JPJC 9758 2023 Prelim P1 Solution

Jurong Pioneer Junior College H2 Mathematics JC2-2023 Prelim Paper 1 Solution 1 (i) 2 12 15 60nnu u n nA−= − ++ 2 21 2 15(2) 60(2) 4 2(2) 60 120 60 uu A AA = − ++ = − + +⇒= − ( ) 2 32 2 15(3) 60 3 60 2(4) 135 180 60 7uu= − + −= − + −= − (ii) Given 1 2u = , thus 24 pqr= −+ ----Equation (1) Given 2 4u = , thus 48 4 p qr=−+ ---Equation (2) Given 3 7u =− , thus 7 16 9p qr−= − + ---Equation (3) Solving, 43, 15, 304p qr= − = −= 2 (i) 1ln ln 1 11 n nu nn  = = − ++  As 1, 0, ln1 01 nnu n→∞ → → =+ (finite value) Hence, sequence is convergent. (ii) Method 1 ( ) ( ) ( ) ( ) ( ) ( ) 11 1 ln ln ln( 1)1 ln1 ln 2 ln 2 ln 3 ln 3 ln 4 ...... ln 2 ln 1 ln 1 ln ln ln 1 ln1 ln 1 ln 1 NN N n nn n nu nnn NN NN NN N N = = = = = −+ + = − +− +− + + −− − + −− +− + =−+ = −+ ∑∑ ∑ Method 2

2 ( ) 11 ln 1 123 1ln ln ln ...... ln ln ln234 1 1 123 1ln ......234 1 1 1ln 1 ln 1 NN n nn nu n NN N N NN NN N N NN N N = = =  + −       =+++ + + +        −+         −       =         −+        =  + = −+ ∑∑ 3 ( ) ( ) 2 2 2 22 22 11 d d At d1 1 ,1d Tangent at 111 2 Shown y x y xx xa y yxa a xa y xa aa ay a a x a ay x a a − = − = = = = − = −+ = − − +=− −= − 2 22 If is parallel to 9 1 d 11 d9 3 (reject as 0) or 3 When 3, 3 3 2(3) 93 l yx y xa aa a yx yx −= = = =−> = −= − −= 4 (i) 2 3 2 xxy x ++= − 223yx y x x− = ++ 2 (1 ) 3 2 0x yx y+− ++ = Consider Discriminant 0≥ ( ) 2 1 4(1)(3 2 ) 0yy−− +≥ 2 10 11 0yy− −≥ 11

3 ( )1 ( 11) 0yy+ −≥ 1y≤− or 11y≥ (ii) (iii) ( ) ( ) 2 22 22 2( 1) ( 2) 3 2x x x x kx+ − + ++ = − 22 22 3( 1) 2 xxxk x  ++++ =  − 222( 1)x yk++= For ( ) ( ) 2 22 22 2( 1) ( 2) 3 2x x x x kx+ − + ++ = − to have at least 1 positive real root, Since 0k > , ( ) ( ) 22 1 0 0 ( 1.5)k > −− + −− 3.25k > 1.803k > (3 d.p) 5 y x O k

4 (i) ( ) ( ) ( ) ( ) ( ) 2 f 45 f 41 f 37 ...... f 1 4 1 3= = === −= (ii) [ ]fR 0, 4= (iii) [ ] ( )fgR 0, 4 D 0,1= ⊄= Hence, gf does not exist. (iv) ( ) 2 g : 1 2 for , 0 1.xx x x − + ∈ << ( )gR 2,3= ( ) ( ) ( ) 22 fg 2 1 2 4 2 1xx x = − + −= − Let ( ) 1 1fg ( ) 2 α − = ( ) 1fg 2α = ( ) 2 121 2α −= ( ) 2 11 4α −= ( ) 111 or 22α −= − 31 or22α = Since ( )fg gD D 0,1= = , 1 2α = x y O 2 4 6 7 −2 −4 4

5 Alternative Method ( ) 2 21yx= − , 01 x<< ( ) 2 1 2 yx−= ( )1 2 yx−= ± 1 2 yx= ± Since 01 x<< , 1 2 yx= − ( ) 1 fg ( ) 1 2 xx − = − ( ) 1 1 11 2fg ( ) 12 22 − = −= 6 (i) Given p: 1 2 12 2  ⋅− = r . Equation of the line AN: 21 12 32 λ    = +−  −  r where λ∈  . Then 2 12 32 ON λ λ λ + = −−+  for a specific value of λ. Since the point N lies on the plane p, 21 1 2 2 12 32 2 λ λ λ +     − ⋅− =    −+   2 2(1 2 ) 2( 3 2 ) 12λλ λ+ − − + −+ = Solving, 2λ = Thus 22 4 14 3 34 1 ON +     = −= −    −+    . The coordinates of the point N is (4, 3, 1)− Let A' be the point of reflection of A. By Ratio Theorem, ' 2 OA OAON +=   

6 ' 426 2 23 1 7 1 35 OA ON OA      = − =−− = −     −       The coordinates of the reflection of A in p is (6, 7, 5)− . (ii) For plane p: 1 2 2 12 33  ⋅− = r Distance from O to the plane p = 12 43 = (since 1 2 3 2  −= ) ∴ The two planes with perpendicular distance 15 to p are 1 2 2 4 15 113  ⋅− = −= − r and 1 2 2 4 15 193  ⋅− = += r Cartesian equations are 2 2 33xyz−+= − and 2 2 57xyz−+= 7 (a) ( ) 3 4 3 4 4 44 2 d1 14 d21 1 ln 12 1 ln 1 since 1 02 x xx x xx xc xc x + = + = ++ = + + +> ∫ ∫ (b) p p1 15 p2 15 O

7 cos 4 sin10 d sin10 cos 4 d 1 sin14 sin 6 d2 1 cos14 cos 6 2 14 6 x xx x xx x xx xx c = = + = − ++ ∫ ∫ ∫ (c) ( ) ( ) ( ) 31 30 2 3 24 30 2 3 24 30 2 3 4 40 34 4 30 34 0 34 0 4 4 0 d 1 tan sec d tan 1 tan sec d sec tan dsec sin cos dcos cos sin d (Shown) cos sin d sin 1 4 16 x x x π π π π π π π θ θθ θ θ θθ θ θ θθ θθ θθ θ θθ θ θθ θ + = + = = = = = = ∫ ∫ ∫ ∫ ∫ ∫ ∫ 8 (i) For the sprinter: 56 + 62 + 68 + … + [56 ( 1)(6)]n+− AP: First term = 56, common difference = 6 The time the sprinter takes to complete n laps = [ ]2(56) ( 1)(6)2 n n+− = ( ) 256 3 3 3 53n n nn+−= + (Shown) (ii) For the marathon runner: 60 + 23 160(1.04) 60(1.04) 60(1.04) ... 60(1.04) n−+ + ++ GP: First term = 60, common ratio = 1.04 The time the marathon runner takes to complete n laps = ( )60 1 1.04 1 1.04 n− − or ( )60 1.04 1 1.04 1 n − − 2 tan d secd 1, 4 0, 0 x x x x θ θθ πθ θ = = = = = =

8 ( )1500 1.04 1n= − (iii) Consider ( )1500 1.04 1 2400n −> Method 1 1.04 1 1.6n −> 1.04 2.6n > Taking lg on both sides: lg2.6 lg1.04n> 24.362 (correct to 5 s.f.)n∴> The marathon runner needs to complete 25 laps. Method 2 n ( )1500 1.04 1n − 24 2345 < 2400 25 2498.8 > 2400 26 2658.7 > 2400 The marathon runner needs to complete 25 laps. (iv) 12000 30400n= = Time taken by the sprinter = ( ) 2 3 30 53(30) 4290+= seconds Time taken by the marathon runner = ( ) 301500 1.04 1 3365.1−= seconds (correct to 5 s.f.) The marathon runner will be the first to complete the race. (v) Consider ( ) 21500 1.04 1 3 53n nn−< + Method 1 n ( )1500 1.04 1n − 23 53nn+ 3 187.3 186 4 254.79 260 5 324.98 340 Method 2 (Sketch ( )1 1500 1.04 1XY = − , 2 2 3 53YX X= + )

9 Since 3.25n≥ (3 s.f.), 4n∴= Hence, the marathon runner first overtakes the sprinter on his 4th lap. 9 (a)(i) Method 1 432( i) 4( i) ( i) 16( i) 0k k ak k b− + − += 43 2 4 i 16 i 0k k ak k b+ − − += ( ) ( ) 42 3 4 16 i 0k ak b k k− ++ − = Comparing real and imaginary parts, 3 2 4 16 0 4 ( 0) 2 kk kk k −= = ≠ =± The two roots are 2i and −2i. 42 2 0 (4) (4) 0 16 4 0 (shown) k ak b ab ab − += − += − += Method 2 Since the coefficients of the polynomial are all real, as izk= is a root, its conjugate izk=− is also a root. 432 2 2 22 4 16 ( i)( i)( ) =( )( ) x x a x xb xkxkx c xd x k x cx d −+−+ = − + + + + ++ Comparing constant, 2 2 bb dk d k= ⇒= …(1) Comparing coefficient of x , 216 ck−= …(2) Comparing coefficient of 2x , 2adk= + … (3) Comparing coefficient of 3x , 4 c−= … (4) Substitute (4) into (2), 216 4 k−= −

10 2 4 2 k k = =± The two roots are 2i and −2i. Substitute (1) and 2 4k = into (3), 44 4 16 16 4 0 (shown) ba ab ab = + = + − += (ii) Sub (0, 20) into ( ) 432f 4 16x x x ax x b= −+−+ , 20b= 16 4 20 0 9 a a −+= = 432 2 22 f( ) 4 9 16 20 = ( 2i)( 2i)( ) ( 4)( 4 5) xx x x x x x x cx d x xx ∴= −+−+ − + ++ = + −+ (b) ( ) ( ) ( ) ( ) ( ) ( ) * * * 2 * zi 3i ( ) i zi 3i ii let i i3 ii 1i 3 1i 3 zi 3i i 1 z zz z zz ab z ab z abab ab ab z zz − = ++ − = + −−= =+⇒=−++ −+ += ++ = −