RVHS 2022 JC1 H2 MA CT Soln w Comments

River Valley High School 1 2022 JC1 H2 MA Common Test Markers Comments 1 Solution [4] MOD Let 2 (2 1)(2 1) (2 1) (2 1) AB r r r r     1 2 2 1(2 1) x A r    1 2 2 1(2 1) x B r    2 1 1 (2 1)(2 1) (2 1) (2 1)r r r r     This part is well -done by almost every student. 11 1 1 1 1 (2 1)(2 1) 2 2 1 2 1 nn rr r r r r        11 13 11 351 2 11 2 3 2 1 11 2 1 2 1 nn nn             ... 1112 2 1 n    The common mistake is missing out the 1 2 but most students recognize it is MOD.

River Valley High School 2 4 1 14 1 3 12 11 1 (2 3)(2 1) 1 (2( 1) 3)(2( 1) 1) 1 (2 1)(2 1) 11 (2 1)(2 1) (2 1)(2 1) 1 1 1 1112 2( 1) 1 2 2(2) 1 1112 n r kn k n k n kk rr kk kk k k k k n                                                      (let r = k+1) = 112( 1) 1 2(2) 1 1 1 1 2 5 2 1 n n         The challenging part of this question is letting 1rk and more imp ortantly considering the starting and ending values of k.

River Valley High School 3 2 Solution [4] SLEs Let b, s and f be the price per cup of butter, sugar and flour respectively. 200b + 50s + 200f = 129 40b + 40s + 40f = 32.4 100b + 50s + 300f = 88 From GC, b = 0.5, s = 0.22, f = 0.09 Therefore, the price of a cup of butter, sugar, and flour is $0.50, $0.22, and $0.09 respectively. Hence, the cost to produce the pastries is $81.25 Very well -done by almost all students

River Valley High School 4 3 Solution [5] Inequality (a) 1 2 1 2 23 23 22 1 21 1 1 1 1 11 1 21 2 2 2 2 2 2 2 21 ... (*)2 2! 3! 1 1 11 ... 22 x xx x x x x x xx                                              Expansion is valid for 22 1 1 2 xxx     2 or 2xx  Since 6 6 2   , 6x is valid for getting an estimate for 6 . 23 2 1 1 11 22 x x x x x      Let 6x 23 4 1 1 116 ( 6) 2( 6) 2( 6)         2 353 4326  8646 353 Most students are able to come out with the series; however some did not read the question clearly and attempt to expand in increasing order instead. Not many students can write down the range of values of x correctly and subsequently give a reason why -6 is a good estimate. There is another approximate value of 3536 144 . 2 353 3 432 2 3 353.3 3 432 6 353 3 432 3536 144    

River Valley High School 5 4 Solution [5] Summation (a) 11 1 1 1 u( ) u( 1) 2 4 2 u( ) u( 1) 2 4 2 u(1) u(0) u(2) u(1) 2 4 2... u( ) u( 1) 111 122u( ) u(0) 4 ( 1) 2 1 21 2 1u( ) 5 1 2 ( 1) 22 1u( ) 6 2 r NN r rr N N N r r r r N N r r r r r r r NN N N N N N N N N N                                             22N N Therefore 21u(r) 6 2 2r r   Question was challenging for most students. For those who attempted, many used the original form u( ) u( 1)2 4 2 rr r r     and kept expanding the RHS with no avail. For some, MOD was used for u( ) u( 1)rr but did not apply the summation to 2 4 2r r  . To apply MOD, students need to recognize that they need the form u( ) u( 1)rr . Thus, the first step is to move u(r – 1) to the LHS. Then take 1 N r  on both sides. Then realise that the RHS can be split into sum of a GP, AP and constant respectively.

River Valley High School 6 5 Solution [7] Inequality (a)   32 32 32 32 2 2 4 5 2 12 4 5 2 102 4 5 2 2 02 44 02 44 02 ( 2) 0 2 x x x x x x x x x x x x x x x x x x x x x xx x                     02 x but 2x also satisfy the inequality Therefore 0 2 or 2xx   __________________________ Alternatively 2( 2) 0 2 xx x   2Since ( 2) 0, x with 2x satisfy the inequality 02 x x  02    Therefore 0 2 or 2xx   Generally ok except for some students missing out on the answer 2x . Also, there were mistakes with the combining of fraction expressions as well as in the operation with inequality signs. ‒2 0 2 ‒ ‒ ‒ +

River Valley High School 7 (b) 32 4 5 2 12 x x x x     Replace with xy  32 4 5 2 12 y y y y      32 4 5 2 12 y y y y     Using earlier result 0 2 or y 2y   . Thus 0 2 or 2xx    2 0 or 2xx    Most students were able to identify the replacement for x and proceed to solve the inequality for this part. Some missed the answer 2x due to the first part.

River Valley High School 8 6 Solution [7] Implicit Differentiation (a) 33 22 22 2 2 4 dd3 3 4 4 dd d 3 4 4 3d d 4 3 d 3 4 x y xy yyx y y x xx y y x y xx y y x x y x           Generally, well done. (b) d d y x  2 2 Let 3 4 0 3 4 yx xy       3 2 3 2 3 6 3 3 33 11 33 33 444 3 34 27 2 0 *64 1281280 or y 1.679894727 3 y y y y y y y yy y or                       2 3 10 or x 128 2.116534712x or The coordinates where the tangent to curve is parallel to the y-axis are (0,0) and   12 33 128 128, 1.68, 2.123 12 or    Many candidates miss out the solution 0y from equation (*). The question did not ask for exact values of the coordinates, candidates could have used GC function to evaluate the roots to (*).

River Valley High School 9 7 Solution [10] Tangent normal Parametric Equation (i) 21xt , 4y t where 1t and 0t  . When 0t , 1x and y , therefore the curve has a vertical asymptote at 1x . When t , x and 0y , therefore the curve has a horizontal asymptote at 0y . To show that a line is an asymptote, we need to describe the be haviour of both x and y variables near the line (eg 1x and y ). In this quest ion, since t is the parameter where both x and y are dependent o n, we must start with the behaviour of t first ( 0t and t ), then talk about the behaviour of x and y. Alternative Method 21xt , 4y t where 1t and 0t  . Sub 1 2 xt  into 4y t , 14 2 xy  8 1y x  where 1x and 1t  since ( 1t and 0t  ) As 1x , y , therefore the curve has a vertical asymptote at 1x . As x and 0y , therefore the curve has a horizontal asymptote at 0y .

River Valley High School 10 (ii) 21xt , 4y t where 1t and 0t  . Most students were unable to get the graph right. Most only had the portion for x > 1 because the default Tmin = 0 in the GC was used. To sketch the other po rtion using the GC, you will need to sketch separately for Tmin = 1 and Tmax = 0. Another note is t hat the point (1, 4) corresponds to t = 1 which not in the range of t, th us this point has to be excluded. (iii) d 2d x t  , 2 d4 d y tt 2 d d d 2 d d d y y t x t x t   Alternatively 8 1y x  2 d8 d ( 1) y xx  When 21xt , 2 d2 d y xt Equation of tangent :  2 42 21y x r rr     At point S : 0y   2 420 2 1 xrrr     41xr Manageable for most students, except for few careless mistakes in 21xr . If 2 d8 d ( 1) y xx  is used, then x must be replaced by 2r + 1 to obtain the