[{"id":"2016-P1-Q1","paper":1,"index":0,"stemLatex":"It is given that the expansion of $(ax+b)^3$ is $8x^3 - px^2 + 18x - 3\\sqrt{3}$, where $a$, $b$ and $p$ are real constants.\n\nWhat is the value of $p$?","options":[{"key":"A","latex":"-12\\sqrt{3}"},{"key":"B","latex":"-6\\sqrt{3}"},{"key":"C","latex":"-4\\sqrt{3}"},{"key":"D","latex":"-\\sqrt{3}"},{"key":"E","latex":"\\sqrt{3}"},{"key":"F","latex":"4\\sqrt{3}"},{"key":"G","latex":"6\\sqrt{3}"},{"key":"H","latex":"12\\sqrt{3}"}],"answerKey":"H","explanationLatex":"Expand $(ax+b)^3$ and compare coefficients with $8x^3-px^2+18x-3\\sqrt{3}$. This gives $a^3=8$, $3a^2b=-p$, $3ab^2=18$, and $b^3=-3\\sqrt{3}$. From $a^3=8$ we get $a=2$. From $b^3=-3\\sqrt{3}$ we get $b=-\\sqrt{3}$, because $(-\\sqrt{3})^3=-3\\sqrt{3}$. Then $3ab^2=3\\cdot2\\cdot3=18$, which is consistent. Finally, $3a^2b=3\\cdot4\\cdot(-\\sqrt{3})=-12\\sqrt{3}=-p$, so $p=12\\sqrt{3}$. Therefore the answer is $H$.","tags":["TMUA","2016","Paper1"],"difficulty":2},{"id":"2016-P1-Q2","paper":1,"index":1,"stemLatex":"The expression $3x^3 + 13x^2 + 8x + a$, where $a$ is a constant, has $(x+2)$ as a factor.\n\nWhich one of the following is a complete factorisation of the expression?","options":[{"key":"A","latex":"(x+2)(x-1)(3x-2)"},{"key":"B","latex":"(x+2)(x+1)(3x-2)"},{"key":"C","latex":"(x+2)(x+1)(3x+2)"},{"key":"D","latex":"(x+2)(x-3)(3x+2)"},{"key":"E","latex":"(x+2)(x+3)(3x-2)"},{"key":"F","latex":"(x+2)(x+3)(3x+2)"}],"answerKey":"E","explanationLatex":"We can use the factor theorem to find the value of a; alternatively we can just divide by (x + 2) and deduce a that way, as we will need to divide by (x+ 2) anyway to complete the factorisation. We show both methods. The factor theorem states that in this case, f(-2) = 0 , where f(x) is the polynomial. So 3(-2)^3 + 13(-2)^2 + 8(-2) + a = 0 , giving -24 + 52 - 16 + a = 0 , so a = -12. We can now divide the polynomial by (x + 2). Alternatively, dividing the given polynomial by (x + 2) by considering the x^3, x^2 and x terms, but ignoring the constant term, gives: 3x3 + 13x2 + 8x + a = (x + 2)(3x2 + 7x - 6) and so a = -12, though that is actually not needed to answer the question. Finally, we can factorise 3x2 + 7x - 6 to determine the answer. Alternatively, we can use the suggested options: only options D and E give a constant term in the quadratic of -6, so it must be one of those. D expands to 3x2 - 7x - 6 and E expands to 3x2 + 7x - 6, so the answer is E.","tags":["TMUA","2016","Paper1"],"difficulty":2},{"id":"2016-P1-Q3","paper":1,"index":2,"stemLatex":"A line is drawn normal to the curve $y = \\frac{2}{x^2}$ at the point on the curve where $x=1$.\n\nThis line cuts the x-axis at $P$ and the y-axis at $Q$.\n\nThe length of $PQ$ is","options":[{"key":"A","latex":"\\frac{3\\sqrt{5}}{2}"},{"key":"B","latex":"\\frac{3\\sqrt{17}}{4}"},{"key":"C","latex":"\\frac{7\\sqrt{17}}{4}"},{"key":"D","latex":"\\frac{35}{4}"},{"key":"E","latex":"\\frac{35\\sqrt{5}}{2}"},{"key":"F","latex":"\\frac{3\\sqrt{17}}{2}"}],"answerKey":"C","explanationLatex":"We need to find the gradient of the curve at (1, 2). We have y = 2x- 2 and so dy dx = -4x- 3. At x = 1, this gives dy dx = -4 and so the normal has gradient 1 4 . As it passes through (1, 2), it has equation y - 2 = 1 4 (x - 1). It cuts the x-axis ( y = 0 ) at P , so at P , we have -2 = 1 4 (x - 1), giving x = -7, and so P has coordinates (-7, 0). It cuts the y-axis ( x = 0 ) at Q, so at Q, we have y - 2 = 1 4 (-1), giving y = 7 4 , and so Q has coordinates (0, 7 4 ). Thus P Q has length $\\\\sqrt{72}$ + ( 7 4 )^2 = 7 $\\\\sqrt{42}$ + 1 = 7 √(where we have written 7 = 7 4 × 4 for the first equality). Hence the answer is C.","tags":["TMUA","2016","Paper1"],"difficulty":2},{"id":"2016-P1-Q4","paper":1,"index":3,"stemLatex":"The sequence $a_n$ is defined by the rule for $n \\geq 1$:\n\n$a_n = (-1)^n - (-1)^{n-1} + (-1)^{n+2}$\n\nFind the value of $\\sum_{n=1}^{39} a_n$","options":[{"key":"A","latex":"-39"},{"key":"B","latex":"-3"},{"key":"C","latex":"-1"},{"key":"D","latex":"0"},{"key":"E","latex":"1"},{"key":"F","latex":"3"},{"key":"G","latex":"39"}],"answerKey":"B","explanationLatex":"Commentary: This looks quite scary! It is unlikely that you have ever seen a sequence looking like this, so a sensible thing to do is to work out the first few values and look for any patterns. We calculate the first few terms of the sequence: a1 = (-1)^1 - (-1)^0 + (-1)^3 = (-1) - 1 + (-1) = -3 a2 = (-1)^2 - (-1)^1 + (-1)^4 = 1 - (-1) + 1 = 3 a3 = (-1)^3 - (-1)^2 + (-1)^5 = (-1) - 1 + (-1) = -3 a4 = (-1)^4 - (-1)^3 + (-1)^6 = 1 - (-1) + 1 = 3 The pattern is now clear (and we could prove it if we wished to): the sequence goes -3, 3, -3, 3, -3, 3 and so on. So the sum of each pair of terms is zero: a1 + a2 = 0 , a3 + a4 = 0 , ..., a37 + a38 = 0. Thus 39∑ n=1 an = (a1 + a2) + (a3 + a4) + · · · + (a37 + a38) + a39 = 0 + 0 + · · · + 0 + (-3) = -3 and the answer is B.","tags":["TMUA","2016","Paper1"],"difficulty":2},{"id":"2016-P1-Q5","paper":1,"index":4,"stemLatex":"What is the total area enclosed between the curve $y=x^2-1$, the x-axis and the lines $x=-2$ and $x=2$?","options":[{"key":"A","latex":"\\frac{4}{3}"},{"key":"B","latex":"\\frac{8}{3}"},{"key":"C","latex":"4"},{"key":"D","latex":"\\frac{16}{3}"},{"key":"E","latex":"12"},{"key":"F","latex":"16"}],"answerKey":"C","explanationLatex":"To answer this question, it is worth drawing a sketch. The graph is of y = x^2 - 1 = ( x + 1)(x - 1), so the parabola intersects the x-axis at (1, 0) and (-1, 0): x y -2 -1 1 2 To find the area enclosed, we integrate over the three separate regions, from -2 to -1, from -1 to 1, and from 1 to 2: ∫ - 1 - 2 x^2 - 1 dx = [ 1 3 x^3 - x ] - 1 - 2 = (- 1 3 + 1) - (- 8 3 + 2) = 4 3∫ 1 - 1 x^2 - 1 dx = [ 1 3 x^3 - x ] 1 - 1 = ( 1 3 - 1) - (- 1 3 + 1) = - 4 3∫ 2 x^2 - 1 dx = [ 1 3 x^3 - x ] 2 = ( 8 3 - 2) - ( 1 3 - 1) = 4 Thus the three areas are each 4 3 , and the total area is 4, so the answer is C.","tags":["TMUA","2016","Paper1"],"difficulty":2},{"id":"2016-P1-Q6","paper":1,"index":5,"stemLatex":"$P$, $Q$ and $R$ are each mixtures of red and white paint.\n\nThe percentage by volume of red paint in $P$ is $30\\%$.\n\nThe percentage by volume of red paint in $Q$ is $20\\%$.\n\nThe mixtures $P$, $Q$ and $R$ are combined in the proportion 12:5:3 respectively.\n\nIf the resulting mixture contains $25\\%$ by volume of red paint, what percentage by volume of mixture $R$ is red paint?","options":[{"key":"A","latex":"25\\%"},{"key":"B","latex":"23\\%"},{"key":"C","latex":"13\\frac{1}{3}\\%"},{"key":"D","latex":"19\\frac{1}{2}\\%"},{"key":"E","latex":"9\\frac{3}{4}\\%"},{"key":"F","latex":"It is impossible to achieve this result."}],"answerKey":"C","explanationLatex":"We can either work in fractions or percentages. This solution works in fractions. The fraction of red paint in P is 3 10 and in Q is 1 5 . Let the fraction of red paint in R be x. The combination is in the ratio 12 : 5 : 3 , with 1 4 red paint, so the composition is as follows: Paint Volume Red P 12 12 × 3 Q 5 5 × 1 R 3 3x total 20 20 × 1 We can now sum the individual red amounts, giving: 12 × 3 10 + 5 × 1 5 + 3x = 20 × 14 so 23 5 + 3x = 5, giving 3x = 2 5 = 40%, so x = 13 1 3 %. The correct answer is C.","tags":["TMUA","2016","Paper1"],"difficulty":2},{"id":"2016-P1-Q7","paper":1,"index":6,"stemLatex":"$60 \\%$ of a sports club's members are women and the remainder are men.\n\nThis sports club offers the opportunity to play tennis or cricket. Every member plays exactly one of the two sports.\n\n$\\frac{2}{5}$ of the male members of the club play cricket;\n\n$\\frac{2}{3}$ of the cricketing members of the club are women.\n\nWhat is the probability that a member of the club, chosen at random, is a woman who plays tennis?","options":[{"key":"A","latex":"\\frac{1}{5}"},{"key":"B","latex":"\\frac{7}{25}"},{"key":"C","latex":"\\frac{1}{3}"},{"key":"D","latex":"\\frac{11}{25}"},{"key":"E","latex":"\\frac{3}{5}"}],"answerKey":"B","explanationLatex":"There are two factors for each member: whether the member is a man or woman, and whether the member plays tennis or cricket. We can either display this situation as a tree diagram or as a two-way table. Since the information given has both the fraction of male members who play cricket and the fraction of the cricketing members who are women, it will be simpler to use a two-way table. (In a tree diagram, we could easily display one of the two, but the other will be hard.) We could also work with fractions or with numbers of people. In this solution, we will work with numbers. The 60% suggests that we work with a multiple of 100 people, and the presence of a fraction 2 3 , suggests that we work with a multiple of 3. So we start with 300 people. This is the information we obtain from the 60% of members are women and 2 5 of the resulting 120 male members play cricket: Men Women Total Tennis 72 Cricket 48 Total 120 180 300 We are also told that 2 3 of the cricketing members are women, so the 48 cricketing men are 1 3 of the cricketers, hence 96 cricketers are women, leaving 84 women to play tennis: Men Women Total Tennis 72 84 (156) Cricket 48 96 (144) Total 120 180 300 (The parenthesised figures are not needed to finish the question.) Therefore the probability that a member of the club, chosen at random, is a woman who plays tennis is 84 300 = 28 100 = 7 25 , and the answer is B.","tags":["TMUA","2016","Paper1"],"difficulty":2},{"id":"2016-P1-Q8","paper":1,"index":7,"stemLatex":"Find the maximum angle $x$ in the range $0^{\\circ} \\leq x \\leq 360^{\\circ}$ which satisfies the equation $\\cos^2(2x) + \\sqrt{3}\\sin(2x) - \\frac{7}{4} = 0$","options":[{"key":"A","latex":"30^{\\circ}"},{"key":"B","latex":"60^{\\circ}"},{"key":"C","latex":"120^{\\circ}"},{"key":"D","latex":"150^{\\circ}"},{"key":"E","latex":"210^{\\circ}"},{"key":"F","latex":"240^{\\circ}"},{"key":"G","latex":"300^{\\circ}"},{"key":"H","latex":"330^{\\circ}"}],"answerKey":"F","explanationLatex":"We note that the only angle involved is 2x, and 0◦ ⩽ x ⩽ 360◦ gives 0◦ ⩽ 2x ⩽ 720◦. We start by writing everything in terms of sin 2x, giving: (1 - sin^2 2x) + $\\\\sqrt{3}$ sin 2x - 7 4 = 0 which rearranges to give sin^2 2x - $\\\\sqrt{3}$ sin 2x + 3 4 = 0. We can apply the quadratic formula to this to obtain sin 2x = $\\\\sqrt{3}$ ± $\\\\sqrt{3}$ - 3 2 = $\\\\sqrt{and}$ hence the possible values of 2x in the range are 2x = 60 ◦, 120◦, 420◦ and 480◦. Thus the largest possible value of x in the range 0◦ ⩽ x ⩽ 360◦ is 240◦, and the answer is F.","tags":["TMUA","2016","Paper1"],"difficulty":2},{"id":"2016-P1-Q9","paper":1,"index":8,"stemLatex":"The line segment joining the points (3, 3) and (7, 5) is a diameter of a circle.\n\nThis circle is translated by 3 units in the negative x-direction, then reflected in the x-axis, and then enlarged by a scale factor of 4 about the centre of the resulting circle.\n\nThe equation of the final circle is","options":[{"key":"A","latex":"(x-2)^{2} + (y-4)^{2} = 320"},{"key":"B","latex":"(x-2)^{2} + (y+4)^{2} = 320"},{"key":"C","latex":"(x-2)^{2} + (y-4)^{2} = 80"},{"key":"D","latex":"(x-2)^{2} + (y+4)^{2} = 80"},{"key":"E","latex":"(x-2)^{2} + (y-4)^{2} = 20"},{"key":"F","latex":"(x-2)^{2} + (y+4)^{2} = 20"}],"answerKey":"D","explanationLatex":"The initial circle looks like this: x y (3, 3) (7, 5) The centre of the circle is at the midpoint of this diameter, which is ( 3 + 7 2 , 3 + 5 ) = (5 , 4), and the radius is therefore √(5 - 3)^2 + (4 - 3)^2 = $\\\\sqrt{5}$. The transformations do the following to the centre and radius of the circle: Centre Radius original (5, 4) $\\\\sqrt{after}$ translation (2, 4) $\\\\sqrt{after}$ reflection (2, -4) $\\\\sqrt{after}$ enlargement (2, -4)^4 $\\\\sqrt{Therefore}$ the final circle has equation (x - 2)^2 + (y + 4)^2 = (4 $\\\\sqrt{5}$)^2 = 80 and the answer is D.","tags":["TMUA","2016","Paper1"],"difficulty":2},{"id":"2016-P1-Q10","paper":1,"index":9,"stemLatex":"How many solutions does the equation $x\\tan x = 1$ have in the interval $-2\\pi \\leq x \\leq 2\\pi$?","options":[{"key":"A","latex":"0"},{"key":"B","latex":"1"},{"key":"C","latex":"2"},{"key":"D","latex":"3"},{"key":"E","latex":"4"},{"key":"F","latex":"5"},{"key":"G","latex":"6"}],"answerKey":"E","explanationLatex":"We can rearrange this equation as tan x = 1 x . (We note that x = 0 is not a solution of the original equation x tan x = 1, so we can divide by x without losing any solutions.) It is impossible to solve this equation exactly, but we are only asked for the number of solutions. Therefore we sketch the graphs of y = tan x and y = 1 x on the same axes, and count the number of intersections. x y 0 π 2π-π-2π There are four points of intersection with -2π ⩽ x ⩽ 2π, so the answer is 4, which is option E.","tags":["TMUA","2016","Paper1"],"difficulty":2},{"id":"2016-P1-Q11","paper":1,"index":10,"stemLatex":"The real roots of the equation $4^{2x} + 12 = 2^{2x+3}$ are $p$ and $q$, where $p > q$.\n\nThe value of $p-q$ can be expressed as","options":[{"key":"A","latex":"\\frac{3}{4}"},{"key":"B","latex":"1"},{"key":"C","latex":"4"},{"key":"D","latex":"-\\frac{1}{2} + \\log_{10}\\frac{3}{2}"},{"key":"E","latex":"\\frac{log_{10}3}{log_{10}4}"},{"key":"F","latex":"\\frac{log_{10}3}{log_{10}2}"}],"answerKey":"E","explanationLatex":"This is a quadratic in 22x, so we write y = 2 2x and solve for y. The quadratic becomes y^2 + 12 = 8 y as 42x = (2 2)2x = 2 4x = (2 2x)^2 and 22x+3 = 2 2x × 23. This rearranges to y^2 - 8y + 12 = 0 , which factorises as (y - 6)(y - 2) = 0 , so y = 6 or y = 2 . Thus 22x = 6 or 22x = 2 , with the value of x being larger in the first than in the second, so 22p = 6 and 22q = 2 . The options are given in terms of log to base 10, so we take log10 of the first equation. Thus 2p log10 2 = log 10 6 and 2q = 1, and hence p = log10 6 2 log10 2 and q = 1 2 . Therefore p - q = log10 6 2 log10 2 - 1 = log10 6 - log10 2 2 log10 2 = log10 3 log10 4 so the answer is option E.","tags":["TMUA","2016","Paper1"],"difficulty":2},{"id":"2016-P1-Q12","paper":1,"index":11,"stemLatex":"A right circular cylinder is contained within a sphere of radius 5 cm in such a way that the whole of the circumferences of both ends of the cylinder are in contact with the sphere. The diagram shows a planar cross section through the centre of the sphere and cylinder. Find, in cubic centimetres, the maximum possible volume of the cylinder.","stemImage":"/stem-fallback/2016-P1-Q12.png","options":[{"key":"A","latex":"250\\pi"},{"key":"B","latex":"500\\pi"},{"key":"C","latex":"1000\\pi"},{"key":"D","latex":"\\frac{250\\sqrt{3}}{3}\\pi"},{"key":"E","latex":"\\frac{500\\sqrt{3}}{9}\\pi"},{"key":"F","latex":"\\frac{1000\\sqrt{3}}{9}\\pi"}],"answerKey":"E","explanationLatex":"Let the radius of the cylinder be $r$ and let half its height be $h$. In the central cross-section of the sphere, we get a right-angled triangle with hypotenuse $5$, so\n\n$$r^2+h^2=25.$$\n\nThe cylinder has height $2h$, so its volume is\n\n$$V=\\pi r^2(2h)=2\\pi h(25-h^2).$$\n\nDifferentiate with respect to $h$:\n\n$$\\frac{dV}{dh}=2\\pi(25-3h^2).$$\n\nSet this equal to $0$:\n\n$$25-3h^2=0 \\quad\\Longrightarrow\\quad h=\\frac{5}{\\sqrt{3}}.$$\n\nNow\n\n$$r^2=25-h^2=25-\\frac{25}{3}=\\frac{50}{3}.$$\n\nSo the maximum volume is\n\n$$V=2\\pi\\left(\\frac{50}{3}\\right)\\left(\\frac{5}{\\sqrt{3}}\\right)=\\frac{500\\pi}{3\\sqrt{3}}=\\frac{500\\sqrt{3}\\pi}{9}.$$\n\nSo the correct answer is $\\mathbf{E}$.","tags":["TMUA","2016","Paper1"],"difficulty":2},{"id":"2016-P1-Q13","paper":1,"index":12,"stemLatex":"How many real roots does the equation $3x^5 - 10x^3 - 120x + 30 = 0$ have?","options":[{"key":"A","latex":"1"},{"key":"B","latex":"2"},{"key":"C","latex":"3"},{"key":"D","latex":"4"},{"key":"E","latex":"5"}],"answerKey":"C","explanationLatex":"Commentary: This is not something that you will have learnt how to do, and in general, this is a very difficult problem indeed. But we can use the tools at our disposal to do something helpful. We don’t know how to solve a quintic ( x^5) equation, but we can at least try to find the turning (stationary) points, and see where that gets us. We attempt to sketch the graph of y = 3x5 - 10x3 - 120x + 30, or at least think about what a sketch might look like. We calculate dy dx = 15x4 - 30x2 - 120 = 15( x^4 - 2x2 - 8) = 15( x^2 - 4)(x^2 + 2) so there are stationary points when dy dx = 0, that is, when x = 2 and when x = -2. We can now calculate the y-coordinates: when x = -2, y = -36 + 80 + 240 + 30 > 0 (we do not need to calculate the exact value), and when x = 2, y = 36 - 80 - 240 + 30 < 0. So the graph must look something like the following: x y (-2, y) (2, y) There are clearly 3 points of intersection of this graph with y = 0, so the original equation has 3 real roots, and the answer is C.","tags":["TMUA","2016","Paper1"],"difficulty":2},{"id":"2016-P1-Q14","paper":1,"index":13,"stemLatex":"The terms of an infinite series $S$ are formed by adding together the corresponding terms in two infinite geometric series, $T$ and $U$.\n\nThe first term of $T$ and the first term of $U$ are each 4.\n\nIn order, the first three terms of the combined series $S$ are 8, 3 and $\\frac{5}{4}$\n\nWhat is the sum to infinity of $S$?","options":[{"key":"A","latex":"\\frac{32}{5}"},{"key":"B","latex":"\\frac{20}{3}"},{"key":"C","latex":"\\frac{64}{5}"},{"key":"D","latex":"\\frac{40}{3}"},{"key":"E","latex":"16"},{"key":"F","latex":"32"}],"answerKey":"D","explanationLatex":"Let us write out the first few terms of the three sequences explicitly. Let the common ratio of T be R and the common ratio of U be r. term 1 2 3 4 5 ... n ... T 4 4R 4R2 4R3 4R4 ... 4Rn- 1 ... U 4 4r 4r2 4r3 4r4 ... 4rn- 1 ... S 8 4R + 4r 4R2 + 4r2 4R3 + 4r3 4R4 + 4r4 ... 4Rn- 1 + 4rn- 1 ... We are given the first three terms of S, so 8 = 8 4R + 4r = 3 4R2 + 4r2 = 5 Therefore R + r = 3 4 and R^2 + r^2 = 5 16 . We can work out the values of R and r by guessing, or we can work them out by solving these equations. We show the second approach here. The first equation gives R = 3 4 - r, which we can substitute into the second equation to give ( 3 4 - r)^2 + r^2 = 5 so 2r2 - 3 2 r + 1 4 = 0 , hence 8r2 - 6r + 1 = 0 , which factorises as (4r - 1)(2r - 1) = 0 , giving r = 1 2 or r = 1 4 and R = 1 4 or R = 1 2 respectively. We can therefore work out the sums to infinity of T and U, taking R = 1 4 and r = 1 2 : sum to infinity of T = 4 1 - 1 = 16 sum to infinity of U = 4 1 - 1 = 8 giving the sum to infinity of S as 16 3 + 8 = 40 3 , since S is the sum of T and U. If we took the values of R and r to be the other way round, we would have the series T and U swapped, but their sum would be unchanged. So the answer is D.","tags":["TMUA","2016","Paper1"],"difficulty":2},{"id":"2016-P1-Q15","paper":1,"index":14,"stemLatex":"The least possible value of the gradient of the curve $y = (2x+a)(x-2a)^2$ at the point where $x=1$, as $a$ varies, is","options":[{"key":"A","latex":"-\\frac{49}{4}"},{"key":"B","latex":"-8"},{"key":"C","latex":"-\\frac{25}{4}"},{"key":"D","latex":"\\frac{7}{4}"},{"key":"E","latex":"\\frac{47}{16}"}],"answerKey":"C","explanationLatex":"We start by expanding the brackets to get y = (2x + a)(x^2 - 4ax + 4a2) = 2 x^3 - 7ax2 + 4a^2x + 8a3 so the derivative is given by dy dx = 6x2 - 14ax + 4a2. Therefore at x = 1 , the gradient is 6 - 14a + 4a2. To find the least possible value of this as a varies, we can either differentiate this expression with respect to a or complete the square. The latter approach gives 4a2 - 14a + 6 = 4( a2 - 7 2 a) + 6 = 4 ( (a - 7 4 )^2 - ( 7 4 )^2) + 6 = 4(a - 7 4 )^2 - 49 4 + 6 = 4(a - 7 4 )^2 - 25 so the minimum value is - 25 4 and the answer is C.","tags":["TMUA","2016","Paper1"],"difficulty":2},{"id":"2016-P1-Q16","paper":1,"index":15,"stemLatex":"Given the simultaneous equations:\n\n$\\log_{10}2 + \\log_{10}(y-1) = 2\\log_{10}x$\n\n$\\log_{10}(y+3-3x) = 0$\n\nthe values of $y$ are","options":[{"key":"A","latex":"\\frac{5}{2} \\pm \\frac{3\\sqrt{5}}{2}"},{"key":"B","latex":"3 \\pm \\sqrt{3}"},{"key":"C","latex":"7 \\pm 3\\sqrt{3}"},{"key":"D","latex":"3, 9"},{"key":"E","latex":"1, 13"}],"answerKey":"C","explanationLatex":"The first equation involves the sum of two logarithms, so we can rewrite it as: log10(2(y - 1)) = log 10 x^2 and so 2(y - 1) = x^2. The second equation becomes y + 3 - 3x = 1, as loga 1 = 0 for any base a. We therefore have two simultaneous equations: 2y - 2 = x^2 y - 3x + 2 = 0 . We want to find the values of y, so we rewrite the second equation to eliminate x: x = y + 2 and so the first equation becomes 2y - 2 = (y + 2)^2 or 18y - 18 = y^2 + 4y + 4 which rearranges to give the quadratic y^2 - 14y + 22 = 0 . The solutions to this quadratic are 14 ± $\\\\sqrt{142}$ - 4 × 22 2 = 7 ± $\\\\sqrt{72}$ - 22 = 7 ± $\\\\sqrt{and}$ so the correct answer is C. It would be good to check whether both of these solutions are valid (though the question does not require us to do so). To be valid, we require both x > 0 and y > 1. Since $\\\\sqrt{27}$ < 6, we see that y > 1. Also, as x = 1 3 (y + 2) and y > 0, it follows that x > 0. So there are, indeed, two valid solutions.","tags":["TMUA","2016","Paper1"],"difficulty":2},{"id":"2016-P1-Q17","paper":1,"index":16,"stemLatex":"It is given that for $0 < x < \\pi$,\n\n$y = (1+2\\cos x)\\cos 2x$.\n\nThe complete set of values of $x$ for which $y$ is negative is","options":[{"key":"A","latex":"0 < x < \\frac{\\pi}{4}, \\frac{2\\pi}{3} < x < \\frac{3\\pi}{4}"},{"key":"B","latex":"0 < x < \\frac{\\pi}{4}, \\frac{3\\pi}{4} < x < \\pi"},{"key":"C","latex":"0 < x < \\frac{2\\pi}{3}, \\frac{3\\pi}{4} < x < \\pi"},{"key":"D","latex":"\\frac{\\pi}{4} < x < \\frac{2\\pi}{3}, \\frac{3\\pi}{4} < x < \\pi"},{"key":"E","latex":"\\frac{\\pi}{4} < x < \\frac{2\\pi}{3}"},{"key":"F","latex":"\\frac{\\pi}{4} < x < \\frac{3\\pi}{4}"}],"answerKey":"D","explanationLatex":"The formula for y is the product of two factors, namely 1 + 2 cos x and cos 2x. The whole expression is negative when one of the two factors is positive and the other is negative. The factors change sign when they cross a point where the factor is zero, and we can find these points: • 1 + 2 cos x = 0 when cos x = - 1 2 , which is when x = 2π 3 (within the range 0 < x < π ). • cos 2x = 0 when x = π 4 and when x = 3π 4 . We can now make a table showing the signs of the two factors in the different parts of the interval 0 < x < π . (This is a useful technique in general.)^0 < x < π 4 x = π π 4 < x < 2π 3 x = 2π 2π 3 < x < 3π 4 x = 3π 3π 4 < x < π 1 + 2 cos x + + + 0 - - - cos 2x + 0 - - - 0 + y + 0 - 0 + 0 - Therefore y is negative when π 4 < x < 2π 3 and when 3π 4 < x < π , and so the answer is D.","tags":["TMUA","2016","Paper1"],"difficulty":2},{"id":"2016-P1-Q18","paper":1,"index":17,"stemLatex":"The function $\\frac{1-x}{\\sqrt[3]{x^2}}$ is defined for all $x \\neq 0$.\n\nThe complete set of values of $x$ for which the function is decreasing is","options":[{"key":"A","latex":"x \\leq -2, x>0"},{"key":"B","latex":"-2 \\leq x < 0"},{"key":"C","latex":"x \\leq 1, x \\neq 0"},{"key":"D","latex":"x \\geq 1"},{"key":"E","latex":"-2 \\leq x \\leq 1, x \\neq 0"},{"key":"F","latex":"x \\leq -2, x \\geq 1"}],"answerKey":"A","explanationLatex":"Write the function as\n\n$$y=\\frac{1-x}{\\sqrt[3]{x^2}}=x^{-2/3}-x^{1/3}.$$\n\nDifferentiate:\n\n$$\\frac{dy}{dx}=-\\frac{2}{3}x^{-5/3}-\\frac{1}{3}x^{-2/3}=-\\frac{1}{3}x^{-5/3}(x+2).$$\n\nNow analyse the sign of the derivative.\n\n- If $x<-2$, then $x^{-5/3}<0$ and $x+2<0$, so $x^{-5/3}(x+2)>0$, hence $\\dfrac{dy}{dx}<0$.\n- If $-2<x<0$, then $x^{-5/3}<0$ and $x+2>0$, so $x^{-5/3}(x+2)<0$, hence $\\dfrac{dy}{dx}>0$.\n- If $x>0$, then $x^{-5/3}>0$ and $x+2>0$, so $x^{-5/3}(x+2)>0$, hence $\\dfrac{dy}{dx}<0$.\n\nTherefore the function is decreasing on\n\n$$(-\\infty,-2)\\cup(0,\\infty).$$\n\nSo the intended answer is $\\mathbf{A}$.","tags":["TMUA","2016","Paper1"],"difficulty":2},{"id":"2016-P1-Q19","paper":1,"index":18,"stemLatex":"The coefficient of $x^3$ in the expansion of $(1+2x+3x^2)^6$ is equal to twice the coefficient of $x^4$ in the expansion of $(1-ax^2)^5$.\n\nFind all possible values of the constant $a$.","options":[{"key":"A","latex":"\\pm 2\\sqrt{2}"},{"key":"B","latex":"\\pm \\sqrt{17}"},{"key":"C","latex":"\\pm \\sqrt{34}"},{"key":"D","latex":"\\pm 2\\sqrt{17}"},{"key":"E","latex":"There are no possible values of $a$."}],"answerKey":"B","explanationLatex":"We expand the first expression up to powers of x^3 to find its coefficient. We can think of the first expression as (1 + (2x + 3x2))^6 and use the binomial theorem to expand it. We note that the first few binomial coefficients are ( 6 ) = 1; ( 6 ) = 6; ( 6 ) = 6 × 5 2! = 15; ( 6 ) = 6 × 5 × 4 3! = 20. We thus have (1 + (2x + 3x2))^6 = 1 + ( 6 ) (2x + 3x2) + ( 6 ) (2x + 3x2)^2 + ( 6 ) (2x + 3x2)^3 + · · · = 1 + 6(2 x + 3x2) + 15((2x)^2 + 2(2x)(3x2) + · · · ) + 20((2x)^3 + · · · ) + · · · where we have stopped when the powers reach 3. We can read off the coefficient of x^3 from this; it is 15 × 2 × 2 × 3 + 20 × 8 = 340 . We can expand (1 - ax2)^5 similarly, and obtain 1 + ( 5 ) (-ax2) + ( 5 ) (-ax2)^2 + · · · = 1 - 5ax2 + 10a^2x4 + · · · so the coefficient of x^4 is 10a2. We are told how these two coefficients relate to each other: we have 340 = 2 × 10a2 so a2 = 17 and a = ± $\\\\sqrt{17}$, giving the answer as option B.","tags":["TMUA","2016","Paper1"],"difficulty":2},{"id":"2016-P1-Q20","paper":1,"index":19,"stemLatex":"The diagram shows a square-based pyramid with base PQRS and vertex O. All the edges of the pyramid are of length 20 metres. Find the shortest distance, in metres, along the outer surface of the pyramid from P to the midpoint of OR.","stemImage":"/stem-fallback/2016-P1-Q20.png","options":[{"key":"A","latex":"10\\sqrt{5-2\\sqrt{3}}"},{"key":"B","latex":"10\\sqrt{3}"},{"key":"C","latex":"10\\sqrt{5}"},{"key":"D","latex":"10\\sqrt{7}"},{"key":"E","latex":"10\\sqrt{5+2\\sqrt{3}}"}],"answerKey":"D","explanationLatex":"Let us label the midpoint of OR as T , so that we can refer to it. We first observe that since all the edges of the pyramid are of length 20m, the triangles are all equilateral. There are two ways to get from P to the midpoint of OR in a short distance: one can either travel along the base to RS and then along the triangular face to T , or one could travel along the triangular face OP S to the edge OS and from there to T . It is not obvious which route is shorter, so we will calculate both. (There are also routes which are mirror-reflections of these going via the edge QR or OQ.) To find these distances, we can “unfold” the tetrahedron, that is, make a net for the tetrahedron. For the first route, we draw just the faces P QRS and ORS: P Q R S O T The height of the triangle OSR is 20 sin 60◦ = 20 × $\\\\sqrt{2}$ = 10 $\\\\sqrt{3}$, so T lies 5 $\\\\sqrt{3}$ above the line SR; T also lies 3 4 × 20 = 15 to the right of S (as it is at the midpoint of OR). We can therefore find the length P T using Pythagoras’s theorem: P T 2 = 15 2 + (20 + 5 $\\\\sqrt{3}$)^2 = 225 + 400 + 200 $\\\\sqrt{3}$ + 75 = 700 + 200 $\\\\sqrt{We}$ now look at the other route, which we can do by drawing the triangles OP S and OSR: O S P R T Using the height of the triangle as calculated above, P lies 10 $\\\\sqrt{3}$ to the left of OS and T lies $\\\\sqrt{3}$ to its right, while T lies 5 above P (being a quarter of OS). Pythagoras’s theorem therefore gives P T 2 = (15 $\\\\sqrt{3}$)^2 + 52 = 225 × 3 + 25 = 700 This is therefore the shorter of the two routes, and its length is $\\\\sqrt{700}$ = 10 $\\\\sqrt{7}$, hence the solution is option D. Our research-based tests provide a fair measure of skills and aptitude to help you make informed decisions. As a trusted partner, we work closely with universities, governments and employers to enhance their selection processes. Cambridge Assessment Admissions Testing The Triangle Building Shaftesbury Road Cambridge CB2 8EA United Kingdom Admissions tests support: admissionstesting.org/help","tags":["TMUA","2016","Paper1"],"difficulty":2},{"id":"2016-P2-Q1","paper":2,"index":0,"stemLatex":"Find the value of $\\int_1^2 \\left(x^2 - \\frac{4}{x^2}\\right)^2\\,dx$","options":[{"key":"A","latex":"\\frac{43}{15}"},{"key":"B","latex":"3"},{"key":"C","latex":"\\frac{97}{15}"},{"key":"D","latex":"\\frac{103}{15}"},{"key":"E","latex":"\\frac{163}{15}"},{"key":"F","latex":"18"}],"answerKey":"A","explanationLatex":"We begin by expanding the brackets, either using the binomial theorem or by multiplying them out by hand, to get ∫ 2 ( x^2 + 4 x^2 )^2 dx = ∫ 2 x^4 - 8 + 16 x^4 dx = ∫ 2 x^4 - 8 + 16x- 4 dx = [ x^5 5 - 8x + 16x- 3 - 3 ] 2 = ( 32 5 - 8 × 2 - 16/8 ) - ( 1 5 - 8 - 16 ) = 31 5 - 8 + 14 = 93 - 120 + 70 = 43 and so the answer is A.","tags":["TMUA","2016","Paper2"],"difficulty":2},{"id":"2016-P2-Q2","paper":2,"index":1,"stemLatex":"$f(x) = \\frac{(x^2+5)(2x)}{\\sqrt[4]{x^3}}$, where $x>0$.\n\nWhich one of the following is equal to $f'(x)$?","options":[{"key":"A","latex":"8x^{\\frac{9}{4}} + \\frac{40}{3}x^{\\frac{1}{4}}"},{"key":"B","latex":"\\frac{9}{2}x^{\\frac{5}{4}} + \\frac{5}{2}x^{-\\frac{3}{4}}"},{"key":"C","latex":"8x^{\\frac{9}{4}} + \\frac{40}{3}x^{-\\frac{1}{4}}"},{"key":"D","latex":"\\frac{8}{13}x^{\\frac{13}{4}} + 8x^{\\frac{5}{4}}"}],"answerKey":"B","explanationLatex":"We begin by expanding the brackets and simplifying the powers of x: f(x) = 2x3 + 10x x = 2x 4 + 10x so f ′(x) = 2 × 9 4 x 4 + 10 × 1 4 x- 3 = 9 2 x 4 + 5 2 x- 3 which is option B.","tags":["TMUA","2016","Paper2"],"difficulty":2},{"id":"2016-P2-Q3","paper":2,"index":2,"stemLatex":"What is the value, in radians, of the largest angle $x$ in the range $0 \\leq x \\leq 2\\pi$ that satisfies the equation $8\\sin^2x + 4\\cos^2x = 7$?","options":[{"key":"A","latex":"\\frac{2\\pi}{3}"},{"key":"B","latex":"\\frac{5\\pi}{6}"},{"key":"C","latex":"\\frac{4\\pi}{3}"},{"key":"D","latex":"\\frac{5\\pi}{3}"},{"key":"E","latex":"\\frac{7\\pi}{4}"},{"key":"F","latex":"\\frac{11\\pi}{6}"}],"answerKey":"D","explanationLatex":"We use the identity sin^2 x + cos^2 x = 1 to write the equation in terms of sin x: 8 sin^2 x + 4 cos^2 x = 4 sin^2 x + 4 = 7 so 4 sin^2 x = 3, giving sin^2 x = 3 4 so sin x = ± $\\\\sqrt{2}$ . The solutions to sin x = $\\\\sqrt{2}$ lie in the interval 0 < x < π , while the solutions to sin x = - $\\\\sqrt{2}$ lie in the interval π < x < 2π, and thus the largest angle in the range 0 < x < 2π which satisfies the original equation is the largest solution to sin x = - $\\\\sqrt{2}$ . This is 2π - π 3 = 5π 3 , which is option D.","tags":["TMUA","2016","Paper2"],"difficulty":2},{"id":"2016-P2-Q4","paper":2,"index":3,"stemLatex":"Five sealed urns, labelled $P$, $Q$, $R$, $S$ and $T$, each contain the same (non-zero) number of balls. The following statements are attached to the urns.\n\nUrn $P$: This urn contains one or four balls.\n\nUrn $Q$: This urn contains two or four balls.\n\nUrn $R$: This urn contains more than two balls and fewer than five balls.\n\nUrn $S$: This urn contains one or two balls.\n\nUrn $T$: This urn contains fewer than three balls.\n\nExactly one of the urns has a true statement attached to it.\n\nWhich urn is it?","options":[{"key":"A","latex":"Urn P"},{"key":"B","latex":"Urn Q"},{"key":"C","latex":"Urn R"},{"key":"D","latex":"Urn S"},{"key":"E","latex":"Urn T"}],"answerKey":"C","explanationLatex":"We work through the possibilities systematically. • Urn P’s statement is true, so the urns each contain one or four balls. As Urn Q has a false statement, there cannot be two or four balls in the urns, and so there is one ball in each urn. But then Urn S has a true statement, which is impossible. • Urn Q’s statement is true, so the urns each contain two or four balls. As Urn P has a false statement, there cannot be one or four balls in each urn, so there are two balls in each urn. But then Urn S has a true statement, which is impossible. • Urn R’s statement is true, so there are three or four balls in each urn. As Urn P’s statement is false, there cannot be four balls in the urn, so there are three balls in each urn. This means that Urn Q, Urn S and Urn T each have a false statement, and therefore this is possible. • Urn S’s statement is true, so the urns each contain one or two balls. As Urn P’s statement is false, there cannot be one ball in each urn, so there are two balls in each urn. But then Urn Q has a true statement, which is impossible. • Urn T’s statement is true, so the urns each contain one or two balls. But then Urn S’s statement is true, which is impossible. Therefore the only urn which has a true statement is Urn R, so option C is correct.","tags":["TMUA","2016","Paper2"],"difficulty":2},{"id":"2016-P2-Q5","paper":2,"index":4,"stemLatex":"Consider the statement:\n\n$(\\ast)$ A whole number $n$ is prime if it is 1 less or 5 less than a multiple of 6.\n\nHow many counterexamples to $(\\ast)$ are there in the range $0 < n < 50$?","options":[{"key":"A","latex":"2"},{"key":"B","latex":"3"},{"key":"C","latex":"4"},{"key":"D","latex":"5"},{"key":"E","latex":"6"}],"answerKey":"C","explanationLatex":"We can rewrite the statement ( ∗) in the form If a whole number n is 1 less or 5 less than a multiple of 6, then n is prime. A counterexample to this statement is a whole number n which makes this if statement false, that is, it is a whole number n that is 1 less or 5 less than a multiple of 6, but which is not prime. So we list these numbers and check whether they are prime: n Prime? 1 no 5 yes 7 yes 11 yes 13 yes 17 yes 19 yes 23 yes 25 no 29 yes 31 yes 35 no 37 yes 41 yes 43 yes 47 yes 49 no Thus there are 4 numbers which are counterexamples in this range, and so the answer is C.","tags":["TMUA","2016","Paper2"],"difficulty":2},{"id":"2016-P2-Q6","paper":2,"index":5,"stemLatex":"The sequence of functions $f_1(x), f_2(x), f_3(x),...$ is defined as follows:\n\n$f_1(x) = x^{10}$\n\n$f_{n+1}(x) = xf'_n(x)$ for $n \\geq 1$\n\nwhere $f'_n(x) = \\frac{df_n(x)}{dx}$.\n\nFind the value of $\\sum_{n=1}^{20} f_n(x)$","options":[{"key":"A","latex":"\\frac{x^{10}(x^{20}-1)}{x-1}"},{"key":"B","latex":"\\frac{x^{10}(x^{21}-1)}{x-1}"},{"key":"C","latex":"(\\frac{10^{20}-1}{9})x^{10}"},{"key":"D","latex":"(\\frac{10^{21} - 1}{9})x^{10}"},{"key":"E","latex":"(\\frac{(10x)^{20}-1}{10x-1})x^{10}"},{"key":"F","latex":"(\\frac{(10x)^{21} - 1}{10x-1})x^{10}"},{"key":"G","latex":"x^{10} + x^9 + x^8 + ... + x + 1"},{"key":"H","latex":"x^{10} + 10x^9 + (10 \\times 9)x^8 + ... + (10 \\times 9 \\times ... \\times 2)x + (10 \\times 9 \\times ... \\times 2 \\times 1)"}],"answerKey":"C","explanationLatex":"We work out the first few functions in the sequence to find a pattern. f1(x) = x10 f^2(x) = x(10x9) = 10 x10 f^3(x) = x(100x9) = 100 x10 f^4(x) = x(1000x9) = 10 3x10 f^5(x) = x(104x9) = 10 4x10 and so the pattern is clear: fn(x) = 10 n- 1x10. The sum expands to 20∑ n=1 fn(x) = x10 + 10x10 + 102x10 + 103x10 + · · ·+ 1019x10 = (1 + 10 + 10 2 + · · ·+ 1019)x10 = ( 1020 - 1 10 - 1 ) x10 where on the final line we have summed the geometric series with a = 1 , r = 10 , n = 20 using the formula Sn = a(rn - 1) r - 1 . The answer is therefore option C.","tags":["TMUA","2016","Paper2"],"difficulty":2},{"id":"2016-P2-Q7","paper":2,"index":6,"stemLatex":"The four real numbers $a$, $b$, $c$, and $d$ are all greater than 1. Suppose that they satisfy the equation $\\log_c d = (\\log_a b)^2$. Use the given lines to construct a proof that $\\log_b d = (\\log_a b)(\\log_a c)$.","stemImage":"/stem-fallback/2016-P2-Q7.png","options":[{"key":"A","latex":"(1). Then (2), so (6), so (8), so (7), and therefore (4), hence $(\\ast)$ as required."},{"key":"B","latex":"(1). Then (2), so (7), so (8), so (6), and therefore (4), hence $(\\ast)$ as required."},{"key":"C","latex":"(1). Then (3), so (5), so (9), so (7), and therefore (4), hence $(\\ast)$ as required."},{"key":"D","latex":"(1). Then (3), so (7), so (9), so (5), and therefore (4), hence $(\\ast)$ as required."},{"key":"E","latex":"(1). Then (4), so (5), so (9), so (7), and therefore (3), hence $(\\ast)$ as required."},{"key":"F","latex":"(1). Then (4), so (6), so (8), so (7), and therefore (2), hence $(\\ast)$ as required."},{"key":"G","latex":"(1). Then (4), so (7), so (8), so (6), and therefore (2), hence $(\\ast)$ as required."},{"key":"H","latex":"(1). Then (4), so (7), so (9), so (5), and therefore (3), hence $(\\ast)$ as required."}],"answerKey":"C","explanationLatex":"The proof certainly begins with (1): Let x = log a b and y = log a c. We are then given a choice of writing d in various ways (line (2), (3) or (4)). We are given logc d = (log a b)^2 so we can rewrite this using x as logc d = x^2. Therefore d = c(x^2), which is line (3). Our next choice is between lines (5) and (7) (these are the only options, as the correct answer must be C or D); these are equivalent, but expressed differently. How can we get from d = c(x^2) to d being a power of a? We must rewrite c in terms of y: since y = log a c, it follows that c = ay, so d = (ay)(x^2), which is line (5). Without doing any more work, the answer must therefore be option C, and the whole proof reads as follows: Suppose that logc d = (log a b)^2. (1) Let x = log a b and y = log a c. Then (3) d = c(x^2), so (5) d = ( ay)(x^2), so (9) d = a(x^2y), so (7) d = (ax)xy, and therefore (4) d = bxy, hence (*) as required.","tags":["TMUA","2016","Paper2"],"difficulty":2},{"id":"2016-P2-Q8","paper":2,"index":7,"stemLatex":"A region is defined by the inequalities $x+y>6$ and $x-y > -4$\n\nConsider the three statements:\n\nI: $x>1$\n\nII: $y > 5$\n\nIII: $(x+y)(x-y) > -24$\n\nWhich of the above statements is/are true for every point in the region?","options":[{"key":"A","latex":"none of them"},{"key":"B","latex":"I only"},{"key":"C","latex":"II only"},{"key":"D","latex":"III only"},{"key":"E","latex":"I and II only"},{"key":"F","latex":"I and III only"},{"key":"G","latex":"II and III only"},{"key":"H","latex":"I, II and III"}],"answerKey":"B","explanationLatex":"A useful way to approach this question is to sketch the graph of these inequalities; we shade the invalid regions. (You might find it helpful to rewrite the inequalities as y > 6 - x and y < x + 4.) x y 0 6 The boundaries of the regions intersect when x + y = 6 and x - y = - 4; adding these equations gives 2x = 2, so x = 1 and y = 5, that is, they intersect at (1, 5). Therefore for every point in the white region, condition 1, x > 1, is satisfied. However, there are clearly points for which condition 2, y > 5, is not satisfied, for example (7, 0). Finally, (x + y)(x - y) > - 24 is not immediately clear. This is obtained by multiplying the two inequalities together, but is this permissible? We could try finding a counterexample to this condition. x + y > 6 means that x + y is always positive, so a counterexample must have x - y < 0 (that is, y > x ) to give a product less than - 24. We could take y = x + 3, giving x - y = - 3, and then if we had x + y > 8, the product would be less than - 24. (Recall that the first inequality is equivalent to x + y > 6, so we may do this.) So we have a counterexample: take x = 5 , y = 8 , giving x + y = 13 , and (x + y)(x - y) = - 39 < - 24. Hence this third condition is not true for every point in the region. Therefore the answer is option B.","tags":["TMUA","2016","Paper2"],"difficulty":2},{"id":"2016-P2-Q9","paper":2,"index":8,"stemLatex":"Triangles ABC and XYZ have the same area. Which of the extra conditions, taken independently, would imply that they are congruent?","stemImage":"/stem-fallback/2016-P2-Q9.png","options":[{"key":"A","latex":"$A$"},{"key":"B","latex":"$B$"},{"key":"C","latex":"$C$"},{"key":"D","latex":"$D$"},{"key":"E","latex":"$E$"},{"key":"F","latex":"$F$"},{"key":"G","latex":"$G$"},{"key":"H","latex":"$H$"}],"answerKey":"D","explanationLatex":"We work through the conditions in the order given. We recall three relevant conditions for these two triangles to be congruent: • side-side-side (SSS): all three corresponding pairs of sides are equal in length • side-angle-side (SAS): two corresponding pairs of sides are equal in length, and the corresponding included angles are equal • angle-side-angle (ASA): two corresponding pairs of angles are equal, and the corresponding sides between these angles are equal in length (1) If we knew that either AC = XZ or the angle B equals the angle Y , then they would certainly be congruent by SSS or SAS. We know that the areas are equal, so we have 2 AB.BC sin B = 1 2 XY.Y Z sin Y , and so sin B = sin Y . This does not imply that B = Y though: we could have Y = 180 ◦ - B, for example: B = Y C = Z A X (2) We now have an angle at the end of the given side, so we can use the equal-area property to work out a second side: the angle ∠ ABC is included between AB and BC , and likewise for XY Z . Therefore, as the areas are 1 2 AB.BC sin(∠ ABC ) and 1 2 XY.Y Z sin(∠ XY Z ), it follows that BC = Y Z. The triangles are therefore congruent by SAS. (3) We now have two pairs of corresponding equal angles, and so the third pair of corresponding angles is also equal and the triangles are similar. Since the areas are equal, they must be congruent. Hence conditions (2) and (3) each imply that the triangles are congruent, and the answer is option D.","tags":["TMUA","2016","Paper2"],"difficulty":2},{"id":"2016-P2-Q10","paper":2,"index":9,"stemLatex":"In this question $x$ and $y$ are non-zero real numbers.\n\nWhich one of the following is sufficient to conclude that $x < y$?","options":[{"key":"A","latex":"x^4 < y^4"},{"key":"B","latex":"y^4 < x^4"},{"key":"C","latex":"x^{-1} < y^{-1}"},{"key":"D","latex":"y^{-1} < x^{-1}"},{"key":"E","latex":"x^{\\frac{3}{5}} < y^{\\frac{3}{5}}"},{"key":"F","latex":"y^{\\frac{3}{5}} < x^{\\frac{3}{5}}"}],"answerKey":"E","explanationLatex":"The only sufficient condition is option $E$.\n\nWhy? Because if\n\n$$x^{3/5}<y^{3/5},$$\n\nthen raising both sides to the power of $5$ gives\n\n$$x^3<y^3.$$\n\nSince the cube function is strictly increasing for all real numbers, this implies\n\n$$x<y.$$\n\nSo $E$ is sufficient.\n\nThe others are not sufficient, because they allow counterexamples with $x\\ge y$.\n\nFor example:\n\n- $A$ is not sufficient: take $x=1$, $y=-2$.\n- $B$ is not sufficient: take $x=2$, $y=1$.\n- $C$ is not sufficient: take $x=2$, $y=1$.\n- $D$ is not sufficient: take $x=1$, $y=-1$.\n- $F$ is not sufficient: from $y^{3/5}<x^{3/5}$ we get $y<x$, not $x<y$.\n\nSo the correct answer is $\\mathbf{E}$.","tags":["TMUA","2016","Paper2"],"difficulty":2},{"id":"2016-P2-Q11","paper":2,"index":10,"stemLatex":"$f(x)$ is a polynomial with real coefficients.\n\nThe equation $f(x) = 0$ has exactly two real roots, $x=-p$ and $x=p$, where $p > 0$.\n\nConsider the following three statements:\n\nI: $f'(x) = 0$ for exactly one value of $x$ between $-p$ and $p$\n\nII: The area between the curve $y=f(x)$, the x-axis and the lines $x=-p$ and $x=p$ is given by $2\\int_0^pf(x)dx$\n\nIII: The graph of $y=-f(-x)$ intersects the x-axis at the points $x=-p$ and $x=p$ only\n\nWhich of the above statements must be true?","options":[{"key":"A","latex":"none of them"},{"key":"B","latex":"I only"},{"key":"C","latex":"II only"},{"key":"D","latex":"III only"},{"key":"E","latex":"I and II only"},{"key":"F","latex":"I and III only"},{"key":"G","latex":"II and III only"},{"key":"H","latex":"I, II and III"}],"answerKey":"D","explanationLatex":"We know that the polynomial intersects the x-axis at x = - p and x = p only. It may touch the x-axis at those points or it may cross it - we are not told whether x = p is a single root or double root, for example. We consider the three statements in turn. 1 It is entirely possible that y = f(x) has multiple stationary points between x = - p and x = p without crossing the x-axis, for example in this sketch: x y p- p Therefore this statement need not be true. 2 This is not necessarily true for (at least) two different reasons: • If the graph of y = f(x) is entirely below the x-axis between x = p and x = - p, as in the above sketch, then the integral ∫ p 0 f(x) dx will be negative, but the area is positive. • If the graph of y = f(x) is not symmetrical about x = 0, then there is no reason that the integal ∫ p - p f(x) dx will be twice ∫ p 0 f(x) dx. An example is this sketch, which is of the graph of y = (x + 1)(x - 1)(x^2 + x + 1); here p = 1, and the graph of y = x^2 + x + 1 is always positive, but it is not symmetrical about x = 0: x y p- p 3 The graph of y = f( - x) is obtained by reflecting the graph of y = f( x) in the y-axis, so this graph intersects the x-axis at x = - p and x = p. The graph of y = - f(- x) is obtained from this by reflecting the graph in the x-axis, which does not change the points of intersection with the x-axis. Therefore this statement must be true. The graph obtained by combining these two transformations is the rotation of the original graph about the origin by 180◦. The answer is therefore option D.","tags":["TMUA","2016","Paper2"],"difficulty":2},{"id":"2016-P2-Q12","paper":2,"index":11,"stemLatex":"The first term of an arithmetic sequence is $a$ and the common difference is $d$.\n\nThe sum of the first $n$ terms is denoted by $S_n$.\n\nIf $S_8 > 3S_6$, what can be deduced about the sign of $a$ and the sign of $d$?","options":[{"key":"A","latex":"both $a$ and $d$ are negative"},{"key":"B","latex":"$a$ is positive, $d$ is negative"},{"key":"C","latex":"$a$ is negative, $d$ is positive"},{"key":"D","latex":"$a$ is negative, but the sign of $d$ cannot be deduced"},{"key":"E","latex":"$d$ is negative, but the sign of $a$ cannot be deduced"},{"key":"F","latex":"neither the sign of $a$ nor the sign of $d$ can be deduced"}],"answerKey":"F","explanationLatex":"The formula for Sn in terms of a and d is Sn = 1 2 n(2a + (n - 1)d). Substituting n = 8 and n = 6 gives S^8 = 4(2a + 7d) S^6 = 3(2a + 5d) so the condition S^8 > 3S6 becomes 4(2a + 7d) > 3 × 3(2a + 5d). Expanding the brackets gives 8a + 28d > 18a + 45d, which on rearranging gives - 8a > 17d, or d < - 8 17 a. If a is positive, then - 8 17 a is negative, so d must be negative. If a is negative, then - 8 17 a is positive, so d could be positive or negative. So a can be positive or negative, as can d. We therefore cannot deduce the sign of either a or d, and so the correct option is F.","tags":["TMUA","2016","Paper2"],"difficulty":2},{"id":"2016-P2-Q13","paper":2,"index":12,"stemLatex":"In this question, a, b, and c are positive integers. The following is an attempted proof of the false statement: If a divides bc, then a divides b or a divides c. Which line contains the error in this proof?","stemImage":"/stem-fallback/2016-P2-Q13.png","options":[{"key":"A","latex":"Line 1"},{"key":"B","latex":"Line 2"},{"key":"C","latex":"Line 3"},{"key":"D","latex":"Line 4"},{"key":"E","latex":"Line 5"},{"key":"F","latex":"Line 6"}],"answerKey":"E","explanationLatex":"Working through the lines of the argument one at a time: 1. This looks like it might be the contrapositive of the statement we are aiming to prove; let us check this. The original statement is: If a divides bc, then a divides b or a divides c so the contrapositive is If not (a divides b or a divides c), then not (a divides bc) which simplifies to If not (a divides b) and not (a divides c), then a does not divide bc or If a does not divide b and a does not divide c, then a does not divide bc So line 1 is correct. 2. “Suppose ...” starts the proof: we suppose the antecedent of the “ if” statement. The part about remainders is correct; this is what it means for a to not divide b. 3. This is an expression of division with remainder: b = ax + r could be read as “ b divided by a is x remainder r. ” So this line is correct. 4. A small amount of algebra shows that this line is correct. 5. This is not clearly correct: what happens if rs > a , for example? We will return to this line in a moment. 6. This is clearly a correct assertion. 7. If line 5 were correct, then this would be correct. Since we are told that the statement is false, the error must lie in line 5. We have already identified one possible problem, but we can do even better by finding a case where the remainder when dividing bc by a is 0, showing that a does divide bc. Let us take, for example, a = 6. Then if r = 2 and s = 3, we have rs = 6 = a, so that the remainder is 0. An explicit example would then be a = 6, b = 2 and c = 3. The correct answer is therefore E.","tags":["TMUA","2016","Paper2"],"difficulty":2},{"id":"2016-P2-Q14","paper":2,"index":13,"stemLatex":"$f(x) = ax^4 + bx^3 + cx^2 + dx + e$, where $a$, $b$, $c$, $d$ and $e$ are real numbers.\n\nSuppose $f(x) = 1$ has $p$ distinct real solutions, $f(x) = 2$ has $q$ distinct real solutions, $f(x) = 3$ has $r$ distinct real solutions, and $f(x)=4$ has $s$ distinct real solutions.\n\nWhich one of the following is not possible?","options":[{"key":"A","latex":"$p=1$, $q=2$, $r=4$ and $s=3$"},{"key":"B","latex":"$p=1$, $q=3$, $r=2$ and $s=4$"},{"key":"C","latex":"$p=1$, $q=4$, $r=3$ and $s=2$"},{"key":"D","latex":"$p=2$, $q=4$, $r=3$ and $s=1$"},{"key":"E","latex":"$p=4$, $q=3$, $r=2$ and $s=1$"}],"answerKey":"B","explanationLatex":"The number of solutions to f(x) = 1 is the number of times the graph of y = f( x) intersects with the line y = 1, and similarly for the other three cases. So we need to think about possible shapes of y = f( x). We assume that a ≠ 0 for now, so that we are dealing with a quartic; this will be enough to answer the question. (Each of the given options has one of p, q, r or s equal to 4, which is impossible if the curve is a cubic, quadratic or linear.) A quartic with a > 0 either has a W-shape, or it is a bit like a quadratic in that it has exactly one local minimum. If a < 0, the possibilities are the reflections of these. We can then try to construct curves with the given conditions. A p = 1 and q = 2 indicates that a > 0 and that y = 1 is tangent to the curve. So something like this would work, where we have drawn dashed lines to show y = 1 , y = 2 , y = 3 and y = 4. (We could have left out the y-axis for clarity.) x y (An accurately-plotted quartic would not look quite like this, as you can see if you use graphdrawing software. Nevertheless, the general shape could be correct, with the stationary points in roughly the locations shown.) We could also draw this schematically as: x y We will use this schematic representation for the rest of the cases. B Again, we require a minimum at y = 1. But as there are 4 crossings at y = 4, the curve must do something like this: x y where the last part is not yet clear. However, to get r = 2 , the last minimum must lie above y = 3 , but then we cannot have q = 3. Hence this is not possible. We look at the other three options for the sake of completeness. C Again, there is a minimum at y = 1, and r = 3 gives a stationary point at y = 3. This sketch does the job: x y D Since s = 1 , we must have a < 0 and y = 4 is a maximum. Likewise, y = 3 must also be a stationary point. The following sketch achieves this: x y E This is similar to the previous option: a < 0 and there is a maximum at y = 4 . As q = 3 , there is a stationary point at y = 2, which gives the following sketch: x y The correct option is B.","tags":["TMUA","2016","Paper2"],"difficulty":2},{"id":"2016-P2-Q15","paper":2,"index":14,"stemLatex":"Consider the quadratic $f(x) = x^2 - 2px + q$ and the statement:\n\n$(\\ast)$ $f(x) = 0$ has two real roots whose difference is greater than 2 and less than 4.\n\nWhich one of the following statements is true if and only if $(\\ast)$ is true?","options":[{"key":"A","latex":"q < p^2 < q+4"},{"key":"B","latex":"\\sqrt{q+1} < p < \\sqrt{q+4}"},{"key":"C","latex":"q-3 \\leq p^2 - 4 \\leq q"},{"key":"D","latex":"q < p^2 - 1 < q+3"},{"key":"E","latex":"q-2 < p^2 - 3 < q+2"}],"answerKey":"D","explanationLatex":"We start by finding the roots of f(x) = 0 . They are given by the quadratic formula, and are r1, r^2 = 2p ± √(- 2p)^2 - 4q 2 = p ± √p^2 - q. The roots are real if and only if the discriminant is non-negative, so if and only if p^2 - q ⩾ 0. The difference between the roots is r^2 - r1 = 2 √p^2 - q, so the condition (∗) is true if and only if p^2 - q ⩾ 0 and 2 < 2 √p^2 - q < 4 so it is true if and only if p^2 - q ⩾ 0 and 1 < √p^2 - q < 2 which is true if and only if p^2 - q ⩾ 0 and 1 < p 2 - q < 4 The first condition is implied by the second condition, so both conditions are true if and only if 1 < p 2- q < 4, which we can rearrange to q+1 < p 2 < q +4. Subtracting 1 gives q < p 2- 1 < q +3, which is condition D. We can also show that the other four statements are not true if and only if (∗) is true, as follows. Statement A fails, because the left inequality q < p 2 is not the same as q < p 2 - 1. Statement A is true if (∗) is true, but (∗) could be true without statement A being true; in other words “statement A is true only if (∗) is true” is false. Statement B fails, because this requires p > 0, whereas we could have p < 0; thus, statement B is true only if (∗) is true, but “statement B is true if (∗) is true” is false. Statement C fails, because this allows for the difference between the roots to be exactly 2 or 4. So statement C is true if (∗) is true, but “statement C is true only if (∗) is true” is false. Finally, statement E fails, because the second inequality is equivalent to p^2 - 1 < q + 4, so similarly to statement A, statement E is true if (∗) is true, but “statement E is true only if (∗) is true” is false. So the correct option is D.","tags":["TMUA","2016","Paper2"],"difficulty":2},{"id":"2016-P2-Q16","paper":2,"index":15,"stemLatex":"The figure shows trapezium PQRS with PQ parallel to SR. The diagonals meet at X. U lies on SP and T lies on RQ such that UT passes through X and is parallel to PQ. The length of PQ is 12 cm and the length of SR is 3 cm.","stemImage":"/stem-fallback/2016-P2-Q16.png","options":[{"key":"A","latex":"4.2"},{"key":"B","latex":"4.5"},{"key":"C","latex":"4.8"},{"key":"D","latex":"5.25"},{"key":"E","latex":"6"}],"answerKey":"C","explanationLatex":"There are several pairs of similar triangles in this diagram. We note that SU X and SP Q are similar, RXT and RP Q are similar, and XSR and XQP are similar. The last pair is the most helpful to start with, as we know the lengths of two corresponding sides: SR = 3 and QP = 12 , so the scale factor is 12 3 = 4 . Thus the height of XQP (from X) is 4 times the height of XSR (again from X). It follows that U P = 4SU , and so SP = 5SU . Since SU X is similar to SP Q, and SP = 5SU , we must have P Q = 5U X, so U X = 1 5 P Q = 12 5 . Likewise, XT = 12 5 , so U T = 24 5 = 4.8 and the correct answer is C. We could alternatively have used the similarity of triangles P U X and P SR, and of triangles QXT and QSR; we would reach the same result.","tags":["TMUA","2016","Paper2"],"difficulty":2},{"id":"2016-P2-Q17","paper":2,"index":16,"stemLatex":"Consider these simultaneous equations, where $c$ is a constant:\n\n$y = 3\\sin x + 2$\n\n$y = x + c$\n\nWhich of the following statements is/are true?\n\nI: For some value of $c$: there is exactly one solution with $0 \\leq x \\leq \\pi$ and there is at least one solution with $-\\pi < x < 0$.\n\nII: For some value of $c$: there is exactly one solution with $0 \\leq x \\leq \\pi$ and there are no solutions with $-\\pi < x < 0$.\n\nIII: For some value of $c$: there is exactly one solution with $0 \\leq x \\leq \\pi$ and there are no solutions with $x > \\pi$.","options":[{"key":"A","latex":"none of them"},{"key":"B","latex":"I only"},{"key":"C","latex":"II only"},{"key":"D","latex":"III only"},{"key":"E","latex":"I and II only"},{"key":"F","latex":"I and III only"},{"key":"G","latex":"II and III only"},{"key":"H","latex":"I, II and III"}],"answerKey":"H","explanationLatex":"The graph of y = 3 sin x + 2 is the standard sine graph stretched vertically by a factor of 3 and then translated in the y-direction by 2. It thus oscillates between 2 - 3 = - 1 and 2 + 3 = 5, and its y-axis intercept is 2. y = x + c has gradient 1 and y-intercept c. We sketch the graph of y = 3 sin x + 2 to help us: x y π- π 2π- 2π - 1 We now consider the statements in turn: 1 To have exactly one solution with 0 ⩽ x ⩽ π and at least one solution with - π < x < 0, we could take c to be just less than 2. So this statement is true. 2 We want to choose c to be negative for this case; if c is such that y = x + c intersects the sine curve at x = π, there will be exactly one solution with 0 ⩽ x ⩽ π. This value of c must satisfy 2 = π + c, so c = 2 - π < - 1, so there are no solutions with x < 0. So this statement is true. 3 The same value of c as in statement 2 will work here: c = 2 - π. At x = 2π, x + c = 2 + π > 5, so there cannot be any solutions with x ⩾ 2π. As the sine graph is below 2 for π < x < 2π, there are no solutions there either. So this statement is true. The correct option is therefore H.","tags":["TMUA","2016","Paper2"],"difficulty":2},{"id":"2016-P2-Q18","paper":2,"index":17,"stemLatex":"Consider this statement about a function $f(x)$:\n\n$(\\ast)$ If $(f(x))^2 \\leq 1$ for all $-1 \\leq x \\leq 1$, then\n\n$\\int_{-1}^1 (f(x))^2\\,dx \\leq \\int_{-1}^1 f(x)\\,dx$.\n\nWhich one of the following functions provides a counterexample to $(\\ast)$?","options":[{"key":"A","latex":"f(x) = x + \\frac{1}{2}"},{"key":"B","latex":"f(x) = x - \\frac{1}{2}"},{"key":"C","latex":"f(x) = x + x^3"},{"key":"D","latex":"f(x) = x - x^3"},{"key":"E","latex":"f(x) = x^2 + x^4"},{"key":"F","latex":"f(x) = x^2 - x^4"}],"answerKey":"D","explanationLatex":"To be a counterexample to the implication, the antecedent must be true but the consequent must be false.\n\nSo we first look for functions satisfying\n\n$$\\bigl(f(x)\\bigr)^2\\le 1 \\qquad \\text{for all } -1\\le x\\le 1.$$\n\nOptions $A$, $B$, $C$ and $E$ fail this immediately, since each takes a value with $|f(x)|>1$ somewhere in the interval.\n\nFor $D$, the function is $f(x)=x-x^3$. Then\n\n$$f'(x)=1-3x^2,$$\n\nso the stationary points occur at\n\n$$x=\\pm \\frac{1}{\\sqrt{3}}.$$\n\nAt these points,\n\n$$f\\!\\left(\\frac{1}{\\sqrt{3}}\\right)=\\frac{1}{\\sqrt{3}}-\\frac{1}{3\\sqrt{3}}=\\frac{2}{3\\sqrt{3}}=\\frac{2\\sqrt{3}}{9},$$\n\nand similarly\n\n$$f\\!\\left(-\\frac{1}{\\sqrt{3}}\\right)=-\\frac{2\\sqrt{3}}{9}.$$\n\nSo $|f(x)|\\le \\frac{2\\sqrt{3}}{9}<1$ on $[-1,1]$, and the antecedent is true for $D$.\n\nNow compare the integrals. Since $x-x^3$ is an odd function,\n\n$$\\int_{-1}^{1}f(x)\\,dx=0.$$\n\nBut $\\bigl(f(x)\\bigr)^2\\ge 0$, and it is not identically zero, so\n\n$$\\int_{-1}^{1}\\bigl(f(x)\\bigr)^2\\,dx>0.$$\n\nHence\n\n$$\\int_{-1}^{1}\\bigl(f(x)\\bigr)^2\\,dx>\\int_{-1}^{1}f(x)\\,dx,$$\n\nso the consequent is false. Therefore $D$ is a counterexample.\n\nFor completeness, $F$ is not a counterexample: there $f(x)=x^2-x^4$, so $0\\le f(x)\\le 1$ on $[-1,1]$, which implies $f(x)^2\\le f(x)$ and therefore\n\n$$\\int_{-1}^{1}f(x)^2\\,dx\\le \\int_{-1}^{1}f(x)\\,dx.$$\n\nSo the correct answer is $\\mathbf{D}$.","tags":["TMUA","2016","Paper2"],"difficulty":2},{"id":"2016-P2-Q19","paper":2,"index":18,"stemLatex":"Some identical unit cubes are glued face-to-face to form a 3D object. Sketches are made from the right-hand side, from the front, and from above. The side elevation, front elevation, and plan view are shown.","stemImage":"/stem-fallback/2016-P2-Q19.png","options":[{"key":"A","latex":"exactly 6"},{"key":"B","latex":"either 6 or 7"},{"key":"C","latex":"exactly 7"},{"key":"D","latex":"either 7 or 8"},{"key":"E","latex":"exactly 8"},{"key":"F","latex":"either 8 or 9"},{"key":"G","latex":"exactly 9"}],"answerKey":"F","explanationLatex":"We first note that the object fits within a 3 × 3 cube. We begin with the plan view of the object: all of the cubes must be within three layers fitting within this L-shape. Let us show the three layers: ? ? ? ? ? bottom layer ? ? ? ? ? middle layer ? ? ? ? ? top layer We have used ‘?’ to indicate that we do not yet know where there are cubes. We do, however, know that there must be a cube in at least one of the three layers for each of the five positions. Next, let us look at the front elevation. This tells us that in the left hand column, we can see two cubes. Since there is only one possible location for the cubes in this column, the first two layers must have a cube and the third not. Likewise, for the middle column, there is only a cube in the bottom layer. In the right hand column, there must be at least one cube in each layer, but it is not yet clear where these would be, as there are three possible positions in each layer. We now know the following: Y Y ? ? ? bottom layer Y x ? ? ? middle layer x x ? ? ? top layer Next, let us consider the side elevation. The left hand column, which corresponds to the bottom row of the plan view layers, has height 2, which means that the bottom right corner of the top layer cannot have a cube. The middle column, which corresponds to the middle row of the plan view layers, has height three, so all three of these cubes must be present (as there is only one possible position in the middle row). Finally, the right column, corresponding to the top row of the plan view layers, has height one, so only the bottom layer has a cube in this position. We thus reach this position: Y Y ? Y Y bottom layer Y x ? Y x middle layer x x x Y x top layer One of the ‘?’ cubes must be present, to give the required plan view, and the other one may or may not be - it makes no difference to any of the views. So there are either 8 or 9 cubes, and the answer is option F. (Note that we would need the corner of the bottom layer to be ‘Y’ if it is a single object.)","tags":["TMUA","2016","Paper2"],"difficulty":2},{"id":"2016-P2-Q20","paper":2,"index":19,"stemLatex":"Each interior angle of a regular polygon with $n$ sides is $\\frac{3}{4}$ of each interior angle of a second regular polygon with $m$ sides.\n\nHow many pairs of positive integers $n$ and $m$ are there for which this statement is true?","options":[{"key":"A","latex":"none"},{"key":"B","latex":"1"},{"key":"C","latex":"2"},{"key":"D","latex":"3"},{"key":"E","latex":"4"},{"key":"F","latex":"5"},{"key":"G","latex":"6"},{"key":"H","latex":"infinitely many"}],"answerKey":"E","explanationLatex":"We can find the interior angle of a regular polygon with n sides as follows. The exterior angle is 360◦ n , so the interior angle is 180◦ - 360◦ n . Therefore the given condition can be written as 180◦ - 360◦ n = 3 ( 180◦ - 360◦ m ) . We start by dividing by 180◦ for simplicity, giving 1 - 2 n = 3 ( 1 - 2 m ) . Now multiplying both sides by 4 and expanding the brackets gives 4 - 8 n = 3 - 6 m , which rearranges to n - 6 m = 1. (∗) We will look at two methods of solving this equation, starting with a case-based approach. We know that n and m must each be at least 3. If n ⩾ 8, then 8 n ⩽ 1, and so this cannot give a solution to ( ∗). So we need only consider n = 3 , 4, 5, 6 and 7. For each of these, we can solve the equation to find m; if m is an integer greater than 2, we have a solution. n m 3 18 4 6 5 10 6 18 7 42 There are four pairs of n and m, and the answer is E. A second approach to solving ( ∗) is to multiply both sides by mn to eliminate the fractions; this gives 8m - 6n = mn. We can rearrange this to give mn + 6n - 8m = 0 , and we can factorise the left hand side by adding or subtracting a constant to get (m + 6)(n - 8) = - 48. So we need to find two integers which multiply to - 48. Since m and n are both positive, we must have n - 8 < 0 and m + 6 ⩾ 9 (as m ⩾ 3), so the possibilities are: m + 6 n - 8 m n 12 - 4 6 4 16 - 3 10 5 24 - 2 18 6 48 - 1 42 7 as before. Our research-based tests provide a fair measure of skills and aptitude to help you make informed decisions. As a trusted partner, we work closely with universities, governments and employers to enhance their selection processes. Cambridge Assessment Admissions Testing The Triangle Building Shaftesbury Road Cambridge CB2 8EA United Kingdom Admissions tests support: admissionstesting.org/help","tags":["TMUA","2016","Paper2"],"difficulty":2}]