2024 Y6 Timed Prac Rev Paper 1 (Soln)

RAFFLES INSTITUTION H2 Mathematics (9758) 2024 Year 6   2024 Year 6 Timed Practice Revision Practice Paper 1 Solutions Source: 2019 Year 6 Term 3 Common Test 1 The curve 1C with equation 22 22 14 xy aa is transformed to curve 2C by a translation of a units in the positive x-direction, followed by a stretch with scale factor 2 parallel to the x-axis, and followed by a reflection in the y-axis, where a is a positive constant. (i) Find the equation of 2.C [3] (ii) Describe the shape of 2C geometrically. [2]   1(i) [3] 22 22 22 22 2 2 22 a translation of units in the po sitive -direction i.e replace with a stretch with scale factor 2 parallel to 14 () 14 1 2 the -axis i.e replace 4 with 2 xy aa xa y aa xy a a axx x a xxx a          22 22 22 22 2 22 a reflection in the -axis i.e r (2 )11 44 (2 ) 1( 2 ) ( epla 2)44 ce with xa y aa xa y xa y aa yx x a           It is good practice to simplify as you go along. Otherwise, you may end up with a very complicated expression which makes simplification much harder. (ii) [2] C 2 is a circle with the centre (2, 0 )a , with a radius of 2a units.

2    2 Do not use a calculator in answering this question. The equation 2 23 4 0zz  has two complex roots 1z and 2 ,z where 10a r g ( ) 2z  . (i) Find 1z and 2z in polar form. [3] (ii) Show that 4 1 2 2 4z z  . [1] (iii) Find the set of possible values of n, ,n  for which 1 13 i nz  is a real number. [3] 2(i) [3] 2 23 4 0 2 3 12 4(4) 2 23 4 2 3i zz z           2 2 12 31 2zz    1 1 1arg( ) tan 3 z    πi 6 1 3i2 e z    and πi 6 2 3i 2 ez     Avoid the tedious method of replacing z by ix y in the equation and comparing real and imaginary parts. (ii) [1]   4πi 6 4 1 22 πi2 6 4 42i π i π66 2 2e 2e 2 e4 e 4 ( S h o w n )2 z z             (iii) [3] 1 1 ππarg arg( ) arg(1 3 i) 6313 i nzn nz      For 1 13 i nz  to be a real number, ππ π,w h e r e 63 n kk   Setting ππ π 62 , 63 n kn k k       Therefore the set of possible values is  |6 2 , nn k k  .

3    3 The function f is defined as follows. 2f: 3 ,xx x  for xa . (i) State the largest value of a for which the function 1f exists. Hence find  1f x and state the domain of 1f for this value of a. [3] In the rest of the question, use 1a . The function g is defined on an interval A as follows.  g: l n 3 2 ,xx  for .xA (ii) A student suggests 2 possible intervals for A as follows. (a) 21, 33  (b) 1 , 02  Determine which of the above intervals will result in the existence of the composite function fg, justifying your answer. In the case(s) that fg exists, find the range of fg. [4]   3(i) [3] From the graph, largest value of 3 2a  .  2 2 Let f 3 39 24 yx xx x     0 3

4    39 24 39 3 3 since .24 2 2 xy yx       Hence  1 39f 24xx   and domain of 1 92 7f, . 44   (ii) [4] For fg to exist, we need gfRD 1 , 1 .  For (a) where   g 21D, 33   ,  gfR, 0 D .     Hence fg does not exists for (a). For (b) where g 1D, 0 2  , gfR ln 2, ln 2 ( 0.693,0.693) D .   Hence fg exists for (b).     g g 22 fg 1D, 0 2 g Rl n 2 , l n 2 f R ln2 3ln 2, ln 2 3ln 2 ( 1.60, 2.56) (to3s.f.)        Take note that f is a decreasing function in this domain.

5    4 (a) A geometric series has first term 22 s i n  and second term . (i) Show that the series is convergent for 2    . [2] (ii) It is given that 3 4   and nS denotes the sum of the first n terms of the series. Find nS and hence determine the exact value of .S [3] (b) (i) Show that 1 1 1 nn nn    , where n   . [1] (ii) Hence find the least possible value of N such that 4 1 100. 1 N n nn   [2]   4(a) (i) [2] Common ratio, sin 2 2sin cos cos 22 s i n 22 s i n 2 r       . Since cos 1 1 cos 1 since cos 1 for 222 2 r           , the series is convergent (shown). (ii) [3] Now, first term, 3122 s i n 22 24 2 a    and 1311 1cos .4222 2 r       Hence 121 2 411.1 321 2 n n nS           As n  , 1 02 n  , thus 4.3 nSS  (b) (i) [1]  11 1 11 1 1 1 1 (Shown). nn nn nn nn nn nn nn         sin 2

6    (ii) [2]  4 4 1 1 1 54 65 76 1 1 12 1 2 100 10403 N n N n nn nn NN NN N NN                      Therefore the least possible value of N is 10404.

7    5 The parametric equations of a curve are 2xt , 21y t t  , where t  , 1 .2t  (i) Sketch the curve, stating the equations of any asymptotes and the coordinates of any points where the curve crosses the axes. [3] (ii) The tangent to the curve at the point 2, 21 pp p    intersects the x-axis at point A and the y-axis at point B. Find, in terms of p, an expression for the area of the triangle OAB. [5]   5(i) [3] Note that you should find the Cartesian equation from the set of parametric equations to derive the asymptotes. You should also indicate the intercepts and equations of the asymptotes clearly on your sketch. (ii) [5] 22 2 d 1d d( 2 1 ) 2 1 d ( 21 ) ( 21 ) dd d 1 dd d ( 2 1 ) x t yt t tt t yy x xtt t        When tp , 2 d1 d( 2 1 ) y xp 

8    2 2 22 1 221 ( 21 ) 12 ( 1 ) (2 1) (2 1) pyx p pp pyx pp             When 0,y 22( 1)xp , therefore  22( 1),0Ap  . When 0,x 2 2 2( 1) (2 1) py p   , therefore 2 2 2( 1)0, (2 1) pB p     . It follows that the area of the triangle OAB is 22 2 2( 1) (2 1) p p   .

9    6 (a) Given a and b are two non-zero and non-parallel vectors and show that the length of projection of c onto a and the length of projection of c onto b have the same magnitude. [3]   (b) The equations of line 1,l planes and are 2 1 1 31:, 22 :2 3 5 , :3 2 , xylz xy z ax y z b         respectively. ( i ) If lies on , find the values of a and b. [2] For the rest of the question, does not lie on and intersects at point F. ( i i ) Find the coordinates of F. [3] ( i i i ) Find, in terms of a, a direction vector of the line of intersection between and . [2] (iv) Find the relationship between a and b if F also lies on . State, in terms of a, a vector equation of . [2]     6(a) [3] Length of projection of c onto a: Length of projection of c onto b: ˆ     2 aca b a+ ab a ba a=+ b aa ba=+ b aa =ba + ba =     2 ˆ    bcb b a + a b b a=ab + bbb ab=ab + b =ab +ab =   Since ba ab and ab = ba , then ˆˆca cb i.e. the length of projection of c onto a and of c onto b have the same magnitude. ,cb aa b 1 2 1l 2 1l 2 1l 1 2l 1 2 2 2l

10    (b)(i) [2] 1 32 :1 2 , 01 l                 r    For line to lie on plane , its dir ection vector must be parallel to (i.e. perpendicular to the normal of ) and the point (3,1,0) on must lie on . Direction vector perpendicular to normal: 2 23 0 12 2a a              Point (3,1,0) lies on the plane : 32 13 02 9 b b                Alternative Method: 32 12 3 2 a b                     32 3 62 33 26 2 0 2 9 aa