SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 2 NOVEMBER 2016

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1 SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY NOVEMBER 06 Mark Scheme: Each part of Question is worth 4 marks which are awarded solely for the correct answer. Each of Questions -7 is worth 5 marks QUESTION : A. Considering the sequence, a = l, a 3 = l, a 4 = l 3, each additional term multiplies the previous term y l. The product of the first 5 terms is equal to l = l 4 5 = l 05. The answer is (d). B. Call the length of one of the sides of the hexagon p, then the side of the square is equal to p + ( p) =. Then as the hexagon side forms a triangle in each corner of the square, using Pythagoras, p = ( p) + ( p). Solving this quadratic results in p = ±, ut as the length must e less than the answer is (). C. We can rewrite the given equation as (x + a ) + (y + a ) = c + a +. For the circle to contain the 4 4 origin, the distance from the centre to the origin must e less than the radius, so a + < c + a The answer is (a). D. cos n (x) + cos n (x) = cos n (x)( + cos n (x)) = 0. For this to e true, if n is even, cos(x) = 0 has two roots, ut when n is odd either cos(x) = 0 or cos(x) =, which is three roots. Hence the answer is (d). E. When x = 0, y = = 0, so we can rule out (d) and (e). To work out the numer of x-axis intersection points, consider (x ) = cos(πx). The shape of these graphs means they cannot intersect 6 times (eliminating ()). The answer cannot e (c), ecause we know there is a crossing point x =, ut that y is positive when x =. So the answer is (a). F. Using the factor theorem, for (x + ) to e a factor, (x + ) = 0, so x =. Then the equation given ecomes (4) n () n ( ) n = 0. This only holds when ( ) n is positive, so the answer is (). G. Considering the first few terms x 0 =, x = x 0 =, x =, x 3 = 4, x 4 = 8, and so on. By oservation, x n = n for n. As this is a geometric progression, we can evaluate the sum of the sequence as The answer is (d). k=0 = x k + k k= = + = 3

2 H. The area ounded y the x-axis and the curve y = f(x), A is equal to A = a a f(x) dx = 4 3 a 3, whilst the area ounded y the x-axis and the curve y = g(x), A is equal to We require an a such that A > A, so and so the answer is (e). 4 a A = g(x) dx = 8 4 a 5 a a 3 > 8 5 a 5 4 0a 6 4 > 4a 5 4 a 4 > 6 5 I. Let ax + y = c, which rearranges to y = ax + c. Given that is positive we can interpret this as achieving the maximum c when the line y = ax + c is moved up the y-axis whilst still intersecting the disc formed y x + y. Hence the line should e tangent to the unit circle. By Pythagoras, ( c ) = + ( a ) and so the answer is (c). J. We are trying to construct counter-examples for each of the statements. Note that 0 x(n) 9. (a) is true since, for example, Π(4) =, ut 4 is not prime. () is false - we don t need to consider even n eyond x n = 4; for this case we know no primes end in a 4, ut for example Π(64) = as 64 = 6. For odd n, x(n) =, Π(n) =, counterexample n = = ; x(n) = 3, Π(n) =, counterexample n = 43 = 3 5 ; x(n) = 5, Π(n) =, counterexample n = 5 = 5 ; x(n) = 7, Π(n) =, counterexample n = 6807 = 7 5 ; x(n) = 9, Π(n) =, counterexample n = 9 = 3. (c), (d), and (e) are all true. The answer is ().

3 . (i) [ mark] We have A(B(x)) = (3x + ) + = 6x + 5; B(A(x)) = 3(x + ) + = 6x + 5. (ii) [3 marks] We note A (x) = (x + ) + = 4x + +, A 3 (x) = (4x + + ) + = 8x , and so more generally A n (x) = n x + n + n = n x + ( n ) using the geometric series formula (pattern spotting sufficient). (iii) [4 marks] As 08 = 3 3 then F can e achieved using two applications of A and three applications of B. As AB = BA then only one such F can e achieved ut the numer of different orders in which A, A, B, B, B might e performed is 5 C = 0. (iv) [3 marks] Note that in each case the constant coefficient is one less than the coefficient of x. We can prove this y noting A(ax + (a )) = (ax + (a )) + = ax + a ; B(ax + (a )) = 3(ax + (a )) + = 3ax + 3a. So c = 07. [Alternatively to find c a student might just determine A B 3.] [Alternative: Commuting argument: By part (i) A and B commute. Therefore we only need to check of the possile configurations. From this calculation we find that c = 07.] (v) [4 marks] As each A m i B n i (x) will have a constant coefficient one less than its x coefficient it follows that k = 4 9 =. However the x coefficient of A m i B n i (x) can never e less than so the sum of such functions cannot have an x coefficient less than 44. [Alternative: Divisile y 6 argument: Each term contains at least an A and at least a B, and so each x coefficient is a multiple of 6. However 4 is not divisile y 6 and hence there exist no positive integers.] 3

4 3. 3. Solution: (i) [ mark] Note that for all x and hence f is ilateral. f(α x) = (α x α) = (α x) = (x α) = f(x) (ii) [ marks] Consider, for example, x = α + where It follows that f is not ilateral. (iii) [ marks] Note that f(α + ) = = f(α ) = f(α (α + )). as required. a [ x x n n+ dx = n + ] a = n+ a n+ n + ( ) a n+ n+ = n + [ ] x n+ a = = n + a x n dx [Alternatively: Some students may show this graphically and argue that area is preserved under reflection] (iv) [3 marks] Since f is a polynomial there is a non-negative integer d and reals c 0,..., c d such that f(x) = c 0 + c x + + c d x d for all x. Integration is linear so y the previous part we have as required. a f(x) dx = d c i x i dx i=0 = a d a c i x i dx = a i=0 f(x) dx (v) [ marks] The first integral is just the signed area under the graph of y = f(x) etween α and t and the second integral is the signed area under the graph of y = f(x) etween α t and α. The second signed area is a reflection of the first, and area is preserved under reflection. Hence the integrals are equal. (vi) [3 marks] For t α we have y the previous two parts that G(t) = t α f(x) dx = α α t f(x) dx = α t α f(x) dx = G(α t). If t α then put u = α t α and note that y what we have just shown The result follows. G(α t) = G(u) = G(α u) = G(t). (vii) [ marks] Since f is a ilateral polynomial we see G(α t) = G(t) for all t. On the other hand since G is ilateral we have G(α t) = G(t) for all t, so G(t) = 0 for all t as required. 4

5 4. (i) [3 marks] Let d e the distance from (0, 0) to the point where C touches the x-axis. Note that the x-axis is tangent to C and hence perpendicular to the radius at this point. So C has centre (d, ). We have a right-angled triangle, with d = tan(α), so d = tan(α). So the centre of C is ( tan(α), ). (ii) [ mark] C has centre (, ) and radius, so has equation tan(α) (x tan(α) ) + (y ) =. (iii) [3 marks] Let d e the distance etween the points where C and C touch the x-axis. Then Pythagoras on the right-angled triangle gives ( + 3) = + d, so d =. Also we have similar triangles (oth have a right angle and share angle α) so 3 = d + d d so d = d. So d = (d ) = 4d, so d = 3 (must have d > 0). So tan(α) = d = 3 so α = 30 (or π 6 ). (iv) [3 marks] Take α = 30. Let C 3 have radius r. Let d 3 e the distance etween the points where C and C 3 touch the x-axis. Then y similar triangles we have So r = d +d 3 +d 3 = 3 + d 3 3. So d 3 = 3(r 3). Also since C and C 3 touch Pythagoras gives r = =. d + d + d 3 d 3 (r + 3) = (r 3) + d 3 = (r 3) + 3(r 3) = 4(r 3), so r + 6r + 9 = 4r 4r + 36, which factorises to (r )(r 9) = 0. We re looking for C 3 larger than C so r = 9. (v) [5 marks] Centres of triangle C and C are (, ) and ( 3, 3) respectively. Area of trapezium is half (ottom plus top) times height, tan(α) tan(α) so: 3 + tan(α) = 4 tan(α). Or reak down as rectangle (area = ) plus triangle (area = ). tan(α) tan(α) Now deduct C sector and C sector from trapezium area. Area of C sector is ( π + α) = π + α. 4 Area of C sector is 3 ( π α) = 9π 9α. 4 So interstitial area is: 4 ( π tan(α) 4 + α ) ( 9π 4 9α ) = 4 tan(α) 5π + 4α = 4 3 π 6 5

6 5. (i) [3 marks] We have s = (A + B) + C = s = 4(A + B) + C = 0 s 3 = 8(3A + B) + C = 34 (ii) [3 marks] Sutracting the first equation from the other two gives 6A + B = 8 A + 6B = 3, whence 4A = 8, so A =, B =, C = and f(n) = (n ) n+ +. (iii) [ marks] We have s k+ = f(k) + (k + ) k+ = (k ) k+ + + (k + ) k+ = k k+ + = f(k + ) as required. (iv) [4 marks] We have t n = (n + n + 4n + + n.n) ( n.n) = n( n+ ) f(n) = n( n+ ) (n ) n+ = n+ n. Now u n = t n / n, so u n = n + n. (v) [3 marks] n n s k = (k k k+ + ) k= = = k= n (k k+ ) k= n ( k+ ) + n k= n (k k ) n n k= = f(n) n n = n+ n n+3 + n + 8 6

7 6. (i) [3 marks] There are no possile arrangements - if A is a, then either B and D are oth s or oth 0s. However, if B and D are oth s then C must also e a - ut that would require all the dancers to e s which is foridden. If B and D are oth 0s then C must also e a 0 otherwise D would not e off-eat. But if C is a 0 they cannot e off-eat. (ii) [3 marks] Assume that A is a and holds hands with F and B, then either F and B are oth s or oth 0s. If oth F and B are s then this pattern must propagate around the circle, forcing everyone to e s, which is foridden. If F and B are oth 0s then C and E must also e 0s, to keep F and B off-eat. However to ensure C, D, and E are off-eat D must e a. Hence the only possile arrangements are those where precisely two dancers on opposite positions on the ring are, and there are 3 such arrangements. (iii) [3 marks] Each person holding hands either requires one of the three dancers to e a or all three to e a. If all three, then this propagates round resulting in all s, which is foridden. Thus for each triplet of dancers one person is a. Then either spot the,0,0 pattern which only repeats when n is a multiple of three, or look at the sum of each local triplet of dancers which must e equal to n and also equal to 3k where k is the numer of dancers who are s. (iv) [ marks] If n is even two separate rings form, however each ring can only e off-eat if the numer of dancers are a multiple of 3, y previous argument. If n is odd, then n must e a multiple of 3 still ecause a ring is still formed (with displaced dancers). (v) [ mark] Either one dancer is a or three dancers are s and one is a 0. There are 8 different ways in total. (vi) [3 marks] There must e at least one dancer who is a. Holding hands with this dancer there must e either no dancers or precisely two dancers who are s. If none of the dancers are s, then the alternating 0, pattern is very ovious. If two dancers are s, then this leads to a situation where all dancers are s, which is still foridden. Hence there are possile ways of arranging off-eat dances. 7

8 7. (Example taken from Graham, Knuth, Patashnik, Concrete Mathematics.) (i) [ marks] Three -spans: (ii) [4 marks] Eight 3-spans: (iii) [3 marks] In a 4-span, the top group may have t =,, 3 or 4 elements, and may e connected to the hu y any of t line segments in each case. If t = 4, that is the end of the story, ut if t < 4 then the remaining tips may form any (4 t)-span. Thus (using the notation of the next part), z 4 =.z 3 +.z + 3.z + 4 = =. (iv) [4 marks] More generally, we have z n =.z n +.z n + + (n ).z + n. It follows that z 5 = = 55. (v) [ marks] z 6 = = 44. 8

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