RI 2023 MATH TP SOLN

RAFFLES INSTITUTION H2 Mathematics (9758) 2023 Year 6 ________________________________________________ 2023 Y6 H2 Math Timed Practice: Solutions with comments Page 1 of 17       2023 Year 6 H2 Mathematics Timed Practice: Solutions with Comments 1 (a) (i) Find 2sin dxx . [2] (ii) Hence or otherwise, show that 22 11 1sin d sin 2 cos 2 .44 8xx xx xx x c   [3] (b) (i) Sketch the graph of 2sinyx x   for 10 π2x , labelling the end points clearly. [2] (ii) Hence, find the exact area of the region bounded by the graph of 2siny xx , the x-axis and the line 1 π2x . [3] Solutions Comments (a)(i) [2]  2 1c o s 2sin d d 2 1 1c o s 2 d2 1s i n 2 22 1s i n 2 24 xxx x xx xxc xxc          (a)(ii) [3] 2 2 22 2 11 11sin d sin 2 sin 2 d24 24 11 1 1 c o s 2sin 224 2 24 2 11 11 sin 2 cos 224 48 11 1 sin 2 cos 2 (shown)44 8 x xx x x x x x x xxxx x c xx x x x c xx x x c                            “Hence” method: Integration by parts: ux   2d sind v xx    d 1d u x    1s i n 2 24 xvx

Raffles Institution H2 Mathematics 2023 Year 6 __________________________________________________________________________________________     ________________________________________________ 2023 Y6 H2 Math Timed Practice: Solutions with comments Page 2 of 17       y x 2sinyx x ππ,22   O 0, 0 π 2x Alternatively, 2 2 2 2 2 1c o s 2sin d d 2 1 cos 2 d42 1 sin 2 sin 2 d42 2 2 sin 2 1 cos 2 44 4 2 11 1 sin 2 cos 2 (shown)44 8 xxx x x x x xx x xx x xx xx x x c xx x x c                    (b)(i) [2] Students needs to zoom to the appropriate scale to see the shape of the graph. Do observe the gradient as well.

Raffles Institution H2 Mathematics 2023 Year 6 __________________________________________________________________________________________     ________________________________________________ 2023 Y6 H2 Math Timed Practic e: Solutions with comments Page 3 of 17       (b)(ii) [3]       π 22 0 π 22 0 2 2 2 2 2 Required area sin d 11 1 sin 2 cos 244 8 1 π 1 ππ 1 πsin 2 cos 242 42 2 8 2 11 100 s i n 2 0 c o s 2 044 8 π 11 116 8 8 π 1 units16 4 xx x xx x x                            Some students made careless mistakes with the signs.

Raffles Institution H2 Mathematics 2023 Year 6 __________________________________________________________________________________________     ________________________________________________ 2023 Y6 H2 Math Timed Practic e: Solutions with comments Page 4 of 17       2 (a) A point R has position vector r = 1 24 35 a a        , where a is a real number. Describe geometrically the set of all possible positions of R as a varies. [2] (b) The angle between the vectors a and b is π 3 radians. (i) Given that the angle between the vectors b and 2ab is a right angle, show that 4ab . [3] (ii) Given also that 2b , find the exact length of projection of  2 ab onto  2ab . [5] Solutions Comments (a) [2] r = 1 24 35 a a         =  02 24 35 a         , a    R lies on the line passing through the point with coordinates (0,2,3) , and the line is parallel to the vector 245ij + k . Many students did not manipulate the expression to have a as a parameter. Equation of a line must take the form: r = c + a d, where c and d are constant vectors. (b)(i) [3] Angle between a and b is π 3  π 1cos 32 ab ab ab … (1) Angle between b and a – 2b is a right angle  (2 ) 0 ba b  20 ba bb  2 2 bb a … (2) Substitute (1) into (2) : 2 1 4ba b Divide throughout by b , we have 4ab . (shown)

Raffles Institution H2 Mathematics 2023 Year 6 __________________________________________________________________________________________     ________________________________________________ 2023 Y6 H2 Math Timed Practic e: Solutions with comments Page 5 of 17       (b)(ii) [5] From (bi), since  2b , 48ab  and 8ab Length of projection of  2 ab onto  2ab = (2 ) ( 2 ) 2   ab a b ab = 22 225 (2 ) (2 )     ab a b ab ab   = 22 176 44 aa b b   = 176 112 = 44 7 units (or 44 7 7 units) Reminder: 4ab DOES NOT mean 4ab The modulus sign is necessary to ensure a positive value is obtained for length of projection. Many students did not consider the dot product to find 2ab . Note that 22 aba b  in general.

Raffles Institution H2 Mathematics 2023 Year 6 __________________________________________________________________________________________     ________________________________________________ 2023 Y6 H2 Math Timed Practic e: Solutions with comments Page 6 of 17       3 Do not use a calculator in answering this question. Two complex numbers p and q are given by 33 i 22 p and 1i3q respectively. (a) Find 2*p q in the form iab , where a and b are exact real values, and *p denotes the conjugate of p. [3] (b) (i) Find the modulus and argument of p and q. [2] (ii) Write down the value of 2*p q in the form  cos isin ,r   where 0r  and ππ . [2] (c) (i) Represent 2*pq   on an Argand diagram, labelling clearly the information found in parts (a) and (bii). [1] (ii) Hence, find the exact value of 1sin π12 . [2] Solutions Comments (a) [3]           22 33*i 1 i 3 22 3 1i1i 23 3 2 3 1i 2i 23 2 3 2i 2 32 i + 2 3 2 32 1 3 i 1 3 32 36 i 32 36 pq                     32 36a   and 32 36b  Many students did not simplify the answer. (b)(i) [2] 22 33 3 22 p          ,    πarg( ) 4p      22 13 2q                ,    πarg( ) 3q    (b)(ii) [2] From (bi), πi 43ep  and πi 32eq Many students did not use the exponential form/properties to find the argument and modulus. It is very difficult to

Raffles Institution H2 Mathematics 2023 Year 6 __________________________________________________________________________________________     ________________________________________________ 2023 Y6 H2 Math Timed Practic e: Solutions with comments Page 7 of 17       2ππ ii2 34 π 2πi 43 5πi12 *3 e 4 e 12e 12e 55 π12 cos isin12 12 pq              Alternatively, 222*3 ( 2 ) 1 2pq p q     2 25arg * 2arg arg 34 1 2pq q p        2 55 π*1 2 c o s i s i n12 12pq      work from the cartesian form since students are not supposed to use GC for this question. (c)(i) [1] Many students did not illustrate the argument proportionally. Note that 5 12  is 75. (c)(ii) [2]   2 2 Re *πsin 12 * 32 36 6 2 12 4 pq pq       Do refer to the diagram in (c)(i).

Raffles Institution H2 Mathematics 2023 Year 6 __________________________________________________________________________________________     ________________________________________________ 2023 Y6 H2 Math Timed Practic e: Solutions with comments Page 8 of 17       Alternatively,  2 2 ππ 5πsin sin12 2 12 5πcos 12 Re * * 32 36 6 2 12 4 pq pq                    π 12   5π 12

Raffles Institution H2 Mathematics 2023 Year 6 __________________________________________________________________________________________     ________________________________________________ 2023 Y6 H2 Math Timed Practice: Solutions with comments Page 9 of 17       4 Two planes 1p and 2p are perpendicular. Plane 1p has equation