2024 Y6 Timed Prac Rev Paper 3 (Soln)

RAFFLES INSTITUTION H2 Mathematics (9758) 2024 Year 6     2024 Year 6 Timed Practice Revision Practice Paper 3 Solution Source: 2021 Year 6 Term 3 Common Test Section A: PURE MATHS (52 Marks) 1 An arithmetic seri es has first term a and common difference d, where a and d are non-zero. If the first, fifth and eleventh terms of the arithmetic seri es are equal to the first, second and third terms of a geometric series respectively, find the exact sum of the first ten terms of the geometric series, in the form ka , where k is a simplified rational number. [4] Solution Comments [4] The first 3 terms of the GP are ,4 , 1 0aa da d    2 22 2 2 41 0 4 41 0 81 6 1 0 16 2 0 (Since 0)8 ad a d aa d ad a a d aa dd a a d da d add           Let the common ratio of the GP be r. 443 8 2 aaadr aa     . Sum of the first 10 terms = 10(1 ) 1 ar r   10 10 3 12 3 12 321 2 58025 512 a a a         The question asked for a simplified rational number, which is a number of the form p q , where p and q are integers, ie, express k as a fraction in lowest terms.

Alternatively, to find the required sum of the geometric series, we just need to find r. We can do so by eliminating d from the equations: 2 4 ( 1 ) 10 (2) ad a r ad a r     (1) (1 ) 4 ard  and thus substituting into (2): we have 2 2 2 (1 )10 4 4 10( 1) 4 since 0 25 3 0 31 or 2 araa r rra rr rr           We reject r = 1, because if r = 1 then (1) (or (2)  d = 0 The required sum of the first 10 terms then follows the same working as in the first solution. Note that the justification that r cannot be 1 has to be clearly stated as it is not given in the question.

2 (i) By writing   2 21 23rr in partial fractions, find an expression for  2 2 21 23 n r rr  in terms of n. [3] (ii) Hence find the exact value of 21 21 21 19 21 21 23 23 25  . [2] Solution Comments (i) [3] Let   2 21 23 21 23 AB rr r r      22 3 2 1Ar Br  When 12,121 3rA    When 32,123 1rB      21 1 21 23 21 23rr r r    22 21 1 21 23 21 23 11 57 11 + 79 11 + 91 1 11 + 21 21 11 + 21 23 11 52 3 nn rr rr r r nn nn n                         Check your partial fractions by either 1) re-combining the 2 fractions into a single one OR 2) using GC to check both sides of the expression give the same graph (both graphs should overlap completely when sketched) Do remember to write down the first 2 cancellations as well as the last one.

(ii) [2]  9 21 21 21 19 21 21 23 23 25 2 21 23r rr           8 22 22 21 23 21 23 11 1 1lim 52 3 52 ( 8 ) 3 11 1 55 1 9 1 19 rr n rr rr n                     Alternative:     9 9 9 21 21 21 19 21 21 23 23 25 2 21 23 2lim 21 23 11lim 21 23 11lim 2(9) 1 2 3 1 19 r n n r n n r n rr rr rr n                            Do take note that the series is not finite. Hence the need to take limits or discuss what happens when n is large. Note:   1 0 as 23 nn  

3 For a curve with equation 23 38xx y y  , find (i) the x-coordinate of the point at which the tangent is parallel to the x-axis, [4] (ii) the equation(s) of the normal(s) at the point(s) where 2.y [3] Solution Comments (i) [4] 23 38xyxy  Differentiate w.r.t x 2 2 dd33 3 0dd d3 2 2 d3 3 yyxy yxx yy x xyx x       Since tangent // x-axis, d 0d y x  320 2 3 y x xy   Sub 2 3 xy into 23 38xyxy  3 2 22 3 32 2238 33 27 82 72 1 6 ) 5 4 4 0 .6329 (using 86 GC 21xx xxxx x xx x               x-coordinate of the point at which the tangent is parallel to the x-axis is 4.63 (3 s.f.) Be careful of sign errors as you manipulate the equation after differentiating. (ii) [3] At 2y  , 23 2 0 or 6 3( 2 ) 2 8 60 x xx x x     At  0, 2 , d 2d 1y x  Gradient of normal = 2 Equation of normal is 22y x  At  6, 2 , d 1d y x  Gradient of normal = 1 Equation of normal is 8yx  Do read the question carefully and remember that the equation of the normal is requested, not the tangent.

4 The polar form of a complex number z is given by iezr  , where 0r and 0 π , and the complex number 13 i.33wz  (i) Find w in exact polar form in terms of r and . [3] (ii) Given that 5 * z w is real and positive, find the possible value(s) of  in exact form, leaving your answer(s) in terms of . [4] Solution Comments (i) [3]  1 π3 itan i 3113 1 1 2i= 1 3i 1 3 e e33 3 3 3       ππ ii i 3322ee e33wr r            Alternative: Let 13 i33v . Then 12 1333v    and   3 1 3 1 3 πarg tan 3v      wv z 12 1333wv z z r    and    πarg arg arg 3wv z      πi 32 e3wr     The argument of a complex number is not found in general by simply taking tangent inverse of the imaginary part over the real part. You should ALWAYS sketch to see which quadrant the point corresponding to the complex number lies in, and work out the basic angle it makes with the real axis before finding the argument. You can also use your GC to check if you have converted to polar form correctly: (ii) [4]   π 3 π 3 55 i 5 i64 * i2 3 e3 e2e zr rw r          OR: Most students realise that there is a need to work in polar form (hence (i)), but are not careful with the conjugate in the denominator.

    5 * 5* To find arg : arg arg 5arg arg π6 3 z w zw zw         Given that 5 * z w is real and positive, ππ π 260 , 2 π,4 π since 0< 6 5 π33 3 3 π 7π 13π 6 , , 33 3 π 7π 13π , , 18 18 18            Remember that for the complex number to be real and positive, the possible arguments are even multiples of π.

5 The function f is defined by 2f : 2 5 3, , .xx x x x a   (i) State the greatest value of a such that the function 1f exists. [1] For the rest of the question, use the value of a found in part (i). (ii) Find 1f in a similar form. [3] (iii) Find the exact solution of 1f. xx  [3] (iv) The function g is defined for specific integer values of x as follows. x 5 4 1 0 2 g(x) 6 4 23 0 3 Find the value of b, where 1fg ( ) 3 .b  [1] Solution Comments (i) [1] 22 2 2 5525 3 2 2 3 44 54 9 2 48 xx x x               The greatest value of 5 4a     Alternatively, the largest value of a is the x-coordinate of the minimum point of the quadratic, which occurs when  5f' 0 4 5 0 4xx x   For students who complete the square, they should check their answers by either 1) re-expanding the completed square form OR 2) using the GC to check both forms give the same graph (both graphs should overlap completely) (ii) [3] 2 54 9Let 2 48yx    2 54 92 48 54 9 8 4 9 42 1 6 1 6 51 8 49 44 xy yyx xy           Since 5 4x , 51 84 944xy   . 1 51 4 9f : 8 49, , 44 8xx x x          It is important to write  before  choosing t