2025 IB DP Mathematics - Analysis and Approaches Practice Paper with Answers

The first three terms of the arithmetic sequence are:
\(u_1 = \ln(a)\)
\(u_2 = \ln(a) + \ln(2) = \ln(2a)\)
\(u_3 = \ln(a) + 2\ln(2) = \ln(4a)\)

The sum of the first three terms is given by:
\(S_3 = u_1 + u_2 + u_3\)
\(S_3 = \ln(a) + \ln(2a) + \ln(4a)\)

Using the properties of logarithms:
\(S_3 = \ln(a \times 2a \times 4a) = \ln(8a^3)\)

Alternatively, using the sum formula:
\(S_3 = \frac{3}{2}(2u_1 + 2d) = 3(u_1 + d) = 3(\ln(a) + \ln(2)) = 3\ln(2a) = \ln((2a)^3) = \ln(8a^3)\)

We are given that \(S_3 = \ln(1728)\):
\(\ln(8a^3) = \ln(1728)\)
\(8a^3 = 1728\)
\(a^3 = 216\)
\(a = 6\)

M1: For expressing the sum of the first three terms in terms of \(a\), e.g., \(\ln(a) + \ln(2a) + \ln(4a)\) or \(3\ln(a) + 3\ln(2)\).
A1: For simplifying the sum expression to \(\ln(8a^3)\) or \(\ln((2a)^3)\).
M1: For equating their simplified expression to \(\ln(1728)\).
A1: For obtaining \(8a^3 = 1728\) (or \(2a = 12\)).
A1.83: For correctly solving to find \(a = 6\).

(a) The horizontal asymptote of a rational function of the form \(f(x) = \frac{ax + 5}{2x - b}\) as \(x \to \pm\infty\) is given by the ratio of the leading coefficients:
\(y = \frac{a}{2}\)

Given that the horizontal asymptote is \(y = 3\):
\(\frac{a}{2} = 3 \implies a = 6\)

(b) Substituting \(a = 6\) into the function gives:
\(f(x) = \frac{6x + 5}{2x - b}\)

Since the graph passes through \((2, -1)\), we substitute \(x = 2\) and \(f(x) = -1\):
\(-1 = \frac{6(2) + 5}{2(2) - b}
\)-1 = \frac{17}{4 - b}

Multiplying both sides by \(4 - b\):
\(-(4 - b) = 17\)
\(b - 4 = 17 \implies b = 21\)

Part (a):
M1: For identifying that the horizontal asymptote is given by \(y = \frac{a}{2}\).
A1: For obtaining \(a = 6\).

Part (b):
M1: For substituting \(a = 6\) and the point \((2, -1)\) into the function equation.
A1: For setting up the equation \(-1 = \frac{17}{4-b}\) (or equivalent).
M1: For attempting to solve for \(b\).
A0.83: For finding \(b = 21\).

Using the Pythagorean identity \(\cos^2(x) = 1 - \sin^2(x)\), substitute this into the equation:
\(2(1 - \sin^2(x)) - \sin(x) - 1 = 0\)
\(2 - 2\sin^2(x) - \sin(x) - 1 = 0\)
\(-2\sin^2(x) - \sin(x) + 1 = 0\)

Multiply the entire equation by \(-1\):
\(2\sin^2(x) + \sin(x) - 1 = 0\)

Let \(y = \sin(x)\). The quadratic equation becomes:
\(2y^2 + y - 1 = 0\)

Factorizing the quadratic:
\((2y - 1)(y + 1) = 0\)

Thus, \(y = \frac{1}{2}\) or \(y = -1\).

This gives two trigonometric equations to solve in the interval \(0 \leq x \leq 2\pi\):

1) \(\sin(x) = \frac{1}{2}\)
This has solutions in the first and second quadrants:
\(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\)

2) \(\sin(x) = -1\)
This has a solution at the boundary:
\(x = \frac{3\pi}{2}\)

Combining these, the complete set of solutions is:
\(x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}\)

M1: For using the identity \(\cos^2(x) = 1 - \sin^2(x)\) to express the equation in terms of \(\sin(x)\).
A1: For obtaining the correct quadratic equation \(2\sin^2(x) + \sin(x) - 1 = 0\) (or equivalent).
M1: For factorizing or solving the quadratic to get \(\sin(x) = \frac{1}{2}\) and \(\sin(x) = -1\).
A1: For finding \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\) from \(\sin(x) = \frac{1}{2}\).
A1: For finding \(x = \frac{3\pi}{2}\) from \(\sin(x) = -1\).
A0.83: For writing all three solutions clearly with no extra invalid solutions in the given domain.

Let \(u = x^2 + 1\). Then, differentiating both sides with respect to \(x\) gives:
\(\frac{du}{dx} = 2x \implies x \, dx = \frac{1}{2} du\)

Next, find the limits of integration in terms of \(u\):
When \(x = 0\), \(u = 0^2 + 1 = 1\).
When \(x = \sqrt{3}\), \(u = (\sqrt{3})^2 + 1 = 4\).

Substitute these into the integral:
\(\int_{0}^{\sqrt{3}} \frac{x}{\sqrt{x^2 + 1}} \, dx = \int_{1}^{4} \frac{\frac{1}{2}}{\sqrt{u}} \, du\)
\(= \frac{1}{2} \int_{1}^{4} u^{-1/2} \, du\)

Integrate using the power rule:
\(= \frac{1}{2} \left[ 2u^{1/2} \right]_{1}^{4}\)
\(= \left[ \sqrt{u} \right]_{1}^{4}\)
\(= \sqrt{4} - \sqrt{1}\)
\(= 2 - 1 = 1\)

M1: For attempting to use substitution, e.g., setting \(u = x^2 + 1\).
A1: For obtaining \(x \, dx = \frac{1}{2} du\) (or equivalent).
A1: For finding the correct limits of integration in terms of \(u\) as \(1\) and \(4\) (or for correctly substituting back to \(x\) later).
M1: For correct integration resulting in \(\sqrt{u}\) (or \(\sqrt{x^2+1}\)).
A1.83: For substituting the limits and showing that the final answer is \(1\).

Let \(P\) be the set of students studying Physics, and \(C\) be the set of students studying Chemistry.
The total number of students in the class is \(U = 30\).

The number of students who study neither subject is \(N((P \cup C)') = 5\), which means the number of students studying at least one of the subjects is:
\(N(P \cup C) = 30 - 5 = 25\)

Using the set union formula:
\(N(P \cup C) = N(P) + N(C) - N(P \cap C)\)
\(25 = 18 + 15 - N(P \cap C)\)
\(25 = 33 - N(P \cap C) \implies N(P \cap C) = 8\)

(a) The probability that a randomly selected student studies both subjects is:
\(P(P \cap C) = \frac{N(P \cap C)}{30} = \frac{8}{30} = \frac{4}{15}\)

(b) We want to find the conditional probability \(P(C' | P)\), where \(C'\) is the set of students who do not study Chemistry.
Using the conditional probability formula:
\(P(C' | P) = \frac{N(P \cap C')}{N(P)}

The number of students who study Physics but do not study Chemistry is:
\)N(P \cap C') = N(P) - N(P \cap C) = 18 - 8 = 10\)

Therefore, the probability is:
\(P(C' | P) = \frac{10}{18} = \frac{5}{9}\)

Part (a):
M1: For setting up an equation to find the number of students in the intersection, e.g., \(18 + 15 - x = 25\).
A1: For finding the number of students who study both is \(8\).
A1: For expressing the probability as \(\frac{8}{30}\) (or \(\frac{4}{15}\)).

Part (b):
M1: For identifying that we need the ratio of students studying Physics-only to the total Physics students, i.e., \(\frac{N(P \cap C')}{N(P)}\).
A1: For finding the number of students studying Physics-only is \(10\).
A0.83: For obtaining the probability \(\frac{10}{18}\) (or \(\frac{5}{9}\)).

First, find the \(y\)-coordinate of the point of tangency by substituting \(x = 0\) into the curve equation:
\(y(0) = e^{2(0)} \cos(0) = 1 \times 1 = 1\)
So, the point of tangency is \((0, 1)\).

Next, find the derivative of \(y = e^{2x} \cos(x)\) using the product rule:
Let \(u = e^{2x} \implies u' = 2e^{2x}\)
Let \(v = \cos(x) \implies v' = -\sin(x)\)

\(\frac{dy}{dx} = u'v + uv' = 2e^{2x}\cos(x) - e^{2x}\sin(x)\)

To find the gradient of the tangent at \(x = 0\), evaluate the derivative \(\frac{dy}{dx}\) at \(x = 0\):
\(m = 2e^{0}\cos(0) - e^{0}\sin(0) = 2(1)(1) - (1)(0) = 2\)

Using the equation of a straight line with gradient \(m = 2\) passing through the point \((0, 1)\):
\(y - y_1 = m(x - x_1)\)
\(y - 1 = 2(x - 0)\)
\(y = 2x + 1\)

M1: For substituting \(x = 0\) to find the correct \(y\)-coordinate \(1\).
M1: For attempting to use the product rule to differentiate \(e^{2x} \cos(x)\).
A1: For obtaining the correct derivative \(\frac{dy}{dx} = 2e^{2x}\cos(x) - e^{2x}\sin(x)\).
A1: For evaluating the derivative at \(x = 0\) to get the gradient \(m = 2\).
M1: For substituting their gradient and point \((0, 1)\) into a straight line equation.
A0.83: For stating the final equation as \(y = 2x + 1\).

(a) First, find the first derivative using the product rule: \( f'(x) = 2x\ln(x) + x^2\left(\frac{1}{x}\right) = x(2\ln(x) + 1) \). Set \( f'(x) = 0 \). Since \( x > 0 \), we have \( 2\ln(x) + 1 = 0 \), which gives \( \ln(x) = -\frac{1}{2} \) or \( x = e^{-1/2} \). Substitute \( x = e^{-1/2} \) back into \( f(x) \): \( f(e^{-1/2}) = (e^{-1/2})^2\ln(e^{-1/2}) = e^{-1}\left(-\frac{1}{2}\right) = -\frac{1}{2e} \). Thus, the coordinates of the local minimum are \( \left(e^{-1/2}, -\frac{1}{2e}\right) \). (b) At \( x = e \), the y-coordinate is \( f(e) = e^2\ln(e) = e^2 \). The gradient of the tangent is \( f'(e) = e(2\ln(e) + 1) = 3e \). Using the point-slope formula: \( y - e^2 = 3e(x - e) \), which simplifies to \( y = 3ex - 2e^2 \). (c) The area is given by the integral \( I = \int_1^e x^2 \ln(x) \, dx \). Use integration by parts with \( u = \ln(x) \) and \( dv = x^2 \, dx \), so \( du = \frac{1}{x} \, dx \) and \( v = \frac{1}{3}x^3 \). Then, \( \int x^2 \ln(x) \, dx = \frac{1}{3}x^3 \ln(x) - \int \frac{1}{3}x^2 \, dx = \frac{1}{3}x^3 \ln(x) - \frac{1}{9}x^3 \). Evaluating this from 1 to \( e \): \( \left[ \frac{1}{3}x^3 \ln(x) - \frac{1}{9}x^3 \right]_1^e = \left( \frac{1}{3}e^3 - \frac{1}{9}e^3 \right) - \left( 0 - \frac{1}{9} \right) = \frac{2}{9}e^3 + \frac{1}{9} = \frac{2e^3 + 1}{9} \). (d) Find the second derivative: \( f''(x) = \frac{d}{dx}(2x\ln(x) + x) = 2\ln(x) + 2 + 1 = 2\ln(x) + 3 \). Set \( f''(x) = 0 \) to get \( 2\ln(x) + 3 = 0 \), which yields \( x = e^{-3/2} \). Since \( f''(x) < 0 \) for \( x < e^{-3/2} \) and \( f''(x) > 0 \) for \( x > e^{-3/2} \), the second derivative changes sign at this point, showing it is a point of inflexion.

(a) M1 for an attempt to use the product rule. A1 for correct first derivative \( f'(x) = x(2\ln(x) + 1) \). M1 for setting \( f'(x) = 0 \) and solving for \( x \). A1 for correct coordinates. (b) A1 for correct y-coordinate and gradient. M1 for attempting to use the linear equation formula. A1 for the correct equation. (c) M1 for setting up the correct definite integral. M1 for identifying parts correctly. A1 for the correct antiderivative. M1 for substituting the limits. A1 for the correct final area. (d) A1 for the correct second derivative. A1 for setting it to 0 and finding \( x = e^{-3/2} \). R1 for showing or stating that a change in sign of the second derivative occurs.

(a) Since \( f(x) \) is a probability density function, the total area under the curve is 1: \( \int_0^4 kx(4-x)\,dx = 1 \). Evaluating the integral: \( k \int_0^4 (4x - x^2)\,dx = k \left[ 2x^2 - \frac{1}{3}x^3 \right]_0^4 = k \left( 32 - \frac{64}{3} \right) = k \left( \frac{32}{3} \right) = 1 \). This yields \( k = \frac{3}{32} \). (b) Since the PDF is a symmetric parabola on the interval \( [0, 4] \), the expected value is at the line of symmetry: \( \text{E}(X) = 2 \). Alternatively, \( \text{E}(X) = \int_0^4 x f(x)\,dx = \frac{3}{32}\int_0^4 (4x^2 - x^3)\,dx = \frac{3}{32}\left[\frac{4}{3}x^3 - \frac{1}{4}x^4\right]_0^4 = \frac{3}{32}\left(\frac{256}{3} - 64\right) = \frac{3}{32}\left(\frac{64}{3}\right) = 2 \). (c) The probability is \( \text{P}(X > 3) = \int_3^4 \frac{3}{32}(4x - x^2)\,dx = \frac{3}{32}\left[ 2x^2 - \frac{1}{3}x^3 \right]_3^4 = \frac{3}{32}\left( \left(32 - \frac{64}{3}\right) - \left(18 - 9\right) \right) = \frac{3}{32}\left( \frac{32}{3} - 9 \right) = \frac{3}{32}\left( \frac{5}{3} \right) = \frac{5}{32} \). (d) This follows a binomial scenario with \( n = 2 \) trials and success probability \( p = \frac{5}{32} \). The probability of exactly one success is \( \text{P}(Y=1) = 2 \times p \times (1 - p) = 2 \left(\frac{5}{32}\right)\left(\frac{27}{32}\right) = \frac{270}{1024} = \frac{135}{512} \). (e) For \( 0 \le x \le 4 \), the CDF is \( F(x) = \int_0^x f(t)\,dt = \frac{3}{32}\int_0^x (4t - t^2)\,dt = \frac{3}{32}\left[ 2t^2 - \frac{1}{3}t^3 \right]_0^x = \frac{3}{32}\left( 2x^2 - \frac{1}{3}x^3 \right) = \frac{6x^2 - x^3}{32} \).

(a) M1 for equating the total integral of the PDF to 1. M1 for attempting to integrate. A1 for obtaining \( k = \frac{3}{32} \). (b) A1 for recognizing symmetry or writing the correct expectation integral. A1 for the correct answer 2. (c) M1 for setting up the correct definite integral. A1 for correct antiderivative. M1 for substituting the limits 3 and 4. A1 for the correct value \( \frac{5}{32} \). (d) M1 for recognizing binomial probability structure. M1 for substituting their value of \( p \). A1 for correct calculation of \( \frac{135}{512} \). (e) M1 for using the definition \( F(x) = \int_0^x f(t)\,dt \). A1 for correct integration. A1 for the correct final expression.

(a) Equating the corresponding components of the two lines gives the system: (1) \( 1 + 2\lambda = 2 + \mu \implies 2\lambda - \mu = 1 \), (2) \( \lambda = -1 - ̅\mu \implies \lambda + \mu = -1 \), (3) \( -1 + 2\lambda = 2 + 3\mu \implies 2\lambda - 3\mu = 3 \). From (2), we get \( \mu = -\lambda - 1 \). Substitute this into (1): \( 2\lambda - (-\lambda - 1) = 1 \implies 3\lambda + 1 = 1 \implies \lambda = 0 \). Substituting \( \lambda = 0 \) back into (2) yields \( \mu = -1 \). We must verify these parameters in (3): \( \text{LHS} = 2(0) - 3(-1) = 3 \), which equals the RHS. Therefore, the lines intersect. Substituting \( \lambda = 0 \) into \( L_1 \) yields the position vector \( \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} \), corresponding to the point \( P(1, 0, -1) \). (b) The direction vectors are \( \mathbf{v}_1 = \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} \) and \( \mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \\ 3 \end{pmatrix} \). Compute the dot product: \( \mathbf{v}_1 \cdot \mathbf{v}_2 = (2)(1) + (1)(-1) + (2)(3) = 2 - 1 + 6 = 7 \). Find the magnitudes: \( |\mathbf{v}_1| = \sqrt{2^2 + 1^2 + 2^2} = 3 \) and \( |\mathbf{v}_2| = \sqrt{1^2 + (-1)^2 + 3^2} = \sqrt{11} \). Using the angle formula: \( \cos \theta = \frac{|\mathbf{v}_1 \cdot \mathbf{v}_2|}{|\mathbf{v}_1||\mathbf{v}_2|} = \frac{7}{3\sqrt{11}} \). (c) The normal vector \( \mathbf{n} \) to the plane is perpendicular to both direction vectors: \( \mathbf{n} = \mathbf{v}_1 \times \mathbf{v}_2 = \begin{pmatrix} (1)(3) - (2)(-1) \\ (2)(1) - (2)(3) \\ (2)(-1) - (1)(1) \end{pmatrix} = \begin{pmatrix} 5 \\ -4 \\ -3 \end{pmatrix} \). The equation of the plane is \( 5x - 4y - 3z = d \). Since the plane contains \( P(1, 0, -1) \), we substitute its coordinates: \( 5(1) - 4(0) - 3(-1) = 8 \), so \( d = 8 \). The Cartesian equation is \( 5x - 4y - 3z = 8 \).

(a) M1 for equating component expressions. M1 for solving equations to find the parameters. A1 for obtaining \( \lambda = 0 \) and \( \mu = -1 \). M1 for checking consistency with the third equation. A1 for concluding with the correct intersection point. (b) M1 for identifying both direction vectors. M1 for computing the scalar product. M1 for computing the magnitudes. A1 for the correct cosine ratio. (c) M1 for recognizing that the normal vector is the cross product. M1 for attempting to calculate the cross product. A1 for the correct normal vector \( \begin{pmatrix} 5 \\ -4 \\ -3 \end{pmatrix} \). M1 for setting up the plane equation \( ax + by + cz = d \). M1 for substituting a point on the plane to find \( d \). A1 for the correct Cartesian equation.

(a) To find the points of intersection, we set \( f(x) = g(x) \):
\( 3\ln(x) = x^2 - 4x + 2 \).

Using a graphic display calculator (GDC) to find the intersection points of the two curves:
\( x_1 \approx 0.8179... \)
\( x_2 \approx 0.558... \)

To 3 significant figures, the \( x \)-coordinates are:
\( x = 0.818 \) and \( x = 4.56 \).

(b) The area of the enclosed region is given by the integral of the upper function minus the lower function between these two intersection points:
\( A = \int_{0.8179...}^{4.558...} (f(x) - g(x)) \, \mathrm{d}x \)
\( A = \int_{0.8179...}^{4.558...} (3\ln(x) - (x^2 - 4x + 2)) \, \mathrm{d}x \).

Using the numerical integration feature on the GDC, we obtain:
\( A \approx 11.392... \)

To 3 significant figures, the area is \( 11.4 \).

(a)
M1 for attempting to solve \( f(x) = g(x) \) (e.g., writing the equation or showing a sketch).
A1 for \( x = 0.818 \) (accept \( 0.8179... \)).
A1 for \( x = 4.56 \) (accept \( 4.558... \)).

(b)
M1 for setting up the correct integral with their limits from part (a).
M1 for integrating \( f(x) - g(x) \).
A1 for \( 11.4 \) (accept \( 11.392... \)).

(a) The particle is at rest when \( v(t) = 0 \).
\( t^2 \cos(t) - 2t = 0 \implies t(t \cos(t) - 2) = 0 \).
For \( t \neq 0 \), we solve \( t \cos(t) - 2 = 0 \) using a GDC in radian mode.
\( t \approx 5.1141... \)
So, \( t = 5.11 \text{ s} \) (to 3 s.f.).

(b) The acceleration \( a(t) \) is the derivative of velocity: \( a(t) = v'(t) \).
Using the GDC to find the numerical derivative of \( v(t) \) at \( t = 3 \):
\( a(3) = v'(3) \approx -9.2113... \)
So, the acceleration is \( -9.21 \text{ m s}^{-2} \) (to 3 s.f.).

(c) The total distance travelled is given by:
\( d = \int_{0}^{6} |v(t)| \, \mathrm{d}t = \int_{0}^{6} |t^2 \cos(t) - 2t| \, \mathrm{d}t \).
Using the GDC numerical integration tool:
\( d \approx 54.757... \)
So, the total distance is \( 54.8 \text{ m} \) (to 3 s.f.).

(a)
M1 for setting \( v(t) = 0 \).
A1 for \( t = 5.11 \) (accept \( 5.114... \)).

(b)
M1 for recognizing that \( a(t) = v'(t) \) (can be shown by writing the derivative formula or using GDC notation).
A1 for \( -9.21 \) (accept \( -9.211... \)).

(c)
M1 for setting up the integral of the absolute value of velocity.
A1 for \( 54.8 \) (accept \( 54.757... \)).

(a) Let us determine the interior angle \( \angle PAB \).
Using parallel lines, the angle from \( A \) to North is parallel to the North line at \( P \).
Thus, the direction from \( A \) back to \( P \) has bearing \( 40^\circ + 180^\circ = 220^\circ \).
Since the bearing to \( B \) is \( 120^\circ \), the interior angle is:
\( \angle PAB = 220^\circ - 120^\circ = 100^\circ \).

Applying the cosine rule to triangle \( PAB \) to find \( AB = x \):
\( PB^2 = PA^2 + AB^2 - 2(PA)(AB)\cos(\angle PAB) \)
\( 16^2 = 10^2 + x^2 - 2(10)(x)\cos(100^\circ) \)
\( 256 = 100 + x^2 - 20x\cos(100^\circ) \)
\( x^2 - 20x\cos(100^\circ) - 156 = 0 \).

Using GDC to solve the quadratic equation:
\( x \approx 10.874... \)
So, \( AB = 10.9\text{ km} \) (to 3 s.f.).

(b) Let \( \theta = \angle APB \).
Using the sine rule in triangle \( PAB \):
\( \frac{\sin(\theta)}{AB} = \frac{\sin(100^\circ)}{PB} \)
\( \sin(\theta) = \frac{10.874... \times \sin(100^\circ)}{16} \)
\( \sin(\theta) \approx 0.6693... \)
\( \theta \approx 42.01^\circ \).

Since \( B \) lies clockwise from \( PA \), the bearing of \( B \) from \( P \) is:
\( \text{Bearing} = 40^\circ + 42.01^\circ = 82.01^\circ \).
To 3 s.f., the bearing is \( 082^\circ \).

(a)
M1 for finding \( \angle PAB = 100^\circ \) (or equivalent angle in a diagram).
M1 for correctly substituting into the cosine rule.
A1 for \( AB = 10.9\text{ km} \) (accept \( 10.874... \)).

(b)
M1 for substituting into the sine rule (or cosine rule) to find \( \angle APB \).
A1 for \( \angle APB \approx 42.0^\circ \).
A1 for bearing \( 082^\circ \) (or \( 082.0^\circ \)).

(a) The terms of the arithmetic sequence are given by \( T_n = a + (n-1)d \).
\( T_3 = a + 2d \)
\( T_7 = a + 6d \)
\( T_{15} = a + 14d \).

Since these are the first three terms of a geometric sequence, the common ratio is constant:
\( \frac{a+6d}{a+2d} = \frac{a+14d}{a+6d} \)
\( (a+6d)^2 = (a+2d)(a+14d) \)
\( a^2 + 12ad + 36d^2 = a^2 + 16ad + 28d^2 \)
\( 8d^2 - 4ad = 0 \).

Since \( d \neq 0 \), we can divide by \( 4d \):
\( 2d - a = 0 \implies a = 2d \). (AG)

(b) Given \( d = 4 \), then \( a = 2(4) = 8 \).
The sum of the first \( n \) terms of an arithmetic sequence is given by:
\( S_n = \frac{n}{2}(2a + (n-1)d) \)
\( 1400 = \frac{n}{2}(2(8) + (n-1)4) \)
\( 1400 = \frac{n}{2}(16 + 4n - 4) \)
\( 1400 = \frac{n}{2}(4n + 12) \)
\( 1400 = 2n^2 + 6n \)
\( 2n^2 + 6n - 1400 = 0 \implies n^2 + 3n - 700 = 0 \).

Using GDC or factoring:
\( (n - 25)(n + 28) = 0 \).
Since \( n > 0 \), \( n = 25 \).

(a)
M1 for setting up the ratio equation \( \frac{a+6d}{a+2d} = \frac{a+14d}{a+6d} \).
M1 for expanding the brackets correctly: \( a^2 + 12ad + 36d^2 = a^2 + 16ad + 28d^2 \).
A1 for simplifying to \( 8d^2 = 4ad \) and correctly concluding \( a = 2d \).

(b)
M1 for substituting \( a = 8 \) and \( d = 4 \) into the sum formula.
M1 for forming the quadratic equation \( 2n^2 + 6n - 1400 = 0 \) (or equivalent).
A1 for \( n = 25 \).

(a) Let \( X \) be the mass of an apple. We are given \( X \sim N(150, \sigma^2) \) and \( P(X > 180) = 0.15 \).
Standardizing the variable:
\( P\left(Z > \frac{180 - 150}{\sigma}\right) = 0.15 \implies P\left(Z < \frac{30}{\sigma}\right) = 0.85 \).

Using the inverse normal cumulative distribution function on GDC:
\( \frac{30}{\sigma} \approx 1.03643... \)
\( \sigma = \frac{30}{1.03643...} \approx 28.9453... \)
So, \( \sigma = 28.9 \) (to 3 s.f.).

(b) We want to find \( P(130 < X < 160) \) with \( \mu = 150 \) and \( \sigma = 28.9453... \).
Using normal cumulative distribution function on GDC (normalcdf with lower limit 130, upper limit 160, mean 150, sd 28.9453):
\( P(130 < X < 160) \approx 0.39034... \)
So, the probability is \( 0.390 \) (to 3 s.f.).

(a)
M1 for standardizing: \( P\left(Z < \frac{30}{\sigma}\right) = 0.85 \).
A1 for finding \( z = 1.036... \).
A1 for \( \sigma = 28.9 \) (accept \( 28.945... \)).

(b)
M1 for setting up the probability expression \( P(130 < X < 160) \).
M1 for substituting their \( \sigma \) value into the normalcdf calculation.
A1 for \( 0.390 \) (accept \( 0.3903... \)).

(a) The volume \( V \) of the solid of revolution is given by:
\( V = \pi \int_{0}^{k} y^2 \, \mathrm{d}x \)
Substitute \( y = \sqrt{x} e^{-x} \):
\( y^2 = (\sqrt{x} e^{-x})^2 = x e^{-2x} \)
\( V = \pi \int_{0}^{k} x e^{-2x} \, \mathrm{d}x \).

(b) Set the volume equation to \( 0.5 \):
\( \pi \int_{0}^{k} x e^{-2x} \, \mathrm{d}x = 0.5 \).

Using GDC numerical solver to find \( k \) such that \( f(k) = \pi \int_{0}^{k} x e^{-2x} \, \mathrm{d}x - 0.5 = 0 \):
\( k \approx 1.0818... \)
So, \( k = 1.08 \) (to 3 s.f.).

(a)
M1 for using \( V = \pi \int y^2 \, \mathrm{d}x \).
A1 for \( V = \pi \int_{0}^{k} x e^{-2x} \, \mathrm{d}x \).

(b)
M1 for setting up the equation \( \pi \int_{0}^{k} x e^{-2x} \, \mathrm{d}x = 0.5 \).
M1 for attempting to solve the equation using a GDC solver or graphical intersection.
A2 for \( k = 1.08 \) (to 3 s.f., accept \( 1.0818... \)). (Award A1 for \( 1.1 \) if 2 s.f. is used).

(a)
We use the standard normal distribution \(Z = \frac{D - \mu}{\sigma}\).
From \(\text{P}(D < 11.8) = 0.115\):
\(\frac{11.8 - \mu}{\sigma} = \Phi^{-1}(0.115) = -1.20035...\)
\(\mu - 1.20035\sigma = 11.8\) (Equation 1)

From \(\text{P}(D > 12.5) = 0.231 \implies \text{P}(D < 12.5) = 0.769\):
\(\frac{12.5 - \mu}{\sigma} = \Phi^{-1}(0.769) = 0.73556...\)
\(\mu + 0.73556\sigma = 12.5\) (Equation 2)

Subtracting Equation 1 from Equation 2:
\(1.93591\sigma = 0.7 \implies \sigma \approx 0.36159\)

Substituting \(\sigma\) back into Equation 2:
\(\mu = 12.5 - 0.73556(0.36159) \approx 12.234\)

To 3 significant figures: \(\mu = 12.2\) mm and \(\sigma = 0.362\) mm.

(b)
Using the values \(\mu = 12.234\) and \(\sigma = 0.3616\) on a GDC:
\(\text{P}(11.9 < D < 12.4) = \text{normalcdf}(11.9, 12.4, 12.234, 0.3616) \approx 0.499\) (to 3 s.f.)

(c)
Let \(X\) be the number of acceptable rods in a box of 20.
\(X \sim \text{B}(20, 0.4991)\)
We want \(\text{P}(X \ge 15) = 1 - \text{P}(X \le 14)\)
Using GDC:
\(\text{P}(X \ge 15) \approx 0.0203\) (to 3 s.f.)

(d)
This follows a geometric distribution with success probability \(p_1 = \text{P}(X \ge 15) \approx 0.02027\).
The expected value is \(E(Y) = \frac{1}{p_1}\)
\(E(Y) = \frac{1}{0.02027} \approx 49.3\) boxes.

(e)
We want to find \(\text{P}(X \ge 15 \mid X \ge 12) = \frac{\text{P}(X \ge 15 \cap X \ge 12)}{\text{P}(X \ge 12)} = \frac{\text{P}(X \ge 15)}{\text{P}(X \ge 12)}\)
Using GDC, find \(\text{P}(X \ge 12) = 1 - \text{P}(X \le 11) \approx 0.2476\)
Thus, \(\text{P}(X \ge 15 \mid X \ge 12) = \frac{0.02027}{0.2476} \approx 0.0819\) (to 3 s.f.)

(a)
- M1 for attempting to find the z-score corresponding to 0.115.
- M1 for attempting to find the z-score corresponding to 0.769 (or 1 - 0.231).
- A1 for obtaining the correct equations or correct values for the z-scores.
- A1 for correct value of \sigma (0.362).
- A1 for correct value of \mu (12.2).

(b)
- M1 for writing down the correct normal integral or GDC parameters.
- A1 for 0.499.

(c)
- M1 for identifying the binomial distribution parameters.
- M1 for writing down 1 - P(X \le 14).
- A1 for 0.0203.

(d)
- M1 for using the geometric expectation formula 1/p.
- A1 for 49.3.

(e)
- M1 for writing down the conditional probability definition.
- A1 for P(X \ge 12) \approx 0.248 (or 0.2476).
- A1 for 0.0819.

(a)
Evaluate \(h(0)\):
\(h(0) = 22 - 18 \cos(0) \text{e}^0 = 22 - 18 = 4\) metres.

(b)
Set \(h(t) = 30\):
\(22 - 18 \cos(0.25 t) \text{e}^{-0.015 t} = 30\)
Using GDC to solve this equation for \(0 \le t \le 120\):
Sketching the line \(y = 30\) and \(y = h(t)\) to find the first intersection point
\(t \approx 8.41\) seconds (to 3 s.f.)

(c)
Using GDC derivative function:
By product rule:
\(h'(t) = -18 \left[ -0.25 \sin(0.25 t) \text{e}^{-0.015 t} - 0.015 \cos(0.25 t) \text{e}^{-0.015 t} \right]\)
Evaluating at \(t = 10\) using GDC:
\(h'(10) \approx 2.13\) m/s (to 3 s.f.)

(d)
(i) The first maximum occurs when \(h'(t) = 0\)
\(0.25 \sin(0.25 t) + 0.015 \cos(0.25 t) = 0 \implies \tan(0.25 t) = -0.06\)
For the first positive solution:
\(0.25 t = \pi + \arctan(-0.06) \approx 3.0817 \implies t_1 \approx 12.3\) seconds.

(ii) Substituting \(t_1 = 12.33\) into \(h(t)\):
\(h(12.33) \approx 36.9\) metres (to 3 s.f.)

(e)
The total distance is given by \(\int_{0}^{60} |h'(t)| \mathrm{d}t\)
Using GDC to compute this definite integral directly:
\(\int_{0}^{60} |h'(t)| \mathrm{d}t \approx 116\) metres (to 3 s.f.)

(a)
- A1 for 4.

(b)
- M1 for setting up the equation h(t) = 30.
- M1 for attempting to solve using GDC (e.g., graphing).
- A1 for 8.41.

(c)
- M1 for indicating the derivative or using d/dt at t=10 on GDC.
- M1 for applying the product/chain rule correctly if done analytically.
- A1 for 2.13.

(d)
- (i) M1 for setting h'(t) = 0.
- M1 for solving for t.
- A1 for 12.3.
- (ii) A1 for 36.9.

(e)
- M1 for expressing distance as the integral of the absolute value of the derivative.
- M2 for correctly evaluating using GDC (or decomposing into intervals).
- A1 for 116.

(a)
Side length of \(S_n = 10 \times \left(\frac{p}{100}\right)^{n-1}\).

(b)
The perimeter of a square with side length \(s\) is \(4s\).
Perimeter of \(S_5 = 4 \times 10 \times \left(\frac{p}{100}\right)^4 = 18.5\)
\(40 \times \left(\frac{p}{100}\right)^4 = 18.5 \implies \left(\frac{p}{100}\right)^4 = 0.4625\)
\(\frac{p}{100} = (0.4625)^{0.25} \approx 0.82444 \implies p \approx 82.4\).

(c)
The perimeters form a geometric sequence with first term \(u_1 = 40\) and common ratio \(r = 0.82444\).
\(S_{12} = \frac{40(1 - r^{12})}{1 - r\)}
Substituting \(r = 0.82444\) (or \(r = 0.824\)):
\(S_{12} = \frac{40(1 - 0.82444^{12})}{1 - 0.82444} \approx 205\) cm (to 3 s.f.).

(d)
The areas of the squares form a geometric sequence with first term \(v_1 = 10^2 = 100\) and common ratio \(R = r^2 = (0.82444)^2 \approx 0.6797\).
Since \(|R| < 1\), the sum to infinity exists:
\(S_\infty = \frac{100}{1 - R}\)
\(S_\infty = \frac{100}{1 - 0.6797} \approx 312\text{ cm}^2\) (to 3 s.f.)
(If \(p=82.4\) is used, \(R = 0.678976\) and \(S_\infty \approx 311\text{ cm}^2\), both 311 and 312 are accepted).

(e)
(i) The area of circle \(C_n\) is given by \(A_n = 100 + (n-1)(-6) = 106 - 6n\).
We require \(106 - 6n > 0 \implies 6n < 106 \implies n < 17.67\)
Thus, the maximum number of circles is 17.

(ii) The total area is the sum of the first 17 terms of the arithmetic sequence:
\(S_{17} = \frac{17}{2} (2 a_1 + 16 d) = \frac{17}{2} [2(100) + 16(-6)]\)
\(S_{17} = 8.5 \times [200 - 96] = 8.5 \times 104 = 884\text{ cm}^2\).

(a)
- A1 for correct formula.

(b)
- M1 for setting up the equation for the perimeter of the 5th square.
- M1 for isolating (p/100)^4.
- A1 for showing p \approx 82.4.

(c)
- M1 for identifying geometric series sum formula.
- M1 for substituting the correct parameters.
- A1 for 205.

(d)
- M1 for identifying the sequence of areas is geometric with ratio R = r^2.
- M1 for using the infinite sum formula.
- A1 for 312 (or 311 if using 82.4).

(e)
- (i) M1 for setting up 106 - 6n > 0.
- A1 for 17.
- (ii) M1 for using the arithmetic sum formula.
- M1 for substituting the values n=17, a_1=100, d=-6.
- A1 for 884.