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