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