9740/01 October/November MATHEMATICS (H2) Paper 1 Suggested Solutions. (ii)

Size: px

Start display at page:

Download "9740/01 October/November MATHEMATICS (H2) Paper 1 Suggested Solutions. (ii)"

Sophie Davis
5 years ago
Views:

1 GCE A Level October/November 9 Suggested Solutions Mathematics H (97/) version. MATHEMATICS (H) Paper Suggested Solutions. Topic: Matrices (i) Given that u n is a quadratic polynomial in n, Let u n an + bn + c, where a, b and c are constants When u, u a + b + c When u, u a + b + c When u 5, u 9a + b + c 5 TI-8 Plus 97/ October/November 9 (ii) n 7 n + 7 > n 7n + > n 7n > () Consider: n 7n n 7±7 +()( () From (), n < 5. or n >.8 n 7 8 or or.8 Since n Z +, n is the set of values of n.. Topic: Definite Integrals By Partial Fractions: x x, B ( x)(+x) A + B x +x A(+ x) + B( x) 5..8 Casio fx-98g Using G.C., a, b 7 and c 7 u n n 7 n + 7 x, A x dx + ( x) (+x) ln +x x ln Alternatively, quote formula from formula list: dx a x a dx ln a+x a x / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 9 ϕ exampaper.com.sg. All rights reserved. /

2 GCE A Level October/November 9 Suggested Solutions Mathematics H (97/) version. p p x dx p p p (px) dx p [sin (px)] p p sin sin () Given dx x From above, ln k p. Topic: Notation (i) + n n n+ k p p p x p k ln π ln dx, n(n+) (n+)(n )+n(n ) n(n )(n+) n +n n ++n n n(n ) n n A d dx [sin f(x)] f (x) [f(x)] (ii) Using (i) and Method of Difference: n r n r r r r r n + r r r r n n n + + n n n + + n n n n n n+ + n n+ n r r r n r r r r + n n+ lim n r lim n + n r + n+ (iii) the series converges to a constant value of as n / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 9 ϕ exampaper.com.sg. All rights reserved. /

3 GCE A Level October/November 9 Suggested Solutions Mathematics H (97/) version.. Topics: Functions, Graphs, Integration Given f(x) 7 x for < x x for < x (i) f(7) f( + ) f() f(9 + ) f(9) f(5) f() f(7) f() () 5 f(5) f( + ) f() f(7) f() f() 7 f(7) + f(5) 5 + and f(x) f(x + ) Graph is a periodic curve that repeats itself at every -unit interval. x, lies in interval < x f(x) x x, lies in interval < x f(x) 7 x (ii) (iii) f( 7) f( 7 + ) f( ) f( + ) f(+) 7 f(x) dx f(x)dx f(x)dx f(x)dx area of trapezium (7 x )dx + x dx (5 + 7)() 7x x + ( + 7)() 8 + Area of trapezium to be subtracted unit / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 9 ϕ exampaper.com.sg. All rights reserved. /

4 GCE A Level October/November 9 Suggested Solutions Mathematics H (97/) version. 5. Topic: Mathematical Induction Let P n be the statement such that n r r n(n + )(n + ) where n When n, L.H.S. r r R.H.S. ()( + )( + ) n rn+ r n r r n r r (n)(n + )(n + ) n(n + )(n + ) n(n + )[8n + n ] n(n + )(7n + ) Using given equation: r n(n + )(n + ) n r ()() L.H.S. R.H.S. P is true Assume P k is true i.e. k r r + k(k + )(k + ), for some k Z To show that P k + is also true i.e. k+ r r (k + )(k + )[(k + ) + ], L.H.S. k+ r r (k + ) th term k r r + (k + ) Bring out (k + ) factor since it is found on RHS. k(k + )(k + ) + (k + ) (k + )[k(k + ) + (k + )] (k + )[k + 7k + ] (k + )(k + )(k + ) (k + )(k + )[(k + ) + ] R. H. S. i.e. P k + is true if P k is true. Since P k is true, by Mathematical Induction, P n is true for all n Z / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 9 ϕ exampaper.com.sg. All rights reserved. /

5 GCE A Level October/November 9 Suggested Solutions Mathematics H (97/) version.. Topic: Graphing Techniques Given C : y x x+ y x+ and C : x + y x + ( ) y ( ) C is an ellipse y a x h + k Vertical asymptote: x h Horizontal asymptote: y k y (x h) a + (y k) b (i) Vertical asymptotes: x Horizontal asymptotes: y C : When x, y When y, x b a (h, k) C : When x, y ± When y, x ± x (ii) Given C : y x.. () C : x + y x+ x + y... () Sub () into (), TI-8 Plus x + x x+ x (x + ) + (x ) (x + ) (x ) (x + ) x (x + ) (x ) ( x )(x + ) (Shown) (iii) Using G.C.: x.55 or x.5 ( sig. fig.) Casio fx-98g / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 9 ϕ exampaper.com.sg. All rights reserved. 5 /

6 GCE A Level October/November 9 Suggested Solutions Mathematics H (97/) version. 7. Topic: Maclaurin s Series cos x (i) Given f(x) e cos x Let y e (ii) dy f() e dx sin x ecos x f () dy dx d y y sin x dy dx sin x y cos x dx By Maclaurin s theorem: f(x) f() + f ()! e + x + e x x + f () x! f () e e ex () a+bx (a + bx ) a + b a x ( ) + b a! a x +.. b a a x +.. b a a x () Given st two non-zero terms of () st two non-zero terms of (), e ex b a a x Comparing constant terms: e a f(x) f() + f () x + f () x! + + f (n) () xn n! + Factorise a to make into general term ( + ax) n ( + ax) n + nax + n(n ) (ax) +! e e e 8. Topic: Geometric Progressions (i) Given instrument A, the bars form a G.P. series. Given T a cm Given T 5 ar 5 cm r r [ ] Total length of n bars a( rn ) r n Since r <, n as n Total length 5. < 57 cm (Shown) a e Comparing coefficients of x : e b a b ea / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 9 ϕ exampaper.com.sg. All rights reserved. /

GCE A Level October/November 9 Suggested Solutions Mathematics H (97/) version. (ii) Given instrument B consists of only 5 bars which are identical to the st 5 bars of instrument A.

7 GCE A Level October/November 9 Suggested Solutions Mathematics H (97/) version. (ii) Given instrument B consists of only 5 bars which are identical to the st 5 bars of instrument A. 5 Total length of 5 bars instrument B, L L 7 cm ( sig. fig.) Length of th bar T T n ar n 9. Topic: Complex Numbers (i) Given z 7 ( + i) z 7 ( + i) z 7 e k+ ki z e 7 k+ ki, where k,,,,,, Hence for k, z e 8 ki z 7 for k, z e 5 8 ki z z n re i(θ+kk) z r ne n (θ+kk)i (iii) Using results from 8(ii), n Given T 5 5 cm, 5 (a + l) where a is value of st term and l is value of last term L 7 cm cm (a + l) 7 () l 5 () Sub () into (), 5 (a + 5) 7 a.7 cm From (), we also have: a + d d 5 d.9 cm Value of d.9 cm Longest bar a.7 cm (exact) length of 5 th bar st bar will be longest because d is negative, hence subsequent bars will get shorter for k, z e 7 8 ki z 5 for k, z e 8 ki z for k, z e 9 8 ki z for k, z e 7 8 ki z for k, z e 5 8 ki z The roots are e 8 πi, e 5 8 πi, e 7 8 πi, e 8 πi, e 9 8 πi, e 7 8 πi and e 5 8 πi / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 9 ϕ exampaper.com.sg. All rights reserved. 7 /

8 GCE A Level October/November 9 Suggested Solutions Mathematics H (97/) version. (ii) Note z z z z z 5 z z 7 and θ 8 π π 8 7 (iii) Given z z z z () for z + i, we have z z for z + i, we have z z (, ) is one of the locus points for () Hence, the locus of all points z s.t. z z z z passes through origin. Since OA OB, OM intersects AOB arg (z ) + θ Gradient OM tan k + k 8 8 tan 5k 8 Cartesian equation of OM: y tan 5π. x 8 y tan 5π (x ) 8. Topic: Vectors p : r. Let n n p : r. Let n n (i) Let θ be the acute angle between p and p Then cos θ n.n n. n Modulus because angle is acute. θ cos ( d.p.) Scalar product of two vectors a and b: a b a b cos θ a θ b / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 9 ϕ exampaper.com.sg. All rights reserved. 8 /

9 GCE A Level October/November 9 Suggested Solutions Mathematics H (97/) version. a (ii) n n b a b a b a b a b a b a 5 b a b a b Let d be the direction vector of l. d // n n, Take d p : r. x + y + z Solve by simultaneous equations by letting x. p : r. x + y + z Using G.C., the point of intersection of p and p is Vector equation of line l: r + λ, λ R () ALTERNATE APPROACH Using rref, TI-8 Plus TI-8 Plus Casio fx-98g Casio fx-98g x + z x z y + z y z x Vector equation of l: r y z z z z + z + λ where λ is a parameter / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 9 ϕ exampaper.com.sg. All rights reserved. 9 /

10 GCE A Level October/November 9 Suggested Solutions Mathematics H (97/) version. (iii) Given p : x + y + z + k( x + y + z ) ( k)x + (k + )y + (k + )z k + x Let r y, k p : r. k + k + () z k + k Let n k + k + k From (), we know lies on line l and. k + k + k + also lies on p. k d n k + k + k + k + (k + ) d n d // plane p Two non-zero vectors a and b are if a b a b cos 9 Since lies in p and d // p, l lies in p for any k. ALTERNATE APPROACH x + y + z + k( x + y + z ) () sub x z and y z into L.H.S. of (), L.H.S. ( z) + ( z) + z + k[z + ( z) + z ] R.H.S. l lies in p and is independent of k. k Given lies on p then. k + k + k k + 5 From (), p : r 5 5 d 5x 5y 5 x y, z R d. Topic: Graphs, Integration by Substitution, Applications of Integration (Volume) Given y f(x) where f(x) xe x (i) TI-8 Plus Casio fx-98g p p l / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 9 ϕ exampaper.com.sg. All rights reserved. /

11 GCE A Level October/November 9 Suggested Solutions Mathematics H (97/) version. (ii) dy dx y xe x e x + xe x ( x) e x [ x ] When dy dx, e x [ x ] Since e x >, x From above graph, e x is always +ve x When x, y e When x, y e x ± The coordinates of the turning points are, e and, e. e n + e n the area of the region between the curve and the positive x-axis n lim n f(x) dx ( ) e n as n unit (iv) Using (i): f(x) dx f(x) dx ( e ) e (v) Volume about x-axis π y dx π x e x dx π(.578).8. unit ( sig. fig.) TI-8 Plus Casio fx-98g (iii) Use u x, du x dx When x, u and when x n, u n n f(x) dx n xe x dx n xe x dx n e u du [ e u ] n / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 9 ϕ exampaper.com.sg. All rights reserved. /

4038/02 October/November 2008

4038/02 October/November 2008 GCE O Level October/November 008 Suggested Solutions Additional Mathematics (408/0) version 1.1 ADDITIONAL MATHEMATICS Paper Suggested Solutions 1. Topic: Exponential Functions (i) Given: V =10000e -pt