SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 2014


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1 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 xxis 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 distncepreserving 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 yxis) 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
2 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 = 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 xyterms 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) = ( ) + 3 ( ) + 4 (1000) = = The nswer is (c).
3 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 xxis (from the (x ) term), nd then performing stretch prllel to the yxis (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 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 yxis 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
4 . (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 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
5 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 n n Then 1 n = b 1 + n 1. 5
6 4. (i) [ mrks] As ABC is isosceles then ABC = π α. So re of tringle ABC = 1 (AB)(BC) sin( ABC) = 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
7 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 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 = = 15. c 3, =.c, + c,1 = = 3 c 4, =.c 3, + c 3,1 = = 7 c 4,3 = 3.c 3,3 + c 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,
8 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
9 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
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