2024 Y6 Timed Prac Rev Paper 4 (Soln)

RAFFLES INSTITUTION 2022 Year 6 H2 Mathematics Common Test Questions and Solutions with comments     Page 1 of 19   1 The curve C has equation 2 24 , xxyx b xb   . (i) Given that y cannot take values between 2 and 6, show that b = 2 using an algebraic method. [3] (ii) Describe a sequence of 2 transforma tions that transform the graph of C onto the graph of 2 54 , 0xxyx x  . [3] (i) [3] Method 1 Since y cannot take values between 2 and 6, then the graph of this rational function will have 2 stationary points, with minimum point at 6y and maximum point at 2y . Consider 2 2 24 2 42 0 xx xb xb     and 2 2 24 6 84 60 xx xb xx b     Since 2y and 6y are tangents to 2 24 ,xxy xb   then the discriminant of the above 2 quadratic equations is equal to 0. 4(4 2 ) 0 2 b b   and 64 4(4 6 ) 0 2 b b   Thus, b = 2. Method 2 2 2 2 24 24 (2 ) ( 4 ) 0 xxy x b yx by x x xx y b y       2 2 Discriminant ( 2 ) 4(1)(4 ) (4 4 ) 12 yb y yy b       Since y cannot take values between 2 and 6, then 26 y  is the solution for 2 (4 4 ) 12 0yy b  . We can rewrite 26 y  as   2 26 0 41 2 0 yy yy    Comparing coefficients of y, b = 2. 6y 2y 2024 Y6 H2 Math Timed Practice Revision Paper 4 Solution

Page 2 of 19 (ii) [3] We can rewrite the 2 equations as 2 24 4 22 xxyx x x      and 2 54 4 5xxyx x x        Replace by 2 Replace by 3 44 22 432 45 xx yy yx yx xx yx x yx x                The sequence of transformation is 1) Translate C in the negative x-direction by 2 units to get 42yx x , 2) followed by a translation of 3 units in the positive y- direction. Note that for such question, you need to describe (in words) the transformation used. Stating “Replace x by x+2” is not a description, but is part of your working. Need to use the proper wordings for describing transformations. In this question, “translate by k units positive/negative x/y-direction”. It is mentioned by the Cambridge report that “along the x-axis”, “in the x-axis” or “on the x-axis” are not accepted.

Page 3 of 19 2 The function g is defined by g: s i n ,x xx for ,x  .x  (i) Show that g0 x  for .x  Hence, or otherwise, show that g has an inverse. [2] (ii) Sketch on the same diagram the graphs of g( )yx  and 1g( )yx  . [3] (iii) Show that the composite function 2g exists. [1] (iv) Find the exact solutions of the equation   2g. x x [2] (i) g( ) sinx xx g( ) c o s 1x x  For ,x  1c o s 1 0c o s 12 0g ( )2 x x x      Thus, g0 x  for .x  Since g is a strictly increasing function on x  , and so g is a one-one function. Hence, g has an inverse. Alternatively, Any horizontal line yk , where  ,k   , cuts the graph of g( )yx at one and only one point. Thus, g is a one-one function and 1g exists. Common mistake: d sin cosd x xx  Quite a few wrote  k  This not correct. For example, consider the line 10y , the horizontal line does cut the graph of g( )yx at any point.  gRk Need to state the range of g explicitly. (ii) There are a few scripts who drew the above graph. Note that  g (0) 2 grad of line yx    can use the GC to draw the graph of g( )yx and yx on the same diagram  the x- and y- scale should be the same (iii)  ggR[ , ] , D ,    Since ggRD , 2g exists. Some use the wrong symbol/notation:  gR[ , ]    ( should be = )  gR[ , ]  ( missing = )  ggRD 1g( )y x g( )y x y x g( )y x y k

Page 4 of 19 (iv)      2 12 1 1 g gg g gg xx x x x x      From the graphs of g( )yx and 1g( )yx  in (ii), they intersect at , 0 , x   The solutions are , 0 , x  

Page 5 of 19 3 (i) Prove by the method of differences that 1 1 11 . 1 N r N rr    [3] (ii) Find 1 4 1 1 N r rr    in term of N. [2] (iii) Using part (i), show that  1 1 21 1 N r N r   [2] (i)  1 1 1 1 1 1 11 11 1 (1 ) 1 21 32 43 . . . 1 1 11 N r N r N r N r rr rr rr rr rr rr rr NN NN N                         Most are able to rewrite the term by rationalizing the denominator, and obtain a difference of two related terms. (ii) 11 3 41 1 111 111 1( 41 ) 2 NN rr r rr rr rr N N            (iii) For 1r ,  11 1 1 1 11 1 11 1 1111 2 1 21 1 NN rr N r N r rr rr rr rr rr rr N r N r                 

Page 6 of 19 4 A curve C has parametric equations ec o s,tx t es i n .tyt  (i) Find the equation of the tangent to C at the point with parameter p. Give your answer in the form ym x c . [5] (ii) The tangent meets the x-axis at point A and the y-axis at point B. Find, in terms of p, the area of the triangle OAB, where O is the origin. [3]   (i)     ec o s d es i n e c o sd e cos sin t tt t xt x ttt tt       es i n d ec o s es i nd e cos sin t tt t yt y ttt tt       2d ed ty x  Equation of tangent:    2 2 2 es i n e e c o s ee s i n e c o s ee s i n c o s pp p pp p pp yp x p yx p p yx p p           Need to use product rule to differentiate x and y with respect to t. (ii) At A, 0y and  2 es i n c o s es i n c o se p p p ppx pp      At B, 0x and es i n c o spyp p Area of triangle OAB    2 2 1 es i n c o s e s i n c o s2 1 sin cos2 1 sin cos2 pp pp pp pp pp        Quite a number of solutions gave the answer as  21 sin cos2 p p , and not realizing that this expression is always negative. Actually either OA or OB is negative, so there is a need to find the absolute value for the lengths when finding the area.

Page 7 of 19 5 The complex number z satisfies the equation f( ) 0z  , where   2f( ) 1 i 2 i 1 0 2 0 iza z z   and a is a real number. It is given that one root is of the form 1i b , where b is an integer . (i) Show that b satisfies the equation   43 2 21 2 4 2 9 0bb b b   . [3] (ii) Hence find a and b, and the other root of f( ) 0z  . [4] (i) [3]        2 2 22 22 1 i 1 i 2i 1 i 10 20i 0 1i 1 2 i 2 i + 21 0 2 0 i 0 1 2 i i 2 i 2 2 10 20i 0 22 9 i 2 2 2 0 ab b ab b b bb a a b a b i b ba b b b a a b                         Comparing 2 2 Real part: 2 2 9 0 92 (1)2 ba b b bba b      Imaginary part: 222 2 0 ( 2 )baa b     Subst (1) into (2),  22 22 2 2 43 2 92 9222 2 0 22 49 2 9 2 4 4 0 21 2 4 2 9 0 bb bbbb b bb b b b b b bb b b                Quite a few are careless in simplifying the equation. There are a few who wrote that 1i b (conjugate pair) is a root of f( ) 0z  . However, not all the coefficients of f( ) 0z  are real numbers. Thus, the theorem cannot be used. (ii) [4] Solving the quartic equation in (i) using GC and since b is an integer, we get 3b . Subst 3b into eqn (1), we get 2a . Method 1 Let the other root be 0z .         2 012 i 2 i 1 02 0 i 12 i 13 izz z z z        Comparing the constant term,      0 0 10 20i 1 2i 1 3i 10 20i 1.8 2.6i, using GC12 i13 i z z        The other root is 1.8 2.6i     There are a few who wrote    2 0 12 i 2 i 1 02 0 i 13 i 12 i zz zz z            This is not wrong, but do note that 0z is not the root of the equation f( ) 0z  . The ro