NYJC_9758_2025_prelim_P2_Solution

2025 NYJC J2 H2 Mathematics Preliminary Exam 9758/2 Marking Guide 1 of 13 Q1 Suggested Answers (a) Method 1: From GC, set required is  : 2 8 .xx   Method 2: ( ) ( ) ( )( ) 22 22 2 2 0 3 12 9 3 1 2 9 54 81 1 4 4 5 50 80 0 10 16 0 28 3 x xx x x x x xx xx xx x − −  − − +  − + − +  −+ −− −   The set required is  : 2 8 .xx   (b) ( ) ( )( ) ( )( ) ( ) ( )( ) ( )( ) ( )( ) 2 2 2 2 7 or 0 2 0 3 5 3 5 0 7 or 3 5 3 5 2 18 15 14 18 5 14 072 64 72 35 72 7 or 2 x xx xx xx x x x xxx xx xx x xx xx x + −+− + + + − +−   −   + − + +   ++ − +− +− +− + − −  + − − Q2 Suggested Answers (a) 2 2 3 3 3 2 i and i where , : i i = 0 50 i 5i 0 Replacing i 5i x b x x x x x a b a x x x       −+ + = − + + = = − + +  − − = Since the coefficients of the equation are all real, the complex roots occur in conjugate pair. (b) ( ) ( ) ( ) 3 2 * * 32 i 5i 0 Since is purely imaginary, : i 5i 0 2 i, 2 i and 1 2 i, 2 i and 1 x x a x b a a a x x a x b x x + + + = =− − − + − + − + = − − = − + = + − − −

2025 NYJC J2 H2 Mathematics Preliminary Exam 9758/2 Marking Guide 2 of 13 Q3 Suggested Answers (a) The width of each of the n rectangles is 2 n . The height of the first rectangle is 2f1 n + , second rectangle is 4f1 n + , …and the n rectangle is 2f1 n n + . The area of the n rectangles is 2 2 4 2f 1 f 1 ... f 1 n n n n n      + + + + + +           . When n→ , the area of the n rectangles tends towards the area under the curve of ( )fyx= from 1x= to 3x= which is ( ) 3 1 f dxx . (b)   ( )       ( ) 1 3 1 33 1 1 33 1 1 33 11 22lim ln 1 ln d 1ln d ln 1 d ln 3ln 3 3 1 3ln 3 2 n n k k nn xx x x x x x x x x x x x → =  + = =− =− =− = − − =−    

2025 NYJC J2 H2 Mathematics Preliminary Exam 9758/2 Marking Guide 3 of 13 Q4 Suggested Answers (a) Differentiating wrt x : 2 22 2 d d d 33 6 0 d d d 6 y y y y xx x y y x x x x y −−+ + + =  = + (b) If tangent is parallel to y-axis, 226 0, ie, 6x y x y+ = =− Point of contact of tangent with C is given by ( ) ( ) 32 2 36 6 2y y y y k− + − + = ie, 6 3 3 6 3216 6 2 216 4 0y y y k y y k− − + =  + + = (shown) For y to be real, 2discriminant 0, ie, 4 4(216)( ) 0 k −  414 216 , ie, 216 54k k k    (c) When 26, 6 6 1x y y=− − =−  = Therefore, 220 or 212k =− −

2025 NYJC J2 H2 Mathematics Preliminary Exam 9758/2 Marking Guide 4 of 13 Q5 Suggested Answers (a) Let  be the acute angle between 1p and 2p . Then 44 13 11 18 3cos 44 18 26 13 13 11                −   = = =               −    . Therefore 2sin 13  = . (b) Since ( )4 1 0 4 8+ + = and ( ) ( )4 1 3 0 4 0+ − = , the point A ( )1 ,0,4 lies on 1p and 2p . 4 4 4 1 1 3 8 4 2 1 1 8 2 −−                = =              −        . The vector equation of the l