2023 ACJC JC1 H1 Promo solution with remarks

ACJC 2023 H1 Math JC1 Promo Solutions 1 ( ) 22(ln ) ln e 7xx − 22(ln ) ln ln 7 0x e x− − −  2(ln ) ln 2ln 7 0x e x− − −  2(ln ) 2ln 8 0xx − −  Let lnux= 2( ) 2 8 0uu − −  ( 4)( 2) 0uu− +  4 or 2uu − ln 4 or ln 2xx − 4 -2 or < ex e x Since x > 0 for lnx to be defined Hence 4 -2 or 0< < ex e x Many students were not able to simplify ( ) 2ln e x as ln e 2ln x+ . For those who did, majority of them wrote ( ) 2ln e x− ln e 2ln x=− + instead, resulting in wrong values of roots. The common misconception to solve ( 4)( 2) 0uu− +  is “splitting” as ( 4) 0u− or ( 2) 0u+ . Very few students can write 20e x − as part of the answer. 2 Method 1 25 5 2xx−+ = 2 25( ) 5xx−+ = 2 21 1 25(( ) ) 2 2 5x − − + = 2135(( ) ) 2 20x−+ = 2135( ) 24x−+ > 0 Method 2 Discriminant of 25 5 2 0xx− + = is 25 4(5)(2) 15 0− =−  (1) Hence 25 5 2y x x= − + does not intersect the x- axis . Also since the coefficient of x2 >0, the curve 25 5 2y x x= − + has a minimum point. Hence 25 5 2xx−+ is positive for all real values of x. Most students use completing the square to show that the expression is positive. A handful of them use discriminant but a few of them did D > 0 instead. 24 (12 8) ( 2) 0+ − − + =x k x k k Discriminant = 22(12 8) 4(4)( 2 )− + +k k k = 22144 64 192 16 32k k k k+ − + + = 2160 160 64kk= − + = 232(5 5 2)kk−+ > 0 for all real values of k from above Almost all students are able to begin with D > 0 in this part. However, only a few students are able to derive that “for all real values of k” as the final answer.

3a ( ) ( ) ( ) 2 2e1ln ln e ln 2 5 225 1 2 ln 2 5 2 x x x x xx − − = − + + =− − + ( ) ( ) d 1 52 ln 2 5 2d 2 2 2 5 5 2 4 10 xxxx x − − + =− − + =− − + 2, 5, 10A B C=− =− = Quite well done. A common error is to differentiate 2eln 25 x x −  + directly, which lead to complicated wrong answers. Do note that product rule is not necessary in H1 Math syllabus. b ( ) 2 3 1/ 2 2 3 2 3 6 dx 4 23 2364 1 (3)2 2 3 1 2 3 4 1 2 3 3 7 3 k k k x x x k k k = +   + =    += + − = += =  Majority were able to integrate. A handful of them could not obtain the answer due to careless mistakes. 4i A common error was that the graph was truncated near x =1. A handful of students wrote y = 2 as the horizontal asymptote as well. Most of the students did not realise that “exact” coordinates is needed. x = 1 ( ) 21 e ,0−+

4ii Rearrange the equation to ln( 1) 2 2 5xx− + = − add the 25yx=− line. From GC, the 2 solutions are 1.01 (3sf) and 4.06 (3sf) Many students obtained 4.06 using the line 25yx=− . Very few students are able to obtain 1.01 as one of the answers. 4iii From GC,