HCI H2 MATH P1 ANSWERS

2015 H2 Mathematics C2 Prelim Paper 1 Solutions Qn Solutions 1(a) Let 21ux 11(1 ) 21 d 1 d 22 ux xx u u     11 1 d22 2 u uu  = 1 1 d4 uu u = 31 221 d4 uu u = 53 2212 2 45 3 uu C    = 53 2211 10 6uu C =   53 22 11 21 2110 6 x xC   1(b) Using MF15 32 PQ x  2 PQ x  6PQ x 2PQ x 28Px => 4Px 2Qx 1sin 3 sin d cos 4 cos 2 d2x xx x x x   1 sin 4 sin 2 24 2 xx C   sin 4 sin 2 84 xx C   2       fl n 1l n 1 c o s , f 0 l n 2x x x    1s i nf' ( ) , f' ( 0 ) 111 c o s xx x x          22 1c o s c o s s i n s i n1f ''( ) + 11 c o s x xx xx xx      22 22 1c o s c o ss i nf' ' ( ) + 11 c o s x xxx xx     2 11 1f ''( ) + , f '' 0 1c o s 21 x xx      32 2s i nf '''( ) + , f ''' 0 2 11 c o s xx xx    Using series formula in MF15:      23 f f 0 f '0 f ' '0 f ' ' '0 . . .2! 3! xxx x     2311f ln 2 ... 43x x x x        11c o sln ln 11 c o s xx x x        2311ln 2 ... 43xx x   

3 Let 1FGh and let 2BC h . Form equation of the ellipse: 22 22 12000 xy a   --- (1) Since the areas of ABCD and EFGH are equal: 211000 1435hh --------- (2) Substitute the point 1 1435,2Gh  into (1): 22 1 22 717.5 12000 h a  --------- (3) Substitute the point 2 1000,2Ch  into (1): 22 2 22 500 12000 h a  --------- (4) Substitute (2) into (4):  22 1 22 1.435500 12000 h a   --- (5) From (3) and (5): 2 764806.25 874.532aa   2 1749.06 1749 mmMN a  4(i) 3sloped length sin  31 3 9total area 8 3 3 24 9cosec cotsin 2 tan 2             4(ii)  2 2 924 9cosec cot 2 d9 9cot cosec cosecd2 92 c o s 1 2 sin A A          Let d 0d A   , 2cos 1  . So, 3   . 91 8 9 9 324 9cosec cot 24 24 16.2 3 s.f.32 3 2 32 3 A        3     3   3     d d A  + 0 – The maximum area is 16.2 m2 .

5(i) Since a line such as 2y  intersects the curve  fyx at 2 points, f is not one-to- one. 1 f does not exist. 5(ii) Largest possible domain for h is [3, ) . 5(iii)  2 ln 2 2yx  2 22e y x   Since 3,x 2 1 2h: 2e , , 2 x xx x       5(iv)   11hh h hx xx  which is always valid since the inverse function exists. Since domain of 1hh  is domain of  1hD2 ,    and domain of 1hh  is domain of  hD3 , . Thus the set of values is  :3xx  . 6(i) As t , sinx t and cosyt . Using trigonometric identity 22sin cos 1tt  , the Cartesian equation of C is 22 1xy . Thus the shape of C is a circle with centre at the origin and with unit length radius. 6(ii) d ec o sd tx tt  , d es i nd ty tt  . de s i n de c o s t t y t x t    At P, gradient of normal is ec o s es i n       .  equation of normal is   ec o sec o s es i n es i nyx            ec o s es i n ec o ses i nyx             ec o s 2ees i nyx          6(iii) Using equation of normal found in (ii), point D is  0, 2e . E is a point on C. From ec o styt , when 0y , ec o st t By inspection, 0t . 0es i n 0 1x   . Hence point E is  1, 0 . 6(iv) The mid-point of DE is 1 ,e2   . Since 1 2x is a fixed value and 0e 1 y   , required locus is a half-line 1 2x , with 1y . y x x = 2 y = 2

7(i) 000 033 0 OC OB BC hh          7(ii) By ratio theorem 2 3 OC OBOP     00 0 1 60 23 2hh h                   7(iii) Select a suitable direction vector parallel to the plane such as BE OE OB      20 2 30 3 0 hh hh                    . Thus  0BE a b ik  2 30 0 ha hb             1 2 a b Since C is on the plane, 01 30 2 2 h h        

1 02 22 2 hxzh      r 7(iv) Given that 3h , 0 2 3 OP    , 6 0 0 OF     n 06 0 0 20 1 86 3 30 1 2 2                    The equation of plane OPF is 0 30 2    r Shortest distance = 222 10 23 22 032              

2 13  units.

8(i)  d1 4000 d 1 dd4 0 0 0 11 d1 d4000 1 4000 4000ln e 4000 e where e 4000e kt kC kt kC kt A kA A ttk A Atk A k At C Akk k A k A k                      Sub   , 0,60000tA  : 4000 400060000 60000 kk   ---- (1) Sub   , 3,69500tA  : 3 400069500 e k k ---- (2) Sub (1) into (2): 34000 400069500 60000 e k kk   Using G.C.,  0.1111343 0.111 3 s.f.k  4000 4000Thus 60000 e ktA kk   0.11124000e 36000tA    with 24000 3 s.f. and 36000 3 s.f. 8(ii) 2 2 dSince , 2 d d d d 4000 4000Thus dd d 2 2 AAr r r rA A k A k r tt r r r       Sub 200r ,      2 0.1111343 200 4000d 7.93 3 s.f.d 2 200 r t    8(iii) Let the rate Mac needs to cut the weeds be n m2 per month.       d 0.1111343d 0 0.1111343 69500 7720 3 s.f. A Ant nn     8(iv) d 0d A t  means that the rate which Mac needs to cut the weeds is equal to the rate the weeds grow. Thus, the area covered in weeds is unchanged. 9(i) Amount after 16 days = 2 11000 250 mg2 

9(ii) Amount of I-131 on Day 49  222 23 1111000 1000 1000 1000 *222 11 11000 1 44 4                        4 11 41000 1328.125 m 1328 mg nearest1 mgg 1 4        9(iii) 1000 1333.33 1334 mg11 4 S    Amount of I-131 will never exceed 1334 mg. 9(iv) Amount of I-125 on Day 121 = 2 11000 250 mg2  I-131 is added on Day 17, …, 113,  total 7 times Amount of I-131 on Day 121 7 26 1111 1 1 41000 1 500 666.626 mg 144 4 2 1 4                   Total amount of radioisotopes  250 666.626 917 mg nearest mg  10(a) (i) It is not necessarily true because to conclude that *i is a root, the coefficients of the equation must be real. 10(a) (ii) Sub iw into 32 24 i 0za z a z  32ii 2 i 4 i 0aa  i2 i 4 i 0aa    (1 2i) 5ia 2ia 10(a) (iii) Let  2 i0bz cz d z  By inspection, 1b  4d  ,  2 4i 0zc z z  Compare z terms: i4 2ca 42 i4 2ic    2Thus 4 0zc z  24 4 ( 1 ) ( 4 ) 2z  121 2  13 i 13 iz or 13 iz

10(b) arg 2i 4z  2tan 4 y x   2yx where 2y  , 0x  --------- (1) From *1 i 2z  i1 i2xy    1i 1 2xy    22 11 4xy --------- (2) Sub (1) into (2): 22 21 21 4xx xx   222x 1x  Since 0x , therefore 1x i1 3 izxy   11(a) (i) 11(a) (ii) 11(a) (iii) 11(b) (i) Since x = –2 is the asymptote, (–2)2 – b = 0  b = 4 Substitute the point 14, 4  into G, 18 5.41 6 4 a a   

11(b) (ii) 11(b) (iii) From G. C For incre Use long Thus 5x Sketch th intersecti C, we find t easing and c division to 2 61 52 4 x x    he equation ions = 3. Th (–1, –1 x = –2 he maximum oncave dow obtain 35x 2 5 204 x x of the line hus there are y 1) um point  wnwards, th 2 2 21 4 4 x x x    51 543   1 53yx  e 3 distinct x 1, 1 . he only rang 7 5 2x x   2x 2 onto th real roots to ge is 2 x  2 61 52 4 x x   he diagram t o the equatio 1 . to obtain nu on. umber of