NYJC 2026 Prelim P1 Ans

2026 NYJC J2 H2 Mathematics Preliminary Examination 9758/1 solutions 1 of 15 Q1 Suggested Answers Substitute (3, 7): 7 (3) 31 137 2 cab ab c = ++ − ⇒ ++ = 2 2 d d ( 1) At 3, 0(3 1) 1 04 yc axx cxa ac = − − = −= − ⇒− = 3 2 3 2 2 d1 ln | 1| 4.5 ln162 cax b x x a x bx c x  ++ − = ++ − = + ⌠⌡ ( )9 3 ln 2 2 2 ln1 4.5 4 ln 22 a bc a bc ++ − ++ = + 5 ln2 4.5 4ln22 abc++ = + 13 7 (1)2ab c++ = − 1 0 (2)4ac−= − 5 ln2 4.5 4ln2 (3)2 abc++ = + − Solving (1), (2), (3) simultaneously, 1, 2, 4ab c= = =

2026 NYJC J2 H2 Mathematics Preliminary Examination 9758/1 solutions 2 of 15 Q2 Suggested Answers (a) 2 2 51 93 51 093 x xx x xx +≤−+ +−≤−+ 5 ( 1)( 3) 0 (*)( 3)( 3)xxxx−+ − ≤−+ 2 2 28 0( 3)( 3) 28 0( 3)( 3) ( 4)( 2) 0( 3)( 3) xx xx xx xx xx xx −++ ≤−+ −− ≥−+ −+⇒≥ −+ + − + − + ̶ 3 2− 3 4 3x<− or 23 x−≤< or 4x≥ (b) 2 5 e1 e 9e 3 x xx +≤−+ Replace with exx , Method 1: Algebraic method e3x <− or 2e 3 x−≤ < or e4x ≥ Since e0x > for all real x, e3x <− has no solution. 0e 3 ln3 2e 3 x x x − ⇒< < < ≤ < ⇒ or 4 e l 4 n x x ≥ ⇒≥ Hence ln3x< or ln4x≥ Method 2: Graphical Method e3x <− ⇒ No solution y (0,1) x O y= −3 y=0

2026 NYJC J2 H2 Mathematics Preliminary Examination 9758/1 solutions 3 of 15 For 2 e 3 or e 4xx−≤ < ≥ , Solution are ln3x< or ln4x≥ Q3 Suggested Answers (a) and ×=⇒⊥ ⊥ab c a c b c and k×= ⇒⊥ ⊥bc a b a c a Therefore, ,ab and c are mutually perpendicular. since and 0 k kk ×= = ⊥> bc a bc a b c since ×= = ⊥ ab c ab c a b Substitute =ab c into :k=bc a 2 2 k = = = bc ab aa b (b) ( ) ( )2 22 2 22 2 (since , 0) 6 −=− − = −+ = + ⊥∴ = = + = bc bc bc bb bc cc b c b c bc b ab     6−=bc (c) ×bc is the area of the square with sides OB and .OC y (0,1) y=3 x O ln 3 y=4 ln 4

2026 NYJC J2 H2 Mathematics Preliminary Examination 9758/1 solutions 4 of 15 Q4 Suggested Answers (a) 1 3iz= + is another root (b) ( )( ) ( ) ( )( ) ( )( ) ( ) ( ) 2 2 2 32 2 Method 1: 1 3i 1 3i 1 9 2 10 is a quadratic factor. Method 2: Sum of roots = 1 + 3i + 1 3i = 2 Product of roots = 1 3i 1 3i 10 1 3i 1 3i 2 10 12 2 10 10 Coefficient z z z zz z z zz qz pz p z q z z z −− −+ = − += − + − + −= −− −+ = − + + + + += − ∴ ++  2 of : 2 10 Coefficient of : 12 10 5 2 12 1010 5 Hence, 0, 2 Third root is 0 qzp qzp qq qp z = −+ + = −+ −+ + = − + = =− = ⇒ (c) ( ) ( ) ( ) Let be the complex number required. Let represent the complex number i 1 3i 3 i Then 3 i 1 3i 4 2i c B c c −= + =++− = −

2026 NYJC J2 H2 Mathematics Preliminary Examination 9758/1 solutions 5 of 15 Q5 Suggested Answers (a) 11 2 634 rnuS nn= = +− ++∑ As n → ∞, 1 03n →+ and 2 04n →+ . Hence 16rnuS= →∑ , which is a constant. The series ru∑ converges Sum to infinity 16= (b) ( )( ) ( )( ) ( )( ) ( )( )( ) ( ) ( ) ( ) ( )( )( ) ( )( )( ) 1 2 22 1 12 1 12 634 623 321 342 3 24 2 23 34 234 3 6 8 2 5 6 7 12 234 234 n nnuSS nn nn nnn nn nn nn nnn nn nn nn nnn n nnn −= −   =+ − −+ −  ++ ++   =−−+++ + +− + +−+ += +++ ++− ++− ++ = +++ = +++ (c) 11 59 k rr rk r uu ∞ = += =∑∑ ( )59 kkSS S∞ −= 5 14 kSS∞ = Method 1 (use Sn formula): 5 11 2146634 kk=+− ++ k 11 2 514 6 3 46kk +− − ++ 3 16− 4 0 5 536 Using GC, k = 4 Method 2 (use un formula): ( )( )( )1 5 146 234 kr rrrr= =  +++∑ k ( )( )( )1 514 234 6kr rrrr=  − +++∑

2026 NYJC J2 H2 Mathematics Preliminary Examination 9758/1 solutions 6 of 15 3 16− 4 0 5 536 Using GC, k = 4

2026 NYJC J2 H2 Mathematics Preliminary Examination 9758/1 solutions 7 of 15 Q6 Suggested Answers (a) 2 2 ln(5 ) (5 ) 1 (5 ) ln(5 ) ( 1)d 5 d (5 ) 1 ln(5 ) (5 ) xy x xxy x xx x x −= − − − − − ×− −= − −+ −= − When 0, 4.yx= = Gradient of tangent is −1 Gradient of normal is 1 Equation of normal: 0 1( 4) 4 yx yx −= − = − (b) ( ) 4 44 0 00 4 442 22 0 00 22 2 ln(5 ) 1 ln(5 )Area d (4)(4) or d ( 4) d (5 ) 2 (5 ) (ln(5 )) (ln(5 )) ( 4)8 or 2 22 11( (ln5) ) 8 or ( (ln5) ) 0 822 18 (ln5)2 xx x xx xxx x xx −−= + −−−−    − −−= − +− −       =−− + =−− − − = + ∫ ∫∫

2026 NYJC J2 H2 Mathematics Preliminary Examination 9758/1 solutions 8 of 15 Q7 Suggested Answers (a) ( )2cos e 1 xy −= − ( ) ( ) ( ) 2 2 22d sin e 1 2e 2e sin e 1d x x xxyx − −−−= − − ×− = − ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 22 2 2 2 2 22 4 2 4 4 4 d 2 2e sin e 1 2e cos e 1 2ed 2 2e sin e 1 2 2e cos e 1 d2 2 2ed d2 2ed d2 2ed xx x x x xx x x x x x y x y yx y yx y yx −− − − − −− −− − − − = − − + × − ×− = −× − −× − = − −× = −+ = − −− Therefore k = −2 (b) 32 44 32 d dd 2 2 e 8ed dd xxy yy yx xx −−= −+ − When x = 0, y = 1, d 0dyx = , 2 2d 4d yx =− , 3 3d 24d yx = ( ) ( ) ( ) 23 2 23 cos e 1 1 4 24 ... 2! 3! 1 2 4 ... x xxy xx −= − =+− + + = −++ (c) Include more terms in the expansion of ( )2cos e 1 x− − (d) 11 33 22 3 1 23 1 11 ...3 ba bx a x a bax a −= −  = −× + Comparing coefficients, a = 1, 1 263 bb− = −⇒=

2026 NYJC J2 H2 Mathematics Preliminary Examination 9758/1 solutions 9 of 15 Q8 Suggested Answers (a) sec 1 d sec tand x x θ θθθ = + = 2 2 tan 2 d secd d sec 1 d sec tan sin y y y x θ θθ θ θθ θ = − = = = When 3,sec 2 1cos 23 x θ πθθ = = =⇒= ± d2When , 3 2,3d 3 2Equation of tangent: ( 3 2) ( 3) 3 2 32 3 yy x yx yx πθ == −= − −= − ⇒= − − d2When , 3 2, 3d 3 2Equation of tangent: 3 2 ( 3) 3 2 32 3 yy x yx yx πθ = − = −− = − + += − − ⇒= − + − Angle between 2 tangents = 11 22tan tan ( ) 98.21 33 −− − −= ° Acute angle = 180 98.21 81.8°− °= ° (b) ( ) ( ) 2 2 22 22 2 2 1 sec ( 2) tan Since tan 1 sec 1 ( 2) 1 x y xy θ θ θθ −= += += −−+ = (c) ( ) 2 2 2 1 ( 2) 1 2 ( 1) 1 xy yx −−+ = = −± − − 33 22 22 22 Volume = ( 2 ( 1) 1) d ( 2 ( 1) 1) d 30.246064 3.26425 26.981 27.0 xx xxππ −− − − − −+ − − = − = ≈ ∫∫

2026 NYJC J2 H2 Mathematics Preliminary Examination 9758/1 solutions 10 of 15 Q9 Suggested Answers (a) Let OA = r ( ) 21 1.2 572 r ≤ 2 57 2 951.2r ×≤= 0 95r<≤ or 0 9.74679345r<≤ 0 9.74r<≤ (b)(i) Area of the 5th sector, u5 = ar4 4 3 97260 or 7.7765 125= = (exact) (b)(ii) 360 1 5 148.4831 5 n nS  −  = ≤ − 3150 1 148.48 05 n −−≤   Method 1 (use GC): n 3150 1 148.485 n −−   8 − 0.999 < 0 9 0.00835 > 0 Using GC, maximum number of sectors to be printed is 8. Method 2 (algebraic method): 3 148.481 5 150 3 148.4815 150 148.48ln 1 150 8.9893ln 5 n n n −≤  ≥− −≤=  Maximum number of sectors to be printed is 8. (c) Given S5 ≤ 4.99, d = −0.15. ( )( )5 5 2 5 1 0.15 4.992SC= +−− ≤