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                                 

6 Solution (a)(i)  1 21 121 21 23 log 3 log 3 3log 3 log 9 which is a constant independent of nn nn aa n a n a du u n         Therefore, the series is an arithmetic series. (a)(ii) Method 1: use 2 nSl n a 2(30) 1 30 30S log 3 log 3 3002 aa     6 3 0 1 15 log 3 300 900log 3 300 log 3 3 27 1 3 a a a a a      Method 2: use 2 21nSa n dn   30 30S 2(log 3) 29 log 9 3002 aa       2 1 3 1 52l o g3 2 9l o g3 3 0 0 900 log 3 300 1log 3 3 3 27 aa a a a a       (b) 2 3 4( 1 ) 7( 2 ) 9( 3 ) bd c r bd c r bd c r    (Eliminate b): 2 32 (2) (1): 3 (3) (2) : 2 cr cr d cr cr d     (Eliminate d):

322 23 cr cr cr cr  Since 0,cr  , we divide both sides by c and r, and rearrange to get 2 )30 o 52 ( s h w nrr  Solving, 21 (rejected ) or 0 3rd r   321 3 cSc   7 Solution (a) Explanation 1 (“Horizontal Line Test”) g is not a one-to-one function as the horizontal line 0y meets the graph of gyx more than once. Explanation 2 (State two inputs with same output) g is not a one-to-one function as there are distinct inputs producing the same output under function g, e.g. g(1) g(2) 0 . (b) Greatest k = 0. (c) Let g( ), 0yx x . Then 1g( )xy  . 2 2g( ) , since 0.1yx x x   2 21 x y 2 2 1x y , 2 1x y  Since 0x , 12 1g ( )x yy    1 2g( ) 1 x x     1 ggDR0 , 2  .

(d) The line in which the graph of g( )y x is reflected to obtain the graph of 1g( )y x is y = x. 8 Solution (a) Method 1: Find d d y x then simplify Differentiating implicitly w.r.t. x, Method A: Consider Chain Rule  11 dd 2dd yy xyx x      Method B: Consider product rule  222 2 ddd12 2 2 2ddd x yx y x x y y yyy x xy yx xx        dd 22 212 dd y yxy xyx x   d12 2 12 2 d yxy xy x      d1 22 d1 22 yx y x xy    Add 1 to both sides,  12 2 12 2d1 d1 2 2 (shown)2 12 2 x yx yy xx y xy      y x y = x y = g-1(x), 0 < x ≤ 2 y = g(x), x ≤ 0 2 2 O 0x 0y

Method 2: Consider adding d1 d y x to both sides Differentiating implicitly w.r.t. x,  11 dd 2dd yy xyx x      Add d1 d y x to both sides,   dd d d12 1 1dd d d d2221 1 d d1 ( s h o w n ) 1 d 2 221xy yy y y xyx xx x yxy x y x                     (b) Diff implicitly w.r.t. x,   2 11 2 2 2 dd 21 2 2 2 2dd d1 d12 2 dd11 d 4 d yy xyx x y xxy yy xx                 32 2 dd (1dd shown)yy xx    (c) 2 2 dd 0 1 0 minimum pointdd yy x x   9 Solution (a)   22 2 Eqn 1 Eqn 2 l n 2e e 2e e 2ex yx x yy          Method 1: implicit differentiation Differentiating Eqn 1 implicitly w.r.t. x, 2de2 ed yx y x  Then  2 from Eqn 2 d 2ee2 ( s h o w n )22 e e 4 ed xy y yyy x      Method 2: direct differentiation

  from Eqn 2 2 2 from Eqn 1 22 ed2 e 4e 2 (shown)d2 e e yx y xy y x         Method 3: make x the subject, implicit differentiation From Eqn 2,  2e2 e 2 l n 2 ex yy x      Differentiating implicitly w.r.t. x, 1d2e 2e d y y y x     Then  22 ed 4e 2 (shown)de y y y y x    (b) Method 1: further differentiation of result in (a) Differentiating d 4e 2d yy x  implicitly w.r.t. x, 2 2 dd d 4e 4 edd d yyyy y xx x     When 0x , 0,y d 2d y x  ,  2 0 2 d 42e 8d y x     22802 2 4 2!yx x x x      Method 2: direct differentiation of 1st derivative in x       22 2 22 22 2 2 22 224e 2 e e ed d 2e 8e 2e x xx x x x x y x            When 0x , 0,y d 2d y x  ,  20 22 0 d8 e 8d 2e y x        22802 2 4 2!yx x x x      (c) From MF26,  2 22 (2)1 2 ... 1 2 2 2e. . .x xxx x      Then

      2 2 22 2 Using the expansion for ln 1 2 f 2 2 2 ln 2 e ln 2 1 2 2 ... ln 1 2 2 ... 22 . . . 2 2 ... ... 2 42 2 ... 2 24 . . . x xy xx x xxx xx x xx xx                     This is the same expression as found part (b) and hence we can conclude that the expansion is correct. 10 Solution (a)     s 1sin 3 cos d 2sin 3 cos d2 1 sin 3 sin 3 d2 1 sin 4 sin 2 d2 1 sin 4 d sin 2 d2 11 1 4sin4 d 2sin2 d24 2 11 1 cos 4 cos 224 2 where is an arbitr s ay 11co 4 r 4c o s 2 co 8 n , xx x xx x xx xxx xx x xx xx xx x x c x x xc xc                      tant. (b) [Since  2d 41 324d xx xx  , we re-write x as 24x Ax B  in order to split the numerator into 2 parts. Compare coefficients to get A and B.]   22 22 2 22 21 d d 41 3 41 3 12 4 1 t d2 d2 41 3 41 3 11 1 2 4 where is l an arbitrary c t ln 4 13 2 d2 (2 )3 12 2 n1 3 t a n ons a