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 line of intersection of 1p and 2p is 11 0 2 , 42 r  −       = +           .

2025 NYJC J2 H2 Mathematics Preliminary Exam 9758/2 Marking Guide 5 of 13 (c) 1 0 1 2 3 2 6 6 0 2 3 2 AB −−           = = − =          −      So AB is perpendicular to l. ( ) 22 0 3 3 3 3 2 3 AB  = = + − =− . By (a), perpendicular distance from B to 2p sin 232 13 AB = =  B(1,3,1) 12 26 = A(1,0,4)  2p (d) Since 3p is perpendicular to both 1p and 2p , the normal vectors of 1p and 2p are both parallel to 3p . From (b), we infer that a normal vector of 3p is 1 2 2 −   . Equation of 3p is therefore 1 1 1 2 3 2 1 6 2 7 2 1 2 r −−           = =− + + =                which in cartesian form is 227x y z− + + = .

2025 NYJC J2 H2 Mathematics Preliminary Exam 9758/2 Marking Guide 6 of 13 Q6 Suggested Answers (a)(i) If two scores are equal (Event C), suppose x and x, sum of two scores will be 2x which is even (which is thus not odd). A and C cannot occur concurrently. So Event A and C are mutually exclusive (a)(ii) A and B are independent. P(A) = P(1 odd and 1 even) = 1 22 112( ( ))2 = P(B) = P( at least 1 of scores >4) = P(1st die =5,6) +P(2nd die = 5,6) – P(1st and 2nd die =5 or 6) = 21 1 1 5 ()3 3 3 9+ =− 8 10P 5 or or or 3( ) 2P(5, 2 4 6) 2P(6 11 3) 6,AB = = = + Alternatively, by using table of outcomes: 1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12 P(B) = 20 36 5 9= 8 10P 3) 1( 5 6AB == (Look for number of ) P(A)P(B) = 1 5 5( ) P( )2 9 18 AB= =  Thus A and B are independent. (b) 0 P( | ) P( ) () P(5,5) P(6,6) P( ) 2 36 5 9 1 1 CB CB PB B = = = += 

2025 NYJC J2 H2 Mathematics Preliminary Exam 9758/2 Marking Guide 7 of 13 Q7 Suggested Answers (a) s =0,1,2,3,4,5 ; f = 5,4,3,2,1,0 d = 5,3,1 D can take values 1, 3 or 5 (b) 1~ B 5, 3S   1 P( 3 P or ( 1) 40 8 2)SS D = = = == (c) d 1 3 5 P(D=d) 40/81 10/27 11/81 E(D2) = ( ) 2Pd D d = = 65/9. (d) 1B 5, 3S   E(S) = 155 33= , Var(S) = 1 2 105 3 3 9  = ( ) 222 10 5 35E Var( ) (E( )) 9 3 9()S S S= + = = + (e) 222 (5 ) 4 20 25D S S S S= − − − + = 22 2 ) E( ) E(4 20 25) 4E( ) 20E( ) 25 35 54( ) 20 25 65 (verified9 93 D S S SS = − + = − + = − +  =

2025 NYJC J2 H2 Mathematics Preliminary Exam 9758/2 Marking Guide 8 of 13 Q8 Suggested Answers (a) 880. 4 .75 11xx n= = = ( ) 22 1 1s n xx=− −  (24.29) 0.5520 ) 1 44 (5sf ) (3s0.552 f 5= = = (b) Let  be the population mean mass of adult trout. To test 0 1 H: H: w w   = at 10% level of significance Under 0H , since n = 45 is large, by the Central Limit Theorem, 2 ~ N , 45X sw  approximately ) / 45 ~ N(0,1wZ X s −= approximately Reject 0H if calc calc 1.6448 or 1.6448zz −  (5 s.f.) calc 45 1.78 s wz −= Since 0H is not rejected, 4 0.55 c 205 cal 5 45 1.6448 1.6448 1.781.6448 1.6448 1.781.6448 1.6448 0.18218 1.78 0.18218 1.5978 1.9622 1.60 1.96 s z w w w w w −   −−   −−   −  −    { 1.60 }: 1.96ww  (c) Since the sample size (n=45) is large, we can approximate the sample mean mass of adult trout with a normal distribution using Central limit theorem.

2025 NYJC J2 H2 Mathematics Preliminary Exam 9758/2 Marking Guide 9 of 13 Q9 Suggested Answers (a) The probability that a plate being faulty is constant. The event that a plate is faulty is independent of the other plates (being faulty). (b) Let X be the number of faulty plates out of 100 plates. ~ B(100, )Xp P(X = 3) 3 100 3100 (1 )3 pp −=− = 3 97161700 (1 ) .pp − (c) Since X = 3 is the mode P(X = 3)> P(X = 2) 3 97 2 98161700 (1 ) 4950 (1 ) 98 13 101 13 3 101 p p p p pp p p −  − −   and P(X = 3)> P(X = 4) 3 97 4 96161700 (1 ) 3921225 (1 ) 97 14 101 14 4 101 p p p p pp p p −  − −   Hence 34 101 101p (d) P(no faulty items)+P(1 faulty plate) +P(1 faulty bowl) = 0.88 2 2 2 2 2 (0.99) (1 ) 2(0.99)(0.01)(1 ) (0.99) (2 )(1 ) 0.8 8 0.9999(1 ) (1.9602) (1 ) 0.88 Using GC, since 0 1, 0.33333 0.333 q q q q q q q q q − + − + − = − + − =  =

2025 NYJC J2 H2 Mathematics Preliminary Exam 9758/2 Marking Guide 10 of 13 Q10 Suggested Answers (a) By GC, 42.8a = and 485 10 kt += . Since the point ( ),at lies on the line 0.8957914 14.16013ta=+ , ( )485 0.8957914 42.8 14.1601310 40 k k + =+ = (b) 21 77 a (c) Time spent by a 60-year old ( )0.8957914 60 14.16013 67.9 + = 68 minutes. (d) By GC, the sum of residuals of the least squares regression line of t on a = 352.1937458