[{"id":"2021-P1-Q1","paper":1,"index":0,"stemLatex":"Two circles have the same radius. \n\n The centre of one circle is (-2, 1). \n\n The centre of the other circle is (3, -2). \n\n The circles intersect at two distinct points. \n\n What is the equation of the straight line through the two points at which the circles intersect?","options":[{"key":"A","latex":"3x-5y=4"},{"key":"B","latex":"3x+5y=-1"},{"key":"C","latex":"5x-3y=-4"},{"key":"D","latex":"5x-3y=-1"},{"key":"E","latex":"5x-3y=1"},{"key":"F","latex":"5x-3y=4"},{"key":"G","latex":"5x+3y=1"}],"answerKey":"F","explanationLatex":"Approach 1: using the equations of the circles We can find the equation by solving the equations of the circles simultaneously. The equations are (x + 2)^2 + (y - 1)^2 = r^2 (x - 3)^2 + (y + 2)^2 = r^2 (We do not know the radius, so we have called it r; we are given that both circles have the same radius.) We can expand the brackets to give x^2 + 4x + 4 + y^2 - 2y + 1 = r^2 x^2 - 6x + 9 + y^2 + 4y + 4 = r^2 Subtracting these equations then gives an equation which must be satisfied by any points lying on both circles: 10x - 6y - 8 = 0 which simplifies to 5 x - 3y = 4, and so the answer is F. Approach 2: using geometry The circles have the same radius, so by symmetry, the straight line through the two points of intersection must be the perpendicular bisector of the line segment joining the two circle centres, as shown in this diagram: In our case, the two centres are at (-2, 1) and (3, -2), so the midpoint is as (1 2 , - 1 2). The gradient of the line joining the centres is -2-1 3-(-2) = - 3 5. Therefore the gradient of the perpendicular bisector is 5 3, and it has equation y - (- 1 2) = 5 3(x - 1 2) Multiplying by 3 gives 3y + 3 2 = 5(x - 1 2) which rearranges to give 5x - 3y = 4 which is option F.","tags":["TMUA","2021","Paper1"],"difficulty":2},{"id":"2021-P1-Q2","paper":1,"index":1,"stemLatex":"The curve $y=x^3-6x+3$ has turning points at $x = \\alpha$ and $x = \\beta$, where $\\beta > \\alpha$. \n\n Find $$\\int_{\\alpha}^{\\beta} x^3-6x+3dx$$","options":[{"key":"A","latex":"-8\\sqrt{2}"},{"key":"B","latex":"-10"},{"key":"C","latex":"-10+6\\sqrt{2}"},{"key":"D","latex":"0"},{"key":"E","latex":"12-8\\sqrt{2}"},{"key":"F","latex":"6\\sqrt{2}"},{"key":"G","latex":"12"}],"answerKey":"F","explanationLatex":"Differentiate to find the turning points:\n\n$$\\frac{dy}{dx}=3x^2-6=0 \\quad\\Longrightarrow\\quad x^2=2.$$\n\nSo\n\n$$\\alpha=-\\sqrt{2},\\qquad \\beta=\\sqrt{2}.$$\n\nNow evaluate the integral:\n\n$$\\int_{\\alpha}^{\\beta}(x^3-6x+3)\\,dx=\\int_{-\\sqrt{2}}^{\\sqrt{2}}(x^3-6x+3)\\,dx.$$\n\nThe terms $x^3$ and $-6x$ are odd, so their integrals cancel over the symmetric interval $[-\\sqrt{2},\\sqrt{2}]$. Only the constant term remains:\n\n$$\\int_{-\\sqrt{2}}^{\\sqrt{2}}3\\,dx=3\\bigl(2\\sqrt{2}\\bigr)=6\\sqrt{2}.$$\n\nSo the correct answer is $\\mathbf{F}$.","tags":["TMUA","2021","Paper1"],"difficulty":2},{"id":"2021-P1-Q3","paper":1,"index":2,"stemLatex":"An arithmetic progression and a convergent geometric progression each have first term $\\frac{1}{2}$ \n\n The sum of the second terms of the two progressions is $\\frac{1}{2}$ \n\n The sum of the third terms of the two progressions is $\\frac{1}{8}$ \n\n What is the sum to infinity of the geometric progression?","options":[{"key":"A","latex":"-2"},{"key":"B","latex":"-1"},{"key":"C","latex":"-\\frac{1}{2}"},{"key":"D","latex":"-\\frac{1}{3}"},{"key":"E","latex":"\\frac{1}{3}"},{"key":"F","latex":"\\frac{1}{2}"},{"key":"G","latex":"1"},{"key":"H","latex":"2"}],"answerKey":"G","explanationLatex":"Let us write the arithmetic progression as 2 , 1 2 + d, 1 2 + 2d, 1 2 + 3d, ... and the geometric progression as 2 , 1 2 r, 1 2 r^2, 1 2 r^3, ... Then the given information about the second terms and third terms can be written algebraically as 2 + d + 1 2 r = 1 2 + 2d + 1 2 r^2 = 1 The first equation simplifies to give d + 1 2 r = 0, so d = - 1 2 r. We can then substitute this into the second equation to give 2 + 1 2 r^2 - r = 1 (We could equivalently have eliminated d by subtracting twice the first equation from the second.) Multiplying by 8 and rearranging slightly gives 4r2 - 8r + 3 = 0. We can solve this by factorising or using the formula, for example; this gives r = 8 ± p (-8)^2 - 4 × 4 × 3 2 × 4 = 1 2 or 3 Since the geometric progression is convergent, we must have |r| < 1, so r = 1 2 and the sum to infinity is S∞ = a 1 - r = ( 1 2)^1 - 1 = 1 so the answer is option G.","tags":["TMUA","2021","Paper1"],"difficulty":2},{"id":"2021-P1-Q4","paper":1,"index":3,"stemLatex":"Find the minimum value of the function $$2^{2x} - 2^{x+3} + 4$$","options":[{"key":"A","latex":"-16"},{"key":"B","latex":"-12"},{"key":"C","latex":"-8"},{"key":"D","latex":"0"},{"key":"E","latex":"4"},{"key":"F","latex":"20"}],"answerKey":"B","explanationLatex":"This function is a quadratic in u = 2x; we can rewrite it as u^2 - 23u + 4 or u^2 - 8u + 4. We can complete the square to express this as (u - 4)^2 - 42 + 4 = (u - 4)^2 - 12. The minimum therefore occurs when u = 4, at which point the function value is 12. We must also check that this value can actually be attained: u = 4 if and only if 2 x = 4, which happens when x = 2, so this minimum is attained, and the correct answer is B. The minimum can also be found using calculus: differentiating u^2 - 8u + 4 gives 2u - 8; this is zero when u = 4, at which point the function value is 4 2 - 8 × 4 + 4 = -12; this is a minimum as the function is a quadratic with positive u^2 coefficient.","tags":["TMUA","2021","Paper1"],"difficulty":2},{"id":"2021-P1-Q5","paper":1,"index":4,"stemLatex":"The function $f$ is such that\n\n$f(mn)=\\begin{cases}f(m)f(n) & \\text{if } mn \\text{ is a multiple of }3\\\\mn & \\text{if } mn \\text{ is not a multiple of }3\\end{cases}$\n\nfor all positive integers $m$ and $n$.\n\nGiven that $f(9)+f(16)-f(24)=0$, what is the value of $f(3)$?","stemImage":"/stem-fallback/2021-P1-Q5.png","options":[{"key":"A","latex":"\\frac{8}{3}"},{"key":"B","latex":"2\\sqrt{2}"},{"key":"C","latex":"3"},{"key":"D","latex":"\\frac{16}{5}"},{"key":"E","latex":"3\\sqrt{2}"},{"key":"F","latex":"4"}],"answerKey":"F","explanationLatex":"We can rewrite f(9), f(16) and f(24) using the information given about the function: f(9) = f(3 × 3) = f(3)f(3) as 9 is a multiple of 3 f(16) = f(16 × 1) = 16 as 16 is not a multiple of 3 f(24) = f(3 × 8) = f(3)f(8) as 24 is a multiple of 3 = f(3)f(8 × 1) = f(3) × 8 as 8 is not a multiple of 3 = 8f(3) Substituting these into the given equation then gives f(3)^2 + 16 - 8f(3) = 0 which is a quadratic in f(3); we can factorise it to give f(3) - 4 2 = 0 which has one (repeated) solution: f(3) = 4. Hence the answer is option F.","tags":["TMUA","2021","Paper1"],"difficulty":2},{"id":"2021-P1-Q6","paper":1,"index":5,"stemLatex":"The function $f$ is given by $$f(x) = \\frac{\\cos x + 3}{7 + 5\\cos x - \\sin^2x}$$ Find the positive difference between the maximum and the minimum values of $f(x)$.","options":[{"key":"A","latex":"0"},{"key":"B","latex":"\\frac{1}{3}"},{"key":"C","latex":"\\frac{1}{2}"},{"key":"D","latex":"\\frac{2}{3}"},{"key":"E","latex":"1"},{"key":"F","latex":"2"}],"answerKey":"D","explanationLatex":"This is almost a function in cos x except for the sin^2 x term, so we rewrite this in terms of cos x: f(x) = cos x + 3 7 + 5 cosx - (1 - cos^2 x) = cos x + 3 6 + 5 cosx + cos^2 x We can then factorise the denominator to give f(x) = cos x + 3 (cos x + 2)(cos x + 3) and the cos x + 3 factor cancels. (We should also check that this can never be zero; if it were, the function would be undefined at some points.) Therefore we can write f(x) = 1 cos x + 2 The maximum value of this function occurs where cosx = -1, giving f(x) = 1, and the minimum value occurs where cos x = +1, giving f( x) = 1 3. The difference between the minimum and maximum values is therefore 1 - 1 3 = 2 3, which is option D.","tags":["TMUA","2021","Paper1"],"difficulty":2},{"id":"2021-P1-Q7","paper":1,"index":6,"stemLatex":"The function $f$ is such that $f(0)=0$, and $xf(x) > 0$ for all non-zero values of $x$. \n\n It is given that $$\\int_{-2}^2 f(x)dx = 4$$ and $$\\int_{-2}^2 |f(x)|dx = 8$$ Evaluate $$\\int_{-2}^0 f(|x|)dx$$","options":[{"key":"A","latex":"-8"},{"key":"B","latex":"-6"},{"key":"C","latex":"-4"},{"key":"D","latex":"-2"},{"key":"E","latex":"2"},{"key":"F","latex":"4"},{"key":"G","latex":"6"},{"key":"H","latex":"8"}],"answerKey":"G","explanationLatex":"Let us start with the condition xf(x) > 0 for all non-zero values of x. When x > 0, this means that f( x) > 0, as we can divide the inequality by x and the direction of the inequality is unchanged. On the other hand, when x < 0, this gives f(x) < 0, since dividing by x now changes the direction of the inequality. In summary: f(x) > 0 when x > 0 f(x) = 0 when x = 0 f(x) < 0 when x < 0 Now let us turn to the integrals. The final question asks about an integral from -2 to 0, so we would do well to split the two given integrals into the ranges -2 to 0 and 0 to 2: Z 2 -2 f(x) dx = Z 0 -2 f(x) dx + Z 2 f(x) dx = 4 Z 2 -2 |f(x)| dx = Z 0 -2 |f(x)| dx + Z 2 |f(x)| dx = Z 0 -2 -f(x) dx + Z 2 f(x) dx = - Z 0 -2 f(x) dx + Z 2 f(x) dx = 8 Therefore Z 0 -2 f(x) dx + Z 2 f(x) dx = 4 - Z 0 -2 f(x) dx + Z 2 f(x) dx = 8 If we now add these two simultaneous equations together, the R 0 -2 f(x) dx terms cancel and we are left with 2 R 2 0 f(x) dx = 12, so R 2 0 f(x) dx = 6 and R 0 -2 f(x) dx = -2. The question asks us to evaluate R 0 -2 f(|x|) dx. Thinking about graph transformations, the graph of y = f( x) is transformed into the graph of y = f( |x|) by removing the left half of the graph (left of the y-axis) and reflecting the right half of the graph in the y-axis (the line x = 0). The integral of this transformed function from -2 to 0 is therefore equal to the integral of the original function from 0 to 2, thus Z 0 -2 f(|x|) dx = Z 2 f(x) dx = 6 and the correct option is G.","tags":["TMUA","2021","Paper1"],"difficulty":2},{"id":"2021-P1-Q8","paper":1,"index":7,"stemLatex":"The line $y=2x+3$ meets the curve $y=x^2+bx+c$ at exactly one point. \n\n The line $y=4x-2$ also meets the curve $y = x^2 + bx + c$ at exactly one point. \n\n What is the value of $b-c$?","options":[{"key":"A","latex":"-9"},{"key":"B","latex":"-5.5"},{"key":"C","latex":"-1"},{"key":"D","latex":"5"},{"key":"E","latex":"6"},{"key":"F","latex":"14"}],"answerKey":"A","explanationLatex":"A straight line meets a quadratic at exactly one point if and only if there is a repeated root to the simultaneous equations. Substituting for y in the first pair gives 2x + 3 = x^2 + bx + c. This rearranges to x^2 + (b - 2)x + (c - 3) = 0. As this has a repeated root, its discriminant must be zero, so (b - 2)^2 - 4(c - 3) = 0. The second line likewise gives ( b - 4)^2 - 4(c + 2) = 0. Expanding and simplifying these two equations gives the following simultaneous equations for b and c: b^2 - 4b - 4c = -16 b^2 - 8b - 4c = -8 Subtracting these two equations gives 4b = -8 so b = -2 and hence c = 7. Therefore b- c = -9, which is option A.","tags":["TMUA","2021","Paper1"],"difficulty":2},{"id":"2021-P1-Q9","paper":1,"index":8,"stemLatex":"Find the area enclosed by the graph of $$|x| + |y| = 1$$","options":[{"key":"A","latex":"\\frac{1}{2}"},{"key":"B","latex":"1"},{"key":"C","latex":"2"},{"key":"D","latex":"4"},{"key":"E","latex":"\\frac{1}{2}\\sqrt{2}"},{"key":"F","latex":"\\sqrt{2}"},{"key":"G","latex":"2\\sqrt{2}"}],"answerKey":"C","explanationLatex":"We first need to work out what the graph looks like. There are four cases for x and y depending on whether there are positive or negative. We list them in a table: quadrant x y |x| | y| equation rearranged to y = · · · 1st + + x y x + y = 1 y = -x + 1 2nd - + -x y -x + y = 1 y = x + 1 3rd - - - x -y -x - y = 1 y = -x - 1 4th + - x -y x - y = 1 y = x - 1 We therefore have (at most) four lines making up this graph, and we can sketch these. We need to recall that we only draw the part of each line that appears within the relevant quadrant; for example, y = x + 1 only appears in the 2nd quadrant. x y 1-1 -1 This shape is a square. We can think of it as made up of 4 right angled triangles, each with base and height equal to 1, so each has area 1 2 × 1 × 1 = 1 2, and hence the square has area 2. Thus the correct option is C.","tags":["TMUA","2021","Paper1"],"difficulty":2},{"id":"2021-P1-Q10","paper":1,"index":9,"stemLatex":"Use the trapezium rule with 3 strips to estimate $$\\int_{\\frac{1}{2}}^2 2\\log_{10}x \\: dx$$","options":[{"key":"A","latex":"\\log_{10}\\frac{\\sqrt{6}}{2}"},{"key":"B","latex":"\\log_{10}\\frac{3}{2}"},{"key":"C","latex":"\\log_{10}\\frac{9}{4}"},{"key":"D","latex":"\\log_{10}3"},{"key":"E","latex":"\\log_{10}\\frac{81}{16}"},{"key":"F","latex":"\\log_{10}\\frac{\\sqrt{23}}{2}"}],"answerKey":"B","explanationLatex":"The interval has width 2 - 1 2 = 3 2, so each strip will have width 1 2. The boundaries of the three strips will be at 1 2, 1, 3 2 and 2. The trapezium rule then gives the estimate (writing f(x) = 2 log10 x): Z 2 f(x) dx ≈ 1 2 × 1 f( 1 2) + 2f(1) + 2f(3 2) + f(2) We can now work out the individual function values: f( 1 2) = 2 log10 2 = -2 log10 2 f(1) = 2 log10 1 = 0 f( 3 2) = 2 log10 2 = 2 log10 3 - 2 log10 2 f(2) = 2 log10 2 Therefore we get the approximation Z 2 f(x) dx ≈ 1 2 × 1 -2 log10 2 + 2 × 0 + 2(2 log10 3 - 2 log10 2) + 2 log10 2 = 1 4(4 log10 3 - 4 log10 2) = log10 3 - log10 2 = log10 which is option B.","tags":["TMUA","2021","Paper1"],"difficulty":2},{"id":"2021-P1-Q11","paper":1,"index":10,"stemLatex":"The function $f$ is given by $$f(x) = x^{\\frac{1}{7}}(x^2 - x + 1)$$ Find the fraction of the interval $0 < x < 1$ for which $f(x)$ is decreasing.","options":[{"key":"A","latex":"\\frac{2}{15}"},{"key":"B","latex":"\\frac{1}{5}"},{"key":"C","latex":"\\frac{1}{3}"},{"key":"D","latex":"\\frac{1}{2}"},{"key":"E","latex":"\\frac{2}{3}"},{"key":"F","latex":"\\frac{4}{5}"},{"key":"G","latex":"\\frac{13}{15}"}],"answerKey":"A","explanationLatex":"The function is decreasing when f ′(x) < 0 (and possibly also at points where f ′(x) = 0). We have f(x) = x15/7 - x^8/7 + x1/7 so f ′(x) = 15 7 x^8/7 - 8 7 x1/7 + 1 7 x-6/7 = 1 7 x-6/7(15x2 - 8x + 1) = 1 7 x-6/7(5x - 1)(3x - 1) This is negative when (5 x - 1)(3x - 1) < 0 (note that we are only considering the region x > 0 here). We note that (5x - 1)(3x - 1) = 0 exactly when x = 1 5 or x = 1 3 and is therefore negative when 5 < x < 1 3. Therefore the function f( x) is decreasing when 1 5 < x < 1 3 (and possibly also at the endpoints of this interval). Thus the fraction of the interval 0 < x < 1 for which f( x) is decreasing is 1 3 - 1 5 = 2 15, which is option A.","tags":["TMUA","2021","Paper1"],"difficulty":2},{"id":"2021-P1-Q12","paper":1,"index":11,"stemLatex":"The minimum value of the function $x^4 - p^2x^2$ is $-9$ \n\n $p$ is a real number. \n\n Find the minimum value of the function $x^2 - px + 6$","options":[{"key":"A","latex":"-3"},{"key":"B","latex":"6-\\frac{3\\sqrt{2}}{2}"},{"key":"C","latex":"\\frac{3}{2}"},{"key":"D","latex":"3"},{"key":"E","latex":"\\frac{9}{2}"},{"key":"F","latex":"6+\\frac{3\\sqrt{2}}{2}"}],"answerKey":"E","explanationLatex":"Complete the square for the first function: $x^4-p^2x^2=(x^2-\\frac{p^2}{2})^2-\\frac{p^4}{4}$. Its minimum value is therefore $-\\frac{p^4}{4}$. We are told this minimum is $-9$, so $\\frac{p^4}{4}=9$, hence $p^4=36$ and therefore $p^2=6$. Now complete the square for the second function: $x^2-px+6=(x-\\frac{p}{2})^2+6-\\frac{p^2}{4}$. Its minimum value is $6-\\frac{p^2}{4}=6-\\frac{6}{4}=\\frac{9}{2}$. So the answer is $E$.","tags":["TMUA","2021","Paper1"],"difficulty":2},{"id":"2021-P1-Q13","paper":1,"index":12,"stemLatex":"The function $f$ is such that, for every integer $n$ $$\\int_n^{n+1} f(x)dx = n + 1$$ Evaluate $$\\sum_{r=1}^8 (\\int_0^r f(x)dx)$$","options":[{"key":"A","latex":"36"},{"key":"B","latex":"84"},{"key":"C","latex":"120"},{"key":"D","latex":"165"},{"key":"E","latex":"204"},{"key":"F","latex":"288"}],"answerKey":"C","explanationLatex":"Let us first work out the value of the integral in the summation (the summand). We have Z r f(x) dx = Z 1 f(x) dx + Z 2 f(x) dx + · · · + Z r r-1 f(x) dx = (0 + 1) + (1 + 1) +· · · + ((r - 1) + 1) = 1 + 2 +· · · + r = 1 2 r(r + 1). We are thus being asked to evaluate 8X r=1 2 r(r + 1). There are algebraic methods to do this (as mentioned below), but they are not on the specification. The most straightforward approach here is to just perform the calculation manually: 8X r=1 2 r(r + 1) = 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 = 120, which is option C. In general, it turns out that nX r=1 2 r(r + 1) = 1 6 n(n + 1)(n + 2). The expressions r(r + 1) and n(n + 1)(n + 2) are known as rising powers, and can be written as r^2 and n^3 respectively. On multiplying by 2, the formula then becomes nX r=1 r^2 = 1 3 n^3, which looks very similar to the calculus formula R x 0 t^2 dt = 1 3 x^3. There is a much longer discussion of sums such as this in Graham, Knuth and Patashnik, Concrete Mathematics.","tags":["TMUA","2021","Paper1"],"difficulty":2},{"id":"2021-P1-Q14","paper":1,"index":13,"stemLatex":"(This question uses radians.) \n\n Find the number of distinct values of $x$ that satisfy the equation $$(x+1)(3-x) = 2(1-\\cos(\\pi x))$$","options":[{"key":"A","latex":"2"},{"key":"B","latex":"3"},{"key":"C","latex":"4"},{"key":"D","latex":"5"},{"key":"E","latex":"6"},{"key":"F","latex":"7"}],"answerKey":"B","explanationLatex":"We do not know a method of solving equations involving both trigonometric functions (on the right-hand side) and quadratics (on the left-hand side), but the question only asks us to count the number of solutions, not to solve the equation. So we will instead sketch the graphs of the left-hand and right-hand sides and see how many times they intersect. The graph of y = (x+1)(3 - x) is a quadratic with roots at x = -1 and x = 3. When expanded, it becomes y = -x^2 + · · ·, so it has a maximum point midway between the roots, at x = 1, y = 4. -4 -3 -2 -1 1 2 3 4 5 6 -4 -2 x y The graph of y = 2(1 - cos(πx)) is a transformed cosine graph. y = cos(πx) has a period of 2: it is the graph of y = cos x scaled by a factor of 1 /π in the x-direction, and looks like this: -4 -3 -2 -1 1 2 3 4 5 6 -1 x y We can then work out the graph of y = 1 - cos(πx) by calculating the values of y at a few key points, or by observing that the graph of y = - cos(πx) is the reflection of y = cos(πx) in the x-axis, and then y = 1 - cos(πx) is this translated by 1 unit in the y-direction. The result is: -4 -3 -2 -1 1 2 3 4 5 6 x y Finally, the graph of y = 2(1 - cos(πx)) is the previous graph scaled by a factor of 2 in the y-direction: -4 -3 -2 -1 1 2 3 4 5 6 x y We can now superimpose the graph of y = (x + 1)(3 - x) on top of y = 2(1 - cos(πx)). We see that the trigonometric graph is never below the x-axis, so the only possible points of intersection are between x = -1 and x = 3. There is one point of intersection at (1 , 4), one between x = -1 and x = 0 and one between x = 2 and x = 3, so there are three points of intersection in total. -4 -3 -2 -1 1 2 3 4 5 6 x y Thus the correct answer is option B. You might wonder about a subtle point: we see that the two graphs touch each other at (1, 4), but how do we know that the trigonometric graph is ‘sharper’ there, so does not also cross the quadratic graph close to x = 1, something like this: This is a plausible possibility, but it turns out not to be the case for reasons that are beyond the scope of the specification.","tags":["TMUA","2021","Paper1"],"difficulty":2},{"id":"2021-P1-Q15","paper":1,"index":14,"stemLatex":"The diagram shows the graph of $y=f(x)$.\n\nThe graph consists of alternating straight-line segments of gradient $1$ and $-1$ and continues in this way for all values of $x$.\n\nThe function $g$ is defined as\n\n$g(x)=\\sum_{r=1}^{10} f(2^{r-1}x).$\n\nFind the value of\n\n$\\int_0^1 g(x)\\,dx.$","stemImage":"/stem-fallback/2021-P1-Q15.png","options":[{"key":"A","latex":"\\frac{1023}{1024}"},{"key":"B","latex":"\\frac{1023}{512}"},{"key":"C","latex":"5"},{"key":"D","latex":"10"},{"key":"E","latex":"\\frac{55}{2}"},{"key":"F","latex":"55"}],"answerKey":"C","explanationLatex":"We can write the integral out in full by substituting the value of g( x): Z 1 g(x) dx = Z 1 10X r=1 f(2r-1x) dx = Z 1 f(20x) + f(21x) + f(22x) + · · · + f(29x) dx = Z 1 f(20x) dx + Z 1 f(21x) dx + Z 1 f(22x) dx + · · · + Z 1 f(29x) dx So we can calculate the integral of g( x) if we can calculate the integral of each f(2 r-1x). Now R 1 0 f(20x) dx = R 1 0 f(x) dx is just the area under the graph of y = f(x) between x = 0 and x = 1, which is 1 2 (it is a right angled triangle with base and height both 1). To work out R 1 0 f(2x) dx, we consider the graph of y = f(2x). This is the graph of y = f(x) scaled by a factor of 1 /2 in the x-direction, so looks like this: x y The area between x = 0 and x = 1 is again 1 2. The graph of y = f(22x) between x = 0 and x = 1 consists of two triangles: x y Helpfully, the area under this graph is still 1 2: this is perhaps easiest to see by noticing that we can draw a rectangle (actually a square) around the graph: x y In each rectangular slice, half of the area is below the graph and half is above, so the total area below the graph is 1 2. This same reasoning will apply for each of the functions f(2 r-1x), and so the total area is 10 × 1 2 = 5, which is option C.","tags":["TMUA","2021","Paper1"],"difficulty":2},{"id":"2021-P1-Q16","paper":1,"index":15,"stemLatex":"Consider the expansion of $$(a+bx)^n$$ The third term, in ascending powers of $x$, is $105x^2$ \n\n The fourth term, in ascending powers of $x$, is $210x^3$ \n\n The fourth term, in descending powers of $x$, is $210x^3$ \n\n Find the value of $(\\frac{a}{b})^2$","options":[{"key":"A","latex":"\\frac{1}{4}"},{"key":"B","latex":"\\frac{4}{9}"},{"key":"C","latex":"\\frac{25}{36}"},{"key":"D","latex":"\\frac{5}{6}"},{"key":"E","latex":"1"}],"answerKey":"B","explanationLatex":"We can expand the binomial expression in ascending powers of x using the binomial theorem: (a + bx)n = an + n an-1bx + n an-2b^2x2 + n an-3b^3x3 + · · ·+ n n - 1 abn-1xn-1 + bnxn The third term in ascending powers of x is n an-2b^2x2 = 105x2. The fourth term in ascending powers of x is n an-3b^3x3 = 210x3. The fourth term in descending powers of x is n n-3 a^3bn-3xn-3 = 210x3. From the last statement, we see that n = 6. Then the first two statements give: 6 a^4b2 = 105 6 a^3b3 = 210 Now 6 = 15 and 6 = 20, so a^4b2 = 7 a^3b3 = 21 Dividing these two equations gives a b = 2 and so a b 2 = 4 9, which is option B.","tags":["TMUA","2021","Paper1"],"difficulty":2},{"id":"2021-P1-Q17","paper":1,"index":16,"stemLatex":"Which of the following sketches shows the graph of\n\n$\\sin(x^2+y^2)=\\frac{1}{2}$\n\nwhere $x^2+y^2 \\leq 8\\pi$?","stemImage":"/stem-fallback/2021-P1-Q17.png","options":[{"key":"A","latex":"$A$"},{"key":"B","latex":"$B$"},{"key":"C","latex":"$C$"},{"key":"D","latex":"$D$"},{"key":"E","latex":"$E$"}],"answerKey":"A","explanationLatex":"We note that x^2 + y^2 ≥ 0 for all x and y, so we need only consider non-negative values of x^2 + y^2 that satisfy this equation. Those with x^2 + y^2 ≤ 8π are x^2 + y^2 = π 6 , 5π 6 , 13π 6 , 17π 6 , 25π 6 , 29π 6 , 37π 6 , 41π (These were obtained by noting that the solutions with x^2 + y^2 ≤ 2π are π 6 and 5π 6 , and then adding multiples of 2 π = 12π 6 to these.) Consider the first value; x^2 + y^2 = π 6 ; this is a circle with radius p π/6. Likewise, the other circles have radii p 5π/6, ... , p 41π/6. Taking out a common factor of p π/6, we see that the circles have radii: ρ r π 6 , ρ = 1, $\\\\sqrt{5}$, $\\\\sqrt{13}$, $\\\\sqrt{17}$, $\\\\sqrt{25}$, $\\\\sqrt{29}$, $\\\\sqrt{37}$, $\\\\sqrt{We}$ can approximate these square roots to get an idea of how they grow: ρ ≈ 1, 2, 3.5, 4, 5, 5.5, 6, 6.5 So the smallest three circles are quite far apart, but after that, they get closer and closer, which is graph A. (We can be more precise than this, and observe that the circles appear in close pairs with larger gaps between the pairs, but that is not needed to answer this question.)","tags":["TMUA","2021","Paper1"],"difficulty":2},{"id":"2021-P1-Q18","paper":1,"index":17,"stemLatex":"The curve with equation $$x=y^2-6y+11$$ is rotated $90^{\\circ}$ clockwise about the point $P$ to give the curve $C$. \n\n $P$ has x-coordinate -2 and y-coordinate 3. \n\n What is the equation of $C$?","options":[{"key":"A","latex":"y = -x^2 - 4x - 3"},{"key":"B","latex":"y = -x^2 - 4x - 5"},{"key":"C","latex":"y = -x^2 - 6x - 7"},{"key":"D","latex":"y = -x^2 - 6x - 11"},{"key":"E","latex":"y = x^2 - 4x + 5"},{"key":"F","latex":"y = x^2 + 4x + 3"},{"key":"G","latex":"y = x^2 - 6x + 11"},{"key":"H","latex":"y = x^2 + 6x + 7"}],"answerKey":"B","explanationLatex":"We should start by sketching the original curve. This is a quadratic in y. It does not factorise, but we can complete the square: x = (y - 3)^2 + 2 It therefore has its vertex at y = 3, x = 2, and x ≥ 2 for all values of y. We can then sketch it: -2 2 4 6 (2, 3)P x y The vertex of the parabola is 4 units to the right of P . Now rotating this by 90◦ clockwise about P moves the vertex to 4 units belowP , that is (-2, -1), giving the following: -4 -2 2 4 6 -4 -2 (2, 3)P x y This curve then has equation y = -(x + 2)^2 - 1, which we can expand to y = -x^2 - 4x - 5 giving option B.","tags":["TMUA","2021","Paper1"],"difficulty":2},{"id":"2021-P1-Q19","paper":1,"index":18,"stemLatex":"The equation $$\\sin^2(4^{\\cos \\theta} \\times 60^{\\circ}) = \\frac{3}{4}$$ has exactly three solutions in the range $0^{\\circ} \\leq \\theta \\leq x^{\\circ}$ \n\n What is the range of all possible values of $x$?","options":[{"key":"A","latex":"90 \\leq x < 120"},{"key":"B","latex":"90 \\leq x < 270"},{"key":"C","latex":"120 \\leq x < 240"},{"key":"D","latex":"270 \\leq x < 300"},{"key":"E","latex":"300 \\leq x < 360"},{"key":"F","latex":"450 \\leq x < 630"}],"answerKey":"B","explanationLatex":"We can start by solving the equation before considering the possible values of x. We have sin 4cos θ × 60◦ = ± $\\\\sqrt{so}$ that 4cos θ × 60◦ = 60◦, 120◦, 240◦, 300◦, ... (noting that the left hand side is always positive, so we do not need to consider negative values for the right hand side). Dividing by 60 ◦ gives 4cos θ = 1, 2, 4, 5, ... and so cos θ = 0, 1 2 or 1; there are no values of cos θ for which 4 cos θ > 4. We can now find the possible values of θ: • cos θ = 0 gives θ = 90◦, 270◦, 450◦, ... . • cos θ = 1 2 gives θ = 60◦, 300◦, 420◦, ... . • cos θ = 1 gives θ = 0◦, 360◦, ... . Collecting these solutions and arranging them in increasing order gives θ = 0◦, 60 ◦, 90 ◦, 270 ◦, 300◦, ... . We are told that there are exactly three solutions in the range 0 ◦ ≤ θ ≤ x◦. Therefore x ≥ 90 to get the third solution, and x < 270 to avoid getting the fourth solution. This gives the answer 90 ≤ x < 270, which is option B.","tags":["TMUA","2021","Paper1"],"difficulty":2},{"id":"2021-P1-Q20","paper":1,"index":19,"stemLatex":"Find the length of the curve with equation $$2\\log_{10}(x-y) = \\log_{10}(2-2x) + \\log_{10}(y+5)$$","options":[{"key":"A","latex":"5"},{"key":"B","latex":"10"},{"key":"C","latex":"15"},{"key":"D","latex":"3\\pi"},{"key":"E","latex":"9\\pi"},{"key":"F","latex":"12\\pi"}],"answerKey":"D","explanationLatex":"We can combine the logarithms on the right and rearrange the left hand side to give log10(x - y)^2 = log10 (2 - 2x)(y + 5) and so (x - y)^2 = (2 - 2x)(y + 5). Expanding the brackets gives x^2 - 2xy + y^2 = 2y + 10 - 2xy - 10x which rearranges to x^2 + y^2 + 10x - 2y = 10. Completing the squares gives (x + 5)^2 + (y - 1)^2 = 36 which is a circle with centre ( -5, 1) and radius 6. The obvious answer is therefore 12π, but we must take care: does every point on the circle satisfy the original equation? The manipulations we have performed make assumptions; in particular, for the logarithms to be defined, we require x - y > 0, 2 - 2x > 0 and y + 5 > 0, which we can rearrange to x > y, x < 1 and y > -5 Let us sketch the circle and these three inequalities. Here is the circle with two points on the circumference marked: x y (-5, 1) (1, 1) (-5, -5) The restriction x < 1 holds for all of the points on the circle except for the single point (1 , 1), and the restriction y > -5 holds everywhere except for ( -5, -5). The restriction x > y is more significant for us. We can draw the line x = y on our sketch: x y (-5, 1) (1, 1) (-5, -5) The region x > y is below this line, and we see that only part of the circle is included in the region. Since the points of intersection of the line x = y with the circle are at (1, 1) and (-5, -5), it is clear that exactly one quarter of the circle lies in the region x > y . Therefore the length of the curve with the given equation is 1 4 of the circumference of the circle, that is, 1 4 × 12π = 3π, and the correct option is D. This document was initially designed for print and as such does not reach accessibility standard WCAG 2.1 in a number of ways including missing text alternatives and missing document structure. If you need this document in a different format please email admissionstesting@cambridgeassessment.org.uk telling us your name, email address and requirements and we will respond within 15 working days. This document contains maths equations and symbols which at present we are not able to produce adequately for accessibility. We are working to improve this. 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","2021","Paper1"],"difficulty":2},{"id":"2021-P2-Q1","paper":2,"index":0,"stemLatex":"Find the value of $$\\int_1^4 (3\\sqrt{x} + \\frac{4}{x^2})dx$$","options":[{"key":"A","latex":"-0.75"},{"key":"B","latex":"7.125"},{"key":"C","latex":"11"},{"key":"D","latex":"17"},{"key":"E","latex":"18"},{"key":"F","latex":"21.875"},{"key":"G","latex":"34.5"}],"answerKey":"D","explanationLatex":"We write the integral as powers of x and then integrate: Z 4 3√x + 4 x^2 dx = Z 4 3x 2 + 4x-2 dx = \" 3x ( 3 2) + 4x-1 -1 #4 = 2x 2 - 4x-1 4 = (2 × 8 - 4 × 1 4) - (2 × 1 - 4 × 1) = 15 - (-2) = 17","tags":["TMUA","2021","Paper2"],"difficulty":2},{"id":"2021-P2-Q2","paper":2,"index":1,"stemLatex":"$A$(0, 2) and $C$(4, 0) are opposite vertices of the square $ABCD$. \n\n What is the equation of the straight line through $B$ and $D$?","options":[{"key":"A","latex":"y=-2x+5"},{"key":"B","latex":"y= -\\frac{1}{2}x - 3"},{"key":"C","latex":"y = -\\frac{1}{2}x + 2"},{"key":"D","latex":"y = x"},{"key":"E","latex":"y = 2x-3"},{"key":"F","latex":"y = 2x+2"}],"answerKey":"E","explanationLatex":"We can sketch the situation: x y A B C D We know that the two diagonals of the square bisect each other at right angles, so this question requires us to find the perpendicular bisector of AC. The midpoint of AC is 0+4 2 , 2+0 = (2, 1), which is the centre of the square. The gradient of AC is 0-2 4-0 = - 1 2, so the gradient of BD is 2. The equation of BD is therefore y - 1 = 2(x - 2), which rearranges to give y = 2x - 3, which is option E.","tags":["TMUA","2021","Paper2"],"difficulty":2},{"id":"2021-P2-Q3","paper":2,"index":2,"stemLatex":"A student is chosen at random from a class. Each student is equally likely to be chosen. \n\n Which of the following conditions is/are necessary for the probability that the student wears glasses to equal $\\frac{4}{15}$? \n\n I: Exactly 11 students in the class do not wear glasses. \n\n II: The number of students in the class is divisible by 3. \n\n III: The class contains 30 students, and 8 of them wear glasses.","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":"C","explanationLatex":"The probability is given as a fraction in its lowest terms. The probability that a student does not wear glasses is 11 15, so the ratio of glasses to non-glasses is 4 : 11, in its lowest terms. Therefore the number of glasses/non-glasses wearers could be 4 and 11 (total 15), or 8 and 22 (total 30), or ... Necessary means that it follows as a consequence. Therefore I and III are not necessary (there could be 22 non-glasses wearers or the class could contain 15 students. The total must be a multiple of 15, and so the number of students is divisible by 3. The correct answer is therefore C.","tags":["TMUA","2021","Paper2"],"difficulty":2},{"id":"2021-P2-Q4","paper":2,"index":3,"stemLatex":"Consider the following claim about positive integers $a$, $b$ and $c$: \n\n If $a$ is a factor of $bc$, then $a$ is a factor of $b$ or $a$ is a factor of $c$ \n\n Which of the following provide(s) a counterexample to this claim? \n\n I: $a = 5$, $b=10$, $c=20$ \n\n II: $a=8$, $b=4$, $c=4$ \n\n III: $a=6$, $b=7$, $c=12$","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":"C","explanationLatex":"I bc = 200, a is a factor of 200, a is a factor of b, so not a counterexample. II bc = 16, a is a factor of 16, a is not a factor of b and a is not a factor of c, so this is a counterexample. III bc = 84, a is a factor of 84,a is not a factor ofb but a is a factor of c, so not a counterexample. The correct option is therefore C.","tags":["TMUA","2021","Paper2"],"difficulty":2},{"id":"2021-P2-Q5","paper":2,"index":4,"stemLatex":"On which line is the first error in the following argument.","options":[{"key":"A","latex":"$\\sin^2x + \\cos^2x = 1$ for all values of $x$."},{"key":"B","latex":"Therefore $\\cos x = \\sqrt{1-\\sin^2x}$ for all values of $x$."},{"key":"C","latex":"Hence $1 + \\cos x = 1 + \\sqrt{1-\\sin^2x}$ for all values of $x$."},{"key":"D","latex":"Thus $(1+\\cos x)^2 = (1 + \\sqrt{1 - \\sin^2x})^2$ for all values of $x$."},{"key":"E","latex":"Substituting $x=\\pi$ gives $0=4$."}],"answerKey":"B","explanationLatex":"Line A is a known identity, so this is fine. Line B rearranges the identity to get cos^2 x = 1 - sin^2 x and then takes the square root of both sides. However, when we take the square root, we always get the positive (or zero) root, but cos x is negative for some values of x. So this is where the error is introduced. Another way of seeing this is to start with the nonsensical 0 = 4 of line E, and substitute x = π in each line to find the first one which fails; it is, indeed, line B, as this gives -1 = 1.","tags":["TMUA","2021","Paper2"],"difficulty":2},{"id":"2021-P2-Q6","paper":2,"index":5,"stemLatex":"Consider the following two statements about the polynomial $f(x)$: \n\n $P$: $f(x) = 0$ for exactly three real values of $x$ \n\n $Q$: $f(x)=0$ for exactly two real values of $x$ \n\n Which one of the following is correct?","options":[{"key":"A","latex":"$P$ is necessary but not sufficient for $Q$."},{"key":"B","latex":"$P$ is sufficient but not necessary for $Q$."},{"key":"C","latex":"$P$ is necessary and sufficient for $Q$."},{"key":"D","latex":"$P$ is not necessary and not sufficient for $Q$."}],"answerKey":"D","explanationLatex":"Statement $P$ says that $f(x)=0$ has exactly three real roots. This does not force the graph to have exactly two stationary points.\n\nA counterexample is\n\n$$f(x)=x^4-x^2.$$\n\nHere the roots are\n\n$$x=-1,\\ 0,\\ 1,$$\n\nso there are exactly three real roots.\n\nBut\n\n$$f'(x)=4x^3-2x=2x(2x^2-1),$$\n\nso the stationary points are at\n\n$$x=0,\\ \\pm\\frac{1}{\\sqrt{2}}.$$\n\nThat is three stationary points, not two. So $P$ is not sufficient for $Q$.\n\nIt is also not necessary. For example,\n\n$$f(x)=x^3-x+1$$\n\nhas exactly two stationary points, but only one real root.\n\nSo $Q$ is neither necessary nor sufficient for $P$.\n\nHence the correct answer is $\\mathbf{D}$.","tags":["TMUA","2021","Paper2"],"difficulty":2},{"id":"2021-P2-Q7","paper":2,"index":6,"stemLatex":"A circle has equation $(x-9)^2 + (y+2)^2 = 4$ \n\n A square has vertices at (1, 0), (1, 2), (-1, 2) and (-1, 0). \n\n A straight line bisects both the area of the circle and the area of the square. \n\n What is the x-coordinate of the point where this straight line meets the x-axis?","options":[{"key":"A","latex":"2"},{"key":"B","latex":"3"},{"key":"C","latex":"4"},{"key":"D","latex":"4.5"},{"key":"E","latex":"5"},{"key":"F","latex":"6"},{"key":"G","latex":"The straight line is not uniquely determined by the information given, so there is more than one possible point of intersection."},{"key":"H","latex":"There is no straight line that bisects both the area of the circle and the area of the square."}],"answerKey":"B","explanationLatex":"We start by drawing the square and the circle. Since the straight line bisects the area of each shape, it must pass through their centres, as shown. x y This straight line therefore passes through the centre of the square at (0 , 1) and the centre of the circle at (9 , -2). The gradient of this line is then -2-1 9-0 = - 1 3, so the line has equation y = - 1 3 x + 1. The line meets the x-axis where y = 0, so 0 = - 1 3 x + 1, giving x = 3. Hence the correct answer is B.","tags":["TMUA","2021","Paper2"],"difficulty":2},{"id":"2021-P2-Q8","paper":2,"index":7,"stemLatex":"Consider the following statement about the polynomial $p(x)$, where $a$ and $b$ are real numbers with $a<b$: \n\n $(\\ast)$ There exists a number $c$ with $a<c<b$ such that $p(c) = 0$. \n\n Which one of the following is true?","options":[{"key":"A","latex":"The condition $p(a) = p(b)$ is necessary and sufficient for $(\\ast)$"},{"key":"B","latex":"The condition $p(a) = p(b)$ is necessary but not sufficient for $(\\ast)$"},{"key":"C","latex":"The condition $p(a) = p(b)$ is sufficient but not necessary for $(\\ast)$"},{"key":"D","latex":"The condition $p(a) = p(b)$ is not necessary and not sufficient for $(\\ast)$"}],"answerKey":"C","explanationLatex":"If p(a) = p(b) = c, say, then if the graph ofy = p(x) (strictly) increases as it leaves (a, c) towards x = b, but then it must decrease again to get back to ( b, c), so there must be a stationary point between x = a and x = b. Similarly, if the graph (strictly) decreases as it leaves ( a, c), it must then increase again, which also results in a stationary point. The only other possibility is that the graph of y = p(x) is constant, in which case p ′(x) = 0 throughout. In all cases, ( ∗) holds, so p(a) = p(b) is sufficient for ( ∗). However, it is not necessary for ( ∗): if p( x) = x^2 and a = -1, b = 2, then this satisfies ( ∗) (with c = 0) but p( a) ≠ p(b). Thus the correct option is C.","tags":["TMUA","2021","Paper2"],"difficulty":2},{"id":"2021-P2-Q9","paper":2,"index":8,"stemLatex":"Consider the following statements about a polynomial $f(x)$:\n\nI. $f(x)=px^3+qx^2+rx+s$, where $p \\neq 0$.\nII. There is a real number $t$ for which $f'(t)=0$.\nIII. There are real numbers $u$ and $v$ for which $f(u)f(v)<0$.\n\nWhich of these statements is/are sufficient for the equation $f(x)=0$ to have a real solution?\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/2021-P2-Q9.png","options":[{"key":"A","latex":"I yes, II yes, III yes"},{"key":"B","latex":"I yes, II yes, III no"},{"key":"C","latex":"I yes, II no, III yes"},{"key":"D","latex":"I yes, II no, III no"},{"key":"E","latex":"I no, II yes, III yes"},{"key":"F","latex":"I no, II yes, III no"},{"key":"G","latex":"I no, II no, III yes"},{"key":"H","latex":"I no, II no, III no"}],"answerKey":"C","explanationLatex":"I Every cubic crosses the x-axis, so this condition is sufficient for f(x) = 0 to have a real root. This also follows from III below if we take u to be a very large positive number and v to be a very large negative number. II Consider, for example, f( x) = x^2 + 1. This quadratic has f ′(x) = 2x so f ′(0) = 0, but there is no real x for which x^2 + 1 = 0. So this condition is not sufficient. III As f( u) f(v) < 0, we must have f( u) < 0 and f( v) > 0 or the other way round; either way, the graph of y = f(x) has values above and below the x-axis. Since f( x) is a polynomial, it must therefore cross the x-axis, and so the equation f( x) = 0 has a real root. (If we did not know that f( x) was a polynomial, this would not be sufficient; consider f( x) = 1 x for example.) The correct option is therefore C.","tags":["TMUA","2021","Paper2"],"difficulty":2},{"id":"2021-P2-Q10","paper":2,"index":9,"stemLatex":"The first seven terms of a sequence of positive integers are: $$u_1 = 15$$ $$u_2 = 21$$ $$u_3=30$$ $$u_4=37$$ $$u_5=44$$ $$u_6=51$$ $$u_7=59$$ Consider the following statement about this sequence: \n\n $(\\ast)$ If $n$ is a prime number, then $u_n$ is a multiple of 3 or $u_n$ is a multiple of 5. \n\n What is the smallest value of $n$ that provides a counterexample to $(\\ast)$?","options":[{"key":"A","latex":"1"},{"key":"B","latex":"2"},{"key":"C","latex":"3"},{"key":"D","latex":"4"},{"key":"E","latex":"5"},{"key":"F","latex":"6"},{"key":"G","latex":"7"}],"answerKey":"E","explanationLatex":"To be a counterexample, we need the antecedent (the ‘if’ part) to be true but the consequent (the ‘then’ part) to be false. n n is prime an multiple of 3 or 5 1 false - 2 true true 3 true true 4 false - 5 true false 6 false - 7 true false Therefore n = 5 and n = 7 are both counterexamples, and the smallest counterexample is n = 5, which is option E.","tags":["TMUA","2021","Paper2"],"difficulty":2},{"id":"2021-P2-Q11","paper":2,"index":10,"stemLatex":"A student attempts to solve the following problem, where $a$ and $b$ are non-zero real numbers:\n\nShow that if $a^2-4b^3 \\geq 0$ then there exist real numbers $x$ and $y$ such that\n$a=xy(x+y)$ and $b=xy$.\n\nConsider the following attempt:\n\n(I) $(x-y)^2 \\geq 0$\n(II) so $x^2+y^2-2xy \\geq 0$\n(III) so $(x+y)^2-4xy \\geq 0$\n(IV) so $x^2y^2(x+y)^2-4x^3y^3 \\geq 0$\n(V) so $a^2-4b^3 \\geq 0$\n\nWhich of the following best describes this attempt?","stemImage":"/stem-fallback/2021-P2-Q11.png","options":[{"key":"A","latex":"It is completely correct."},{"key":"B","latex":"It is incorrect, but it would be correct if written in the reverse order."},{"key":"C","latex":"It is incorrect, but the student has correctly proved the converse."},{"key":"D","latex":"It is incorrect because there is an error in line (II)."},{"key":"E","latex":"It is incorrect because there is an error in the line (III)."},{"key":"F","latex":"It is incorrect because there is an error in line (IV)."}],"answerKey":"C","explanationLatex":"We start by checking that each step works. Line (II) follows from line (I) by expanding the brackets. The algebraic expression in line (III) expands as x^2 + 2xy + y^2 - 4xy, so it follows from line (II). Line (IV) follows from line (III) by multiplying by x^2y2. Since x^2y2 ≥ 0, the inequality keeps the same direction, so this is a valid deduction. However, how does line (V) follow from line (IV)? It assumes that a = xy(x + y) and b = xy in order to make the deduction. This argument has therefore shown that if a = xy(x + y) and b = xy, then a2 - 4b3 ≥ 0, or even more explicitly: if there exist real numbers x and y such that a = xy(x + y) and b = xy, then a2 - 4b3 ≥ 0. However, the problem posed is the converse: ‘show that if a2 - 4b3 ≥ 0, then there exist real numbers x and y such that a = xy(x + y) and b = xy. So this attempt has proved the converse, though the argument would have been slightly better presented if the student had explicitly stated ‘assume that x and y are real numbers with a = xy(x + y) and b = xy’. The correct option is therefore C.","tags":["TMUA","2021","Paper2"],"difficulty":2},{"id":"2021-P2-Q12","paper":2,"index":11,"stemLatex":"Which of the following statements about polynomials $f$ and $g$ is/are true? \n\n I: If $f(x) \\geq g(x)$ for all $x \\geq 0$, then $\\int_0^x f(t)\\,dt \\geq \\int_0^x g(t)\\,dt$ for all $x \\geq 0$. \n\n II: If $f(x) \\geq g(x)$ for all $x \\geq 0$, then $f'(x) \\geq g'(x)$ for all $x \\geq 0$. \n\n III: If $f'(x) \\geq g'(x)$ for all $x \\geq 0$, then $f(x) \\geq g(x)$ for all $x \\geq 0$.","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":"I We can think of the integral R x 0 f(t) dt as being the (signed) area under the graph of y = f(x) from 0 to x. Since f( x) ≥ g(x) for all x ≥ 0, the area under the graph of y = f(x) will also be greater than that under the graph of y = g(x), so this statement is true. You might be concerned about values of x for which g( x) is negative. We could therefore demonstrate this algebraically as well. We have Z x f(t) dt - Z x g(t) dt = Z x (f(t) - g(t)) dt. Since f( t) - g(t) ≥ 0 for all x ≥ 0, we are integrating a non-negative function, and so the integral must be non-negative, that is to say, we must have R x 0 f(t) dt ≥ R x 0 g(t) dt. II It is not clear why f( x) ≥ g(x) should imply that the gradient (derivative) of f( x) is greater than that of g(x), and it is relatively straightforward to find a counterexample. Here is one: f(x) = (x - 1)^2 g(x) = 0 x y In this example, f( x) ≥ g(x) for all x, but f ′(0) < 0 while g ′(x) = 0 for all x. III This is not necessarily true, as we could add a constant to g( x) without changing g′(x). For example, we could have f ′(x) = g ′(x) = 1 for all x, but f( x) = x while g( x) = x + 1, so f(x) < g(x) for all x. The correct option is therefore B.","tags":["TMUA","2021","Paper2"],"difficulty":2},{"id":"2021-P2-Q13","paper":2,"index":12,"stemLatex":"A region $R$ in the (x, y)-plane is defined by the simultaneous inequalities $$y - x < 3$$ $$y - x^2 < 1$$ Which of the following statements is/are true for every point in $R$? \n\n I: $-1 < x < 2$ \n\n II: $(y-x)(y-x^2) < 3$ \n\n III: $y<5$","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":"Manipulating inequalities algebraically requires care. For example, we cannot (in general) multiply two inequalities together and obtain a valid inequality (which is what statement II has done). Neither can we subtract two inequalities (which is where statement I comes from). In this case, one approach is to sketch the region R described by the pair of inequalities and then determine whether the inequalities in statements I-III are satisfied in the whole of R. Another approach is to search for counterexamples for each statement, that is, values of x and y that do not satisfy the statement but do satisfy the original inequalities. This, however, might be harder to do, and if the inequalities are true, it may take a while to discover this. So let us sketch the given inequalities. The inequality y - x < 3 can be rearranged as y < x + 3, so it is the region below y = x + 3. Similarly, y - x^2 < 1 can be rearranged as y < x 2 + 1, so it is the region below the quadratic y = x^2 + 1. The region R is therefore the shaded area shown here (excluding the boundaries): x y We can now consider the three statements. I It is clear that this is not true in general, as the region R extends in all directions. For example, the point (10 , 0) lies in R (it is easy to check that this satisfies the original inequalities: 0 - 10 < 3 and 0 - 102 < 1) but x = 10 does not satisfy -1 < x < 2. II If we take x = 0, the inequality becomes y^2 < 3, but (0, -2) is in R and (-2)^2 = 4. So this statement is not true for all points in R. III We could take, for example, (6 , 6) as a point in R with y ≥ 5, so this statement is not true in general. The correct option is therefore A.","tags":["TMUA","2021","Paper2"],"difficulty":2},{"id":"2021-P2-Q14","paper":2,"index":13,"stemLatex":"Consider the following simultaneous equations, where $p$ is a real number:\n\n$$p\\cdot 2^x + \\log_2 y = 2$$\n$$2^x + \\log_2 y = 1$$\n\nWhat is the complete range of $p$ for which these simultaneous equations have a real solution $(x, y)$?","options":[{"key":"A","latex":"p<1"},{"key":"B","latex":"p=1"},{"key":"C","latex":"p>1"},{"key":"D","latex":"$p < 1$ or $p>2$"},{"key":"E","latex":"$p = 1$ and $p < 2$"},{"key":"F","latex":"$p > 1$ and $p < 2$"},{"key":"G","latex":"p > 2"},{"key":"H","latex":"All real values of $p$"}],"answerKey":"C","explanationLatex":"Writing X = 2x and Y = log2 y gives the simultaneous equations pX + Y = 2 X + Y = 1 Subtracting gives (p - 1)X = 1, so as long as p ≠ 1, we have X = 1 p-1 and so Y = p-2 p-1. Now the equation log 2 y = p-2 p-1 gives a real value for y for any value of p ≠ 1 (specifically y = 2(p-2)/(p-1)), but the equation 2 x = 1 p-1 has a solution if and only if 1 p-1 > 0, which is if and only if p > 1. Therefore the original simultaneous equations have a real solution ( x, y) if and only if p > 1, which is option C.","tags":["TMUA","2021","Paper2"],"difficulty":2},{"id":"2021-P2-Q15","paper":2,"index":14,"stemLatex":"A circle has equation $$x^2 + ax + y^2 + by + c = 0$$ where $a$, $b$ and $c$ are non-zero real constants. \n\n Which one of the following is a necessary and sufficient condition for the circle to be tangent to the y-axis?","options":[{"key":"A","latex":"a^2 = 4c"},{"key":"B","latex":"b^2 = 4c"},{"key":"C","latex":"\\frac{a}{2} = \\sqrt{\\frac{a^2 + b^2}{4}-c}"},{"key":"D","latex":"\\frac{b}{2} = \\sqrt{\\frac{a^2 + b^2}{4}-c}"},{"key":"E","latex":"-\\frac{a}{2} = \\sqrt{\\frac{a^2 + b^2}{4}-c}"},{"key":"F","latex":"-\\frac{b}{2} = \\sqrt{\\frac{a^2 + b^2}{4}-c}"}],"answerKey":"B","explanationLatex":"Approach 1 We start by completing the square to find the centre and radius of the circle: (x + 1 2 a)^2 - ( 1 2 a)^2 + (y + 1 2 b)^2 - ( 1 2 b)^2 + c = 0 which gives (x + 1 2 a)^2 + (y + 1 2 b)^2 = ( 1 2 a)^2 + (1 2 b)^2 - c so the circle has centre ( - 1 2 a, - 1 2 b) and radius q ( 1 2 a)^2 + (1 2 b)^2 - c. Now let us sketch a circle tangent to the y-axis: x y We can see that the circle is tangent to the y-axis if and only if its radius is equal to the distance of its centre from the y-axis. This is not the same as saying the radius is equal to the x-coordinate of the centre, as the centre may have a negative x-coordinate. As an equation, this condition is then q ( 1 2 a)^2 + (1 2 b)^2 - c = - 1 2 a which, on squaring, is equivalent to ( 1 2 a)^2 + (1 2 b)^2 - c = ( 1 2 a)^2 (This is equivalent to the previous equation because both sides of the first equation are nonnegative.) Subtracting ( 1 2 a)^2 from both sides gives 1 4 b^2 - c = 0 or b^2 = 4c, so this condition is necessary and sufficient for the circle to be tangent to the y-axis, which is option B. Approach 2 The circle is tangent to the y-axis, x = 0, if and only if the pair of simultaneous equations x^2 + ax + y^2 + by + c = 0 and x = 0 has a repeated root. Substituting x = 0 into the circle equation gives y^2 + by + c = 0, which has discriminant b^2 - 4c. So there is a repeated root if and only if b^2 - 4c = 0, which is equivalent to option B.","tags":["TMUA","2021","Paper2"],"difficulty":2},{"id":"2021-P2-Q16","paper":2,"index":15,"stemLatex":"$p$ and $q$ are real numbers, and the equation $$x|x| = px + q$$ has exactly $k$ distinct real solutions for $x$. \n\n Which one of the following is the complete list of possible values for $k$?","options":[{"key":"A","latex":"0, 1, 2"},{"key":"B","latex":"0, 1, 2, 3"},{"key":"C","latex":"0, 1, 2, 3, 4"},{"key":"D","latex":"0, 2, 4"},{"key":"E","latex":"1, 2, 3"},{"key":"F","latex":"1, 2, 3, 4"}],"answerKey":"E","explanationLatex":"Approach 1: graph sketching We are looking for the intersections of the graphs of y = x |x| and y = px + q. The latter is just a straight line graph, while the former is y = x^2 when x ≥ 0 and y = -x^2 when x < 0: x y This graph looks quite similar to the graph of y = x^3: it has a point of inflexion at the origin. Any straight line will meet this graph at least once: if q > 0, then it will meet it for some x > 0, and if q < 0, then it will meet it for some x < 0. A straight line can meet the graph twice if it is tangent to the curve somewhere, and it can meet it three times, for example y = x. But a straight line cannot meet the curve four times. It can meet the curve at most twice for x ≥ 0, and at most twice for x < 0 (as each half of the curve is a quadratic, and a straight line and quadratic can meet at most twice). But if a straight line meets the quadratic twice in the region x < 0, then it must have q > 0, for example: x y It can then meet the curve only once in the region x > 0. The same applies if it meets the curve twice in the region x > 0. And if the line passes through the origin, then it can meet the curve in at most two other locations. So the correct answer is option E. Approach 2: algebraic If x is a solution, then either x ≤ 0 and -x^2 = px + q or x > 0 and x^2 = px + q; we can rewrite these as x^2 + px + q = 0 and x^2 - px - q = 0 respectively. If q < 0, then the quadratic equations both have positive discriminant. The first quadratic has roots of opposite signs (as p p^2 - 4q > p ), so there is exactly one negative solution. The second quadratic (corresponding to x > 0) may have 0, 1 or 2 positive solutions, giving 1, 2 or 3 solutions in total. If q > 0, exactly the opposite is true, and we still end up with 1, 2 or 3 solutions. If q = 0, then x = 0 is a root of both quadratics, and there are 0, 1 or 2 other roots. Thus the possible values are 1, 2 and 3. Examples with these numbers are: p = q = 0; p = -1, q = 0, and p = -2, q = 1 respectively.","tags":["TMUA","2021","Paper2"],"difficulty":2},{"id":"2021-P2-Q17","paper":2,"index":16,"stemLatex":"Consider the following functions defined for $x>1$: $$f(x) = \\log_2(\\log_2\\sqrt{x})$$ $$g(x) = \\log_2{(\\sqrt{\\log_2x})}$$ Which one of the following is true for all values of $x > 1$?","options":[{"key":"A","latex":"$0 \\leq f(x) \\leq g(x)$ or $g(x) \\leq f(x) \\leq 0$"},{"key":"B","latex":"$0 \\leq g(x) \\leq f(x)$ or $f(x) \\leq g(x) \\leq 0$"},{"key":"C","latex":"$\\frac{1}{2} \\leq f(x) \\leq g(x)$ or $g(x) \\leq f(x) \\leq \\frac{1}{2}$"},{"key":"D","latex":"$\\frac{1}{2} \\leq g(x) \\leq f(x)$ or $f(x) \\leq g(x) \\leq \\frac{1}{2}$"},{"key":"E","latex":"$1 \\leq f(x) \\leq g(x)$ or $g(x) \\leq f(x) \\leq 1$"},{"key":"F","latex":"$1 \\leq g(x) \\leq f(x)$ or $f(x) \\leq g(x) \\leq 1$"}],"answerKey":"F","explanationLatex":"The expressions for f( x) and g(x) look quite complicated, so we start by simplifying them using the laws of logarithms: f(x) = log2 log2 $\\\\sqrt{x}$ = log2( 1 2 log2 x) = -1 + log2(log2 x) g(x) = log2 p log2 x = 1 2 log2(log2 x) We now see that both f( x) and g( x) can be written as a function of log 2(log2 x). As x > 1, log2 x > 0 and log 2(log2 x) can take any real value. Let us write z = log 2(log2 x), so that we have f( x) = z - 1 and g( x) = 1 2 z. We can sketch a graph showing these two functions of z: z g(x) = 1 2 z f(x) = z - 1 0 2 We can calculate the intersection of y = z - 1 and y = 1 2 z, obtaining z = 2, y = 1. We can then see from the graph that f(x) ≥ g(x) when both are at least 1, and f( x) ≤ g(x) when both are at most 1, so either f( x) ≥ g(x) ≥ 1 or f( x) ≤ g(x) ≤ 1 (or both), which is option F. (Note that the ‘or’ in the options is inclusive ‘or’, so we are allowed to have both of the inequalities being true.)","tags":["TMUA","2021","Paper2"],"difficulty":2},{"id":"2021-P2-Q18","paper":2,"index":17,"stemLatex":"A student chooses two distinct real numbers $x$ and $y$ with $ 0 < x < y < 1 $. \n\n The student then attempts to draw a triangle $ABC$ with: $$AB = 1$$ $$\\sin A = x$$ $$\\sin B = y$$ Which of the following statements is/are correct? \n\n I: For some choice of $x$ and $y$, there is exactly one triangle the student could draw. \n\n II: For some choice of $x$ and $y$, there are exactly two different triangles the student could draw. \n\n III: For some choice of $x$ and $y$, there are exactly three different triangles the student could draw. \n\n (Note that congruent triangles are considered to be the same.)","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":"C","explanationLatex":"Since we are only given a condition on sin A, the angle A could be acute or obtuse, and similarly for angle B. Let us write θ = sin-1 x and φ = sin-1 y for the acute angles; the respective obtuse angles are then 180◦ - θ and 180◦ - φ. As we are given x < y , we have θ < φ . We can now consider the four possible cases. • Case 1: both A and B are acute. Then we have this situation: 1A B θ φ We therefore know the three angles of the triangle ABC (as given two, we automatically know the third) and one side, and this determines the triangle uniquely (regarding congruent triangles as the same). So we always obtain exactly one triangle from this case. • Case 2: A is obtuse and B is acute. Then A = 180◦ - θ and we have this situation: 1A B θ φ Since θ < φ , the two lines (rays) from A and B shown in this diagram will never meet. Alternatively, we could note that A + B = 180◦ - θ + φ, and since φ - θ > 0, this means that the angles A and B sum to more than 180 ◦. Therefore this case can never produce a triangle. • Case 3: A is acute and B is obtuse. Then B = 180◦ - φ and we have this situation: 1A B θ φ Since θ < φ , the two lines (rays) from A and B shown will eventually meet, and so we get a unique triangle. Alternatively, we have A + B = θ + 180◦ - φ = 180◦ - (φ - θ), and since φ - θ > 0, this means that the angles A and B sum to less than 180 ◦, so they can be two angles of a triangle. So we always obtain exactly one triangle from this case. • Case 4: A and B are both obtuse. This is impossible in a triangle. Therefore this case can never produce a triangle. In summary, cases 1 and 3 always produce a unique triangle, and cases 2 and 4 never produce a triangle, so the student can always draw exactly two different triangles, regardless of the values of x and y. Therefore the correct answer is C.","tags":["TMUA","2021","Paper2"],"difficulty":2},{"id":"2021-P2-Q19","paper":2,"index":18,"stemLatex":"The angle $\\theta$ can take any of the values $1^{\\circ}, 2^{\\circ}, 3^{\\circ}, ...,359^{\\circ}, 360^{\\circ} $. \n\n For how many of these values of $\\theta$ is it true that $$\\sin \\theta \\sqrt{1 + \\sin \\theta}\\sqrt{1 - \\sin \\theta} + \\cos \\theta \\sqrt{1 + \\cos \\theta}\\sqrt{1 - \\cos \\theta} = 0$$","options":[{"key":"A","latex":"0"},{"key":"B","latex":"1"},{"key":"C","latex":"2"},{"key":"D","latex":"4"},{"key":"E","latex":"93"},{"key":"F","latex":"182"},{"key":"G","latex":"271"},{"key":"H","latex":"360"}],"answerKey":"F","explanationLatex":"We can rewrite the given equation as sin θ p (1 + sinθ)(1 - sin θ) + cosθ p (1 + cosθ)(1 - cos θ) = 0 and then multiply out the brackets to give sin θ p 1 - sin^2 θ + cos θ p 1 - cos^2 θ = 0 which simplifies to sin θ √cos^2 θ + cos θ p sin^2 θ = 0. It would be easy at this point to make the mistake of simplifying √cos^2 θ to cos θ, but this would only work in the case that cos θ ≥ 0 as square roots are always positive. We can, though, write |cos θ| for this square root: we always have √x^2 = |x|. Therefore we can rewrite the equation as sin θ |cos θ| + cos θ |sin θ| = 0. If both sin θ and cos θ are positive, then this becomes 2 sin θ cos θ = 0, which cannot be true. Likewise, if both sin θ and cos θ are negative, this becomes -2 sinθ cos θ = 0, which again cannot be true. However, if one is positive and one is negative, say sin θ > 0 and cos θ < 0, this becomes - sin θ cos θ + sin θ cos θ = 0, which is certainly true. The same holds if sin θ < 0 and cos θ > 0. Finally, if either sin θ = 0 or cos θ = 0, the equation is also satisfied. We therefore need to count the number of given angles for which sin θ and cos θ have opposite signs or one of them is zero. A quick sketch helps here: 90 180 270 360 -1 x y We see that the relevant ranges are 90 ◦ ≤ θ ≤ 180◦ and 270 ◦ ≤ θ ≤ 360◦; x = 0 ◦ would also work, but is not one of the angles in the given list. There are 91 angles in each of the two ranges (remembering to include both endpoints), so there are 182 angles in total, which is option F.","tags":["TMUA","2021","Paper2"],"difficulty":2},{"id":"2021-P2-Q20","paper":2,"index":19,"stemLatex":"A sequence of functions $f_1, f_2, f_3, ...$ is defined by $$f_1(x) = |x|$$ $$f_{n+1}(x) = |f_n(x) + x|$$ for $n \\geq 1$ \n\n Find the value of $$\\int_{-1}^1 f_{99}(x)dx$$","options":[{"key":"A","latex":"0"},{"key":"B","latex":"0.5"},{"key":"C","latex":"1"},{"key":"D","latex":"49.5"},{"key":"E","latex":"50"},{"key":"F","latex":"99"},{"key":"G","latex":"99.5"},{"key":"H","latex":"100"}],"answerKey":"E","explanationLatex":"It is helpful to get a feeling for what this sequence of functions looks like, so we will sketch the first few. We are only interested in the range from -1 to 1 (because that is what the integral asks for). The first function, f 1(x) = |x| is standard: -1 1 y = f1(x) x y Note that when x < 0, this is the graph of y = -x, and when x ≥ 0, this is the graph of y = x. We next have f 2(x) = |f1(x) + x|. We can calculate this on the two regions x < 0 and x ≥ 0 separately: when x < 0, we get f^2(x) = |-x+x| = 0, and when x > 0, we get f^2(x) = |x+x| = 2x, giving this graph: -1 1 y = f^2(x) x y We can determine f 3(x) and f 4(x) in a similar way: -1 1 y = f^3(x) x y -1 1 y = f^4(x) x y Here, the graphs for x ≥ 0 are y = 3x and y = 4x respectively. We can see a pattern forming: when n is odd, the graph of y = fn(x) looks like y = f^3(x), with y = -x for x < 0 and y = nx for x ≥ 0, and when n is even, it looks like y = f^4(x), with y = 0 for x < 0 and y = nx for x ≥ 0. We are interested in f 99(x), which looks like y = f^3(x): it is y = -x for x < 0 and y = 99x for x ≥ 0. We can calculate the integral R 1 -1 f99(x) dx by finding the area under the graph, which is just the sum of two triangles. The area between x = -1 and x = 0 is 1 2 × 1 × 1 and the area between x = 0 and x = 1 is 2 ×1 ×99, so the integral equals 1 2 ×1 + 1 2 ×99 = 1 2 ×100 = 50; the correct option is therefore E. This document was initially designed for print and as such does not reach accessibility standard WCAG 2.1 in a number of ways including missing text alternatives and missing document structure. If you need this document in a different format please email admissionstesting@cambridgeassessment.org.uk telling us your name, email address and requirements and we will respond within 15 working days. This document contains maths equations and symbols which at present we are not able to produce adequately for accessibility. We are working to improve this. 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","2021","Paper2"],"difficulty":2}]