2024 Cambridge IGCSE International Mathematics (0607) 模擬試題連答案詳解

(a) Using a graphic display calculator (GDC) or algebraic differentiation: \(f'(x) = 3x^2 - 6x - 9\). Setting \(f'(x) = 0 \implies 3(x - 3)(x + 1) = 0\), which gives critical values at \(x = -1\) and \(x = 3\). Evaluating the function at these points: \(f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = 10\) and \(f(3) = (3)^3 - 3(3)^2 - 9(3) + 5 = -22\). Thus, the local maximum is at \((-1, 10)\) and the local minimum is at \((3, -22)\).
(b) The intersections with the \(x\)-axis occur where \(f(x) = 0\). Using the GDC's root-finding tool on \(x^3 - 3x^2 - 9x + 5 = 0\), we get three solutions: \(x \approx -1.91\), \(x \approx 0.483\), and \(x \approx 4.43\).
(c) The horizontal line \(y = k\) intersects the curve \(y = f(x)\) exactly three times when \(k\) lies strictly between the local minimum \(y\)-value and the local maximum \(y\)-value. Therefore, \(-22 < k < 10\).

(a) M1 for finding derivative or using GDC features. A1 for local maximum at \((-1, 10)\). A1 for local minimum at \((3, -22)\). B1 for correct distinction between max and min.
(b) B1 for each correct root: \(x \approx -1.91\), \(x \approx 0.483\), and \(x \approx 4.43\).
(c) M1 for identifying y-values of turning points. A2 for correct inequality \(-22 < k < 10\) (A1 if boundary values are correct but incorrect inequality sign used, e.g., \(-22 \le k \le 10\)).

(a) The area \(A\) is given by: \(A = \text{length} \times \text{width} = (2x + 5)(x - 1)\). Expanding the brackets: \(A = 2x(x) - 2x(1) + 5(x) - 5(1) = 2x^2 - 2x + 5x - 5 = 2x^2 + 3x - 5\).
(b)(i) Since \(A = 30\), we have: \(2x^2 + 3x - 5 = 30\). Subtracting 30 from both sides gives: \(2x^2 + 3x - 35 = 0\).
(ii) Factoring the quadratic: \(2x^2 + 10x - 7x - 35 = 0 \implies 2x(x + 5) - 7(x + 5) = 0 \implies (2x - 7)(x + 5) = 0\). This gives \(x = 3.5\) or \(x = -5\). The solution \(x = -5\) is not suitable because a length or width cannot be negative (e.g., width \(x - 1 = -6\) m is impossible). Hence, \(x = 3.5\).
(c) For \(x = 3.5\): Width \(= 3.5 - 1 = 2.5\) m, Length \(= 2(3.5) + 5 = 12\) m. Perimeter \(= 2 \times (\text{length} + \text{width}) = 2 \times (12 + 2.5) = 29\) m.

(a) M1 for writing area as product of the two linear expressions. A1 for correct expansion to get the given formula.
(b)(i) M1 for setting the area formula equal to 30. A1 for correct rearrangement to show \(2x^2 + 3x - 35 = 0\).
(ii) M1 for a valid method to solve the quadratic. A1 for finding both solutions \(x = 3.5\) and \(x = -5\). A1 for selecting \(x = 3.5\) and explaining why \(x = -5\) is rejected.
(c) M1 for calculating correct length (12) and width (2.5) using \(x = 3.5\). M1 for using the formula for perimeter of a rectangle. A1 for 29.

(a)(i) The graph of \(y = f(x) - 4\) is a translation of \(y = f(x)\) by 4 units down, represented by the vector \(\begin{pmatrix} 0 \\ -4 \end{pmatrix}\).
(ii) The graph of \(y = f(x - 3)\) is a translation of \(y = f(x)\) by 3 units right, represented by the vector \(\begin{pmatrix} 3 \\ 0 \end{pmatrix}\).
(iii) The graph of \(y = 3f(x)\) is a stretch with scale factor 3, parallel to the y-axis.
(b) Translating by vector \(\begin{pmatrix} -2 \\ 5 \end{pmatrix}\) replaces \(x\) with \(x + 2\) and adds 5 to the function. Thus: \(g(x) = f(x + 2) + 5 = \frac{1}{(x + 2)^2 + 1} + 5 = \frac{1}{x^2 + 4x + 4 + 1} + 5 = \frac{1}{x^2 + 4x + 5} + 5\). Putting this over a single denominator: \(g(x) = \frac{1 + 5(x^2 + 4x + 5)}{x^2 + 4x + 5} = \frac{5x^2 + 20x + 26}{x^2 + 4x + 5}\).

(a)(i) B1 for 'translation', B1 for vector \(\begin{pmatrix} 0 \\ -4 \end{pmatrix}\) or '4 units down'.
(ii) B1 for 'translation', B1 for vector \(\begin{pmatrix} 3 \\ 0 \end{pmatrix}\) or '3 units right'.
(iii) B1 for 'stretch', B1 for 'scale factor 3 parallel to the y-axis' (or vertical stretch scale factor 3).
(b) M1 for replacing \(x\) with \(x + 2\). M1 for adding 5. A1 for expanding the denominator \((x + 2)^2 + 1 = x^2 + 4x + 5\). A1 for correct simplified form, either \(\frac{1}{x^2+4x+5} + 5\) or \(\frac{5x^2+20x+26}{x^2+4x+5}\).

(a) On the first draw, \(P(R) = \frac{7}{12}\) and \(P(B) = \frac{5}{12}\). Since drawing is without replacement, there are 11 counters left for the second draw. If the first was red: \(P(R|R) = \frac{6}{11}\) and \(P(B|R) = \frac{5}{11}\). If the first was blue: \(P(R|B) = \frac{7}{11}\) and \(P(B|B) = \frac{4}{11}\).
(b)(i) \(P(R, R) = \frac{7}{12} \times \frac{6}{11} = \frac{42}{132} = \frac{7}{22} \approx 0.318\).
(ii) Different colours means either \((R, B)\) or \((B, R)\): \(P(\text{different}) = \left(\frac{7}{12} \times \frac{5}{11}\right) + \left(\frac{5}{12} \times \frac{7}{11}\right) = \frac{35}{132} + \frac{35}{132} = \frac{70}{132} = \frac{35}{66} \approx 0.530\).
(iii) At least one blue counter is the complement of getting no blue counters (both red): \(P(\text{at least one blue}) = 1 - P(R, R) = 1 - \frac{7}{22} = \frac{15}{22} \approx 0.682\).

(a) B1 for first branch probabilities correct. B2 for second branch probabilities correct (B1 if only one branch set is correct).
(b)(i) M1 for \(\frac{7}{12} \times \frac{6}{11}\). A1 for \(\frac{7}{22}\) or 0.318.
(ii) M1 for identifying \(P(R, B)\) and \(P(B, R)\). M1 for adding \(\frac{35}{132} + \frac{35}{132}\). A1 for \(\frac{35}{66}\) or 0.530.
(iii) M1 for \(1 - P(R, R)\) or summing three valid paths. A1 for \(\frac{15}{22}\) or 0.682.

(a) The total volume is the sum of the volume of the cylinder and the hemisphere: \(V = \pi r^2 h + \frac{2}{3}\pi r^3\).
(b)(i) The total height is \(h + r = 15\). Since \(r = 4\), \(h = 15 - 4 = 11\) cm.
(ii) Substituting \(r = 4\) and \(h = 11\): \(V = \pi (4)^2 (11) + \frac{2}{3}\pi (4)^3 = 176\pi + \frac{128}{3}\pi = \frac{528}{3}\pi + \frac{128}{3}\pi = \frac{656}{3}\pi\) \(\text{cm}^3\).
(iii) The total surface area consists of the curved surface of the hemisphere, the curved surface of the cylinder, and the single circular base at the bottom: \(A = 2\pi r^2 + 2\pi r h + \pi r^2 = 3\pi r^2 + 2\pi r h\). Substituting \(r = 4\) and \(h = 11\): \(A = 3\pi(4)^2 + 2\pi(4)(11) = 48\pi + 88\pi = 136\pi \approx 427.26\) \(\text{cm}^2\). To 3 s.f., this is 427 \(\text{cm}^2\).

(a) B1 for cylinder volume term \(\pi r^2 h\). B1 for hemisphere volume term \(\frac{2}{3}\pi r^3\).
(b)(i) B1 for indicating that the height of the cylinder is total height minus hemisphere radius.
(ii) M1 for substituting 4 and 11 into their volume expression. A1 for correct cylinder volume \(176\pi\) or hemisphere volume \(\frac{128}{3}\pi\). A1 for \(\frac{656}{3}\pi\) (or equivalent simplified fractional multiple of \(\pi\)).
(iii) M1 for identifying correct components of surface area: curved hemisphere \(2\pi r^2\), curved cylinder \(2\pi r h\), and one flat circular base \(\pi r^2\). M1 for substituting \(r = 4, h = 11\) into their surface area expression. A1 for \(136\pi\) or 427.26... A1 for 427 (accept 427 to 428).

(a) Gradient \(m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 2}{5 - (-3)} = \frac{6}{8} = \frac{3}{4}\).
(b) Using the point-slope form: \(y - 2 = \frac{3}{4}(x + 3)\). Multiply by 4: \(4y - 8 = 3x + 9 \implies 3x - 4y + 17 = 0\).
(c) Midpoint \(M = \left(\frac{-3 + 5}{2}, \frac{2 + 8}{2}\right) = (1, 5)\).
(d) The gradient of the perpendicular line is \(m_{\perp} = -\frac{1}{m_{AB}} = -\frac{4}{3}\). It passes through the midpoint \((1, 5)\). Using point-slope form: \(y - 5 = -\frac{4}{3}(x - 1) \implies y = -\frac{4}{3}x + \frac{4}{3} + 5 \implies y = -\frac{4}{3}x + \frac{19}{3}\).

(a) M1 for attempting to use the gradient formula. A1 for \(\frac{3}{4}\) (or 0.75).
(b) M1 for using their gradient and point in point-slope or slope-intercept form. M1 for multiplying to clear fractions. A1 for \(3x - 4y + 17 = 0\) (or \(-3x + 4y - 17 = 0\) or any integer multiple).
(c) M1 for attempting to find midpoint coordinates. A1 for \((1, 5)\).
(d) M1 for finding the perpendicular gradient: \(-\frac{4}{3}\). M1 for substituting their perpendicular gradient and midpoint into a line equation. A1 for \(y = -\frac{4}{3}x + \frac{19}{3}\) (accept decimal equivalent \(y = -1.33x + 6.33\) with coefficients to 3 s.f.).

Equating the two expressions for \(y\):
\(2x^2 - 3x - 4 = x + 2\)

Rearranging into standard quadratic form:
\(2x^2 - 4x - 6 = 0\)

Divide the entire equation by 2:
\(x^2 - 2x - 3 = 0\)

Factorise the quadratic expression:
\((x - 3)(x + 1) = 0\)

This gives the two \(x\)-values:
\(x = 3\) or \(x = -1\)

Now, find the corresponding \(y\)-values using \(y = x + 2\):
For \(x = 3\): \(y = 3 + 2 = 5\)
For \(x = -1\): \(y = -1 + 2 = 1\)

The solutions are \(x = 3, y = 5\) and \(x = -1, y = 1\).

M1: Equating the two equations to form an equation in one variable: \(2x^2 - 3x - 4 = x + 2\)
M1: Rearranging to get a correct 3-term quadratic equation, e.g., \(2x^2 - 4x - 6 = 0\) or \(x^2 - 2x - 3 = 0\)
M1: Correct method to solve their quadratic equation (by factorisation or formula)
A1: Correct \(x\) values: \(x = 3\) and \(x = -1\)
M1: Attempt to find corresponding \(y\) values by substituting at least one of their \(x\) values
A1: Correct matching pairs: \(x = 3, y = 5\) and \(x = -1, y = 1\)

Let the three consecutive odd integers be represented as \(2n - 1\), \(2n + 1\), and \(2n + 3\), where \(n\) is an integer.

Write the sum of their squares:
\(S = (2n - 1)^2 + (2n + 1)^2 + (2n + 3)^2\)

Expand each term:
\((2n - 1)^2 = 4n^2 - 4n + 1\)
\((2n + 1)^2 = 4n^2 + 4n + 1\)
\((2n + 3)^2 = 4n^2 + 12n + 9\)

Sum the expanded expressions:
\(S = (4n^2 - 4n + 1) + (4n^2 + 4n + 1) + (4n^2 + 12n + 9)\)
\(S = 12n^2 + 12n + 11\)

Factorise out 12 from the first two terms:
\(S = 12(n^2 + n) + 11\)

Since \(n\) is an integer, \(n^2 + n\) must also be an integer. Let \(k = n^2 + n\).

\(S = 12k + 11\), which is 11 more than a multiple of 12. This completes the proof.

M1: Define three consecutive odd integers algebraically (e.g., \(2n-1, 2n+1, 2n+3\))
M1: Express the sum of their squares mathematically
M2: Correctly expand all three squared binomials (1 mark for at least two correct, 2 marks for all three correct)
M1: Simplify the expression to a 3-term quadratic in terms of \(n\), such as \(12n^2 + 12n + 11\)
A1: Factorise the expression into the form \(12(n^2 + n) + 11\) and write a concluding statement

Let the length of the garden be \(x\) and the width be \(y\).

The perimeter is given by:
\(2x + 2y = 34 \implies x + y = 17\) (Equation 1)

The area is given by:
\(xy = 60\) (Equation 2)

From Equation 1, we express \(y\) in terms of \(x\):
\(y = 17 - x\)

Substitute this into Equation 2:
\(x(17 - x) = 60\)
\(17x - x^2 = 60\)

Rearrange into standard quadratic form:
\(x^2 - 17x + 60 = 0\)

Factorise the quadratic:
\((x - 12)(x - 5) = 0\)

So, \(x = 12\) or \(x = 5\).

If \(x = 12\), then \(y = 17 - 12 = 5\).

If \(x = 5\), then \(y = 17 - 5 = 12\).

Thus, the dimensions of the garden are \(5\text{ m}\) and \(12\text{ m}\).

M1: Write down correct equations for perimeter and area: \(2x + 2y = 34\) (or \(x + y = 17\)) and \(xy = 60\)
M1: Rearrange linear equation to make one variable the subject, e.g., \(y = 17 - x\)
M1: Substitute into the area equation to form a quadratic equation, e.g., \(x(17 - x) = 60\)
A1: Correct standard form quadratic equation: \(x^2 - 17x + 60 = 0\) (or equivalent)
M1: Solve the quadratic equation by factorisation or formula to find the values of one dimension (e.g., 5 or 12)
A1: Correct final dimensions: \(5\text{ m}\) and \(12\text{ m}\)

Let the two consecutive even integers be represented as \(2n\) and \(2n + 2\), where \(n\) is an integer.

The odd integer that lies directly between them is \(2n + 1\).

Write the expression for the difference of their squares (larger square minus smaller square):
\(D = (2n + 2)^2 - (2n)^2\)

Expand the squared terms:
\((2n + 2)^2 = 4n^2 + 8n + 4\)
\((2n)^2 = 4n^2\)

Subtract the two expressions:
\(D = (4n^2 + 8n + 4) - 4n^2\)
\(D = 8n + 4\)

Factorise the result to relate it to the odd intermediate integer:
\(D = 4(2n + 1)\)

Since \(2n + 1\) is the odd integer between \(2n\) and \(2n + 2\), the difference is indeed four times the odd integer between them. This completes the proof.

M1: Define two consecutive even integers algebraically, e.g., \(2n\) and \(2n + 2\)
M1: State that the odd integer between them is \(2n + 1\)
M1: Write the subtraction of the squares of the two even integers: \((2n + 2)^2 - (2n)^2\)
M1: Expand \((2n + 2)^2\) correctly to obtain \(4n^2 + 8n + 4\)
A1: Simplify the expression to \(8n + 4\)
A1: Factorise \(8n + 4\) to \(4(2n + 1)\) and conclude the proof

Let \(u = \frac{1}{x}\) and \(v = \frac{1}{y}\).

Rewrite the simultaneous equations in terms of \(u\) and \(v\):
\(4u + 3v = 1\) (Equation 1)
\(u - 2v = 14\) (Equation 2)

From Equation 2, express \(u\) in terms of \(v\):
\(u = 2v + 14\)

Substitute this expression for \(u\) into Equation 1:
\(4(2v + 14) + 3v = 1\)
\(8v + 56 + 3v = 1\)
\(11v + 56 = 1\)
\(11v = -55 \implies v = -5\)

Substitute \(v = -5\) back into the expression for \(u\):
\(u = 2(-5) + 14 = -10 + 14 = 4\)

Now find \(x\) and \(y\) using \(u = \frac{1}{x}\) and \(v = \frac{1}{y}\):
\(\frac{1}{x} = 4 \implies x = \frac{1}{4} = 0.25\)
\(\frac{1}{y} = -5 \implies y = -\frac{1}{5} = -0.2\)

The final solution is \(x = 0.25, y = -0.2\).

M1: Use substitution (e.g., \(u = 1/x, v = 1/y\)) or elimination method to eliminate one variable
M1: Form a correct linear equation in one variable, e.g., \(4(2v + 14) + 3v = 1\)
A1: Find the value of one of the intermediate variables: \(u = 4\) or \(v = -5\)
A1: Find the value of the other intermediate variable: the other of \(u = 4\) or \(v = -5\)
M1: Use the reciprocal relation to find \(x\) and \(y\) from their intermediate values
A1: Correct final values: \(x = 0.25\) (or \(1/4\)) and \(y = -0.2\) (or \(-1/5\))

1. **Find the area of the regular hexagonal base**: The area of a regular hexagon with side length \(s = 6\text{ cm}\) is given by:
\(A_{\text{base}} = \frac{3\sqrt{3}}{2} s^2 = \frac{3\sqrt{3}}{2} \times 36 = 54\sqrt{3} \approx 93.531\text{ cm}^2\).

2. **Find the lateral surface area**:
\(A_{\text{lateral}} = 210 - 54\sqrt{3} \approx 116.469\text{ cm}^2\).

3. **Find the slant height \(L\)**: The lateral area consists of 6 congruent triangles, each with base \(s = 6\text{ cm}\) and height \(L\):
\(6 \times \left(\frac{1}{2} \times 6 \times L\right) = 116.469 \implies 18 L = 116.469 \implies L \approx 6.4705\text{ cm}\).

4. **Find the height of the pyramid \(h\)**: Let \(M\) be the midpoint of one of the base sides. The apothem \(OM\) is the perpendicular distance from the center \(O\) to the side, which is:
\(OM = 6 \cos(30^\circ) = 3\sqrt{3} \approx 5.1962\text{ cm}\).
In the right-angled triangle \(VOM\):
\(h = \sqrt{L^2 - OM^2} = \sqrt{6.4705^2 - (3\sqrt{3})^2} = \sqrt{41.8676 - 27} = \sqrt{14.8676} \approx 3.8558\text{ cm}\).

5. **Find the angle \(\theta\) between the slant edge \(VA\) and the base**: In a regular hexagon, the distance from the center \(O\) to any vertex \(A\) is equal to the side length, so \(OA = 6\text{ cm}\). In the right-angled triangle \(VOA\):
\( \tan(\theta) = \frac{h}{OA} = \frac{3.8558}{6} \approx 0.64263\).
\(\theta = \arctan(0.64263) \approx 32.723^\circ\).
Rounding to 3 significant figures, we get \(32.7^\circ\).

M1 for expressing the base area as \(54\sqrt{3}\) or \(93.5\)
M1 for subtracting the base area from 210 to get the lateral area \(116.5\)
M1 for setting up the equation \(18L = 116.469\) to find the slant height \(L\)
A1 for finding \(L \approx 6.47\)
M1 for finding the apothem \(OM = 3\sqrt{3}\) or \(5.20\)
M1 for applying Pythagoras' theorem to find the pyramid height \(h\)
A1 for \(h \approx 3.86\)
M1 for identifying that the distance \(OA = 6\) and using \(\tan(\theta) = \frac{h}{OA}\)
M1 for applying the inverse tangent function
A1 for the final angle \(32.7^\circ\) (accept answers in the range 32.6 to 32.8)

1. **Find the area of the base triangle \(ABC\)**:
\(A_{\text{base}} = \frac{1}{2} a c \sin B = \frac{1}{2} \times 9 \times 7 \times \sin(64^\circ) = 31.5 \sin(64^\circ) \approx 28.312\text{ cm}^2\).

2. **Find the height of the pyramid \(VA\)**: Since \(V\) is vertically above \(A\), \(VA\) is perpendicular to the base, and its length is the height \(h\) of the pyramid.
\(V = \frac{1}{3} A_{\text{base}} \times h \implies 110 = \frac{1}{3} \times 28.312 \times h \implies h = \frac{330}{28.312} \approx 11.656\text{ cm}\).

3. **Find the length of the base line \(AC\)**: Use the cosine rule on triangle \(ABC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\)
\(AC^2 = 7^2 + 9^2 - 2(7)(9)\cos(64^\circ) = 49 + 81 - 126\cos(64^\circ) \approx 130 - 55.235 = 74.765\)
\(AC = \sqrt{74.765} \approx 8.6467\text{ cm}\).

4. **Calculate the angle between \(VC\) and the base**: Since \(VA\) is perpendicular to the base, the projection of \(VC\) onto the base is \(AC\). Therefore, the angle is \(\angle VCA\) in the right-angled triangle \(VAC\):
\(\tan(\angle VCA) = \frac{VA}{AC} = \frac{11.656}{8.6467} \approx 1.3480\).
\(\angle VCA = \arctan(1.3480) \approx 53.43^\circ\).
Rounding to 3 significant figures, the angle is \(53.4^\circ\).

M1 for the base area formula \(\frac{1}{2} \times 7 \times 9 \times \sin(64^\circ)\)
A1 for the base area \(28.3\) or \(28.312\)
M1 for the volume formula \(V = \frac{1}{3} A h\)
A1 for the height \(VA \approx 11.7\) or \(11.656\)
M1 for using the cosine rule: \(7^2 + 9^2 - 2(7)(9)\cos(64^\circ)\)
A1 for \(AC^2 \approx 74.8\)
A1 for \(AC \approx 8.65\)
M1 for recognizing the required angle is \(\angle VCA\) and using \(\tan(\theta) = \frac{VA}{AC}\)
M1 for inverse tangent calculation
A1 for the final answer \(53.4^\circ\) (accept answers in the range 53.3 to 53.5)

1. **Express the volume of the solid in terms of \(r\)**:
- Volume of the hemisphere: \(V_{\text{hemi}} = \frac{2}{3}\pi r^3\).
- Volume of the cone: \(V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 (2r) = \frac{2}{3}\pi r^3\).
- Total volume: \(V_{\text{total}} = \frac{2}{3}\pi r^3 + \frac{2}{3}\pi r^3 = \frac{4}{3}\pi r^3\).

2. **Solve for \(r\)**:
\(\frac{4}{3}\pi r^3 = 350 \implies r^3 = \frac{1050}{4\pi} \approx 83.556\)
\(r = (83.556)^{1/3} \approx 4.3718\text{ cm}\).

3. **Determine the slant height of the cone \(l\)**:
\(l = \sqrt{r^2 + h^2} = \sqrt{r^2 + (2r)^2} = \sqrt{5r^2} = r\sqrt{5}\)
\(l \approx 4.3718 \times 2.2361 \approx 9.7757\text{ cm}\).

4. **Calculate the total surface area**:
The total surface area consists of the curved surface area of the hemisphere and the curved surface area of the cone:
- Curved area of hemisphere: \(A_{\text{hemi}} = 2\pi r^2\).
- Curved area of cone: \(A_{\text{cone}} = \pi r l = \pi r (r\sqrt{5}) = \sqrt{5}\pi r^2\).
- Total area: \(A_{\text{total}} = (2 + \sqrt{5})\pi r^2\).
Using \(r^2 \approx 19.113\):
\(A_{\text{total}} \approx (2 + 2.2361) \times 3.1416 \times 19.113 \approx 4.2361 \times 60.046 \approx 254.35\text{ cm}^2\).
Rounding to 3 significant figures, we get \(254\text{ cm}^2\).

M1 for expressing volume of hemisphere as \(\frac{2}{3}\pi r^3\)
M1 for expressing volume of cone as \(\frac{2}{3}\pi r^3\) (using \(h=2r\))
A1 for total volume formula \(\frac{4}{3}\pi r^3 = 350\)
A1 for finding \(r \approx 4.37\) or \(r^3 \approx 83.6\)
M1 for the slant height formula \(l = \sqrt{r^2 + h^2}\)
A1 for expressing \(l = r\sqrt{5}\) or finding \(l \approx 9.78\)
M1 for formula for curved surface area of hemisphere \(2\pi r^2\)
M1 for formula for curved surface area of cone \(\pi r l\)
A1 for sum of areas expression \((2+\sqrt{5})\pi r^2\) or substituting values
A1 for the final answer \(254\) (accept 253 to 255)