2. Topic: Summation of Series (Mathematical Induction) When n = 1, L.H.S. = S 1 = u 1 = 3 R.H.S. = 1 (1)(1+1)(4+5) = 3

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1 GCE A Level Otober/November 008 Suggested Solutions Mthemtis H (970/0) version. MATHEMATICS (H) Pper Suggested Solutions. Topi: Definite Integrls From the digrm: Are A = y dx = x Are B = x dy = y dy dx () = y dy () Equting () & (): x dx () 3 = y dy x3 = 3 3 y3 ()3 3 3 = 3 ()3 3 ()3 8 = = 9 = ( 9 ) 3.73 (3 sig. fig.) 970/0 Otober/November 008 x n dx = xn+ n + +. Topi: Summtion of Series (Mthemtil Indution) Let P n denotes the sttement S n = n(n + )(n + 5), n Z+. When n =, L.H.S. = S = u = 3 R.H.S. = ()(+)(+5) = 3 L.H.S. = R.H.S. P is true. Assume P k is true i.e. S k = k(k + )(k + 5), for some k Z+ To show tht P k+ is lso true i.e. S k+ = (k + )(k + )(k + 9), Bring out the (k + ) ftor sine it s found on the R.H.S. L.H.S. = S k+ = S k + u k+ = = k (k + ) (k + 5) + (k + ) [(k+) + ] (k + ) [k (k + 5) + (k + 3)] = (k + ) (k + 5k + k + 8) = (k + ) (k + 7k + 8) = (k + ) (k + ) (k + 9) = R. H. S. P k+ is lso true if P k is true. S k = k(k + )(k + 5) u k+ = (k + ) [(k+) + ] Sine P is true nd P k+ is true if P k is true, by mthemtil indution, P n is true n Z + For tuition, exm ppers & Lst-Minute Buddh Foot Hugging Syndrome tretment / missloi@exmpper.om.sg febook.om/jossstikstuition twitter.om/missloi Unuthorized opying, resle or distribution prohibited. Copyright 008 ϕ exmpper.om.sg. All rights reserved. / 9

2 GCE A Level Otober/November 008 Suggested Solutions Mthemtis H (970/0) version. 3. Topi: Vetors (i) OA = OA + AA = OA + OO 5 = = 3 3 (ii) os AOB = OA OB = OA OB Equl Vetors: OO = AA sme mgnitude & diretion ( 3) 5 +( ) +0 O B A Slr Produt of two vetors nd b: b = b os θ b P. Topis: Differentition, Differentil Equtions (i) dy = 3x dx x + dy = 3x x + dx y = 3 x x + dx = 3 ln x + + (ii) Sub y =, x = 0 into y = 3 ln x + +, = 3 ln = y = 3 ln x + + Express in f (x) f(x) f (x) dx = ln f(x) + f(x) = ()(5)+()( )+( 3)(0) = AOB = os ( ) = (3 sig. fig.) θ (iii) As x +, dy dx 0+ As x, dy dx 0 the grdient of every solution urve tends to be horizontl s x ±. (iv) (iii) Are of prllelogrm OAPB = OA OO 5 i j k = 3 = 3 3 i j + k = 3i 5j k 3 = 5 = ( 3) + ( 5) + ( ) = 75 = 5 3 units N.B. Finl nswer expressed in surd form s question sks for ext re. TI-8 Plus Csio fx-980g For tuition, exm ppers & Lst-Minute Buddh Foot Hugging Syndrome tretment / missloi@exmpper.om.sg febook.om/jossstikstuition twitter.om/missloi Unuthorized opying, resle or distribution prohibited. Copyright 008 ϕ exmpper.om.sg. All rights reserved. / 9

3 GCE A Level Otober/November 008 Suggested Solutions Mthemtis H (970/0) version. 5. Topi: Integrtion (i) x dx = 3 0 +(3x) 3 0 +(3x) = 3 3 dx dx = 3 [tn (3x)] 0 3 = 3 tn ( 3 3 ) tn (0) = 3 π 3 0 = π 9 e (ii) x n ln x dx = ln x xn+ n+ ILATE/LIATE Rule: Sub u = ln x (L) du dx = x Sub dv = dx xn (A) v = xn+ n + e x n+ e n+ x dx = ln e en+ e xn ln n+ n+ n+ = en+ xn+ e n+ n+ n+ = en+ n+ (n+) (en+ n+ ) = [(n + (n+) )en+ e n+ + ] = nen+ + (n+) Express in f (x) +[f(x)] f (x) + [f(x)] dx = tn [f(x)] + Integrtion by prts: u dv du dx = uu v dx dx dx n+ dx. Topi: Mlurin s Series () By Cosine Rule, os ABC = AB + BC AC (AB)(BC) os θ = + 3 AC ()(3) Mlurin s expnsion: ( + x) n = + nn + x term i.e. 3 θ & bove ignored sine θ is smll 0 AC = AC = 0 os θ 0 ( θ ) 0 + 3θ + 3θ AC + 3θ (Shown) + 3 θ [ + 3 θ + ] [ θ + ] + 3 θ =, b = 3 (b) f(x) = tn (x + π) f(0) = tn ( π) = f (x) = se (x + π) f (0) = se ( π) = f (x) = se(x + π) [ se(x + π) tn (x + π)] = 8 se (x + π) tn (x + π) f (0) = 8()() = f(x) = f(0) + xf (0) + x f (0) +! = + x() + x +! = + x + 8x + A C θ 3 B Smll Angle Approximtion: os θ θ Express in ( + x) n sine n is not positive integer d dx sen [f(x)] = n se n d [f(x)] dx se[f(x)] d dx se[f(x)] = f (x) se[f(x)] tn[f(x)] For tuition, exm ppers & Lst-Minute Buddh Foot Hugging Syndrome tretment / missloi@exmpper.om.sg febook.om/jossstikstuition twitter.om/missloi Unuthorized opying, resle or distribution prohibited. Copyright 008 ϕ exmpper.om.sg. All rights reserved. 3 / 9

4 GCE A Level Otober/November 008 Suggested Solutions Mthemtis H (970/0) version. 7. Topi: Differentition Totl time tken: 80 = (y + x)(3) + π x (9) 0 = y + x + π x (3) Totl time tken = y = 0 x 3πx length of stright prt 3 hrs/m + length of semiirulr prt 9 hrs/m y = 30 x 3πx () Totl re of retngulr & semiirulr prts: Sub y expression from () A = xy + π x = x 30 x 3πx + π x = 30x x 3πx + πx 8 = 30x x 5πx 8 When A is mximum, da = 0. dx 30 x 0πx 8 x + 0πx 8 Seond derivtive test: = 0 = 30 x = π (3 sig. fig.) d A dx Sub x.0889 into (): y = π = < 0 A is mximum when x = (3 sig. fig.) 3π(.0889) x =.09 & y =. gives the flower-bed mximum re. 8. Topi: Complex Numbers (i) z 3 = + 3i 3 Alterntive Method = + 3 3i + 3 3i + 3i 3 z = + 3i = e iπ 3 = + 3 3i 3(3) 3 3i z 3 = e iπ 3 3 = 3 e iπ 3 = 8 = 8 (ii) Given tht + 3i is root, sub z = + 3i into z 3 + z + bz + = 0: z 3 = 8 from ( 8) + + 3i + b + 3i + = i 3 + b + b 3i + = 0 + 3i + b + b 3i + = 0 + b + 3i + b 3i = 0 Compring rel prts: + b = 0 + b = b = + () Compring imginry prts: 3 + b 3 = 0 b =. () Equting () & (): + = = = 3 b = ( 3) = (iii) Sine the eqution z 3 3z + z + = 0 hs rel oeffiients, Given + 3i is root 3i is lso root. A qudrti ftor of the eqution is z + 3iz 3i = z z + Binomil Expnsion: ( + x) 3 = + 3x + 3x + x 3 Non-rel roots our in onjugte pirs in polynomil equtions with rel oeffiients. z 3 3z + z + (z z + )(z + ) = 0 z = + 3i, 3i, By inspetion or long division. For tuition, exm ppers & Lst-Minute Buddh Foot Hugging Syndrome tretment / missloi@exmpper.om.sg febook.om/jossstikstuition twitter.om/missloi Unuthorized opying, resle or distribution prohibited. Copyright 008 ϕ exmpper.om.sg. All rights reserved. / 9

5 GCE A Level Otober/November 008 Suggested Solutions Mthemtis H (970/0) version. ALTERNATE APPROACH (ii) Sine the eqution z 3 + z + bz + = 0 hs rel oeffiients, Given + 3i is root 3i is lso root. A qudrti ftor of the eqution is z + 3iz 3i = z z + z 3 + z + bz + (z z + )(Az O) where A, B R Compring oeffiients of z 3, A = Compring oeffiients of onstnt, = B B = Compring oeffiients of z, = B A = ( ) () = 3 Compring oeffiients of z, b = ( B) + A = [ ( )] + () = (iii) Sub = 3, b =, A =, B = in (ii), z 3 3z + z + z + 3iz 3i(z + ) = 0 z = + 3i, 3i, Im (z) 3 z = + 3i Non-rel roots our in onjugte pirs in polynomil equtions with rel oeffiients. 9. Topi: Grphing Tehniques (i) f(x) = x+b x+d (x + d)() (x + b)() f (x) = u (x + d) v x + d x b = (x + d) d b = 0, x d b 0 (Given) (x + d) By differentition, f (x) 0, x the grph of y = f(x) hs no turning points. (ii) f(x) = d b + x+d = + b d (x+d) = d b (x+d) When d b = 0, y = f(x) =, x d Quotient Rule: = vu uu v By long division: x + d x + b x + d d b z 3 = 0 3 Re (z) z = 3i The grph is horizontl line t y =, but undefined t x = d y (iii) y = 3x 7 x+ Sub = 3, b = 7, =, d = into f (x) in (i): dy 3() ( 7)() = = 7 > 0 dx (x+) (x+) d the grph hs positive grdient t ll points sine (x + ) > 0, x. + b y = + d, = bb x For tuition, exm ppers & Lst-Minute Buddh Foot Hugging Syndrome tretment / missloi@exmpper.om.sg febook.om/jossstikstuition twitter.om/missloi Unuthorized opying, resle or distribution prohibited. Copyright 008 ϕ exmpper.om.sg. All rights reserved. 5 / 9

6 GCE A Level Otober/November 008 Suggested Solutions Mthemtis H (970/0) version. (iv) TI-8 Plus () x = (b) x = y Csio fx-980g y = 3 (0, 7) y y = 3 7 3, 0 y = 3 y = 3x 7 x + 7 3, 0 y = 3x 7 x + x x 0. Topi: Arithmeti & Geometri Series Sum of A. P.: S n (i) Sine the student sves $3 more thn the previous month in eh subsequent month, the monthly svings = 0, 3,, 9, Arithmeti series with the st term, = 0, nd ommon differene, d = 3. = n [ + (n )d] To sve over $000 in totl, S n > 000 n [ + (n )d] > 000 Sub = 0, d = 3 into S n : n [(0) + (n )(3)] > 000 n [7 + 3n] > 000 7n + 3n > 000 3n + 7n 000 > 0 Csio fx-980g n > or n < 39. (rejet) n = 3 (erliest no. of months sine Jn 009) she will first hve sved over $000 on the st of Otober 0. (ii) () Tking into ount only the originl $0 deposit: Month Blne t Strt of Month ($) Interest Erned End of Month ($) 0 (0.0)(0) 0+(0.0)(0) = (.0)(0) (0.0)(.0)(0) 3 (.0)(0)+(0.0)(.0)(0) (0.0)(.0)(.0)(0) = (.0)(.0)(0) n (.0) n (0) (0.0)(0)(.0) n Totl ompound interest erned fter n months from the originl $0 = (0) (0.0) + (0) (0.0) [ n- ] = (0)(0.0) + (0)(0.0).0.0n.0 = (0)(0.0) +.0n = 0(.0 n ) Sum of G.P. where r > : S n = (rn ) r For tuition, exm ppers & Lst-Minute Buddh Foot Hugging Syndrome tretment / missloi@exmpper.om.sg febook.om/jossstikstuition twitter.om/missloi Unuthorized opying, resle or distribution prohibited. Copyright 008 ϕ exmpper.om.sg. All rights reserved. / 9

0 (0) +.0(0) n (.0 +.0 + +.0 n )0 = 0.0(.0n ).0 = 50(.0 n ) () Totl in ount t end of yers ( months) = 50(.0 ) Sub n = into () 30.30997 $30 (3 sig. fig.

7 GCE A Level Otober/November 008 Suggested Solutions Mthemtis H (970/0) version. totl interest erned in yers ( months) from originl $0 = 0(.0 ) $.08 (3 sig. fig.) (b) Aount blne inluding $0 deposited monthly: Month Blne t Strt of Month ($) Blne t End of Month ($) 0 (.0)(0) (.0)(0) + 0.0[.0(0) + 0] =.0 (0) +.0(0) 3.0 (0) +.0(0) + 0.0[.0 (0) +.0(0) + 0] =.0 3 (0) +.0 (0) +.0(0) n ( n )0 = 0.0(.0n ).0 = 50(.0 n ) () Totl in ount t end of yers ( months) = 50(.0 ) Sub n = into () $30 (3 sig. fig.) () Let n be the number of omplete months tken for the totl in the ount to first exeed $000. Using () from (b)(ii), 50(.0 n ) > n >.95 n ln.0 > ln.95 n > n = 8 months.. Topi: Three-Dimensionl Geometry x Let ybe the point of intersetion of p, p, p 3. Solving for p, p nd p 3, z Crtesin eqution of plne x y = = x n r n = y n = d: z z n 3 n x + n y + n 3 z = d TI-8 Plus For tuition, exm ppers & Lst-Minute Buddh Foot Hugging Syndrome tretment / missloi@exmpper.om.sg febook.om/jossstikstuition twitter.om/missloi Unuthorized opying, resle or distribution prohibited. Copyright 008 ϕ exmpper.om.sg. All rights reserved. 7 / 9

8 GCE A Level Otober/November 008 Suggested Solutions Mthemtis H (970/0) version. (i) Sine l lies on p nd p solve for p nd p, x 5y + 3z = 3 () 3x + y 5z = 5 () Crtesin eqution of () + () 5: Eliminte y b 9x 9z = 9 z = x + line + s b : () 3 () : Eliminte x 3 b 3 9y + 9z = 9 z = y + x = y + = z 3 x+ = y+ = z b b b 3 vetor eqution of l is: r = + s, s R 0 ALTERNATE APPROACH Diretion vetor of l, d l = n p n p (where n p nd n p re the norml vetors of p nd p respetively) 3 = 5 = Using the ommon point obtined erlier, vetor eqution of l: r = TI-8 Plus x z = y z = + s, s R 7 i j k d l = = i j + k = 9i + 9j + 9k d l = (ii) Given tht l lies on p 3 ll points on l lies on p 3. Piking two points on l from the vetor eqution of l in (i), 5 Let s = 0: r = λ = μ λ = μ () Let s = : r = 0 0 λ = μ 7 μ = 7. () Sub () into (): 5 λ = 7 λ = λ =, µ = 7 (iii) For the three plnes to hve no point in ommon,. p 3 must be prllel to l n p3 d l diretion vetor of l 5 λ = λ + 7 = 0 λ = b. p 3 must not ontin l l n p3 μ 5 μ 0 7 μ 7 b b = 0 l : Pik point on l obtined from (i) Given tht the three plnes hve no point in ommon, λ = nd µ n be ny rel number but not 7. n p3 d l p p p 3 For tuition, exm ppers & Lst-Minute Buddh Foot Hugging Syndrome tretment / missloi@exmpper.om.sg febook.om/jossstikstuition twitter.om/missloi Unuthorized opying, resle or distribution prohibited. Copyright 008 ϕ exmpper.om.sg. All rights reserved. 8 / 9

9 GCE A Level Otober/November 008 Suggested Solutions Mthemtis H (970/0) version. (iv) Let p nd q be two distint diretion vetors tht lie on the plne (required to get the eqution of the plne). n q (, -, 3) p l (-, -, 0) O Diretion vetor of l, d Sine plne ontins l p = l obtined in (i). Sine plne ontins the point (,, 3) q = l 3 Position vetor of point on l, = = 0 l obtined in (i) Norml vetor to the plne, n = p q = 0 = 3 3 Vetor eqution of plne: r n = 3 x 3 y = z Crtesin eqution of plne: 3x y z = Crtesin eqution of plne x n r n = y n = d: z n 3 n x + n y + n 3 z = d For tuition, exm ppers & Lst-Minute Buddh Foot Hugging Syndrome tretment / missloi@exmpper.om.sg febook.om/jossstikstuition twitter.om/missloi Unuthorized opying, resle or distribution prohibited. Copyright 008 ϕ exmpper.om.sg. All rights reserved. 9 / 9

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

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.