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, area = 6.14 units2 Quite well done. 5i 2 2 2 2 0 0 or k x kx k x kx x x k − =− + −= == Sub into equation to find y-coordinates. 2 (0, )Ak and ( ,0)Bk Mostly well done. 5ii ( ) ( ) ( ) 2 2 2 0 2 2 2 0 23 0 33 3 23 23 6 k k k k x kx k dx k x kx k dx kx x kk k − − − + = − + − =− =− =   Most students are able to begin with the first line for either methods. Some errors in integrating 2k . Many mistaken k as x in integrating with respect to x.

Alternative method ( ) ( )( ) 22 0 3 22 0 33 3 area of triangle 1 32 2 32 6 k k k x dx xk x k k kk k −− = − − =− =  6i 2 3 12212 12xy x x x −+= = + ( ) 3122 12 32 3122 2 3 62 36Let 0 2 2 6 3 4 2, 2(na) dy xxdx dy xdx x xx x x − =− =  = =  =  = − Alternative method: Sub x = 2 into 3312 2 233 6 2 6(2) 022 dy xxdx −− = − = − = 2 12 4 12 16 22 xy x ++= = = Stationary point is 162 , 2   Pretty well done as most of the students are aware that the expression requires some simplification before differentiating. A handful of them made some errors with the indices rules thus not able to obtain the correct answers.

31223 62 dy xxdx − =− 1st derivative test: x 1.99 2 2.01 2.01 dy dx −0.0213 0 +0.0211 0.0211 Alternative method: 2 5122 2 3 9 2.12 04 dy xxdx −− = + =  Therefore 162 , 2   is a minimum turning point. A handful of them are not aware that computing values of d d y x is necessary in first derivative test. ii. From GC, gradient 4.50=− , Coordinate is (1, 13) Equation of the tangent is 13 4.50( 1)yx− =− − 4.5 17.5yx=− + (2dp) Most students are aware that they need to find equation of a straight line. Some of the students obtained the wrong d d y x value due to the error made in the earlier parts.

7i 50002500 2 0.4 −=+ +S t 2 0.4 0 5tt+ =  =− When 0, 0St== Horizontal asymptotes 2500S = Or Horizontal asymptote is 1000 25000.4S == Majority of the students are aware that the horizontal asymptote is S = 2500. A handful of them labelled as y = 2500. A number of them did not realise that 0t  . ii d d S t = 2 2000 (2 0.4 )t+ When 2t = , 22 2000 2000 255(2 0.4 ) (2 0.8) dS dt t= = =++ (3 sf) The sales of the company is changing at the rate of $255000 per year in the year 2012. Many students did not include the word “rate”. Many students mistaken that the value is meant for 2010 to 2012. In fact, it should just be 2012. iii In the long run the sales of the company increases and tends towards $2 500 000. Not many students use the similar phrasing. iv 4 2 1000 d2 0.4+ t tt = 4 2 5000(2500 )d2 0.4− + tt = 4 2 5000[2500 ln(2 0.4 )]0.4−+tt = 5000[2500(4) ln(2 0.4(4)]0.4−+ 5000[2500(2) ln(2 0.4(2)]0.4− − + = 2.8(2500(2) 12500ln ) 3.6+ = 1859 (to the nearest integer) It represents the total sales in thousand dollars for the jewellery store from 2012 to 2014. A handful of them use G.C to integrate. Students are not aware that “Find algebraically” means G.C is not allowed in solving. S=2500 S t

v 32 12 ( 36)C t t k t= − + + 2d 3 24 ( 36)d C t t kt = − + + > 0 for all real t since x is an increasing function of t Method 1: Discriminant = (24)2 – 4(3) (k+36) < 0  k > 12 Method 2: 2 2 23 24 36 3 8 12 3 ( 4) 16 12 0 33 kkt t k t t t    − + + = − + + = − − + +        for all real t if 16 12 03 k− + +   43 k   k > 12 Quite badly done as most students did D > 0 instead.