[{"id":"2018-P1-Q1","paper":1,"index":0,"stemLatex":"Find the value of $\\int_1^4 \\frac{3-2x}{x\\sqrt{x}}\\:dx$.","options":[{"key":"A","latex":"-\\frac{13}{2}"},{"key":"B","latex":"-\\frac{85}{16}"},{"key":"C","latex":"-\\frac{13}{8}"},{"key":"D","latex":"-1"},{"key":"E","latex":"-\\frac{1}{4}"},{"key":"F","latex":"\\frac{7}{4}"},{"key":"G","latex":"7"}],"answerKey":"D","explanationLatex":"We write the integrand (the thing being integrated) in terms of powers of x and split up the fraction in order to be able to compute the integral: ∫ 4 3- 2x x√x dx = ∫ 4 3- 2x x dx = ∫ 4 3x- 3 2- 2x- 1 2 dx = [ 3x- 1 (- 1 2)- 2x ]4 = [ -6x- 1 2- 4x ]4 = (-6× 1 2- 4× 2)- (-6× 1- 4× 1) =-11- (-10) =-1 so the correct answer is D.","tags":["TMUA","2018","Paper1"],"difficulty":2},{"id":"2018-P1-Q2","paper":1,"index":1,"stemLatex":"An arithmetic progression has first term $a$ and common difference $d$. \n\n The sum of the first 5 terms is equal to the sum of the first 8 terms. \n\n Which one of the following expresses the relationship between $a$ and $d$?","options":[{"key":"A","latex":"a = -\\frac{38}{3}d"},{"key":"B","latex":"a = -7d"},{"key":"C","latex":"a = -6d"},{"key":"D","latex":"a = 6d"},{"key":"E","latex":"a = 7d"},{"key":"F","latex":"a = \\frac{38}{3}d"}],"answerKey":"C","explanationLatex":"One approach is to use the formulae for the sum of the first 5 and the first 8 terms (that is, Sn = 1 2n(2a + (n- 1)d)). We have S^5 = 1 2× 5(2a + 4d) = 5 2(2a + 4d) S^8 = 1 2× 8(2a + 7d) = 4(2a + 7d) These are equal, so we have 2(2a + 4d) = 4(2a + 7d) We want to rearrange this to get a formula in the form a =··· , so we multiply both sides by 2, expand the brackets and collect like terms: 5(2a + 4d) = 8(2a + 7d) ⇐⇒ 10a + 20d = 16a + 56d ⇐⇒ -6a = 36d ⇐⇒ a =-6d so the answer is C. An alternative approach is to note that the difference between the sum of the first 5 and the sum of the first 8 terms equals the sum of the 6th, 7th and 8th terms. (This is because S^5 = u1 +u^2 +··· +u^5 while S^8 = u1 +u^2 +··· +u^5 +u^6 +u^7 +u^8, where we have written the sequence as u1, u^2, ... .) Since the two sums S^5 and S^8 are equal, the sum of the 6th, 7th and 8th terms must be zero. This gives (a + 5d) + (a+ 6d) + (a+ 7d) = 0 Expanding and collecting like terms gives 3a + 18d = 0, so a =-6d, as before.","tags":["TMUA","2018","Paper1"],"difficulty":2},{"id":"2018-P1-Q3","paper":1,"index":2,"stemLatex":"Find the shortest distance between the two circles with equations:\n$(x+2)^2 + (y-3)^2 = 18$\n$(x-7)^2 + (y+6)^2 = 2$.","options":[{"key":"A","latex":"0"},{"key":"B","latex":"4"},{"key":"C","latex":"16"},{"key":"D","latex":"2\\sqrt{2}"},{"key":"E","latex":"5\\sqrt{2}"}],"answerKey":"E","explanationLatex":"The first circle has centre (-2, 3) and radius $\\\\sqrt{18}$ = 3 $\\\\sqrt{The}$ second circle has centre (7,-6) and radius $\\\\sqrt{It}$ would help to draw a quick sketch of the situation. x y (-2, 3) (7,-6) It seems clear from the sketch that to find the shortest distance between the circles, we should draw a line between their centres and find the length of the segment which lies outside both of them: x y (-2, 3) (7,-6) The distance between the centres is √((-2)- 7)^2 + (3- (-6))^2 = $\\\\sqrt{81}$ + 81 = 9 $\\\\sqrt{and}$ if we subtract the two radii (3 $\\\\sqrt{2}$ and $\\\\sqrt{2}$) for the parts of the line which lie within the circles, we are left with 9 √2- 3 √2- $\\\\sqrt{2}$ = 5 $\\\\sqrt{2}$. The correct answer is E. This can also be proved by taking an arbitrary point on each of the circles, finding the distance between them and then minimising this, but that is far more complicated than needed.","tags":["TMUA","2018","Paper1"],"difficulty":2},{"id":"2018-P1-Q4","paper":1,"index":3,"stemLatex":"Consider the simultaneous equations $$3x^2 + 2xy = 4$$ $$x + y = a$$ where $a$ is a real constant. \n\n Find the complete set of values of $a$ for which the equations have two distinct real solutions for $x$.","options":[{"key":"A","latex":"There are no values of $a$."},{"key":"B","latex":"-2 < a < 2"},{"key":"C","latex":"-1 < a < 1"},{"key":"D","latex":"a = 0"},{"key":"E","latex":"$a<-1$ or $a>1$"},{"key":"F","latex":"$a<-2$ or $a>2$"},{"key":"G","latex":"All real values of $a$."}],"answerKey":"G","explanationLatex":"We can solve the equations to find x by substituting for y. The second equation gives y =a-x, so the first equation becomes 3x2 + 2x(a-x) = 4 This rearranges to x^2 + 2ax- 4 = 0 and the discriminant of this equation is (2a)^2- 4× 1× (-4) = 4a2 + 16 This is always positive, as 4a 2 ⩾ 0 for all real values of a, so there are always two distinct real roots, and the correct answer is G.","tags":["TMUA","2018","Paper1"],"difficulty":2},{"id":"2018-P1-Q5","paper":1,"index":4,"stemLatex":"The function $f$ is defined by $f(x) = x^3 + ax^2 + bx + c$ \n\n $a$, $b$ and $c$ take the values 1, 2, and 3 with no two of them being equal and not necessarily in this order. \n\n The remainder when $f(x)$ is divided by $(x+2)$ is $R$. \n\n The remainder when $f(x)$ is divided by $(x+3)$ is $S$. \n\n What is the largest possible value of $R-S$?","options":[{"key":"A","latex":"-26"},{"key":"B","latex":"5"},{"key":"C","latex":"7"},{"key":"D","latex":"17"},{"key":"E","latex":"29"}],"answerKey":"D","explanationLatex":"For this question, we make use of the remainder theorem: the remainder when f( x) is divided by (x-a) is f(a). In this case, we take a =-2 and a =-3, so f(-2) =R f(-3) =S Substituting these into the given formula for f( x) gives f(-2) =-8 + 4a - 2b +c =R f(-3) =-27 + 9a- 3b +c =S We want the largest possible value of R-S, so we subtract these two equations to get 19- 5a +b =R-S To get the largest possible value, we want a to be as small as possible and b to be as large as possible, so we take a = 1, b = 3 (and then c = 2), giving R-S = 19- 5 + 3 = 17 Thus the correct answer is D.","tags":["TMUA","2018","Paper1"],"difficulty":2},{"id":"2018-P1-Q6","paper":1,"index":5,"stemLatex":"Find the number of solutions of the equation $$x\\sin(2x) = \\cos(2x)$$ with $0 \\leq x \\leq 2\\pi$.","options":[{"key":"A","latex":"0"},{"key":"B","latex":"1"},{"key":"C","latex":"2"},{"key":"D","latex":"3"},{"key":"E","latex":"4"}],"answerKey":"E","explanationLatex":"This equation has trigonometric functions and anx on its own. Our best bet is to first rearrange it to bring all of the trigonometric functions together. We have a sin 2x on one side and a cos 2x on the other, which suggests that we should divide by cos 2x to obtain tan 2x. This gives x tan 2x = 1 We should check, though, that we cannot have cos 2x = 0, so that we are allowed to divide by it. If cos 2x = 0, this would require x sin 2x = 0, so either sin 2x = 0, which is impossible if cos 2x = 0, or x = 0. But if x = 0, then cos 2x = 1, so this is also impossible. So we cannot have cos 2x = 0. Returning to x tan 2x = 1, we can divide by x to get tan 2x = 1 x This equation is not possible to solve exactly. But the question (thankfully) only asks us how many solutions there are in the range 0 ⩽x ⩽ 2π So we can sketch the graphs of y = tan 2x and y = 1 x and see how many times they intersect. x y 0 π π 3π 2π -2 -1 W e see that the graph of y = 1 x in tersects every positive branch of y = tan 2x once. Since there are four branches of y = tan 2x in the range 0 ⩽x ⩽ 2π, there are four solutions, so the correct answer is E.","tags":["TMUA","2018","Paper1"],"difficulty":2},{"id":"2018-P1-Q7","paper":1,"index":6,"stemLatex":"The non-zero constant $k$ is chosen so that the coefficients of $x^6$ in the expansions of $(1+kx^2)^7$ and $(k+x)^{10}$ are equal. \n\n What is the value of $k$?","options":[{"key":"A","latex":"\\frac{1}{6}"},{"key":"B","latex":"6"},{"key":"C","latex":"\\frac{\\sqrt{6}}{6}"},{"key":"D","latex":"\\sqrt{6}"},{"key":"E","latex":"\\frac{\\sqrt{30}}{30}"},{"key":"F","latex":"\\sqrt{30}"}],"answerKey":"A","explanationLatex":"We can expand these two expressions using the binomial theorem. We look just at the term involvingx6: (1 +kx2)^7 =··· + (7 ) 14(kx2)^3 +··· =··· + 7× 6× 5 3× 2× 1k^3x6 +··· =··· + 35k^3x6 +··· and (k +x)10 =··· + (10 ) k^4x6 +··· =··· + (10 ) k^4x6 +··· =··· + 10× 9× 8× 7 4× 3× 2× 1 k^4x6 +··· =··· + 210k^4x6 +··· We therefore need 35k3 = 210k4 so k = 35 210 = 5 30 = 1 6. Thus the correct answer is A.","tags":["TMUA","2018","Paper1"],"difficulty":2},{"id":"2018-P1-Q8","paper":1,"index":7,"stemLatex":"The sum to infinity of a geometric progression is 6. \n\n The sum to infinity of the squares of each term in the progression is 12. \n\n Find the sum to infinity of the cubes of each term in the progression.","options":[{"key":"A","latex":"8"},{"key":"B","latex":"18"},{"key":"C","latex":"24"},{"key":"D","latex":"\\frac{216}{7}"},{"key":"E","latex":"72"},{"key":"F","latex":"216"}],"answerKey":"D","explanationLatex":"Let us write the terms of the GP (geometric progression) as a, ar, ar2 and so on. Then we are told that S∞ =a +ar +ar2 +··· = a 1-r = 6 (1) We can rearrange this final equation to a = 6- 6r if we wish to. The squares of each term are a2, a^2r2, a^2r4, a^2r6 and so on. These themselves form a GP with first term a2 and common ratio r^2. Therefore, given that their sum is 12, we have for this GP S∞ =a2 +a^2r2 +a^2r4 +··· = a2 1-r^2 = 12 We can write this last equation as a2 (1 +r)(1-r) = 12 But we can now substitute in equation (1) to give ( a 1 +r ) = 12 so that a 1 +r = 2, or a = 2 + 2r Thus we have simultaneous equations for a and r: a = 6- 6r a = 2 + 2r Subtracting these gives 4- 8r = 0, so r = 1 2 and a = 3 We can now work out the sum to infinity of the cubes: the sequence of terms becomes a3,a^3r3, a^3r6 and so on, which as before is a GP; this time the first term is a3 and the common ratio is r^3. Hence the sum to infinity is S∞ = a3 1-r^3 = 33 1- ( 1 2)^3 = 27 ( 7 8) = 216 and the correct answer is D.","tags":["TMUA","2018","Paper1"],"difficulty":2},{"id":"2018-P1-Q9","paper":1,"index":8,"stemLatex":"Find the complete set of values of the constant $c$ for which the cubic equation $$2x^3 - 3x^2 - 12x + c = 0$$ has three distinct real solutions.","options":[{"key":"A","latex":"-20 < c < 7"},{"key":"B","latex":"-7 < c < 20"},{"key":"C","latex":"c > 7"},{"key":"D","latex":"c > -7"},{"key":"E","latex":"c < 20"},{"key":"F","latex":"c < -20"}],"answerKey":"B","explanationLatex":"Commentary: We want the cubic equation^2x3-3x2-12x+c = 0 to have three distinct real solutions. We don’t know any techniques for finding the roots of a cubic equation, so we will have to be creative. (Actually, there is a formula, often known as Cardano’s formula, for solving a cubic equation, but you are not expected to know this.) What we do know how to do is to sketch a cubic graph by finding its turning points and y-axis intercept, even if we don’t know the x-axis intercepts, so we will do this. We can find the stationary points on the graph of y = 2x3- 3x2- 12x +c by differentiating. We have dy dx = 6x2- 6x- 12 = 6(x^2-x- 2) = 6(x- 2)(x + 1) so there are stationary points at x = 2 and x =-1 (where dy dx = 0) These have coordinates (-1, 7 +c) and (2, -20 +c) For the equation 2x 3- 3x2- 12x +c = 0 to have three distinct real solutions, we need the two turning points of the cubic to be on opposite sides of the x-axis, one above it and one below it. It is clear that -20 +c <7 +c, so we need -20 +c <0 and 7 + c >0. The first inequality becomes c< 20 and the second c> -7. Combining these, there are three distinct real solutions if and only if -7 < c <20, which is option B.","tags":["TMUA","2018","Paper1"],"difficulty":2},{"id":"2018-P1-Q10","paper":1,"index":9,"stemLatex":"$x$ and $y$ satisfy $|2-x| \\leq 6$ and $|y+2| \\leq 4$. \n\n What is the greatest possible value of $|xy|$?","options":[{"key":"A","latex":"16"},{"key":"B","latex":"24"},{"key":"C","latex":"32"},{"key":"D","latex":"40"},{"key":"E","latex":"48"},{"key":"F","latex":"There is no greatest possible value."}],"answerKey":"E","explanationLatex":"We notice that the first inequality involves x only and the second involves y only, so we can solve each inequality separately. For the first one,|2-x| ⩽ 6, this means that x is at distance at most 6 from 2, so -4 ⩽x ⩽ 8 Likewise, the second inequality, |y + 2| ⩽ 4 means that y is at most 4 from -2 (as |y + 2| = |y- (-2)|, so it measures the distance of y from-2). So -6 ⩽y ⩽ 2 We can now obtain the greatest possible value of |xy| =|x|.|y| by taking the greatest possible value of|x|, which is 8, and the greatest possible value of|y|, which is 6, so the greatest possible value of|xy| is 8× 6 = 48, which is option E.","tags":["TMUA","2018","Paper1"],"difficulty":2},{"id":"2018-P1-Q11","paper":1,"index":10,"stemLatex":"The line $y=mx+5$, where $m > 0$, is normal to the curve $y = 10 -x^2$ at the point ($p$, $q$). \n\n What is the value of $p$?","options":[{"key":"A","latex":"\\frac{\\sqrt{2}}{6}"},{"key":"B","latex":"-\\frac{\\sqrt{2}}{6}"},{"key":"C","latex":"\\frac{3\\sqrt{2}}{2}"},{"key":"D","latex":"-\\frac{3\\sqrt{2}}{2}"},{"key":"E","latex":"\\sqrt{5}"},{"key":"F","latex":"-\\sqrt{5}"}],"answerKey":"C","explanationLatex":"Differentiate the curve:\n\n$$y=10-x^2 \\quad \\Longrightarrow \\quad \\frac{dy}{dx}=-2x.$$\n\nAt the point $(p,q)$, the tangent has gradient $-2p$, so the normal has gradient\n\n$$\\frac{1}{2p}.$$\n\nAlso,\n\n$$q=10-p^2.$$\n\nSince the normal line is $y=mx+5$, it must pass through $(p,10-p^2)$, so\n\n$$10-p^2=mp+5.$$\n\nBut the gradient of the normal is also\n\n$$m=\\frac{1}{2p}.$$\n\nSubstitute this into the line equation:\n\n$$10-p^2=\\frac{1}{2}+5.$$\n\nSo\n\n$$p^2=\\frac{9}{2},$$\n\nand hence\n\n$$p=\\pm\\frac{3}{\\sqrt{2}}=\\pm\\frac{3\\sqrt{2}}{2}.$$\n\nSince $m=\\frac{1}{2p}>0$, we need $p>0$. Therefore\n\n$$p=\\frac{3\\sqrt{2}}{2}.$$\n\nSo the correct answer is $\\mathbf{C}$.","tags":["TMUA","2018","Paper1"],"difficulty":2},{"id":"2018-P1-Q12","paper":1,"index":11,"stemLatex":"A curve has equation $y = f(x)$, where $$f(x) = x(x-p)(x-q)(r-x)$$ with $0 < p < q < r$. \n\n You are given that: $$\\int_0^r f(x)\\:dx = 0$$ $$\\int_0^q f(x)\\:dx = -2$$ $$\\int_p^r f(x)\\:dx = -3$$ What is the total area enclosed by the curve and the x-axis for $0 \\leq x \\leq r$?","options":[{"key":"A","latex":"0"},{"key":"B","latex":"1"},{"key":"C","latex":"4"},{"key":"D","latex":"5"},{"key":"E","latex":"6"},{"key":"F","latex":"10"}],"answerKey":"F","explanationLatex":"This curve intercepts the x-axis at 0, p, q and r. As the leading power of x is-x^4, the curve tends to-∞ asx gets very large (both positive and negative), and so the curve looks something like this: x y 0 p q r A B C We have labelled the three areas as A, B andC, and we want to know the value of A +B +C. We can then write the three integrals we have been given in terms of these, referring to the sketch: ∫ r f(x) dx =A-B +C = 0 (1) ∫ q f(x) dx =A-B =-2 (2) ∫ r p f(x) dx =-B +C =-3 (3) Subtracting (3) from (1) gives A = 3, from which (2) gives B = 5 and so C = 2. Therefore A +B +C = 10, hence the correct option is F. Commentary: Although our sketch turns out to be not very accurate, it was sufficient to help us understand what was going on and to solve the problem. This is what we seek from sketches: they should be good enough to give us insight, but they do not need to be perfect. Incidentally, there do exist values ofp,q andr which make the statements of this question true, namely p = 1.25, q = 3.16 and r = 4.25, correct to two decimal places.","tags":["TMUA","2018","Paper1"],"difficulty":2},{"id":"2018-P1-Q13","paper":1,"index":12,"stemLatex":"The function $f(x)$ has derivative $f'(x)$.\n\nThe diagram below shows the graph of $y = f'(x)$.\n\nWhich point corresponds to a local minimum of $f(x)$?","stemImage":"/stem-fallback/2018-P1-Q13.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":"$F$"}],"answerKey":"C","explanationLatex":"A local minimum of f(x) occurs where the gradient is 0 at the point, is negative immediately to the left of the point and is positive immediately to its right. So on the graph of f ′(x), this will be at a point where f ′(x) = 0, and f ′(x) is increasing from negative to positive at that point. The only such point on the graph given is C. The behaviour of f(x) at all of the labelled points is as follows: A Changing from positive to negative gradient, so a local maximum. B A local minimum of the gradient, so a non-stationary point of inflection. If walking along the curve, one would stop turning to the right and start turning to the left. C Changing from negative to positive gradient, so a local minimum. D This point, where x = 0, has no particular relevance to the shape of f( x). E A local maximum of the gradient, so a non-stationary point of inflection. If walking along the curve, one would stop turning to the left and start turning to the right. F A local minimum of the gradient at a point of zero gradient, so a stationary point of inflection. If walking along the curve, one would stop turning to the right and start turning to the left.","tags":["TMUA","2018","Paper1"],"difficulty":2},{"id":"2018-P1-Q14","paper":1,"index":13,"stemLatex":"The line $y = mx+4$ passes through the points ($3$, $\\log_2{p}$) and ($\\log_2{p}, 4$). \n\n What are the possible values of $p$?","options":[{"key":"A","latex":"$p = 1$ and $p = 4$"},{"key":"B","latex":"$p = 1$ and $p = 16$"},{"key":"C","latex":"$p = \\frac{1}{4}$ and $p = 4$"},{"key":"D","latex":"$p = \\frac{1}{4}$ and $p = 64$"},{"key":"E","latex":"$p = \\frac{1}{64}$ and $p = 4$"},{"key":"F","latex":"$p = \\frac{1}{64}$ and $p = 16$"}],"answerKey":"B","explanationLatex":"There are a variety of different ways to approach this question. The following is probably the simplest. We can substitute the coordinates of the points into the equation of the line to obtain log^2p = 3m + 4 4 =m log^2p + 4 The second equation gives m log^2p = 0, so either m = 0 or log 2p = 0, so either m = 0 or p = 1. If m = 0, then the first equation gives log 2p = 4, so p = 16 If log^2p = 0, so p = 1, then the first equation gives 3 m + 4 = 0, which has a solution. So there are two possible values of p, namely 1 and 16. Hence the answer is B. Commentary: You might find it helpful to write q (for example) in place of log^2p, so that the equations become q = 3m + 4 4 =mq + 4 This makes their structure much more obvious.","tags":["TMUA","2018","Paper1"],"difficulty":2},{"id":"2018-P1-Q15","paper":1,"index":14,"stemLatex":"Find the sum of the real solutions of the equation: $$3^x - (\\sqrt{3})^{x+4} + 20 = 0$$","options":[{"key":"A","latex":"1"},{"key":"B","latex":"4"},{"key":"C","latex":"9"},{"key":"D","latex":"\\log_3{20}"},{"key":"E","latex":"2\\log_3{20}"},{"key":"F","latex":"4\\log_3{20}"}],"answerKey":"E","explanationLatex":"Let\n\n$$y=(\\sqrt{3})^x.$$\n\nThen\n\n$$3^x=y^2$$\n\nand\n\n$$(\\sqrt{3})^{x+4}=9y.$$\n\nSo the equation becomes\n\n$$y^2-9y+20=0.$$\n\nFactorising,\n\n$$y^2-9y+20=(y-4)(y-5)=0,$$\n\nso\n\n$$y=4 \\quad \\text{or} \\quad y=5.$$\n\nThat means\n\n$$(\\sqrt{3})^x=4 \\quad \\text{or} \\quad (\\sqrt{3})^x=5.$$\n\nSince $(\\sqrt{3})^x=3^{x/2}$, we get\n\n$$3^{x/2}=4 \\quad \\text{or} \\quad 3^{x/2}=5.$$\n\nTherefore\n\n$$\\frac{x}{2}=\\log_3 4 \\quad \\text{or} \\quad \\frac{x}{2}=\\log_3 5,$$\n\nso the two real solutions are\n\n$$x=2\\log_3 4 \\quad \\text{and} \\quad x=2\\log_3 5.$$\n\nTheir sum is\n\n$$2\\log_3 4+2\\log_3 5=2\\log_3(20).$$\n\nSo the correct answer is $\\mathbf{E}$.","tags":["TMUA","2018","Paper1"],"difficulty":2},{"id":"2018-P1-Q16","paper":1,"index":15,"stemLatex":"The curve $C$ has equation $y = x^2 + bx + 2$, where $b \\geq 0$. \n\n Find the value of $b$ that minimises the distance between the origin and the stationary point of the curve C.","options":[{"key":"A","latex":"b = 0"},{"key":"B","latex":"b = 1"},{"key":"C","latex":"b = 2"},{"key":"D","latex":"b = \\frac{\\sqrt{6}}{2}"},{"key":"E","latex":"b = \\sqrt{2}"},{"key":"F","latex":"b = \\sqrt{6}"}],"answerKey":"F","explanationLatex":"We first need to find the stationary point of the curveC. We can do this either by differentiating or, as the equation is a quadratic, by completing the square. By differentiating, we have dy dx = 2x +b so dy dx = 0 if and only if x =- 1 2b, and so y = (- 1 2b)^2 +b(- 1 2b) + 2 = 2- 1 4b2 Alternatively, we can complete the square to give y = (x + 1 2b)^2- 1 4b2 + 2, and so the stationary point is at (- 1 2b, 2- 1 4b2). The distance of this point from the origin is then given by √(- 1 2b)^2 + (2- 1 4b2)^2 = $\\\\sqrt{4b2}$ + 4- 4( 1 4b2) + 1 16b4 = √16b4- 3 4b2 + 4 To minimise this square root, we can minimise the expression inside the square root. This is a function of B =b^2, namely 1 16B2- 3 4B + 4, so we can minimise this. Commentary: If a quartic is actually a quadratic in the variable squared, it is usually much easier to work with the quadratic than the quartic. We can minimise this by completing the square or differentiating, as before. We can start by taking out a factor of 1 16 so we are trying to minimise 1 16(B^2- 12B) + 4. (There’s no need to think about the 4 at the end: this additional term does not change the value of b orB at which the function is minimised.) We therefore wish to minimise B^2- 12B = (B- 6)^2- 36 so B = 6 gives the minimum value, so b^2 = 6 giving b = $\\\\sqrt{6}$ (as b ⩾ 0), hence the answer is F. Commentary: We could have gone ahead and differentiated the whole square root expression for distance, if we know the chain rule. But that’s more effort than needed. Similarly, we could have differentiated the expression as a quartic, but again that is more complex than required. It is almost always worth spending a moment thinking how to make the problem a little easier before diving into algebra and calculations.","tags":["TMUA","2018","Paper1"],"difficulty":2},{"id":"2018-P1-Q17","paper":1,"index":16,"stemLatex":"There are two sets of data: the mean of the first set is 15, and the mean of the second set is 20. \n\n One of the pieces of data from the first set is exchanged with one of the pieces of data from the second set. \n\n As a result, the mean of the first set of data increases from 15 to 16, and the mean of the second set of data decreases from 20 to 17. \n\n What is the mean of the set made by combining all the data?","options":[{"key":"A","latex":"16\\tfrac{1}{4}"},{"key":"B","latex":"16\\tfrac{1}{3}"},{"key":"C","latex":"16\\tfrac{1}{2}"},{"key":"D","latex":"16\\tfrac{2}{3}"},{"key":"E","latex":"16\\tfrac{3}{4}"}],"answerKey":"A","explanationLatex":"It is not obvious how to begin this question - there are lots of wordy descriptions. We also don’t know how many pieces of data there are in each of the two sets. So let’s try naming things. Let the first set of data have n pieces of data, and the second set have m pieces of data. Since the mean of the first set of data is 15, the sum of the data is 15 n. Likewise, the sum of the second set of data is 20 m. When we exchange one piece of data from the first set with one piece of data from the second set, the overall total stays the same, but the totals of the two sets change. We are told what happens to the means, so the total of the first set is now 16 n and of the second set is 17m. Since the overall total is the same, we have 15n + 20m = 16n + 17m We can simplify this to give n = 3m Therefore the overall total is 15n + 20m = 45m + 20m = 65m and there are n +m = 4m pieces of data. Thus the mean of the combined set of data is 65m 4m = 65 4 = 16 1 4 and the answer is A. Commentary: At the end of this, we are still no wiser about how many pieces of data there were to begin with. All we know is that the first set contains three times as many as the second set.","tags":["TMUA","2018","Paper1"],"difficulty":2},{"id":"2018-P1-Q18","paper":1,"index":17,"stemLatex":"What is the smallest positive value of $a$ for which the line $x = a$ is a line of symmetry of the graph $y = \\sin(2x - \\frac{4\\pi}{3})$?","options":[{"key":"A","latex":"\\frac{\\pi}{12}"},{"key":"B","latex":"\\frac{5\\pi}{12}"},{"key":"C","latex":"\\frac{7\\pi}{12}"},{"key":"D","latex":"\\frac{11\\pi}{12}"},{"key":"E","latex":"\\frac{19\\pi}{12}"}],"answerKey":"B","explanationLatex":"The lines of symmetry of a sine or cosine graph (that is, something of the form sin( kx +θ) +c or cos(kx +θ) +c) are at the x-values where the sine or cosine equals ±1. So in this case, if we solve sin(2x- 4π 3 ) =±1 we will find the lines of symmetry. We solve each of them separately. We have sinθ = 1 when θ = π 2 + 2nπ (for any integer n), so in this case, this gives 2x- 4π 3 = π 2 + 2nπ ⇐⇒ 2x = 11π 6 + 2nπ ⇐⇒ x = 11π 12 +nπ So the smallest positive value of x with sin(2x- 4π 3 ) = 1 is x = 11π In the second case, we have sin θ =-1 when θ =-π 2 + 2nπ (for any integer n), so in this case, this gives 2x- 4π 3 =-π 2 + 2nπ ⇐⇒ 2x = 5π 6 + 2nπ ⇐⇒ x = 5π 12 +nπ So the smallest positive value of x with sin(2x- 4π 3 ) =-1 is x = 5π This is smaller than 11π 12 so the answer is 5π 12 , which is option B.","tags":["TMUA","2018","Paper1"],"difficulty":2},{"id":"2018-P1-Q19","paper":1,"index":18,"stemLatex":"A triangle $ABC$ is to be drawn with $AB = 10$cm, $BC = 7$cm and the angle at $A$ equal to $\\theta$, where $\\theta$ is a certain specified angle. \n\n Of the two possible triangles that could be drawn, the larger triangle has three times the area of the smaller one. \n\n What is the value of $\\cos{\\theta}$?","options":[{"key":"A","latex":"\\frac{5}{7}"},{"key":"B","latex":"\\frac{151}{200}"},{"key":"C","latex":"\\frac{7\\sqrt{2}}{5}"},{"key":"D","latex":"\\frac{\\sqrt{17}}{5}"},{"key":"E","latex":"\\frac{\\sqrt{51}}{8}"},{"key":"F","latex":"\\frac{\\sqrt{34}}{8}"}],"answerKey":"D","explanationLatex":"This questions calls for a sketch. We draw AB = 10, a line from A at an angle θ toAB (though we do not know what θ is), and a circular arc centred on B with radius 7. 10A B θ C^2 C1 The sketch shows the two triangles which could be formed subject to these conditions, with the third vertex being at C1 or C^2. One approach, hinted at by the instruction to find cos θ, is to use the cosine rule. In the triangle ABC1, we have 72 = (AC1)^2 + 102- 2(AC1)10 cosθ (1) while in the triangle ABC2, we have 72 = (AC2)^2 + 102- 2(AC2)10 cosθ (2) Unfortunately, we do not know either AC1 or AC2. We do, however, know that the larger triangle has three times the area of the smaller one. Taking AC1 or AC2 as the base of the triangle, the perpendicular height is from B to the line AC2, and is the same for both triangles. So AC2 = 3(AC1) We can now substitute this into equation (2) to get 72 = 9(AC1)^2 + 102- 6(AC1)10 cosθ Subtracting equation (1) then gives 0 = 8(AC1)^2- 40(AC1) cosθ so dividing by 8AC 1, we have AC1 = 5 cosθ. We can now substitute this back into equation (1) to get 72 = (5 cosθ)^2 + 102- 20(5 cosθ) cosθ so 49 = 25 cos^2θ + 100- 100 cos^2θ which simplifies to 75 cos^2θ = 51 so cos^2θ = 51 75 = 17 25 giving cosθ = $\\\\sqrt{5}$ (as cosθ> 0) There is an alternative approach which does not use the cosine rule. If we drop a perpendicular from B to the line AC2, and let x =AC1, our diagram becomes: 10A B θ C1 C^2 x D NowD is midway betweenC1 andC2 by symmetry (or by using the “RHS” congruency rule on the triangles BDC 1 and BDC 2), so C1D =DC2. Also, the area of the triangle ABC1 is 1 2(AC1)(BD) and the area of the triangle ABC2 is 2(AC2)(BD) Since the larger triangle has three times the area of the smaller one, AC2 = 3(AC1) (as before). So AC2 = 3x, hence C1C2 = 2x, and C1D =DC2 =x We can now apply Pythagoras to BDC 1 and BDA. We obtain, respectively, x^2 + (BD)^2 = 72 (2x)^2 + (BD)^2 = 102 Subtracting gives 3x2 = 51, so x^2 = 17 Since cosθ = AD AB = 2x 10 we get cosθ = $\\\\sqrt{5}$ as before. This also easily shows that BD = $\\\\sqrt{32}$ = 4 $\\\\sqrt{2}$ and that the area of the two triangles ABC1 and ABC2 are 2 $\\\\sqrt{34}$ and 6 $\\\\sqrt{34}$.","tags":["TMUA","2018","Paper1"],"difficulty":2},{"id":"2018-P1-Q20","paper":1,"index":19,"stemLatex":"Find the value of $$\\sin^2{0^{\\circ}} + \\sin^2{1^{\\circ}} + \\sin^2{2^{\\circ}} + \\sin^2{3^{\\circ}} + \\: ... \\: + \\sin^2{87^{\\circ}} + \\sin^2{88^{\\circ}} + \\sin^2{89^{\\circ}} + \\sin^2{90^{\\circ}}$$","options":[{"key":"A","latex":"0.5"},{"key":"B","latex":"1"},{"key":"C","latex":"1.5"},{"key":"D","latex":"45"},{"key":"E","latex":"45.5"},{"key":"F","latex":"46"}],"answerKey":"E","explanationLatex":"This reminds us of adding the terms of an arithmetic series. One way to do so is to write out the series twice, once forwards and once backwards, and then add corresponding pairs of terms together. If we do that here, we obtain S = sin^2 0◦ + sin^2 1◦ + sin^2 2◦ +··· + sin^2 88◦ + sin^2 89◦ + sin^2 90◦ S = sin^2 90◦ + sin^2 89◦ + sin^2 88◦ +··· + sin^2 2◦ + sin^2 1◦ + sin^2 0◦ so that 2S = (sin^2 0◦ + sin^2 90◦) + (sin^2 1◦ + sin^2 89◦) +··· + (sin^2 90◦ + sin^2 0◦) Now we know that sin 2x + cos^2x = 1 for all x, and since sin(90◦-x) = cosx (by, for example, thinking about a right-angled triangle or by looking at the graphs of the functions), we can rewrite this sum as 2S = (sin^2 0◦ + cos^2 0◦) + (sin^2 1◦ + cos^2 1◦) +··· + (sin^2 90◦ + cos^2 90◦) = 1 + 1 +··· + 1 = 91 (since there are 91 terms), therefore S = 45.5, which is option E. 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","2018","Paper1"],"difficulty":2},{"id":"2018-P2-Q1","paper":2,"index":0,"stemLatex":"The function $f$ is given, for $x>0$, by $$f(x) = \\frac{x^3 -4x}{2\\sqrt{x}}$$ Find the value of $f'(4)$.","options":[{"key":"A","latex":"3"},{"key":"B","latex":"9"},{"key":"C","latex":"9.5"},{"key":"D","latex":"12"},{"key":"E","latex":"39.5"},{"key":"F","latex":"88"}],"answerKey":"C","explanationLatex":"We first simplify the fraction using index laws: f(x) = x^3- 4x 2√x = x^3- 4x 2x = 1 2x 2- 2x so f′(x) = 5 4x 2-x- 1 Substituting x = 4 then gives f′(4) = 5 4× 8- 1 2 = 10- 1 2 = 9.5 and so the answer is C.","tags":["TMUA","2018","Paper2"],"difficulty":2},{"id":"2018-P2-Q2","paper":2,"index":1,"stemLatex":"Find the value of the constant term in the expansion of $$(x^6 - \\frac{1}{x^2})^{12}$$","options":[{"key":"A","latex":"-495"},{"key":"B","latex":"-220"},{"key":"C","latex":"-66"},{"key":"D","latex":"66"},{"key":"E","latex":"220"},{"key":"F","latex":"495"}],"answerKey":"B","explanationLatex":"The constant term is the one with x0. The expansion is (x^6)12 + (12 ) (x^6)11 ( - 1 x^2 ) +··· + (12 k ) (x^6)12-k ( - 1 x^2 )k +··· + ( - 1 x^2 )12 (This notation for binomial coefficients is much clearer than the older nCr notation, and is recommended.) The kth term is therefore (12 k ) x^6(12-k)(-1)kx-2k = (12 k ) (-1)kx6(12-k)-2k = (12 k ) (-1)kx72-8k The value of k which makes the power zero is therefore k = 9, so the constant term is (12 ) (-1)^9 =- (12 ) =-12× 11× 10 3× 2× 1 =-2× 11× 10 =-220 where we have used (n k ) = ( n n-k ) to simplify the calculation. Thus the answer is B. Commentary: We have used k here (beginning our counting with k = 0). You might prefer to use another letter such as r.","tags":["TMUA","2018","Paper2"],"difficulty":2},{"id":"2018-P2-Q3","paper":2,"index":2,"stemLatex":"A car journey consists of two parts. In the first part, the average speed is $u$km/h. In the second part, the average speed is $v$km/h. Hence the average speed for the whole journey is $\\frac{1}{2}(u + v)$km/h. \n\n Which of the following examples of car journeys provide(s) a counterexample to the statement? \n\n I: In the first part of the journey, the car travels at a constant speed of 50 km/h for 100 km. In the second part of the journey, the car travels at a constant speed of 40 km/h for 100 km. \n\n II: In the first part of the journey, the car travels at a constant speed of 50 km/h for one hour. In the second part of the journey, the car travels at a constant speed of 40 km/h for one hour. \n\n III: In the first part of the journey, the car travels at a constant speed of 50 km/h for 80 km. In the second part of the journey, the car travels at a constant speed of 40 km/h for 100 km.","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":"F","explanationLatex":"A counterexample to this statement is an example which shows that it is not true. So we consider the journeys described, and determine whether or not the average speed for the whole journey is, indeed, 1 2(u +v). Perhaps the simplest way to do this is to make a table. We know that average speed is distance travelled divided by time, so we will need these for each part of the journey. Filling in the information given, we end up with the following. We write d1, d^2 and D for the distances (in km) in the first part, second part and total, respectively. Likewise, t stands for time (in h), and we use V for the average speed for the whole journey (in km /h). First part Second part Overall d1 t1 u d 2 t^2 v D T V I 100 50 100 40 II 1 50 1 40 III 80 50 100 40 We can now use u = d1/t1 and so on to fill in the rest of the first and second part of the journey, and then sum the distances and times for the overall journey. We can finally calculate the average speed. This gives: First part Second part Overall d1 t1 u d 2 t^2 v D T V I 100 2 50 100 2.5 40 200 4.5 400 II 50 1 50 40 1 40 90 2 45 III 80 1.6 50 100 2.5 40 180 4.1 1800 In every case, 1 2(u +v) = 45 so both I and III are counterexamples (as the fractions cannot be simplified, so they cannot equal 45), and the answer is F.","tags":["TMUA","2018","Paper2"],"difficulty":2},{"id":"2018-P2-Q4","paper":2,"index":3,"stemLatex":"The non-zero real number $c$ is such that the equation $\\cos x = c$ has two solutions for $0 < x < \\frac{3}{2} \\pi$. \n\n How many solutions of the equation $\\cos^22x = c^2$ are there in the range $0 < x < \\frac{3}{2}\\pi$?","options":[{"key":"A","latex":"2"},{"key":"B","latex":"3"},{"key":"C","latex":"4"},{"key":"D","latex":"6"},{"key":"E","latex":"7"},{"key":"F","latex":"8"}],"answerKey":"D","explanationLatex":"It is worth sketching a graph of y = cosx to understand the meaning of the stated condition. Two lines have been drawn corresponding to different values of c. x y 0 1 2π π 3 2π 2π For there to be two solutions for 0 <x< 3 2π, we see from the sketch that we need -1<c< 0 We now consider the equation cos^2 2x =c^2, so cos 2x =±c There are two straightforward approaches. The first is to sketch the graph of y = cos 2x and look at 0 <x< 3 2π; the second is to note that the range can be written as 0 < 2x< 3π, so we could write u = 2x and look for solutions to cos u =±c in the range 0 < u <3π. We will use the first approach here; the second is very similar. Here is a sketch of y = cos 2x: x y 0 1 2π π 3 2π 2π -c c We immediately see that, for 0 <x< 3 2π, there are six solutions to our equation, so the correct answer is D.","tags":["TMUA","2018","Paper2"],"difficulty":2},{"id":"2018-P2-Q5","paper":2,"index":4,"stemLatex":"The two diagonals of the quadrilateral $Q$ are perpendicular. \n\n Consider the following statements: \n\n I: One of the diagonals of $Q$ is a line of symmetry of $Q$. \n\n II: The midpoints of the sides of $Q$ are the vertices of a square. \n\n Which of these statements is/are necessarily true for the quadrilateral $Q$?","options":[{"key":"A","latex":"neither of them"},{"key":"B","latex":"I only"},{"key":"C","latex":"II only"},{"key":"D","latex":"I and II"}],"answerKey":"A","explanationLatex":"Commentary: With a question such as this, it is very wise to try drawing a sketch. We are told that the two diagonals of Q are perpendicular, so it makes sense to start by drawing the two diagonals. If statement I is necessarily true for Q, it means that it is always true, no matter how we choose to draw Q (subject to the condition that it has perpendicular diagonals), and likewise for statement II. We could therefore either try proving these statements or start by looking for counterexamples. Following the above comments, we start by drawing two perpendicular lines, and mark vertices on them with the aim of finding a counterexample to each statement. This turns out not to be that difficult after some playing around. To make a diagonal not be a line of symmetry, we simply need the off-diagonal vertices to be different distances from the diagonal. So this will do it: For II, to make the midpoints of the sides not be the vertices of a square, we can make the quadrilateral be “long and skinny”: So neither I nor II is necessarily true, so the correct answer is A. Commentary: It turns out that the midpoints of the sides ofQ are always the vertices of a rectangle, and for a general quadrilateral, they are always the vertices of a parallelogram, with sides parallel to the diagonals of the original quadrilateral.","tags":["TMUA","2018","Paper2"],"difficulty":2},{"id":"2018-P2-Q6","paper":2,"index":5,"stemLatex":"Which one of the following functions provides a counterexample to the statement: \n\n if $f'(x) > 0$ for all real $x$, then $f(x) > 0$ for all real $x$.","options":[{"key":"A","latex":"f(x) = x^2+1"},{"key":"B","latex":"f(x) = x^2 - 1"},{"key":"C","latex":"f(x) = x^3 + x + 1"},{"key":"D","latex":"f(x) = 1 - x"},{"key":"E","latex":"f(x) = 2^x"}],"answerKey":"C","explanationLatex":"A counterexample to an ‘ if P then Q’ statement is an example where the statement is false, that is where P (the antecedent) is true but Q (the consequent) is false. We can now look through the given functions to see which of them satisfies these requirements. We give our results as a table and make comments on the working below. f(x) f ′(x) f ′(x)> 0 f( x)> 0 for all x for all x A x^2 + 1 2 x false B x^2- 1 2 x false C x^3 +x + 1 3 x^2 + 1 true false D 1 -x -1 false E 2 x ? true true Once we know that f′(x) > 0 for all x is false, there is no need to go any further, so for A, B and D, we stop at this point. For C, 3x2 ⩾ 0 for all x, so 3x2 + 1 ⩾ 1> 0 for all x, and we know that cubics ( x^3 +x + 1 in this case) always take both positive and negative values, so f( x)> 0 for all x is false. For E, we don’t know what the derivative of 2x is (or at least, knowledge of it is not required for this test), but we do know that 2 x is increasing for all x, so the derivative is positive for all x. Finally, only C has the antecedent true and the consequent false, so C provides a counterexample.","tags":["TMUA","2018","Paper2"],"difficulty":2},{"id":"2018-P2-Q7","paper":2,"index":6,"stemLatex":"Sequence 1 is an arithmetic progression with first term 11 and common difference 3. \n\n Sequence 2 is an arithmetic progression with first term 2 and common difference 5. \n\n Some numbers that appear in Sequence 1 also appear in Sequence 2. Let $N$ be the 20th such number. \n\n What is the remainder when $N$ is divided by 7?","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":"B","explanationLatex":"We can list the terms of the two sequences to get a feeling for what is going on: Sequence 1: 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, ... Sequence 2: 2, 7, 12, 17, 22, 27, 32, 37, 42, ... We see that the first two common terms are 17 and 32; the next will be 47, and the common terms are forming an arithmetic progression with first term 17 and common difference 15. (Note that 15 is the lowest common multiple of 3 and 5.) Therefore the 20th such number, N is 17 + 19× 15 = 17 + 285 = 302. Finally, dividing 302 by 7 gives 43 remainder 1, so the answer is B.","tags":["TMUA","2018","Paper2"],"difficulty":2},{"id":"2018-P2-Q8","paper":2,"index":7,"stemLatex":"The diagram shows an example of a mountain profile.\n\nThis consists of upstrokes which go upwards from left to right, and downstrokes which go downwards from left to right. The example shown has six upstrokes and six downstrokes. The horizontal line at the bottom is known as sea level.\n\nA mountain profile of order $n$ consists of $n$ upstrokes and $n$ downstrokes, with the condition that the profile begins and ends at sea level and never goes below sea level (although it might reach sea level at any point). So the example shown is a mountain profile of order 6.\n\nMountain profiles can be coded by using U to indicate an upstroke and D to indicate a downstroke. The example shown has the code UDUUUDUDDUDD. A sequence of U's and D's obtained from a mountain profile in this way is known as a valid code.\n\nWhich of the following statements is/are true?\n\nI. If a valid code is written in reverse order, the result is always a valid code.\nII. If each U in a valid code is replaced by D and each D by U, the result is always a valid code.\nIII. If U is added at the beginning of a valid code and D is added at the end of the code, the result is always a valid code.","stemImage":"/stem-fallback/2018-P2-Q8.png","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":"I Every valid code begins with a ‘U’ and ends with a ‘D’, so the reverse code begins with a ‘D’, meaning it immediately dips below sea level. So it is never a valid code, and this statement is false. II As with I, the first letter of this substituted code will always be ‘D’, which will never give a valid code, so this statement is false. III The new ‘U’ at the beginning of the code and ‘D’ at the end mean that the whole middle part of the profile is lifted one unit higher than the original profile. So this will never dip below sea level, and will always be a valid code. So this statement is true. Therefore the correct answer is D.","tags":["TMUA","2018","Paper2"],"difficulty":2},{"id":"2018-P2-Q9","paper":2,"index":8,"stemLatex":"Consider the following attempt to solve the equation $4x\\sqrt{2x - 1} = 10x - 5$:\n\n(I) $4x\\sqrt{2x - 1} = 10x - 5$\n(II) $4x\\sqrt{2x - 1} = 5(2x - 1)$\n(III) $16x^2(2x - 1) = 25(2x - 1)^2$\n(IV) $16x^2 = 25(2x - 1)$\n(V) $16x^2 - 50x + 25 = 0$\n(VI) $(8x - 5)(2x - 5) = 0$\n\nThe solutions of the original equation are $x = \\frac{5}{8}$ and $x = \\frac{5}{2}$.\n\nWhich one of the following is true?","stemImage":"/stem-fallback/2018-P2-Q9.png","options":[{"key":"A","latex":"The solution is correct."},{"key":"B","latex":"Only one of $x=\\frac{5}{8}$ and $x=\\frac{5}{2}$ is correct and the error arises as a result of step (II)."},{"key":"C","latex":"Only one of $x=\\frac{5}{8}$ and $x=\\frac{5}{2}$ is correct and the error arises as a result of step (III)."},{"key":"D","latex":"Only one of $x=\\frac{5}{8}$ and $x=\\frac{5}{2}$ is correct and the error arises as a result of step (IV)."},{"key":"E","latex":"There is another value of $x$ that satisfies the original equation and the error arises as a result of step (II)."},{"key":"F","latex":"There is another value of $x$ that satisfies the original equation and the error arises as a result of step (III)."},{"key":"G","latex":"There is another value of $x$ that satisfies the original equation and the error arises as a result of step (IV)."}],"answerKey":"F","explanationLatex":"Given the choices, we first check to see what the proposed solutions x = 5 8 and x = 5 2 work in the original equation. Whenx = 5 8 the left side is 4× 5 8× √2× 5 8- 1 = 5 4 and the right side is 10× 5 8- 5 = 5 4 so this is a valid solution. When x = 5 2 the left side is 4 × 5 2× √2× 5 2- 1 = 20 and the right side is 10 × 5 2- 5 = 20 so this is also a valid solution. This therefore eliminates options B, C and D. There may, though, be additional solutions. Let us work through the argument step-by-step, looking for steps which are not reversible; these are often the cause of errors when attempting to solve equations. Step (I) is just factorising the right hand side, so is fine. Step (II) involves squaring. In general, squaring an equation is not reversible, as we do not know whether to take the positive or negative square root when going backwards. However, if we do have a solution to the original equation, it will still be a solution when we square it; the worst that can happen is that we introduce new solutions. Since the two solutions found do work, this step is fine, as is the conclusion (though the author of the attempt should have checked that this is the case). Step (III) involves dividing by 2 x- 1. But what if this is zero? In that case, we would have x = 1 2. Trying this in the original equation makes the left hand side 4 × 1 2× √2× 1 2- 1 = 0, while the right side is 10× 1 2- 5 = 0. So x = 1 2 is another solution. The correct option is therefore F. (The remaining steps are rearrangements, and are all valid.)","tags":["TMUA","2018","Paper2"],"difficulty":2},{"id":"2018-P2-Q10","paper":2,"index":9,"stemLatex":"The function $f(x)$ is defined for all real numbers.\n\nConsider the following three conditions, where $a$ is a real constant:\nI. $f(a - x) = f(a + x)$ for all real $x$.\nII. $f(2a - x) = f(x)$ for all real $x$.\nIII. $f(a - x) = f(x)$ for all real $x$.\n\nWhich of these conditions is/are necessary and sufficient for the graph of $y = f(x)$ to have reflection symmetry in the line $x = a$?\n\nRows A-H correspond to this table:\nA: I yes, II yes, III yes\nB: I yes, II yes, III no\nC: I yes, II no, III yes\nD: I yes, II no, III no\nE: I no, II yes, III yes\nF: I no, II yes, III no\nG: I no, II no, III yes\nH: I no, II no, III no","stemImage":"/stem-fallback/2018-P2-Q10.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":"B","explanationLatex":"It would help to sketch a graph which has reflection symmetry in x =a. We should pick a value of a which is not zero. Let us pick a positive value of a. We can draw a quadratic with vertex at x =a, for example. x y a We now consider the three conditions and see which are certainly notnecessary for the reflection symmetry by looking at this example. We mark on x, a±x and 2a-x for a value of x which is not equal to a, where we have chosen x≈ 1.5a: x y a x a +xa-x^2a-x It seems from this sketch that f(x) and f(2a - x) are likely to be equal, as are f(a + x) and f(a - x). So we reject condition III, leaving us with conditions I and II to consider. For a graph y = f(x) to be symmetrical about x = a, we need the value of the function to be equal whenever we go the same distance to the right or left of x = a. Looking at this sketch, let’s take the distance to be x. Then we need f(a + x) = f(a - x) for every choice of x if the graph is symmetrical about x = a, so condition I is necessary. But it is also sufficient, for how could a graph not be symmetrical? Only if for some x, the function values are different at a + x and a - x. For condition II, we can do the same. This time, the values we want to consider are 2a - x and x. Now 2a - x is x - a to the left of a (assuming that x > a) and x is x - a to the right of a, so these are the same distance from a, and again this condition is necessary. Since every distance can be written as x - a for some choice of x, the condition is also sufficient. So both conditions I and II are necessary and sufficient, and the correct option is B.","tags":["TMUA","2018","Paper2"],"difficulty":2},{"id":"2018-P2-Q11","paper":2,"index":10,"stemLatex":"Consider the equation $2^x = mx + c$, where $m$ and $c$ are real constants. \n\n Which of the following statements is/are true? \n\n I: The equation has a negative real solution only if $c > 1$. \n\n II: The equation has two distinct real solutions if $c > 1$. \n\n III: The equation has two distinct positive real solutions if and only if $c \\leq 1$.","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":"A","explanationLatex":"We sketch the graph of y = 2x. We know that y = mx +c is a straight line with gradient m and a y-intercept ofc, and we will imagine this line for different values of m andc. Solutions to the equation 2x =mx +c correspond to the x-coordinates where these two graphs meet. x y I Recall that ‘ P only if Q’ is the same as saying ‘if P thenQ’, so in this case, the statement becomes ‘if the equation 2x =mx +c has a negative real solution, thenc> 1’. But we see that the line y =-x, with c = 0, has a solution with negative real x, so this statement is false. II We can rewrite this statement as ‘ if c> 1 then the equation has two distinct real solutions’. So we imagine straight lines passing through some point above (0 , 1), and we see that if this straight line has a negative gradient, then it can only intercept the curve once. So this statement is false. III If c ⩽ 1, then a line with a negative gradient will still only intercept the curve once, so this is also false. Therefore none of the statements is true, and the correct option is A.","tags":["TMUA","2018","Paper2"],"difficulty":2},{"id":"2018-P2-Q12","paper":2,"index":11,"stemLatex":"Consider the following statement: \n\n For any positive integer $N$ there is a positive integer $K$ such that $N(Km + 1) - 1$ is not prime for any positive integer $m$. \n\n Which one of the following is the negation of this statement?","options":[{"key":"A","latex":"For any positive integer $N$ there is a positive integer $K$ such that there is a positive integer $m$ for which $N(Km + 1) - 1$ is prime."},{"key":"B","latex":"For any positive integer $N$ there is a positive integer $K$ such that there is a positive integer $m$ for which $N(Km + 1) - 1$ is not prime."},{"key":"C","latex":"For any positive integer $N$ there is a positive integer $K$ such that for any positive integer $m$, $N(Km + 1) - 1$ is not prime."},{"key":"D","latex":"For any positive integer $N$ any positive integer $K$ and any positive integer $m$, $N(Km + 1) - 1$ is not prime."},{"key":"E","latex":"There is a positive integer $N$ such that for any positive integer $K$ there is a positive integer $m$ for which $N(Km + 1) - 1$ is not prime."},{"key":"F","latex":"There is a positive integer $N$ such that for any positive integer $K$ there is a positive integer $m$ for which $N(Km + 1) - 1$ is prime."},{"key":"G","latex":"There is a positive integer $N$ such that for any positive integer $K$ and any positive integer $m$, $N(Km + 1) - 1$ is prime."},{"key":"H","latex":"There is a positive integer $N$ such that for any positive integer $K$ for which there is no positive integer $m$ for which $N(Km + 1) - 1$ is prime."}],"answerKey":"F","explanationLatex":"The negation of ‘ for any N, P ’ is ‘there exists N such that not P ’. So none of options A-D can be a negation of the original statement, but all of E-H still have the potential to be. In this case, the statement P is ‘there is a postiive integer K such that N(Km + 1)- 1 is not prime for any positive integer m’; this is a statement of the form ‘ there exists’, and we wish to negate it. The negation of ‘ there exists K such that Q’ is ‘ for all K, not Q’, so we are looking for a statement which continues with ‘for all positive integers K’, and this eliminates option H. (‘For any’ and ‘for all’ are equivalent ways of expressing this.) So we are now looking at options E, F and G. The statement Q that we wish to negate is ‘ N(Km + 1)- 1 is not prime for any positive integerm’, which we can rewrite with the quantifier (‘for any’) at the start of the statement to give ‘for any positive integer m, N(Km + 1)- 1 is not prime’. And the negation of ‘ for all m, R’ is ‘there exists m such that not R’, which in this case is ‘there exists a positive integer m such that it is not the case that N(Km + 1)- 1 is not prime’, so option G is not correct. Now ‘not not S’ is equivalent to S, so this becomes ‘there exists a positive integer m such that N(Km + 1)- 1 is prime’. The only option which matches this is option F, and this is therefore the correct answer.","tags":["TMUA","2018","Paper2"],"difficulty":2},{"id":"2018-P2-Q13","paper":2,"index":12,"stemLatex":"The following is an attempted proof of the conjecture:\n\nif $\\tan\\theta > 0$, then $\\sin\\theta + \\cos\\theta > 1$.\n\nSuppose $\\tan\\theta > 0$, so in particular $\\cos\\theta \\neq 0$.\n\n(I) Since $\\tan\\theta = \\frac{\\sin\\theta}{\\cos\\theta}$, then $\\sin\\theta\\cos\\theta = \\tan\\theta\\cos^2\\theta > 0$.\n(II) It follows that $1 + 2\\sin\\theta\\cos\\theta > 1$.\n(III) Therefore $\\sin^2\\theta + 2\\sin\\theta\\cos\\theta + \\cos^2\\theta > 1$.\n(IV) which factorises to give $(\\sin\\theta + \\cos\\theta)^2 > 1$.\n(V) Therefore $\\sin\\theta + \\cos\\theta > 1$.\n\nWhich one of the following is the case?","stemImage":"/stem-fallback/2018-P2-Q13.png","options":[{"key":"A","latex":"The proof is correct."},{"key":"B","latex":"The proof is incorrect, and the first error occurs in line (I)."},{"key":"C","latex":"The proof is incorrect, and the first error occurs in line (II)."},{"key":"D","latex":"The proof is incorrect, and the first error occurs in line (III)."},{"key":"E","latex":"The proof is incorrect, and the first error occurs in line (IV)."},{"key":"F","latex":"The proof is incorrect, and the first error occurs in line (V)."}],"answerKey":"F","explanationLatex":"A good way to test a conjecture like this is to look for a counterexample.\n\nSince $\\tan\\theta>0$ in the first and third quadrants, the most promising place to look is the third quadrant, where both $\\sin\\theta$ and $\\cos\\theta$ are negative.\n\nTake\n\n$$\\theta=\\frac{5\\pi}{4}.$$\n\nThen\n\n$$\\tan\\theta=1>0,$$\n\nbut\n\n$$\\sin\\theta+\\cos\\theta=-\\frac{1}{\\sqrt{2}}-\\frac{1}{\\sqrt{2}}=-\\sqrt{2}<1.$$\n\nSo the conjecture is false.\n\nNow test the attempted proof line by line at this value of $\\theta$.\n\n- Line (I) is correct.\n- Line (II) is correct.\n- Line (III) is correct.\n- Line (IV) is correct.\n- Line (V) is the problem.\n\nUp to line (IV) we have\n\n$$(\\sin\\theta+\\cos\\theta)^2>1.$$\n\nBut from this we cannot conclude that\n\n$$\\sin\\theta+\\cos\\theta>1,$$\n\nbecause a number can have square greater than $1$ while still being negative. Here,\n\n$$\\sin\\theta+\\cos\\theta=-\\sqrt{2},$$\n\nand indeed\n\n$$(-\\sqrt{2})^2=2>1.$$\n\nSo the first incorrect line is line $(V)$, and the correct answer is $\\mathbf{F}$.","tags":["TMUA","2018","Paper2"],"difficulty":2},{"id":"2018-P2-Q14","paper":2,"index":13,"stemLatex":"In the triangle $PQR$, $PR = 2$, $QR = p$ and $\\angle RPQ = 30^{\\circ}$. \n\n What is the set of all the values of $p$ for which this information uniquely determines the length of $PQ$?","options":[{"key":"A","latex":"p = 1"},{"key":"B","latex":"p = \\sqrt{3}"},{"key":"C","latex":"1 \\leq p < 2"},{"key":"D","latex":"\\sqrt{3} \\leq p < 2"},{"key":"E","latex":"$p = 1$ or $p \\geq 2$"},{"key":"F","latex":"$p = \\sqrt{}3$ or $p \\geq 2$"},{"key":"G","latex":"p < 2"},{"key":"H","latex":"p \\geq 2"}],"answerKey":"E","explanationLatex":"We sketch the situation using the given information. This sketch shows a case when there is not a unique solution. The length of PQ is uniquely determined when there is a unique triangle satisfying the conditions. 2P R 30◦ p Q^2 p Q1 Ifp is such that the circle centred at R with radiusp touches the line from P at angle 30◦, there will be a unique solution. If p is smaller than this, then there will be no intersections, and hence no solutions. If p is larger than this there will be two solutions, which suggests that the answer is A or B. We can work out the value ofp which makes the circle tangent to the line through P : RQP will be a right-angled triangle with the right angle at Q in this case, and so p = 2 sin 30◦ = 1 This means that p = 1 gives a unique solution, and the solution is either A or possibly E. What happens when p = 2? Then the circle centred on R passes throughP , and so there is only one possible location for Q. And then we see that if p> 2, the point Q1 will no longer exist: it will be beyond P , on the extension of the line “below” PR . So for any p ⩾ 2, there is also only one possible triangle. Hence the answer is option E.","tags":["TMUA","2018","Paper2"],"difficulty":2},{"id":"2018-P2-Q15","paper":2,"index":14,"stemLatex":"It is given that $f(x) = x^3 + 3qx^2 + 2$, where $q$ is a real constant. \n\n The equation $f(x) = 0$ has 3 distinct real roots. \n\n Which of the following statements is/are necessarily true? \n\n I: The equation $f(x) + 1 = 0$ has $3$ distinct real roots. \n\n II: The equation $f(x + 1) = 0$ has $3$ distinct real roots. \n\n III: The equation $f(-x) - 1 = 0$ has $3$ distinct real roots.","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":"G","explanationLatex":"This question appears to be related to graph transformations, so we will start with this approach and see where it leads. The equation f(x) = 0 has 3 distinct real roots, so y = f(x) intersects the x-axis 3 times. I The graph of y = f(x) + 1 is a translation of y = f(x) in the y-direction by 1 unit, which may or may not intersect the x-axis 3 times. II The graph of y = f(x + 1) is a translation of y = f(x) in the negative x-direction by 1 unit, so it will still intersect the x-axis 3 times, and hence f(x + 1) = 0 has 3 distinct roots. III The graph of y = f(-x) is a reflection of y = f(x) in the y-axis, which still intersects the x-axis 3 times. Then y = f(-x)- 1 is a translation in the negative y-direction by 1 unit, which may or may not intersect the x-axis 3 times. It is still not clear whether or not I or III are necessarily true. We will have to identify where the turning points of the graph are to determine this. We have f(x) =x^3 + 3qx2 + 2 so f′(x) = 3x2 + 6qx so f′(x) = 0 when x = 0 or x ==-2q, so the stationary points are at (0 , 2) and (-2q, 2 + 4q3). As the graph crosses the x-axis 3 times, the stationary points must lie either side of the x-axis, so 2 + 4q3< 0. Therefore when the graph is translated up by 1 unit, we may have the new stationary point at (-2q, 2 + 4q3 + 1) with 2 + 4q3 + 1> 0, but when it is translated down by 1 unit, they will be at (0, 1) and (-2q, 2 + 4q3- 1), and 2 + 4q3- 1< 0. So there will still be 3 intersections. Therefore II and III are necessarily true, and the answer is G.","tags":["TMUA","2018","Paper2"],"difficulty":2},{"id":"2018-P2-Q16","paper":2,"index":15,"stemLatex":"In this question, $x_1, x_2,x_3$, ... is an arithmetic progression, all of whose terms are integers. \n\n Let $n$ be a positive integer. If the median of the first $n$ terms of the sequence is an integer, which of the following three statements must be true? \n\n I: The median of the first $n+2$ terms is an integer. \n\n II: The median of the first $2n$ terms is an integer. \n\n III: The median of $x_2, x_4, x_6, ... ,x_{2n}$ is an integer.","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":"F","explanationLatex":"Finding the median of the first n terms depends on whether n is even or odd. If n is odd, then the median of the first n terms is the middle term, which is x(n+1)/2, so it is certainly an integer, whatever the sequence. We can consider the three given statements in this case: I n+2 is also odd, so the median of the first n+2 terms is a term of the sequence, specifically x(n+3)/2, so it is an integer. II 2 n is even, so the median of the first 2 n terms is midway between the nth and (n + 1)th terms. This may or may not be an integer. For example, if the sequence is 1, 2, 3, ... and n = 3, then the median of the first 2n = 6 terms is midway between 3 and 4, so is 3.5. This is not an integer. (On the other hand, it might be: if the sequence were 2, 4, 6, ... , with n = 3 still, then the median of the first 2 n = 6 terms is midway between 6 and 8, which is 7; this is an integer.) III There are n terms listed here, so the median of them is the middle one, which is certainly an integer. Thus in the case that n is odd, I and III must be true, but II need not be. Next we consider the case of n being even. In this case, the median of the first n terms is midway betweenx 1 2n andx 1 2n+1, and we are told that this is an integer. If we call the common difference of this arithmetic progression d, then midway between these is x 1 2n + 1 2d, so 1 2d is an integer. (Alternatively, note that both the middle two terms are even or both the middle two terms are odd, so either all the terms of the arithmetic progression are even or all of them are odd. This is equivalent to saying that the common difference is even.) We again consider the three statements: I Now n+2 is even, and the median is midway between two terms (namelyx 1 2n+1 andx 1 2n+2), so it is an integer since the common difference is even. (Alternatively, it is an integer because either all of the terms are even or all of the terms are odd.) II We do not need to consider this statement, as we know it is not always true from before. (Nevertheless, this statement is always true for this case, using the same reasoning as for I.) III This time, we have n terms, and the common difference between consecutive pairs of them is 2d. The median is midway between central two terms (as n is even), so is an integer as d is an integer. Alternatively, the median is midway between xn and xn+2, so is xn+1, which is an integer. Therefore all three statements are true in the case that n is even. Putting this together, statements I and III must be true, whatever the value of n, but statement II is not necessarily true. Hence the answer is F.","tags":["TMUA","2018","Paper2"],"difficulty":2},{"id":"2018-P2-Q17","paper":2,"index":16,"stemLatex":"A positive integer is called a squaresum if and only if it can be written as the sum of the squares of two integers. For example, 61 and 9 are both squaresums since $61 = 5^2 + 6^2$ and $9 = 3^2 + 0^2$. \n\n A prime number is called awkward if and only if it has a remainder of 3 when divided by 4. For example, 23 is awkward since $23 = 5 \\times 4 + 3$. \n\n A (true) theorem due to Fermat states that: \n\n A positive integer is a squaresum if and only if each of its awkward prime factors occurs to an even power in its prime factorisation. \n\n It follows that $5 \\times 23^2$ is a squaresum, since 23 occurs to the power 2, but $5 \\times 23^3$ is not, since 23 occurs to the power 3. \n\n Which one of the following statements is not true?","options":[{"key":"A","latex":"Every square number is a squaresum."},{"key":"B","latex":"If $N$ and $M$ are squaresums, then so is $NM$."},{"key":"C","latex":"If $NM$ is a squaresum, then $N$ and $M$ are squaresums."},{"key":"D","latex":"If $N$ is not a squaresum, the $kN$ is a squaresum for some number $k$ which is a product of awkward primes."}],"answerKey":"C","explanationLatex":"We work through the given statements in turn. A This is a statement about numbers of the form N 2. If we write N as a product of prime factors, thenN 2 will have the same prime factors, but each to double the power. For example, ifN = 3× 173× 414 thenN 2 = 32× 176× 418. So each prime factor occurs to an even power in its prime factorisation, so it is a squaresum. Alternatively, we can ignore the theorem completely, and note that N 2 =N 2 + 02, so N 2 is a squaresum. So this statement is true. B If N is a squaresum, then each of its awkward prime factors occurs to an even power in its prime factorisation, by the theorem. Likewise, each of the awkward prime factors of M occurs to an even power in its prime factorisation. The prime factorisation of NM is the product of the prime factorisations of M and N, so each awkward prime occurs to an even power in the prime factorisation of NM , and therefore NM is a squaresum. Thus this statement is also true. C If NM is a squaresum, then each of its awkward prime factors occurs to an even power in its prime factorisation. But it is not clear why that would mean that each of the awkward prime factors of M and N each occur to an even power in the factorisations of M and N. And we can find a counterexample to this statement: 23 is awkward, so if N =M = 23 then NM = 232 is a squaresum, but neither N nor M is. So this statement is false. D As N is not a squaresum, some awkward prime factor occurs to an odd power in its prime factorisation. If we multiply N by this awkward prime factor, then the result will have this awkward prime factor to an even power. But N might have more than one awkward prime factor occurring to an odd power, so we multiply by all such awkward prime factors. The result will be a number with every awkward prime factor occurring to an even power. Thus lettingk be the product of all the awkward prime factors of N which occur to an odd power, kN is a squaresum. Hence this statement is true. The only false statement is C.","tags":["TMUA","2018","Paper2"],"difficulty":2},{"id":"2018-P2-Q18","paper":2,"index":17,"stemLatex":"$f(x)$ is a polynomial function defined for all real $x$. \n\n Which of the following is a necessary condition for the inequality $$\\frac{f(a) + f(b)}{2} \\geq f(\\frac{a + b}{2})$$ to be true for all real numbers $a$ and $b$ with $a < b$?","options":[{"key":"A","latex":"$f(x) \\geq 0$ for all real $x$"},{"key":"B","latex":"$f'(x) \\geq 0$ for all real $x$"},{"key":"C","latex":"$f''(x) \\geq 0$ for all real $x$"},{"key":"D","latex":"$f(x) \\leq 0$ for all real $x$"},{"key":"E","latex":"$f'(x) \\leq 0$ for all real $x$"},{"key":"F","latex":"$f''(x) \\leq 0$ for all real $x$"}],"answerKey":"C","explanationLatex":"It is helpful to draw a sketch of y = f(x) to understand what is going on here. Note first that f(a) + f(b)^2 is the mean of the f( a) and f(b), so if we draw a line segment from (a, f(a)) to (b, f(b)) on the graph of y = f(x), then the point (a +b 2 , f(a) + f(b) ) is the midpoint of this line segment. Likewise, (a +b 2 , f (a +b )) is the point on the graph with this x-coordinate. The inequality states that the line segment lies above the graph at this point, so the graph could look something like this: x y a ba +b f (a +b ) f(a) + f(b) We see the for the inequality to be true in this case, the curve needs to “bend up”, so that it is below the chord connecting (a, f(a)) with (b, f(b)). This means that the second derivative, f ′′(x) must be positive. So it would seem likely that the answer is C. Without too much difficulty we can see that the other options don’t work, just by looking at the above sketch. This shows that none of D, E or F is necessary (for the curve could bend back upwards for more negative values ofx, as in a quadratic). If we translate the graph downwards, then it will still have this property, but A is no longer the case, so A is not necessary. And if we continue this graph to more negative values of x, it will have f′(x)< 0 if we go far enough, so B is not necessary either. Therefore by elimination, C must be necessary. We can also prove directly that C is necessary. To do so, we will prove the contrapositive: if f′′(x)< 0 for some real x, then there are real numbers a and b with a<b for which f(a) + f(b)^2 < f (a +b ) (∗) Now, f′′(x)< 0 means that the curve is “bending downwards” at this point, which means that a sketch of the graph will look like a reflection of the above sketch in a horizontal line (at least near the point where f ′′(x) < 0). And then the curve will lie above the chord, so ( ∗) is true. Hence C is necessary. Commentary: An interesting question to wonder about is whether there are any examples of such polynomial functions. Clearly, constant polynomials (f(x) = c) satisfy the condition, as do linear polynomials (f(x) = ax +b) and quadratics with a positive coefficient of x^2 (f(x) =ax2 +bx +c with a> 0) For higher degree polynomials, if f(x) = axn +··· , then f′′(x) = an(n- 1)xn-2 +··· . If n is odd, then this polynomial takes all real values, including negative ones, so this cannot satisfy the condition. If n is even, then it might be the case that f′′(x) ⩾ 0 for allx, which will be the case if f′′(x) ⩾ 0 at all of its stationary points. For example, if we take f′′(x) = 3 x1 + 1, then the corresponding f(x) will have this condition. A function satisfying this condition is called a convex function.","tags":["TMUA","2018","Paper2"],"difficulty":2},{"id":"2018-P2-Q19","paper":2,"index":18,"stemLatex":"Three real numbers $x$, $y$ and $z$ satisfy $x > y > z > 1$. \n\n Which one of the following statements must be true?","options":[{"key":"A","latex":"\\frac{2^{z+1}}{2^x} > \\frac{2^x + 2^z}{2^y}"},{"key":"B","latex":"2 > \\frac{3^x + 3^z}{3^y}"},{"key":"C","latex":"\\frac{2 \\times 5^x}{5^z} > \\frac{5^x + 5^z}{5^y}"},{"key":"D","latex":"2 < \\frac{7^x + 7^z}{7^y}"}],"answerKey":"C","explanationLatex":"There are different ways of approaching this question. Given that the question asks which one of the statements must be true, it is probably simplest to try to find values which “break” each of the inequalities. For each one, we can either multiply through by the denominators first, or leave the fractions as fractions. A We can write this as 2z-x+1> 2x-y + 2z-y To make this false, we want the left side small and right side large. But x > y > z, so the left hand side is less than 2. If we then make x-y >1, the right hand side will be at least 2, and so this statement is not necessarily true. For example, x = 10, y = 5, z = 3 will make the inequality false. B This becomes 2 > 3x-y + 3z-y. If we take x = 10 and y = 2, then the first term on the right side is 3 8 so this inequality is not always true. C Multiplying by 5 z^5y gives 2× 5x+y > 5x+z + 52z Now 5x+y > 5x+z and 5x+y > 52z so this must be true. D This is 2< 7x-y + 7z-y If x and y are very close, then 7 x-y is just greater than 1, and then if z is much smaller than y, 7z-y is close to 0, so the sum will be just greater than 1. So for example, x = 1000, y = 999.99, z = 1.1 will do to show that this statement is not always true. Therefore the correct option is C.","tags":["TMUA","2018","Paper2"],"difficulty":2},{"id":"2018-P2-Q20","paper":2,"index":19,"stemLatex":"It is given that the equation $\\sqrt{x+p} + \\sqrt{x} = p$ has at least one real solution for $x$, where $p$ is a real constant. \n\n What is the complete set of possible values for $p$?","options":[{"key":"A","latex":"$p=1$ or $p=0$"},{"key":"B","latex":"$p=0$ or $p\\geq 1$"},{"key":"C","latex":"p \\geq -x"},{"key":"D","latex":"p \\geq \\sqrt{x}"},{"key":"E","latex":"p \\geq 0"},{"key":"F","latex":"p \\geq 1"}],"answerKey":"B","explanationLatex":"An algebraic approach We can manipulate the equation algebraically. We could square it immediately, but it is probably simpler to first move one of the square roots to the other side so that there is only one square root on each side, that is $\\\\sqrt{x}$ +p =p-$\\\\sqrt{x}$ We can now square this to obtain x +p =p^2- 2p√x +x so 2p√x =p^2-p If p≠ 0, we can divide by p to obtain 2√x =p- 1. Since the left side of this must be positive, there will be a solution as long as p ⩾ 1. We also need to check that whenp ⩾ 1, the solution obtained does satisfy the original equation. We have√x = 1 2(p- 1) so x = 1 4(p- 1)^2. Then the left side of the original equation becomes √4(p- 1)^2 +p + √4(p- 1)^2 = √4(p + 1)^2 + √4(p- 1)^2 = 1 2(p + 1) + 1 2(p- 1) =p and so this is a real solution. Finally, whenp = 0 (which we excluded earlier), the equation becomes√x +$\\\\sqrt{x}$ = 0, which has the solution x = 0. Thus the correct option is B. A graphical approach We can sketch the graphy =$\\\\sqrt{x}$ +p +$\\\\sqrt{x}$. We know that√x is an increasing function, defined for x ⩾ 0, taking only non-negative values. It follows that y =$\\\\sqrt{x}$ +p is also an increasing function, defined for x ⩾-p, being a translation of y =$\\\\sqrt{x}$ byp in the negative x-direction. Here is a sketch of the two graphs for some p> 0: x y Clearly, there cannot be a solution for p< 0, as the left hand side is never negative. For p ⩾ 0, the function on the left hand side is defined for all x ⩾ 0, and is increasing towards infinity. So the smallest value the function takes is when x = 0, which is $\\\\sqrt{p}$, and it takes on every value larger than this. So there will be a solution to the original equation whenever $\\\\sqrt{p}$ ⩽p, which is whenever p ⩽p^2, that is p^2-p ⩾ 0 or p(p- 1) ⩾ 0. This holds for p ⩽ 0 and p ⩾ 1. Since we also require p ⩾ 0, there is a solution for p = 0 and for all p ⩾ 1. Hence the correct option is B.","tags":["TMUA","2018","Paper2"],"difficulty":2}]