[{"id":"2022-P1-Q1","paper":1,"index":0,"stemLatex":"How many real solutions are there to the equation $$2\\cos^4\\theta - 5\\cos^2\\theta + 3 = 0$$ in the interval $0 \\leq \\theta \\leq 2\\pi$?","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"},{"key":"H","latex":"8"}],"answerKey":"C","explanationLatex":"This equation is a quadratic in cos^2θ, and it factorises as (2 cos^2θ- 3)(cos^2θ- 1) = 0. Therefore 2 cos^4θ- 5 cos^2θ + 3 = 0 if and only if cos^2θ = 3 2 or cos^2θ = 1. The first equation has no real solutions, so the original equation holds if and only if cos^2θ = 1, or cosθ =±1. cosθ = 1 has the solutions θ = 0 and θ = 2π in the interval 0≤θ≤ 2π, and cosθ =-1 has the solution θ =π in this interval. Therefore the original equation has three real solutions in the interval, and the correct option is C.","tags":["TMUA","2022","Paper1"],"difficulty":2},{"id":"2022-P1-Q2","paper":1,"index":1,"stemLatex":"Find the complete set of values of $p$ for which the equation $$x^2 -2px + y^2 -6y -p^2 + 8p + 9 = 0$$ describes a circle in the $xy$-plane.","options":[{"key":"A","latex":"p < -\\frac{9}{4}"},{"key":"B","latex":"0 < p < 4"},{"key":"C","latex":"-1 < p < 9"},{"key":"D","latex":"$p > 0$ or $p > 4$"},{"key":"E","latex":"$p < -1$ or $p > 9$"},{"key":"F","latex":"all real values of $p$"}],"answerKey":"D","explanationLatex":"We rewrite the equation in completed square form: starting from x^2- 2px +y^2- 6y-p^2 + 8p + 9 = 0, completing the square for x and y gives (x-p)^2-p^2 + (y- 3)^2- 9-p^2 + 8p + 9 = 0, which simplifies to (x-p)^2 + (y- 3)^2- 2p2 + 8p = 0 or (x-p)^2 + (y- 3)^2 = 2p2- 8p. This is a circle with squared radius 2 p^2- 8p, as long as this is positive. So the complete set of values ofp for which the original equation describes a circle in the xy-plane are those for which 2p2- 8p> 0, or p^2- 4p> 0. This factorises as p(p- 4) > 0, which has roots p = 0, p = 4, so the complete set of values in which the inequality holds is p< 0 or p> 4, which is option D.","tags":["TMUA","2022","Paper1"],"difficulty":2},{"id":"2022-P1-Q3","paper":1,"index":2,"stemLatex":"Given the following statements about a function $f$ \n\n • $f''(x) = a$ for all $x$ \n\n • $f(0) = 1$, $f(1) = 2$ \n\n • $\\int_0^1 f(x) \\: dx = 1$ \n\n find the value of $a$.","options":[{"key":"A","latex":"-6"},{"key":"B","latex":"-3"},{"key":"C","latex":"-2"},{"key":"D","latex":"2"},{"key":"E","latex":"3"},{"key":"F","latex":"6"}],"answerKey":"F","explanationLatex":"Using the first statement, f ′′(x) = a for all x, we can integrate to get f ′(x) = ax +b for some constantb, and a further integration gives f(x) = 1 2ax2 +bx +c for some constant c. The second statement then yields f(0) = c = 1 and f(1) = 1 2a +b + 1 = 2, so a + 2b = 2. Finally, we can now use the third statement: ∫ 1 f(x) dx = [1 6ax3 + 1 2bx2 +x ]1 = 1 6a + 1 2b + 1 = 1 so that a + 3b = 0. Combining the simultaneous equations a + 2b = 2 a + 3b = 0 givesb =-2 and a = 6, so the correct answer is option F.","tags":["TMUA","2022","Paper1"],"difficulty":2},{"id":"2022-P1-Q4","paper":1,"index":3,"stemLatex":"These sectors of circles are similar.\n\nThe arc length of the smaller sector is 6.\n\nThe difference between the areas of the sectors is 21.\n\nFind the positive difference between the perimeters of the sectors.","stemImage":"/stem-fallback/2022-P1-Q4.png","options":[{"key":"A","latex":"4.5"},{"key":"B","latex":"7"},{"key":"C","latex":"8"},{"key":"D","latex":"9"},{"key":"E","latex":"10.5"},{"key":"F","latex":"14"},{"key":"G","latex":"15"}],"answerKey":"C","explanationLatex":"Suppose that the angle of the sector is θ (in radians). Then the arc length of the smaller sector is rθ and the area is 1 2r2θ. For the smaller sector, this gives rθ = 6, so θ = 6/r. Therefore the area is 1 2r2(6/r) = 3r. Likewise, since the larger sector has the same angle as the smaller one, its arc length is (r+3)(6/r) and its area is 1 2(r + 3)^2(6/r) = 3(r + 3)^2/r. We are told the difference between the areas, so we have 3(r + 3)^2 r - 3r = 21. Multiplying this by r and rearranging gives 3(r + 3)^2- 3r2- 21r = 0 which we can expand and simplify to give -3r + 27 = 0, so r = 9. We can now calculate the arc length of the larger sector; it is ( r + 3)(6/r) = 8. Therefore, the perimeter of the first sector is 2r + 6 = 24 and the perimeter of the second sector is 2(r + 3) + 8 = 32. The positive difference is 8, which is option C.","tags":["TMUA","2022","Paper1"],"difficulty":2},{"id":"2022-P1-Q5","paper":1,"index":4,"stemLatex":"The terms $x_n$ of a sequence follow the rule $$x_{n+1} = \\frac{x_n + p}{x_n + q}$$ where $p$ and $q$ are real numbers. \n\n Given that $x_1 = 3, \\: x_2 = 5, \\: x_3 = 7$, find the value of $x_4$","options":[{"key":"A","latex":"-5"},{"key":"B","latex":"5"},{"key":"C","latex":"\\frac{51}{7}"},{"key":"D","latex":"\\frac{15}{2}"},{"key":"E","latex":"\\frac{23}{3}"},{"key":"F","latex":"9"},{"key":"G","latex":"11"},{"key":"H","latex":"13"}],"answerKey":"H","explanationLatex":"We write out the formulae forx2,x^3 andx4 in terms of p andq, using the values x1 = 3,x^2 = 5 and x^3 = 7: n = 1 : 5 = 3 +p 3 +q n = 2 : 7 = 5 +p 5 +q n = 3 : x^4 = 7 +p 7 +q Multiplying the first equation by 3 + q and the second by 5 + q gives 15 + 5q = 3 +p 35 + 7q = 5 +p These are simultaneous equations for p and q; subtracting gives 20 + 2 q = 2 so q =-9 and therefore p =-33. Substituting these into the expression for x^4 gives x^4 = 7- 33 7- 9 = 13 so the answer is option H.","tags":["TMUA","2022","Paper1"],"difficulty":2},{"id":"2022-P1-Q6","paper":1,"index":5,"stemLatex":"Given that $$\\int_{\\log_2 5}^{\\log_2 20} x \\, dx = \\log_2 M$$ what is the value of $M$?","options":[{"key":"A","latex":"4"},{"key":"B","latex":"15"},{"key":"C","latex":"16"},{"key":"D","latex":"20"},{"key":"E","latex":"25"},{"key":"F","latex":"100"},{"key":"G","latex":"10000"}],"answerKey":"F","explanationLatex":"We calculate the integral and use the laws of logarithms; we obtain ∫ log2 20 log2 5 x dx = [1 2x2]log2 20 log2 5 = 1 2(log2 20)^2- 1 2(log2 5)^2 = 1 ( (log2 20)^2- (log2 5)^2) = 1 ( (log2 4 + log2 5)^2- (log2 5)^2) = 1 ( (2 + log2 5)^2- (log2 5)^2) = 1 ( 4 + 4 log2 5 + (log2 5)^2- (log2 5)^2) = 1 2(4 + 4 log2 5) = 2 + 2 log2 5 = log2 4 + log2 25 = log2 100 so the answer is option F.","tags":["TMUA","2022","Paper1"],"difficulty":2},{"id":"2022-P1-Q7","paper":1,"index":6,"stemLatex":"Find the finite area enclosed between the line $y=0$ and the curve $y = x^2 - 4|x|-12$","options":[{"key":"A","latex":"\\frac{128}{3}"},{"key":"B","latex":"\\frac{176}{3}"},{"key":"C","latex":"\\frac{256}{3}"},{"key":"D","latex":"108"},{"key":"E","latex":"144"},{"key":"F","latex":"288"}],"answerKey":"E","explanationLatex":"The line y = 0 is the x-axis; the curve y =x^2- 4|x|- 12 is y =x^2- 4x- 12 when x≥ 0 and y =x^2 + 4x- 12 when x< 0, so we sketch these two graphs. The curve y = x^2- 4x- 12 = (x- 6)(x + 2) intersects the x-axis at x = 6 and x =-2 and has a minimum half way between them, at x = 2. Since we are only considering x≥ 0, we only observe one point of intersection with the x-axis. Similarly, the curvey =x^2 + 4x- 12 = (x + 6)(x- 2) intersects the x-axis at x =-6 and x = 2 and has a minimum half way between them, at x =-2. Since we are only considering x <0, we only observe one point of intersection with the x-axis. We can now sketch the graph, where we have shown the parts of the quadratics we are not using as dashed lines: -6 -2 2 6 The finite area enclosed between thex-axis and the graphy =x^2-4|x|- 12 is therefore between x =-6 and x = 6. We could calculate the integrals of the two quadratics, one between x =-6 and x = 0 and the other between x = 0 and x = 6. But we could also notice that the graph is symmetrical about the y-axis, so we only need to find one of these and double the answer. We have ∫ 6 x^2- 4x- 12 dx = [x^3 3 - 2x2- 12x ]6 = 63 3 - 2× 62- 12× 6- 0 =-72 so ∫ 6 -6x2- 4|x|- 12 dx =-144 and the (unsigned) area required is 144, which is option E.","tags":["TMUA","2022","Paper1"],"difficulty":2},{"id":"2022-P1-Q8","paper":1,"index":7,"stemLatex":"A geometric sequence has first term $a$ and a common ratio $r$, where $a$ and $r$ are positive integers and $r$ is greater than $1$. \n\n The sum of the first $n$ terms of this sequence is denoted by $S_n$ \n\n It is given that the terms of the sequence satisfy $$S_{30} - S_{20} = kS_{10}$$ for some positive integer $k$. \n\n What is the smallest possible value of $k$?","options":[{"key":"A","latex":"2^{10}"},{"key":"B","latex":"2^{20}"},{"key":"C","latex":"2^{30}"},{"key":"D","latex":"\\frac{2^{10}}{2^{10} - 1}"},{"key":"E","latex":"2^{10}(2^{10} - 1)"}],"answerKey":"B","explanationLatex":"We use the formula Sn = a(rn- 1) r- 1 and substitute into the given equation: a(r30- 1) r- 1 - a(r20- 1) r- 1 =ka(r10- 1) r- 1 . We can multiply this by r- 1 and divide by a to get (r30- 1)- (r20- 1) =k(r10- 1), which simplifies to r30-r20 =k(r10- 1). Factorising the left hand side then gives r20(r10- 1) =k(r10- 1); as r> 1, we can divide by r10- 1≠ 0 to obtain r20 =k. Therefore k is a 20th power, and the smallest possible value is 2 20, which is option B.","tags":["TMUA","2022","Paper1"],"difficulty":2},{"id":"2022-P1-Q9","paper":1,"index":8,"stemLatex":"This question is about pairs of functions $f$ and $g$ that satisfy $$f(x) - g(x) =2\\sin x$$ $$f(x)g(x) = \\cos^2 x$$ for all real numbers $x$. \n\n Across all solutions for $f(x)$, what is the minimum value that $f(x)$ attains for any $x$?","options":[{"key":"A","latex":"1 - \\sqrt{2}"},{"key":"B","latex":"- 1 - \\sqrt{2}"},{"key":"C","latex":"0"},{"key":"D","latex":"-1"},{"key":"E","latex":"-2"},{"key":"F","latex":"-3"},{"key":"G","latex":"-4"}],"answerKey":"E","explanationLatex":"It may well require reading this question a couple of times to understand what it is asking! It seems like we need to find all possible functions f(x) and g(x), then for each one, to find the minimum value that f(x) takes, and then to find the minimum of all the minima. There might, of course, be some shortcuts! Let us try to solve the equations to find f( x). The first equation rearranges to give g( x) = f(x)- 2 sinx, so we can substitute this into the second equation to give f(x)(f(x)- 2 sinx) = cos^2x. This is a quadratic in f(x), so we can rearrange it into the standard form and use the quadratic formula: (f(x))^2- (2 sinx)f(x)- cos^2x = 0, hence f(x) = 2 sinx± $\\\\sqrt{4}$ sin^2x + 4 cos^2x = sinx± 1. So there are two possible functions f( x), namely f( x) = sin x + 1 and f(x) = sin x- 1. The minimum value taken by these for any x is 0 for the first function and -2 for the second, so the minimum possible value is-2, which is option E.","tags":["TMUA","2022","Paper1"],"difficulty":2},{"id":"2022-P1-Q10","paper":1,"index":9,"stemLatex":"A sequence of translations is applied to the graph of $y = x^3$ \n\n Which of the following graphs could be the result of this sequence of translations? \n\n I $\\quad y = x^3 - 3x^2 + 9x - 27$ \n\n II $\\quad y = x^3 - 9x^2 + 27x - 3$ \n\n III $\\quad y = 27x^3 - 9x^2 + x - 3$","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":"Any sequence of translations is equivalent to a translation in the x-direction followed by a translation in the y-direction (where the translations could be zero). Translatingy =x^3 bya in the x-direction gives the graph y = (x-a)^3 =x^3- 3ax2 + 3a^2x-a3. Then translating by b in the y-direction gives the graph y =x^3- 3ax2 + 3a^2x-a3 +b. Now considering the three graphs: I The x^2 term is -3x2, which means we require a = 1. Substituting this into the above general equation gives y =x^3- 3x2 + 3x- 1 +b, which does not match the given equation. So this is not a translation of y =x^3. II The x^2 term is-9x2, so we require a = 3. Substituting gives y =x^3- 9x2 + 27x- 27 +b, which matches the equation with b = 24, so this is a translation of y =x^3. III The x^3 term does not match here, so this is not a translation of y =x^3. Therefore the answer is option C, II only.","tags":["TMUA","2022","Paper1"],"difficulty":2},{"id":"2022-P1-Q11","paper":1,"index":10,"stemLatex":"Evaluate $$\\sum_{n = 1}^{100} \\log_{10}(3^{1-n})$$","options":[{"key":"A","latex":"-4950\\log_{10}3"},{"key":"B","latex":"4950\\log_{10}3"},{"key":"C","latex":"-5050\\log_{10}3"},{"key":"D","latex":"5050\\log_{10}3"},{"key":"E","latex":"1 - 4950\\log_{10}3"},{"key":"F","latex":"1 + 4950\\log_{10}3"},{"key":"G","latex":"1 - 5050\\log_{10}3"},{"key":"H","latex":"1 + 5050\\log_{10}3"}],"answerKey":"A","explanationLatex":"We have log10(31-n) = (1-n) log10 3, so the sum becomes 100∑ n=1 log10(31-n) = 100∑ n=1 (1-n) log10 3 = (log10 3) 100∑ n=1 (1-n). Now the sum is actually an arithmetic series: expanding it gives 100∑ n=1 (1-n) = 0 + (-1) + (-2) +··· + (-99) =-(0 + 1 + 2 +··· + 99) =- 1 2× 100(0 + 99× 1) =-4950 using the formula Sn = 1 2n(2a + (n- 1)d). (Alternatively, we could have used the formula for the nth triangular number.) Therefore the original sum evaluates to -4950 log10 3, which is option A.","tags":["TMUA","2022","Paper1"],"difficulty":2},{"id":"2022-P1-Q12","paper":1,"index":11,"stemLatex":"A family of quadratic curves is given by $$y_k = 2(x - \\frac{k}{2})^2 + \\frac{k^2}{2}+4k+3$$ where $k$ is any real number and $y_k$ is a function of $x$. \n\n All these curves are sketched, and the point with the lowest $y$-coordinate among all the curves $y_k$ is $(a, b)$. \n\n Find the value of $a + b$","options":[{"key":"A","latex":"-1"},{"key":"B","latex":"-3"},{"key":"C","latex":"-5"},{"key":"D","latex":"-7"},{"key":"E","latex":"-9"}],"answerKey":"D","explanationLatex":"The formula foryk is given in completed square form. The minimum of the graphy =a(x-b)^2+c, where a > 0, is at x = b, and the minimum value is c. Therefore the minimum of yk is at (k 2, k^2 2 + 4k + 3). The minimum of the minimum y-coordinates is therefore the minimum of k^2 2 + 4k + 3. We can find this minimum either by completing the square or by differentiating this with respect to k. Let us use the latter method here as it seems simpler. We have d dk (k^2 2 + 4k + 3 ) =k + 4 so the minimum occurs at k =-4, giving (k 2,k^2 2 + 4k + 3 ) = (-2,-5) so that a +b =-7, which is option D.","tags":["TMUA","2022","Paper1"],"difficulty":2},{"id":"2022-P1-Q13","paper":1,"index":12,"stemLatex":"Given that $$(a^3 + \\frac{2}{b^3})(\\frac{2}{a^3} - b^3) = \\sqrt{2}$$ where $a$ and $b$ are real numbers, what is the least value of $ab$?","options":[{"key":"A","latex":"-\\sqrt{2}"},{"key":"B","latex":"\\sqrt{2}"},{"key":"C","latex":"-2\\sqrt{2}"},{"key":"D","latex":"2\\sqrt{2}"},{"key":"E","latex":"-\\frac{\\sqrt{2}}{2}"},{"key":"F","latex":"\\frac{\\sqrt{2}}{2}"},{"key":"G","latex":"-2^{\\frac{1}{6}}"},{"key":"H","latex":"2^{\\frac{1}{6}}"}],"answerKey":"A","explanationLatex":"Expanding the brackets gives 2-a^3b3 + 4 a^3b3- 2 = $\\\\sqrt{so}$ 4 a^3b3-a^3b3 = $\\\\sqrt{2}$. We are interested in the least possible value of ab, so we can find the least possible value of a^3b3 and take the cube root. Let us simplify the notation by writing x =a^3b3, so 4 x-x = $\\\\sqrt{2}$. Multiplying byx gives a quadratic: 4-x^2 = $\\\\sqrt{2x}$, which we can rearrange to x^2 + √2x- 4 = 0 and then solve using the quadratic formula: x =- √2±√2 + 16 2 =- √2± 3 $\\\\sqrt{2}$ =-2 $\\\\sqrt{2}$ or $\\\\sqrt{2}$. Since we want the least possible value of x, we take x = a^3b3 =-2 $\\\\sqrt{2}$ = -2 2 , and so the minimum possible value of ab is (-2 2 )^3 =-2 2 =- $\\\\sqrt{2}$, which is option A.","tags":["TMUA","2022","Paper1"],"difficulty":2},{"id":"2022-P1-Q14","paper":1,"index":13,"stemLatex":"A circle has center $O$ and radius $6$. \n\n $P$, $Q$ and $R$ are points on the circumference with angle $POQ \\geq \\frac{\\pi}{2}$ \n\n The area of the triangle $POQ$ is $9\\sqrt{3}$ \n\n What is the greatest possible area of the triangle $PRQ$?","options":[{"key":"A","latex":"18 + 9\\sqrt{3}"},{"key":"B","latex":"18\\sqrt{3}"},{"key":"C","latex":"27 + 9\\sqrt{3}"},{"key":"D","latex":"27\\sqrt{3}"},{"key":"E","latex":"36 + 9\\sqrt{3}"},{"key":"F","latex":"36\\sqrt{3}"}],"answerKey":"D","explanationLatex":"Let us start by drawing the triangle P OQ in the circle: O P Q 6 62θ (We have called the angle 2 θ as we will shortly want to consider half of the angle.) We are told that the area of P OQ is 9 $\\\\sqrt{3}$, so we can use the formula for the area of a triangle given two sides and the included angle to determine 2θ: the area formula is A = 1 2 ab sin C, which in this case gives 2 × 6 × 6 × sin^2θ = 9 $\\\\sqrt{so}$ sin^2θ = $\\\\sqrt{2}$ . Since 2 θ ≥ π 2 , we must have 2 θ = 2 3 π (or 120◦). We can now work out the length P Q by dropping a perpendicular from O to P Q: O P QS 6 6θ θ We have P S = SQ = 6 sinθ = 6 sin 1 3 π = 3 $\\\\sqrt{3}$, so P Q = 6 $\\\\sqrt{3}$, and we also have OS = 6 cosθ = 6 cos 1 3 π = 3. We can now work out the maximum possible area of the triangle P QR. The triangle P QR looks like this: O P Q R $\\\\sqrt{Fixing}$ P Q and allowing R to move, we see that the largest possible area of P QR occurs when R is as far as possible from P Q, as the area of the triangle is 1 2 × base × height. When R is at the top of the circle (in our diagram), we have this configuration: O P Q R S $\\\\sqrt{We}$ have RS = RO + OS. Now RO = 6 (it is a radius of the circle) and OS = 3 as we calculated earlier, so RS = 9. Therefore the maximum possible area of P QR is 2 × P Q × RS = 1 2 × 6 $\\\\sqrt{3}$ × 9 = 27 $\\\\sqrt{3}$, which is option D.","tags":["TMUA","2022","Paper1"],"difficulty":2},{"id":"2022-P1-Q15","paper":1,"index":14,"stemLatex":"A rectangle is drawn in the region enclosed by the curves $p$ and $q$, where $$p(x) = 8 - 2x^2$$ $$q(x) = x^2 - 2$$ such that the sides of the rectangle are parallel to the $x$- and $y$-axes. \n\n What is the maximum possible area of the rectangle?","options":[{"key":"A","latex":"\\frac{26}{9}"},{"key":"B","latex":"\\frac{52}{9}"},{"key":"C","latex":"\\frac{4\\sqrt{6}}{3}"},{"key":"D","latex":"\\frac{8\\sqrt{6}}{3}"},{"key":"E","latex":"4\\sqrt{2}"},{"key":"F","latex":"8\\sqrt{2}"},{"key":"G","latex":"\\frac{20\\sqrt{10}}{9}"},{"key":"H","latex":"\\frac{40\\sqrt{10}}{9}"}],"answerKey":"H","explanationLatex":"We start by sketching the two curves to understand what is happening. They are both quadratics. The graph of p has a maximum at (0 , 8) and the graph of q has a minimum at (0,-2). They are both symmetrical about x = 0 (the y-axis). It might be helpful to work out where they intersect: 8 - 2x2 = x^2- 2 implies 3x2 = 10, so x = √10/3, which is between 1.5 and 2. The graph of these therefore looks like this, where we have also sketched a possible rectangle. For the rectangle to have the maximum possible area, it must meet the graphs at all four of its vertices. Let us suppose that the vertices of the rectangle are at x =±a, with 0 < a < √10/3. Then the width of the rectangle is 2a and the height is (8- 2a2)- (a2- 2) = 10- 3a2. Therefore the area is A = 2a(10- 3a2) = 20a- 6a3. To maximise this area, we differentiate with respect to a to find the stationary points. We have dA da = 20- 18a2 so there is a stationary point when this is zero, at a =± √20/18 =± √10/9 =± 1 $\\\\sqrt{10}$. We are only interested in the positive value, and this is less than √10/3, so it is a valid value for our problem. We therefore obtain the maximum possible area: it is A = 2a(10- 3a2) = 2 √10(10- 1 3× 10) = 2 √10× 2 3× 10 = 40 $\\\\sqrt{which}$ is option H.","tags":["TMUA","2022","Paper1"],"difficulty":2},{"id":"2022-P1-Q16","paper":1,"index":15,"stemLatex":"The solutions to $7x^4 - 6x^2 + 1 = 0$ are $\\pm \\cos \\theta$ and $\\pm \\cos \\beta$. \n\n Which one of the following equations has solutions $\\pm \\sin \\theta$ and $\\pm \\sin \\beta$?","options":[{"key":"A","latex":"7x^4 - 8x^2 - 5 = 0"},{"key":"B","latex":"7x^4 - 8x^2 + 2 = 0"},{"key":"C","latex":"7x^4 - 6x^2 - 2 = 0"},{"key":"D","latex":"7x^4 - 6x^2 + 1 = 0"},{"key":"E","latex":"7x^4 + 6x^2 - 1 = 0"},{"key":"F","latex":"7x^4 + 6x^2 + 5 = 0"}],"answerKey":"B","explanationLatex":"Treat the quartic as a quadratic in $x^2$:\n\n$$7x^4-6x^2+1=0 \\quad \\Longrightarrow \\quad 7u^2-6u+1=0,$$\n\nwhere $u=x^2$.\n\nSolving,\n\n$$u=\\frac{6\\pm\\sqrt{36-28}}{14}=\\frac{6\\pm 2\\sqrt{2}}{14}=\\frac{3\\pm\\sqrt{2}}{7}.$$\n\nSo the two values of $x^2$ are\n\n$$\\cos^2\\theta,\\ \\cos^2\\beta=\\frac{3\\pm\\sqrt{2}}{7}.$$\n\nHence\n\n$$\\sin^2\\theta,\\ \\sin^2\\beta=1-\\frac{3\\pm\\sqrt{2}}{7}=\\frac{4\\mp\\sqrt{2}}{7}.$$\n\nTherefore the equation with roots $\\pm\\sin\\theta$ and $\\pm\\sin\\beta$ is\n\n$$\\left(x^2-\\frac{4+\\sqrt{2}}{7}\\right)\\left(x^2-\\frac{4-\\sqrt{2}}{7}\\right)=0.$$\n\nExpanding,\n\n$$x^4-\\frac{8}{7}x^2+\\frac{(4+\\sqrt{2})(4-\\sqrt{2})}{49}=0,$$\n\nso\n\n$$x^4-\\frac{8}{7}x^2+\\frac{14}{49}=0,$$\n\nthat is,\n\n$$x^4-\\frac{8}{7}x^2+\\frac{2}{7}=0.$$\n\nMultiplying by $7$ gives\n\n$$7x^4-8x^2+2=0.$$\n\nSo the correct answer is $\\mathbf{B}$.","tags":["TMUA","2022","Paper1"],"difficulty":2},{"id":"2022-P1-Q17","paper":1,"index":16,"stemLatex":"Find the complete set of values of $x$ for which there are two non-congruent triangles with the side lengths and angle shown in the diagram.","stemImage":"/stem-fallback/2022-P1-Q17.png","options":[{"key":"A","latex":"1 < x < 3"},{"key":"B","latex":"1 < x < 4"},{"key":"C","latex":"1 < x < 5"},{"key":"D","latex":"3 < x < 4"},{"key":"E","latex":"3 < x < 5"},{"key":"F","latex":"4 < x < 5"}],"answerKey":"D","explanationLatex":"We first require that the lengths shown are both positive; x- 1 > 0 if and only if x >1, and -x^2 + 6x- 5 =-(x- 5)(x- 1)> 0 if and only if 1 <x< 5. Therefore we require 1 <x< 5 to ensure both lengths are positive. There are two non-congruent triangles when the side marked x- 1 can be drawn in two configurations: x- 1 -x^2 + 6x- 5 30◦ This is possible when the following conditions are satisfied: • x- 1<-x^2 + 6x- 5 (otherwise one of the possibilities will lie to the right of the 30◦ angle in the above diagram) • x- 1> (-x^2 + 6x- 5) sin 30◦ (so that the distance from the end of the -x^2 + 6x- 5 side to the base is at least x- 1, making it possible to form a triangle; if there is equality here, then the dashed circle will be tangent to the base of the triangle, and there will only be one possibility) The first inequality rearranges as x^2- 5x + 4< 0, which factorises as (x- 1)(x- 4)< 0, giving 1<x< 4. The second equality, using sin 30◦ = 1 2, becomes 2x- 2>-x^2 + 6x- 5, so x^2- 4x + 3> 0. This factorises as (x- 1)(x- 3)> 0, so x< 1 or x> 3. Requiring both of these inequalities to hold thus gives 3 <x< 4; this also satifies the positivity requirement we started with (1 <x< 5), and so the answer is option D.","tags":["TMUA","2022","Paper1"],"difficulty":2},{"id":"2022-P1-Q18","paper":1,"index":17,"stemLatex":"It is given that $$f(x) = x^2(x - 1)^2(x - 2)$$ $$g(x) = -p(x - q)^2(x - r)^2$$ where $p$, $q$ and $r$ are positive and $q < r$ \n\n Find the set of values of $q$ and $r$ that guarantees the greatest number of distinct real solutions of the equation $f(x) = g(x)$ for all $p$.","options":[{"key":"A","latex":"$q < 1$ and $r < 1$"},{"key":"B","latex":"$q < 1$ and $1 < r < 2$"},{"key":"C","latex":"$q < 1$ and $r > 2$"},{"key":"D","latex":"$1 < q < 2$ and $1 < r < 2$"},{"key":"E","latex":"$1 < q < 2$ and $r > 2$"},{"key":"F","latex":"$q > 2$ and $r > 2$"}],"answerKey":"B","explanationLatex":"The equation f(x) = g(x) is a polynomial equation with degree 5, so has at most 5 distinct real solutions; we wish there to be 5 such solutions for all values of p. We would have fewer than 5 distinct real solutions if we either have a square factor of the form (x-a)^2 for some a, or if the graph of y = f(x)- g(x) has fewer than four turning (stationary) points, or if this graph has a minimum point above the x-axis or a maximum point below it, for then it would not cross the x-axis 5 times. Let us write h(x) = f(x)- g(x) =x^2(x- 1)^2(x- 2) +p(x-q)^2(x-r)^2 We can try expanding and differentiating this, but that looks horrible. Let us think about the above description, though, remembering that there need to be 5 distinct real solutions for all values of p. The graph of y = f(x) =x^2(x- 1)^2(x- 2) intersects the x-axis at x = 0 (where it is tangent to the axis, because it is a double root), x = 1 (likewise) and x = 2, and it is positive when x> 2 and negative or zero when x< 2, so the graph looks like this: 1 2 The graph of y = g(x) =-p(x-q)^2(x-r)^2 has two points of intersection with the x-axis, one at x = q and one at x = r, and at both of these points, the graph is tangent to the axis. (In the case q =r, there is just one point of tangency.) The graph is entirely below or touching the x-axis as we are given that p> 0. It looks like this: q r If r >2, then the graph of y = g(x) will entirely miss one of the “valleys” of y = f(x) if p is large, as it will if r < 1; in this case, there would be at most three points of intersection for large p. So we must have 1 < r <2. If 1 < q < r, then it will miss the left valley for large enough p, so we must have 0 < q <1 and 1 < r <2. In this case, the graph of y = g(x) must intersect both valleys twice as g(1) < 0: This sketch shows four points of intersection, and we are looking for five. But the graph of y = f(x) tends to -∞ as x tends to-∞ far faster that y = g(x), regardless of the value of p, as y = f(x) is a quintic but y = g(x) is only a quartic. So there will always be another point of intersection with x< 0. Hence the correct answer is option B. Note that there is actually a small simplification in the above argument. We showed that we cannot have 1<q <r and deduced that we must have 0<q < 1. But what about the possibility q = 1? In that case, there would be a double root at x = 1, and so there would not be five distinct roots, hence we cannot have q = 1. But what about r? We showed that we cannot have r> 2 or r <1. If r = 1, we have the same problem, but what about r = 2 and 0<q < 1? In this case, it is still true that the equation has five distinct real solutions for all positive p, as x = 2 is only a single root of f(x) = 0. So the full set is actually q <1 and 1<r ≤ 2.","tags":["TMUA","2022","Paper1"],"difficulty":2},{"id":"2022-P1-Q19","paper":1,"index":18,"stemLatex":"Circle $C_1$ is defined as $x^2 + y^2 = 25$ \n\n A second circle $C_2$ has radius $4$ and centre $(a, b)$ where $$-2 \\leq a \\leq 2 \\quad and \\quad -3 \\leq b \\leq 3$$ If the centre of $C_2$ is equally likely to be located anywhere within the given range, what is the probability that $C_2$ intersects $C_1$?","options":[{"key":"A","latex":"\\frac{1}{25}"},{"key":"B","latex":"\\frac{9}{25}"},{"key":"C","latex":"\\frac{16}{25}"},{"key":"D","latex":"\\frac{6 - \\pi}{6}"},{"key":"E","latex":"\\frac{16 - \\pi}{24}"},{"key":"F","latex":"\\frac{24 - \\pi}{24}"}],"answerKey":"F","explanationLatex":"Note first that the region where the centre of C^2 may lie is entirely within the circle C1, as 22 + 32 = 13< 25. As the radius of C1 is 5, the circle C^2 will intersect C1 if and only if the centre of C^2 is at least a distance of 1 from the origin (the centre of C1). This is because in this case, a line from the centre of C1 passing through the centre of C^2 will reach the circle C1 before (or as) it reaches the circle C^2, but a line from the centre of C1 in the opposite direction will reach C^2 first. If, though, the centre of C^2 is at a distance of less than 1 from the origin, the furthest point on C^2 from the origin will be at a distance of less than 1 + 4 = 5 from the origin, so C^2 lies entirely inside C1. The area of the allowed ( a,b ) region is 4 × 6 = 24. The ( a,b ) region where the circles do not intersect is a circle of radius 1, with area π, lying entirely within the allowed region, so the region where the circles do intersect has area 24 -π. Thus the probability of intersection is 24-π 24 , which is option F.","tags":["TMUA","2022","Paper1"],"difficulty":2},{"id":"2022-P1-Q20","paper":1,"index":19,"stemLatex":"$n$ is the number of points of intersection of the graphs $$y = |x^2 - a^2| \\quad y = a^2|x - 1|$$ where $a$ is a real number. \n\n What is the smallest value of $n$ that is not possible?","options":[{"key":"A","latex":"n = 1"},{"key":"B","latex":"n = 2"},{"key":"C","latex":"n = 3"},{"key":"D","latex":"n = 4"},{"key":"E","latex":"n = 5"}],"answerKey":"B","explanationLatex":"The graph y =|x^2-a2| intersects the x-axis at ±a. The graph y = a2|x- 1| intersects the x-axis at x = 1. Replacing a with-a does not change the equations, so we may as well assume that a≥ 0. Here is a sketch of the graphs with a = 2; we note that the graphs always intersect at x = 0 (and y =a2 there): -2 1 2 We immediately see that whena> 1, there will be 4 points of intersection: the graphy =a2|x-1| is below the “hill” at (1, 0), so intersects it twice asx moves away fromx = 1, and then intersects the quadratic y =x^2-a2 for some x< -a and some x>a as the quadratic grows faster than the linear y =-a2(x- 1) or y =a2(x- 1) in these regions. What happens when a = 1? In this case, the equations become y =|x^2- 1| andy =x- 1. The derivative of the quadratic y =x^2- 1 is 2x, so the gradient is just over 2 when x is just over 1. (There is no gradient at the point x = 1.) Therefore the graph looks like this: -1 1 We see that we have lost one point of intersection: there are now only three points of intersection. What happens if 0 <a< 1? Now the vertex of y =a2|x- 1| will be to the right of x =a, where the quadratic meets the axis, and so there will be no points of intersection there - the quadratic grows too quickly to meet the linear function for x >1. (To be more precise: the gradient of the quadratic to the right of x =a is 2x> 2a>a 2, so the quadratic will be growing faster than the linear function.) On the left of x = a, we have one point of intersection to the left of x =-a as before. There is also a point of intersection at (0 ,a 2). Since this is the vertex of the quadratic, the linear function must go below the “hill”, giving another two points of intersection, yielding four points of intersection in total: -1 1 There is one remaining case to consider: a = 0. In this case, the graphs become y =|x^2| =x^2 and y = 0, and these graphs have a single point of intersection at the origin. Therefore the possible numbers of points of intersection are 1, 3 and 4, and the smallest number not possible is n = 2, option B.","tags":["TMUA","2022","Paper1"],"difficulty":2},{"id":"2022-P2-Q1","paper":2,"index":0,"stemLatex":"Determine the number of stationary points on the curve with equation $$y = 3x^4 + 4x^3 + 6x^2 - 5$$","options":[{"key":"A","latex":"0"},{"key":"B","latex":"1"},{"key":"C","latex":"2"},{"key":"D","latex":"3"},{"key":"E","latex":"4"}],"answerKey":"B","explanationLatex":"We differentiate to obtain dy dx = 12x3 + 12x2 + 12x. We solve the equation dy dx = 0 to find the number of stationary points: 12x3 + 12x2 + 12x = 0 if and only if 12 x(x^2 +x + 1) = 0. So either x = 0 or x^2 +x + 1 = 0. But this quadratic has discriminant 1 2- 4× 1 =-3< 0, so it has no real solutions. Thus there is only one stationary point, at x = 0, and the answer is option B.","tags":["TMUA","2022","Paper2"],"difficulty":2},{"id":"2022-P2-Q2","paper":2,"index":1,"stemLatex":"Find the coefficient of the $x^5$ term in the expansion of $$(1 + x)^5 \\times \\sum_{i=0}^5x^i$$","options":[{"key":"A","latex":"1"},{"key":"B","latex":"5"},{"key":"C","latex":"16"},{"key":"D","latex":"25"},{"key":"E","latex":"32"}],"answerKey":"E","explanationLatex":"We expand the brackets in the first term using the binomial theorem and write out the sum in full to get (1 + 5x + 10x2 + 10x3 + 5x4 +x^5)× (1 +x +x^2 +x^3 +x^4 +x^5). Thex5 term in the product is obtained by taking all possible pairs of a term in the first bracket and a term in the second whose product is cx5 for some number c, and then adding these together. Therefore the x^5 term in the product is 1·x^5 + 5x·x^4 + 10x2·x^3 + 10x3·x^2 + 5x4·x +x5· 1 = (1 + 5 + 10 + 10 + 5 + 1)x^5 = 32x5. Therefore the correct answer is option E.","tags":["TMUA","2022","Paper2"],"difficulty":2},{"id":"2022-P2-Q3","paper":2,"index":2,"stemLatex":"Consider the following statement about the positive integer $n$ \n\n 'If $n$ is prime, then $n^2 + 2$ is not prime' \n\n Which of the following is a counterexample to this statement \n\n I $\\quad n = 2$ \n\n II $\\quad n = 3$ \n\n III $\\quad n = 4$","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 n = 2 is prime and n^2 + 2 = 6 is not prime, so this case satisfies the statement and is not a counterexample. II n = 3 is prime but n^2 + 2 = 11 is prime, so this is a counterexample as it does not satisfy the statement. III n = 4 is not prime, so it satisfies the statement and is not a counterexample. Therefore only II provides a counterexample, which is option C.","tags":["TMUA","2022","Paper2"],"difficulty":2},{"id":"2022-P2-Q4","paper":2,"index":3,"stemLatex":"The point $P$ has coordinates $(p, q)$, and the equation of a circle is $$x^2 + 2fx + y^2 + 2gy + h = 0$$ where $f$, $g$, $h$, $p$ and $q$ are all real constants. \n\n Let $L$ be the distance between the centre of the circle and the point $P$. \n\n Which one of the following is sufficient on its own to be able to calculate $L$?","options":[{"key":"A","latex":"the values of $f$, $g$ and $h$"},{"key":"B","latex":"the values of $f$, $g$, $p$ and $q$"},{"key":"C","latex":"the values of $f$, $h$, $p$ and $q$"},{"key":"D","latex":"the values of $g$, $h$, $p$ and $q$"},{"key":"E","latex":"none of the options A-D is sufficient on its own"}],"answerKey":"B","explanationLatex":"We can rewrite the equation of circle by completing the square to find its centre. We get: (x +f)^2-f 2 + (y +g)^2-g^2 +h = 0 which we can rearrange as (x +f)^2 + (y +g)^2 =f 2 +g^2-h. Therefore the centre is at (-f,-g) (and the radius is $\\\\sqrt{f}$ 2 +g^2-h, but that is not relevant to us here). Then the distance between the centre of the circle and the point P is given by L^2 = (p +f)^2 + (q +g)^2 so we need the values of f, g, p and q. Hence the correct answer is option B.","tags":["TMUA","2022","Paper2"],"difficulty":2},{"id":"2022-P2-Q5","paper":2,"index":4,"stemLatex":"A straight line $L$ passes through $(1, 2)$. \n\n Let $P$ be the statement \n\n 'if the $y$-intercept of $L$ is negative, then the $x$-intercept of $L$ is positive.' \n\n Which one of the following statements must be true? \n\n I $\\quad P$ \n\n II $\\quad$ the converse of $P$ \n\n III $\\quad$ the contrapositive of $P$","options":[{"key":"A","latex":"none of them"},{"key":"B","latex":"I only"},{"key":"C","latex":"II only"},{"key":"D","latex":"III only"},{"key":"E","latex":"I and II only"},{"key":"F","latex":"I and III only"},{"key":"G","latex":"II and III only"},{"key":"H","latex":"I, II and III"}],"answerKey":"F","explanationLatex":"I Let us sketch a line through (1 , 2) with a negative y-intercept: x y It is clear from this sketch that if the line has a negative y-intercept, it must have a positive x-intercept. (A more formal proof goes as follows: since y <0 when x = 0 and y >0 when x = 1, there must be some value ofx with 0<x< 1 wherey = 0. Therefore the x-intercept must lie between 0 and 1.) II The converse of P reads: If the x-intercept of L is positive, then the y-intercept of L is negative. This is the case if the x-intercept is between 0 and 1, as in the previous sketch, but if the x-intercept is greater than 1, this is no longer true: x y So the converse of P is false. III The contrapositive of P has the same truth value as P itself, so the contrapositive of P is true. Therefore only I and III are true, and the correct answer is option F.","tags":["TMUA","2022","Paper2"],"difficulty":2},{"id":"2022-P2-Q6","paper":2,"index":5,"stemLatex":"A list consists of $n$ integers. \n\n Consider the following statements: \n\n P: $\\quad$ $n$ is odd. \n\n Q: $\\quad$ The median of the list is one of the numbers in the list. \n\n Which one of the following is true?","options":[{"key":"A","latex":"P is necessary and sufficient for Q"},{"key":"B","latex":"P is necessary but not sufficient for Q"},{"key":"C","latex":"P is sufficient but not necessary for Q"},{"key":"D","latex":"P is not necessary and not sufficient for Q"}],"answerKey":"C","explanationLatex":"Suppose P is true. Since there are an odd number of integers in the list, there is a unique middle one (when they are written in increasing order), so this is the median. Therefore Q is true, and P is sufficient for Q. Now suppose that Q is true. It does not follow that n is odd; here is a counterexample: if the list is 2, 2, then n = 2 and the median is 2, so Q is true, but P is false. (In general, any list with n even and the middle pair of numbers equal is a counterexample.) Therefore P is not necessary for Q. The correct answer is option C.","tags":["TMUA","2022","Paper2"],"difficulty":2},{"id":"2022-P2-Q7","paper":2,"index":6,"stemLatex":"Consider the following claim: \n\n 'The difference between two consecutive positive cube numbers is always prime.' \n\n Here is an attempted proof of this claim: \n\n I $\\quad (x + 1)^3 = x^3 + 3x^2 + 3x + 1$ \n\n II $\\quad$ Taking $x$ to be a positive integer, the difference between two consecutive cube numbers can be expressed as $(x+1)^3 - x^3 = 3x^2 + 3x + 1$ \n\n III $\\quad$ It is impossible to factorise $3x^2 + 3x + 1$ into two linear factors with integer coefficients because its discriminant is negative. \n\n IV $\\quad$ Therefore for every positive integer value of $x$ the integer $3x^2 + 3x + 1$ cannot be factorised. \n\n V $\\quad$ Hence, the difference between two consecutive cube numbers will always be prime. \n\n Which of the following best describes this proof?","options":[{"key":"A","latex":"The proof is completely correct, and the claim is true."},{"key":"B","latex":"The proof is completely correct, but there are counterexamples to the claim."},{"key":"C","latex":"The proof is wrong, and the first error occurs on line I."},{"key":"D","latex":"The proof is wrong, and the first error occurs on line II."},{"key":"E","latex":"The proof is wrong, and the first error occurs on line III."},{"key":"F","latex":"The proof is wrong, and the first error occurs on line IV."},{"key":"G","latex":"The proof is wrong, and the first error occurs on line V."}],"answerKey":"F","explanationLatex":"Searching for small counterexamples to the claim does not find any; the first few differences are all prime. So we do have to check the proof in detail, as it might be correct. The algebraic expansion on line I is correct. The expression in line II is also correct, using the expansion from line I. For line III, if we could write 3x2 + 3x + 1 = (ax +b)(cx +d) with real coefficients a,b,c andd, then we would have real roots of the equation 3 x^3 + 3x + 1 = 0, namely x =-b a and x =-d c . But the discriminant is 3 2- 4× 3× 1 =-3< 0, so this is impossible. So this statement is true. Line IV is more tricky: just because the algebraic expression is not factorisable into algebraic factors does not mean that the integer it represents for any particular value ofx is not factorisable in integers. As a simpler example, the algebraic expression x^2 + 1 is not factorisable into real algebraic factors, yet substituting integers for x shows that the integer value can sometimes be factorised: taking x = 3 gives 3 2 + 1 = 10 = 2× 5, for example. So this line of the argument is not valid. Line V does follow from lines II and IV. Therefore the proof is wrong, and the first (and only) error occurs on line IV (option F). It is worth noting that the ‘opposite’ argument does work: if an algebraic expression does factorise as (ax +b)(cx +d) with a, b, c and d integers, then the value of original expression always factorises (as an integer) when x is an integer, just by substituting the value of x into the factorised form. (It is possible, though, that one of the factors is 1, so the integer may be prime for some value(s) of x.) It also turns out that in this case, if we search far enough, we do find a counterexample to the claim: 6 3- 53 = 216- 125 = 91 = 7× 13. (It might have been the case that the claim was true even though the attempted proof had an error.)","tags":["TMUA","2022","Paper2"],"difficulty":2},{"id":"2022-P2-Q8","paper":2,"index":7,"stemLatex":"A selection, $S$, of $n$ terms is taken from the arithmetic sequence $1, \\: 4, \\: 7, \\: 10, \\: ..., \\: 70$. \n\n Consider the following statement: \n\n $(*)$ There are two distinct terms in $S$ whose sum is $74$. \n\n What is the smallest value of $n$ for which $(*)$ is necessarily true?","options":[{"key":"A","latex":"12"},{"key":"B","latex":"13"},{"key":"C","latex":"14"},{"key":"D","latex":"21"},{"key":"E","latex":"22"},{"key":"F","latex":"23"}],"answerKey":"C","explanationLatex":"We first determine the number of terms in the whole sequence: there are 70-1 3 + 1 = 23 + 1 = 24 terms. We next determine the number of pairs that sum to 74. The pairs are: 4 + 70 7 + 67 ... 34 + 40 [37 + 37] The final pair is excluded because the two terms have to be distinct to fit the rule in ( ∗). The number of valid pairs is therefore 34-4 3 + 1 = 10 + 1 = 11. If a selection S does not satisfy the rule in (∗), then it can have at most one term from each of these pairs, so it has to leave out at least 11 terms. So it can have at most 24 - 11 = 13 terms. Therefore if S has at least 14 terms, it must satisfy (∗). We also need to show that if S only has 13 terms, it might not satisfy (∗). We can include 1, 37 and one term from each listed pair, giving 2 + 11 = 13 terms. These 13 terms do not contain a distinct pair summing to 37. Hence the smallest value ofn which forces (∗) to be true is 14, and the correct answer is option C..","tags":["TMUA","2022","Paper2"],"difficulty":2},{"id":"2022-P2-Q9","paper":2,"index":8,"stemLatex":"Consider the following statement: \n\n $(*)$ For all real numbers $x$, if $x < k$ then $x^2 < k$ \n\n What is the complete set of values of $k$ for which $(*)$ is true?","options":[{"key":"A","latex":"no real numbers"},{"key":"B","latex":"k > 0"},{"key":"C","latex":"k < 1"},{"key":"D","latex":"k \\leq 1"},{"key":"E","latex":"0 < k < 1"},{"key":"F","latex":"0 < k \\leq 1"},{"key":"G","latex":"all real numbers"}],"answerKey":"A","explanationLatex":"The statement (∗) does not restrict us to positive real numbers x, so whatever the value of k, we can always find a large negative number x with x<k and x^2≥k. We can even be explicit about this (though that is not required to answer this question): if k≤ 0, we can take x =k- 1 in which case x^2 > 0≥k, and if k >0, we can take x =- $\\\\sqrt{k}$ so x< 0<k and x^2 =k≥k. Therefore (∗) is true for no real values of k, and the correct option is A.","tags":["TMUA","2022","Paper2"],"difficulty":2},{"id":"2022-P2-Q10","paper":2,"index":9,"stemLatex":"Which of the following statements is/are true? \n\n I $\\quad$ For all real numbers $x$ and for all positive integers $n$, $x < n$ \n\n II $\\quad$ For all real numbers $x$, there exists a positive integer $n$ such that $x < n$ \n\n III $\\quad$ There exists a real number $x$ such that for all positive integers $n$, $x < n$","options":[{"key":"A","latex":"none of them"},{"key":"B","latex":"I only"},{"key":"C","latex":"II only"},{"key":"D","latex":"III only"},{"key":"E","latex":"I and II only"},{"key":"F","latex":"I and III only"},{"key":"G","latex":"II and III only"},{"key":"H","latex":"I, II and III"}],"answerKey":"G","explanationLatex":"I We can find a counterexample to this statement: taking x = 2 and n = 1 shows that this statement is false. II This is true; if x≤ 0, we can take n = 1, and if x >0, we can take n to be the smallest integer greater than x. III This is true; we can take x = 0. (If, though, we replaced ‘positive integers’ with ‘integers’, the resulting statement would be false.) Therefore II and III are true, and the correct option is G.","tags":["TMUA","2022","Paper2"],"difficulty":2},{"id":"2022-P2-Q11","paper":2,"index":10,"stemLatex":"The diagram shows a kite $PQRS$ whose diagonals meet at $O$.\n\n$OP=x$\n$OQ=y$\n$OR=x$\n$OS=z$\n\nWhich of the following is necessary and sufficient for angle $SPQ$ to be a right angle?","stemImage":"/stem-fallback/2022-P2-Q11.png","options":[{"key":"A","latex":"x=y=z"},{"key":"B","latex":"2x=y+z"},{"key":"C","latex":"x^2 = yz"},{"key":"D","latex":"y=z"},{"key":"E","latex":"y^2=x^2 + z^2"}],"answerKey":"C","explanationLatex":"The angle SPQ is a right angle if and only if the triangle SPQ satisfies Pythagoras’s theorem, that is, if and only if SP 2 +PQ 2 =SQ2. Since PQRS is a kite, its diagonals intersect at right angles, so both SOP and POQ are rightangled triangles. Therefore, again by Pythagoras, SP 2 =x^2 +z^2 PQ 2 =x^2 +y^2. Substituting this, along with SQ =y +z, into the above equation gives the following necessary and sufficient condition for angle SPQ to be a right angle: x^2 +z^2 +x^2 +y^2 = (y +z)^2. Expanding and simplifying, this is equivalent to 2x2 = 2yz or x^2 =yz. Therefore option C gives a necessary and sufficient condition. (One should also check that none of the others are necessary and sufficient; we could take x = 2, y = 4, z = 1: this satisfies option C but none of the others.)","tags":["TMUA","2022","Paper2"],"difficulty":2},{"id":"2022-P2-Q12","paper":2,"index":11,"stemLatex":"Place the following integrals in order of size, starting with the smallest. $$P = \\int_0^1 2^{\\sqrt{x}} \\: dx$$ $$Q = \\int_0^1 2^x \\: dx$$ $$R = \\int_0^1 (\\sqrt{2})^x \\: dx$$","options":[{"key":"A","latex":"P < Q < R"},{"key":"B","latex":"P < R < Q"},{"key":"C","latex":"Q < P < R"},{"key":"D","latex":"Q < R < P"},{"key":"E","latex":"R < P < Q"},{"key":"F","latex":"R < Q < P"}],"answerKey":"F","explanationLatex":"We compare the integrands on the interval $0\\le x\\le 1$.\n\nFirst,\n\n$$R=(\\sqrt{2})^x=2^{x/2}.$$\n\nSo the three integrands are:\n\n$$2^{x/2},\\qquad 2^x,\\qquad 2^{\\sqrt{x}}.$$\n\nOn $0<x<1$ we have\n\n$$\\frac{x}{2}<x<\\sqrt{x}.$$\n\nSince the function $2^t$ is increasing, this gives\n\n$$2^{x/2}<2^x<2^{\\sqrt{x}}.$$\n\nTherefore the integrands satisfy\n\n$$R<Q<P$$\n\nthroughout the interval, and so the integrals satisfy the same ordering:\n\n$$R<Q<P.$$\n\nSo the correct option is $\\mathbf{F}$.","tags":["TMUA","2022","Paper2"],"difficulty":2},{"id":"2022-P2-Q13","paper":2,"index":12,"stemLatex":"Consider the statement $(*)$ about a real number $x$: \n\n $(*)$ There exists a real number $y$ such that $x - xy + y$ is negative. \n\n For how many real values of $x$ is $(*)$ true?","options":[{"key":"A","latex":"no values of $x$"},{"key":"B","latex":"exactly one value of $x$"},{"key":"C","latex":"exactly two values of $x$"},{"key":"D","latex":"all except exactly two values of $x$"},{"key":"E","latex":"all except exactly one value of $x$"},{"key":"F","latex":"all values of $x$"}],"answerKey":"E","explanationLatex":"For each fixed value of x, we can think about this expression as being a function of y. We can write the expression as (1 -x)y +x, which is just my +c where m = 1-x and c =x. This is the equation of a straight line graph, so it takes every real value, both positive and negative, as long as m≠ 0. The casem = 0 occurs exactly whenx = 1, and in this case, the expression becomes (1-1)y+1 = 1, so it is always 1, for every value of y. Therefore (∗) is true for all except exactly one value of x, namely x = 1, and the correct answer is option E.","tags":["TMUA","2022","Paper2"],"difficulty":2},{"id":"2022-P2-Q14","paper":2,"index":13,"stemLatex":"Consider the two inequalities: $$|x + 5| < |x + 11|$$ $$|x + 11| < |x + 1|$$ Which one of the following is correct?","options":[{"key":"A","latex":"There is no real number for which both inequalities are true."},{"key":"B","latex":"There is exactly one real number for which both inequalities are true."},{"key":"C","latex":"The real numbers for which both inequalities are true form an interval of length $1$."},{"key":"D","latex":"The real numbers for which both inequalities are true form an interval of length $2$."},{"key":"E","latex":"The real numbers for which both inequalities are true form an interval of length $3$."},{"key":"F","latex":"The real numbers for which both inequalities are true form an interval of length $4$."},{"key":"G","latex":"The real numbers for which both inequalities are true form an interval of length $5$."}],"answerKey":"D","explanationLatex":"To solve this question, we use the fact that|x-a| can be understood as the distance of x froma. The first inequality is true if and only if x is closer to-5 than to-11, i.e., if and only if x> -8. The second inequality is true if and only if x is closer to -11 than to -1, i.e., if and only if x< -6. Therefore both inequalities are true if and only if -8< x <-6, which is an interval of length 2, which is option D. One could also do this question by sketching the graphs of y =|x + 5| and y =|x + 11| and working out where they intersect, and so on. But that requires a lot more work than than the approach presented here.","tags":["TMUA","2022","Paper2"],"difficulty":2},{"id":"2022-P2-Q15","paper":2,"index":14,"stemLatex":"The real numbers $x$, $y$ and $z$ are all greater than $1$, and satisfy the equations $$\\log_x y = z \\quad \\text{and} \\quad \\log_y z = x$$ Which one of the following equations for $\\log_z x$ must be true?","options":[{"key":"A","latex":"\\log_zx = y"},{"key":"B","latex":"\\log_zx = \\frac{1}{y}"},{"key":"C","latex":"\\log_zx = xy"},{"key":"D","latex":"\\log_zx = \\frac{1}{xy}"},{"key":"E","latex":"\\log_zx = xz"},{"key":"F","latex":"\\log_zx = \\frac{1}{xz}"},{"key":"G","latex":"\\log_zx = yz"},{"key":"H","latex":"\\log_zx = \\frac{1}{yz}"}],"answerKey":"F","explanationLatex":"Since logxy =z, we have y =xz. Likewise z =yx. Combining these gives z =yx = (xz)x =xxz. Therefore raising the equation to the power of 1 xz givesz1/xz =x, hence logzx = 1 xz , which is option F.","tags":["TMUA","2022","Paper2"],"difficulty":2},{"id":"2022-P2-Q16","paper":2,"index":15,"stemLatex":"In this question, $a_1$, ..., $\\: a_{100}$ and $b_1$, ..., $\\: b_{100}$ and $c_1$, ..., $\\: c_{100}$ are three sequences of integers such that $$a_n \\leq b_n + c_n$$ for each $n$. \n\n Which of the following statements must be true ? \n\n I $\\quad$ (minimum of $a_1$, ..., $\\: a_{100}$) $\\leq$ (minimum of $b_1$, ..., $\\: b_{100}$) + (minimum of $c_1$, ..., $\\: c_{100}$) \n\n II $\\quad$ (minimum of $a_1$, ..., $\\: a_{100}$) $\\geq$ (minimum of $b_1$, ..., $\\: b_{100}$) + (minimum of $c_1$, ..., $\\: c_{100}$) \n\n III $\\quad$ (maximum of $a_1$, ..., $\\: a_{100}$) $\\leq$ (maximum of $b_1$, ..., $\\: b_{100}$) + (maximum of $c_1$, ..., $\\: c_{100}$)","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":"For each of these, we can try to construct a counterexample, and if that doesn’t work, we may understand why the statement must be true. For simplicity in the explanations below, let’s write Amin = minimum of a1, ... ,a100 Bmin = minimum of b1, ... ,b100 Cmin = minimum of c1, ... ,c100 and similarly for Amax etc. I Let’s suppose that Bmin = 0 and Cmin = 0. Can we then make the minimum of the ans greater than zero? Yes, we can: if we take b1 = 0 and the rest of the bns to be 100, and we take c100 = 0 and the rest of the cns to be 100, then we can have an = 100 for every n, so the minimum of the ans is 100. Therefore this statement is not necessarily true. II If we try the same example as in I, we find that Amin = 100, which is greater than Bmin + Cmin, so the inequality holds in this case. But to get a counterexample, we want to make Amin small. Helpfully, the condition an≤ bn +cn allows us to make each an as small as we like. So if we take an = 0, bn = 1 and cn = 1 for all n, then the condition will be satisfied for each n, but we will have Amin = 0 and Bmin =Cmin = 1, so Amin<B min +Cmin in this case. Therefore this statement is not necessarily true either. III Both of the counterexamples we used for I and II satisfy this statement. It is not obvious how to construct a counterexample: every simple example satisfies this statement. So let us instead try to prove it. Each an satisfies the condition an≤ bn +cn. Now bn≤ Bmax and cn≤ Cmax for each n, so an≤Bmax +Cmax for each n. But this means that the maximum of all of the ans also satisfies this condition, that is Amax≤Bmax +Cmax, so the given statement must be true. Therefore only statement III must be true, and the correct answer is option D.","tags":["TMUA","2022","Paper2"],"difficulty":2},{"id":"2022-P2-Q17","paper":2,"index":16,"stemLatex":"A student answered the following question:\n\n$a$ and $b$ are non-zero real numbers.\n\nProve that the equation $x^3+ax^2+b=0$ has three distinct real roots if\n$27b\\left(b+\\frac{4a^3}{27}\\right)<0$.\n\nHere is the student's solution:\n\nI. We differentiate $y=x^3+ax^2+b$ to get $\\frac{dy}{dx}=3x^2+2ax=x(3x+2a)$.\n\nSolving $\\frac{dy}{dx}=0$ shows that the stationary points are at $\\left(0,b\\right)$ and $\\left(-\\frac{2a}{3}, b+\\frac{4a^3}{27}\\right)$.\n\nII. If $27b\\left(b+\\frac{4a^3}{27}\\right)<0$, then $b$ and $b+\\frac{4a^3}{27}$ must have opposite signs, and so one of the stationary points is above the x-axis and one is below.\n\nIII. If the cubic has three distinct real roots, then one of the stationary points is above the x-axis and one is below.\n\nIV. Hence if $27b\\left(b+\\frac{4a^3}{27}\\right)<0$, then the equation has three distinct real roots.\n\nWhich one of the following options best describes the student's solution?","stemImage":"/stem-fallback/2022-P2-Q17.png","options":[{"key":"A","latex":"It is a completely correct solution."},{"key":"B","latex":"The student has instead proved the converse of the statement in the question."},{"key":"C","latex":"The solution is wrong, because the student should have stated step II after step III."},{"key":"D","latex":"The solution is wrong, because the student should have shown the converse of the result in step II."},{"key":"E","latex":"The solution is wrong, because the student should have shown the converse of the result in step III."}],"answerKey":"E","explanationLatex":"Considering the offered options, the focus is on the order of the steps and whether the steps prove what they should be proving; there is no requirement for us to check the algebraic calculations themselves. The task is to prove that if 27b ( b + 4a3 ) < 0, then x^3 +ax2 +b = 0 has three distinct real roots. In step I, the student finds the stationary points of y =x^3 +ax2 +b. In step II, the student assumes that 27 b ( b + 4a3 ) < 0, which is the correct thing to do to prove an ‘if ... then’ statement. In step III, the student says ‘if the cubic has three distinct real roots then ... ’, which is correct but is not useful: we know that one of the stationary points is above the x-axis and the other is below and we wish to deduce that the cubic has three distinct real roots. In step IV, the student uses the result of II and the result ‘ if one of the stationary points is above thex-axis and one is below, then the equation has three distinct real roots’, which is the converse of the statement in step III. Therefore the correct option is E: the student should have shown the converse of the result in step III.","tags":["TMUA","2022","Paper2"],"difficulty":2},{"id":"2022-P2-Q18","paper":2,"index":17,"stemLatex":"$P$, $Q$, $R$ and $S$ show the graphs of\n$y=(\\cos x)^{\\cos x}$, $y=(\\sin x)^{\\sin x}$, $y=(\\cos x)^{\\sin x}$ and $y=(\\sin x)^{\\cos x}$\nfor $0<x<\\frac{\\pi}{2}$ in some order.\n\nWhich row in the following table correctly identifies the graphs?\n\nRows A-H map in this column order:\n$y=(\\cos x)^{\\cos x}$, $y=(\\sin x)^{\\sin x}$, $y=(\\cos x)^{\\sin x}$, $y=(\\sin x)^{\\cos x}$\n\nA: P, Q, R, S\nB: P, Q, S, R\nC: Q, P, R, S\nD: Q, P, S, R\nE: R, S, P, Q\nF: R, S, Q, P\nG: S, R, P, Q\nH: S, R, Q, P","stemImage":"/stem-fallback/2022-P2-Q18.png","options":[{"key":"A","latex":"P, Q, R, S"},{"key":"B","latex":"P, Q, S, R"},{"key":"C","latex":"Q, P, R, S"},{"key":"D","latex":"Q, P, S, R"},{"key":"E","latex":"R, S, P, Q"},{"key":"F","latex":"R, S, Q, P"},{"key":"G","latex":"S, R, P, Q"},{"key":"H","latex":"S, R, Q, P"}],"answerKey":"E","explanationLatex":"We identify the graphs by checking convenient values of $x$.\n\nAt $x=0$:\n\n$$ (\\cos x)^{\\cos x}=1^1=1, $$\n$$ (\\sin x)^{\\sin x}\\to 1, $$\n$$ (\\cos x)^{\\sin x}=1^0=1, $$\n$$ (\\sin x)^{\\cos x}=0^1=0. $$\n\nSo the graph that starts at $0$ is\n\n$$y=(\\sin x)^{\\cos x},$$\n\nwhich is graph $Q$.\n\nAt $x=\\frac{\\pi}{2}$:\n\n$$ (\\cos x)^{\\sin x}=0^1=0. $$\n\nSo the graph that ends at $0$ is\n\n$$y=(\\cos x)^{\\sin x},$$\n\nwhich is graph $P$.\n\nThat leaves graphs $R$ and $S$ for\n\n$$y=(\\cos x)^{\\cos x} \\quad \\text{and} \\quad y=(\\sin x)^{\\sin x}.$$\n\nNow test $x=\\frac{\\pi}{6}$.\n\nFor $y=(\\sin x)^{\\sin x}$,\n\n$$\\left(\\frac12\\right)^{1/2}=\\frac{1}{\\sqrt{2}}\\approx 0.707.$$\n\nThis matches graph $S$, so graph $S$ is\n\n$$y=(\\sin x)^{\\sin x}.$$\n\nHence graph $R$ must be\n\n$$y=(\\cos x)^{\\cos x}.$$\n\nSo the matching is:\n\n- $(\\cos x)^{\\cos x}$ : $R$\n- $(\\sin x)^{\\sin x}$ : $S$\n- $(\\cos x)^{\\sin x}$ : $P$\n- $(\\sin x)^{\\cos x}$ : $Q$\n\nThis is option $\\mathbf{E}$.","tags":["TMUA","2022","Paper2"],"difficulty":2},{"id":"2022-P2-Q19","paper":2,"index":18,"stemLatex":"A polygon has $n$ vertices, where $n \\geq 3$. It has the following properties: \n\n • Every vertex of the polygon lies on the circumference of a circle $C$. \n\n • The centre of the circle $C$ is inside the polygon. \n\n • The radii from the centre of the circle $C$ to the vertices of the polygon cut the polygon into $n$ triangles of equal area. \n\n For which values of $n$ are these properties sufficient to deduce that the polygon is regular?","options":[{"key":"A","latex":"no values of $n$"},{"key":"B","latex":"$n = 3$ only"},{"key":"C","latex":"$n = 3$ and $n = 4$ only"},{"key":"D","latex":"$n = 3$ and $n \\geq 5$ only"},{"key":"E","latex":"all values of $n$"}],"answerKey":"B","explanationLatex":"We sketch a part of the polygon with the triangles described in the question: θ Suppose the angle of one of the triangles is θ, as shown. Then the area of the triangle is given by 1 2r2 sinθ where r is the radius of the circle (using the formula A = 1 2ab sinC). Since r is the same for each of the triangles, all the triangles will have equal area if and only if sinθ is the same for each triangle. This will certainly be the case if the polygon is regular, but we have to determine if this can be true in any other situation. If two of the triangles have angles at the centre of the circle of θ andφ, then sinφ = sinθ if and only if φ = θ or φ = π-θ. Can we have the latter possibility? If so, it would mean we have some angles being θ and some being π-θ. Suppose then that θ is acute and π-θ is obtuse. (We cannot have a reflex angle as the centre of the circle lies inside the polygon, and if θ is a right angle, then π-θ =θ.) Suppose further thatk of the angles of the polygon triangles equal π-θ and the remaining n-k equalθ. Then the sum of these triangle angles is given by (n-k)θ +k(π-θ) = 2π since they form a whole circle. We can rearrange this to ( n- 2k)θ = (2-k)π and hence θ = (2-k)π n- 2k . There can be at most three obtuse angled triangles (as four would be more than a whole circle), so let’s try k = 1, k = 2 and k = 3 in turn. If k = 1, we have θ = π n-2. It looks as though this will work for every n, but we have to be a little careful as we need θ to be acute. When n = 3, we get θ = π, which is not allowed. When n = 4, we get θ = π 2 , which is again not allowed. When n> 4, θ <π 2 , so this does work. Therefore when n≥ 5, it is possible for the conditions to be satisfied but for the polygon not to be regular. Now consider k = 2, giving θ = 0 n-4. We only need to consider n = 3 and n = 4, as we have dealt withn≥ 5 above. This is zero when n = 3, so it does not work for a triangle. When n = 4, this formula is not meaningful (it gives 0 0). We can instead go back to the original equation (n- 2k)θ = (2-k)π, which becomes 0 = 0 in this case, meaning that it works for any value of θ. And indeed, this gives two angles of θ and two of π-θ, so we get a non-regular polygon with equal area triangles. The simplest example is a non-square rectangle. We are left with the case k = 3 to consider. When n = 3, there would be no acute angles and three obtuse ones, and the formula gives θ = -π -3 = π 3 . And indeed, a triangle with three obtuse angles of π 3 is an equilateral triangle satisfying the conditions. Therefore the only triangle (n = 3) satisfying the conditions is regular. We have thus been able to construct non-regular polygons satisfying the conditions for every n≥ 4, but not for n = 3. Hence the correct answer is option B.","tags":["TMUA","2022","Paper2"],"difficulty":2},{"id":"2022-P2-Q20","paper":2,"index":19,"stemLatex":"The functions $f_1$ to $f_5$ are defined on the real numbers by $$f_1(x) = \\cos x$$ $$f_2(x) = \\sin(\\cos x)$$ $$f_3(x) = \\cos(\\sin(\\cos x))$$ $$f_4(x) = \\sin(\\cos(\\sin(\\cos x)))$$ $$f_5(x) = \\cos(\\sin(\\cos(\\sin(\\cos x))))$$ where all numbers are taken to be in radians. \n\n These functions have maximum values $m_1$, $m_2$, $m_3$, $m_4$ and $m_5$, respectively. \n\n Which one of the following statements is true?","options":[{"key":"A","latex":"$m_1$, $m_2$, $m_3$, $m_4$ and $m_5$ are all equal to $1$"},{"key":"B","latex":"0 < m_5 < m_4 < m_3 < m_2 < m_1 = 1"},{"key":"C","latex":"$m_1 = m_3 = m_5 = 1$ and $0 < m_2 = m_4 < 1$"},{"key":"D","latex":"$m_1 = m_3 = m_5 = 1$ and $0 < m_4 < m_2 < 1$"},{"key":"E","latex":"$m_1 = m_3 = 1$ and $0 < m_2 = m_4 < 1$ and $0 < m_5 < 1$"},{"key":"F","latex":"$m_1 = m_3 = 1$ and $0 < m_4 < m_2 < 1$ and $0 < m_5 < 1$"}],"answerKey":"E","explanationLatex":"We note that we can rewrite each of the functions f 2 to f^5 in terms of the previous function: f1(x) = cosx f^2(x) = sin f1(x) f^3(x) = cos f^2(x) f^4(x) = sin f^3(x) f^5(x) = cos f^4(x) Since f1(x) = cosx is periodic with period 2π, each of the other functions also repeats every 2 π (though it may have a shortest period less than this), and so we only need to consider 0≤x≤ 2π. (We could actually restrict ourselves to 0≤x≤π, as the values of cosx in the rangeπ <x≤ 2π are a repeat of the values in the range 0 ≤x≤π, but we will draw a complete period anyway.) Sketching the graph of each of these functions is the most straightforward way to proceed. We note first a result that we will need later: π≈ 3.14 so π 2≈ 1.6 and hence sin 1 < sin π 2 = 1. A sketch of f 1(x) = cosx is straightforward: -1 x y We see that m1 = 1, and the minimum value taken is -1. Now f 2(x) = sin f1(x), so this goes between- sin 1 and sin 1, with roots where f 1(x) = 0; recall that sin 1 < 1: -1 x y We see that m^2 = sin 1< 1, and the minimum value taken is - sin 1. Now f^3(x) = cos f^2(x). When f 2(x) = 0, f^3(x) = 1, so f 3(x) has a maximum of 1 at x = π 2 and at x = 3π 2 . But f 3(x) never reaches 0 as f 2(x) never reaches π 2≈ 1.6 or-π 2≈- 1.6. Therefore the graph looks something like this: -1 x y Therefore m^3 = 1, and the minimum value taken is cos(sin 1). Next, f^4(x) = sin f^3(x), so this will be an oscillating curve going between sin(cos(sin 1)) and sin 1: -1 x y Thusm4 = sin 1. Finally, f 5(x) = cos f 4(x), so this will be an oscillating curve going between cos(sin 1) and cos(sin(cos(sin 1))): -1 x y To summarise, we have m1 =m^3 = 1, m^2 =m^4 = sin 1 and m^5 = cos(sin(cos(sin 1))), and the correct option is E. It is also interesting to note from these accurate graphs that the subsequent functions are getting flatter and flatter and seem to be tending to a straight line (though the limiting value is different in the functions fn where n is even and those where n is odd).","tags":["TMUA","2022","Paper2"],"difficulty":2}]