NYJC H2 MATH P1 Answer

2015 NYJC JC2 Prelim Exam 9740/1 Solutions Page 1 of 11 Qn Solutions 1 2 2 2 2 2 2 2 4 24 4 024 ( 2 4) 4 024 ( 2 ) 0( 1) 5 ( 2) 0 ( 1 5)( 1 5) x xxx x xxx x x x x xx x x x x xx xx                1 5 2 or 1 5 and 0x x x       2 (i)   2 32 d4 d 1 x t t     2 d2 d 1 x At t    2 1x At Bt   When 0t , 3x 32 B  1B Hence 2 11x Att   (ii) 2 x 0 1 5 1 5 + + + − − x 1x At , 0A  0,3 1x , 0A t 1x At , 0A

2015 NYJC JC2 Prelim Exam 9740/1 Solutions Page 2 of 11 Qn Solutions 3(i) sin c cos os OA AP r Ox r Qr       By Pythagoras Theorem,   2 2 2 2 2 2 2 2 2 2 2 sin ( cos ) cos 1 16 (since )6 sin cos 0 cos 16 sin AQ PQ r x r x r r r xr AP r xr                     Alternative solution Using Cosine rule,   2 224 2 cosr r x rx    22 2 cos 15x rx r    2 2 2 2cos cos 15x r r r      2 2 2 2cos 15 cosx r r r        2 2 2 2cos 15 1 sinx r r r         2 22cos 16 sinx r r    2cos 16 sinx r r    (Since cos 0xr   )   2cos 16 sinxr     3(ii)     1 2 2 2 d d d d d d 1dsin 16 sin 2sin cos2d cos dsin 1 d16 sin xx tt r t r t                     O P Q A  r 4r

2015 NYJC JC2 Prelim Exam 9740/1 Solutions Page 3 of 11 Qn Solutions Given that d 0.5 rad/sdt   and when 2 3   , 1 d 3 1 21 0.378 cm/sd 2 2 316 4 x rrt             P is moving towards O at a rate of 0.378 cm/sr . 4 (a) a b b a       a b b a 0 2  a b 0 ab is the zero vector. (a) Since the vector perpendicular to both a ( OA ) and b ( OB ) is also perpendicular to c OC , ( ).a b c 0 implies that the four points are coplanar. Vector normal to plane ABC = 5 2 1   Equation of plane ABC is 5 0 5 2 1 2 0 1 2 1                               r Cartesian equation is 5 2 0x y z   Foot of perpendicular of P to plane ABC is given by 55 22 511 1 2230 30 1 OP FP OP                        nn 2 2 5 3 1 2 1 3 1 2 OQ OP PQ OP PF                                 

2015 NYJC JC2 Prelim Exam 9740/1 Solutions Page 4 of 11 Qn Solutions 5(i) 11f ( 2) 2 4 1 2 3r rr            1 1 1 3 4f ( 2) f ( ) 2 3 2 1 2 3 2 1 2 3 2 1rr r r r r r r            5(ii)   1 2 1 1 1 1 f ( 2) f ( )1 5 3 7 5 9 4 n r rr                                    f 0 f 2 f 1 f 3 f 2 f 4 f 3 f 1 f 2 f f 1 f 1 1 4 nn nn nn                    1 f (0) f (1) f ( ) f ( 1)4 nn     1 1 1 3 2 1 2 1 14 3nn           1 1 2 3 2 1 3 4 2 1 2 3 nn nn          11 3 2 1 2 3 n nn   5(iii) As ,n    1 2 1 2 3 0n nn    , therefore the series converges. The sum to infinity = 1 3 5(iv) 2 2 2 3 3 3 3 3 3 1 311 5 3 7 5 9 35 7 9              6(i)   222R h R r   2 2 2 2 2R h hR R r    22 2 (shown)r hR h  

2015 NYJC JC2 Prelim Exam 9740/1 Solutions Page 5 of 11 Qn Solutions 6(ii) Let V be the volume of the cone. 21 3V r h    2 2 31 2 123 3 3 Rh h h Rh h      2d4 d3 V Rh hh  For max. volume, set d 0.d V h  2d4 0d3 V Rh hh    4 03h R h  40 (rejected) or 3h h R When 4 3hR , 2 2 d4 2d3 4 4 4 2 03 3 3 V Rhh R R R           The volume of the cone is a maximum when 4 3hR . 6(iii) 22 2 2 2 4 4 8 =2 = 3 3 9 r hR h R R R R            22 ( 0) 3r R r   Ratio 22 3 2 4 2 3 R r h R     7(i)      f " 0 f ''' 0f ' 0 2, and 2! 3!ab     2 2 dd1 2 2 0 dd yyx xx    ---(1) When 0x , from (1), we have

2015 NYJC JC2 Prelim Exam 9740/1 Solutions Page 6 of 11 Qn Solutions    f " 0 2f ' 0 0    f " 0 2f ' 0 2(2) 4    4 2 (shown) 2!a   Differentiate (1) w.r.t. x:   3 2 2 3 2 2 d d d1 2 2 2 0 d d d yyyx x x x       32 32 dd1 2 4 0 dd yyx xx    ---(2) When 0x , from (1), we have    f "' 0 4f '' 0 0    f "' 0 4f '' 0 4( 4) 16     16 8 3! 3b   7(ii) 23 3 822 3 8 x x x x     1 23 3 82 2 8 3x x x x       1 1 323 3 82 2 8 1 38 xx x x               2 23 14 1 8 1 332 2 1 ...2 3 3 8 2! 8 xxx x x                                    2 3 2 4 1 1 1 ...3 24 288x x x x x            2 3 2 3 31 1 1 4 24 288 24 3x x x x x x      2325 397 24 288x x x   8 (i)  6 6 3 2 2 12xx    3A , 12B

2015 NYJC JC2 Prelim Exam 9740/1 Solutions Page 7 of 11 Qn Solutions     11 22 00 11 22 200 1 21 0 11 1 3 2 2 1266 dd2 5 2 5 2 2 13 d 12 d 25 12 13ln 2 5 6 tan 2 13ln 8 6 tan 1 3ln 5 6 tan 2 813 ln 2 tan5 2 2 xx xxx x x x x xxxx x xxx                               (ii) Required area 1 21 66 d25 x xxx   1 20 1 662d 25 816 ln 2 tan5 2 2 x xxx        9(a) (i) 5x ( 21x ) = 4096    4 20 4096ar ar   2 24 4096 (1)ar  12 4096 64ar  25 4 1 log k k x   = 4 1 4 2 4 3 4 25log log log ... logx x x x    =  4 1 2 3 25log ...x x x x =       2 24 4log ...a ar ar ar =   25 1 2 ... 24 4log ar    =  25 24 25 2 4log ar   =   25 25 12 4log (2)ar  1 –1 2 66 25 xy xx  

2015 NYJC JC2 Prelim Exam 9740/1 Solutions Page 8 of 11 Qn Solutions     2512 12 44log 25logar ar    4 3 425log 64 25log 4 25 3 75   (ii) 1 4 4 1 4 1 log ( ) log ( ) log n n n n n n xxy x xy      1 4 2log n n ar ar   4log r is a constant free from n. Hence, { ny } is an arithmetic sequence. (b) Let the amount of oil mined in the first year be a. 23 The maximum total amount of oil mined 0.94 0.94 0.94 ....    a a a a 16.66 171 0.94 a aa Let n be the number of year at which the mine will be in operation. (1 0.94 ) 161 0.94 na a  0.94 0.04n lg 0.04 52.02lg 0.94n   Smallest n is 53. (1 0.94 ) 161 0.94 n  n (1 0.94 ) 1 0.94 n  52 15.999096 53 16.03915 The mine will be closed in the 53th year. Therefore, the mine will be closed in 2049. 10 x y O yx yx x ( 1,0) x (1,0) x

2015 NYJC JC2 Prelim Exam 9740/1 Solutions Page 9 of 11 Qn Solutions (i)(a) (i)(b) (ii)     by 2 1 2 2 2 2 2 2 2 2 2 by 2 1 2 by 6 1 2 2 1 ( 2) 1 ( 2) 1 ( 6)( 2) 1 x y y T S T x y x y yx yx                  11(i) 8 16 324 1 2 3 4 , , , i i ii w w w we e e e        11(ii) Note that 1 4   and 1a 1rg ) arg( )2( nnw w  . Thus 1 1 2 nn  . Since 1 1 2 n n     for all 1n , thus n is a geometric sequence with common ratio 1 2 . 1 1