2025 Edexcel IAS-Level Mathematics (XMA01) 模擬試題連答案詳解

Equating the line and the curve: \(x^2 + 5x + 1 = kx - 3\). Rearranging into standard quadratic form: \(x^2 + (5 - k)x + 4 = 0\). For the line and curve not to intersect, the quadratic equation must have no real roots, so the discriminant must be less than zero: \((5 - k)^2 - 4(1)(4) < 0\). Expanding and simplifying: \(25 - 10k + k^2 - 16 < 0\), which gives \(k^2 - 10k + 9 < 0\). Factorising the quadratic: \((k - 1)(k - 9) < 0\). Therefore, the range of values for \(k\) is \(1 < k < 9\).

M1: Equates the equations of the line and curve and rearranges to form a quadratic. A1: Correct quadratic equation \(x^2 + (5 - k)x + 4 = 0\) or equivalent. M1: Recognises the condition for no intersection is \(b^2 - 4ac < 0\) and substitutes their coefficients. M1: Solves the quadratic inequality to find critical values \(k = 1\) and \(k = 9\). A1.33: Correct final inequality \(1 < k < 9\) or equivalent notation.

First rewrite the curve equation in a form suitable for differentiation: \(y = \frac{4x^3}{2x^{1/2}} - \frac{5}{2x^{1/2}} = 2x^{5/2} - \frac{5}{2}x^{-1/2}\). Differentiating with respect to \(x\): \(\frac{dy}{dx} = 2 \left(\frac{5}{2}\right)x^{3/2} - \frac{5}{2}\left(-\frac{1}{2}\right)x^{-3/2} = 5x^{3/2} + \frac{5}{4}x^{-3/2}\). At the point \(P\) where \(x = 1\), the gradient of the tangent is: \(\frac{dy}{dx} = 5(1)^{3/2} + \frac{5}{4}(1)^{-3/2} = 5 + \frac{5}{4} = \frac{25}{4}\). The gradient of the normal is the negative reciprocal of the tangent gradient: \(m = -\frac{4}{25}\). Using the equation of a straight line with gradient \(m = -\frac{4}{25}\) and point \((1, -1/2)\): \(y - (-1/2) = -\frac{4}{25}(x - 1)\). Multiplying the entire equation by 50 to clear fractions: \(50(y + 1/2) = -8(x - 1) \implies 50y + 25 = -8x + 8\). Rearranging into the form \(ax + by + c = 0\) gives: \(8x + 50y + 17 = 0\).

M1: Rewrites the curve equation as \(A x^p + B x^q\) with at least one correct index. A1: Correctly differentiates to obtain \(\frac{dy}{dx} = 5x^{3/2} + \frac{5}{4}x^{-3/2}\) or equivalent. M1: Substitutes \(x = 1\) into their \(\frac{dy}{dx}\) to find the tangent gradient, and takes the negative reciprocal to find the normal gradient. M1: Uses their normal gradient and point \((1, -0.5)\) to form the equation of a line. A1.33: Correct equation in the required form, \(8x + 50y + 17 = 0\) or any non-zero integer multiple.

To find \(f(x)\), integrate \(f'(x)\) with respect to \(x\): \(f(x) = \int \left(6x^2 - 4x + 3x^{-2}\right) dx\). Integrating term by term: \(f(x) = \frac{6x^3}{3} - \frac{4x^2}{2} + \frac{3x^{-1}}{-1} + C = 2x^3 - 2x^2 - \frac{3}{x} + C\), where \(C\) is the constant of integration. Since the curve passes through \((1, 5)\), substitute \(x = 1\) and \(f(1) = 5\): \(5 = 2(1)^3 - 2(1)^2 - \frac{3}{1} + C \implies 5 = 2 - 2 - 3 + C \implies 5 = -3 + C \implies C = 8\). Therefore, \(f(x) = 2x^3 - 2x^2 - \frac{3}{x} + 8\).

M1: Integrates at least one term of \(f'(x)\) correctly (power increases by 1). A1: Correctly integrates at least two terms. A1: Fully correct integration with constant of integration, \(2x^3 - 2x^2 - \frac{3}{x} + C\). M1: Substitutes \(x = 1\) and \(f(x) = 5\) into their integrated expression to find \(C\). A1.33: Fully correct simplified expression \(f(x) = 2x^3 - 2x^2 - \frac{3}{x} + 8\) or \(y = 2x^3 - 2x^2 - \frac{3}{x} + 8\).

(a) Equating the expressions for the curve \( C \) and the line \( L \):
\[ k(x^2 + 1) + 8x = 6 \]
\[ kx^2 + 8x + k - 6 = 0 \]
Since the curve and line do not intersect, this quadratic equation has no real roots. Therefore, its discriminant must be strictly negative:
\[ \Delta = b^2 - 4ac < 0 \]
\[ 8^2 - 4(k)(k - 6) < 0 \]
\[ 64 - 4k^2 + 24k < 0 \]
Dividing the entire inequality by \(-4\) (and reversing the inequality sign):
\[ k^2 - 6k - 16 > 0 \]

(b) Factoring the quadratic expression:
\[ (k - 8)(k + 2) > 0 \]
The critical values are \( k = 8 \) and \( k = -2 \).
Since we require the expression to be greater than zero, the solution is:
\[ k > 8 \quad \text{or} \quad k < -2 \]

Part (a):
- M1: Equates the equations of the curve and the line to form a single quadratic equation in \( x \).
- A1: Rearranges the equation into the correct form \( kx^2 + 8x + (k - 6) = 0 \).
- M1: Attempts to use the discriminant condition \( b^2 - 4ac < 0 \) for no real roots.
- A1: Obtains \( 64 - 4k(k - 6) < 0 \) or equivalent.
- A1*: Completes the proof rigorously to reach the given inequality \( k^2 - 6k - 16 > 0 \).

Part (b):
- M1: Attempts to find the critical values by factorising or using the quadratic formula on \( k^2 - 6k - 16 = 0 \).
- A1: Obtains critical values of \( 8 \) and \( -2 \).
- A1.43: Correctly identifies the outside regions, writing \( k > 8 \text{ or } k < -2 \) (allow set notation).

(a) First, find the midpoint \( M \) of \( AB \):
\[ M = \left( \frac{1 + 5}{2}, \frac{7 - 1}{2} \right) = (3, 3) \]

Next, find the gradient of the line segment \( AB \):
\[ m_{AB} = \frac{-1 - 7}{5 - 1} = \frac{-8}{4} = -2 \]

The gradient of the perpendicular bisector is the negative reciprocal of \( m_{AB} \):
\[ m = -\frac{1}{-2} = \frac{1}{2} \]

The equation of the perpendicular bisector passing through \( M(3, 3) \) is:
\[ y - 3 = \frac{1}{2}(x - 3) \]
\[ 2y - 6 = x - 3 \]
\[ x - 2y + 3 = 0 \]

(b) The perpendicular bisector intersects the \( y \)-axis at \( C \), so we set \( x = 0 \):
\[ 0 - 2y + 3 = 0 \implies y = 1.5 \]
Thus, the coordinates of \( C \) are \( (0, 1.5) \).

To find the area of triangle \( ABC \), we can use the length of the base \( AB \) and the perpendicular height from \( C \) to \( AB \), which is the distance \( MC \):
\[ AB = \sqrt{(5 - 1)^2 + (-1 - 7)^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5} \]
\[ MC = \sqrt{(3 - 0)^2 + (3 - 1.5)^2} = \sqrt{9 + 2.25} = \sqrt{11.25} = \sqrt{\frac{45}{4}} = \frac{3\sqrt{5}}{2} \]

Therefore, the area of triangle \( ABC \) is:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4\sqrt{5} \times \frac{3\sqrt{5}}{2} = 15 \]

Part (a):
- M1: Correct attempt to find the midpoint of \( AB \).
- A1: Midpoint is \( (3, 3) \).
- M1: Correct attempt to find the gradient of \( AB \) and use negative reciprocal to find perpendicular gradient.
- A1.43: Formulates the correct equation in the specified form \( x - 2y + 3 = 0 \) (or any integer multiple thereof).

Part (b):
- M1: Sets \( x = 0 \) to find the coordinates of point \( C \).
- A1: Correct coordinates \( C(0, 1.5) \).
- M1: Correct method to find the area of triangle \( ABC \) (e.g., using \( \frac{1}{2} \times AB \times MC \) or Shoelace formula).
- A1: Obtains the correct final answer of \( 15 \).

(a) Using the trigonometric identity \( \cos^2(2\theta) = 1 - \sin^2(2\theta) \):
\[ 6(1 - \sin^2(2\theta)) - \sin(2\theta) - 5 = 0 \]
\[ 6 - 6\sin^2(2\theta) - \sin(2\theta) - 5 = 0 \]
\[ -6\sin^2(2\theta) - \sin(2\theta) + 1 = 0 \]
Multiplying the entire equation by \(-1\):
\[ 6\sin^2(2\theta) + \sin(2\theta) - 1 = 0 \]

(b) Let \( y = \sin(2\theta) \). The quadratic equation becomes:
\[ 6y^2 + y - 1 = 0 \]
\[ (3y - 1)(2y + 1) = 0 \]
This gives:
\[ \sin(2\theta) = \frac{1}{3} \quad \text{or} \quad \sin(2\theta) = -\frac{1}{2} \]

Since \( 0 \le \theta < \pi \), the interval for \( 2\theta \) is \( 0 \le 2\theta < 2\pi \).

Case 1: \( \sin(2\theta) = \frac{1}{3} \)
\[ 2\theta = \arcsin\left(\frac{1}{3}\right) \approx 0.3398 \text{ rad} \implies \theta \approx 0.170 \text{ rad} \]
\[ 2\theta = \pi - 0.3398 \approx 2.8018 \text{ rad} \implies \theta \approx 1.40 \text{ rad} \]

Case 2: \( \sin(2\theta) = -\frac{1}{2} \)
\[ 2\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \approx 3.6652 \text{ rad} \implies \theta = \frac{7\pi}{12} \approx 1.83 \text{ rad} \]
\[ 2\theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6} \approx 5.7596 \text{ rad} \implies \theta = \frac{11\pi}{12} \approx 2.88 \text{ rad} \]

Part (a):
- M1: Uses \( \cos^2(2\theta) = 1 - \sin^2(2\theta) \).
- A1*: Completes steps to show the given equation with no errors.

Part (b):
- M1: Factorises or solves the quadratic equation to find the two values of \( \sin(2\theta) \).
- M1: Finds at least one correct principal value for \( 2\theta \) in radians.
- A1: Obtains \( \theta \approx 0.170 \).
- A1: Obtains \( \theta \approx 1.40 \).
- A1: Obtains \( \theta \approx 1.83 \) (or exact \( \frac{7\pi}{12} \)).
- A1.43: Obtains \( \theta \approx 2.88 \) (or exact \( \frac{11\pi}{12} \)).

(a) Differentiating \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 2 \left( \frac{3}{2}x^{1/2} \right) - 8 \left( \frac{1}{2}x^{-1/2} \right) \]
\[ \frac{dy}{dx} = 3x^{1/2} - 4x^{-1/2} \]

(b) Set \( \frac{dy}{dx} = 4 \):
\[ 3x^{1/2} - 4x^{-1/2} = 4 \]
Multiply through by \( x^{1/2} \):
\[ 3x - 4 = 4x^{1/2} \]
\[ 3x - 4x^{1/2} - 4 = 0 \]
Letting \( u = x^{1/2} \), this is a quadratic equation:
\[ 3u^2 - 4u - 4 = 0 \]
\[ (3u + 2)(u - 2) = 0 \]
Since \( x > 0 \), we must have \( u = x^{1/2} > 0 \), so \( u = 2 \).
\[ x^{1/2} = 2 \implies x = 4 \]
Substitute \( x = 4 \) back into the original curve equation to find \( y \):
\[ y = 2(4)^{3/2} - 8(4)^{1/2} + 5 = 2(8) - 8(2) + 5 = 16 - 16 + 5 = 5 \]
So the coordinates of \( P \) are \( (4, 5) \).

(c) The tangent has gradient \( 4 \) and passes through \( (4, 5) \):
\[ y - 5 = 4(x - 4) \]
\[ y - 5 = 4x - 16 \]
\[ y = 4x - 11 \]

Part (a):
- M1: Power decreased by 1 on at least one fractional term.
- A1.43: Correct derivative \( 3x^{1/2} - 4x^{-1/2} \) (or equivalent).

Part (b):
- M1: Sets their \( \frac{dy}{dx} = 4 \) and forms a quadratic in \( x^{1/2} \) or equivalent.
- M1: Solves the quadratic to find a positive value for \( x^{1/2} \).
- A1: Obtains \( x = 4 \).
- A1: Obtains \( y = 5 \), so \( P = (4, 5) \).

Part (c):
- M1: Uses their point \( P \) and gradient \( 4 \) to write a straight line equation.
- A1: Obtains the correct final equation \( y = 4x - 11 \).

(a) Simplifying the fraction term-by-term:
\[ f'(x) = \frac{2x^2}{x} - \frac{5x^{1/2}}{x} \]
\[ f'(x) = 2x - 5x^{-1/2} \]
So \( A = 2 \), \( B = -5 \), and \( k = -0.5 \).

(b) Integrate \( f'(x) \) to find \( f(x) \):
\[ f(x) = \int (2x - 5x^{-1/2}) \, dx \]
\[ f(x) = x^2 - \frac{5x^{1/2}}{1/2} + C \]
\[ f(x) = x^2 - 10x^{1/2} + C \]
Since the curve passes through \( (9, -2) \):
\[ f(9) = -2 \]
\[ (9)^2 - 10(9)^{1/2} + C = -2 \]
\[ 81 - 10(3) + C = -2 \]
\[ 81 - 30 + C = -2 \]
\[ 51 + C = -2 \implies C = -53 \]
Thus, the equation of the curve is:
\[ f(x) = x^2 - 10\sqrt{x} - 53 \]

Part (a):
- M1: Attempts to divide at least one term of the numerator by \( x \).
- A1.43: Obtains the correct expression \( 2x - 5x^{-1/2} \).

Part (b):
- M1: Attempts to integrate \( f'(x) \) (at least one power increased by 1).
- A1: Integrates at least one term correctly.
- A1: Fully correct integration including the constant of integration \( + C \).
- M1: Substitutes \( x = 9 \) and \( y = -2 \) to find \( C \).
- A1: Obtains \( C = -53 \).
- A1: Formulates the final expression \( f(x) = x^2 - 10\sqrt{x} - 53 \).

(a) The curve \( y = (x - 2)^2(x + 3) \) is a positive cubic.
- It touches the \( x \)-axis at \( (2, 0) \).
- It crosses the \( x \)-axis at \( (-3, 0) \).
- Substituting \( x = 0 \) gives the \( y \)-intercept:
\[ y = (0 - 2)^2(0 + 3) = 4 \times 3 = 12 \]
So the \( y \)-intercept is \( (0, 12) \).
Sketch shows a standard positive cubic curve touching at \( x = 2 \), crossing at \( x = -3 \), and crossing \( y \)-axis at \( 12 \).

(b) A translation by the vector \( \begin{pmatrix} 2 \\ -4 \end{pmatrix} \) replaces \( x \) with \( x - 2 \) and subtracts \( 4 \) from the equation:
\[ y = ((x - 2) - 2)^2((x - 2) + 3) - 4 \]
\[ y = (x - 4)^2(x + 1) - 4 \]
Expanding the brackets:
\[ y = (x^2 - 8x + 16)(x + 1) - 4 \]
\[ y = x^3 + x^2 - 8x^2 - 8x + 16x + 16 - 4 \]
\[ y = x^3 - 7x^2 + 8x + 12 \]
So \( p = -7 \), \( q = 8 \), and \( r = 12 \).

Part (a):
- M1: Correct shape of a positive cubic.
- A1: Clearly touches the \( x \)-axis at \( (2, 0) \).
- A1: Clearly crosses the \( x \)-axis at \( (-3, 0) \).
- A1.43: Crosses the \( y \)-axis at \( (0, 12) \).

Part (b):
- M1: Correctly substitutes \( x \to x - 2 \) and subtracts \( 4 \) to reflect the translation.
- M1: Attempts to expand \( (x-4)^2(x+1) \).
- A1: Obtains expanded part \( x^3 - 7x^2 + 8x + 16 \).
- A1: Final correct equation \( y = x^3 - 7x^2 + 8x + 12 \).

(a) Let the height of the box be \( h \) cm.
Since the box is open-topped, the surface area consists of the base and 4 vertical sides:
- Base area: \( 3x \times x = 3x^2 \)
- Two sides of area: \( 3x \times h = 6xh \)
- Two sides of area: \( x \times h = 2xh \)

Thus, the total surface area \( S \) is:
\[ S = 3x^2 + 8xh = 576 \]
Rearrange to express \( h \) in terms of \( x \):
\[ 8xh = 576 - 3x^2 \implies h = \frac{576 - 3x^2}{8x} \]

The volume \( V \) of the box is:
\[ V = \text{length} \times \text{width} \times \text{height} = 3x \times x \times h = 3x^2 h \]
Substitute the expression for \( h \):
\[ V = 3x^2 \left( \frac{576 - 3x^2}{8x} \right) \]
\[ V = \frac{3x}{8}(576 - 3x^2) \]
\[ V = 216x - \frac{9}{8}x^3 \]

(b) To find the maximum volume, differentiate \( V \) with respect to \( x \):
\[ \frac{dV}{dx} = 216 - \frac{27}{8}x^2 \]
Set \( \frac{dV}{dx} = 0 \):
\[ 216 - \frac{27}{8}x^2 = 0 \]
\[ \frac{27}{8}x^2 = 216 \implies x^2 = \frac{216 \times 8}{27} = 64 \]
Since \( x > 0 \), we have \( x = 8 \).

Substitute \( x = 8 \) back into the volume formula to find the maximum volume:
\[ V = 216(8) - \frac{9}{8}(8)^3 = 1728 - 576 = 1152 \text{ cm}^3 \]

To justify that this is a maximum, find the second derivative:
\[ \frac{d^2V}{dx^2} = -\frac{27}{4}x \]
At \( x = 8 \):
\[ \frac{d^2V}{dx^2} = -\frac{27}{4}(8) = -54 \]
Since \( \frac{d^2V}{dx^2} < 0 \), the value is indeed a maximum.

Part (a):
- M1: Formulates the surface area of the open box as \( S = 3x^2 + 8xh \).
- A1: Rearranges correctly to find \( h = \frac{576 - 3x^2}{8x} \).
- M1: Substitutes their expression for \( h \) into the volume formula \( V = 3x^2h \).
- A1*: Shows convincing algebraic steps to arrive at \( V = 216x - \frac{9}{8}x^3 \).

Part (b):
- M1: Differentiates \( V \) to obtain \( \frac{dV}{dx} = 216 - kx^2 \).
- A1: Finds the correct critical value \( x = 8 \).
- M1: Substitutes their \( x = 8 \) into \( V \) to calculate the maximum volume.
- M1: Finds the second derivative and evaluates it at their \( x \) value.
- A0.43: Obtains the volume \( 1152 \) and successfully shows \( \frac{d^2V}{dx^2} < 0 \) to justify it is a maximum.

Using the formula for the sum of the first \(n\) terms of a geometric series, the sum of the first two terms is given by \(S_2 = a + ar = a(1 + r) = 15\) (Equation 1). Using the formula for the sum to infinity of a geometric series, \(S_{\infty} = \frac{a}{1 - r} = 27\) (Equation 2). From Equation 2, we can express \(a\) in terms of \(r\) as \(a = 27(1 - r)\). Substituting this expression into Equation 1 gives \(27(1 - r)(1 + r) = 15\), which simplifies to \(27(1 - r^2) = 15\). Dividing both sides by 27 gives \(1 - r^2 = \frac{5}{9}\), so \(r^2 = \frac{4}{9}\). Since \(r > 0\), we take the positive square root to find \(r = \frac{2}{3}\). Substituting this back into the expression for \(a\) gives \(a = 27\left(1 - \frac{2}{3}\right) = 9\). Thus, the values are \(a = 9\) and \(r = \frac{2}{3}\).

M1: Writes a correct equation for the sum of the first two terms, e.g., \(a + ar = 15\) or \(a(1+r) = 15\). M1: Writes a correct equation for the sum to infinity, e.g., \(\frac{a}{1-r} = 27\). M1: Eliminates \(a\) to obtain an equation in terms of \(r\) only, e.g., \(27(1-r)(1+r) = 15\). A1: Solves to find \(r = \frac{2}{3}\) (accept 0.67 or better; must reject negative root). A1: Finds \(a = 9\).

Using the laws of logarithms, we can rewrite the first term as \(\log_4((x+3)^2)\). This gives \(\log_4((x+3)^2) - \log_4(x) = 2\). Using the division law, we combine the logarithms to get \(\log_4\left(\frac{(x+3)^2}{x}\right) = 2\). Removing the logarithm by raising the base 4 to the power of both sides yields \(\frac{(x+3)^2}{x} = 4^2\), which simplifies to \(\frac{x^2 + 6x + 9}{x} = 16\). Multiplying by \(x\) gives \(x^2 + 6x + 9 = 16x\). Rearranging this quadratic equation yields \(x^2 - 10x + 9 = 0\). Factoring the quadratic gives \((x - 1)(x - 9) = 0\), which has solutions \(x = 1\) and \(x = 9\). Since both values are greater than 0, they are both valid solutions.

M1: Applies the power law of logarithms to write \(2\log_4(x+3)\) as \(\log_4(x+3)^2\). M1: Applies the division law of logarithms to obtain a single logarithm, e.g., \(\log_4\left(\frac{(x+3)^2}{x}\right) = 2\). M1: Removes logarithms correctly to form a quadratic equation, e.g., \(\frac{(x+3)^2}{x} = 16\). A1: Obtains the correct quadratic equation \(x^2 - 10x + 9 = 0\). A1: Correctly solves the quadratic to find \(x = 1\) and \(x = 9\) (both required).

(a) The sum of the first two terms is given by \(a + ar = 15 \implies a(1+r) = 15\). The sum to infinity is given by \(S_\infty = \frac{a}{1-r} = 27 \implies a = 27(1-r)\). Substituting \(a\) into the first equation: \(27(1-r)(1+r) = 15 \implies 27(1-r^2) = 15 \implies 27 - 27r^2 = 15 \implies 27r^2 = 12\). This simplifies to \(r^2 = \frac{12}{27} = \frac{4}{9}\), giving \(r = \pm \frac{2}{3}\). (b) Since \(r > 0\), we have \(r = \frac{2}{3}\). Substituting this into \(a = 27(1-r)\) gives \(a = 27(1 - \frac{2}{3}) = 9\). The 4th term is \(u_4 = ar^3 = 9 \times \left(\frac{2}{3}\right)^3 = 9 \times \frac{8}{27} = \frac{8}{3}\). (c) The sum of the first 5 terms is \(S_5 = \frac{a(1-r^5)}{1-r} = \frac{9(1 - (2/3)^5)}{1 - 2/3} = 27\left(1 - \frac{32}{243}\right) = 27 - \frac{32}{9} = \frac{211}{9} \approx 23.44\).

(a) M1: Formulates two equations in \(a\) and \(r\). A1: Eliminates \(a\) to obtain a correct equation in \(r\). M1: Clearly demonstrates \(27r^2 = 12\). A1: Finds both solutions \(r = \pm \frac{2}{3}\). (b) M1: Correctly identifies \(r = \frac{2}{3}\) to calculate \(a = 9\). A1: Calculates the exact 4th term as \(\frac{8}{3}\). (c) M1: Attempts to use the sum formula with \(n=5\). A1: Obtains \(\frac{211}{9}\) or equivalent. A1.28: Rounds correctly to 2 decimal places to get 23.44.

(a) Using the trigonometric identity \(\sin^2 \theta = 1 - \cos^2 \theta\), we substitute this into the equation: \(4(1 - \cos^2 \theta) - 7\cos \theta - 2 = 0 \implies 4 - 4\cos^2 \theta - 7\cos \theta - 2 = 0 \implies -4\cos^2 \theta - 7\cos \theta + 2 = 0\). Multiplying the entire equation by \(-1\) gives \(4\cos^2 \theta + 7\cos \theta - 2 = 0\). (b) Let \(y = \cos \theta\), then the equation becomes \(4y^2 + 7y - 2 = 0 \implies (4y - 1)(y + 2) = 0\). This gives \(\cos \theta = \frac{1}{4}\) or \(\cos \theta = -2\) (which has no real solutions since \(-1 \le \cos \theta \le 1\)). For \(\cos \theta = 0.25\), the principal value is \(\theta = \arccos(0.25) \approx 75.5^\circ\). The other solution in the range is \(360^\circ - 75.52^\circ \approx 284.5^\circ\). (c) Replacing \(\theta\) with \(2\phi\), we have \(\cos(2\phi) = 0.25\). In radians, the solutions for \(2\phi\) in the interval \(0 \le 2\phi < 4\pi\) are \(2\phi \approx 1.3181, 2\pi - 1.3181 = 4.9651, 2\pi + 1.3181 = 7.6013, 4\pi - 1.3181 = 11.2483\). Dividing by 2 gives \(\phi \approx 0.659, 2.48, 3.80, 5.62\) radians.

(a) M1: Uses identity \(\sin^2 \theta = 1 - \cos^2 \theta\). A1: Fully completes the proof showing all steps. (b) M1: Solves the quadratic equation to find a value for \(\cos \theta\). A1: Correctly identifies \(\cos \theta = 0.25\) as the only valid branch. M1: Finds one correct angle. A1: Finds both angles \(75.5^\circ\) and \(284.5^\circ\). (c) M1: Solves for \(2\phi\) in radians. A1: Identifies at least two correct values for \(2\phi\). A1.28: Provides all four correct solutions \(0.659, 2.48, 3.80, 5.62\) rounded to 3 significant figures.

(a) Completing the square for both \(x\) and \(y\): \((x-3)^2 - 9 + (y+4)^2 - 16 - 11 = 0 \implies (x-3)^2 + (y+4)^2 = 36\). Therefore, the coordinates of the centre are \((3, -4)\) and the radius is \(\sqrt{36} = 6\). (b) Substituting \(y = 2x + k\) into the circle equation \((x-3)^2 + (y+4)^2 = 36\): \((x-3)^2 + (2x + k + 4)^2 = 36 \implies x^2 - 6x + 9 + 4x^2 + 4x(k+4) + (k+4)^2 - 36 = 0 \implies 5x^2 + (4k + 10)x + (k^2 + 8k - 11) = 0\). Since \(l\) is a tangent, the discriminant of this quadratic equation in \(x\) must be zero: \(B^2 - 4AC = 0 \implies (4k+10)^2 - 4(5)(k^2 + 8k - 11) = 0 \implies 16k^2 + 80k + 100 - 20(k^2 + 8k - 11) = 0 \implies -4k^2 - 80k + 320 = 0 \implies k^2 + 20k - 80 = 0\). Solving this quadratic using the quadratic formula: \(k = \frac{-20 \pm \sqrt{400 - 4(1)(-80)}}{2} = \frac{-20 \pm \sqrt{720}}{2} = -10 \pm 6\sqrt{5}\).

(a) M1: Attempts to complete the square for both \(x\) and \(y\). A1: Coordinates of centre \((3, -4)\) correct. A1: Radius \(6\) correct. (b) M1: Substitutes \(y = 2x + k\) into the circle equation. A1: Obtains a correct 3-term quadratic in \(x\). M1: Uses the discriminant condition \(b^2 - 4ac = 0\). A1: Simplifies to the quadratic equation \(k^2 + 20k - 80 = 0\). M1: Solves the quadratic equation for \(k\). A1.28: Yields exact solutions \(k = -10 \pm 6\sqrt{5}\) (or equivalent simplified surd form).

(a) Using logarithm laws: \(2\log_3(x-2) = \log_3(x-2)^2\). Then, the subtraction law yields: \
\log_3\left(\frac{(x-2)^2}{x+4}\right) = 1\). Removing the logarithm: \(\frac{(x-2)^2}{x+4} = 3^1 \implies x^2 - 4x + 4 = 3x + 12 \implies x^2 - 7x - 8 = 0\). Factoring the quadratic: \((x-8)(x+1) = 0\), which gives \(x = 8\) or \(x = -1\). However, for the terms \(\log_3(x-2)\) and \(\log_3(x+4)\) to be defined, we require \(x > 2\). Therefore, \(x = 8\) is the only valid solution. (b) Let \(u = 3^y\). The equation becomes: \(u^2 - 10u + 9 = 0 \implies (u-9)(u-1) = 0\), giving \(u = 9\) or \(u = 1\). Substituting back \(3^y = u\): If \(3^y = 9 \implies y = 2\). If \(3^y = 1 \implies y = 0\). Both solutions are valid.

(a) M1: Uses power law \(k\log x = \log x^k\). M1: Uses division law \(\log a - \log b = \log(a/b)\). M1: Removes logarithms correctly by setting base to power 1. A1: Solves quadratic to find both roots. A1: Rejects \(x = -1\) with justification and states \(x = 8\). (b) M1: Converts the equation into a quadratic in terms of \(3^y\). A1: Solves the quadratic to find \(3^y = 1\) and \(3^y = 9\). M1: Solves exponential equations for \(y\). A1.28: Correctly states both solutions \(y = 0\) and \(y = 2\).

(a) Equating the two equations to find the points of intersection: \(10 + 3x - x^2 = x + 7 \implies x^2 - 2x - 3 = 0\). Factoring gives \((x-3)(x+1) = 0 \implies x = 3\) or \(x = -1\). When \(x = 3\), \(y = 3 + 7 = 10\). When \(x = -1\), \(y = -1 + 7 = 6\). So the points of intersection are \((-1, 6)\) and \((3, 10)\). (b) The area of the region \(R\) is given by: \(\int_{-1}^{3} (y_C - y_l) \, dx = \int_{-1}^{3} ((10 + 3x - x^2) - (x + 7)) \, dx = \int_{-1}^{3} (3 + 2x - x^2) \, dx\). Integrating term-by-term: \(\left[ 3x + x^2 - \frac{x^3}{3} \right]_{-1}^{3}\). Substituting the upper limit \(3\): \(3(3) + 3^2 - \frac{27}{3} = 9 + 9 - 9 = 9\). Substituting the lower limit \(-1\): \(3(-1) + (-1)^2 - \frac{(-1)^3}{3} = -3 + 1 + \frac{1}{3} = -\frac{5}{3}\). Area \(= 9 - \left(-\frac{5}{3}\right) = 9 + \frac{5}{3} = \frac{32}{3}\).

(a) M1: Equates equations of curve and line. A1: Solves quadratic to obtain correct x-coordinates \(-1\) and \(3\). A1: Correctly gives the coordinates of both points of intersection. (b) M1: Formulates a correct definite integral with appropriate limits. A1: Integrates correctly to obtain \(3x + x^2 - \frac{x^3}{3}\). M1: Substitutes the upper limit of 3. M1: Substitutes the lower limit of -1. A1: Evaluates both boundary values correctly. A1.28: Subtracts and provides exact area \(\frac{32}{3}\) (or \(10\frac{2}{3}\)).

(a) Using the factor theorem, \(f(2) = 0 \implies 2(8) + a(4) + b(2) - 10 = 0 \implies 4a + 2b = -6 \implies 2a + b = -3\). Using the remainder theorem, \(f(-1) = -24 \implies 2(-1) + a(1) + b(-1) - 10 = -24 \implies a - b = -12\). Adding the two equations: \((2a+b) + (a-b) = -3 - 12 \implies 3a = -15 \implies a = -5\). Substituting \(a = -5\) into \(a - b = -12\) gives \(-5 - b = -12 \implies b = 7\). (b) Since \(f(x) = 2x^3 - 5x^2 + 7x - 10\), we can divide \(f(x)\) by \((x-2)\) using algebraic long division or matching coefficients: \((x-2)(2x^2 + px + 5) = 2x^3 + (p - 4)x^2 + (5 - 2p)x - 10\). Comparing \(x^2\) terms: \(p - 4 = -5 \implies p = -1\). Hence, \(f(x) = (x-2)(2x^2 - x + 5)\). (c) Setting \(f(x) = 0\) gives \(x - 2 = 0 \implies x = 2\) or \(2x^2 - x + 5 = 0\). For the quadratic part, the discriminant is \(b^2 - 4ac = (-1)^2 - 4(2)(5) = 1 - 40 = -39\). Since the discriminant is negative (\(-39 < 0\)), the quadratic equation has no real roots. Thus, \(x = 2\) is the only real root.

(a) M1: Applies factor theorem to write \(f(2) = 0\). A1: Correct simplified equation \(2a + b = -3\). M1: Applies remainder theorem to write \(f(-1) = -24\). A1: Correct simplified equation \(a - b = -12\). A1: Solves simultaneously to find \(a = -5\) and \(b = 7\). (b) M1: Attempts to divide \(f(x)\) by \(x-2\) or match coefficients. A1.28: Fully shows the steps leading to the correct quotient \(2x^2 - x + 5\). (c) M1: Calculates the discriminant of the quadratic factor \(2x^2 - x + 5\). A1: Explicitly states the discriminant is negative and draws the conclusion that there are no other real roots.

(a) The volume \(V\) of the box is given by \(V = x \times 2x \times h = 2x^2 h = 360 \implies h = \frac{180}{x^2}\). The total external surface area \(A\) of the open-topped box is \(A = \text{base area} + 2 \times \text{front area} + 2 \times \text{side area} = 2x^2 + 2(2xh) + 2(xh) = 2x^2 + 6xh\). Substituting \(h\): \(A = 2x^2 + 6x\left(\frac{180}{x^2}\right) = 2x^2 + \frac{1080}{x}\). (b) Differentiating \(A\) with respect to \(x\): \(\frac{dA}{dx} = 4x - \frac{1080}{x^2}\). For a stationary point, \(\frac{dA}{dx} = 0 \implies 4x - \frac{1080}{x^2} = 0 \implies 4x^3 = 1080 \implies x^3 = 270 \implies x = \sqrt[3]{270} \approx 6.4633\). Substituting \(x\) back into \(A\): \(A = 2(270)^{2/3} + \frac{1080}{270^{1/3}} \approx 83.55 + 167.10 = 250.65\). Rounded to the nearest integer, the minimum surface area is \(251\text{ cm}^2\). (c) The second derivative is \(\frac{d^2A}{dx^2} = 4 + \frac{2160}{x^3}\). Since \(x \approx 6.46 > 0\), we have \(\frac{d^2A}{dx^2} = 4 + \frac{2160}{270} = 12 > 0\). Since the second derivative is positive, the value is a minimum.

(a) M1: Uses volume \(2x^2h = 360\) to express \(h\) in terms of \(x\). A1: Correct expression for \(h\). M1: Formulates surface area equation \(A = 2x^2 + 6xh\). A1: Achieves the given formula \(A = 2x^2 + \frac{1080}{x}\) with no errors. (b) M1: Differentiates \(A\) to obtain a form \(px + qx^{-2}\). A1: Correct derivative \(4x - \frac{1080}{x^2}\). M1: Sets derivative to 0 and solves for \(x^3\) or \(x\). A1.28: Finds minimum area to the nearest integer (251). (c) B1: Correctly evaluates \(\frac{d^2A}{dx^2}\) at the stationary point and concludes it is a minimum because \(\frac{d^2A}{dx^2} > 0\).

(a)
There are \(n = 15\) observations.
Median \(Q_2\) is the \(\frac{15+1}{2} = 8\)th value:
\(Q_2 = 140\) g.

Lower quartile \(Q_1\) is the \(\frac{15+1}{4} = 4\)th value:
\(Q_1 = 130\) g.

Upper quartile \(Q_3\) is the \(3 \times \frac{15+1}{4} = 12\)th value:
\(Q_3 = 149\) g.

(b)
Interquartile Range (\(\text{IQR}\)) = \(Q_3 - Q_1 = 149 - 130 = 19\).

Lower outlier boundary = \(Q_1 - 1.5 \times \text{IQR} = 130 - 1.5 \times 19 = 130 - 28.5 = 101.5\).
Upper outlier boundary = \(Q_3 + 1.5 \times \text{IQR} = 149 + 1.5 \times 19 = 149 + 28.5 = 177.5\).

Since \(182 > 177.5\), 182 is an outlier.
No data value is less than 101.5 (minimum value is 115), and no other data value is greater than 177.5 (the next highest is 165).
Therefore, 182 is the only outlier.

(c)
Using quartiles:
\(Q_3 - Q_2 = 149 - 140 = 9\)
\(Q_2 - Q_1 = 140 - 130 = 10\)
Since \(Q_3 - Q_2 < Q_2 - Q_1\) (i.e. \(9 < 10\)), the data is negatively skewed.

Alternatively, using the mean:
\(\sum x = 2124 \implies \text{Mean} = \frac{2124}{15} = 141.6\).
Since \(\text{Mean} > \text{Median}\) (\(141.6 > 140\)), this suggests positive skew.
Accept either justification if supported by correct calculations.

(a)
B1: For Median = 140
B1: For \(Q_1\) = 130
B1: For \(Q_3\) = 149

(b)
M1: For a correct attempt to find IQR and use the \(1.5 \times \text{IQR}\) rule
A1: For correct boundaries: 101.5 and 177.5
A1: For fully justifying why 182 is the only outlier (reference to other data points required)

(c)
M1: For a valid comparison test (e.g., comparing \(Q_3 - Q_2\) and \(Q_2 - Q_1\) or Mean and Median)
A1: Correct comparison of values
A1ft: Correct conclusion of skewness based on their comparison

(a)
\(S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 2410 - \frac{136^2}{8} = 2410 - 2312 = 98\)

\(S_{xy} = \sum xy - \frac{\sum x \sum y}{n} = 25160 - \frac{136 \times 1420}{8} = 25160 - 24140 = 1020\)

(b)
\(b = \frac{S_{xy}}{S_{xx}} = \frac{1020}{98} \approx 10.408... \approx 10.4\) (to 3 s.f.)

\(\bar{x} = \frac{136}{8} = 17\)
\(\bar{y} = \frac{1420}{8} = 177.5\)

\(a = \bar{y} - b\bar{x} = 177.5 - 10.408... \times 17 = 177.5 - 176.938... = 0.5612... \approx 0.561\) (to 3 s.f.)

So the equation of the regression line is:
\(y = 0.561 + 10.4x\)

(c)
For each 1 °C increase in the daily average temperature, the daily sales of ice cream increase by approximately £10.40 (or £10.41 using the unrounded value of \(b\)).

(d)
When \(x = 18\):
\(y = 0.5612... + 10.408... \times 18 = 187.91\)
So the estimated daily sales of ice cream are £188 (to 3 s.f.).

(a)
M1: For a correct formula applied to calculate \(S_{xx}\) or \(S_{xy}\)
A1: \(S_{xx} = 98\)
A1: \(S_{xy} = 1020\)

(b)
M1: For a correct method to find the gradient \(b\)
M1: For a correct method to find the intercept \(a\)
A1: For the equation \(y = 0.561 + 10.4x\) with coefficients to 3 s.f. (Accept awrt 0.561 and 10.4)

(c)
B1: For a correct contextual interpretation linking 1 °C increase to approximately £10.40 increase in sales.

(d)
M1: For substituting \(x = 18\) into their regression equation.
A1: awrt 188 (or 187.9)

(a)
Using the addition law of probability:
\(\text{P}(A \cup B) = \text{P}(A) + \text{P}(B) - \text{P}(A \cap B)\)
\(0.80 = 0.45 + 0.60 - \text{P}(A \cap B)\)
\(\text{P}(A \cap B) = 1.05 - 0.80 = 0.25\)

(b)
The Venn diagram should show two intersecting circles labelled \(A\) and \(B\) inside a box representing the sample space \(S\):
- Probability in the intersection \(A \cap B\) is 0.25.
- Probability in \(A\) only is \(0.45 - 0.25 = 0.20\).
- Probability in \(B\) only is \(0.60 - 0.25 = 0.35\).
- Probability outside both circles is \(1 - 0.80 = 0.20\).

(c)
\(\text{P}(A' \cap B) = \text{P}(B) - \text{P}(A \cap B) = 0.60 - 0.25 = 0.35\).

(d)
\(\text{P}(A' | B') = \frac{\text{P}(A' \cap B')}{\text{P}(B')} = \frac{1 - \text{P}(A \cup B)}{1 - \text{P}(B)}\)
\(\text{P}(A' | B') = \frac{0.20}{1 - 0.60} = \frac{0.20}{0.40} = 0.5\).

(a)
M1: Correct use of the addition formula to write an equation with \(\text{P}(A \cap B)\)
A1: 0.25

(b)
M1: Venn diagram template with two overlapping circles inside a rectangle.
A1: Correct circle probabilities: 0.20, 0.25, and 0.35.
A1ft: Correct probability of 0.20 in the region outside both circles (must sum to 1).

(c)
B1: 0.35 (or ft from diagram)

(d)
M1: Correct formula for conditional probability \(\frac{\text{P}(A' \cap B')}{\text{P}(B')}\)
A1: Correct fraction or calculation \(\frac{0.20}{0.40}\)
A1: 0.5

(a)
First draw branches:
- Red (\(R_1\)): \(\frac{5}{8}\)
- Blue (\(B_1\)): \(\frac{3}{8}\)

Second draw branches (without replacement):
- If \(R_1\) is chosen first:
- Red (\(R_2\)): \(\frac{4}{7}\)
- Blue (\(B_2\)): \(\frac{3}{7}\)
- If \(B_1\) is chosen first:
- Red (\(R_2\)): \(\frac{5}{7}\)
- Blue (\(B_2\)): \(\frac{2}{7}\)

(b)
\(\text{P}(R_2) = \text{P}(R_1 \cap R_2) + \text{P}(B_1 \cap R_2)\)
\(\text{P}(R_2) = \left(\frac{5}{8} \times \frac{4}{7}\right) + \left(\frac{3}{8} \times \frac{5}{7}\right)\)
\(\text{P}(R_2) = \frac{20}{56} + \frac{15}{56} = \frac{35}{56} = \frac{5}{8} = 0.625\)

(c)
Using Bayes' theorem / conditional probability:
\(\text{P}(R_1 | R_2) = \frac{\text{P}(R_1 \cap R_2)}{\text{P}(R_2)}\)
\(\text{P}(R_1 | R_2) = \frac{\frac{5}{8} \times \frac{4}{7}}{\frac{5}{8}} = \frac{4}{7} \approx 0.571\) (to 3 s.f.)

(a)
M1: Correct structure of tree diagram with 2 stages
A1: Correct first stage probabilities (\(5/8\) and \(3/8\))
A1: Correct second stage probabilities (\(4/7, 3/7, 5/7, 2/7\))

(b)
M1: Attempting to sum the two correct paths \(\text{P}(R_1 \cap R_2)\) and \(\text{P}(B_1 \cap R_2)\)
A1: Correct substitution of values
A1: \(\frac{5}{8}\) or 0.625

(c)
M1: Correct conditional probability statement \(\frac{\text{P}(R_1 \cap R_2)}{\text{P}(R_2)}\)
A1ft: Correct numerical values used from their (b)
A1: \(\frac{4}{7}\) or awrt 0.571

(a)
Since the sum of the probabilities in a probability distribution is 1:
\(\sum \text{P}(X=x) = 1 \implies k + 2k + 3k + 4k = 1\)
\(10k = 1\)
\(k = 0.1\) (as required).

(b)
Using the formula for expectation:
\(\text{E}(X) = \sum x \cdot \text{P}(X=x)\)
\(\text{E}(X) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4)\)
\(\text{E}(X) = 0.1 + 0.4 + 0.9 + 1.6 = 3.0\).

(c)
First, find \(\text{E}(X^2)\):
\(\text{E}(X^2) = \sum x^2 \cdot \text{P}(X=x)\)
\(\text{E}(X^2) = 1^2(0.1) + 2^2(0.2) + 3^2(0.3) + 4^2(0.4)\)
\(\text{E}(X^2) = 1(0.1) + 4(0.2) + 9(0.3) + 16(0.4)\)
\(\text{E}(X^2) = 0.1 + 0.8 + 2.7 + 6.4 = 10.0\).

Now, calculate variance:
\(\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2\)
\(\text{Var}(X) = 10.0 - (3.0)^2 = 10.0 - 9.0 = 1.0\).

(a)
M1: Setting the sum of the probabilities equal to 1 to form a linear equation in \(k\)
A1: Fully correct completion to show \(k = 0.1\)

(b)
M1: Attempt to use \(\sum x \text{P}(X=x)\) with at least 3 correct terms
A1: Correct values substituted
A1: 3.0 (or 3)

(c)
M1: Attempt to find \(\text{E}(X^2)\) with at least 3 terms of form \(x^2 \text{P}(X=x)\)
A1: Correct value of \(\text{E}(X^2) = 10.0\)
M1: Correct use of the formula \(\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2\)
A1: 1.0

(a)
Using the properties of expectation:
\(\text{E}(W) = \text{E}(4Y - 3) = 4\text{E}(Y) - 3\)
\(\text{E}(W) = 4(4.5) - 3 = 18 - 3 = 15\).

(b)
Using the properties of variance:
\(\text{Var}(W) = \text{Var}(4Y - 3) = 4^2 \text{Var}(Y)\)
\(\text{Var}(W) = 16 \times 1.25 = 20\).

The standard deviation is:
\(\text{SD}(W) = \sqrt{\text{Var}(W)} = \sqrt{20} \approx 4.47\) (to 3 s.f.).

(c)
Using the formula for variance:
\(\text{Var}(W) = \text{E}(W^2) - [\text{E}(W)]^2\)
\(20 = \text{E}(W^2) - 15^2\)
\(20 = \text{E}(W^2) - 225\)
\(\text{E}(W^2) = 225 + 20 = 245\).

(a)
M1: Correctly applying expectation properties \(4\text{E}(Y) - 3\)
A1: 15

(b)
M1: Correctly applying variance coding property \(16\text{Var}(Y)\)
A1: \(\text{Var}(W) = 20\)
M1: Attempting to square root their variance
A1: awrt 4.47

(c)
M1: Correct formula relating \(\text{Var}(W)\), \(\text{E}(W^2)\) and \(\text{E}(W)\)
A1ft: Correct substitution of their values
A1: 245

(a)
Standardising \(L < 80\):
\(\text{P}\left(Z < \frac{80 - \mu}{\sigma}\right) = 0.0668\)
Since the probability is less than 0.5, the z-value must be negative.
From the standard normal table, \(\Phi(1.5) = 1 - 0.0668 = 0.9332\).
Therefore:
\(\frac{80 - \mu}{\sigma} = -1.5 \implies 80 - \mu = -1.5\sigma \implies \mu - 1.5\sigma = 80\) (Equation 1)

Standardising \(L > 120\):
\(\text{P}\left(Z > \frac{120 - \mu}{\sigma}\right) = 0.1587 \implies \text{P}\left(Z < \frac{120 - \mu}{\sigma}\right) = 0.8413\)
From the standard normal table, \(\Phi(1.0) = 0.8413\).
Therefore:
\(\frac{120 - \mu}{\sigma} = 1.0 \implies 120 - \mu = 1.0\sigma \implies \mu + \sigma = 120\) (Equation 2)

(b)
Subtract Equation 1 from Equation 2:
\((\mu + \sigma) - (\mu - 1.5\sigma) = 120 - 80\)
\(2.5\sigma = 40\)
\(\sigma = 16\)

Substitute \(\sigma = 16\) into Equation 2:
\(\mu + 16 = 120 \implies \mu = 104\).

(a)
M1: Attempt to standardise either probability expression
B1: For finding \(z = -1.5\) (or awrt -1.5) associated with 0.0668
B1: For finding \(z = 1.0\) (or awrt 1.0) associated with 0.1587
A1: Correctly formed equation: \(\mu - 1.5\sigma = 80\)
A1: Correctly formed equation: \(\mu + \sigma = 120\)

(b)
M1: A valid method to solve the two simultaneous equations to find \(\mu\) or \(\sigma\)
A1: \(\sigma = 16\)
A1: \(\mu = 104\)

(a)
We want to find \(\text{P}(V < 500)\) where \(V \sim \text{N}(505, 4^2)\).
Standardising:
\(\text{P}\left(Z < \frac{500 - 505}{4}\right) = \text{P}(Z < -1.25)\)
\(\text{P}(Z < -1.25) = 1 - \Phi(1.25)\)
Using standard normal tables, \(\Phi(1.25) = 0.8944\).
Therefore, \(\text{P}(V < 500) = 1 - 0.8944 = 0.1056\).

(b)
We are given \(\text{P}(V > v) = 0.05\).
Standardising:
\(\text{P}\left(Z > \frac{v - 505}{4}\right) = 0.05\)
From the standard normal percentage points table, the z-value corresponding to a right-tail probability of 0.05 is 1.6449.
Therefore:
\(\frac{v - 505}{4} = 1.6449\)
\(v = 505 + 4 \times 1.6449\)
\(v = 505 + 6.5796 = 511.58\) ml (approx. 511.6 ml or 512 ml to 3 s.f.)

(c)
The probability that a single bottle contains more than 500 ml is:
\(1 - \text{P}(V < 500) = 1 - 0.1056 = 0.8944\).

Since the three bottles are independent, the probability that all 3 contain more than 500 ml is:
\(0.8944^3 \approx 0.7154... \approx 0.715\) (to 3 s.f.).

(a)
M1: Attempting to standardise with 500, 505, and 4.
A1: Correct standardisation to \(Z = -1.25\).
A1: 0.1056 (or awrt 0.106)

(b)
M1: Setting up equation of standardisation equated to a z-value.
B1: Correct z-value of 1.6449 (or 1.645) used.
A1: awrt 511.6 or 512

(c)
M1: Identifying that the probability for more than 500 ml is \(1 - \text{their (a)}\).
M1: Attempting to cube \((1 - \text{their (a)})\).
A1: awrt 0.715

(a) The velocity-time graph is a trapezium starting at \((0,0)\), accelerating to a maximum speed of \(v = 2.5T\) at time \(t = T\), remaining constant at this speed until \(t = 5T\), and then decelerating to \(0\) at \(t = 5T + 6\).

(b) The distance travelled is the area under the velocity-time graph:
\text{Area} = \frac{1}{2}(T)(2.5T) + (4T)(2.5T) + \frac{1}{2}(6)(2.5T) = 450
\(1.25T^2 + 10T^2 + 7.5T = 450\)
\(11.25T^2 + 7.5T - 450 = 0\)
Multiplying by 4 to clear decimals:
\(45T^2 + 30T - 1800 = 0\)
Dividing by 15:
\(3T^2 + 2T - 120 = 0\)
Factoring the quadratic:
\((3T + 20)(T - 6) = 0\)
Since \(T > 0\), we have \(T = 6\).

(c) The maximum speed reached is \(v = 2.5 \times 6 = 15 \text{ m s}^{-1}\).
During the final \(6 \text{ s}\), the motorcyclist decelerates from \(15 \text{ m s}^{-1}\) to rest.
\text{Deceleration} = \frac{15}{6} = 2.5 \text{ m s}^{-2}\.

(a)
- M1: Standard trapezium shape starting at origin and ending on the time axis.
- A1: Correct labels showing time intervals \(T\), \(5T\), and \(5T+6\) (or equivalent intervals) and peak velocity \(2.5T\).

(b)
- M1: Attempting to find the total distance using areas of three sections or using the trapezium formula.
- A1: Correct expression for total area: \(1.25T^2 + 10T^2 + 7.5T\) (or equivalent).
- A1: Formulates the quadratic equation: \(3T^2 + 2T - 120 = 0\) (or equivalent).
- M1: Solves the quadratic equation for a positive value of \(T\).
- A1: Obtain \(T = 6\).

(c)
- M1: Calculates the maximum velocity \(v = 2.5 \times 6 = 15\).
- A1.71: Correctly calculates deceleration as \(2.5 \text{ m s}^{-2}\).

Given \(\tan\alpha = \frac{3}{4}\), we have \(\sin\alpha = 0.6\) and \(\cos\alpha = 0.8\).

(a) For particle \(P\) on the inclined plane:
Normal reaction: \(R = mg\cos\alpha = 0.8mg\).
Since \(Q\) is much heavier, it will accelerate downwards, and \(P\) will accelerate up the plane.
Therefore, the friction force \(F\) acts down the plane:
\(F = \mu R = 0.25 \times 0.8mg = 0.2mg\).

Equation of motion for \(P\) moving up the plane:
\(T - mg\sin\alpha - F = ma\)
\(T - 0.6mg - 0.2mg = ma\)
\(T - 0.8mg = ma\) --- (Equation 1)

Equation of motion for \(Q\) moving downwards:
\(3mg - T = 3ma\) --- (Equation 2)

Adding Equation 1 and Equation 2:
\(3mg - 0.8mg = 4ma\)
\(2.2mg = 4ma\)
\(a = 0.55g\).

(b) Substituting \(a = 0.55g\) into Equation 2:
\(T = 3mg - 3m(0.55g)\)
\(T = 3mg - 1.65mg = 1.35mg\).

(a)
- B1: Correctly deduces \(\sin\alpha = 0.6\) and \(\cos\alpha = 0.8\).
- M1: Resolves forces perpendicular to the plane to find \(R = mg\cos\alpha\).
- A1: Correct expression for friction: \(F = 0.2mg\).
- M1: Formulates the equation of motion for \(P\) with three terms: \(T\), gravity component, and friction.
- A1: Correct equation for \(P\): \(T - 0.8mg = ma\).
- M1: Formulates the equation of motion for \(Q\): \(3mg - T = 3ma\).
- A1: Obtains \(a = 0.55g\) (or \(\frac{11}{20}g\)).

(b)
- M1: Sustitutes their value of \(a\) back into one of the motion equations to solve for \(T\).
- A2.71: Correctly obtains \(T = 1.35mg\) (or \(\frac{27}{20}mg\)).

(a) Let the initial direction of \(A\)'s motion be positive.
Initial velocities:
\(u_A = 4 \text{ m s}^{-1}\)
\(u_B = -3 \text{ m s}^{-1}\)

After collision, directions are reversed:
\(v_A = -v\) (where \(v\) is the speed of \(A\))
\(v_B = 1 \text{ m s}^{-1}\)

By conservation of linear momentum:
\(m_A u_A + m_B u_B = m_A v_A + m_B v_B\)
\(0.4(4) + 0.6(-3) = 0.4(-v) + 0.6(1)\)
\(1.6 - 1.8 = -0.4v + 0.6\)
\(-0.2 = -0.4v + 0.6\)
\(0.4v = 0.8\)
\(v = 2 \text{ m s}^{-1}\).
So the speed of \(A\) immediately after the collision is \(2 \text{ m s}^{-1}\).

(b) Impulse on \(A\):
\(I_A = m_A(v_A - u_A) = 0.4(-2 - 4) = -2.4 \text{ N s}\).
Magnitude of impulse is \(2.4 \text{ N s}\).

(c) Before colliding with the wall, \(B\) is moving with velocity \(1 \text{ m s}^{-1}\).
After rebounding, its velocity is \(-0.8 \text{ m s}^{-1}\).
Impulse exerted by the wall on \(B\):
\(I_{\text{wall}} = m_B(v_{\text{new}} - v_{\text{old}}) = 0.6(-0.8 - 1) = -1.08 \text{ N s}\).
Magnitude of the impulse is \(1.08 \text{ N s}\).

(a)
- M1: Applies conservation of linear momentum with correct signs for opposite directions.
- A1: Correctly substitutes values into the momentum equation: \(0.4(4) - 0.6(3) = 0.6(1) - 0.4v\).
- M1: Solves the equation for \(v\).
- A1: Obtains \(2 \text{ m s}^{-1}\).

(b)
- M1: Applies impulse formula \(I = m(v - u)\) for particle \(A\).
- A1: Correct substitution of values: \(0.4 \times (2 - (-4))\) or equivalent.
- A1: State magnitude is \(2.4 \text{ N s}\).

(c)
- M1: Identifies the velocities of \(B\) before and after impact with the wall with opposite signs (e.g., \(+1\) and \(-0.8\)).
- M1: Applies impulse formula \(I = 0.6(-0.8 - 1)\).
- A1.71: Obtains \(1.08 \text{ N s}\).

We resolve forces perpendicular to the plane:
\(R = mg\cos 25^\circ = 5 \times 9.8 \times \cos 25^\circ = 49\cos 25^\circ \approx 44.41 \text{ N}\).
The maximum frictional force is:
\(F_{\max} = \mu R = 0.3 \times 44.41 = 13.32 \text{ N}\).
The component of weight acting down the plane is:
\(W_{\parallel} = mg\sin 25^\circ = 5 \times 9.8 \times \sin 25^\circ = 49\sin 25^\circ \approx 20.71 \text{ N}\).

(a) For the minimum value of \(P\), the particle is on the point of slipping down the plane, so friction \(F_{\max}\) acts up the plane:
\(P_{\min} + F_{\max} = mg\sin 25^\circ\)
\(P_{\min} = 20.71 - 13.32 = 7.39 \text{ N}\) (to 3 sig figs).

(b) For the maximum value of \(P\), the particle is on the point of slipping up the plane, so friction \(F_{\max}\) acts down the plane:
\(P_{\max} = mg\sin 25^\circ + F_{\max}\)
\(P_{\max} = 20.71 + 13.32 = 34.0 \text{ N}\) (to 3 sig figs).

(a)
- M1: Resolves forces perpendicular to the plane to find \(R = 5g\cos 25^\circ\).
- A1: Correct calculation of \(R \approx 44.4 \text{ N}\).
- M1: Uses \(F = \mu R\) to find friction \(F \approx 13.3 \text{ N}\).
- M1: Formulates the equilibrium equation along the slope for the minimum force: \(P + F = mg\sin 25^\circ\).
- A1: Correct minimum value of \(P = 7.39 \text{ N}\) (accept \(7.4 \text{ N}\)).

(b)
- M1: Identifies that for maximum force, friction acts down the plane.
- M1: Formulates the equilibrium equation along the slope: \(P = mg\sin 25^\circ + F\).
- A1: Correct terms substitution: \(P = 20.71 + 13.32\).
- A1.71: Correct maximum value of \(P = 34.0 \text{ N}\) (accept \(34 \text{ N}\)).

(a) When the plank is on the point of tilting about \(C\), the reaction force at support \(D\) is zero: \(R_D = 0\).
We take moments about the support \(C\):
\(15g \times (x - 1) = 20g \times 1\)
\(15(x - 1) = 20\)
\(x - 1 = \frac{20}{15} = \frac{4}{3}\)
\(x = \frac{7}{3} \approx 2.33 \text{ m}\).

(b) Assuming the block is a particle means its weight acts at a single point (specifically at the end \(A\)), rather than being distributed over a width.

(c) The block is placed at \(B\) and the plank is on the point of tilting about \(D\), so the reaction force at \(C\) is zero: \(R_C = 0\).
Distance of \(D\) from \(A\) is \(6 - 1.5 = 4.5 \text{ m\).
Distance of the center of mass of the plank from \(D\) is \(4.5 - x = 4.5 - \frac{7}{3} = \frac{13}{6} \text{ m}\).
We take moments about the support \(D\):
\(15g \times (4.5 - x) = Mg \times 1.5\)
\(15 \times \frac{13}{6} = 1.5M\)
\(\frac{65}{2} = 1.5M\)
\(32.5 = 1.5M \implies M = \frac{65}{3} \approx 21.7 \text{ kg}\).

(a)
- B1: States or uses \(R_D = 0\).
- M1: Formulates a moment equation about \(C\).
- A2: Correct equation: \(15g(x - 1) = 20g(1)\) (A1 for one side correct, A2 for both sides correct).
- M1: Solves the equation for \(x\).
- A1: Obtains \(x = \frac{7}{3}\) or \(2.33\).

(b)
- B1: Explains that the weight of the block acts at a single point (at the end \(A\)).

(c)
- B1: States or uses \(R_C = 0\).
- M1: Formulates a moment equation about \(D\) using their value of \(x\).
- A1: Correct equation: \(15(4.5 - \frac{7}{3}) = 1.5M\) (or equivalent).
- A0.71: Obtains \(M = 21.7\) or \(\frac{65}{3}\).

(a) Using \(\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t\) for constant velocity:
\(\mathbf{r}_P = (-10 + 4t)\mathbf{i} + (12 + t)\mathbf{j}\).

(b) Using \(\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\) for constant acceleration:
\(\mathbf{r}_Q = \mathbf{0} + (u\mathbf{i} - 3\mathbf{j})t + \frac{1}{2}(0.5\mathbf{i} + 2\mathbf{j})t^2\)
\(\mathbf{r}_Q = (ut + 0.25t^2)\mathbf{i} + (-3t + t^2)\mathbf{j}\).

(c) At the time of collision \(t = T\), we have \(\mathbf{r}_P = \mathbf{r}_Q\).
Equating the \(\mathbf{j}\) components:
\(12 + T = -3T + T^2\)
\(T^2 - 4T - 12 = 0\)
\((T - 6)(T + 2) = 0\)
Since \(T > 0\), the collision occurs at \(T = 6\) seconds.

Equating the \(\mathbf{i}\) components at \(T = 6\):
\(-10 + 4(6) = u(6) + 0.25(6)^2\)
\(14 = 6u + 9\)
\(6u = 5\)
\(u = \frac{5}{6} \approx 0.833\).

(a)
- M1: Uses constant velocity position vector formula.
- A1: Correctly writes \(\mathbf{r}_P = (-10 + 4t)\mathbf{i} + (12 + t)\mathbf{j}\).

(b)
- M1: Uses constant acceleration position vector formula.
- A1: Correctly writes \(\mathbf{r}_Q = (ut + 0.25t^2)\mathbf{i} + (-3t + t^2)\mathbf{j}\).

(c)
- M1: Equates the \(\mathbf{j}\) components of \(\mathbf{r}_P\) and \(\mathbf{r}_Q\).
- A1: Obtains the correct quadratic: \(T^2 - 4T - 12 = 0\).
- M1: Solves the quadratic to find a positive value of \(T\).
- A1: Obtains \(T = 6\).
- M1: Equates the \(\mathbf{i}\) components with their value of \(T\).
- A1.71: Obtains \(u = \frac{5}{6}\) (or \(0.833\)).

(a) Let the acceleration of the lift be \(a \text{ m s}^{-2}\).
We consider the system of the lift and the crate together.
Total mass \(M = 800 + 120 = 920 \text{ kg}\).
Using \(F = Ma\) vertically upwards:
\(T - Mg = Ma\)
\(11000 - 920 \times 9.8 = 920a\)
\(11000 - 9016 = 920a\)
\(1984 = 920a\)
\(a = \frac{1984}{920} \approx 2.16 \text{ m s}^{-2}\) (to 3 sig figs).

(b) Consider the crate alone of mass \(m = 120 \text{ kg}\).
Let \(R\) be the reaction force exerted by the floor of the lift on the crate.
Using \(F = ma\) vertically upwards:
\(R - mg = ma\)
\(R = m(g + a) = 120(9.8 + 2.1565)\)
\(R = 120 \times 11.9565 \approx 1430 \text{ N}\) (to 3 sig figs).

(c) The deceleration is \(1.5 \text{ m s}^{-2}\) upwards, so the acceleration is \(-1.5 \text{ m s}^{-2}\).
Using \(F = Ma\) on the combined system:
\(T - Mg = M(-1.5)\)
\(T = 920(9.8 - 1.5)\)
\(T = 920 \times 8.3 = 7636 \approx 7640 \text{ N}\) (to 3 sig figs).

(a)
- B1: Identifies total mass is \(920 \text{ kg}\).
- M1: Formulates equation of motion for combined system: \(T - Mg = Ma\).
- A1: Substitutes correct values: \(11000 - 920 \times 9.8 = 920a\).
- A1: Correctly calculates \(a \approx 2.16 \text{ m s}^{-2}\) (or \(2.2 \text{ m s}^{-2}\)).

(b)
- M1: Formulates equation of motion for the crate: \(R - mg = ma\).
- A1: Correctly substitutes values into equation of motion.
- M1: Solves for \(R\) using their value of \(a\).
- A1: Correctly calculates \(R \approx 1430 \text{ N}\) (or \(1400 \text{ N}\)).

(c)
- M1: Recognizes acceleration is now \(-1.5 \text{ m s}^{-2}\) and sets up the combined system equation.
- A1.71: Correctly calculates tension \(T = 7640 \text{ N}\) (accept \(7600 \text{ N}\)).