2024 Y6 Timed Prac Rev Paper 2 (Soln)

RAFFLES INSTITUTION H2 Mathematics (9758) 2024 Year 6 2024 Year 6 Timed Practice Revision Practice Paper 2 Solutions Source: 2020 Year 6 Term 3 Timed Practice Section A: Pure Mathematics (60 marks) 1 The complex number z is given by ixy , where x and y are real numbers. (i) Express y in terms of x if  2arg 2z  . [2] (ii) State the values of x and y if  Re 0z  and 2z  . [1] Using these values of x and y, find the smallest positive integer n for which * n z z is a negative real number. [2] (i) [2] i 22 i ( 2 ) e e zr zr       2 2   2,kk    ,4 kk      tan 1 y x    yx ,0x Alternative Given izxy  2arg 2 32a rg or 22 3arg or is in the 2nd or 4th quadrant44 , 0 z z zz yx x           Represent 2z on an argand diagram! (ii) [1] 2, 2xy 

2      (iii) [2] 2i2z  arg 4z     i* 1 i e e n n nn zr rz r      i1 e n     1n     2,k k     (1 ) ( 21 ) ,4 1 21 ,4 nk k n kk              83 ,nk k   The smallest positive integer n is 3. Alternative 0, ( 1) , 3 , 5 ...4nn    The smallest positive integer n is 3. For * n z z to be a negative real number means that * arg n z z   2 (a) By writing  2 2 1rr  in partial fractions, find an expression for  2 2 2 . 1 n r rr  [ 3 ] (b) A geometric series has first term a and common ratio r, where a and r are non- zero and 1r  . The 3rd and 9th terms of the series are 448 and 7 respectively. Given also that the sum of the first n terms is 1197, find the values of a, r and n. [ 4 ]   (a) [3] Let   2 2 111 A BC rr rrr      2( 1 ) ( 1 ) ( 1 ) ( 1 )Ar r B rr C rr       Let 0, 2 Let 1, 1 Let 1, 1 rA rC rB         2 22 1 1 111 rr rrr        2 22 21 2 1 111 nn rr rr rrr       To apply MOD write in “order” 121 11rr r  .

3      121 123 121 234 121 345 121 456 121 321 nnn       12 1 21 121 11 nn n nn n       2 2 21 1 1 211 n r nnrr   (b) [4] 28 448, 7ar ar   6 1 64r 1 , 2r  1792a Sum of first n terms  1792 1 11971 nr r      11792 1 21if , 1197 12 2 n r       1 1 11971( ) 2 2 1792 13 4 1 25 1 2 n n      0.586n (rej as n is an integer) 11792 1 21if , 1197 12 1 2 n r       1 3 11971( ) 2 2 1792 n  

4      11() 25 1 2 n  Note that n must be an odd integer, 11()25 1 2 n    11ln ln2 512n    19, 1792, 2na r      Alternative Observe that since every term is positive, and the first term 1792 sum of terms =1197,an 1 2r  otherwise the sum can only get larger from 1792. Sum of n terms 11792 1 2 119711 2 n     1 3 11971( ) 2 2 1792 11() 25 1 2 n n     Note that n must be an odd integer, 11()25 1 2 n    11ln ln2 512 9 n n     3 The functions f and g are defined by      f: 1 6 4 , , 4xx x x  αϒ ,      2g: , xx x αϒ . (i) Sketch on the same diagram, the graphs of f and 1f , giving the coordinates of all points of intersection. [4] (ii) Explain why the composite function fg does not exist. [1] (iii) Find gf in similar form and state its range. [2]

5      (i) Clearly (0, 4), (4, 0) are solutions to     1ff x x Another solution to   1ff xx   lies on yx . 16 4 xx . By GC, x = 2.47 (3sf) The coordinates of points of intersections are (0, 4), (4, 0) and (2.47, 2.47). The graphs of f and 1f should appear as “reflections about the line y = x”. There are 3 points of intersection. Note that 16 4 22 5 xx x      (ii) fD(, 4 ]  gR[ 0 , ) Since gfRD , hence fg does not exist. (iii) gf : 16 4 , , 4xx x x     gfR0 , “similar form” means arrow notation which includes “domain” 4 A curve C has equation  41, , 1 1yk x x x x      , where k is a constant, 02 k . (i) Sketch C, labelling clearly the axial intercept(s), the coordinates of the turning points and equations of the asymptotes. [4] The graph of C is transformed by a reflection in the x-axis, followed by a translation of 1 unit in the positive x-direction, followed by a stretch with scale factor 1 2 parallel to the y-axis. (ii) Find the equation of the resulting curve in the form  f.yx [3] y x

6      (i)  41 1yk x x   When 0x , 4yk coordinates are 0, 4k   When 0y ,    2 410 1 41 1 41 kx x kx x x k         Not applicable since 0k  . The graph does not cut the x-axis.  2 d4 2 01d 1 y kxx kx        When 21x k   , 2411 4 211 yk k k k          When 21x k   , 2411 4 211 yk k k k         Note that 02 k so 210 k   Turning points are in quadrant 1 and 3. (ii) Applying transformations to C :  41 1yk x x   Reflection in the x-axis (replace by yy  )   41 1 41 1 yk x x yk x x        Translate 1 unit in the positive x-direction (replace by 1xx  )              x y

7        411 11 4 yk x x yk x x        Scale with a factor 1 2 parallel to the y-axis (replace by 2yy ) 42 2 2 yk x x kxy x     5 The curve C has equation  21 2 e. x y   (i) Sketch C, labelling clearly the coordinates of the axial intercept(s) and turning point(s), if any. [2] (ii) Show that the equation of the tangent to C at the point where x p  can be expressed as 21 22(2 1) e 2 1 p px yp p     . Hence find the equations of the tangents to C which passes through the origin. [4] (iii) The straight line y = mx intersects C at two distinct points. State the range of values of m. [2]   (i)                   No x-intercepts, y-intercept at 1 40, e   and min turning point at 1 ,12   . (ii)  21 2 e x y       21 21 2 d 2ed xy xx   y x 1 ,12    1 4e

8      Gradient of tangent at xp is   21 21 2 d 2ed p xp y px      Equation of tangent is  2211 22e( 2 1 ) e pp yp x p         2 2 1 22 1 22 1( 2 1 ) 2e (2 1) e 2 1 (shown) p p pxp py px yp p           When tangent passes through the origin, 0, 0xy    221 0 21 1 0 pp pp       1 or 12p p   At 1d, 2 e2d yp x  Equation of tangent: 2eyx At 1 4d1, ed yp x Equation of tangent: 1 4eyx           (iii) For a straight line passing through the origin to intersect the curve at two points, the straight line must cut the curve in the region bounded by the 2 tangent lines pasing through the origin, 1 4eyx  and 2eyx . 1 4e or 2emm     y x

9      6 A frigate is stationed at position  1, 2, 0F . Two submarines 1S and 2S are under the sea surface. Submarine 1S is at position  2, 1, 1A   and travelling in a path parallel to vector –3 i + 2 j – k. An enemy submarine 2S is detected at position 3, 2, 2B  travelling in a path parallel to vector –2i – 3j + k.  (i) Determine if the paths of the submarines will intersect each other. [3] (ii) The enemy submarine 1S will launch a torpedo at the frigate when it is at a point P in its path that is closest to F. Find the co-ordinates of P. [4] (iii) Find a cartesian equation of the plane  that contains F and the path of 2S . Calculate the acute angle between  and the x-y plane. [4] (iv) A depth charge is a countermeasure used against submarines. The frigate releases