[{"id":"2019-P1-Q1","paper":1,"index":0,"stemLatex":"$f(x)$ is a quadratic function in $x$. \n\n The graph of $y = f(x)$ passes through the point (1, -1) and has a turning point at (-1, 3). \n\n Find an expression for $f(x)$.","options":[{"key":"A","latex":"-x^2 - 2x + 2"},{"key":"B","latex":"-x^2 + 2x + 3"},{"key":"C","latex":"x^2 - 2x"},{"key":"D","latex":"x^2 + 2x - 4"},{"key":"E","latex":"2x^2 + 4x + 1"},{"key":"F","latex":"-2x^2 - 4x + 5"}],"answerKey":"A","explanationLatex":"We could try substituting in (1 ,-1) and (-1, 3) into the six given expressions: when we substitute in x = 1, only A, C, D and F give -1, so the answer must be one of these. When we substitute in x =-1 into these four, only A and C give 3, so the answer must be one of these two. Finally, we need a turning point at x =-1, so we differentiate these two: A gives-2x- 2, which is zero at x =-1, which C gives 2x- 2, which is -4 at x =-1. So the correct answer is A. Alternatively, we could work out the answer without reference to the given options. Let f(x) =ax2 +bx +c. We are given f(1) = -1 and f(-1) = 3, so a +b +c =-1 a-b +c = 3 Adding these gives 2a + 2c = 2, so a +c = 1 and hence b =-2. We then differentiate to get f′(x) = 2ax +b, so as f′(-1) = 0, we require-2a +b = 0, soa =-1 and c = 2. Hence the answer is A.","tags":["TMUA","2019","Paper1"],"difficulty":2},{"id":"2019-P1-Q2","paper":1,"index":1,"stemLatex":"Find the complete set of values of the real constant $k$ for which the expression $$x^2 + kx + 2x + 1 - 2k$$ is positive for all real values of $x$.","options":[{"key":"A","latex":"-12 < k < 0"},{"key":"B","latex":"$k < -12$ or $k > 0$"},{"key":"C","latex":"-\\sqrt{6} - 3 < k < \\sqrt{6} - 3"},{"key":"D","latex":"$k < -\\sqrt{6} - 3$ or $k > \\sqrt{6} - 3$"},{"key":"E","latex":"-2 < k < \\frac{1}{2}"},{"key":"F","latex":"$k < -2$ or $k > \\frac{1}{2}$"},{"key":"G","latex":"0 < k < 4"},{"key":"H","latex":"$k < 0$ or $k > 4$"}],"answerKey":"A","explanationLatex":"Let us write the expression asx2 +(k+2)x+(1-2k) so that the coefficient ofx and the constant are clear. The expression is positive for large values of x, and so is positive for all values of x if the discriminant is negative. Hence we require (k + 2)^2- 4(1- 2k)< 0. Expanding and simplifying this gives k^2 + 12k< 0, so k(k + 12)< 0. If we sketch a graph of the function k(k + 12), we obtain k y -12 We see that k(k + 12)< 0 for-12<k< 0, so the correct option is A.","tags":["TMUA","2019","Paper1"],"difficulty":2},{"id":"2019-P1-Q3","paper":1,"index":2,"stemLatex":"Find the coefficient of $x$ in the expression: $$(1 + x)^0 + (1 + x)^1 + (1 + x)^2 + (1 + x)^3 +... + (1 + x)^{79} + (1 + x)^{80}$$","options":[{"key":"A","latex":"80"},{"key":"B","latex":"81"},{"key":"C","latex":"324"},{"key":"D","latex":"628"},{"key":"E","latex":"3240"},{"key":"F","latex":"3321"},{"key":"G","latex":"6480"},{"key":"H","latex":"6642"}],"answerKey":"E","explanationLatex":"The coefficient of x in the expansion of (1 + x)n is (n ) =n, so the coefficient of x in this whole expression is 0 + 1 + 2 + 3 +··· + 79 + 80 = 1 2× 80× 81 using the formula for the sum of the firstn positive integers, 1 2n(n+1). This gives 40×81 = 3240, so the answer is E.","tags":["TMUA","2019","Paper1"],"difficulty":2},{"id":"2019-P1-Q4","paper":1,"index":3,"stemLatex":"The sequence $x_n$ is given by: $$x_1 = 10$$ $$x_{n+1} = \\sqrt{x_n} \\text{ for } n \\geq 1$$ What is the value of $x_{100}$? \n\n [Note that $a^{b^c}$ means $a^{(b^c)}$]","options":[{"key":"A","latex":"10^{2^{99}}"},{"key":"B","latex":"10^{2^{100}}"},{"key":"C","latex":"10^{2^{-99}}"},{"key":"D","latex":"10^{2^{-100}}"},{"key":"E","latex":"10^{-2^{99}}"},{"key":"F","latex":"10^{-2^{100}}"},{"key":"G","latex":"10^{-2^{-99}}"},{"key":"H","latex":"10^{-2^{-100}}"}],"answerKey":"C","explanationLatex":"We work out the first few terms, writing them in the form of the given options: x1 = 10 x^2 = $\\\\sqrt{10}$ = 10 2 = 102-1 x^3 = √102-1 = ( 102-1)^1 2 = 102-1× 1 2 = 102-2 As this pattern continues, we will end up with x100 = 102-99 , which is option C.","tags":["TMUA","2019","Paper1"],"difficulty":2},{"id":"2019-P1-Q5","paper":1,"index":4,"stemLatex":"$S$ is a geometric sequence. \n\n The sum of the first 6 terms of $S$ is equal to 9 times the sum of the first 3 terms of $S$. \n\n The 7th term of $S$ is 360. \n\n Find the 1st term of $S$.","options":[{"key":"A","latex":"\\frac{40}{27}"},{"key":"B","latex":"\\frac{40}{9}"},{"key":"C","latex":"\\frac{40}{3}"},{"key":"D","latex":"\\frac{45}{16}"},{"key":"E","latex":"\\frac{45}{8}"},{"key":"F","latex":"\\frac{45}{4}"}],"answerKey":"E","explanationLatex":"Let the first term of S be a and the common ratio r, as usual. The sum of the first n terms of S is a(rn- 1) r- 1 , so the second line of the question can be written as a(r^6- 1) r- 1 = 9a(r^3- 1) r- 1 which can be multiplied by (r- 1) and divided by a to give r^6- 1 = 9(r^3- 1). Writing R = r^3, this becomes R^2- 1 = 9 R- 9, so R^2- 9R + 8 = 0. This factorises to (R- 1)(R- 8) = 0, so R = 1 or R = 8, and hence r = 1 or r = 2. However, ifr = 1, then the original equation is not valid; in that case, the sequence is constant, the sum of the first 6 terms is 6 a and the sum of the first 3 terms is 3 a. We could only have 6a = 9× 3a ifa = 0, but that is impossible as the 7th term of S is 360. So we must have r = 2. The 7th term of S is ar6, so 360 = 2 6a = 64a, and hence a = 360 64 = 45 8 and so the answer is E.","tags":["TMUA","2019","Paper1"],"difficulty":2},{"id":"2019-P1-Q6","paper":1,"index":5,"stemLatex":"The circles with equations $$(x+4)^2 + (y+1)^2 = 64$$ $$(x-8)^2 + (y-4)^2 = r^2$$ have exactly one point in common. \n\n Find the difference between the two possible values of $r$.","options":[{"key":"A","latex":"4"},{"key":"B","latex":"10"},{"key":"C","latex":"16"},{"key":"D","latex":"26"},{"key":"E","latex":"50"}],"answerKey":"C","explanationLatex":"We could try to solve this algebraically or geometrically. An algebraic approach looks difficult, as we have two quadratic equations, and one of them has an unknown coefficient (that is, r^2). So we use a geometric approach instead. The first circle has centre ( -4,-1) and radius 8, while the second circle has centre (8 , 4) and radiusr. The point (8, 4) does not lie inside the first circle (it is more than 8 - (-4) = 12 units away), so the two circles will have exactly one point in common when the first circle is tangent to the second circle externally or when it lies inside the second circle and touches it internally. Here is a sketch of these two situations: x y x y In the first case, where r =r1, we see that r1 + 8 is the distance between the centres of the two circles. In the second case, where r =r^2, we see that r^2- 8 is the distance between the centres. Therefore r1 + 8 =r^2- 8, so r^2-r1 = 16, and the answer is C. Note that we did not even need to work out the actual values of r1 and r^2. If we had done so, the distance between the centres is $\\\\sqrt{122}$ + 52 = 13, so r1 = 13- 8 = 5 and r^2 = 13 + 8 = 21.","tags":["TMUA","2019","Paper1"],"difficulty":2},{"id":"2019-P1-Q7","paper":1,"index":6,"stemLatex":"A curve has equation $$y = (2q - x^2)(2qx + 3)$$ The gradient of the curve at $x=-1$ is a function of $q$. \n\n Find the value of $q$ that minimises the gradient of the curve at $x=-1$.","options":[{"key":"A","latex":"-1"},{"key":"B","latex":"-\\frac{3}{4}"},{"key":"C","latex":"-\\frac{1}{2}"},{"key":"D","latex":"0"},{"key":"E","latex":"\\frac{1}{2}"},{"key":"F","latex":"\\frac{3}{4}"},{"key":"G","latex":"1"}],"answerKey":"F","explanationLatex":"We expand the given expression to get y = 4q^2x- 2qx3 + 6q- 3x2 Differentiating gives dy dx = 4q2- 6qx2- 6x so the gradient of the curve at x =-1 is 4q2- 6q + 6. We can minimise this either by using calculus or by completing the square. Using calculus, we have d dq (4q2- 6q + 6) = 8q- 6, and 8q- 6 = 0 when q = 6 8 = 3 4, so the answer is F. Alternatively, completing the square, we have 4q2- 6q + 6 = 4(q^2- 3 2q) + 6 = 4 ( (q- 3 4)^2- ( 3 4)^2) + 6 so the minimum occurs when q = 3 4, as before.","tags":["TMUA","2019","Paper1"],"difficulty":2},{"id":"2019-P1-Q8","paper":1,"index":7,"stemLatex":"A function $f$ is such that $0 < f(x) < 1$ for $0 \\leq x \\leq 1$. \n\n The trapezium rule with $n$ equal intervals is used to estimate $\\int_0^1 f(x)dx$ and produces an underestimate. \n\n Using the same number of equal intervals, for which one of the following does the trapezium rule produce an overestimate?","options":[{"key":"A","latex":"\\int_0^1 (f(x) + 1)dx"},{"key":"B","latex":"\\int_0^1 2f(x)dx"},{"key":"C","latex":"\\int_{-1}^0 f(x+1)dx"},{"key":"D","latex":"\\int_{-1}^0 f(-x)dx"},{"key":"E","latex":"\\int_0^1 (1-f(x))dx"}],"answerKey":"E","explanationLatex":"A This translates the graph upwards (in the positive y-direction) by 1 unit, so the trapezia will just be 1 unit taller. The approximation will therefore still be an underestimate. B This stretches the graph by a factor of 2 in the y-direction, and the trapezia with it, so the approximation will still be an underestimate. C This translates the graph by 1 unit leftwards (in the negative x-direction), and correspondingly calculates the area between -1 and 0 rather than between 0 and 1; the trapezia are correspondingly translated leftwards by 1 unit, and so we will still get an underestimate. D This reflects the graph in the y-axis, and correspondingly calculates the area between -1 and 0. The trapezia are correspondingly reflected, so we will still get an underestimate. E The graph is reflected in the x-axis and then translated upwards by 1 unit. Whether the approximation using the trapezium rule is an over- or underestimate is not affected by the vertical translation, as we noted in A, so we may as well not translate the graph at all, giving just ∫ 1 0 (-f(x)) dx. Let us sketch an example using just two intervals: x y x y The original (blue) trapezia give an underestimate for the integral. The red trapezia give an underestimate for the area of the region, but since the integral is negative (the graph lies below the x-axis), the trapezium rule gives an overestimate for the integral (for example, it may estimate-0.6 instead of the correct -0.65, and-0.6>-0.65). Hence the correct answer is E.","tags":["TMUA","2019","Paper1"],"difficulty":2},{"id":"2019-P1-Q9","paper":1,"index":8,"stemLatex":"$p$ is a positive constant. \n\n Find the area enclosed between the curves $y = p\\sqrt{x}$ and $x = p\\sqrt{y}$","options":[{"key":"A","latex":"\\frac{2}{3}p^{\\frac{5}{2}} - \\frac{1}{2}p^2"},{"key":"B","latex":"\\frac{4}{3}p^{\\frac{5}{2}} - p^2"},{"key":"C","latex":"\\frac{p^4}{6}"},{"key":"D","latex":"\\frac{p^4}{3}"},{"key":"E","latex":"\\frac{2}{3}p^3 - \\frac{1}{2}p^4"},{"key":"F","latex":"\\frac{4}{3}p^3 - p^4"},{"key":"G","latex":"2p^4"}],"answerKey":"D","explanationLatex":"We can rearrange x =p√y to give√y = x p , so y = x^2 p^2 (for x≥ 0). We can then sketch the two curves to see what is intended: x y We need to find where these two curves intersect. We have p√x = x^2 p^2 so x 2 =p^3, giving x 2 =p, so x =p^2 and thus y =p^2 as well; the point of intersection thus lies on the line y =x and the graph is symmetrical about this line. We can now find the area between the curves by integrating the difference fromx = 0 tox =p^2: area = ∫ p^2 p√x- x^2 p^2 dx = [ px ( 3 2)- x^3 3p2 ]p^2 = p.p^3 ( 3 2)- p^6 3p2- 0 = 2 3p4- 1 3p4 = 1 3p4 so the answer is D.","tags":["TMUA","2019","Paper1"],"difficulty":2},{"id":"2019-P1-Q10","paper":1,"index":9,"stemLatex":"Evaluate $$\\int_{-1}^3 |x|(1-x)dx$$","options":[{"key":"A","latex":"\\frac{17}{3}"},{"key":"B","latex":"-\\frac{17}{3}"},{"key":"C","latex":"\\frac{16}{3}"},{"key":"D","latex":"-\\frac{16}{3}"},{"key":"E","latex":"\\frac{11}{3}"},{"key":"F","latex":"-\\frac{11}{3}"}],"answerKey":"F","explanationLatex":"We do not have a rule for integrating |x|, so we will split this into two separate integrals, one from x =-1 to x = 0, where|x| =-x, and one from x = 0 to x = 3, where|x| =x. This gives ∫ 3 -1 |x|(1-x) dx = ∫ 0 -1 |x|(1-x) dx + ∫ 3 |x|(1-x) dx = ∫ 0 -1 (-x)(1-x) dx + ∫ 3 x(1-x) dx = ∫ 0 -1 -x +x^2 dx + ∫ 3 x-x^2 dx = [ -x^2 2 + x^3 ]0 -1 + [x^2 2 - x^3 ]3 =-(- 1 2- 1 3) + (9 2- 9) = 5 6- 9 =- 22 =- 11 so the correct answer is F.","tags":["TMUA","2019","Paper1"],"difficulty":2},{"id":"2019-P1-Q11","paper":1,"index":10,"stemLatex":"Find the sum of the real values of $x$ that satisfy the simultaneous equations: $$\\log_3{(xy^2)} = 1$$ $$(\\log_3x)(\\log_3y) = -3$$","options":[{"key":"A","latex":"\\frac{1}{3}"},{"key":"B","latex":"1"},{"key":"C","latex":"3"},{"key":"D","latex":"3\\tfrac{1}{9}"},{"key":"E","latex":"9\\tfrac{1}{27}"},{"key":"F","latex":"9\\tfrac{1}{3}"},{"key":"G","latex":"27"},{"key":"H","latex":"27\\tfrac{1}{9}"}],"answerKey":"H","explanationLatex":"We can write the first equation as log^3x+2 log^3y = 1. Then writing X = log^3x andY = log^3y, the two equations become X + 2Y = 1 XY =-3 We are interested in the values of X, so we rearrange the second equation to get Y =-3/X, and substitute this into the first equation to give X- 6 X = 1. Multiplying this by X gives the quadratic X 2- 6 = X, or X 2-X- 6 = 0. This factorises to (X- 3)(X + 2) = 0, so X = 3 or X =-2, so x = 3X is either 27 or 1 9, and their sum is 27 1 9, which is option H. We could also have solved this problem by writingY in terms of X using the first equation, and then substituting this into the second; it would have given the same result.","tags":["TMUA","2019","Paper1"],"difficulty":2},{"id":"2019-P1-Q12","paper":1,"index":11,"stemLatex":"It is given that for $t \\geq 1$ $$\\frac{dV}{dt} = \\frac{24\\pi (t-1)}{(1 + \\sqrt{t})}$$ and $V = 7$ when $t = 1$. \n\n Find the value of $V$ when $t=9$.","options":[{"key":"A","latex":"208\\pi + 7"},{"key":"B","latex":"216\\pi + 7"},{"key":"C","latex":"224\\pi + 7"},{"key":"D","latex":"416\\pi + 7"},{"key":"E","latex":"608\\pi + 7"},{"key":"F","latex":"744\\pi + 7"}],"answerKey":"C","explanationLatex":"To find the value of V when t = 9, we need to integrate this expression. There is no general rule for integrating quotients such as these, but we might notice that t- 1 looks somewhat similar to the denominator $\\\\sqrt{t}$ + 1. In fact, we can use the difference of two squares to write t- 1 = ( $\\\\sqrt{t}$ + 1)( √t- 1), so we can rewrite the fraction to give dV dt = 24π( √t- 1) and this is easy to integrate; we get V = 24πt ( 3 2) - 24πt +c = 16πt 2- 24πt +c Substituting in V = 7 and t = 1 gives 7 = 16π- 24π +c so c = 8π + 7. Then at t = 9, we have V = 16π× 9 2- 24π× 9 + 8π + 7 = (16× 27- 24× 9 + 8)π + 7 = 224π + 7 so the answer is option C. Commentary: The final calculation can be simplified somewhat by writing 16× 27 = 16× 3× 9 = 48× 9, so the brackets become 24× 9 + 8. Then 24× 9 = 240- 24 = 216, giving the final answer of 224π + 7.","tags":["TMUA","2019","Paper1"],"difficulty":2},{"id":"2019-P1-Q13","paper":1,"index":12,"stemLatex":"Find the maximum value of $$4^{\\sin x} - 4 \\times 2^{\\sin x} + \\frac{17}{4}$$ for real $x$.","options":[{"key":"A","latex":"\\frac{1}{4}"},{"key":"B","latex":"\\frac{5}{2}"},{"key":"C","latex":"\\frac{13}{2}"},{"key":"D","latex":"\\frac{21}{2}"},{"key":"E","latex":"\\frac{65}{4}"},{"key":"F","latex":"There is no maximum value."}],"answerKey":"B","explanationLatex":"We will not use calculus for this question, as the specification does not expect knowledge of differentiating exponentials or trigonometric functions. (And it also turns out that using calculus would make the solution far more complicated.) Instead, we note that this is a quadratic expression in u = 2sinx, and we can write the function as u^2- 4u + 17 4. We also note that as-1≤ sinx≤ 1, 1 2≤u≤ 2 for all real x (andu can take every value in this interval). Now we can complete the square on the quadratic to understand its behaviour; we get (u- 2)^2- 22 + 17 4 = (u- 2)^2 + 1 Thus the minimum value of the function occurs whenu = 2, and foru in the interval 1 2≤u≤ 2, the maximum value occurs when u = 1 2. In this case, we get (- 3 2)^2 + 1 4 = 10 4 = 5 2, so the correct answer is B.","tags":["TMUA","2019","Paper1"],"difficulty":2},{"id":"2019-P1-Q14","paper":1,"index":13,"stemLatex":"$x$ satisfies the simultaneous equations $$\\sin{2x} + \\sqrt{3}\\cos{2x} = -1$$ and $$\\sqrt{3}\\sin{2x} - \\cos{2x} = \\sqrt{3}$$ where $0^{\\circ} \\leq x \\leq 360^{\\circ}$. \n\n Find the sum of the possible values of $x$.","options":[{"key":"A","latex":"210^{\\circ}"},{"key":"B","latex":"330^{\\circ}"},{"key":"C","latex":"390^{\\circ}"},{"key":"D","latex":"660^{\\circ}"},{"key":"E","latex":"780^{\\circ}"},{"key":"F","latex":"930^{\\circ}"}],"answerKey":"B","explanationLatex":"We writeS = sin 2x and C = cos 2x, so the simultaneous equations become S + $\\\\sqrt{3C}$ =-1 √3S- C = $\\\\sqrt{Multiplying}$ the second equation by $\\\\sqrt{3}$ and then adding to the first equation gives 4S = 2 so S = 1 2 and hence C =- $\\\\sqrt{2}$ . The solutions to sin 2x = 1 2 for 0◦≤x≤ 360◦, or 0◦≤ 2x≤ 720◦, are given by 2x = 30◦, 150◦, 390◦ and 510◦. The solutions to cos 2x =- $\\\\sqrt{2}$ in the same range are given by 2 x = 150◦, 210◦, 510◦ and 570◦, so the values of 2x for which both are true are 150◦ and 510◦. Thus the solutions forx in this range are 75◦ and 255◦, and their sum is 330◦. Hence the correct answer is option B.","tags":["TMUA","2019","Paper1"],"difficulty":2},{"id":"2019-P1-Q15","paper":1,"index":14,"stemLatex":"Find the real non-zero solution to the equation $$\\frac{2^{(9^x)}}{8^{(3^x)}} = \\frac{1}{4}$$","options":[{"key":"A","latex":"\\log_32"},{"key":"B","latex":"2\\log_32"},{"key":"C","latex":"1"},{"key":"D","latex":"2"},{"key":"E","latex":"\\log_23"},{"key":"F","latex":"2\\log_23"}],"answerKey":"A","explanationLatex":"We look to see how we can simplify this problem. First, the powers are 9x and 3x. We can write 9x = (32)x = 32x = (3x)^2, but it is not clear how this will help. But without a better idea, let us go with it, and write u = 3x, so the question becomes 2(u^2) 8u = 1 4. Next, let us write everything as powers of 2, giving 2(u^2) 23u = 1 22. Therefore 2u2-3u = 2-2 so we require u^2- 3u =-2. We can rearrange this quadratic to give u^2- 3u + 2 = 0, so (u- 1)(u- 2) = 0 giving u = 1 or u = 2. Thus 3x = 1 or 3x = 2, giving two possibilities: x = 0 or x = log3 2. The question explicitly says “find the real non-zero solution”, so the correct answer is option A.","tags":["TMUA","2019","Paper1"],"difficulty":2},{"id":"2019-P1-Q16","paper":1,"index":15,"stemLatex":"Given that $$2\\int_0^1 f(x)dx + 5\\int_1^2 f(x)dx = 14$$ and $$\\int_0^1 f(x+1)dx = 6$$ find the value of $$\\int_0^2 f(x)dx$$","options":[{"key":"A","latex":"-8"},{"key":"B","latex":"-4"},{"key":"C","latex":"-2"},{"key":"D","latex":"2"},{"key":"E","latex":"4"},{"key":"F","latex":"\\frac{29}{5}"},{"key":"G","latex":"\\frac{32}{5}"},{"key":"H","latex":"14"}],"answerKey":"C","explanationLatex":"Let A = ∫ 1 0 f(x) dx, which is the (signed) area under the graph of y = f(x) between x = 0 and x = 1 (taking signs into account, as usual: parts of the graph under the x-axis are counted as negative areas). Likewise, let B = ∫ 2 1 f(x) dx, which is the (signed) area between x = 1 and x = 2. The integral ∫ 1 0 f(x+ 1) dx gives the (signed) area under the graph of y = f(x+ 1) betweenx = 0 and x = 1. This graph is the graph of y = f(x) translated left by 1 unit, so this equals the area under the original graph between x = 1 and x = 2, which is B. Hence B = 6. The first equation gives 2A + 5B = 14, and as B = 6, we get A =-8. The final integral ∫ 2 0 f(x) dx is the (signed) area between x = 0 and x = 2, which is A +B = -8 + 6 =-2. Hence the answer is C.","tags":["TMUA","2019","Paper1"],"difficulty":2},{"id":"2019-P1-Q17","paper":1,"index":16,"stemLatex":"Find the fraction of the interval $0 \\leq \\theta \\leq \\pi$ for which the inequality $$(\\sin{(2\\theta)} - \\frac{1}{2})(\\sin{\\theta} - \\cos{\\theta}) \\geq 0$$ is satisfied.","options":[{"key":"A","latex":"\\frac{1}{12}"},{"key":"B","latex":"\\frac{1}{6}"},{"key":"C","latex":"\\frac{1}{4}"},{"key":"D","latex":"\\frac{5}{12}"},{"key":"E","latex":"\\frac{7}{12}"},{"key":"F","latex":"\\frac{3}{4}"},{"key":"G","latex":"\\frac{5}{6}"},{"key":"H","latex":"\\frac{11}{12}"}],"answerKey":"C","explanationLatex":"A product of two terms is at least 0 if both factors are positive or both terms are negative. So we look at the signs of the two factors. We start by finding the values of θ for which they are zero. We have sin^2θ = 1 2 when 2θ = π 6 or 5π 6 , so for θ = π 12 or 5π 12 . Thinking about a sketch of the graph of y = sin^2θ, we see that sin^2θ- 1 2 is positive for θ between these values and negative elsewhere. For the second factor, we have sin θ = cosθ when tanθ = 1, so θ = π 4 . Thinking about the graphs of y = sinθ andy = cosθ, we see that sin θ- cosθ is negative for θ< π 4 and positive for θ> π 4 . For consistency, we write π 4 = 3π 12 in what follows. We thus have the following table: 0≤θ< π 12 θ = π π 12 <θ< 3π 12 θ = 3π 3π 12 <θ< 5π 12 θ = 5π 5π 12 <θ ≤π sin^2θ- 1 2 - 0 + + + 0 - sinθ- cosθ - - - 0 + + + product + 0 - 0 + 0 - The lengths of the subintervals where the product is positive (or zero) is ( π 12- 0) + (5π 12- 3π 12 ) = 3π 12 = π The whole interval has lengthπ, so the fraction of the interval for which the inequality is satisfied is 1 4, and the answer is option C.","tags":["TMUA","2019","Paper1"],"difficulty":2},{"id":"2019-P1-Q18","paper":1,"index":17,"stemLatex":"Find the shortest distance between the curve $y = x^2 + 4$ and the line $y = 2x -2$.","options":[{"key":"A","latex":"2"},{"key":"B","latex":"\\sqrt{5}"},{"key":"C","latex":"\\frac{6\\sqrt{5}}{5}"},{"key":"D","latex":"3"},{"key":"E","latex":"\\frac{5\\sqrt{5}}{3}"},{"key":"F","latex":"5"},{"key":"G","latex":"6"}],"answerKey":"B","explanationLatex":"From the given options, it is clear that the curve and line do not intersect. One way to do this would be to take a point on the curve, say (a,a 2 +4), and a point on the line, say (b, 2b- 2), find the (square of) the distance between them, and then try to find some way of minimising it. That, though, seems quite challenging, as there are two independent variables involved. Another way is to draw a sketch of the situation and to think about what the question means geometrically. x y The shortest distance from any point on the curve to the straight line is found by drawing a perpendicular from the curve to the straight line, like this: x y This may give us a few different ideas; any of the following would work: • We can use this to find the distance of any point ( a,a 2 + 4) to the line; we then find the smallest possible such distance. • The smallest distance will be when the perpendicular to the line is also perpendicular to the curve; in the above sketch, it is clear that if we move the joining line closer to the origin, its length will become shorter. We can find this position by finding the gradient of the normal to the curve and finding a point at which this is perpendicular to the line. • Taking this one step further, if we translate the line (without changing its gradient), we shorten the minimum distance of the curve from the line, but as long as we don’t cross the curve, the point which is closest to the line will remain so. Therefore the closest point to the line is the point on the curve where the curve’s gradient equals the line’s gradient, for when the line just touches the curve, this point will be at zero distance from the curve. We will use the final approach to solve this problem. The gradient of the line is 2, and the curve has gradient dy dx = 2x, so dy dx = 2 when x = 1, so (1, 5) is the closest point on the curve to the line. The line through (1, 5) perpendicular to the line y = 2x- 2 has equation y- 5 =- 1 2(x- 1). This intersects the line y = 2x- 2 when (2x- 2)- 5 =- 1 2(x- 1) so 2x- 7 =- 1 2(x- 1). Multiplying this by 2 gives 4x- 14 =-x + 1 so 5x = 15, giving x = 3, y = 4. The distance between (1, 5) and (3, 4) is √(3- 1)^2 + (4- 5)^2 = $\\\\sqrt{5}$, so the correct answer is B. Commentary: Though we dismissed the original two-variable idea, it turns out that we could have used it in a similar way to the first idea listed above: if we fix a value for a, we can find the value of b which gives the smallest distance from that point to the line. This gives us a formula for the smallest distance from a given point to the line in terms of a; we can then find the value of a which makes this as small as possible. This changes the problem from one two-variable problem to two one-variable problems, which we know how to solve.","tags":["TMUA","2019","Paper1"],"difficulty":2},{"id":"2019-P1-Q19","paper":1,"index":18,"stemLatex":"Find the value of $$\\sum_{k = 0}^{90} \\sin ({10 + 90k})^{\\circ}$$","options":[{"key":"A","latex":"0"},{"key":"B","latex":"\\sin 10^{\\circ}"},{"key":"C","latex":"\\sin 100^{\\circ}"},{"key":"D","latex":"\\sin 190^{\\circ}"},{"key":"E","latex":"\\sin 280^{\\circ}"},{"key":"F","latex":"1"}],"answerKey":"C","explanationLatex":"We start by writing out the first few terms of the summation, and writing each of them in terms of sin 10◦ or sin 100◦: k = 0 : sin 10 ◦ k = 1 : sin 100 ◦ k = 2 : sin 190 ◦ =- sin 10◦ k = 3 : sin 280 ◦ =- sin 100◦ k = 4 : sin 370 ◦ = sin 10◦ k = 5 : sin 460 ◦ = sin 100◦ k = 6 : sin 550 ◦ =- sin 10◦ k = 7 : sin 640 ◦ =- sin 100◦ and we see that we repeat every four terms. The first four terms sum to zero, as do the next four, and so on. In total, there are 91 terms, and 91 = 22× 4 + 3, so the only terms which contribute to the final sum are the last three. Therefore the whole sum is that from k = 88,k = 89 and k = 90, giving sin 10◦ + sin 100◦ + (- sin 10◦) = sin 100◦. Hence the answer is C.","tags":["TMUA","2019","Paper1"],"difficulty":2},{"id":"2019-P1-Q20","paper":1,"index":19,"stemLatex":"What is the complete range of values of $k$ for which the curves with equations $$y = x^3 - 12x$$ and $$y = k - (x-2)^2$$ intersect at three distinct points, of which exactly two have positive x-coordinates?","options":[{"key":"A","latex":"-4 < k < 0"},{"key":"B","latex":"-4 < k < 4"},{"key":"C","latex":"-4 < k < 16"},{"key":"D","latex":"-16 < k < 0"},{"key":"E","latex":"-16 < k < 4"},{"key":"F","latex":"-16 < k < 16"}],"answerKey":"E","explanationLatex":"Finding the values of x at which the two curves intersect is the same as finding the values of x at which the y-values are equal, that is, x^3- 12x =k- (x- 2)^2 or x^3- 12x-k + (x- 2)^2 = 0. Thus we are finding out something about the roots of a cubic. We can expand the brackets in the final equation to give x^3 +x^2- 16x-k + 4 = 0, and we write f(x) =x^3 +x^2- 16x-k + 4. We can determine something about the locations of the roots of f( x) by finding the stationary points. We have f′(x) = 3x2 + 2x- 16, so f′(x) = 0 when x =-2± $\\\\sqrt{22}$ + 4× 3× 16 6 =-2± $\\\\sqrt{6}$ =-2± 14 6 = 2 or-8 3. (We could have found these roots by factorising, but that did not look particularly easy.) Therefore there are stationary points at x = 2 and x =- 8 3. We have f(2) =-16-k, but calculating f(- 8 3) by hand would be painful, so let us try to avoid doing so if we can! We note that f(0) = 4-k, and that f(x) gets big as x gets big, so if there are exactly two roots with positive x-coordinates, we need the graph to look something like this: x y (2,-16-k) (0, 4-k) In particular, we require f(0) > 0, so we must have k< 4. It must also be the case that the stationary point at x = 2 lies below the x-axis, so we need -16-k< 0, that is,-16<k . Once we have met these two conditions, it becomes clear that there are three distinct roots, exactly one of which has x < 0. Therefore, the complete range of values of k is given by -16<k< 4, which is option E.","tags":["TMUA","2019","Paper1"],"difficulty":2},{"id":"2019-P2-Q1","paper":2,"index":0,"stemLatex":"Find the coefficient of the $x^4$ term in the expansion $$x^2(2x + \\frac{1}{x})^6$$","options":[{"key":"A","latex":"15"},{"key":"B","latex":"30"},{"key":"C","latex":"60"},{"key":"D","latex":"120"},{"key":"E","latex":"240"}],"answerKey":"E","explanationLatex":"We need the coefficient of x^2 in the bracketed term to get the term in x^4 in the whole expansion. We expand this using the binomial theorem to get x^2 ( 2x + 1 x )^6 = x^2 ( (2x)^6 + (6 ) (2x)^5 (1 x ) + (6 ) (2x)^4 (1 x )^2 +··· ) . We stop here as the power of x in this last term is x^4-2 = x^2, as we require. This term is then x^2 (6 ) (2x)^4 (1 x )^2 = 15x2.16x4.x-2 = 240x4 so the answer is E.","tags":["TMUA","2019","Paper2"],"difficulty":2},{"id":"2019-P2-Q2","paper":2,"index":1,"stemLatex":"$(2x+1)$ and $(x-2)$ are factors of $2x^3 + px^2 + q$. \n\n What is the value of $2p+q$?","options":[{"key":"A","latex":"-10"},{"key":"B","latex":"-\\frac{38}{5}"},{"key":"C","latex":"-\\frac{22}{3}"},{"key":"D","latex":"\\frac{22}{3}"},{"key":"E","latex":"\\frac{38}{5}"},{"key":"F","latex":"10"}],"answerKey":"C","explanationLatex":"Since 2x + 1 = 2(x + 1 2) is a factor of f( x) = 2x3 + px2 + q, it follows that f(- 1 2) = 0. Similarly, f(2) = 0. Substituting these values in to f( x) gives 2(- 1 2)^3 + p(- 1 2)^2 + q = 0 2(2)^3 + p(22) + q = 0 so - 1 4 + 1 4 p + q = 0 16 + 4p + q = 0 Multiplying the first equation by 16 gives -4 + 4p + 16q = 0. Subtracting 16 + 4p + q = 0 from this gives-20 + 15q = 0, so q = 4 3, hence 4p =-16- q =- 52 3 . Therefore 2p + q =- 26 3 + 4 3 =- 22 3 , which is option C.","tags":["TMUA","2019","Paper2"],"difficulty":2},{"id":"2019-P2-Q3","paper":2,"index":2,"stemLatex":"$a$, $b$ and $c$ are real numbers. \n\n Given that $ab = ac$, which of the following statements must be true? \n\n I: $a=0$ \n\n II: $b=0$ or $c=0$ \n\n III: $b=c$","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":"It might be tempting to divide by a to get b = c, but a might be zero, and we cannot divide by zero. Instead, we can rearrange and factorise to get ab- ac = 0, so a(b- c) = 0. From this, we can deduce that a = 0 or b- c = 0 (or both). I This might not be true; we could have b = c instead. II This might not be true; we could have b = c = 1, for example. III This might not be true; we could have a = 0 instead. So the correct answer is A.","tags":["TMUA","2019","Paper2"],"difficulty":2},{"id":"2019-P2-Q4","paper":2,"index":3,"stemLatex":"Consider the following conjecture: \n\n If $N$ is a positive integer that consists of the digit 1 followed by an odd number of 0 digits and then a final digit 1, then N is a prime number. \n\n Here are three numbers: \n\n I: $N = 101$ (which is prime) \n\n II: $N = 1001$ (which equals $7 \\times 11 \\times 13$) \n\n III: $N=10001$ (which equals $73 \\times 137$) \n\n Which of these provide(s) a counterexample to the conjecture?","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":"A counterexample with have the antecedent (“N is a positive integer that consists of the digit 1 followed by an odd number of 0 digits and then a final digit 1”) true, and the consequent (“ N is a prime number”) false. We can now check each of the three offered values of N: I The form of N is correct: there is 1 zero, and 1 is odd, so the antecedent is true. However, the consequent is also true, so this is not a counterexample. II The form of N is not correct, as there are 2 zeros, so the antecedent is false. Hence this is not a counterexample. III The form of N is correct, as there are 3 zeros, and the consequent is false, as N is not prime. Hence this is a counterexample. The correct solution is D.","tags":["TMUA","2019","Paper2"],"difficulty":2},{"id":"2019-P2-Q5","paper":2,"index":4,"stemLatex":"Consider the following statement about the positive integers $a$, $b$ and $n$: \n\n $(\\ast)$: $ab$ is divisible by $n$ \n\n The condition 'either $a$ or $b$ is divisible by $n$' is:","options":[{"key":"A","latex":"necessary but not sufficient for $(\\ast)$"},{"key":"B","latex":"sufficient but not necessary for $(\\ast)$"},{"key":"C","latex":"necessary and sufficient for $(\\ast)$"},{"key":"D","latex":"not necessary and not sufficient for $(\\ast)$"}],"answerKey":"B","explanationLatex":"Saying that ‘either a or b is divisible by n’ is necessary for (∗) is the same as saying that if ab is divisibly by n, then either a or b is divisible by n. We can look for a counterexample or proof for this assertion. On the surface of it, it seems surprising: when a product is divisible by a number, why should one of the two factors also be? We know something similar to this for a product which equals zero, and we may know that if a product is divisible by a prime, then one of the factors must be. But there is no mention of zero or primes here. So let us try a simple example where n is not prime, such as n = 4 (the smallest non-prime other than 1). Then the statement becomes: if ab is divisible by 4, then either a or b is divisible by 4. A small amount of exploration finds counterexamples, such as a = 2, b = 2 or a = 6, b = 2. So the condition is not necessary for (∗). What about sufficiency? Saying that ‘either a or b is divisible by n’ is sufficient for (∗) is the same as saying that if either a or b is divisible by n, then ab is divisibly by n. We can again look for a counterexample or proof for this assertion. This time, though, a proof is clear: if a is divisible by n, then a is a multiple of n, so ab is also a multiple of n (as b is an integer). This means that ab is divisible by n. The same is the case if b is divisible by n, so the condition is sufficient. The correct answer is therefore B.","tags":["TMUA","2019","Paper2"],"difficulty":2},{"id":"2019-P2-Q6","paper":2,"index":5,"stemLatex":"A student attempts to solve the equation\n\n$\\cos x + \\sin x \\tan x = 2\\sin x - 1$\n\nin the range $0 \\leq x \\leq 2\\pi$.\n\nThe student's attempt is as follows:\n\n$\\cos x + \\sin x \\tan x = 2\\sin x - 1$\n\nSo $\\cos x - \\sin x + \\sin x \\tan x - \\sin x = -1$ (I)\n\nSo $(\\sin x - \\cos x)(\\tan x - 1) = -1$ (II)\n\nSo $\\sin x - \\cos x = -1$ or $\\tan x - 1 = -1$ (III)\n\nSo $(\\sin x - \\cos x)^2 = 1$ or $\\tan x = 0$ (IV)\n\nSo $2\\sin x \\cos x = 0$ or $\\tan x = 0$ (V)\n\nSo $x = 0, \\frac{\\pi}{2}, \\pi, \\frac{3\\pi}{2}, 2\\pi$ (VI)\n\nWhich of the following best describes this attempt?","stemImage":"/stem-fallback/2019-P2-Q6.png","options":[{"key":"A","latex":"It is completely correct"},{"key":"B","latex":"It is incorrect, and the first error occurs on line (I)"},{"key":"C","latex":"It is incorrect, and the first error occurs on line (II)"},{"key":"D","latex":"It is incorrect, and the first error occurs on line (III)"},{"key":"E","latex":"It is incorrect, and the first error is that extra solutions were introduced on line (IV)"},{"key":"F","latex":"It is incorrect, and the first error is that extra solutions were introduced on line (V)"},{"key":"G","latex":"It is incorrect, and the first error is not eliminating the values where $\\tan x$ is undefined on line (VI)"}],"answerKey":"D","explanationLatex":"The following is just one approach to checking an argument. We start by checking the final solutions on line (VI).x = 0 makes the LHS (left hand side) equal 1 but the RHS (right hand side) equal -1, so the attempt cannot be correct. Furthermore, this is not a value of x for which tan x is undefined, so G is not the correct answer. We must then identify where the error arose. We can now test the incorrect solution x = 0 at various stages of the argument to see where things first go wrong. Rather than working line-by-line, we start about half way through to see whether the mistake happens in the first half of the argument or the second half; this is a little more efficient. Let us try the incorrect solution x = 0 in line (III) (which is midway between (I) and (V)): this gives-1 =-1 or-1 =-1, so x = 0 is valid at this point. This tells us that the error must occur on line (I), (II) or (III). On line (I), x = 0 gives 1 =-1, while on line (II), x = 0 gives (-1)× (-1) =-1. So it is most likely that the first error occurs on line (III). We check the algebra on line (I): all that has happened is that 2 sin x has been subtracted from both sides, so this is correct. We likewise check the algebra on line (II): by multiplying out the brackets, we get the equation on line (I), so this is correct. (We might be concerned about the case cos x = 0 here, but this is not possible on either line (I) or line (II) because of the tan x terms on each line.) Finally, we are left with line (III). The implication here is that because two numbers multiply to give-1, one of them must be -1. This would be true if we were dealing with integers, but we are not, so this is an invalid implication. The correct solution is therefore D: the first error occurs on line (III).","tags":["TMUA","2019","Paper2"],"difficulty":2},{"id":"2019-P2-Q7","paper":2,"index":6,"stemLatex":"For which one of the following statements can the fact that $12^2 + 16^2 = 20^2$ be used to produce a counterexample?","options":[{"key":"A","latex":"If $a$, $b$ and $c$ are positive integers which satisfy the equation $a^2 + b^2 = c^2$, and the three numbers have no common divisor, then two of them are odd and the other is even."},{"key":"B","latex":"The equation $a^4 + b^2 = c^2$ has no solutions for which $a$, $b$ and $c$ are positive integers."},{"key":"C","latex":"The equation $a^4 + b^4 = c^4$ has no solutions for which $a$, $b$ and $c$ are positive integers."},{"key":"D","latex":"If $a$, $b$ and $c$ are positive integers which satisfy the equation $a^2 + b^2 = c^2$, then one is the arithmetic mean of the other two."}],"answerKey":"B","explanationLatex":"We check each statement in turn. A: The triple $(12,16,20)$ has a common divisor of $4$, so it does not satisfy the antecedent. Dividing by $4$ gives the primitive triple $(3,4,5)$, which has two odd terms and one even term, so this is not a counterexample. B: Since $16=4^2$, the identity $12^2+16^2=20^2$ can be rewritten as $4^4+12^2=20^2$. This is a positive-integer solution to $a^4+b^2=c^2$, so it is a counterexample. C: Neither $12$ nor $20$ is a perfect square, so the identity cannot be rewritten with all three terms as fourth powers. D: Although $12^2+16^2=20^2$, the number $16$ is the arithmetic mean of $12$ and $20$, so this triple is not a counterexample. Therefore the correct answer is B.","tags":["TMUA","2019","Paper2"],"difficulty":2},{"id":"2019-P2-Q8","paper":2,"index":7,"stemLatex":"$a$, $b$ and $c$ are real numbers with $a < b < c < 0$ \n\n Which of the following statements must be true? \n\n I: $ac < ab < a^2$ \n\n II: $b(c+a) > 0$ \n\n III: $\\frac{c}{b} > \\frac{a}{b}$","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":"E","explanationLatex":"When thinking about inequalities involving negative numbers, we must take care to reverse the sign of any inequality when multiplying or dividing by a negative number. In this case, all three of a, b and c are negative. I We start with a < b < c and multiply by a < 0 to obtain a2 > ab > ac so this statement must be true. II Since a < 0 and c < 0, a+ c < 0. Since b < 0, it follows that b(a+ c) > 0. So this statement must be true. III We start with c > a (since a < c ). Dividing by b < 0 then gives c b < a b and so this statement must be false. The correct answer is E.","tags":["TMUA","2019","Paper2"],"difficulty":2},{"id":"2019-P2-Q9","paper":2,"index":8,"stemLatex":"A large circular table has 40 chairs round it. \n\n What is the smallest number of people who can be sitting at the table already such that the next person to sit down must sit next to someone?","options":[{"key":"A","latex":"9"},{"key":"B","latex":"10"},{"key":"C","latex":"13"},{"key":"D","latex":"14"},{"key":"E","latex":"19"},{"key":"F","latex":"20"}],"answerKey":"D","explanationLatex":"If the next person must sit next to someone, there cannot be three spaces between one person and the next. But if there are only two spaces between them, someone sitting down in one of those spaces must sit next to someone. So the smallest number of people will be achieved when all of the gaps are exactly two spaces, except perhaps for the final person to the first person. So counting the seats around the table as 1, 2, 3, ... , we place people in seats 1, 4, 7, 10, ... , 37, 40. (The final person is sitting immediately next to the first person, but that does not matter.) There are 14 seats in this list, so the answer is D.","tags":["TMUA","2019","Paper2"],"difficulty":2},{"id":"2019-P2-Q10","paper":2,"index":9,"stemLatex":"$PQRS$ is a quadrilateral, labelled anticlockwise. \n\n Which one of the following is a necessary but not sufficient condition for $PQRS$ to be a parallelogram?","options":[{"key":"A","latex":"$PQ = SR$ and $PS$ is parallel to $QR$"},{"key":"B","latex":"$PQ = SR$ and $PQ$ is parallel to $SR$"},{"key":"C","latex":"$PQ = QR = SR = PS$"},{"key":"D","latex":"$PR = QS$"}],"answerKey":"A","explanationLatex":"We sketch a quadrilateral for ease of reference. P Q R S A The sides P Q and SR are equal in length, which is certainly necessary if P QRS is a parallelogram, and P S being parallel to QR is also necessary. If these are both the case, then on first glance it may seem that P QRS must be a parallelogram. However, if we draw two parallel lines and ask how we can join them with two equal-length sides. We discover there are two ways: P Q RS P Q RS So this condition is necessary but not sufficient. B As before, P Q = SR is necessary, and P Q being parallel to SR is obviously necessary, so the whole condition is necessary. Unlike in the previous part, though, this is also sufficient: if two sides of the quadrilateral are both equal and parallel, the remaining two sides must also be equal and parallel. (This may seem obvious, and it can be proven using geometric or vector methods; see below for details.) C This condition is not necessary, as not all four sides of a parallelogram need to be the same length. It does mean that the parallelogram is a rhombus, so it is a sufficient condition. D The diagonals of a parallelogram need not be equal in length, so this is not a necessary condition. It is also not sufficient. The correct answer is therefore A. We now give two proofs that condition B is sufficient, that is, if P Q = SR and P Q is parallel to SR, then P QRS is a parallelogram. We first give a vector proof. Since P Q = SR and they are parallel, - - →P Q =-→SR; we call this vector a. Let -→P S = b, as shown in this diagram: P Q RS a a b Then - - →QR =- - →QP +-→P S +-→SR = (-a) + b + a = b so- - →QR =-→P S, showing that P S and QR are equal and parallel, and thus P QRS is a parallelogram. And here is a geometric approach. We are given that P Q = RS and that they are parallel. Draw in the diagonal QS, as shown. P Q RS Then ∠P QS = ∠RSQ as they are alternating angles. Therefore triangles P QS and RSQ are congruent by side-angle-side, as P Q = RS (given), QS = SQ (common side) and ∠P QS = ∠RSQ. Thus ∠QSP = ∠SQR, so SP is parallel to QR (by alternating angles) and P S = RQ. Hence P QRS is a parallelogram.","tags":["TMUA","2019","Paper2"],"difficulty":2},{"id":"2019-P2-Q11","paper":2,"index":10,"stemLatex":"An arithmetic series has $n$ terms, all of which are integers. \n\n The sum of the series is 20. \n\n Which of the following statements must be true? \n\n I: The first term of the series is even. \n\n II: $n$ is even. \n\n III: The common difference is even.","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":"Let the first term of the series be a and the common difference be d, as usual. Therefore we have 2 n(2a + (n- 1)d) = 20. Multiplying by 2 gives n(2a + (n- 1)d) = 40, so we need to find integers n, a and d which satisfy this. We list values of n which are factors of 40 and the corresponding value of 2 a + (n- 1)d. n 2a + (n- 1)d 1 40 = 2a 2 20 = 2a + d 4 10 = 2a + 3d 5 8 = 2a + 4d 8 5 = 2a + 7d 10 4 = 2a + 9d 20 2 = 2a + 19d 40 1 = 2a + 39d With this table to hand, we consider the possibilities for each of the statements. I n = 1 gives a = 20, which is even. n = 2 gives 2a + d = 20. If we take d = 2, this gives a = 9, and indeed 9 + 11 = 20, so this statement need not be true. II The table shows that we could have n = 1 and a = 20, or n = 5, a = 2, d = 1, giving 2 + 3 + 4 + 5 + 6 = 20. So this statement need not be true. III The example in the previous part, 2 + 3 + 4 + 5 + 6 = 20 shows that this is not necessarily true. Therefore the correct answer is A.","tags":["TMUA","2019","Paper2"],"difficulty":2},{"id":"2019-P2-Q12","paper":2,"index":11,"stemLatex":"Most students in a large college study Mathematics. A teacher chooses three different students at random, one after the other. \n\n Consider the three probabilities: \n\n $R = P$(At least one of the students chosen studies Mathematics) \n\n $S = P$(The second student chosen studies Mathematics) \n\n $T = P$(All three of the students chosen study Mathematics) \n\n Which of the following is true?","options":[{"key":"A","latex":"R < S < T"},{"key":"B","latex":"R < T < S"},{"key":"C","latex":"S < R < T"},{"key":"D","latex":"S < T < R"},{"key":"E","latex":"T < R < S"},{"key":"F","latex":"T < S < R"}],"answerKey":"F","explanationLatex":"One way to answer this question is to draw a tree diagram and to work out the probabilities of the three events. Another way is to reason as follows. If all three students study Mathematics, then certainly the second student chosen studies Mathematics and at least one of the students chosen studies Mathematics. Therefore T < S and T < R . Likewise, if the second student chosen studies Mathematics, certainly at least one of the students chosen studies Mathematics. Therefore S < R . Putting these together gives T < S < R , and the answer is F.","tags":["TMUA","2019","Paper2"],"difficulty":2},{"id":"2019-P2-Q13","paper":2,"index":12,"stemLatex":"A student approximates the integral $\\int_a^b \\sin^2x$ dx using the trapezium rule with 4 strips. The resulting approximation is an overestimate. \n\n Which of the following is/are necessarily true? \n\n I: If the student approximates $\\int_{-b}^{-a} \\sin^2x$ dx in the same way, the result will be an overestimate. \n\n II: If the student approximates $\\int_{a}^{b} \\cos^2x$ dx in the same way, the result will be an underestimate.","options":[{"key":"A","latex":"neither of them"},{"key":"B","latex":"I only"},{"key":"C","latex":"II only"},{"key":"D","latex":"I and II"}],"answerKey":"D","explanationLatex":"It will help us to draw a sketch of this situation, remembering that we do not know that values of a and b. The exact shape of the graph does not matter; the only thing that is important for this question is that it is symmetrical about the y-axis, since sin^2(-x) = (- sin x)^2 = sin^2 x. x y a b I We can now add to our sketch the trapezium rule calculation for the integral ∫-a -b sin^2 x dx: x y a b-a-b We see that the strips are just a reflection of the original 4 strips. Since the original approximation was an overestimate, this approximation is also an overestimate. II We note that cos^2 x is related to sin^2 x by the identity sin^2 x + cos^2 x = 1, so cos^2 x = 1- sin^2 x. Thus the graph of y = cos^2 x is a reflection of the graph of y = sin^2 x in the line y = 1 2 (why?). Therefore the sketch of the graph and trapezia for ∫ b a cos^2 x dx is: x y a b Since cos^2 x = 1- sin^2 x, we can also draw on the original trapezia, coming down from the line y = 1: x y a b We can see from this sketch that the blue and red trapezia exactly sum to ∫ b a sin^2 x dx +∫ b a cos^2 x dx; since the blue trapezia give an overestimate, the red ones must give an underestimate. Hence this statement is also necessarily true. Therefore the correct answer is D.","tags":["TMUA","2019","Paper2"],"difficulty":2},{"id":"2019-P2-Q14","paper":2,"index":13,"stemLatex":"Consider the following statements about the polynomial $p(x)$, where $a < b$: \n\n I: $p(a) \\leq p(b)$ \n\n II: $p'(a) \\leq p'(b)$ \n\n III: $p''(a) \\leq p''(b)$ \n\n Which of these statements is a necessary condition for $p(x)$ to be increasing for $a \\leq x \\leq b$?","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 necessary condition is one which must follow. To show that a condition is not necessary, we need to exhibit a counterexample, while to show that a condition is necessary, we need to give a justification. I If p( x) is increasing for a≤ x≤ b, the value of the polynomial is not decreasing in this interval, so we must have p(a)≤ p(b), and this is therefore a necessary condition. II A quadratic such as p( x) = 9-x^2 provides a counterexample: it is increasing for-3≤ x≤ 0, but p(-3) > 0 while p(0) = 0. Hence this is not a necessary condition. III It is not clear why this should be the case, and we need to look beyond quadratics (as p ′′(x) is constant for every quadratic). Let us see if we can construct a counterexample. Let us take p′′(x) =-x, so that it is decreasing. Then p′(x) =- 1 2 x^2 + c, giving p(x) =- 1 6 x^3 + cx. (There is a further constant of integration here, but we can choose it to be zero, as it has no impact on where p( x) is increasing.) If we now choose some c > 1, then p′(x) will be positive for 0 ≤ x≤ 1, so p( x) will be increasing there, yet p′′(0) < p′′(1). Therefore this condition is not necessary. Another way of thinking about this is: to say that p( x) is increasing for a≤ x≤ b only tells us about the function values; it does not tell us anything about the derivatives, so neither II nor III could possibly be necessary conditions. Either way, the correct answer is B.","tags":["TMUA","2019","Paper2"],"difficulty":2},{"id":"2019-P2-Q15","paper":2,"index":14,"stemLatex":"The numbers $a$, $b$ and $c$ are each greater than 1. \n\n The following logarithms are all the same base: $$\\log(ab^2c) = 7$$ $$\\log(a^2bc^2) = 11$$ $$\\log(a^2b^2c^3) = 15$$ What is this base?","options":[{"key":"A","latex":"a"},{"key":"B","latex":"b"},{"key":"C","latex":"c"},{"key":"D","latex":"It is possible to determine the base, but the base is not $a$, $b$ or $c$."},{"key":"E","latex":"There is insufficient information given to determine the base."}],"answerKey":"B","explanationLatex":"We expand the expressions using the logarithm rules, namely log(ab) = log a+log b and log an = n log a when a > 1. This gives log a + 2 logb + log c = 7 2 loga + log b + 2 logc = 11 2 loga + 2 logb + 3 logc = 15 This is a set of simultaneous equations! If we write A = log a, B = log b and C = log c, they become A + 2B + C = 7 (1) 2A + B + 2C = 11 (2) 2A + 2B + 3C = 15 (3) Subtracting (3)- (2) and (3)- 2× (1) to eliminate A gives B + C = 4 (4) -2B + C = 1 (5) Subtracting these then gives 3B = 3 so B = 1 (and thus C = 3, A = 2). Therefore log b = 1, so the base of logarithms is b (and also a = b^2, c = b^3). The correct answer is option B.","tags":["TMUA","2019","Paper2"],"difficulty":2},{"id":"2019-P2-Q16","paper":2,"index":15,"stemLatex":"The graph of the quadratic\n\n$y = px^2 + qx + p$\n\nwhere $p > 0$, intersects the x-axis at two distinct points.\n\nIn which one of the following graphs does the shaded region show the complete set of possible values that $p$ and $q$ could take?","stemImage":"/stem-fallback/2019-P2-Q16.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":"F","explanationLatex":"For the quadratic to intersect the x-axis at two distinct points, we require the discriminant to be positive. In this case, this gives q^2- 4p2 > 0. We can either deduce the correct graph by solving this inequality, or decide between the given options by checking various points. We show both approaches here. First, we can write the inequality as ( q- 2p)(q + 2p) > 0. Therefore, this inequality holds when either q- 2p > 0 and q + 2p > 0, or when q- 2p < 0 and q + 2p < 0. The first case gives q > 2p and q >-2p. As p > 0, this simplifies to q > 2p. The second case gives q < 2p and q <-2p, which simplifies to q <-2p. So the solution is q > 2p or q <-2p, and this is shown in graph F. Alternatively, we can check values of p and q. When p = 1, q = 0, q^2- 4p2 =-4 < 0, so this point cannot lie in the region. This eliminates options A, C, E and G. Next, we test a point with q < 0, say p = 1, q =-4. This gives q^2- 4p2 = 12 > 0, so this point is in the region. This eliminates options B and D, so we are left with F or H. Finally, we can check a point which distinguishes the two, say p = 2, q = 3, which gives q^2- 4p2 =-7 < 0, so this point is not in the region. Therefore the correct answer is graph F.","tags":["TMUA","2019","Paper2"],"difficulty":2},{"id":"2019-P2-Q17","paper":2,"index":16,"stemLatex":"A multiple-choice test question offered the following four options relating to a certain statement: \n\n A: The statement is true if and only if $x>1$ \n\n B: The statement is true if $x > 1$ \n\n C: The statement is true if and only if $x>2$ \n\n D: The statement is true if $x>2$ \n\n Given that exactly one of these options was correct, which one was it?","options":[{"key":"A","latex":"A"},{"key":"B","latex":"B"},{"key":"C","latex":"C"},{"key":"D","latex":"D"}],"answerKey":"D","explanationLatex":"As exactly one of the options was correct, if option X is implied by option Y, then option Y cannot be correct (as then option X would also be correct). Option A implies option B: if A is correct, then the statement is true if and only if x > 1, which means the statement is true if x > 1 and the statement is true only if x > 1, so option B is also correct. Therefore option A cannot be the correct option. For the same reason, option C cannot be correct, as it implies option D. We must therefore consider options B and D. Let us assume that option B is correct and ask whether option D follows from it. Option B tells us that whenever x > 1, the statement is true, so certainly whenever x > 2, the statement is true. Therefore option D follows from option B, and option B cannot be the correct option. The opposite direction does not follow, though: the statement could be true for all x > 2, but be false for x = 3 2, for example. Therefore the correct option must have been option D.","tags":["TMUA","2019","Paper2"],"difficulty":2},{"id":"2019-P2-Q18","paper":2,"index":17,"stemLatex":"Consider the following inequality: $$(\\ast): a|x| + 1 \\leq |x-2|$$ where $a$ is a real constant. \n\n Which one of the following describes the complete set of values of $a$ such that $(\\ast)$ is true for all real $x$?","options":[{"key":"A","latex":"a \\leq \\frac{3}{2}"},{"key":"B","latex":"a \\leq 1"},{"key":"C","latex":"a \\leq \\frac{1}{2}"},{"key":"D","latex":"a \\leq 0"},{"key":"E","latex":"a \\leq -\\frac{1}{2}"},{"key":"F","latex":"a \\leq -1"},{"key":"G","latex":"a \\leq -\\frac{3}{2}"},{"key":"H","latex":"There are no such values of $a$."}],"answerKey":"E","explanationLatex":"We would do well to sketch a graph for this question! There are two functions involved, y = a|x| + 1 and y =|x- 2|, so we sketch these for different values of a. The graph of y = a|x|+1 is the graph of y =|x| first stretched by a factor of a in the y-direction, and then translated by 1 in the positive y-direction. The graph of y =|x- 2| is the graph of y =|x| translated by 2 in the positive x-direction. This gives us the following sketch: x y a = 2 a =-2 What we can see from this sketch is that y = a|x| + 1 will be below y =|x- 2| when a is sufficiently negative; when a is positive, it will certainly intersect the graph of y =|x- 2|. We can find the value of a at which the behaviour changes by sketching the graph at this point and then doing some algebra: x y We can work out the value of a by noting that the graph passes through (2 , 0): a|2| + 1 = 0 therefore a =- 1 2, and hence the correct answer is a≤- 1 2, which is option E.","tags":["TMUA","2019","Paper2"],"difficulty":2},{"id":"2019-P2-Q19","paper":2,"index":18,"stemLatex":"Find the value of the expression $$\\sqrt{8 - 4\\sqrt{2} + 1} + \\sqrt{9 - 12\\sqrt{2} + 8}$$","options":[{"key":"A","latex":"\\sqrt{26 - 16\\sqrt{2}}"},{"key":"B","latex":"4\\sqrt{2} - 4"},{"key":"C","latex":"-2"},{"key":"D","latex":"4 - 4\\sqrt{2}"},{"key":"E","latex":"2"},{"key":"F","latex":"\\sqrt{26} - 4\\sqrt{2}"},{"key":"G","latex":"1"}],"answerKey":"E","explanationLatex":"First rewrite the expression as $\\sqrt{9-4\\sqrt{2}}+\\sqrt{17-12\\sqrt{2}}$. Now recognise each term as a perfect square. Since $(2\\sqrt{2}-1)^2=8+1-4\\sqrt{2}=9-4\\sqrt{2}$, we have $\\sqrt{9-4\\sqrt{2}}=2\\sqrt{2}-1$. Also, $(3-2\\sqrt{2})^2=9+8-12\\sqrt{2}=17-12\\sqrt{2}$, so $\\sqrt{17-12\\sqrt{2}}=3-2\\sqrt{2}$. Therefore the total is $(2\\sqrt{2}-1)+(3-2\\sqrt{2})=2$, so the answer is $E$.","tags":["TMUA","2019","Paper2"],"difficulty":2},{"id":"2019-P2-Q20","paper":2,"index":19,"stemLatex":"When the graph of the function $y = f(x)$, defined on the real numbers, is reflected in the y-axis and then translated by 2 units in the negative x-direction, the result is the graph of the function $y = g(x)$. \n\n When the graph of the same function $y = f(x)$ is translated by 2 units in the negative x-direction and then reflected in the y-axis, the result is the graph of the function $y = h(x)$. \n\n Which one of the following conditions on $y = f(x)$ is necessary and sufficient for the functions $g(x)$ and $h(x)$ to be identical?","options":[{"key":"A","latex":"$f(x) = f(x+2)$ for all $x$"},{"key":"B","latex":"$f(x) = f(x+4)$ for all $x$"},{"key":"C","latex":"$f(x) = f(x+8)$ for all $x$"},{"key":"D","latex":"$f(x) = f(-x)$ for all $x$"},{"key":"E","latex":"$f(x) = f(2-x)$ for all $x$"},{"key":"F","latex":"$f(x) = f(4-x)$ for all $x$"},{"key":"G","latex":"$f(x) = f(8-x)$ for all $x$"}],"answerKey":"B","explanationLatex":"Commentary: It could be helpful for this question to take a piece of paper and physically move it around following the given instructions to get a feel for what’s going on. This suggests that we may be looking for a periodic function. We could then try to see what happens with a sine or cosine function, but these turn out to have too much symmetry, and could lead us to the wrong answer. A tan graph is periodic, but does not have line symmetry, so is likely to be better for our needs. With all of this, though, we might still get ourselves in quite a muddle, so beginning with an algebraic approach is likely to be more successful. When we reflect the graph of y = f(x) in the y-axis, we obtain the graph of y = f(-x). When we translate this by 2 units in the negative x-direction, we obtain the graph of y = f(-(x + 2)), so g(x) = f(-(x + 2)) = f(-x- 2). Similarly, when we translate the graph of y = f( x) by 2 units in the negative x-direction, we obtain the graph of y = f( x + 2). When we reflect this in the y-axis, we obtain the graph of y = f(-x + 2), so h(x) = f(-x + 2). If the functions g( x) and h( x) are identical, then we require f(-x- 2) = f(-x + 2) for all x. This is quite close to what we want, so let us let u =-x- 2, so that-x + 2 = u + 4. Then the above becomes f(u) = f(u + 4) for all u, which is option B, just with x replaced by u. This means that the function is periodic with a period of 4 (though it may have a smaller period such as 2 or 1 7, as long as it repeats every 4).","tags":["TMUA","2019","Paper2"],"difficulty":2}]