x = 0\). Rearranging to group \(\frac{\mathrm{d}y}{\mathrm{d}x}\) terms: \((3y^2 + 3x)\frac{\mathrm{d}y}{\mathrm{d}x} = 2x - 3y \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2x - 3y}{3y^2 + 3x}\). (b) Substituting \(P(1, 2)\) into the equation: \(2^3 + 3(1)(2) - 1^2 = 8 + 6 - 1 = 13\). Since this equals the right-hand side, \(P(1, 2)\) lies on the curve. Substitute \(x = 1\) and \(y = 2\) into the gradient formula: \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2(1) - 3(2)}{3(2)^2 + 3(1)} = \frac{2 - 6}{12 + 3} = -\frac{4}{15}\). The gradient of the normal is the negative reciprocal of the tangent gradient: \(m_{\text{normal}} = -\frac{1}{-4/15} = \frac{15}{4}\). The equation of the normal at \((1, 2)\) is: \(y - 2 = \frac{15}{4}(x - 1) \Rightarrow y = \frac{15}{4}x - \frac{15}{4} + 2 \Rightarrow y = \frac{15}{4}x - \frac{7}{4}\).

(a) M1: Differentiates \(y^3\) to get \(3y^2 \frac{\mathrm{d}y}{\mathrm{d}x}\). M1: Applies product rule correctly to differentiate \(3xy\). A1: Correct differentiation of the entire equation: \(3y^2 \frac{\mathrm{d}y}{\mathrm{d}x} + 3y + 3x\frac{\mathrm{d}y}{\mathrm{d}x} - 2x = 0\). A1: Correctly rearranges to make \(\frac{\mathrm{d}y}{\mathrm{d}x}\) the subject. (b) B1: Verifies that \(P(1, 2)\) satisfies the equation of the curve. M1: Substitutes \(x=1\) and \(y=2\) into their expression to find the tangent gradient. M1: Finds the negative reciprocal gradient for the normal and forms the equation of a line. A1: Correct final equation of the normal: \(y = \frac{15}{4}x - \frac{7}{4}\).

(a) Separating the variables: \(\int \frac{1}{180 - \theta} \, \mathrm{d}\theta = \int 0.1 \, \mathrm{d}t\). Integrating both sides: \(-\ln|180 - \theta| = 0.1t + C\) (where \(C\) is the constant of integration). \(\ln|180 - \theta| = -0.1t - C \Rightarrow 180 - \theta = A\mathrm{e}^{-0.1t}\) (where \(A = \mathrm{e}^{-C}\)). Given \(\theta = 20\) when \(t = 0\): \(180 - 20 = A\mathrm{e}^{0} \Rightarrow A = 160\). Therefore, \(180 - \theta = 160\mathrm{e}^{-0.1t} \Rightarrow \theta = 180 - 160\mathrm{e}^{-0.1t}\). (b) Set \(\theta = 100\): \(100 = 180 - 160\mathrm{e}^{-0.1t} \Rightarrow 160\mathrm{e}^{-0.1t} = 80\) \(\mathrm{e}^{-0.1t} = 0.5 \Rightarrow -0.1t = \ln(0.5) \Rightarrow 0.1t = \ln(2)\) \(t = 10\ln(2) \approx 6.9315 \approx 6.93\) minutes.

(a) M1: Separates variables correctly to set up the integral equation. A1: Correct integration of both sides to get \(-\ln(180 - \theta) = 0.1t + C\) (or equivalent). M1: Substitutes the boundary condition \(t = 0\), \(\theta = 20\) to find the constant. M1: Correct algebraic manipulation to make \(\theta\) the subject. A1: Shows the given formula clearly and convincingly. (b) M1: Substitutes \(\theta = 100\) into the given equation. M1: Rearranges and applies logarithms correctly to solve for \(t\). A1: Obtains \(t \approx 6.93\) (allow awrt 6.93).

(a) Using the formula for the volume of revolution about the \(x\)-axis: \(V = \pi \int_{a}^{b} y^2 \, \mathrm{d}x\). Here, \(y^2 = \left(\frac{x}{\sqrt{x^2 + 4}}\right)^2 = \frac{x^2}{x^2 + 4}\). With limits \(x=0\) and \(x=2\), we get: \(V = \pi \int_{0}^{2} \frac{x^2}{x^2 + 4} \, \mathrm{d}x\). (b) Let \(x = 2\tan\theta\), then \(\mathrm{d}x = 2\sec^2\theta \, \mathrm{d}\theta\). Changing the limits: When \(x = 0\), \(2\tan\theta = 0 \Rightarrow \theta = 0\). When \(x = 2\), \(2\tan\theta = 2 \Rightarrow \tan\theta = 1 \Rightarrow \theta = \frac{\pi}{4}\). Substitute these into the integral: \(V = \pi \int_{0}^{\pi/4} \frac{4\tan^2\theta}{4\tan^2\theta + 4} (2\sec^2\theta) \, \mathrm{d}\theta = \pi \int_{0}^{\pi/4} \frac{4\tan^2\theta}{4\sec^2\theta} (2\sec^2\theta) \, \mathrm{d}\theta = \pi \int_{0}^{\pi/4} 2\tan^2\theta \, \mathrm{d}\theta\). Using the identity \(\tan^2\theta = \sec^2\theta - 1\): \(V = 2\pi \int_{0}^{\pi/4} (\sec^2\theta - 1) \, \mathrm{d}\theta = 2\pi \left[ \tan\theta - \theta \right]_{0}^{\pi/4} = 2\pi \left( \tan\left(\frac{\pi}{4}\right) - \frac{\pi}{4} - (0 - 0) \right) = 2\pi \left( 1 - \frac{\pi}{4} \right) = 2\pi - \frac{\pi^2}{2}\).

(a) B1: Recalls volume formula \(V = \pi \int y^2 \, \mathrm{d}x\). B1: Substitutes the boundaries and correctly squares \(y\) to show the given expression. (b) M1: Employs the substitution \(x = 2\tan\theta\) with correct derivative \(\mathrm{d}x = 2\sec^2\theta \, \mathrm{d}\theta\). A1: Correctly updates the limits of integration to \(0\) and \(\frac{\pi}{4}\). M1: Performs the substitution and uses the identity \(1 + \tan^2\theta = \sec^2\theta\) to simplify the integrand to \(2\tan^2\theta\). M1: Uses identity \(\tan^2\theta = \sec^2\theta - 1\) and integrates correctly. A1: Obtains the correct exact value of \(2\pi - \frac{\pi^2}{2}\) (or equivalent form like \(\frac{\pi(4 - \pi)}{2}\)).

Suppose that \(\sqrt{3}\) is rational. Then we can write \(\sqrt{3} = \frac{a}{b}\) where \(a\) and \(b\) are positive integers and the fraction \(\frac{a}{b}\) is in its simplest form, meaning \(a\) and \(b\) share no common factors other than 1. Squaring both sides: \(3 = \frac{a^2}{b^2} \Rightarrow a^2 = 3b^2\). Since \(b^2\) is an integer, \(a^2\) must be a multiple of 3. By the given assumption, since \(a^2\) is a multiple of 3, \(a\) must also be a multiple of 3. Therefore, we can write \(a = 3k\) for some integer \(k\). Substituting this into \(a^2 = 3b^2\) gives: \((3k)^2 = 3b^2 \Rightarrow 9k^2 = 3b^2 \Rightarrow b^2 = 3k^2\). Since \(k^2\) is an integer, \(b^2\) must be a multiple of 3, and hence \(b\) must also be a multiple of 3. This means both \(a\) and \(b\) have 3 as a common factor, which directly contradicts the initial assumption that \(\frac{a}{b}\) was in its simplest form. Thus, our assumption that \(\sqrt{3}\) is rational must be false, meaning \(\sqrt{3}\) is irrational.

M1: Begins proof by contradiction by assuming \(\sqrt{3}\) is rational and setting \(\sqrt{3} = \frac{a}{b}\) where \(a\) and \(b\) have no common factors. A1: Correctly squares and rearranges to get \(a^2 = 3b^2\). M1: Concludes that \(a^2\) is a multiple of 3, hence \(a\) is a multiple of 3. A1: Expresses \(a = 3k\) and substitutes back into the squared equation. M1: Simplifies to find \(b^2 = 3k^2\). A1: Concludes that \(b\) is a multiple of 3. M1: Clear explanation of the contradiction that both \(a\) and \(b\) share the common factor 3. A1: Fully correct, rigorous proof ending with the conclusion that \(\sqrt{3}\) must be irrational.

(a) Separating the variables:
\[\int \frac{1}{100 - \theta} \,\mathrm{d}\theta = \int k\cos\left(\frac{t}{10}\right) \,\mathrm{d}t\]

Integrating both sides gives:
\[-\ln|100 - \theta| = 10k\sin\left(\frac{t}{10}\right) + C\]

Using the initial condition \(t = 0\), \( \theta = 20 \):
\[-\ln|100 - 20| = 10k\sin(0) + C \implies C = -\ln(80)\]

Substituting \(C\) back into the equation:
\[-\ln(100 - \theta) = 10k\sin\left(\frac{t}{10}\right) - \ln(80)\]

\[\ln(80) - \ln(100 - \theta) = 10k\sin\left(\frac{t}{10}\right)\]

Using the laws of logarithms:
\[\ln\left(\frac{80}{100 - \theta}\right) = 10k\sin\left(\frac{t}{10}\right) \quad \text{(as required)}\]

(b) Substituting \(t = 10\) and \(\theta = 50\):
\[\ln\left(\frac{80}{100 - 50}\right) = 10k\sin\left(\frac{10}{10}\right)\]

\[\ln\left(\frac{80}{50}\right) = 10k\sin(1)\]

\[\ln(1.6) = 10k\sin(1)\]

\[k = \frac{\ln(1.6)}{10\sin(1)}\]

Evaluating this (using radians for the trigonometric function):
\[k \approx \frac{0.4700036}{10 \times 0.8414710} \approx 0.055855\]

\[k \approx 0.0559 \text{ (to 3 s.f.)}\]

(c) Substituting \(t = 5\) and the unrounded value of \(k \approx 0.055855\) (or \(k \approx 0.0559\)):
\[\ln\left(\frac{80}{100 - \theta}\right) = 10(0.055855)\sin(0.5)\]

\[\ln\left(\frac{80}{100 - \theta}\right) \approx 0.55855 \times 0.479426 \approx 0.26778\]

\[\frac{80}{100 - \theta} = e^{0.26778} \approx 1.3071\]

\[100 - \theta = \frac{80}{1.3071} \approx 61.206\]

\[\theta \approx 100 - 61.206 = 38.794 \approx 39{}^\circ\text{C}\]

(a)
- M1: Separates variables and attempts to integrate both sides, obtaining \(\pm\ln(100-\theta)\) and \(A\sin(t/10)\).
- A1: Fully correct integration including a constant of integration, e.g., \(-\ln|100 - \theta| = 10k\sin\left(\frac{t}{10}\right) + C\).
- M1: Uses the boundary condition \(t = 0, \theta = 20\) to find their constant \(C\).
- A1*: Correctly derives the given printed expression with no algebraic errors, showing clear logical steps.

(b)
- M1: Substitutes \(t = 10\) and \(\theta = 50\) into the given or shown equation.
- M1: Rearranges the equation to make \(k\) the subject, evaluating the trigonometric function in radians.
- A1: \(k \approx 0.0559\) (accept \(0.056\) or \(0.0559\)).

(c)
- B1: \(\theta \approx 39{}^\circ\text{C}\) (accept 39, or 38.8 from using \(k = 0.0559\)).

(a) For the first \(T\) seconds, \(Q\) accelerates from rest: \(s_1 = \frac{1}{2} \times 1.5 \times T^2 = 0.75 T^2\). The velocity reached is \(V = 1.5 T\). For the next \(10\text{ s}\), \(Q\) travels at constant speed: \(s_2 = 10 \times 1.5 T = 15 T\). The total distance is \(s_Q = 0.75 T^2 + 15 T\). (b) In \(T + 10\) seconds, \(P\) travels \(s_P = 8(T + 10)\). Since \(s_Q - s_P = 65\), we have \(0.75 T^2 + 15 T - (8T + 80) = 65\), which simplifies to \(0.75 T^2 + 7T - 145 = 0\), or \(3 T^2 + 28 T - 580 = 0\). Factoring gives \((3T + 58)(T - 10) = 0\). Since \(T > 0\), we have \(T = 10\). (c) Using the expression for velocity, \(V = 1.5 T = 1.5 \times 10 = 15\text{ m s}^{-1}\).

(a) M1: Attempts to find the distance during the acceleration stage or constant speed stage. A1: Correct expressions for both distances. A1: Correct total distance expression. (b) M1: Expresses the distance of \(P\) in terms of \(T\). M1: Sets up the equation \(s_Q - s_P = 65\). A1: Obtains a correct quadratic equation. M1: Solves their quadratic equation. A1: \(T = 10\). (c) M1: Uses \(V = 1.5 T\). A1ft: \(V = 15\).

(a) The speed-time graph starts at \((0,0)\), increases linearly to \((30, V)\), remains horizontal to \((30+T, V)\), and then decreases linearly to \((30+T+T_d, 0)\). (b) For the acceleration phase: \(V = u + at = 0 + 0.4 \times 30 = 12\text{ m s}^{-1}\). (c) Deceleration distance is the area of the final triangle: \(s_d = \frac{1}{2} \times V \times T_d = 180\), so \(6 T_d = 180 \implies T_d = 30\text{ s}\). The deceleration is \(a_d = \frac{V}{T_d} = \frac{12}{30} = 0.4\text{ m s}^{-2}\). (d) Total distance is the sum of the three stages: \(s_{\text{total}} = s_a + s_c + s_d = 1440\). Since \(s_a = \frac{1}{2} \times 30 \times 12 = 180\text{ m}\) and \(s_c = 12T\), we have \(180 + 12T + 180 = 1440 \implies 360 + 12T = 1440 \implies 12T = 1080 \implies T = 90\).

(a) B1: Correct general trapezoidal shape. B1: Correctly labelled axes and values. (b) M1: Use of \(v = u + at\) or equivalent. A1: \(V = 12\). (c) M1: Area for deceleration stage set to 180. A1: Correctly showing \(T_d = 30\). M1: Deceleration calculation. A1: \(0.4\text{ m s}^{-2}\). (d) M1: Total area expression set to 1440. A1: Correct equation. A1: \(T = 90\).

(a) Since \(\tan \alpha = 0.75\), \(\sin \alpha = 0.6\) and \(\cos \alpha = 0.8\). For \(P\), the normal reaction is \(R = 2mg \cos \alpha = 1.6mg\). The maximum friction force is \(F = \mu R = 0.25 \times 1.6mg = 0.4mg\). (i) For \(P\) moving up the plane: \(T - 2mg \sin \alpha - F = 2ma \implies T - 1.2mg - 0.4mg = 2ma \implies T - 1.6mg = 2ma\). (ii) For \(Q\) hanging vertically and moving down: \(3mg - T = 3ma\). (b) Adding the two equations of motion: \(1.4mg = 5ma \implies a = 0.28g\). With \(g = 9.8\text{ m s}^{-2}\), \(a = 0.28 \times 9.8 = 2.744 \approx 2.7\text{ m s}^{-2}\) (to 2 s.f.). (c) From the equation of motion for \(Q\): \(T = 3mg - 3ma = 3m(g - 0.28g) = 2.16mg\).

(a) B1: States \(\sin \alpha = 0.6\) and \(\cos \alpha = 0.8\). M1: Calculates normal reaction \(R\) and friction \(F\). A1: Correct equation for \(P\). A1: Correct equation for \(Q\). (b) M1: Eliminates \(T\) to find acceleration in terms of \(g\). A1: \(a = 0.28g\). M1: Substitutes \(g = 9.8\). A1: \(a = 2.7\text{ m s}^{-2}\) (accept \(2.74\text{ m s}^{-2}\)). (c) M1: Solves for \(T\) using their \(a\). A1: \(T = 2.16mg\).

(a) Using \(v^2 = u^2 + 2as\): \(0 = 1.5^2 + 2 a (2.25) \implies 4.5a = -2.25 \implies a = -0.5\text{ m s}^{-2}\). The deceleration is \(0.5\text{ m s}^{-2}\). (b) Since the crate is decelerating downwards, the acceleration is upwards with magnitude \(0.5\text{ m s}^{-2}\). Using \(F = ma\) upwards: \(T - mg = ma \implies T - 80 \times 9.8 = 80 \times 0.5 \implies T - 784 = 40 \implies T = 824\text{ N}\). (c) (i) For the combined mass during ascent: \(T - (80+M)g = (80+M)a \implies 1100 = (80+M)(9.8 + 1.2) \implies 1100 = 11(80+M) \implies 80+M = 100 \implies M = 20\text{ kg}\). (ii) For the box of mass \(20\text{ kg}\): \(R - Mg = Ma \implies R = M(g+a) = 20(9.8 + 1.2) = 220\text{ N}\). By Newton's third law, the force exerted by the box on the floor of the crate is \(220\text{ N}\).

(a) M1: Uses \(v^2 = u^2 + 2as\). A1: Correct substitution. A1: Deceleration = \(0.5\text{ m s}^{-2}\). (b) M1: Resolves vertically upwards. A1: Correct equation setup. M1: Substituting correct values. A1: \(T = 824\text{ N}\). (c) M1: Equation of motion for combined mass. A1: \(M = 20\). M1: Equation of motion for the box. A1: \(220\text{ N}\).

(a) When the beam is on the point of tilting about \(C\), the tension at \(D\) is zero. Taking moments about \(C\): \(30g \times CG = 15g \times AC\). Since \(CG = x - 1.5\) and \(AC = 1.5\): \(30g(x - 1.5) = 15g(1.5) \implies 30(x - 1.5) = 22.5 \implies x - 1.5 = 0.75 \implies x = 2.25\). (b) When the container is removed and the person of mass \(W\) stands at \(B\), the beam is on the point of tilting about \(D\), so the tension at \(C\) is zero. Taking moments about \(D\): \(Wg \times DB = 30g \times GD\). Since \(DB = 1\text{ m}\), \(AD = 6 - 1 = 5\text{ m}\), and \(GD = 5 - x = 5 - 2.25 = 2.75\text{ m}\): \(W \times 1 = 30 \times 2.75 \implies W = 82.5\).

(a) M1: Identifies tension at \(D\) is zero. M1: Takes moments about \(C\). A1: Correct moment equation. M1: Solves for \(x\). A1: \(x = 2.25\). (b) M1: Identifies tension at \(C\) is zero. M1: Identifies distance \(GD = 2.75\text{ m}\). M1: Takes moments about \(D\). A1: Correct moment equation. A1: \(W = 82.5\).

(a) Using \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\): \(6\mathbf{i} - 9\mathbf{j} = (-2\mathbf{i} + 3\mathbf{j}) + 4\mathbf{a} \implies 4\mathbf{a} = 8\mathbf{i} - 12\mathbf{j} \implies \mathbf{a} = (2\mathbf{i} - 3\mathbf{j})\text{ m s}^{-2}\). (b) Using \(\mathbf{F} = m\mathbf{a} = 0.5(2\mathbf{i} - 3\mathbf{j}) = (\mathbf{i} - 1.5\mathbf{j})\text{ N}\). Its magnitude is \(|\mathbf{F}| = \sqrt{1^2 + (-1.5)^2} = \sqrt{3.25} \approx 1.80\text{ N}\). (c) Using \(\mathbf{r}(t) = \mathbf{r}_0 + \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\): \(\mathbf{r}(t) = (5\mathbf{i} + 2\mathbf{j}) + (-2\mathbf{i} + 3\mathbf{j})t + \frac{1}{2}(2\mathbf{i} - 3\mathbf{j})t^2 = (5 - 2t + t^2)\mathbf{i} + (2 + 3t - 1.5t^2)\mathbf{j}\). At \(t = 6\text{ s}\), \(\mathbf{r}(6) = (5 - 12 + 36)\mathbf{i} + (2 + 18 - 54)\mathbf{j} = 29\mathbf{i} - 34\mathbf{j}\). The distance is \(d = \sqrt{29^2 + (-34)^2} = \sqrt{1997} \approx 44.7\text{ m}\) (to 3 s.f.).

(a) M1: Uses \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\). A1: Correct setup. A1: \(\mathbf{a} = 2\mathbf{i} - 3\mathbf{j}\). (b) M1: Uses \(\mathbf{F} = m\mathbf{a}\). A1: Correct force vector. M1: Correct magnitude method. A1: Magnitude is \(1.80\text{ N}\). (c) M1: Correct formula for position vector. A1: Correct expression in terms of \(t\). M1: Substitutes \(t = 6\) and finds magnitude. A1: Distance is \(44.7\text{ m}\).

(a) Since \(\sin \theta = 0.6\), \(\cos \theta = 0.8\). Resolving perpendicular to the plane: \(R = mg \cos \theta + P \sin \theta = 4g(0.8) + P(0.6) = 3.2g + 0.6P\). (b) When \(P\) is at its maximum, the particle is on the point of slipping up the plane, so the friction force \(F\) acts down the plane: \(F = F_{\max} = \mu R = 0.5(3.2g + 0.6P) = 1.6g + 0.3P\). Resolving parallel to the plane (upwards): \(P \cos \theta - mg \sin \theta - F = 0 \implies 0.8P - 4g(0.6) - (1.6g + 0.3P) = 0 \implies 0.5P - 4g = 0 \implies 0.5P = 4g\). Using \(g = 9.8\text{ m s}^{-2}\): \(P = 8 \times 9.8 = 78.4\text{ N}\).

(a) M1: Identifies \(\cos \theta = 0.8\). M1: Resolves perpendicular to the plane. A1: Correctly shows the given formula. (b) M1: Identifies that friction acts down the plane and uses \(F = \mu R\). A1: Correct expression for friction. M1: Resolves parallel to the plane. A1: Correct equation of equilibrium parallel to the plane. M1: Substitutes \(F\) and solves for \(P\). A1: \(0.5P = 4g\) or equivalent. A1: \(P = 78.4\text{ N}\) (or \(78\text{ N}\)).

(a) For the class \(12 \le L < 14\):
- Class width \(= 14 - 12 = 2\) units, which is represented by 1 cm.
So 1 unit of length is represented by \(0.5\text{ cm}\).
- Frequency density \(= \frac{\text{Frequency}}{\text{Class Width}} = \frac{18}{2} = 9\).
This frequency density of 9 is represented by a height of 4.5 cm.
So 1 unit of frequency density is represented by \(0.5\text{ cm}\).

For the class \(14 \le L < 17\):
- Class width \(= 17 - 14 = 3\) units.
- Represented width \(= 3 \times 0.5 = 1.5\text{ cm}\).
- Frequency density \(= \frac{24}{3} = 8\).
- Represented height \(= 8 \times 0.5 = 4\text{ cm}\).

(b) The median is the \(40\)th value.
Since there are 12 values in the first class and 18 in the second, the cumulative frequency at the upper boundary 14 is 30.
Therefore, the median lies in the class \(14 \le L < 17\).
\[
\text{Median} = 14 + \left(\frac{40 - 30}{24}\right) \times 3 = 14 + 1.25 = 15.25\text{ mm}
\]

(c) Mean \(\bar{L} = \frac{\sum fL}{n} = \frac{1251}{80} = 15.6375 \approx 15.6\text{ mm}\) (to 3 s.f.).
Standard deviation \(\sigma = \sqrt{\frac{\sum fL^2}{n} - \bar{L}^2} = \sqrt{\frac{20470.5}{80} - 15.6375^2} = \sqrt{255.88125 - 244.531406} = \sqrt{11.349844} \approx 3.37\text{ mm}\) (to 3 s.f.).

(a)
M1: For finding the scaling factor for class width (0.5 cm per unit) or frequency density (0.5 cm per unit).
A1: For correct width of 1.5 cm.
M1: For finding the frequency density of 8 and applying the height scale.
A1: For correct height of 4 cm.

(b)
M1: For identifying the correct class \(14 \le L < 17\) and the fraction \(\frac{40 - 30}{24}\).
M1: For a correct interpolation expression using their values.
A1: For 15.25.

(c)
B1: For mean = 15.6 (accept 15.6375).
M1: For a correct standard deviation formula substituted with their values.
A1: For standard deviation = 3.37 (accept 3.3689...).

(a) Since the sum of probabilities must equal 1:
\[
a + b + b + a + 0.1 = 1 \implies 2a + 2b = 0.9 \implies a + b = 0.45
\]
Next, we calculate the expected value of \(X\):
\[
\text{E}(X) = -1(a) + 0(b) + 1(b) + 2(a + 0.1) = -a + b + 2a + 0.2 = a + b + 0.2
\]
Since \(a + b = 0.45\), we have:
\[
\text{E}(X) = 0.45 + 0.2 = 0.65
\]
Now, we find \(\text{E}(X^2)\):
\[
\text{E}(X^2) = (-1)^2(a) + 0^2(b) + 1^2(b) + 2^2(a + 0.1) = a + b + 4(a + 0.1) = 5a + b + 0.4
\]
Since \(\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2\):
\[
0.9275 = 5a + b + 0.4 - (0.65)^2 \implies 0.9275 = 5a + b + 0.4 - 0.4225 \implies 5a + b = 0.95
\]
We now have the simultaneous equations:
1) \(a + b = 0.45\)
2) \(5a + b = 0.95\)

Subtracting (1) from (2):
\[
4a = 0.5 \implies a = 0.125
\]
Substituting back into (1):
\[
b = 0.45 - 0.125 = 0.325
\]

(b)
\[
\text{P}(-1.5 < X \le 1) = \text{P}(X = -1) + \text{P}(X = 0) + \text{P}(X = 1) = a + b + b = a + 2b
\]
Substituting \(a = 0.125\) and \(b = 0.325\):
\[
\text{P}(-1.5 < X \le 1) = 0.125 + 2(0.325) = 0.775
\]

(c) The possible outcomes where \(X_1 + X_2 > 2\) are:
- \(X_1 = 1, X_2 = 2\) with probability \(0.325 \times 0.225 = 0.073125\)
- \(X_1 = 2, X_2 = 1\) with probability \(0.225 \times 0.325 = 0.073125\)
- \(X_1 = 2, X_2 = 2\) with probability \(0.225 \times 0.225 = 0.050625\)

Summing these probabilities:
\[
\text{P}(X_1 + X_2 > 2) = 0.073125 + 0.073125 + 0.050625 = 0.196875 \approx 0.197 \text{ (to 3 s.f.)}
\]

(a)
M1: Formulates the equation \(2a + 2b + 0.1 = 1\).
M1: Finds \(\text{E}(X) = 0.65\).
M1: Obtains \(\text{E}(X^2) = 5a + b + 0.4\).
M1: Correctly sets up simultaneous equations by substituting into \(\text{Var}(X)\).
A1: For both \(a = 0.125\) and \(b = 0.325\).

(b)
M1: Identifies the sum is \(\text{P}(X=-1) + \text{P}(X=0) + \text{P}(X=1)\).
A1: For 0.775.

(c)
M1: Lists the correct pairs: \((1,2)\), \((2,1)\), and \((2,2)\).
M1: Computes the probability of at least one of these pairs using multiplication for independent events.
M1: Adds the probabilities of all valid pairs.
A1: For \(0.197\) (accept \(0.196875\) or \(63/320\)).

(a)
\[
S_{tt} = \sum t^2 - \frac{(\sum t)^2}{n} = 2184 - \frac{142^2}{10} = 2184 - 2016.4 = 167.6
\]
\[
S_{dd} = \sum d^2 - \frac{(\sum d)^2}{n} = 432500 - \frac{2050^2}{10} = 432500 - 420250 = 12250
\]
\[
S_{td} = \sum td - \frac{(\sum t)(\sum d)}{n} = 27900 - \frac{142 \times 2050}{10} = 27900 - 29110 = -1210
\]

(b)
\[
r = \frac{S_{td}}{\sqrt{S_{tt} S_{dd}}} = \frac{-1210}{\sqrt{167.6 \times 12250}} = \frac{-1210}{\sqrt{2053100}} = \frac{-1210}{1432.864} \approx -0.8444 \approx -0.844 \text{ (to 3 s.f.)}
\]

(c) Yes, a linear regression model is suitable because the product moment correlation coefficient, \(r = -0.844\), is close to \(-1\), indicating a strong negative linear correlation between temperature and hot drink sales.

(d)
\[
b = \frac{S_{td}}{S_{tt}} = \frac{-1210}{167.6} \approx -7.21957 \approx -7.22 \text{ (to 3 s.f.)}
\]
\[
a = \bar{d} - b\bar{t} = \frac{2050}{10} - (-7.21957) \left(\frac{142}{10}\right) = 205 - (-7.21957 \times 14.2) = 205 + 102.518 = 307.518 \approx 307.5 \text{ (to 3 s.f.)}
\]
So the regression equation of \(d\) on \(t\) is:
\[
d = 307.5 - 7.22t
\]
(Accept \(d = 308 - 7.22t\))

(e) Substituting \(t = 18\):
\[
d = 307.518 - 7.21957(18) = 177.566 \approx 178 \text{ hot drinks (nearest integer)}
\]
This estimate is likely to be reliable because \(t = 18\) is close to the range of data (interpolation, with the mean temperature being \(14.2\) °C) and there is a strong correlation.

(a)
M1: For any one correct formula or calculation of \(S_{tt}\), \(S_{dd}\), or \(S_{td}\).
A1: For two correct values.
A1: For all three correct: \(S_{tt} = 167.6\), \(S_{dd} = 12250\), \(S_{td} = -1210\).

(b)
M1: Correct substitution of their values into the PMCC formula.
A1: For \(r \approx -0.844\).

(c)
B1: For stating "suitable" with a valid reason referencing the strong negative correlation.

(d)
M1: Correct formula for \(b = S_{td} / S_{tt}\).
A1: For \(b \approx -7.22\).
A1: For \(a \approx 307.5\) or \(308\).

(e)
M1: Substitutes \(t = 18\) into their regression equation.
A1: For 178 (accept 177 or 177.6) and comments that it is reliable due to interpolation/strong correlation.

(a) Since \(B\) and \(C\) are mutually exclusive, there is no overlap between circle \(B\) and circle \(C\) (i.e., \(B \cap C = \emptyset\)). Therefore, \(A \cap B \cap C = \emptyset\).
- \(A \cap B = 0.15\) (this region is inside \(A\) and \(B\) but outside \(C\)).
- \(A \cap C = 0.12\) (this region is inside \(A\) and \(C\) but outside \(B\)).
- The region representing "\(A\) only" is \(0.45 - 0.15 - 0.12 = 0.18\).
- The region representing "\(B\) only" is \(0.35 - 0.15 = 0.20\).
- The region representing "\(C\) only" is \(0.30 - 0.12 = 0.18\).
- The probability of being outside all circles is:
\[
1 - (0.18 + 0.15 + 0.12 + 0.20 + 0.18) = 1 - 0.83 = 0.17
\]

(b) Since \(B\) and \(C\) are mutually exclusive, \(B \cap C' = B\).
Therefore, we need to find:
\[
\text{P}(A \cup B) = \text{P}(A) + \text{P}(B) - \text{P}(A \cap B) = 0.45 + 0.35 - 0.15 = 0.65
\]

(c) Using the definition of conditional probability:
\[
\text{P}(A' \cap C' | B') = \frac{\text{P}(A' \cap C' \cap B')}{\text{P}(B')}
\]
- \(\text{P}(B') = 1 - \text{P}(B) = 1 - 0.35 = 0.65\).
- \(A' \cap B' \cap C'\) represents the region outside all three circles, which is \(0.17\).
\[
\text{P}(A' \cap C' | B') = \frac{0.17}{0.65} = \frac{17}{65} \approx 0.262 \text{ (to 3 s.f.)}
\]

(d) Two events \(A\) and \(B\) are independent if and only if \(\text{P}(A \cap B) = \text{P}(A) \times \text{P}(B)\).
\[
\text{P}(A) \times \text{P}(B) = 0.45 \times 0.35 = 0.1575
\]
Since \(\text{P}(A \cap B) = 0.15 \ne 0.1575\), the events \(A\) and \(B\) are not independent.

(a)
M1: Draw three circles with B and C completely separate (no intersection).
M1: Correctly places 0.15 in \(A \cap B\) and 0.12 in \(A \cap C\).
A1: Correct values for only A (0.18), only B (0.20), only C (0.18).
A1: Correct value outside (0.17).

(b)
M1: Identifies \(B \cap C'\) is \(B\) or uses Venn diagram to sum correct regions.
A1: For 0.65.

(c)
M1: Writes down a correct conditional probability formula/ratio.
M1: Identifies \(\text{P}(A' \cap B' \cap C') = 0.17\).
A1: For \(17/65\) or \(0.262\).

(d)
M1: Computes \(\text{P}(A) \times \text{P}(B)\).
A1: Compares \(\text{P}(A \cap B)\) with \(\text{P}(A) \times \text{P}(B)\) and states that they are not independent with a valid quantitative reason.

(a) Standardizing the first equation:
\[
\text{P}\left(Z < \frac{120 - \mu}{\sigma}\right) = 0.1587
\]
Since \(0.1587 < 0.5\), the z-score must be negative. From normal tables, \(\Phi(-1.00) = 0.1587\).
\[
\frac{120 - \mu}{\sigma} = -1.00 \implies \mu - \sigma = 120 \quad \text{(Equation 1)}
\]

Standardizing the second equation:
\[
\text{P}\left(Z > \frac{164 - \mu}{\sigma}\right) = 0.1151 \implies \text{P}\left(Z < \frac{164 - \mu}{\sigma}\right) = 0.8849
\]
From normal tables, \(\Phi(1.20) = 0.8849\).
\[
\frac{164 - \mu}{\sigma} = 1.20 \implies \mu + 1.2\sigma = 164 \quad \text{(Equation 2)}
\]

(b) Subtracting Equation 1 from Equation 2:
\[
(\mu + 1.2\sigma) - (\mu - \sigma) = 164 - 120
\]
\[
2.2\sigma = 44 \implies \sigma = 20
\]
Substituting \(\sigma = 20\) into Equation 1:
\[
\mu - 20 = 120 \implies \mu = 140
\]

(c) Find the probability that a single apple is small (\(Y < 130\)):
\[
\text{P}(Y < 130) = \text{P}\left(Z < \frac{130 - 140}{20}\right) = \text{P}(Z < -0.50)
\]
\[
\text{P}(Z < -0.50) = 1 - \Phi(0.50) = 1 - 0.6915 = 0.3085
\]
Let \(p = 0.3085\). For a sample of 3 apples, the number of small apples follows a binomial distribution with \(n=3\) and \(p=0.3085\).
\[
\text{P}(\text{Exactly 1}) = \binom{3}{1} p^1 (1 - p)^2 = 3 \times 0.3085 \times (1 - 0.3085)^2
\]
\[
= 3 \times 0.3085 \times 0.6915^2 = 3 \times 0.3085 \times 0.478172 = 0.44254 \approx 0.443 \text{ (to 3 s.f.)}
\]

(a)
M1: Standardizing either equation with \(\mu\) and \(\sigma\).
B1: Correct z-values of \(-1.00\) and \(1.20\) (or better) from tables.
A1: Correct first equation: \(\mu - \sigma = 120\).
A1: Correct second equation: \(\mu + 1.2\sigma = 164\).

(b)
M1: Eliminates one variable from their equations to solve for \(\mu\) or \(\sigma\).
A1: \(\sigma = 20\).
A1: \(\mu = 140\).

(c)
M1: Standardizes \(130\) using their \(\mu\) and \(\sigma\).
A1: Obtains \(p = 0.3085\) (accept \(0.309\)).
M1: Uses the binomial formula \(\binom{3}{1} p (1-p)^2\).
A1: For \(0.443\) (accept in range \(0.442 - 0.444\)).

(a) We are given \(\text{P}(X=x) = k x^2\) for \(x = 1, 2, 3, 4\).
Since the sum of probabilities must equal 1:
\[
\sum_{x=1}^{4} k x^2 = 1 \implies k(1^2 + 2^2 + 3^2 + 4^2) = 1
\]
\[
k(1 + 4 + 9 + 16) = 1 \implies 30k = 1 \implies k = \frac{1}{30}
\]
Thus, \(\text{P}(X=x) = \frac{x^2}{30}\).

(b)
\[
\text{E}(X) = \sum x \text{P}(X=x) = 1\left(\frac{1}{30}\right) + 2\left(\frac{4}{30}\right) + 3\left(\frac{9}{30}\right) + 4\left(\frac{16}{30}\right)
\]
\[
= \frac{1 + 8 + 27 + 64}{30} = \frac{100}{30} = \frac{10}{3} \approx 3.33
\]
Next, we find \(\text{E}(X^2)\):
\[
\text{E}(X^2) = \sum x^2 \text{P}(X=x) = 1^2\left(\frac{1}{30}\right) + 2^2\left(\frac{4}{30}\right) + 3^2\left(\frac{9}{30}\right) + 4^2\left(\frac{16}{30}\right)
\]
\[
= \frac{1(1) + 4(4) + 9(9) + 16(16)}{30} = \frac{1 + 16 + 81 + 256}{30} = \frac{354}{30} = 11.8
\]
\[
\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2 = 11.8 - \left(\frac{10}{3}\right)^2 = 11.8 - \frac{100}{9} = \frac{106.2 - 100}{9} = \frac{6.2}{9} = \frac{31}{45} \approx 0.689
\]

(c) Using coding properties:
\[
\text{E}(Y) = \text{E}(3X - 2) = 3\text{E}(X) - 2 = 3\left(\frac{10}{3}\right) - 2 = 10 - 2 = 8
\]
\[
\text{Var}(Y) = \text{Var}(3X - 2) = 3^2 \text{Var}(X) = 9 \times \frac{31}{45} = \frac{31}{5} = 6.2
\]

(a)
M1: Sets up the equation \(\sum k x^2 = 1\).
M1: Simplifies the sum of squares to 30.
A1: Concludes with \(k = 1/30\) and states the probability distribution correctly.

(b)
M1: Correct method for \(\text{E}(X)\).
A1: \(\text{E}(X) = 10/3\) (or exact equivalent fraction/decimal).
M1: Correct method for \(\text{E}(X^2)\) and applies \(\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2\).
A1: \(\text{Var}(X) = 31/45\) (or exact equivalent, or approx \(0.689\)).

(c)
M1: Applies linear coding to find \(\text{E}(Y)\).
A1: \(\text{E}(Y) = 8\).
B1: \(\text{Var}(Y) = 6.2\) (or equivalent fraction \(31/5\)).

(a) The tree diagram should have:
- First level branches: Machine \(A\) (probability \(0.60\)) and Machine \(B\) (probability \(0.40\)).
- Second level branches from \(A\): Defective \(D\) (probability \(0.03\)) and Not Defective \(D'\) (probability \(0.97\)).
- Second level branches from \(B\): Defective \(D\) (probability \(0.05\)) and Not Defective \(D'\) (probability \(0.95\)).

(b) Using the law of total probability:
\[
\text{P}(D) = \text{P}(A \cap D) + \text{P}(B \cap D) = (0.60 \times 0.03) + (0.40 \times 0.05)
\]
\[
= 0.018 + 0.020 = 0.038
\]

(c) Using Bayes' theorem / conditional probability:
\[
\text{P}(A | D) = \frac{\text{P}(A \cap D)}{\text{P}(D)} = \frac{0.018}{0.038} = \frac{18}{38} = \frac{9}{19} \approx 0.474 \text{ (to 3 s.f.)}
\]

(d) The probability of selecting a non-defective bolt is \(1 - 0.038 = 0.962\).
\[
\text{P}(\text{at least one defective}) = 1 - \text{P}(\text{both non-defective}) = 1 - (0.962)^2
\]
\[
= 1 - 0.925444 = 0.074556 \approx 0.0746 \text{ (to 3 s.f.)}
\]

(a)
M1: Draws two-stage tree diagram representing Machine and Defect branches.
A1: Correct probabilities on first-level branches (0.60, 0.40).
A1: Correct probabilities on second-level branches (0.03, 0.97, 0.05, 0.95).

(b)
M1: Identifies the two pathways for defective bolts: \(A \cap D\) and \(B \cap D\).
M1: Multiplies and sums these pathways.
A1: For 0.038.

(c)
M1: Uses the conditional probability formula \(\text{P}(A|D) = \text{P}(A \cap D) / \text{P}(D)\).
M1: Substitutes their values from (b) into the formula.
A1: For \(9/19\) or \(0.474\).

(d)
M1: Recognizes the method of subtraction from 1: \(1 - \text{P}(D')^2\).
A1: For \(0.0746\) (accept \(0.074556\)).