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