EJC H2 Promo 2023 (Solutions)

2023 EJC H2 Math Promo Solutions 1 Solution At  2,1 ,    23 12 2 2 4 2 9ab c a b c         --- (1) At 2, 3 ,    32 32 2 2 4 2 1 1ab c a b c       --- (2) Since 2,1 lies on f1yx , Method 1: consider that 3,1 lies on fyx :    32 13 3 3 9 3 2 6ab c a b c       --- (3) Method 2: use   32 f1 1 1 1 acyx x x x b       32 12 1 2 1 2 1 9 3 2 6ab c a b c       --- (3) Solving (1), (2) and (3), 5, 72,abc    [or 32f( ) 2 5 7xx x x  ] 2 Solution (a) y x O

(b) 3 Solution (a) 13 122 xy xx   . Horizontal Asymptote: 1y . Vertical Asymptote: 2x For ln 1 2 x : 021 2 x x  , i.e. 2x is a Vertical Asymptote. Graph exist when 021 2 x x (b) From the graph, 02 x y x O y x O

4 Solution (a)         LHS 0 and RHS (shown)                             rr q rr rq rr rrq q rq rq r r a b b a qqr q p pp pp p p r p pp pp q r qr r r  (b)     11 22 1 2 1 2 OR OP OR OQ PR QR         pp qqrr r r pq             1 2 qqrrpp represents the area of PQR . Alternative 1 2 qqrrpp represents half the area of a parallelogram with PR and QR as adjacent sides. (c) Given , ( ) ( ) (from result in ) PR QR        qqrrp0 r p pr q 0 ( a ) 0    , , are distinct points, , and , , i.e. , , are collinear points. PQR PR QR PR QR P Q R    00         Given also that 3PR QR , and PQ PR ,  Point R divides PQ internally in the ratio 3:1, i.e. :3 : 1PR RQ  . By the ratio theorem, position vector 3 4  pqr . Alternative (to show collinear) . Given , Area of 1 (by p e a i.e. is a d gene re rt )2 1 2 0 ate triangl PQ PQR R          qqrrp0 qqrrp ( bp) 0 p i.e. , , are collinear points.PQR O 4 P 3 R 1 Q

5 Solution (a)         2 ) 32 1 1 1321 1! ! ( 1 ) ! 1! 31 1 (verified! rr r rr r r rr r        (b)  2 1 31 1! n r rr r   1 321 1! ! ( 1 ) ! n r rr r      321 2! 1! 0! 321 3! 2! 1! 321 4! 3! 2! 321 5! 4! 3! 321 ( 1)! ( 2)! ( 3)! 32 1 !( 1 ) !( 2 ) ! 321 1! ! ( 1 ) ! nn n nn n nn n             31 41! !nn  (c ) Method 1: change of variable 22 1 31 3 21 2 21 1 3( 1 ) ( 1 ) 3 !( 1 ) ! 31 (1 ) ! 31 3 (1 ) ! 2 31 3 4!( 1 ) ! 2 31 5 !( 1 ) ! 2 nr n rr n r n r rr r r rr rr r rr r nn nn                              Method 2: Listing     22 2 3 21 2 21 1 33 3 3 3 ...!3 ! ! 11 3 1! 31 3 (1 ) ! 2 31 4!1 ! 3 2 3 2 15 !( 1 ) ! n r n r n r rr nn rn rr r rr r nn nn                