RVHS 9758 2025 Prelim P2 Solutions

River Valley High School 1 2025 H2MA Prelim Paper 2 1 Solution [5] Sequences (a) (i) For 4N = , the sequence takes on a constant value with 4nu = for all 1n . (a) (ii) For 3N = , 1 3u = 2 3(3) 8 1u = − = 3 3( 5) 8 23u = − − =− … The sequence decreases and approaches negative infinity. (b) Method 1 5 409u = Using 1 38nnuu+ =− , 443 8 409 139uu− =  = 333 8 139 49uu− =  = 223 8 49 19uu− =  = 113 8 19 9u u N− =  = = Method 2 1 2 3 4 5 38 9 32 27 104 81 320 uN uN uN uN uN = =− =− =− =− 81 320 409 9 N N −= = (c) From part (a), when 1 4u = , 4nu = for 1n ( ) ( )1 n nnv u k= − + ( )1 4vk=− + ( )2 4vk=+ ( )3 4vk=− +

River Valley High School 2 For ( ) ( )1 n nnv u k= − + to be convergent, then 4k =− .

River Valley High School 3 2 Solution [5] Complex Numbers (a) The line segment AC can be obtained by rotating the line segment AB , 90 in a clockwise direction about .A ( )iC A B Az z z z− =− − ( )iC A B Az z z z= − − ( ) ( ) ( ) ( )   5 6i i 9 3i 5 6i 5 6i i 4 3i 2 2i Cz = + − + − + = + − − =+ (b) Method 1 ABDC is a parallelogram Midpoint of AD = Midpoint of BC 22 BCAD zzzz ++ = ( ) ( ) ( )5 6i 9 3i 2 2iDz+ + = + + + ( )11 5i 5 6iDz = + − + 6iDz =− Method 2 BD AC= D B C Az z z z− = − D B C Az z z z= + − 6iDz =− Method 3 BD can be obtained by rotating the line segment BA , 90 in an anti-clockwise direction about B.

River Valley High School 4 ( ) ( ) ( ) ( ) i i i 5 6i 9 3i 9 3i i 3i 4 9 3i 9 3i i 3i 4 6i BA BD B A D B D D D D zz z z z z z z z z = − = − + − − = − − − = − − = + + − =−

River Valley High School 5 3 Solution [8] Volume of revolution (a) Volume required 25 2 12 3 π 1 d π 5 d 1.95 units y y y y= − + − =  (b) x-coordinate of point of intersection between 2 2 47 y xx = −+ and 2yx=− is 3x= (from GC)

River Valley High School 6 Volume required ( ) ( ) ( ) ( ) ( ) ( ) 2 33 2 212 2 3 2 21 33 2 212 333 2 21 2 333 2 21 2 2 3 1 1 2π d π 2 d 47 21OR π d π 1 1 347 4π d π 4 4 d47 14π d π 2 4 32 4 7 14π d π 2 4 323 1 2 14π tan π 333 x x x xx x xx x x x xxx xx x x x xx x x x x− = − − −+  − −+ = − − +−+ = − − + − − +  = − − + −+  −=−        11 23 4 1 1 1π tan tan π 33 3 3 4 π π 1ππ 6 6 33 4 π1ππ 333 41 3π π units93 41, 93ab −− −   = − −        = + − =−  =− = =−

River Valley High School 7 4 Solution [10] AP GP (a) Let un be the dose on the nth day. ( ) ( ) ( )( )( ) ( ) 2 2 2 0.75 0.5 1 2 0.75 1 0.52 10 1 -----(*)22 11 1042 2 40 1 41 1 41 or 1 41 (rej as no. of days cannot be -ve) 5.40312 n n un nSn nn nn nn n nn n = + − = + − +  + + + − + − −  Therefore, it takes 6 days. Alternatively, When 5n= , 211 8.75 1042nn+ =  When 6n= , 211 12 1042nn+ =  Therefore, it takes 6 days. ( )6 0.75 0.5 5 3.25u = + = The dose on the 6th day is 3.25 units.

River Valley High School 8 (b) (i) ( ) ( ) 1 11 12 11 12 1 1 1 1 2 1 2 1 Increase on day 3 5 4 3 5 4 20 5 20 5 3 5 4 5 4 20 5 5 3 1 4 5 4 5 420 5 5 3 1 1 420 5 5 34 20 5 th nn n n n n nn n n n n nn n n n n n n n n n SS − −− −− −− −− − − − − − − − − =−    −−=−         −−=−    = − − +       =      =  which is of the form arn-1, where 3 20a= and 4 5r = which is a GP. Alternative: Consider 11 1 1 34 420 5 534 20 5 n n n n u u +− + − ==  which is a constant Since consecutive terms have a common ratio 4 5r = , the increase in dosage follows a GP. (b) (ii) Method 1 1 Long term dosage 3 5 4lim 20 5 320 41 5 3 4 nn nn −→  −=  = − = In the long term, the total dosage approaches 0.75 units. Note: After n days, the total dosage of the medicine is given by

River Valley High School 9 1 3 5 4 20 5 nn n−  −  . Let nS denote the total dosage of the new medicine after n days. Then 1 3 5 4 20 5 nn n nS −  −=  . Let nv the change in dosage on nth day found in part (b)(i) where 1n n nv S S −=− Since nv is a geometric progression, we can use 1 aS r  = − , the sum to infinity formula to find the long term total dosage. Method 2 ( ) ( ) 1 3 5 4 20 5 3 5 4520 5 34 5120 5 nn n nn n n −  −   −=   =−    As n→ , 4 05 n  → Total dosage in the long term ( )( )3 5 1 020 3 4 =− = (c) If the dosage increases as an AP, then it will continue to increase to (positive) infinity.

River Valley High School 10 5 Solution [12] 3D Vectors (a) Since 12 1 3 2 3 5 0 51 −       − • =− − + =          , the satellite’s path is parallel to Earth’s Orbital plane. Since 31 2 1 3 2 15 16 0 35        • − = − + =           , the satellite does not sit on Earth’s Orbital plane. Therefore, the satellite does not cross Earth’s Orbital Plane. Alternatively Earth: 1 1 0 (1) 5  • − = −−− r Satellite: 32 2 3 , (2) 31  −       = +  −−−          r Sub (2) into (1): 3 2 1 2 3 1 0 (3) 3 1 5   −            + • − = −−−                16 0 0+= There is no consistent value of  for eqn (3) Therefore, the satellite does not cross Earth’s Orbital Plane.