2019 JC1 RVHS H2 Maths CT Solutions (w Markers Comments)

1 JC1 H2 Mathematics 2019 Common Test Solutions 1 Solution [5] Markers’ Comments Based on the given information, we have the following: 0 - - - - - (1)y x z x y z= +  − + = 29 21 124 295x y z+ + = - - - - - - (2) 24 27 118 295 3 24 27 118 298x y z x y z+ + = +  + + = - - (3) Using GC to solve the above system of equation, we have: 3, 4 and 1x y z= = = . Thus, the scores for the various units are as follow: Unit A: 298 points Unit B: 295 points Unit C: 26 3 25 4 121 1 299 +  +  = points Thus, Unit C is the champion of the competition. Generally quite well done across the level; some students tend to solve the system of linear equations manually instead of using the GC. 2 Solution [7] i Method 1 ( ) 22 2 2 2 1 2 1 112 22 x x x x x − + = − + = − + Since 2 1 02x − , 2 1120 22x − +  for all real x. 22 2 1xx − + is always positive for all real x. _______________________________________________ Method 2 Since discriminant = 2( 2) 4(2)(1) 4 0− − =−  and coefficient of 2x is positive, 22 2 1xx − + is always positive for all real x. For method 1, many students made mistakes in their argument, e.g. writing 2 1 02x − instead of 2 1 02x − etc. Some completed the square wrongly. For method 2, a complete argument needs to have both “discriminant < 0” and “coefficient of 2x > 0”. Having just the former might imply the quadratic expression is always negative instead of positive.

2 ii ( )( ) 32 2 2 2 1 0, , 1 22 1 1 x x x xx xx −+  −  +− ( ) ( )( ) 2 2 2 2 1 0 2 1 1 x x x xx −+  +− Since 22 2 1 0xx− +  for all real x, ( )( ) 2 0 2 1 1 x xx  +− Using number line, 1 2x− or 0x , 1x It would be good to note from the start that 1 2x− and 1x . Students need to show proper working and explain why ( ) 22 2 1xx−+ need not be considered in the number-line investigation. Students need to take note of when “=” is included. Some students used other analytical approaches which considered cases how each sub -expression’s sign should be etc. It is tedious to do that and students have to take care of many details in the argument. iii ( ) ( ) ( )( ) 32 2 2 ln 2 ln ln 0 2ln 1 1 ln x x x xx −+  +− Replace x by ln x : 1ln 2x− or ln 0x , ln 1x 1 20e x −  or 1x , ex Almost all students know to make use of part (ii)’s answer and replace x with lnx. Quite many students thought that 1ln 2x− cannot be solved (that ln x has to be 0 ). For those who attempted solving it, many did not realise that “ 0x ”, else ln x is undefined.

3 3 Solution [7] (i) Note: the point (0,1) should be above the point (4, ln2) Generally quite badly done. Some common mistakes are: • Missing axial intercepts; • Missing ending points (need to be obvious) indicated and labeled; • Need to know that ln 2 1 and hence there is a need to be indicated correctly on the graph (ii) g 90, 8R =  Since    gf 90, 0, 4 \ 28RD =  =  , The composite function fg exist. Very badly attempted as students did not look at the graph of g to determine the range. The correct domain of f should also be  ) ( 0, 2 2, 4 . A lot of students did not read the question carefully to determine the domain of f. (iii) ( ) 2 9 2 16f 0 1, f 92 0 8 7 2 8 = = = = −  − By mapping method,  ) gf 9 160,1 0, 1, 87    ⎯⎯ → ⎯⎯ →     fg 161, 7R =  Alternatively, Quite a number of students simply ignore all workings and simply write out the wrong answer and this resulted in a lot of marks lost. Students should show clear workings or thoughts. Some common errors include fg gRR= or fg fRR= y x y = f(x) O

4 ( ) ( ) ( ) 2 2 2 f g f 2 1 2 2 2 1 2 21 x x x xx xx = − + + = − − + + = −+ (since g 90, 8R =  ) Dfg = Dg =  )0,1 ( fg 1, 2.29R = (to 3 s.f.)

5 4 Solution [8] i Since C has an oblique asymptote at 1yx=− with coefficient of x equals to 1, a = 1 Since C has a vertical asymptote at 2x= , 2c= Since C passes through the point P 30, 2  − , 3 322 b b− =  =− (shown) Most students largely have an issue with explaining the reason why a = 1. Many left their answer as ‘Because 1yx=− is the asymptote, therefore a = 1.” ii 2 33 2 xxy x −+= − iii 2 3 3 1 122 xxyx xx −+= = − +−− ( ) 2 d1 1d 2 y x x =− − When x = 0, d3 d4 y x = The equation of normal is thus, ( )31 032 4 yx − − =− − 43 32yx =− − O

6 5 Solution [7] A 3 unit  Applying cosine rule: B 4 units 2 2 2 2 2 3 4 2 3 4cos 25 24 1 , since is small2 1 12 BC     = + −    − −  =+ C 21 12 (shown)BC   + Then ( ) 1 2 21 12BC + ( ) ( )( ) 11 22222 2 4 4 111 12 12 .....2 2! 1 6 18 (up to term)     −= + + +  + − where 6, 18pq= =− . (shown) Sub 1 4 = into the above result: 2 2 4 24 1 1 11 12 1 6 184 4 4 7 1 11 6 184 4 4      +  + −               + −       Thus, 24 1 1 1677 4 1 6 18 4 4 64       + − =         As 1 10 is closer to 0 compared to 1 4 , the substitution 1 10 = will give a better estimate of the value of 7 . Most students are aware that they need to use cosine rule and small angle approximation, and were able to show the correct result, except for some who fumbled with applying binomial expansion to obtain the correct final answer. Overall a huge majority lost presentation mark in this question due to the careless use of “=” and “  ” interchangeably. This part was not well done with a significant number of students only substituting ¼ to the LHS of the expression, some did not obtain the correct answer due to the error in the previous part. There are many who did not provide proper explanation for this part e.g. some used “closer to 1” and some others gave the correct reason but stated that ¼ is better

7 6 Solution [10] (i) When 3a , any horizontal line y = k where k will intersect the graph of ( )fyx= , where  ),xa , at most once. Thus f is one-to-one and 1f− exist. Also, the turning point 26y x x=− occurs when 3x= .  Least value of a is 3. Most students were able to sketch the curve with the domain [3, ), with some not labelling the max pt and x-intercept. For the 2 nd part, students need to mention about the max pt (3, 9). Otherwise it is not sufficient to explain that the function is not 1 -1 for 3a as we still need to ascertain that the function is 1-1 for 3a . Also, attention should be paid to the correct use of term l ike ‘at most 1 point’ or ‘more than 1 point’ in the explanation of horizontal line test for 1-1 property. (ii) Let 26y x x=− ( ) ( ) ( ) 2 2 2 2 2 6 6 3 3 39 y x x xx x =− − =− − + − =− − + ( ) 2 39xy− = − 39xy− = − 39xy=  − Since 3x , 39xy= + − ( 1 ff ,9DR− = = − 1f : 3 9 , 9x x x− + −  Some students were not able to perform completing square or use the quadratic root formula in trying to express x in terms of y. Some students also did not explain on the rejection of the expression 39xy= − − . Other errors include not expressing the answer for 1f− in similar form and not stating the domain for 1f− . y = f(x) O

8 (iii) Many students did not take note of the domain of all 3 graphs for this part while sketching. There were many instances where students did not label the end points and relevant axial intercepts Also there should be equal scaling on both axes so as to