An original Thinka practice paper modelled on the structure and difficulty of the Nov 2024 (V2) Cambridge International A Level International Mathematics (0607) paper. Not affiliated with or reproduced from Cambridge.
Paper 22 (Extended)
Answer all questions. Calculators must not be used.
18 Question · 39.519999999999996 marks
Question 1 · Short Answer
2.22 marks
Simplify \(\frac{4}{\sqrt{5} - 1}\) by rationalising the denominator.
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Worked solution
Multiply the numerator and denominator by the conjugate \(\sqrt{5} + 1\): \(\frac{4(\sqrt{5} + 1)}{(\sqrt{5} - 1)(\sqrt{5} + 1)} = \frac{4(\sqrt{5} + 1)}{5 - 1} = \frac{4(\sqrt{5} + 1)}{4} = \sqrt{5} + 1\).
Marking scheme
M1 for multiplying numerator and denominator by \(\sqrt{5} + 1\) and A1 for correct simplified answer \(\sqrt{5} + 1\)
Question 2 · Short Answer
2.22 marks
Express as a single fraction in its simplest form: \(\frac{3}{x-2} - \frac{2}{x+3}\)
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Worked solution
Find a common denominator: \(\frac{3(x+3) - 2(x-2)}{(x-2)(x+3)} = \frac{3x + 9 - 2x + 4}{(x-2)(x+3)} = \frac{x + 13}{(x-2)(x+3)}\).
Marking scheme
M1 for expressing over a common denominator with correct expansion of terms and A1 for \(\frac{x+13}{(x-2)(x+3)}\)
Question 3 · Short Answer
2.22 marks
Solve the equation: \(\log_2(x) + \log_2(x-3) = 2\)
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Worked solution
Using the laws of logarithms: \(\log_2(x(x-3)) = 2\) which gives \(x(x-3) = 2^2\), so \(x^2 - 3x - 4 = 0\). Factoring this gives \((x-4)(x+1) = 0\), so \(x = 4\) or \(x = -1\). Since we cannot take the logarithm of a negative number, \(x = 4\) is the only valid solution.
Marking scheme
M1 for combining logarithms to get \(x(x-3) = 4\) and A1 for solving the quadratic to find \(x = 4\)
Question 4 · Short Answer
2.22 marks
Find the \(n\)-th term of the sequence: 5, 12, 23, 38, 57, ...
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Worked solution
Find the differences between consecutive terms. First differences: 7, 11, 15, 19. Second differences: 4, 4, 4. Since the second differences are constant, the sequence is quadratic with coefficient of \(n^2\) as \(4/2 = 2\). Subtracting \(2n^2\) from the terms leaves 3, 4, 5, 6, which has the linear form \(n + 2\). Thus, the \(n\)-th term is \(2n^2 + n + 2\).
Marking scheme
M1 for identifying second differences are constant and coefficient of \(n^2\) is 2, M1 for finding the linear sequence \(n + 2\), and A1 for \(2n^2 + n + 2\)
Question 5 · Short Answer
2.22 marks
In a right-angled triangle, the hypotenuse is \(10\text{ cm}\) and one of the acute angles is \(30^\circ\). Find the length of the side opposite to this angle, in cm.
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Worked solution
Using the sine ratio: \(\sin(30^\circ) = \frac{\text{Opposite}}{\text{Hypotenuse}}\). Thus, \(\text{Opposite} = 10 \times \sin(30^\circ)\). Since \(\sin(30^\circ) = 0.5\), the opposite side is \(10 \times 0.5 = 5\text{ cm}\).
Marking scheme
M1 for setting up the equation \(\sin(30^\circ) = \frac{x}{10}\) and A1 for 5
Question 6 · Short Answer
2.22 marks
Solve the inequality: \(x^2 - 3x - 10 < 0\)
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Worked solution
First, solve the equation \(x^2 - 3x - 10 = 0\) to find the critical values: \((x-5)(x+2) = 0\) gives \(x = 5\) and \(x = -2\). Since the inequality is less than zero, the solution is the interval between these roots, which is \(-2 < x < 5\).
Marking scheme
M1 for finding critical values \(x = 5\) and \(x = -2\) and A1 for writing the correct interval inequality \(-2 < x < 5\)
Question 7 · Short Answer
2.22 marks
Calculate \((3 \times 10^5) \times (4 \times 10^{-8})\), giving your answer in standard form.
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Worked solution
Multiply the coefficients and the powers of 10: \((3 \times 4) \times (10^5 \times 10^{-8}) = 12 \times 10^{-3}\). In standard form, this is adjusted to \(1.2 \times 10^{-2}\).
Marking scheme
M1 for calculating \(12 \times 10^{-3}\) or \(0.012\) and A1 for writing in correct standard form \(1.2 \times 10^{-2}\)
Question 8 · Short Answer
2.22 marks
Points \(A\), \(B\), and \(C\) lie on a circle with centre \(O\). \(AC\) is a diameter of the circle. Point \(B\) lies on the circumference such that angle \(BAC = 35^\circ\). Find the size of angle \(BCA\).
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Worked solution
Since \(AC\) is a diameter of the circle, the angle subtended by the diameter at the circumference is a right angle, meaning angle \(ABC = 90^\circ\). In triangle \(ABC\), the angles sum to \(180^\circ\), so angle \(BCA = 180^\circ - 90^\circ - 35^\circ = 55^\circ\).
Marking scheme
M1 for identifying that angle \(ABC = 90^\circ\) and A1 for \(55^\circ\) or 55
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Worked solution
First, use the logarithmic product rule to combine the terms: \(\log_2((x+3)(x-3)) = 4\). This simplifies to \(\log_2(x^2 - 9) = 4\). Convert the logarithmic equation into its exponential form: \(x^2 - 9 = 2^4\), which gives \(x^2 - 9 = 16\). Solving for \(x^2\) gives \(x^2 = 25\). This yields two potential solutions, \(x = 5\) or \(x = -5\). Since the argument of a logarithm must be strictly positive, \(x + 3 > 0\) and \(x - 3 > 0\), which means \(x > 3\). Therefore, we reject \(x = -5\) and the only valid solution is \(x = 5\).
Marking scheme
M1 for applying the logarithmic addition law to get \(\log_2(x^2 - 9) = 4\). M1 for converting to exponential form \(x^2 - 9 = 16\). A1 for the final answer of 5, explicitly or implicitly rejecting -5.
Question 10 · Short Answer
2.22 marks
Express \(\frac{6}{\sqrt{5} - \sqrt{2}}\) in the form \(a\sqrt{5} + b\sqrt{2}\), where \(a\) and \(b\) are integers.
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Worked solution
To rationalise the denominator, multiply the numerator and the denominator by the conjugate of the denominator, which is \(\sqrt{5} + \sqrt{2}\). This gives: \(\frac{6(\sqrt{5} + \sqrt{2})}{(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})} = \frac{6(\sqrt{5} + \sqrt{2})}{5 - 2} = \frac{6(\sqrt{5} + \sqrt{2})}{3} = 2(\sqrt{5} + \sqrt{2}) = 2\sqrt{5} + 2\sqrt{2}\).
Marking scheme
M1 for multiplying both numerator and denominator by the conjugate \(\sqrt{5} + \sqrt{2}\). M1 for correctly simplifying the denominator to 3. A1 for obtaining \(2\sqrt{5} + 2\sqrt{2}\) or equivalent.
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Worked solution
First, factorise the numerator: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\). Next, factorise the denominator as a difference of two squares: \(x^2 - 9 = (x - 3)(x + 3)\). Substitute these factored forms back into the fraction: \(\frac{(2x + 1)(x - 3)}{(x - 3)(x + 3)}\). Cancel out the common factor of \((x - 3)\) to get the simplified fraction: \(\frac{2x + 1}{x + 3}\).
Marking scheme
M1 for factorising the quadratic numerator into \((2x + 1)(x - 3)\). M1 for factorising the denominator into \((x - 3)(x + 3)\). A1 for the fully simplified fraction \(\frac{2x + 1}{x + 3}\).
Question 12 · Short Answer
2.22 marks
The graph of \(y = f(x)\) is mapped onto the graph of \(y = g(x)\) by a stretch, parallel to the \(y\)-axis, with scale factor 3, followed by a translation of \(\begin{pmatrix} 0 \\ -5 \end{pmatrix}\). Write down an expression for \(g(x)\) in terms of \(f(x)\).
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Worked solution
A stretch parallel to the \(y\)-axis with scale factor 3 transforms the function \(f(x)\) into \(3f(x)\). A subsequent translation represented by the vector \(\begin{pmatrix} 0 \\ -5 \end{pmatrix}\) shifts the graph vertically downwards by 5 units, resulting in \(3f(x) - 5\). Therefore, \(g(x) = 3f(x) - 5\).
Marking scheme
M1 for identifying the stretch as \(3f(x)\). A1 for completing the transformation to get \(3f(x) - 5\).
Question 13 · Short Answer
2.22 marks
Find the value of \(\cos 120^\circ \times \sin 300^\circ\).
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Worked solution
First, find the exact values of each trigonometric function. For \(\cos 120^\circ\), the reference angle is \(180^\circ - 120^\circ = 60^\circ\). Since \(120^\circ\) is in the second quadrant where cosine is negative, \(\cos 120^\circ = -\cos 60^\circ = -\frac{1}{2}\). For \(\sin 300^\circ\), the reference angle is \(360^\circ - 300^\circ = 60^\circ\). Since \(300^\circ\) is in the fourth quadrant where sine is negative, \(\sin 300^\circ = -\sin 60^\circ = -\frac{\sqrt{3}}{2}\). Multiply these values together: \((-\frac{1}{2}) \times (-\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{4}\).
Marking scheme
M1 for finding \(\cos 120^\circ = -1/2\) or \(\sin 300^\circ = -\sqrt{3}/2\). A1 for the final simplified exact product of \(\frac{\sqrt{3}}{4}\).
Question 14 · Short Answer
2.22 marks
Find the \(n\)-th term of the sequence: \(2, 9, 20, 35, 54, \dots\)
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Worked solution
Find the first and second differences of the sequence. First differences: \(9-2=7\), \(20-9=11\), \(35-20=15\), \(54-35=19\). Second differences: \(11-7=4\), \(15-11=4\), \(19-15=4\). Since the second difference is constant and equals 4, the sequence has a quadratic term of \(an^2\) where \(a = 4 / 2 = 2\). Subtract \(2n^2\) from the original terms: for \(n=1: 2-2(1)=0\); for \(n=2: 9-2(4)=1\); for \(n=3: 20-2(9)=2\); for \(n=4: 35-2(16)=3\). The resulting sequence is \(0, 1, 2, 3, \dots\), which has the linear term \(n-1\). Combining these, the \(n\)-th term is \(2n^2 + n - 1\).
Marking scheme
M1 for finding constant second differences of 4. M1 for identifying the linear adjustment \(n-1\) after subtracting \(2n^2\). A1 for \(2n^2 + n - 1\).
Question 15 · Short Answer
2.22 marks
y is inversely proportional to the square of x. When x = 3, y = 4. Find y when x = 6.
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Worked solution
We can write the relationship as \(y = \frac{k}{x^2}\), where \(k\) is the constant of proportionality. Substitute the given values to find \(k\): \(4 = \frac{k}{3^2}\), which gives \(4 = \frac{k}{9}\), so \(k = 36\). Our equation is \(y = \frac{36}{x^2}\). Substitute \(x = 6\) into this equation: \(y = \frac{36}{6^2} = \frac{36}{36} = 1\).
Marking scheme
M1 for setting up the proportion equation \(y = \frac{k}{x^2}\) and solving for \(k = 36\). A1 for obtaining \(y = 1\).
Question 16 · Short Answer
2.22 marks
Work out \((3 \times 10^7) \times (8 \times 10^{-4})\), giving your answer in standard form.
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Worked solution
Multiply the coefficients and the powers of 10 separately: \((3 \times 8) \times (10^7 \times 10^{-4}) = 24 \times 10^{7 + (-4)} = 24 \times 10^3\). To write this in standard scientific notation, express \(24\) as \(2.4 \times 10^1\). Thus, the expression becomes \(2.4 \times 10^1 \times 10^3 = 2.4 \times 10^4\).
Marking scheme
M1 for calculating \(24 \times 10^3\) or \(24000\). A1 for writing the final answer correctly in standard form as \(2.4 \times 10^4\).
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Worked solution
First, apply the addition law of logarithms to combine the terms on the left-hand side: \(\log_{3}((x + 1)(x - 5)) = 3\). Next, convert the logarithmic equation to its exponential form: \((x + 1)(x - 5) = 3^3\). Expanding the left side gives \(x^2 - 4x - 5 = 27\). Rearranging this into standard quadratic form yields \(x^2 - 4x - 32 = 0\). Factoring the quadratic expression gives \((x - 8)(x + 4) = 0\), which has solutions \(x = 8\) and \(x = -4\). Because the terms in the original logarithmic equation require the arguments to be positive (i.e., \(x + 1 > 0\) and \(x - 5 > 0\), meaning \(x > 5\)), the solution \(x = -4\) is invalid. Thus, the only valid solution is \(x = 8\).
Marking scheme
M1 for applying logarithm laws to obtain \((x + 1)(x - 5) = 3^3\) or simplifying to the quadratic equation \(x^2 - 4x - 32 = 0\) A1 for the final answer 8 (with the negative root rejected)
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Worked solution
To simplify the expression, first rationalize the denominator of the fraction by multiplying both the numerator and denominator by the conjugate of the denominator, which is \(3 + \sqrt{3}\): \(\frac{12}{3 - \sqrt{3}} \times \frac{3 + \sqrt{3}}{3 + \sqrt{3}} = \frac{12(3 + \sqrt{3})}{3^2 - (\sqrt{3})^2}\). This simplifies to \ rac{12(3 + \sqrt{3})}{9 - 3} = \frac{12(3 + \sqrt{3})}{6}\). Dividing the numerator by the denominator gives \(2(3 + \sqrt{3}) = 6 + 2\sqrt{3}\). Finally, subtract the remaining term in the original expression: \(6 + 2\sqrt{3} - 2\sqrt{3} = 6\).
Marking scheme
M1 for rationalizing the denominator of the fraction, showing the step \(\frac{12(3 + \sqrt{3})}{9 - 3}\) or obtaining \(6 + 2\sqrt{3}\) A1 for the final simplified integer answer 6
Paper 42 (Extended)
Answer all questions. Graphic display calculators should be used where appropriate.
(a) Find an expression, in terms of \(n\), for the \(n\)-th term of: (i) Sequence A (ii) Sequence B (iii) Sequence C
(b) The \(k\)-th term of Sequence B is 819. Find the value of \(k\).
(c) Find the term in Sequence C that has a value of 12288.
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Worked solution
(a)(i) Sequence A is an arithmetic sequence with a common difference of \(6\). The first term is \(5\). Thus, the \(n\)-th term is \(5 + (n-1)6 = 6n - 1\). (ii) Sequence B has first differences of \(7, 11, 15, 19, \dots\) and second differences of \(4, 4, 4, \dots\). Since the second difference is constant, the sequence is quadratic of the form \(an^2 + bn + c\), where \(a = 4 / 2 = 2\). Subtracting \(2n^2\) from the terms yields the sequence \(0, 1, 2, 3, 4, \dots\), which has the \(n\)-th term of \(n-1\). Therefore, the overall \(n\)-th term is \(2n^2 + n - 1\). (iii) Sequence C is a geometric sequence with first term \(a = 3\) and common ratio \(r = 2\). The \(n\)-th term is \(3 \times 2^{n-1}\).
(b) Set the expression for Sequence B equal to 819: \(2k^2 + k - 1 = 819 \implies 2k^2 + k - 820 = 0\). Factorizing this quadratic equation gives \((2k + 41)(k - 20) = 0\). Since \(k\) must be a positive integer, \(k = 20\).
(c) Set the expression for Sequence C equal to 12288: \(3 \times 2^{m-1} = 12288 \implies 2^{m-1} = 4096\). Since \(2^{12} = 4096\), we have \(m - 1 = 12 \implies m = 13\). Thus, it is the 13th term.
Marking scheme
Part (a): - (i) [2 marks]: M1 for recognizing common difference is 6 or seeing \(6n + c\), A1 for \(6n - 1\). - (ii) [3 marks]: M1 for finding second differences are 4, M1 for setting up \(2n^2 + bn + c\), A1 for \(2n^2 + n - 1\). - (iii) [2 marks]: M1 for identifying geometric sequence with ratio 2, A1 for \(3 \times 2^{n-1}\).
Part (b) [2 marks]: - M1 for setting up \(2k^2 + k - 820 = 0\) and attempting to solve. - A1 for \(k = 20\).
Part (c) [2 marks]: - M1 for setting up \(2^{m-1} = 4096\). - A1 for 13 or 13th term.
Question 2 · Structured
11 marks
A solid toy is made from a hemisphere of radius \(r\) cm surmounting a cone of base radius \(r\) cm and height \(h\) cm. The total volume of the toy is \(360\pi\) cm\(^3\). The height of the cone, \(h\), is equal to three times its radius, \(r\).
(a) Show that \(r = 6\).
(b) Calculate the total surface area of the toy. Give your answer correct to 1 decimal place.
(c) The toy is made of wood with a density of \(0.85\) g/cm\(^3\). Calculate the mass of the toy, giving your answer to the nearest gram.
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Worked solution
(a) The volume of a hemisphere is \(V_{\text{hemi}} = \frac{2}{3}\pi r^3\). The volume of a cone is \(V_{\text{cone}} = \frac{1}{3}\pi r^2 h\). Since \(h = 3r\), we substitute to get \(V_{\text{cone}} = \frac{1}{3}\pi r^2 (3r) = \pi r^3\). The total volume is: \(V_{\text{total}} = \frac{2}{3}\pi r^3 + \pi r^3 = \frac{5}{3}\pi r^3\). We are given that \(V_{\text{total}} = 360\pi\), so: \[ \frac{5}{3}\pi r^3 = 360\pi \implies \frac{5}{3}r^3 = 360 \implies r^3 = 216 \implies r = 6 \text{ cm}. \\]
(b) The total surface area consists of the curved surface area of the hemisphere and the curved surface area of the cone. Hemisphere CSA: \(2\pi r^2 = 2\pi (6^2) = 72\pi\) cm\(^2\). Cone CSA: \(\pi r l\), where \(l\) is the slant height \(l = \sqrt{r^2 + h^2}\). Since \(h = 3(6) = 18\), \(l = \sqrt{6^2 + 18^2} = \sqrt{360} = 6\sqrt{10}\) cm. Cone CSA: \(\pi (6)(6\sqrt{10}) = 36\sqrt{10}\pi\) cm\(^2\). Total Surface Area = \(72\pi + 36\sqrt{10}\pi \approx 226.195 + 357.643 = 583.838\) cm\(^2\). To 1 decimal place, this is \(583.8\) cm\(^2\).
(c) The total volume is \(360\pi \approx 1130.973\) cm\(^3\). Mass = Volume \(\times\) Density = \(1130.973 \times 0.85 = 961.327\) g. To the nearest gram, the mass is \(961\) g.
Marking scheme
Part (a) [4 marks]: - M1 for writing volume of hemisphere as \(\frac{2}{3}\pi r^3\). - M1 for substituting \(h = 3r\) into cone volume to get \(\pi r^3\). - M1 for equating the sum to \(360\pi\) and solving for \(r^3\). - A1 for showing clearly that \(r^3 = 216\) leads to \(r = 6\).
Part (b) [5 marks]: - M1 for calculating the slant height of the cone, \(l = \sqrt{6^2 + 18^2} = \sqrt{360}\). - M1 for formula of hemisphere CSA (\(2\pi r^2\)) and evaluating as \(72\pi\) or \(226.2\). - M1 for formula of cone curved surface area (\(\pi r l\)) and evaluating as \(36\sqrt{10}\pi\) or \(357.6\). - M1 for adding both areas. - A1 for \(583.8\) (accept \(584\)).
Part (c) [2 marks]: - M1 for multiplying their volume by \(0.85\). - A1 for \(961\) (accept \(961.3\)).
Question 3 · Structured
11 marks
Consider the function \(f(x) = x + \frac{4}{x-1}\) for \(x \neq 1\).
(a) Write down the equation of the vertical asymptote of the graph of \(y = f(x)\).
(b) Find the coordinates of the point where the graph of \(y = f(x)\) crosses the \(y\)-axis.
(c) Find the coordinates of: (i) the local minimum point (ii) the local maximum point
(d) Write down the set of values of \(k\) for which the equation \(f(x) = k\) has no real solutions.
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Worked solution
(a) The vertical asymptote occurs where the denominator of the fraction is zero, which is \(x - 1 = 0 \implies x = 1\).
(b) The graph crosses the \(y\)-axis when \(x = 0\): \(f(0) = 0 + \frac{4}{0 - 1} = -4\). So the coordinates are \((0, -4)\).
(c)(i) To find the turning points, we take the derivative of \(f(x) = x + 4(x-1)^{-1}\): \(f'(x) = 1 - \frac{4}{(x-1)^2}\). Setting \(f'(x) = 0\) to find the stationary points: \(1 = \frac{4}{(x-1)^2} \implies (x-1)^2 = 4 \implies x - 1 = \pm 2\). Thus, \(x = 3\) or \(x = -1\). For \(x = 3\): \(f(3) = 3 + \frac{4}{3-1} = 5\). This is the local minimum point because for \(x > 1\) the function opens upwards. So the local minimum is \((3, 5)\).
(ii) For \(x = -1\): \(f(-1) = -1 + \frac{4}{-1-1} = -3\). This is the local maximum point: \((-1, -3)\).
(d) From the coordinates of the local minimum \((3, 5)\) and local maximum \((-1, -3)\), the graph consists of two branches. One branch goes from \(-\infty\) to a maximum of \(-3\), and the other branch goes from a minimum of \(5\) to \(\infty\). Therefore, there are no real solutions to \(f(x) = k\) when \(k\) is strictly between the local maximum value and local minimum value, which is \(-3 < k < 5\).
Marking scheme
Part (a) [1 mark]: - A1 for \(x = 1\).
Part (b) [2 marks]: - M1 for substituting \(x = 0\) into the function. - A1 for \((0, -4)\).
Part (c)(i) [3 marks]: - M1 for differentiating \(f(x)\) correctly to get \(1 - \frac{4}{(x-1)^2}\). - M1 for setting their derivative to 0 and finding \(x = 3\). - A1 for \((3, 5)\).
Part (c)(ii) [3 marks]: - M1 for finding the other root \(x = -1\). - A1 for \(y = -3\). - A1 for \((-1, -3)\).
Part (d) [2 marks]: - M1 for recognizing the range is between the y-values of the turning points. - A1 for \(-3 < k < 5\) (accept equivalent inequality or interval notation).
Question 4 · Structured
11 marks
Three ports, \(A\), \(B\), and \(C\), are situated such that \(B\) is \(15\) km from \(A\) on a bearing of \(065^\circ\). \(C\) is \(22\) km from \(B\) on a bearing of \(145^\circ\).
(a) Find the distance \(AC\).
(b) Find the bearing of \(C\) from \(A\), correct to the nearest degree.
(c) Calculate the area of triangle \(ABC\).
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Worked solution
(a) Let \(N_A\) be North at \(A\) and \(N_B\) be North at \(B\). The bearing of \(B\) from \(A\) is \(065^\circ\), so \(\angle N_A AB = 65^\circ\). Since the North lines are parallel, the co-interior angle \(\angle A B N_B = 180^\circ - 65^\circ = 115^\circ\). The bearing of \(C\) from \(B\) is \(145^\circ\), which is \(\angle N_B B C = 145^\circ\). Therefore, the interior angle of the triangle at \(B\) is: \(\angle ABC = 360^\circ - (115^\circ + 145^\circ) = 360^\circ - 260^\circ = 100^\circ\). Using the Cosine Rule on triangle \(ABC\) to find \(AC\): \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\) \(AC^2 = 15^2 + 22^2 - 2(15)(22)\cos(100^\circ)\) \(AC^2 = 225 + 484 - 660(-0.173648)\) \(AC^2 = 709 + 114.607 = 823.607\) \(AC = \sqrt{823.607} \approx 28.698\) km. To 3 significant figures, \(AC = 28.7\) km.
(b) To find the bearing of \(C\) from \(A\), we first need to find the angle \(\angle BAC\). Using the Sine Rule: \(\frac{\sin(\angle BAC)}{22} = \frac{\sin(100^\circ)}{28.698}\) \(\sin(\angle BAC) = \frac{22 \sin(100^\circ)}{28.698} \approx 0.75494\) \(\angle BAC = \arcsin(0.75494) \approx 49.02^\circ\). The bearing of \(C\) from \(A\) is the bearing of \(B\) from \(A\) plus \(\angle BAC\): \(\text{Bearing} = 65^\circ + 49.02^\circ = 114.02^\circ\). To the nearest degree, the bearing is \(114^\circ\).
(c) The area of triangle \(ABC\) is given by: \(\text{Area} = \frac{1}{2} \times AB \times BC \times \sin(\angle ABC)\) \(\text{Area} = \frac{1}{2} \times 15 \times 22 \times \sin(100^\circ) = 165 \sin(100^\circ) \approx 162.49\) km\(^2\). To 3 significant figures, the area is \(162\) km\(^2\).
Marking scheme
Part (a) [4 marks]: - M1 for finding the angle \(\angle ABC = 100^\circ\) (with working showing co-interior angles). - M1 for writing the correct Cosine Rule formula with values substituted. - A1 for \(AC^2 \approx 823.6\). - A1 for \(28.7\) (or \(28.69\) to \(28.70\)).
Part (b) [4 marks]: - M1 for setting up the correct Sine Rule ratio to find \(\angle BAC\). - A1 for \(\angle BAC \approx 49.0^\circ\). - M1 for adding their angle to \(65^\circ\). - A1 for \(114^\circ\) (accept \(114.0^\circ\)).
Part (c) [3 marks]: - M1 for area formula \(\frac{1}{2}ab\sin(C)\) with values. - A1 for \(165\sin(100)\). - A1 for \(162\) (accept \(162.5\)).
Question 5 · Structured
11 marks
An online retailer sells wireless headphones. In one week, they sell \(x\) headphones for a total revenue of \(\$5500\).
(a) Write down an expression in terms of \(x\) for the selling price of one headphone.
(b) In the following week, they reduce the selling price of each headphone by \(\$10\). As a result, they sell \(5\) more headphones than the previous week, and the total revenue is still \(\$5500\). Form an equation in \(x\) and show that it simplifies to \(x^2 + 5x - 2750 = 0\).
(c) Solve the equation \(x^2 + 5x - 2750 = 0\) to find the number of headphones sold in the first week.
(d) Find the reduced selling price of a headphone.
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Worked solution
(a) The selling price of one headphone is \(\frac{5500}{x}\).
(b) In the second week, the selling price of each headphone is \(\frac{5500}{x} - 10\). The number of headphones sold is \(x + 5\). The total revenue is still \(\$5500\), so we can write: \((x + 5)\left(\frac{5500}{x} - 10\right) = 5500\). Expanding the brackets: \(5500 - 10x + \frac{27500}{x} - 50 = 5500\). Subtracting 5500 from both sides: \(-10x - 50 + \frac{27500}{x} = 0\). Divide the entire equation by \(-10\): \(x + 5 - \frac{2750}{x} = 0\). Multiplying by \(x\) to clear the fraction: \(x^2 + 5x - 2750 = 0\). (As required)
(c) To solve \(x^2 + 5x - 2750 = 0\), we can factorize the quadratic equation: We need two numbers that multiply to \(-2750\) and add up to \(5\). These numbers are \(55\) and \(-50\). \((x + 55)(x - 50) = 0\). Thus, \(x = -55\) or \(x = 50\). Since the number of headphones sold must be positive, \(x = 50\).
(d) The original selling price is \(\frac{5500}{50} = \$110\). The reduced selling price of a headphone is \(110 - 10 = \$100\).
Marking scheme
Part (a) [1 mark]: - A1 for \(\frac{5500}{x}\).
Part (b) [4 marks]: - M1 for writing the new price as \(\frac{5500}{x} - 10\) and new quantity as \(x + 5\). - M1 for setting up the product \((x+5)(\frac{5500}{x} - 10) = 5500\). - M1 for expanding the brackets correctly. - A1 for executing algebraic steps systematically to arrive at \(x^2 + 5x - 2750 = 0\).
Part (c) [3 marks]: - M1 for attempting to factorize or using quadratic formula. - A1 for \(x = 50\) and \(x = -55\). - A1 for choosing \(x = 50\) as the final answer.
Part (d) [3 marks]: - M1 for calculating original price \(5500 / 50\). - M1 for subtracting 10. - A1 for \(\$100\).
(b) Solve the quadratic equation \(2x^2 - 5x - 12 = 0\) to find critical values: \(2x^2 - 5x - 12 = 0 \implies (2x + 3)(x - 4) = 0\). The critical values are \(x = -1.5\) and \(x = 4\). Since the inequality is \(> 0\), we look for the regions outside of these roots. Thus, \(x < -1.5\) or \(x > 4\).
(c) Solve the first inequality: \(2y - 3 < 11 \implies 2y < 14 \implies y < 7\).
Solve the second inequality: \(y^2 - 4y - 5 \le 0 \implies (y - 5)(y + 1) \le 0\). The solution to this quadratic inequality is \(-1 \le y \le 5\).
We need the values of \(y\) that satisfy both conditions: \(y < 7\) and \(-1 \le y \le 5\). The intersection is \(-1 \le y \le 5\). Since \(y\) must be an integer, the values are: \(-1, 0, 1, 2, 3, 4, 5\).
Marking scheme
Part (a) [3 marks]: - M1 for expanding LHS to \(6x - 15\). - M1 for isolating terms in \(x\) on one side. - A1 for \(x \le 11\).
Part (b) [4 marks]: - M1 for attempting to factorize \(2x^2 - 5x - 12 = 0\) or use the quadratic formula. - A1 for critical values \(-1.5\) and \(4\). - M1 for identifying the correct regions (outside the interval). - A1 for \(x < -1.5\) or \(x > 4\).
Part (c) [4 marks]: - M1 for solving \(2y - 3 < 11\) to get \(y < 7\). - M1 for solving \(y^2 - 4y - 5 \le 0\) to get \(-1 \le y \le 5\). - M1 for intersecting both intervals to find \(-1 \le y \le 5\). - A1 for listing the integers \(-1, 0, 1, 2, 3, 4, 5\).
Question 7 · Structured
11 marks
Let \(f(x) = 3x - 5\), \(g(x) = \frac{2}{x+1}\) for \(x \neq -1\), and \(h(x) = x^2 - 2\).
(a) Find: (i) \(f(g(3))\) (ii) \(h(f(x))\) in its simplest expanded form.
(b) Find \(g^{-1}(x)\).
(c) Solve the equation \(f(x) = h(x) - 5\). Give your answers correct to 3 significant figures.
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Part (a): - (i) [2 marks]: M1 for finding \(g(3) = 0.5\), A1 for \(-3.5\). - (ii) [2 marks]: M1 for expanding \((3x-5)^2\) correctly, A1 for \(9x^2 - 30x + 23\).
Part (b) [3 marks]: - M1 for putting \(y = \frac{2}{x+1}\) and swapping or isolating variables. - M1 for correct steps to isolate \(x\). - A1 for \(g^{-1}(x) = \frac{2}{x} - 1\) (or equivalent).
Part (c) [4 marks]: - M1 for setting up the equation \(3x - 5 = x^2 - 7\). - M1 for simplifying to quadratic form \(x^2 - 3x - 2 = 0\). - M1 for using GDC or quadratic formula to find roots. - A1 for \(x = 3.56\) and \(x = -0.562\).
Question 8 · Structured
11 marks
A group of 40 students took a mathematics test. The table below shows the distribution of their marks.
| Mark (\(x\)) | Frequency | | :--- | :--- | | \(10 < x \le 20\) | \(4\) | | \(20 < x \le 30\) | \(8\) | | \(30 < x \le 40\) | \(p\) | | \(40 < x \le 50\) | \(q\) | | \(50 < x \le 60\) | \(5\) |
(a) Show that \(p + q = 23\).
(b) Using the mid-interval values, the estimated mean mark of the 40 students is \(36.75\). (i) Show that \(35p + 45q = 935\). (ii) Find the value of \(p\) and the value of \(q\).
(c) Using the values of \(p\) and \(q\) found in part (b)(ii), calculate the estimated standard deviation of the marks.
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Worked solution
(a) The total number of students is 40. Therefore, the sum of frequencies is: \(4 + 8 + p + q + 5 = 40 \implies 17 + p + q = 40 \implies p + q = 23\).
(b)(i) The mid-interval values are: \(15\) for \(10 < x \le 20\) \(25\) for \(20 < x \le 30\) \(35\) for \(30 < x \le 40\) \(45\) for \(40 < x \le 50\) \(55\) for \(50 < x \le 60\)
The estimated mean is given by: \(\text{Mean} = \frac{\sum f \cdot m}{\sum f}\) \(36.75 = \frac{4(15) + 8(25) + p(35) + q(45) + 5(55)}{40}\) Multiply both sides by 40: \(1470 = 60 + 200 + 35p + 45q + 275\) \(1470 = 535 + 35p + 45q\) \(35p + 45q = 1470 - 535 = 935\). (As required)
(ii) Solve the simultaneous equations: 1) \(p + q = 23 \implies 35p + 35q = 805\) 2) \(35p + 45q = 935\) Subtract equation 1 from equation 2: \(10q = 130 \implies q = 13\). Substitute \(q = 13\) back into equation 1: \(p + 13 = 23 \implies p = 10\).
(c) Using \(p = 10\) and \(q = 13\), we calculate the standard deviation. The formula is \(\sigma = \sqrt{\frac{\sum f \cdot m^2}{N} - \bar{x}^2}\). Calculate \(\sum f \cdot m^2\): \(4(15^2) + 8(25^2) + 10(35^2) + 13(45^2) + 5(55^2)\) \(= 4(225) + 8(625) + 10(1225) + 13(2025) + 5(3025)\) \(= 900 + 5000 + 12250 + 26325 + 15125 = 59600\). Now calculate variance: \(\sigma^2 = \frac{59600}{40} - 36.75^2 = 1490 - 1350.5625 = 139.4375\). Thus, the standard deviation is: \(\sigma = \sqrt{139.4375} \approx 11.808\). Correct to 3 significant figures, the estimated standard deviation is \(11.8\).
Marking scheme
Part (a) [1 mark]: - A1 for clear equation showing sum equal to 40 and simplifying.
Part (b): - (i) [4 marks]: M1 for identifying correct mid-interval values (15, 25, 35, 45, 55). M1 for writing the correct mean equation using their midpoints. M1 for multiplying by 40 to get 1470. A1 for leading to the given expression correctly. - (ii) [3 marks]: M1 for attempting to solve the simultaneous equations (elimination/substitution). A1 for \(q = 13\), A1 for \(p = 10\).
Part (c) [3 marks]: - M1 for attempting to calculate \(\sum f \cdot m^2\). - M1 for substituting their values into the variance/standard deviation formula. - A1 for \(11.8\) (accept \(11.81\)).
Question 9 · Structured
11 marks
The first five terms of three sequences, \(P\), \(Q\), and \(R\), are shown in the table below.
(a) Find the 6th term of: (i) Sequence \(P\) (ii) Sequence \(Q\) (iii) Sequence \(R\)
(b) Find an expression, in terms of \(n\), for the \(n\)-th term of Sequence \(P\).
(c) Find an expression, in terms of \(n\), for the \(n\)-th term of Sequence \(Q\).
(d) Using your answers to part (b) and part (c), write down the \(n\)-th term of Sequence \(R\).
(e) Find which term in Sequence \(P\) has a value of 1083.
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Worked solution
(a) (i) For Sequence \(P\), the differences between consecutive terms are: \(12-5=7\), \(23-12=11\), \(38-23=15\), \(57-38=19\). The next difference is \(19+4=23\). Thus, the 6th term is \(57+23 = 80\). (ii) Sequence \(Q\) is geometric with a first term of 3 and a common ratio of 2. The 6th term is \(48 \times 2 = 96\). (iii) Sequence \(R\) is the sum of Sequence \(P\) and Sequence \(Q\). The 6th term is \(80 + 96 = 176\).
(b) The second differences of Sequence \(P\) are constant: \(11-7=4\), \(15-11=4\), \(19-15=4\). Since the second difference is 4, the coefficient of the \(n^2\) term is \(\frac{4}{2} = 2\). Let the \(n\)-th term of Sequence \(P\) be \(2n^2 + bn + c\). For \(n=1\): \(2(1)^2 + b(1) + c = 5 \Rightarrow b + c = 3\). For \(n=2\): \(2(2)^2 + b(2) + c = 12 \Rightarrow 2b + c = 4\). Subtracting the first equation from the second gives \(b = 1\), which means \(c = 2\). So, the \(n\)-th term of Sequence \(P\) is \(2n^2 + n + 2\).
(c) Sequence \(Q\) is geometric with first term \(a = 3\) and common ratio \(r = 2\). The formula for the \(n\)-th term is \(a r^{n-1} = 3 \times 2^{n-1}\).
(d) Since each term of Sequence \(R\) is the sum of the corresponding terms of Sequence \(P\) and Sequence \(Q\), the \(n\)-th term of Sequence \(R\) is \(2n^2 + n + 2 + 3 \times 2^{n-1}\).
(e) Set the \(n\)-th term of Sequence \(P\) equal to 1083: \(2n^2 + n + 2 = 1083\) \(2n^2 + n - 1081 = 0\) Using the quadratic formula: \(n = \frac{-1 \pm \sqrt{1^2 - 4(2)(-1081)}}{2(2)}\) \(n = \frac{-1 \pm \sqrt{1 + 8648}}{4}\) \(n = \frac{-1 \pm \sqrt{8649}}{4}\) \(n = \frac{-1 \pm 93}{4}\) Since \(n\) must be positive, \(n = \frac{92}{4} = 23\).
Marking scheme
(a) (i) B1 for 80 (ii) B1 for 96 (iii) B1 for 176 (or FT from (i) + (ii))
(b) M1 for realizing it is a quadratic sequence and finding second differences equal to 4. M1 for setting up equations to find \(b\) and \(c\), e.g., \(b+c=3\) and \(2b+c=4\). A1 for \(2n^2 + n + 2\).
(c) M1 for finding the common ratio of 2 or form \(k \times 2^{n-1}\). A1 for \(3 \times 2^{n-1}\).
(d) B1 for \(2n^2 + n + 2 + 3 \times 2^{n-1}\) (or FT from (b) + (c)).
(e) M1 for setting up \(2n^2 + n + 2 = 1083\) and attempting to solve the quadratic equation. A1 for \(23\).
Question 10 · Structured
11 marks
A solid metal ornament is made of a cylinder of radius \(r\) cm and height \(4r\) cm. A solid cone of radius \(r\) cm and height \(3r\) cm is attached to one of the flat circular ends of the cylinder. A hemisphere of radius \(r\) cm is carved out from the other flat circular end of the cylinder.
(a) Show that the total volume, \(V\text{ cm}^3\), of the ornament is given by the formula: \(V = \frac{13}{3}\pi r^3\)
(b) The total volume of the ornament is \(350\text{ cm}^3\). Find the value of \(r\), giving your answer correct to 3 significant figures.
(c) (i) Find the slant height of the cone, giving your answer in the form \(r\sqrt{k}\) where \(k\) is an integer. (ii) Find the total external surface area of the ornament, giving your answer in the form \((a + \sqrt{b})\pi r^2\) where \(a\) and \(b\) are integers.
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Worked solution
(a) The total volume \(V\) is the sum of the volume of the cylinder and the volume of the cone, minus the volume of the carved-out hemisphere: \(V = V_{\text{cylinder}} + V_{\text{cone}} - V_{\text{hemisphere}}\)
Substitute these into the equation: \(V = 4\pi r^3 + \pi r^3 - \frac{2}{3}\pi r^3 = 5\pi r^3 - \frac{2}{3}\pi r^3 = \frac{13}{3}\pi r^3\).
(b) Set the volume equal to 350: \(\frac{13}{3}\pi r^3 = 350\) \(r^3 = \frac{350 \times 3}{13\pi} = \frac{1050}{13\pi}\) \(r^3 \approx 25.7096\) \(r = \sqrt[3]{25.7096} \approx 2.9515\text{ cm}\) Correct to 3 significant figures, \(r = 2.95\).
(c) (i) The cone has base radius \(r\) and height \(3r\). The slant height \(l\) is given by Pythagoras' theorem: \(l = \sqrt{r^2 + (3r)^2} = \sqrt{r^2 + 9r^2} = \sqrt{10r^2} = r\sqrt{10}\). Here, \(k = 10\).
(ii) The total external surface area consists of: - Curved surface area of the cone: \(\pi r l = \pi r (r\sqrt{10}) = \sqrt{10}\pi r^2\) - Curved surface area of the cylinder: \(2\pi r h_{\text{cylinder}} = 2\pi r (4r) = 8\pi r^2\) - Curved inner surface area of the carved-out hemisphere: \(2\pi r^2\) (Note: The flat circular rim is completely replaced by the hemisphere, so there is no flat area at that end).
(a) M1 for cylinder volume \(4\pi r^3\) or cone volume \(\pi r^3\). M1 for hemisphere volume \(\frac{2}{3}\pi r^3\) and correct combination of terms: \(4\pi r^3 + \pi r^3 - \frac{2}{3}\pi r^3\). A1 for fully correct simplification to \(\frac{13}{3}\pi r^3\).
(b) M1 for setting up \(\frac{13}{3}\pi r^3 = 350\). M1 for isolating \(r^3\) or \(r = \sqrt[3]{\frac{1050}{13\pi}}\). A1 for \(2.95\) (accept 2.95 or 2.951 to 2.952; do not accept 3).
(c) (i) M1 for using Pythagoras' theorem: \(r^2 + (3r)^2\). A1 for \(r\sqrt{10}\) or \(\sqrt{10}r\). (ii) M1 for curved surface area of cone: \(\sqrt{10}\pi r^2\) or curved surface area of cylinder: \(8\pi r^2\). M1 for adding all three components: \(\sqrt{10}\pi r^2 + 8\pi r^2 + 2\pi r^2\). A1 for \((10 + \sqrt{10})\pi r^2\).
Question 11 · Structured
11 marks
Let \(f(x) = x^3 - 4x + \frac{2}{x}\) for \(x \ne 0\).
(a) Find the \(x\)-coordinates of the four points where the graph of \(y = f(x)\) crosses the \(x\)-axis. Give your answers correct to 3 significant figures.
(b) Find the coordinates of: (i) the local maximum point, giving your coordinates correct to 3 significant figures. (ii) the local minimum point, giving your coordinates correct to 3 significant figures.
(c) Solve the equation \(f(x) = 2x - 1\). Give your answers correct to 3 significant figures.
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Worked solution
(a) To find where \(y = f(x)\) crosses the \(x\)-axis, set \(f(x) = 0\): \(x^3 - 4x + \frac{2}{x} = 0\) Multiplying by \(x\) (since \(x \ne 0\)): \(x^4 - 4x^2 + 2 = 0\) Let \(u = x^2\). Then: \(u^2 - 4u + 2 = 0\) Using the quadratic formula for \(u\): \(u = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(2)}}{2} = \frac{4 \pm \sqrt{16 - 8}}{2} = 2 \pm \sqrt{2}\) So, \(x^2 = 2 + \sqrt{2}\) or \(x^2 = 2 - \sqrt{2}\). This yields four values for \(x\): - \(x = \pm \sqrt{2 + \sqrt{2}} \approx \pm 1.8478\) - \(x = \pm \sqrt{2 - \sqrt{2}} \approx \pm 0.7654\)
Correct to 3 significant figures, the \(x\)-coordinates are \(-1.85\), \(-0.765\), \(0.765\), and \(1.85\).
(b) To find the turning points, find the derivative \(f'(x)\) and set it to 0: \(f'(x) = 3x^2 - 4 - \frac{2}{x^2} = 0\) Multiplying by \(x^2\): \(3x^4 - 4x^2 - 2 = 0\) Let \(v = x^2\). Then: \(3v^2 - 4v - 2 = 0\) Using the quadratic formula for \(v\): \(v = \frac{4 \pm \sqrt{16 - 4(3)(-2)}}{6} = \frac{4 \pm \sqrt{40}}{6} = \frac{2 \pm \sqrt{10}}{3}\) Since \(x^2 = v > 0\), we must have: \(x^2 = \frac{2 + \sqrt{10}}{3} \approx 1.7208\) Thus, \(x = \pm \sqrt{1.7208} \approx \pm 1.3118\).
(i) The local maximum occurs at the negative turning point: \(x \approx -1.31\) \(f(-1.3118) = (-1.3118)^3 - 4(-1.3118) + \frac{2}{-1.3118} \approx -2.258 + 5.247 - 1.525 = 1.464 \approx 1.47\) So, the coordinates of the local maximum are \((-1.31, 1.47)\).
(ii) Since \(f(x)\) is an odd function, the local minimum is symmetric about the origin: \(x \approx 1.31\) \(f(1.3118) \approx -1.47\) So, the coordinates of the local minimum are \((1.31, -1.47)\).
(c) To solve \(f(x) = 2x - 1\): \(x^3 - 4x + \frac{2}{x} = 2x - 1\) \(x^3 - 6x + 1 + \frac{2}{x} = 0\) Multiplying by \(x\): \(x^4 - 6x^2 + x + 2 = 0\) Using a graphic display calculator to find the roots of this polynomial: - \(x \approx -2.466\) - \(x \approx -0.5096\) - \(x \approx 0.7001\) - \(x \approx 2.275\)
Correct to 3 significant figures, the solutions are \(x = -2.47\), \(x = -0.510\), \(x = 0.700\), and \(x = 2.28\).
Marking scheme
(a) M1 for clearing fractions: \(x^4 - 4x^2 + 2 = 0\). M1 for solving the quadratic in \(x^2\) to get \(x^2 = 2 \pm \sqrt{2}\). A2 for all four correct roots: \(-1.85\), \(-0.765\), \(0.765\), \(1.85\) (A1 for two or three correct roots).
(b) (i) M1 for finding the derivative \(f'(x) = 3x^2 - 4 - \frac{2}{x^2}\) and attempting to solve \(f'(x) = 0\). A1 for \((-1.31, 1.47)\) (accept \(x = -1.31, y = 1.47\)). (ii) B1 for \((1.31, -1.47)\) (accept \(x = 1.31, y = -1.47\); or FT from (i)).
(c) M1 for setting up the equation \(x^4 - 6x^2 + x + 2 = 0\) or plotting both functions. A2 for all four solutions: \(-2.47\), \(-0.510\) (or \(-0.51\)), \(0.700\) (or \(0.7\)), \(2.28\). (A1 if two or three correct solutions are found).
Paper 62 (Extended)
Answer both Part A (Investigation) and Part B (Modelling).
2 Question · 60 marks
Question 1 · Structured Tasks
30 marks
### Part A (Investigation): Triangle Grids
This investigation is about geometric patterns of nested equilateral triangles. A large equilateral triangle of side length \(n\) units is built using small, unit equilateral triangles of side length 1. The dots represent the vertices where the sides of the small triangles meet. The lines represent the unit segments that form the sides of the small triangles.
For \(n = 1\), the pattern has 1 small triangle, 3 dots, and 3 lines. For \(n = 2\), the pattern has 4 small triangles, 6 dots, and 9 lines. For \(n = 3\), the pattern has 9 small triangles, 10 dots, and 18 lines.
#### Part 1 (a) Complete the table for \(n = 1, 2, 3, 4, 5\):
| \(n\) | 1 | 2 | 3 | 4 | 5 | | :--- | :---: | :---: | :---: | :---: | :---: | | Number of small triangles, \(T\) | 1 | 4 | 9 | | | | Number of dots, \(V\) | 3 | 6 | 10 | | | | Number of lines, \(E\) | 3 | 9 | 18 | | |
(b) Find an expression, in terms of \(n\), for: (i) the number of small triangles, \(T\). (ii) the number of dots, \(V\). (iii) the number of lines, \(E\).
#### Part 2 A dot is on the boundary if it lies on the outer perimeter of the large triangle. Otherwise, it is an internal dot. Let \(B\) be the number of boundary dots and \(I\) be the number of internal dots. For \(n = 1\): \(B = 3\), \(I = 0\). For \(n = 2\): \(B = 6\), \(I = 0\). For \(n = 3\): \(B = 9\), \(I = 1\).
(b) Write down an expression for \(B\) in terms of \(n\). (c) Show that the expression for the number of internal dots, \(I\), is: \[I = \frac{(n-1)(n-2)}{2}\]
#### Part 3 Find the value of \(n\) for which the number of internal dots is 2 times the number of boundary dots plus 1, i.e., \(I = 2B + 1\).
#### Part 4 A line is a boundary line if it lies on the outer perimeter of the large triangle. Otherwise, it is an internal line. Let \(L_B\) be the number of boundary lines and \(L_I\) be the number of internal lines. (a) Explain why \(L_B = 3n\). (b) Find an expression for \(L_I\) in terms of \(n\) in its simplest form. (c) Find the value of \(n\) for which the number of internal lines is 10 times the number of boundary lines.
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Worked solution
#### Part 1 (a) Continuing the sequences: - \(T\): squares of \(n\), so for \(n=4\), \(16\); for \(n=5\), \(25\). - \(V\): triangular numbers shifted, so for \(n=4\), \(15\); for \(n=5\), \(21\). - \(E\): multiples of 3 of triangular numbers, so for \(n=4\), \(30\); for \(n=5\), \(45\).
#### Part 3 Set up the equation: \[\frac{(n-1)(n-2)}{2} = 2(3n) + 1\] \[\frac{n^2 - 3n + 2}{2} = 6n + 1\] \[n^2 - 3n + 2 = 12n + 2\] \[n^2 - 15n = 0\] Since \(n > 0\), \(n = 15\).
#### Part 4 (a) The outer perimeter of the large triangle consists of 3 sides of length \(n\). Since each unit of length has exactly 1 boundary line, the total is \(3 \times n = 3n\). (b) \(L_I = E - L_B = \frac{3n(n+1)}{2} - 3n = \frac{3n^2 + 3n - 6n}{2} = \frac{3n^2 - 3n}{2} = \frac{3n(n-1)}{2}\). (c) Set \(L_I = 10 L_B\): \[\frac{3n(n-1)}{2} = 10(3n)\] Divide both sides by \(3n\) (since \(n \neq 0\)): \[\frac{n-1}{2} = 10 \implies n-1 = 20 \implies n = 21\].
Marking scheme
Part 1: - (a) [3 marks] 0.5 marks for each completed table cell. - (b)(i) [2 marks] B2 for \(n^2\) (B1 for identifying square number pattern). - (b)(ii) [3 marks] M1 for establishing a quadratic pattern, M1 for working to solve for coefficients, A1 for \(\frac{(n+1)(n+2)}{2}\) or equivalent. - (b)(iii) [3 marks] M1 for establishing a quadratic pattern, M1 for finding coefficient of \(n^2\) is 1.5, A1 for \(\frac{3n(n+1)}{2}\) or equivalent.
Part 2: - (a) [2 marks] 0.5 marks for each completed table cell. - (b) [2 marks] B2 for \(3n\) (B1 for recognizing multiples of 3). - (c) [3 marks] M1 for writing \(I = V - B\), M1 for algebraic substitution of their expressions, A1 for completing steps to show \(\frac{(n-1)(n-2)}{2}\).
Part 3: - [4 marks] M1 for setting up \(\frac{(n-1)(n-2)}{2} = 2(3n) + 1\), M1 for expanding to quadratic form \(n^2 - 15n = 0\), M1 for factoring, A1 for \(n = 15\).
Part 4: - (a) [2 marks] B2 for a clear explanation referring to 3 perimeter sides of length \(n\). - (b) [3 marks] M1 for \(L_I = E - L_B\), M1 for substituting expressions, A1 for \(\frac{3n(n-1)}{2}\). - (c) [3 marks] M1 for setting up equation, M1 for simplifying (e.g., dividing by \(3n\)), A1 for \(n = 21\).
Question 2 · Structured Tasks
30 marks
### Part B (Modelling): Optimizing a Compartmentalized Storage Box
An open-top storage box has a square base of side length \(x\) cm and vertical height \(h\) cm. Inside the box, there are two vertical partitions parallel to one of the sides, dividing the box into three equal compartments. The total volume of the box is fixed at \(V = 12000\text{ cm}^3\).
#### Part 1 (a) Find an expression for \(h\) in terms of \(x\). (b) Show that the total surface area, \(A\text{ cm}^2\), of the sheet metal used to construct the box and the partitions is given by: \[A = x^2 + \frac{72000}{x}\]
#### Part 2 (a) Complete the table of values for \(A = x^2 + \frac{72000}{x}\), giving your answers to the nearest 10:
(b) Describe the general shape of the graph of \(A\) against \(x\) for \(10 \le x \le 50\). (c) Find the value of \(x\) that minimizes the surface area \(A\), and find this minimum surface area. Give your answers to 3 significant figures.
#### Part 3 A lid is now added to cover only the middle compartment of the box. The lid has dimensions \(x\) cm by \(\frac{x}{3}\) cm. (a) Show that the new total surface area, \(A_{\text{new}}\text{ cm}^2\), of the material used (including the partitions and the lid) is given by: \[A_{\text{new}} = \frac{4}{3}x^2 + \frac{72000}{x}\] (b) Find the value of \(x\) that minimizes \(A_{\text{new}}\), and find this new minimum surface area.
#### Part 4 The base of the box is made of a heavy-duty material costing 3 times as much per \(\text{cm}^2\) as the material used for the sides, partitions, and lid. Let the cost factor \(F\) be defined as: \[F = 3 \times (\text{area of base}) + 1 \times (\text{area of sides, partitions, and lid})\] Consider the original design (without the lid) from Part 1. (a) Show that \(F = 3x^2 + \frac{72000}{x}\). (b) Find the value of \(x\) that minimizes the cost factor \(F\). Give your answer to 3 significant figures.
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Worked solution
#### Part 1 (a) Volume \(V = x^2 h = 12000 \implies h = \frac{12000}{x^2}\). (b) The box is open at the top. The surface area consists of: - 1 base of area \(x^2\) - 4 outer sides of height \(h\) and width \(x\) (total area \(4xh\)) - 2 internal partitions of height \(h\) and width \(x\) (total area \(2xh\)) Total Area \(A = x^2 + 4xh + 2xh = x^2 + 6xh\). Substitute \(h = \frac{12000}{x^2}\): \[A = x^2 + 6x\left(\frac{12000}{x^2}\right) = x^2 + \frac{72000}{x}\]
(b) The graph starts very high at \(x = 10\) (7300), decreases steadily to a minimum value near \(x = 33\), and then begins to rise gradually up to 3940 at \(x = 50\). It forms a smooth U-shaped curve.
#### Part 3 (a) The lid has dimensions \(x\) by \(\frac{x}{3}\), giving an area of \(\frac{1}{3}x^2\). \[A_{\text{new}} = A + \frac{1}{3}x^2 = x^2 + \frac{72000}{x} + \frac{1}{3}x^2 = \frac{4}{3}x^2 + \frac{72000}{x}\] (b) Differentiate \(A_{\text{new}}\) with respect to \(x\): \[\frac{d A_{\text{new}}}{dx} = \frac{8}{3}x - \frac{72000}{x^2} = 0 \implies \frac{8}{3}x^3 = 72000 \implies x^3 = 27000 \implies x = 30\text{ cm}\] Minimum surface area: \[A_{\text{new}}(30) = \frac{4}{3}(30^2) + \frac{72000}{30} = 1200 + 2400 = 3600\text{ cm}^2\]
#### Part 4 (a) The base area is \(x^2\), so its weighted cost contribution is \(3x^2\). The total area of sides and partitions is \(6xh = \frac{72000}{x}\), so its weighted cost contribution is \(1 \times \frac{72000}{x}\). Thus, \(F = 3x^2 + \frac{72000}{x}\).
Part 1: - (a) [2 marks] M1 for \(x^2 h = 12000\), A1 for \(h = \frac{12000}{x^2}\). - (b) [3 marks] M1 for identifying component areas (base \(x^2\), sides \(4xh\), partitions \(2xh\)), M1 for substituting \(h\), A1 for obtaining correct simplified form.
Part 2: - (a) [4 marks] 1 mark for each correct value (7300, 4000, 3400, 3940). - (b) [3 marks] B1 for starting high at \(x=10\), B1 for minimum in range \([30, 40]\), B1 for rising curve to \(x=50\). - (c) [4 marks] M1 for differentiating or using GDC, M1 for setting derivative to 0, A1 for \(x \approx 33.0\) (3 s.f.), A1 for \(A \approx 3270\) (3 s.f.).
Part 3: - (a) [3 marks] M1 for area of lid \(\frac{1}{3}x^2\), M1 for adding area to \(A\), A1 for establishing correct expression. - (b) [4 marks] M1 for differentiating or GDC, M1 for setting derivative to 0 (or equating to local minimum), A1 for \(x = 30\), A1 for \(A_{\text{new}} = 3600\).
Part 4: - (a) [3 marks] M1 for multiplying base by 3, M1 for multiplying sides/partitions by 1, A1 for showing the resulting expression. - (b) [4 marks] M1 for differentiating or GDC, M1 for setting to 0, A1 for \(x^3 = 12000\), A1 for \(x \approx 22.9\).
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