SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 014 Mrk Scheme: Ech prt of Question 1 is worth four mrks which re wrded solely for the correct nswer. Ech of Questions -7 is worth 15 mrks QUESTION 1: A. We cn rewrite the inequlity s which rerrnges to x 4 8x 9 < 0 (x 9)(x + 1) < 0. As x + 1 > 0 for ll vlues of x then we cn divide by this term nd see tht the inequlity is equivlent to x < 9. The nswer is (). B. We cn complete the squre in the qudrtic to get (x 1) + 1. So substituting x = 1, we know tht the grph meets the x-xis t (1,0), since log 10 (1) = 0. This elimintes grphs (), (b), (c), nd (d). The nswer is (e). C. The first derivtive is given by dy dx = 3kx (k + )x + ( k) which t x = 1 tkes the vlue 3k k + k = 0. So the grdient is zero for ll vlues of k. However for this to be minimum we lso need tht the second derivtive is positive. The second derivtive is equl to d y = 6kx (k + 1) dx which equls 4k when x = 1. This is positive when k > 1 nd so the nswer is (c). D. One might note tht (d) is the only vector tht is unit vector for ll vlues of m; s reflection is distnce-preserving then (d) is the only possible nswer. Alterntively we might consider the cse when m becomes lrge (the line y = mx begins to verge upon the y-xis) nd (d) is the only nswer tht grees with (, 0) in this extreme. Or one might clculte the imge using geometry. The line through (1, 0) nd perpendiculr to y = mx hs grdient /m nd so hs eqution y = (1 x)/m. This norml intersects y = mx t (1/(m + 1), m/(m + 1)) which is displcement of ( ) m m + 1, m m + 1 wy from the originl point (1, 0). So, using vectors, the reflected imge is t ( ) ( ) m (1, 0) + m + 1, m 1 m = m + 1 m + 1, m. m + 1 Hence the nswer is (d). 1
E. Consider first the extremes of the expression in the brcket. If we use the identity sin x = 1 cos x then we see 4 sin x + 4 cos x + 1 = 5 + 4 cos x 4 cos x = 6 (1 cos x). Thus the expression in the brcket cn tke vlues from 3, when cos x =, to 6, when cos x = 1/. Hence the gretest the squre of the expression cn be is 6 = 36 nd the nswer is (b). F. Whilst t = 1 nd s = 8 (in tht order) do led to the function 8 x there re clerly other wys of chieving this function e.g. t = 3 nd s = 8 in tht order. So we cnnot deduce (). As S is trnsltion unit to the right nd T is reflection in the origin then fter some selection of the two functions we will hve trnslted the rel line to the right or left by n integer (wherever the origin hs moved to) nd it will be pointing in the sme direction or in the reverse direction; more precisely we will hve yielded function n ± x where n is n integer. If t is even we will hve +x nd if t is odd we will hve x. In our exmple we know tht the origin hs moved to the point 8. The order in which the vrious S nd T re performed ffects the position of the origin s ST (x) = 1 x nd T S(x) = x but s these two possibilities differ in how they move the origin by, whilst we cnnot conclude tht s = 8 we cn conclude tht s is even. So the nswer is (c). G. Recll from the Binomil Theorem for ( + b) n tht the sum of the exponents of nd b in ech term lwys equls n, so n expnsion of (xy + y ) k would led solely to terms of totl exponent k (in x nd y together). As x 3 y 5 hs totl exponent 8 then we need to focus on the k = 4 term. The correct term in the binomil expnsion of (1 + xy + y ) n is ( ) n (xy + y ) 4. 4 nd s x 3 y 5 = (xy) 3 y we need to choose 3 xy-terms nd 1 y term, so nd there re ( 4 1) = 4 wys of doing this. The nswer is (d). H. Between 1 nd 6000 there re 3000 vlues of n which re divisible by nd 000 vlues of n which re divisible by 3. Even multiples of 3 re divisible by, nd there re 1000 such vlues of n. Thus f(6000) = (3000 1000) + 3 (000 1000) + 4 (1000) = 4000 + 3000 + 4000 = 11000. The nswer is (c).
I. Completing the squre rerrnges the exponent to (x ) 1. This mens tht x 4x+3 = (x ) = 1 (x ). We re therefore trnslting the grph prllel to the x-xis (from the (x ) term), nd then performing stretch prllel to the y-xis (from the 1 multiplier), nd so the nswer is (b). J. We re interested in the integrl between nd 1 of f(x). If we integrte the originl identity between nd 1, we find 1 + 1 1 f(x) du = f( x) du + [ x 3] 1 1 f(x) du. Note tht 1 f( x) du = 1 f(x)du s the grph of y = f( x) is just reflection in the y-xis of the grph of y = f(x). Substituting A = 1 f(x) du mens we re left with so tht A = 4 nd the nswer is (). 1 + A = A + A, 3
. (i) [5 mrks] As x = 1 is root then which rerrnges (on completing the squre) to 1 + b b = 0 + () =. As 0 then () = 1 b 1 +. (ii) [5 mrks] Substituting = 1 + b b into ( ) nd fctorizing it we get 0 = x 3 + bx + (b )x b = (x 1)(x + (b + 1)x + b ). If ( ) hs x = 1 s repeted root, then x = 1 is lso root of the qudrtic fctor nd so which is impossible for rel b. 1 + b + 1 + b = 1 + (b + 1) = 0 Alterntive pproch: Write the cubic s (x 1) (x γ) to gin equtions in b, γ (nd ); by compring coefficients we rrive t contrdiction. (iii) [5 mrks] As x = 1 cn never be repeted root, then ( ) cn only hve repeted root by the qudrtic fctor hving repeted root. This hppens when the discriminnt is zero tht is, when When b = /4 then (b + 1) = 4 1 b = 1 + 4b = 0 = b = /4. x + (b + 1)x + b = x + x/ + 1/16 = (x + 1/4) nd so the repeted root is lso /4. As the repeted root is less thn the single root, then (from knowledge of cubic s shpe) we know there is locl mximum t the repeted root. Alterntive pproch: Write the cubic s (x 1)(x γ) nd compre coefficients to to gin equtions in b nd γ. We see b = γ 1 nd b = γ, so tht b = /4. Then follow s bove to find x, nd either by observtion or differentition know tht there is locl mximum t the repeted root. 4
3. (i) [ mrks] Setting x = y = 0 in (A) we get As f(0) > 0 by (C) then f(0) = 1. (ii) [3 mrks] By property (B) we hve I = 1 0 f(x) dx = 1 0 f(0) = f(0)f(0). df dx dx = [f(x)]1 0 = f(1) f(0) = 1. (iii) [6 mrks] The trpezium rule estimtes the re s I n = 1/n [ ( ) ( ) ( ) 1 n 1 f(0) + f + f + f n n n ] + f(1). Now b = f(1/n) nd by (B) we hve f(/n) = f(1/n)f(1/n) = b, f(3/n) = b 3, etc. prticulr tht b n = f(n/n) = f(1) =. Hence I n = 1 [ 1 + b + b + b 3 + b n + b n] n = 1 [ ] 1 + b(bn 1) + b n n = 1 [ ] b n b + b n+1 b n + n = 1 [ ] b n+1 + b n n = 1 n = 1 n [ ] (b + 1)(b n 1) ( b + 1 ) ( 1). Note in (iv) [4 mrks] We re given tht I n I for lrge n nd hence ( ) 1 b + 1 ( 1) 1. (+) n Then b + 1 1 + n n Then 1 n = b 1 + n 1. 5
4. (i) [ mrks] As ABC is isosceles then ABC = π α. So re of tringle ABC = 1 (AB)(BC) sin( ABC) = 1 1 1 sin(π α) = 1 sin α. (ii) [3 mrks] When β = α then F = 1 nd when β = 0 then F = 0. Further F is decresing function of β nd hence ech vlue of 0 k 1 is uniquely ttined by F. (iii) [3 mrks] AXB will hve hlf the re of ABC when AXB is right ngle. Hence ABX = π/ α. But we lso hve tht ABX = π β then it follows tht π β = π/ α = β = π/4 + α/. (iv) [4 mrks] As ABX = π β nd AXB = β α then, by the sine rule, we hve AX sin(π β) = 1 sin(β α). Hence Thus re of ABX = 1 sin(π β) sin α (AX)(AB) sin α = sin(β α) F = sin β sin α / sin α sin(β α) = = sin β sin α sin(β α.) sin α. sin β sin α sin(β α). (v) [3 mrks] When 0 < β < α π/, we cn swp the roles of α nd β nd hence re of tringle ABX = sin α sin β sin(α β). Hence F = sin α sin β / sin α sin(α β) = sin β sin(α β). 6
5. (i) [1 mrk] AAA, AAB, ABA, ABB, ABC. (ii) [ mrks] c n,1 = 1: the rhyming scheme must contin just n As. c n,n = 1: the only rhyming scheme is ABC... K, contining the first k letters of the lphbet. (iii) [5 mrks] If the lst symbol mtches n erlier symbol then the first n 1 chrcters hve k different symbols, so there re c n,k wys of choosing the first n 1 symbols of the rhyming scheme; nd there re k wys of choosing the lst symbol totl kc n,k. If the lst symbol does not mtch n erlier one, then the first n chrcters hve k different symbols, so there re c n,k wys of choosing the first n 1 symbols; the finl symbol must be the kth symbol of the lphbet totl c n,k. Adding the bove gives the totl. (iv) [4 mrks] r n = n k=1 c n,k Here s tble of c n,k vlues. n\k 1 3 4 1 1 1 1 3 1 3 1 4 1 7 6 1 The first column nd min digonl come from the bse cses. The other entries re clculted from the recursive eqution s follows Hence r 4 = 1 + 7 + 6 + 1 = 15. c 3, =.c, + c,1 =.1 + 1 = 3 c 4, =.c 3, + c 3,1 =.3 + 1 = 7 c 4,3 = 3.c 3,3 + c 3, = 3.1 + 3 = 6 (v) [3 mrks] c n, = n. The first symbol is A; the next n re either A or B ( n possibilities), but not ll A s (so subtrct 1). Alterntive pproch: Tke k = in the formul of prt (iii): c n, = c n, + c n,1 = c n, + 1. 7
6. (i) [ mrks] We must hve tht the product hs unique fctoristion, so x = 1 nd y is prime or 1. If the product is not prime it would hve more thn one fctoristion, so Pm would not know x nd y. (ii) [ mrks] If we hd x = 1, y = 3, Pm would know x nd y, s in the previous prt. Thus x = nd y =. (iii) [3 mrks] If we hd x = nd y = then Sm would know x nd y, s in the previous prt. Thus x = 1 nd y = 4. (iv) [4 mrks] Sm knows tht the sum is 9, nd so knows tht Pm didn t originlly know x nd y (since 9 is not 1 + prime). If we hd x = nd y = 4 then Sm would know the sum is 6, nd so would not initilly know tht Pm does not know x nd y: from Sm s point of view, it could be tht x = 1 nd y = 5, in which cse Pm would know x nd y. Hence x = 1 nd y = 8. (v) [4 mrks] Pm knows the product is 6, so sys no t her first sttement. Sm knows the sum if 5, so t her first sttement, she considers it possible tht x = 1 nd y = 4. At Pm s second sttement, she still considers it possible tht x = 1 nd y = 6 (in this cse, Sm would know the sum is 7, nd would still consider x =, y = 5 or x = 3, y = 4 possible). If we hd x = 1 nd y = 4 then Pm would know the product is 4. In this cse, Sm s first sttement would tell Pm tht we don t hve x = y =, by prt (ii). Hence Pm would know x nd y t the point of her second sttement contrdiction. Hence x = nd y = 3. 8
7. (i) [ mrks], b, bb,.... All words composed of lternting s nd b s, strting nd ending with n. (ii) [ mrks] All words except those in prt (i). (iii) [ mrks] Automton such s the one illustrted below. s 0 s 1 b b (iv) [3 mrks] Automton such s the one illustrted below. s 0 s 1 b b b b s s 3 (v) [6 mrks] From the stte reched fter i, following the pth lbelled b i leds to n ccepting stte, since i b i is in L. However, from the stte reched fter j, following the pth lbelled b i does not led to n ccepting stte, since j b i is not in L. Hence these re different sttes. Hence ech word i (for i = 0, 1,,...) leds to different stte, so there must be n infinite number of sttes. But this contrdicts the fct tht there re finite number of sttes. 9