2025 IB DP Mathematics - Analysis and Approaches 模擬試題連答案詳解

To find the value of \( k \), we first integrate the function with respect to \( x \): \\int \\frac{6}{3x - 1} \\, dx . Using the rule \\int \\frac{1}{ax + b} \\, dx = \\frac{1}{a} \\ln|ax + b| + C, we have: \\int \\frac{6}{3x - 1} \\, dx = 6 \\cdot \\frac{1}{3} \\ln|3x - 1| = 2 \\ln|3x - 1| . Now we apply the limits of integration from \( 1 \) to \( k \): \\left[ 2 \\ln|3x - 1| \\right]_{1}^{k} = \\ln(16) . Since \( k > 1 \), we have \( 3x - 1 > 0 \) over the interval. Substituting the limits gives: 2 \\ln(3k - 1) - 2 \\ln(3(1) - 1) = \\ln(16) \\implies 2 \\ln(3k - 1) - 2 \\ln(2) = \\ln(16) . Divide the entire equation by 2: \\ln(3k - 1) - \\ln(2) = \\frac{1}{2} \\ln(16) . Using the laws of logarithms: \\ln\\left( \\frac{3k - 1}{2} \\right) = \\ln(16^{1/2}) \\implies \\ln\\left( \\frac{3k - 1}{2} \\right) = \\ln(4) . Taking the exponential of both sides: \\frac{3k - 1}{2} = 4 \\implies 3k - 1 = 8 \\implies 3k = 9 \\implies k = 3 .

**M1** for an attempt to integrate of the form \( A \ln(3x - 1) \). **A1** for correct integration resulting in \( 2 \ln(3x - 1) \) (or equivalent). **M1** for correctly substituting the limits and setting up an equation to solve for \( k \) (e.g., \( 2\ln(3k-1) - 2\ln(2) = \ln(16) \)). **A1** for \( k = 3 \). *Note: Award at most **M1A1M1A0** if \( k = 3 \) is found but other incorrect solutions like \( k = -7/3 \) are also kept (since \( k > 1 \) is given in the question).*

Using the definitions of arithmetic and geometric sequences, we write expressions for the terms: \( u_2 = 4 + d \), \( u_3 = 4 + 2d \), \( v_2 = 4r \), and \( v_3 = 4r^2 \). From the first relation, we have: \( u_2 = v_2 + 2 \Rightarrow 4 + d = 4r + 2 \Rightarrow d = 4r - 2 \). From the second relation, we have: \( u_3 = v_3 + 3 \Rightarrow 4 + 2d = 4r^2 + 3 \). Substituting the expression for \( d \) into this equation gives: \( 4 + 2(4r - 2) = 4r^2 + 3 \). Simplifying the equation: \( 4 + 8r - 4 = 4r^2 + 3 \Rightarrow 8r = 4r^2 + 3 \Rightarrow 4r^2 - 8r + 3 = 0 \). Factoring the quadratic: \( (2r - 1)(2r - 3) = 0 \). Thus, we find the possible values of \( r \) are \( r = \frac{1}{2} \) and \( r = \frac{3}{2} \).

M1: Attempt to write expressions for the terms in terms of \( d \) and \( r \). A1: Correctly establishing the relation \( d = 4r - 2 \) (or equivalent). M1: Substituting the expression for \( d \) into the equation for the third terms. A1: Obtaining the correct quadratic equation, e.g., \( 4r^2 - 8r + 3 = 0 \). A1: Finding both correct values: \( r = \frac{1}{2} \) and \( r = \frac{3}{2} \).

To find \( f(x) \), we integrate the gradient function: \( f(x) = \int 3x \sqrt{x^2 + 5} \, dx \). We use integration by substitution. Let \( u = x^2 + 5 \), which gives \( \frac{du}{dx} = 2x \), or \( x \, dx = \frac{1}{2} \, du \). Substituting these into the integral: \( f(x) = \int 3 \sqrt{u} \cdot \frac{1}{2} \, du = \frac{3}{2} \int u^{\frac{1}{2}} \, du \). Integrating with respect to \( u \): \( f(x) = \frac{3}{2} \left( \frac{2}{3} u^{\frac{3}{2}} \right) + C = u^{\frac{3}{2}} + C \). Substituting back \( u = x^2 + 5 \): \( f(x) = (x^2 + 5)^{\frac{3}{2}} + C \). We use the point \( (2, 10) \) to find the constant \( C \): \( 10 = (2^2 + 5)^{\frac{3}{2}} + C \Rightarrow 10 = (9)^{\frac{3}{2}} + C \Rightarrow 10 = 27 + C \Rightarrow C = -17 \). Therefore, the equation of the curve is \( f(x) = (x^2 + 5)^{\frac{3}{2}} - 17 \).

M1: Realising integration is required and attempting integration by substitution. A1: Correct substituted integral, e.g., \( \frac{3}{2} \int u^{\frac{1}{2}} \, du \). A1: Correct integration yielding \( (x^2 + 5)^{\frac{3}{2}} + C \). M1: Substituting \( x = 2 \) and \( y = 10 \) into their integrated expression to find \( C \). A1: Correct final function \( f(x) = (x^2 + 5)^{\frac{3}{2}} - 17 \).

To find where the tangent is parallel to the line \( y = x \), we set the derivative of \( f(x) \) equal to the gradient of the line, which is \( 1 \). First, find \( f'(x) \) using the chain rule: \( f'(x) = \frac{2x - 3}{x^2 - 3x + 3} \). Set \( f'(x) = 1 \): \( \frac{2x - 3}{x^2 - 3x + 3} = 1 \). Multiply both sides by \( x^2 - 3x + 3 \) (which is always positive since the discriminant is negative): \( 2x - 3 = x^2 - 3x + 3 \). Rearrange into a quadratic equation: \( x^2 - 5x + 6 = 0 \). Factor the quadratic: \( (x - 2)(x - 3) = 0 \). So, \( x = 2 \) or \( x = 3 \). Find the corresponding \( y \)-coordinates: For \( x = 2 \), \( f(2) = \ln(2^2 - 3(2) + 3) = \ln(1) = 0 \). For \( x = 3 \), \( f(3) = \ln(3^2 - 3(3) + 3) = \ln(3) \). Thus, the coordinates of the points are \( (2, 0) \) and \( (3, \ln 3) \).

M1 for attempting to use the chain rule to differentiate. A1 for correct derivative: \( f'(x) = \frac{2x - 3}{x^2 - 3x + 3} \). M1 for setting their \( f'(x) = 1 \). M1 for attempting to solve the resulting equation to form a quadratic equation. A1 for finding \( x = 2 \) and \( x = 3 \). A1 for \( y = 0 \). A1 for \( y = \ln 3 \).

(a) Since the sum of probabilities must equal 1: \( a + b + 0.2 + 0.1 = 1 \implies a + b = 0.7 \). Using the expected value \( \text{E}(X) = 2.2 \): \( 1(a) + 2(b) + 3(0.2) + 4(0.1) = 2.2 \implies a + 2b + 1.0 = 2.2 \implies a + 2b = 1.2 \). Subtracting the first equation from the second gives: \( b = 0.5 \). Substituting \( b = 0.5 \) back gives: \( a = 0.2 \). (b) To find the variance, first find \( \text{E}(X^2) \): \( \text{E}(X^2) = 1^2(0.2) + 2^2(0.5) + 3^2(0.2) + 4^2(0.1) = 0.2 + 2.0 + 1.8 + 1.6 = 5.6 \). Now, use the variance formula: \( \text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2 = 5.6 - (2.2)^2 = 5.6 - 4.84 = 0.76 \).

M1 for using sum of probabilities is 1 to get \( a + b = 0.7 \). M1 for using expected value formula to get \( a + 2b = 1.2 \). A1 for \( b = 0.5 \). A1 for \( a = 0.2 \). M1 for a valid attempt to find \( \text{E}(X^2) \). A1 for \( \text{E}(X^2) = 5.6 \). A1 for \( \text{Var}(X) = 0.76 \).

Use the identity \( \cos^2(x) = 1 - \sin^2(x) \) to rewrite the equation in terms of sine: \( 2(1 - \sin^2(x)) + 3\sin(x) - 3 = 0 \). Simplify the expression: \( 2 - 2\sin^2(x) + 3\sin(x) - 3 = 0 \implies 2\sin^2(x) - 3\sin(x) + 1 = 0 \). Factor the quadratic equation: \( (2\sin(x) - 1)(\sin(x) - 1) = 0 \). This gives two possible cases: \( \sin(x) = \frac{1}{2} \) or \( \sin(x) = 1 \). Solve each equation in the interval \( 0 \le x \le 2\pi \): For \( \sin(x) = \frac{1}{2} \), we find \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \). For \( \sin(x) = 1 \), we find \( x = \frac{\pi}{2} \). Thus, the full set of solutions is \( x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6} \).

M1 for substituting \( \cos^2(x) = 1 - \sin^2(x) \). A1 for obtaining a correct quadratic equation in terms of sine: \( 2\sin^2(x) - 3\sin(x) + 1 = 0 \). M1 for attempting to factorize or solve their quadratic equation. A1 for finding both \( \sin(x) = \frac{1}{2} \) and \( \sin(x) = 1 \). A1 for \( x = \frac{\pi}{6} \). A1 for \( x = \frac{5\pi}{6} \). A1 for \( x = \frac{\pi}{2} \).

(a) Write \(f(x) = (\sin(x))^2 + \cos(x)\).
Using the chain rule to differentiate \(\sin^2(x)\):
\[\frac{d}{dx}(\sin^2(x)) = 2\sin(x)\cos(x)\]
Using the standard derivative for \(\cos(x)\):
\[\frac{d}{dx}(\cos(x)) = -\sin(x)\]
Thus, \(f'(x) = 2\sin(x)\cos(x) - \sin(x)\).
Factorizing out \(\sin(x)\):
\[f'(x) = \sin(x)(2\cos(x) - 1)\]

(b) To find the stationary points, we set \(f'(x) = 0\):
\[\sin(x)(2\cos(x) - 1) = 0\]
This gives two cases:
1. \(\sin(x) = 0\)
Since \(0 \le x \le \pi\), this yields \(x = 0\) and \(x = \pi\).
2. \(2\cos(x) - 1 = 0 \implies \cos(x) = \frac{1}{2}\)
Since \(0 \le x \le \pi\), this yields \(x = \frac{\pi}{3}\).
Therefore, the \(x\)-coordinates of the stationary points are \(x = 0\), \(x = \frac{\pi}{3}\), and \(x = \pi\).

(c) We can use the second derivative test. First, write the derivative as:
\[f'(x) = 2\sin(x)\cos(x) - \sin(x) = \sin(2x) - \sin(x)\]
Differentiating again with respect to \(x\):
\[f''(x) = 2\cos(2x) - \cos(x)\]
Now we test each stationary point:
- For \(x = 0\):
\[f''(0) = 2\cos(0) - \cos(0) = 2(1) - 1 = 1 > 0\]
Since \(f''(0) > 0\), the point at \(x = 0\) is a local minimum.
- For \(x = \frac{\pi}{3}\):
\[f''\left(\frac{\pi}{3}\right) = 2\cos\left(\frac{2\pi}{3}\right) - \cos\left(\frac{\pi}{3}\right) = 2\left(-\frac{1}{2}\right) - \frac{1}{2} = -1 - \frac{1}{2} = -\frac{3}{2} < 0\]
Since \(f''\left(\frac{\pi}{3}\right) < 0\), the point at \(x = \frac{\pi}{3}\) is a local maximum.
- For \(x = \pi\):
\[f''(\pi) = 2\cos(2\pi) - \cos(\pi) = 2(1) - (-1) = 3 > 0\]
Since \(f''(\pi) > 0\), the point at \(x = \pi\) is a local minimum.

(d) The area \(A\) is given by the definite integral:
\[A = \int_{0}^{\frac{\pi}{2}} f(x) \, dx = \int_{0}^{\frac{\pi}{2}} (\sin^2(x) + \cos(x)) \, dx\]
Using the double angle identity \(\sin^2(x) = \frac{1 - \cos(2x)}{2}\):
\[A = \int_{0}^{\frac{\pi}{2}} \left(\frac{1}{2} - \frac{1}{2}\cos(2x) + \cos(x)\right) \, dx\]
Integrating term-by-term:
\[A = \left[ \frac{1}{2}x - \frac{1}{4\vphantom{2}}\sin(2x) + \sin(x) \right]_{0}^{\frac{\pi}{2}}\]
Substitute the upper limit \(x = \frac{\pi}{2}\):
\[\left(\frac{1}{2}\left(\frac{\pi}{2}\right) - \frac{1}{4}\sin(\pi) + \sin\left(\frac{\pi}{2}\right)\right) = \frac{\pi}{4} - 0 + 1 = \frac{\pi}{4} + 1\]
Substitute the lower limit \(x = 0\):
\[\left(0 - 0 + 0\right) = 0\]
Thus, the area is \(\frac{\pi}{4} + 1\).

(a)
M1 for attempting to use the chain rule on \(\sin^2(x)\).
A1 for correct differentiation of both terms, leading to \(2\sin(x)\cos(x) - \sin(x)\).
A1 for factorizing to show the required form \(\sin(x)(2\cos(x) - 1)\).
[3 marks]

(b)
M1 for setting \(f'(x) = 0\).
A1 for finding \(x = 0\) and \(x = \pi\).
A1 for finding \(x = \frac{\pi}{3}\).
[3 marks]

(c)
M1 for attempting to find the second derivative or analyzing the sign of the first derivative around the stationary points.
\(f''(x) = 2\cos(2x) - \cos(x)\)
A1 for correctly determining that \(x = \frac{\pi}{3}\) is a local maximum with supporting work (e.g., \(f''(\frac{\pi}{3}) = -\frac{3}{2} < 0\)).
A1 for correctly determining that \(x = 0\) and \(x = \pi\) are local minima with supporting work (e.g., \(f''(0) = 1 > 0\) and \(f''(\pi) = 3 > 0\)).
Note: Accept sign charts with clear explanation of derivative changing sign.
[3 marks]

(d)
M1 for setting up the correct definite integral \(\int_{0}^{\frac{\pi}{2}} (\sin^2(x) + \cos(x)) \, dx\).
M1 for substituting the identity \(\sin^2(x) = \frac{1 - \cos(2x)}{2}\).
A1 for correct integration: \(\frac{1}{2}x - \frac{1}{4}\sin(2x) + \sin(x)\).
A1 for substituting limits and obtaining the final exact area of \(\frac{\pi}{4} + 1\).
[4 marks]

(a) To find the \( x \)-intercepts, we solve \( f(x) = 0 \):

\( e^{2x} - 5e^x + 4 = 0 \)

Let \( u = e^x \). The equation becomes:

\( u^2 - 5u + 4 = 0 \implies (u-1)(u-4) = 0 \)

Thus, \( u = 1 \) or \( u = 4 \).

Since \( e^x = u \):

\( e^x = 1 \implies x = 0 \)

\( e^x = 4 \implies x = \ln 4 \) (or \( 2\ln 2 \))

So, the \( x \)-intercepts are at \( x = 0 \) and \( x = \ln 4 \).

(b) To find the coordinates of the local minimum, we first find the first derivative \( f'(x) \):

\( f'(x) = 2e^{2x} - 5e^x \)

Set \( f'(x) = 0 \) to find the stationary points:

\( e^x(2e^x - 5) = 0 \)

Since \( e^x > 0 \) for all \( x \in \mathbb{R} \), we have:

\( 2e^x - 5 = 0 \implies e^x = \frac{5}{2} \implies x = \ln\left(\frac{5}{2}\right) \)

To find the corresponding \( y \)-coordinate, substitute \( x = \ln\left(\frac{5}{2}\right) \) into \( f(x) \):

\( f\left(\ln\left(\frac{5}{2}\right)\right) = \left(\frac{5}{2}\right)^2 - 5\left(\frac{5}{2}\right) + 4 = \frac{25}{4} - \frac{25}{2} + 4 = -\frac{9}{4} \)

To verify that this point is indeed a local minimum, we find the second derivative \( f''(x) \):

\( f''(x) = 4e^{2x} - 5e^x \)

Evaluating \( f''(x) \) at \( x = \ln\left(\frac{5}{2}\right) \):

\( f''\left(\ln\left(\frac{5}{2}\right)\right) = 4\left(\frac{25}{4}\right) - 5\left(\frac{5}{2}\right) = 25 - \frac{25}{2} = \frac{25}{2} > 0 \)

Since \( f''\left(\ln\left(\frac{5}{2}\right)\right) > 0 \), the graph is concave up at this point, verifying a local minimum.

Thus, the coordinates of the local minimum are \( \left(\ln\left(\frac{5}{2}\right), -\frac{9}{4}\right) \).

(c) For a point of inflection, we set \( f''(x) = 0 \):

\( 4e^{2x} - 5e^x = 0 \implies e^x(4e^x - 5) = 0 \)

Since \( e^x > 0 \), we have:

\( 4e^x - 5 = 0 \implies e^x = \frac{5}{4} \implies x = \ln\left(\frac{5}{4}\right) \)

To find the corresponding \( y \)-coordinate, substitute \( x = \ln\left(\frac{5}{4}\right) \) into \( f(x) \):

\( f\left(\ln\left(\frac{5}{4}\right)\right) = \left(\frac{5}{4}\right)^2 - 5\left(\frac{5}{4}\right) + 4 = \frac{25}{16} - \frac{25}{4} + 4 = \frac{25 - 100 + 64}{16} = -\frac{11}{16} \)

Since \( f''(x) \) changes sign around \( x = \ln\left(\frac{5}{4}\right) \), this is a point of inflection.

The coordinates of the point of inflection are \( \left(\ln\left(\frac{5}{4}\right), -\frac{11}{16}\right) \).

(d) The region \( R \) is bounded by the graph of \( f \) and the \( x \)-axis. This region lies below the \( x \)-axis between the two \( x \)-intercepts found in part (a), \( x = 0 \) and \( x = \ln 4 \).

Therefore, the area of \( R \) is given by:

\( \text{Area} = -\int_{0}^{\ln 4} f(x) \, dx = \int_{0}^{\ln 4} (5e^x - e^{2x} - 4) \, dx \)

Now, we find the antiderivative:

\( \int (5e^x - e^{2x} - 4) \, dx = 5e^x - \frac{1}{2}e^{2x} - 4x \)

Evaluating this antiderivative at the upper limit \( x = \ln 4 \):

\( 5e^{\ln 4} - \frac{1}{2}e^{2\ln 4} - 4\ln 4 = 5(4) - \frac{1}{2}(16) - 4\ln(2^2) = 20 - 8 - 8\ln 2 = 12 - 8\ln 2 \)

Evaluating this antiderivative at the lower limit \( x = 0 \):

\( 5e^0 - \frac{1}{2}e^0 - 4(0) = 5 - \frac{1}{2} = \frac{9}{2} \)

Subtracting the lower limit value from the upper limit value:

\( \text{Area} = (12 - 8\ln 2) - \frac{9}{2} = \frac{24}{2} - \frac{9}{2} - 8\ln 2 = \frac{15}{2} - 8\ln 2 \)

Thus, the area of \( R \) is indeed \( \frac{15}{2} - 8\ln 2 \).

Part (a)
* **M1**: For setting \( f(x) = 0 \) to find the x-intercepts.
* **A1**: Finding the correct values \( e^x = 1 \) and \( e^x = 4 \).
* **A1**: Finding the correct x-coordinates: \( x = 0, \ln 4 \) (or \( 2\ln 2 \)). Accept coordinates written as \( (0, 0) \) and \( (\ln 4, 0) \).

Part (b)
* **M1**: For differentiating \( f(x) \) to find \( f'(x) = 2e^{2x} - 5e^x \).
* **A1**: For setting \( f'(x) = 0 \) and solving for \( x = \ln\left(\frac{5}{2}\right) \).
* **A1**: For finding the correct y-coordinate \( y = -\frac{9}{4} \).
* **R1**: For justifying that this is a local minimum, e.g., showing \( f''\left(\ln(5/2)\right) = 12.5 > 0 \) (or using a sign diagram for the first derivative).

Part (c)
* **M1**: For setting \( f''(x) = 0 \) to find the point of inflection, where \( f''(x) = 4e^{2x} - 5e^x \).
* **A1**: For finding \( x = \ln\left(\frac{5}{4}\right) \).
* **A1**: For finding the correct y-coordinate \( y = -\frac{11}{16} \). (Must be written as a coordinate pair \( \left(\ln\left(\frac{5}{4}\right), -\frac{11}{16}\right) \)).

Part (d)
* **M1**: For formulating the integral for the area with correct limits, \( -\int_{0}^{\ln 4} f(x) \, dx \) or \( \int_{0}^{\ln 4} (5e^x - e^{2x} - 4) \, dx \).
* **A1**: For the correct antiderivative, e.g., \( 5e^x - \frac{1}{2}e^{2x} - 4x \).
* **M1**: For correctly substituting the limits of integration \( 0 \) and \( \ln 4 \).
* **A1**: For evaluating the upper and lower limits correctly as \( 12 - 8\ln 2 \) and \( \frac{9}{2} \).
* **AG**: For showing all intermediate algebraic steps clearly to obtain the final requested expression: \( \frac{15}{2} - 8\ln 2 \).

(a) (i) Using the product rule on \( f(x) = x \ln(x) \):
\( f'(x) = 1 \cdot \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1 \).

(ii) To find the local minimum, we set \( f'(x) = 0 \):
\( \ln(x) + 1 = 0 \implies \ln(x) = -1 \implies x = e^{-1} = \frac{1}{e} \).

Substituting this back into \( f(x) \) to find the \( y \)-coordinate:
\( f\left(\frac{1}{e}\right) = \frac{1}{e} \ln\left(\frac{1}{e}\right) = -\frac{1}{e} \).

Thus, the coordinates of the local minimum point are \( \left(\frac{1}{e}, -\frac{1}{e}\right) \).

(b) At point \( P(e, e) \), the gradient of the tangent is given by \( f'(e) \):
\( f'(e) = \ln(e) + 1 = 2 \).

Since the normal is perpendicular to the tangent, the gradient of the normal is:
\( m = -\frac{1}{2} \).

Using the point-slope formula for the equation of the normal line through \( (e, e) \):
\( y - e = -\frac{1}{2}(x - e) \)
\( 2y - 2e = -x + e \)
\( x + 2y - 3e = 0 \).

(c) Differentiating \( f'(x) = \ln(x) + 1 \):
\( f''(x) = \frac{1}{x} \).

For a point of inflexion, the second derivative must equal zero or be undefined, and change sign. However, since the domain of \( f \) is \( x > 0 \), we have \( f''(x) = \frac{1}{x} > 0 \) for all \( x \) in the domain. Since \( f''(x) \neq 0 \) and does not change sign (the graph remains concave up), there are no points of inflexion.

(d) Since \( f(x) \ge 0 \) on the interval \( [1, e] \), the area of \( R \) is given by:
\( \text{Area} = \int_1^e x \ln(x) \, dx \).

We use integration by parts: \( \int u \, dv = uv - \int v \, du \).
Let \( u = \ln(x) \) and \( dv = x \, dx \).
Then \( du = \frac{1}{x} \, dx \) and \( v = \frac{x^2}{2} \).

Applying the formula:
\( \int x \ln(x) \, dx = \frac{x^2}{2} \ln(x) - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx \)
\( = \frac{x^2}{2} \ln(x) - \frac{1}{2} \int x \, dx \)
\( = \frac{x^2}{2} \ln(x) - \frac{x^2}{4} \).

Now evaluate the definite integral from \( 1 \) to \( e \):
\( \left[ \frac{x^2}{2} \ln(x) - \frac{x^2}{4} \right]_1^e = \left( \frac{e^2}{2} \ln(e) - \frac{e^2}{4} \right) - \left( \frac{1^2}{2} \ln(1) - \frac{1^2}{4} \right) \)
\( = \left( \frac{e^2}{2} - \frac{e^2}{4} \right) - \left( 0 - \frac{1}{4} \right) \)
\( = \frac{e^2}{4} + \frac{1}{4} = \frac{e^2 + 1}{4} \).

(a) (i)
- M1 for attempting to use the product rule
- A1 for correct derivative: \( f'(x) = \ln(x) + 1 \)
(ii)
- M1 for setting \( f'(x) = 0 \) and solving for \( x \)
- A1 for the correct coordinates \( \left(\frac{1}{e}, -\frac{1}{e}\right) \)
[4 marks]

(b)
- A1 for finding gradient of the tangent: \( f'(e) = 2 \)
- M1 for finding gradient of the normal: \( m = -\frac{1}{2} \)
- M1 for writing down a line equation using their gradient and point \( (e, e) \)
- A1 for correct equation in required form: \( x + 2y - 3e = 0 \) (or any equivalent integer multiple form)
[4 marks]

(c)
- A1 for \( f''(x) = \frac{1}{x} \)
- M1 for explaining that a point of inflexion requires \( f''(x) = 0 \) or a change in sign
- R1 for concluding there are no points of inflexion because \( f''(x) > 0 \) for all \( x > 0 \)
[3 marks]

(d)
- A1 for setting up the correct integral: \( \int_1^e x \ln(x) \, dx \)
- M1 for attempting integration by parts with correct choice of \( u \) and \( dv \)
- A1 for finding the correct antiderivative: \( \frac{x^2}{2} \ln(x) - \frac{x^2}{4} \)
- M1 for substituting the limits \( e \) and \( 1 \) into their antiderivative
- A1 for correct evaluation of upper limit terms and lower limit terms
- A1 for the final exact answer: \( \frac{e^2 + 1}{4} \)
[6 marks]

(a) Let \(X\) be the mass of a cat, so \(X \sim N(4.2, \sigma^2)\).
We are given \(\text{P}(X > 5.1) = 0.15\), which implies \(\text{P}(X \le 5.1) = 0.85\).
Using the standard normal distribution, we convert this to a \(z\)-score:
\(\frac{5.1 - 4.2}{\sigma} = z_{0.85}\)
Using the GDC inverse normal function:
\(z_{0.85} \approx 1.03643\)
\(\frac{0.9}{\sigma} = 1.03643\)
\̂\sigma \approx 0.86836\dots \approx 0.868\text{ kg}

(b) We now need to find \(\text{P}(3.5 \le X \le 4.5)\) where \(X \sim N(4.2, 0.86836^2)\).
Using the GDC normal cumulative distribution function (normalcdf):
\(\text{P}(3.5 \le X \le 4.5) \approx 0.42505\dots \approx 0.425\)

(a)
**M1** for attempting to standardise and setting equal to a standard normal \(z\)-value (e.g., \(\frac{0.9}{\sigma} = 1.036\dots\) or \(\Phi\left(\frac{0.9}{\sigma}\right) = 0.85\)).
**A1** for \(\sigma \approx 0.868\) (or \(0.86836\dots\)).

(b)
**M1** for attempting to calculate the interval probability \(\text{P}(3.5 < X < 4.5)\) using their standard deviation from (a).
**A1** for \(0.425\) (accept \(0.42505\dots\)).

Let \(W \sim N(\mu, \sigma^2)\) and \(Z \sim N(0, 1)\) be the standard normal variable.

We are given:
\(P(W > 180) = 0.15 \implies P(W < 180) = 0.85\)
\(P(W < 120) = 0.06\)

Using the inverse normal cumulative distribution function on a GDC, we find the \(z\)-scores:
For \(P(Z < z_1) = 0.85\):
\(z_1 \approx 1.03643\)

For \(P(Z < z_2) = 0.06\):
\(z_2 \approx -1.55477\)

Using the standardization formula \(z = \frac{x - \mu}{\sigma}\), we set up the following equations:
\(180 - \mu = 1.03643\sigma\)
\(120 - \mu = -1.55477\sigma\)

Subtracting the second equation from the first equation:
\(60 = (1.03643 - (-1.55477))\sigma\)
\(60 = 2.59120\sigma\)
\(120 - \mu = -1.55477\sigma\)

Solving for \(\sigma\):
\(\sigma \approx 23.1553...\)

Substituting \(\sigma\) back to find \(\mu\):
\(\mu = 180 - 1.03643(23.1553...) \approx 155.999...\)

Rounding to 3 significant figures:
\(\mu \approx 156\)
\(\sigma \approx 23.2\)

**[1 Mark]** for finding the correct standard normal value for the upper tail: \(z_1 \approx 1.0364...\) (accept \(1.04\))
**[1 Mark]** for finding the correct standard normal value for the lower tail: \(z_2 \approx -1.5548...\) (accept \(-1.55\))
**[1 Mark]** for setting up at least one correct standardization equation involving \(\mu\) and \(\sigma\), e.g., \(\frac{180 - \mu}{\sigma} = 1.0364...\) or \(\frac{120 - \mu}{\sigma} = -1.5548...\)
**[1 Mark]** for correctly solving the system of equations for one of the variables (e.g., \(\sigma \approx 23.2\))
**[1 Mark]** for finding the other variable (e.g., \(\mu \approx 156\))

*Note:* If 3 s.f. values of \(z\) are used (\(1.04\) and \(-1.55\)), the equations are:
\(180 - \mu = 1.04\sigma\)
\(120 - \mu = -1.55\sigma\)
Subtracting gives \(60 = 2.59\sigma \implies \sigma \approx 23.166... \approx 23.2\).
Then \(\mu = 180 - 1.04(23.166...) \approx 155.90... \approx 156\). Accept these rounded results.

Let \(X \sim N(150, 8^2)\). (a) Using a graphic display calculator (GDC), \(P(X > 155) \approx 0.265985...\). To 3 significant figures, \(P(X > 155) = 0.266\). (b) Let \(Y\) be the number of chocolate bars with a mass greater than 155 g. \(Y \sim B(5, 0.265985...)\). Using binomial probability on a GDC, \(P(Y = 2) = \binom{5}{2} (0.265985...)^2 (1 - 0.265985...)^3 \approx 0.279791...\). To 3 significant figures, \(P(Y = 2) = 0.280\). (c) We want to find the conditional probability \(P(X < 155 \mid X > 145) = \frac{P(145 < X < 155)}{P(X > 145)}\). Using the GDC, \(P(145 < X < 155) \approx 0.468028...\) and \(P(X > 145) \approx 0.734014...\). Thus, \(\frac{0.468028...}{0.734014...} \approx 0.637628...\). To 3 significant figures, the probability is 0.638.

(a) (M1) for attempting to use normal CDF on a GDC, (A1) for 0.266. (b) (M1) for identifying binomial distribution with \(n=5\) and \(p \approx 0.266\), (A1) for 0.280. (c) (M1) for setting up the conditional probability: \(\frac{P(145 < X < 155)}{P(X > 145)}\), (A1) for 0.638.

(a) The area of the sector is \(\frac{1}{2} r^2 \theta = 32\theta\). The area of the triangle is \(\frac{1}{2} r^2 \sin\theta = 32\sin\theta\). Since the sector area is twice the triangle area, \(32\theta = 2(32\sin\theta) \implies \theta = 2\sin\theta\). (b) Using a GDC to solve \(\theta = 2\sin\theta\) in the interval \(0 < \theta < \pi\), we get \(\theta \approx 1.89549...\). To 3 significant figures, \(\theta = 1.90\) radians. (c) The area of the segment is the area of the sector minus the area of the triangle: \(\text{Area} = \frac{1}{2} r^2 \theta - \frac{1}{2} r^2 \sin\theta = 32(\theta - \sin\theta)\). Using \(\theta \approx 1.89549...\), \(\text{Area} = 32(1.89549... - \sin(1.89549...)) \approx 32(0.947743...) \approx 30.3278...\). To 3 significant figures, the area is 30.3.

(a) (A1) for a correct equation, e.g., \(\frac{1}{2}(8)^2\theta = 2 \left(\frac{1}{2}(8)^2\sin\theta\right)\) or \(\theta = 2\sin\theta\). (b) (M1) for attempting to solve the equation using GDC, (A1) for \(\theta \approx 1.90\). (c) (M1) for expressing the area of the segment as \(\frac{1}{2}r^2(\theta - \target)\) or \(32(\theta - \sin\theta)\), (A1) for substituting the values correctly, (A1) for 30.3.

(a) To find the points of intersection, we set \(f(x) = g(x) \implies \mathrm{e}^{-0.5x^2} = x^2 - 2\). Using a GDC to find the intersections of the two curves, we get \(x \approx -1.52138... \) and \(x \approx 1.52138...\). To 3 significant figures, the \(x\)-coordinates are \(x = -1.52\) and \(x = 1.52\). (b) The area of the region \(R\) is given by \(\text{Area} = \int_{-1.52138...}^{1.52138...} (f(x) - g(x)) \mathrm{d}x = \int_{-1.52138...}^{1.52138...} \left(\mathrm{e}^{-0.5x^2} - (x^2 - 2)\right) \mathrm{d}x\). Using the numerical integration function on a GDC, we get \(\text{Area} \approx 5.92438...\). To 3 significant figures, the area is 5.92.

(a) (M1) for setting up the equation \(\mathrm{e}^{-0.5x^2} = x^2 - 2\), (A1) for \(x \approx -1.52\), (A1) for \(x \approx 1.52\). (b) (M1) for setting up the integral for the area with their limits from part (a), (A1) for correct substitution of limits and functions: \(\int_{-1.52}^{1.52} \left(\mathrm{e}^{-0.5x^2} - x^2 + 2\right) \mathrm{d}x\), (A1) for 5.92.

(a) Let \(X\) be the mass of a randomly selected apple, where \(X \sim \text{N}(150, 12^2)\).
We want to find \(P(X > 165)\).
Using a GDC with lower bound \(165\), upper bound \(\infty\), \(\mu = 150\), and \(\sigma = 12\):
\(P(X > 165) \approx 0.105650... \approx 0.106\).

(b) "Standard" apples have a mass between \(135\text{ g}\) and \(165\text{ g}\).
We want to find the conditional probability \(P(X > 145 \mid 135 \le X \le 165)\).
Using the conditional probability formula:
\(P(X > 145 \mid 135 \le X \le 165) = \frac{P(145 < X \le 165)}{P(135 \le X \le 165)}\).
Using a GDC to find the individual probabilities:
\(P(145 < X \le 165) \approx 0.555889...\)
\(P(135 \le X \le 165) \approx 0.788700...\)
Now, calculate the ratio:
\(\frac{0.555889...}{0.788700...} \approx 0.704817... \approx 0.705\).

(c) Let \(Y\) be the number of apples in the box that weigh more than \(135\text{ g}\).
First, we find the probability of a single apple weighing more than \(135\text{ g}\):
\(p = P(X > 135) = 1 - P(X \le 135)\).
Since \(P(X \le 135) \approx 0.105650...\), we have \(p \approx 1 - 0.105650 = 0.894350...\).
Since the apples are randomly and independently selected, \(Y\) follows a binomial distribution:
\(Y \sim \text{B}(10, 0.894350...)\).
We want to find \(P(Y \ge 8) = P(Y = 8) + P(Y = 9) + P(Y = 10)\).
Using a GDC for binomial cumulative probability:
\(P(Y \ge 8) = 1 - P(Y \le 7) \approx 1 - 0.083259... \approx 0.916741... \approx 0.917\).

(a)
M1: For attempting to find \(P(X > 165)\) using the normal distribution with \(\mu = 150\) and \(\sigma = 12\).
A1: \(0.106\) (accept \(0.105\))
[2 marks]

(b)
M1: For a correct formulation of the conditional probability, i.e., \(\frac{P(145 < X \le 165)}{P(135 \le X \le 165)}\).
M1: For finding \(P(145 < X \le 165) \approx 0.556\) and \(P(135 \le X \le 165) \approx 0.789\).
A1: \(0.705\) (accept \(0.704\))
[3 marks]

(c)
M1: For recognizing the binomial scenario with \(n = 10\) and probability of success \(p = P(X > 135) \approx 0.894\).
A1: \(0.917\) (accept \(0.916\))
[2 marks]

(a) Let \(W\) be the weight of a loaf of bread, where \(W \sim N(\mu, \sigma^2)\).
Using the standard normal distribution \(Z \sim N(0, 1)\):
\(P(W < 385) = 0.10 \implies P\left(Z < \frac{385 - \mu}{\sigma}\right) = 0.10\)
Using a GDC, the \(z\)-score for the 10th percentile is \(-1.28155...\)
So, \(\frac{385 - \mu}{\sigma} = -1.28155 \implies \mu - 1.28155\sigma = 385\) (Equation 1)

Similarly,
\(P(W > 415) = 0.05 \implies P\left(Z < \frac{415 - \mu}{\sigma}\right) = 0.95\)
Using a GDC, the \(z\)-score for the 95th percentile is \(1.64485...\)
So, \(\frac{415 - \mu}{\sigma} = 1.64485 \implies \mu + 1.64485\sigma = 415\) (Equation 2)

Subtracting Equation 1 from Equation 2:
\((\mu + 1.64485\sigma) - (\mu - 1.28155\sigma) = 415 - 385\)
\(2.92640\sigma = 30\)
\(\sigma \approx 10.2515...\)
\(\sigma = 10.3\text{ g}\) (to 3 s.f.)

Substituting \(\sigma\) back to find \(\mu\):
\(\mu = 415 - 1.64485(10.2515) \approx 398.138...\)
\(\mu = 398\text{ g}\) (to 3 s.f.)

(b) We want to find \(P(390 < W < 410)\) where \(W \sim N(398.138, 10.2515^2)\).
Using the normal CDF function on a GDC with lower limit \(390\), upper limit \(410\), \(\mu = 398.138\), and \(\sigma = 10.2515\):
\(P(390 < W < 410) \approx 0.66275...\)
\(P(390 < W < 410) = 0.663\) (to 3 s.f.)

(c) (i) We want to find \(P(W > 402)\).
Using the normal CDF function on a GDC with lower limit \(402\), upper limit \(\infty\):
\(P(W > 402) \approx 0.42795...\)
\(P(W > 402) = 0.428\) (to 3 s.f.)

(ii) Let \(X\) be the number of loaves in a bag of 6 that weigh more than \(402\text{ g}\).
Then \(X\) follows a binomial distribution: \(X \sim B(6, p)\) where \(p = 0.42795...\)
We want to find \(P(X \ge 4)\).
\(P(X \ge 4) = 1 - P(X \le 3)\)
Using the binomial CDF function on a GDC with \(n = 6\), \(p = 0.42795\), and \(x = 3\):
\(P(X \le 3) \approx 0.78000...\)
\(P(X \ge 4) = 1 - 0.78000... \approx 0.22000...\)
\(P(X \ge 4) = 0.220\) (to 3 s.f.)

(d) We want to find the conditional probability \(P(X = 6 \mid X \ge 4)\).
\(P(X = 6 \mid X \ge 4) = \frac{P(X = 6 \cap X \ge 4)}{P(X \ge 4)} = \frac{P(X = 6)}{P(X \ge 4)}\)
Using binomial PDF on a GDC:
\(P(X = 6) = (0.42795)^6 \approx 0.0061418...\)
\(P(X = 6 \mid X \ge 4) = \frac{0.0061418}{0.22000} \approx 0.027917...\)
\(P(X = 6 \mid X \ge 4) = 0.0279\) (to 3 s.f.)

**Part (a)**
* **M1**: Evidence of standardizing and setting up a probability equation (e.g., \(z = -1.282\) or \(1.645\)).
* **M1**: Another standardizing equation for the second percentage.
* **A1**: Both standard normal equations correct: \(\frac{385-\mu}{\sigma} = -1.28155...\) and \(\frac{415-\mu}{\sigma} = 1.64485...\) (or equivalent).
* **M1**: Evidence of attempting to solve the system of equations for \(\mu\) and \(\sigma\).
* **A1**: Correct values for both: \(\mu = 398\) (accept \(398.138...\)), \(\sigma = 10.3\) (accept \(10.2515...\)).
*Note: Accept answers that round to 3 s.f. directly from GDC equations.*

**Part (b)**
* **M1**: Evidence of normal CDF calculation on GDC (such as writing down the lower and upper bounds with their mean and standard deviation).
* **A1**: Correct probability: \(0.663\) (accept \(0.662\) to \(0.664\) depending on rounding of \(\mu\) and \(\sigma\)).

**Part (c)(i)**
* **M1**: Attempting to calculate \(P(W > 402)\) using their \(\mu\) and \(\sigma\).
* **A1**: Correct probability: \(0.428\) (accept \(0.42795...\)).

**Part (c)(ii)**
* **M1**: Recognizing binomial distribution with parameters \(n = 6\) and \(p = \text{their answer to (c)(i)}\).
* **M1**: Evidence of calculating \(P(X \ge 4)\), e.g., writing \(1 - P(X \le 3)\) or \(P(X=4) + P(X=5) + P(X=6)\).
* **A1**: Correct probability: \(0.220\) (accept \(0.22000...\)).

**Part (d)**
* **M1**: Writing down a correct expression for conditional probability, \(\frac{P(X = 6)}{P(X \ge 4)}\).
* **A1**: Correct final probability: \(0.0279\) (accept \(0.027917...\) or \(0.0280\) if using rounded 3 s.f. values).

**(a)**
(i) Substitute \(t = 0\) into the model: \(R(0) = 2 + 8 \sin(0) e^0 = 2\) thousand cubic meters per hour.

(ii) Substitute \(t = 2\) into the model: \(R(2) = 2 + 8 \sin\left(\frac{2\pi}{4}\right) e^{-0.12 \times 2} = 2 + 8(1)e^{-0.24} \approx 8.29\) thousand cubic meters per hour.

**(b)**
We set \(R(t) = 0\):
\(2 + 8 \sin\left(\frac{\pi t}{4}\right) e^{-0.12t} = 0\)
Using a graphic display calculator (GDC) to find the first positive root, we obtain:
\(t \approx 4.57\) hours (or \(4.569...\) hours).

**(c)**
Using the GDC maximum finder tool on the graph of \(R(t)\) for \(0 \le t \le 12\), we find a local maximum occurs at:
\(t \approx 1.81\) hours (or \(1.807...\) hours) with a maximum rate of flow of \(R(1.81) \approx 8.37\) thousand cubic meters per hour (or \(8.366...\)).

To justify that this is the global maximum on the interval \(0 \le t \le 12\), we compare it with the boundary values and other local extrema:
- At the boundaries: \(R(0) = 2\) and \(R(12) = 2 + 8 \sin(3\pi) e^{-1.44} = 2\).
- The next local maximum occurs when \(\frac{\pi t}{4} \approx 1.419 + 2\pi \implies t \approx 9.81\) hours, where \(R(9.81) \approx 4.44\).
Since \(8.37 > 4.44\) and \(8.37 > 2\), the local maximum at \(t \approx 1.81\) is the global maximum on the domain.

**(d)**
The volume of water at \(t = 8\) is given by:
\(V(8) = V(0) + \int_0^8 R(t) \, \mathrm{d}t\)
\(V(8) = 50 + \int_0^8 \left( 2 + 8 \sin\left(\frac{\pi t}{4}\right) e^{-0.12t} \right) \mathrm{d}t\)
Using the numerical integration feature of the GDC, we evaluate the definite integral:
\(\int_0^8 R(t) \, \mathrm{d}t \approx 22.143...\)
Therefore, \(V(8) \approx 50 + 22.143 = 72.1\) thousand cubic meters (or \(72143\text{ m}^3\)).

**(e)**
The average net rate of flow is given by:
\(\text{Average} = \frac{1}{8-0} \int_0^8 R(t) \, \mathrm{d}t\)
\(\text{Average} = \frac{22.143...}{8} \approx 2.77\) thousand cubic meters per hour.

**(a)**
(i) \(R(0) = 2\) *(A1)*
(ii) \(R(2) = 8.29\) *(A1)*
*[2 Marks]*

**(b)**
Setting \(R(t) = 0\) *(M1)*
Evidence of using GDC to solve (e.g. sketching or writing down equation) *(M1)*
\(t = 4.57\) (accept \(4.569...\)) *(A1)*
*[3 Marks]*

**(c)**
Method to find local maximum (e.g., setting \(R'(t) = 0\) or using GDC maximum finder) *(M1)*
\(t = 1.81\) (accept \(1.807...\)) *(A1)*
Maximum net rate of flow \(= 8.37\) (accept \(8.366...\)) *(A1)*
Comparing the local maximum value with boundaries and/or other local extrema to justify it is the global maximum *(R1)*
*[4 Marks]*

**(d)**
Recognizing that volume is the integral of the rate plus initial volume *(M1)*
\(V(8) = 50 + \int_0^8 R(t) \, \mathrm{d}t\) *(A1)*
Evaluating the integral to obtain \(\approx 22.143...\) (or \(22.1\)) *(A1)*
\(V(8) = 72.1\) (thousand cubic meters) (accept \(72.143...\) or \(72143\)) *(A1)*
*[4 Marks]*

**(e)**
Correct average value formula *(M1)*
\(\text{Average} = 2.77\) (thousand cubic meters per hour) (accept \(2.767...\)) *(A1)*
*[2 Marks]*

**(a)**
To find the local maximum, we set the first derivative of \(f(x)\) to zero:
\(f'(x) = 2\cos(x) - 0.5 = 0\)
\(\cos(x) = 0.25\)
For \(0 \le x \le 6\), the local maximum occurs at \(x \approx 1.31812... \approx 1.32\).
Substitute \(x\) back into \(f(x)\):
\(f(1.31812...) = 5 + 2\sin(1.31812...) - 0.5(1.31812...) \approx 6.2774...\)
So, the coordinates of the local maximum are \((1.32, 6.28)\).

**(b)**
The volume is given by the formula:
\(V = \pi \int_{0}^{6} [f(x)]^2 \, dx\)
\(V = \pi \int_{0}^{6} (5 + 2\sin(x) - 0.5x)^2 \, dx\)
Using a GDC to evaluate the definite integral:
\(V \approx 103.414... \times \pi \approx 324.881... \approx 325\text{ cm}^3\).

**(c) (i)**
First, we evaluate \(f(x)\) at the given \(x\) values:
- At \(x = 0\), \(f(0) = 5\).
- At \(x = 3\), \(f(3) = 5 + 2\sin(3) - 1.5 \approx 3.78224...\).
- At \(x = 6\), \(f(6) = 5 + 2\sin(6) - 3 \approx 1.44117...\).

Since the curve of \(g(x)\) passes through these points:
- \(g(0) = c = 5\)
- \(g(3) = 9a + 3b + 5 = 3.78224...\)
- \(g(6) = 36a + 6b + 5 = 1.44117...\)

Solving this linear system using a GDC or algebraic elimination:
\(9a + 3b = -1.21776... \implies 3a + b = -0.40592...\)
\(36a + 6b = -3.55883... \implies 6a + b = -0.59314...\)
Subtracting the first simplified equation from the second:
\(3a = -0.18722... \implies a \approx -0.062407... \approx -0.0624\)
Then, \(b = -0.40592... - 3(-0.062407...) \approx -0.21870... \approx -0.219\).
So, \(a \approx -0.0624\), \(b \approx -0.219\), \(c = 5\).

**(c) (ii)**
The volume is given by:
\(V = \pi \int_{0}^{6} (ax^2 + bx + c)^2 \, dx\)
Substituting the exact values of \(a, b, c\):
\(V = \pi \int_{0}^{6} (-0.062407...x^2 - 0.21870...x + 5)^2 \, dx\)
Using a GDC to evaluate this integral:
\(V \approx 84.0458... \times \pi \approx 264.04... \approx 264\text{ cm}^3\).

**(d) (i)**
We compare the predicted volumes with the actual volume of \(290\text{ cm}^3\):
- Error in Model 1: \(|324.88... - 290| \approx 34.9\text{ cm}^3\)
- Error in Model 2: \(|264.04... - 290| \approx 26.0\text{ cm}^3\)
Model 2 is more accurate because its predicted volume is closer to the actual volume of \(290\text{ cm}^3\).

**(d) (ii)**
The percentage error is:
\(\text{Percentage Error} = \frac{|\text{Predicted} - \text{Actual}|}{\text{Actual}} \times 100\%\)
\(\text{Percentage Error} = \frac{|264.04... - 290|}{290} \times 100\% \approx 8.95\%\) (accept \(8.96\%\) or \(8.97\%\) if using rounded values from previous steps).

**(a)**
- **(M1)** for attempting to find the local maximum (e.g. setting \(f'(x) = 0\) or finding the peak on a GDC sketch).
- **(A1)** for \((1.32, 6.28)\) (must have both coordinates correct to 3 sf).

**(b)**
- **(M1)** for setting up the volume of revolution integral \(\pi \int y^2 \, dx\).
- **(A1)** for the correct substitution of \(f(x)\) and limits: \(\pi \int_{0}^{6} (5 + 2\sin(x) - 0.5x)^2 \, dx\).
- **(A1)** for \(325\text{ cm}^3\) (or \(103\pi\)).

**(c) (i)**
- **(M1)** for finding the values of \(f(0)\), \(f(3)\), and \(f(6)\) (at least two correct).
- **(A1)** for \(c = 5\).
- **(M1)** for setting up a correct system of equations in terms of \(a\) and \(b\).
- **(A1)** for \(a \approx -0.0624\) and \(b \approx -0.219\).

**(c) (ii)**
- **(M1)** for setting up the volume integral with their quadratic expression: \(\pi \int_{0}^{6} (g(x))^2 \, dx\).
- **(A1)(FT)** for the correct substitution of their \(a, b, c\).
- **(A1)(FT)** for \(264\text{ cm}^3\) (or \(84.0\pi\)).

**(d) (i)**
- **(M1)** for finding the absolute difference from the actual volume for both models (\(34.9\) vs \(26.0\)).
- **(R1)** for concluding Model 2 is more accurate because \(26.0 < 34.9\).

**(d) (ii)**
- **(M1)** for setting up the percentage error formula.
- **(A1)** for substituting the correct values: \(\frac{|264.04... - 290|}{290} \times 100\%\).
- **(A1)(FT)** for \(8.95\%\) (accept \(8.96\%\) or \(8.97\%\). If Model 1 was chosen in (d)(i), award **(A1)(FT)** for \(12.0\%\)).