2024 H2 MATH P2 with Solutions

2024 GCE A-Level H2 Mathematics 9758 Paper 2 Page 1 of 23 Section A: Pure Mathematics [40 marks] 1 Do not use a calculator in answering this question. The complex number 2i−+ is denoted by z. (a) Find the real numbers a and b such that *z az b=+ . [2] Solutions 2iz=− + *z az b=+ 2 i ( 2 i) 2i ab aba − + = − − + =− + − Comparing imaginary and real parts respectively: 1a=− 22 ab− =− +  2 2( 1) 4b=− + − =− The complex number 1 3i− is denoted by w. (b) Without using a calculator, evaluate wwz z− . Give your answer in the form icd+ , where c and d are real numbers. [4] Solutions 1 3i(1 3i)( 2 i) 2i 1 3i 2 i2 i 6i 3 2 i 2 i 2 i 6i 31 7i 41 5 5i1 7i 5 1 7i 1 i 2 6i wwz z −− = − − + − −+ − − −=− + + + − − + − − − − + −= + − + −+= + − = + + − =+ (i.e. 2c= , 6d = )

2024 GCE A-Level H2 Mathematics 9758 Paper 2 Solutions Page 2 of 23 2 A gardener designs a flower bed ABCDE in the shape of a rectangle with an equilateral triangle on one of the shorter sides. Side AE is of length a m and side ED is of length b m (see diagram). The total perimeter of the flower bed is 20 m. Find the maximum possible area of the flower bed, showing that it is a maximum value. Give your answer correct to 4 significant figures. [6] Solutions Perimeter 3 2 20ab= + =  310 2 ab=− Let area of flower bed be y 2m . ( ) 2 2 1 ( )( )sin 602 3310 2 2 2 3610 4 y ab a a aaa aa = +  = − +   −=+ ( )d 3610d2 y a a −=+ When y is maximum, d 0d y a = ( )3610 0 2 a−+= ( )63 102 a− = 20 63 a= − 2 2 d 36 2.13 02d y a −= −   The area y 2m is a maximum when 20 63 a= − . Maximum area ( ) 2 3620 2010 46 3 6 3     −=+    −−    223.43 m= (4 s.f.)

2024 GCE A-Level H2 Mathematics 9758 Paper 2 Solutions Page 3 of 23 Solutions Comment: From what was computed above, a = 4.68, b = 2.97 (3 s.f.) But question says that the equilateral triangle is on the shorter side. In fact, based on how the question is phrased, there is no maximum value for the area. Non-Calculus Method ( ) 23610 4 aya −=+ Let 63 ( 1.067)4k −= 2 2 22 2 10 10 55 5 25 y a ka k a a k k a k kk ka kk =− =− −    =− − +       =− − + The area y is a quadratic expression in a with a negative coefficient for 2a . Hence it has a maximum turning point at 5 25, kk  . Hence maximum possible area 2 425 63 23.43 m (to 4 s.f.) =  − = 3 The function f is such that 2f ( ) 4 7x x x=− + + , for x . (a) Find the range of f. [2] Solutions 2 2 2 f ( ) 4 7 ( 2) 4 7 ( 2) 11 11 x x x x x =− + + =− − + + =− − +   Range of f is f ( , 11]R = −

2024 GCE A-Level H2 Mathematics 9758 Paper 2 Solutions Page 4 of 23 (b) Sketch the graph of f, giving the exact coordinates of the points where the curve crosses the axes. [2] Solutions 2f ( ) 4 7y x x x= =− + + Let 0:x= 7y=  y-intercept (0, 7)= Let 0y= : 2( 2) 11 0x− − + = 2 11x− = 2 11x=  x-intercepts (2 11, 0)=− and (2 11, 0)+ The function g is such that 2g( ) 1x x=+ , for 1 10x . (c) Explain how you know 1g− exists. [1] Solutions Any horizontal line yk= , where k , cuts the graph of g( )yx= at most once. Hence g is a one-to-one function, and hence 1g− exists. Alternative 2 2g ( ) 0x x  =−  for 1 10x This shows g is a decreasing function, hence it is a one-to-one function and 1g− exists. x y O x y O

2024 GCE A-Level H2 Mathematics 9758 Paper 2 Solutions Page 5 of 23 (d) Find the value of 1fg (1.5)− . [3] Solutions Let 2g( ) 1yx x= = +  12 g ( )1xy y −== −  1 2g ( ) 1x x − = − for 1.2 3 x 1 2 2fg (1.5) f 1.5 1 f (4) (4 2) 11 7 − =  − = =− − + = 4 The sum of the first n terms of the series T is given by 322 8 4n n n−− . (a) Find an expression for nt , the nth term of series T. [2] Solutions ( ) ( ) 3 2 3 2 3 2 3 2 2 2 2 2 2 8 4 2( 1) 8( 1) 4( 1) 2 8 4 2( 3 3 1) 8( 2 1) 4 4 2( 3 3 1) 8 2 1 4 6 6 2 16 8 4 6 22 6 nt n n n n n n n n n n n n n n n n n n n n n nn = − − − − − − − − = − − − − + − + − + + − =− − + − + − + − = − + − + − = − + (Check: When 1n= , 322(1) 8(1) 4(1) 10− − =− and 6(1) 22(1) 6 10− + =−  Expression for nt holds for 1n ) The nth term of the series U is given by 50 204nun=− . (b) Find the values of n for which nnut = . [2] Solutions nnut = 250 204 6 22 6n n n− = − + 26 72 210 0nn− + = 6( 5)( 7) 0nn− − =  5 or 7n=

2024 GCE A-Level H2 Mathematics 9758 Paper 2 Solutions Page 6 of 23 4 [Continued] The nth term of the series V is given by 3 16nvn=+ . (c) Find the smallest number greater than 100 that is in both series U and series V. [2] Solutions Method 1 50 204 100nun=−  304 6.0850n=  100nu  for all 7n 3 16 100nvn= +   84 283n=  100nv  for all 29n Using GC, the smallest number greater than 100 that is in both series U and series V is 196 ( 8 60 196uv== ). Method 2 Series U is an arithmetic series with common difference 50. Series V is an arithmetic series with common difference 3. From GC, 5 10 46uv== is the smallest number that exists in both series U and series V. To get the next number that exists in both series, we add the lowest common multiple of 50 and 3, i.e. 150.  46 150 196+= is the smallest number greater than 100 that exists in both series. The nth term of the series W is given by 23 5 7nw n n= − + . (d) (i) Explain why all the terms in series W are odd numbers. [2] Solutions 23 5 7nw n n= − + When n is odd, 2 oddeven (3 7) 5 is oddnw n n= + − . When n is even, 2 evenodd (3 7) 5 is oddnw n n= + − . Hence all terms in series W are odd numbers.

2024 GCE A-Level H2 Mathematics 9758 Paper 2 Solutions Page 7 of 23 Solutions Alternative Method 1 Case 1: n is even (i.e. 2nk= where k + ) 2 2 2 2 3(2 ) 5(2 ) 7 12 10 7 2(6 5 3) 1 21 k k w k k kk kk K = − + = − + = − + + =+ where 26 5 3kK k k= − + is an integer. Hence 2kw is an odd number. Case 2: n is odd (i.e. 21nk=− where k + ) 2 21 2 2 2 3(2 1) 5(2 1) 7 12 12 3 10 5 7 12 22 15 2(6 11 7) 1 21 k k w k k k k k kk kk L − = − − − + = − + − + + = − + = − + + =+ where 26 11 7kL k k= − + is an integer. Hence 21kw − is an odd number. Hence all terms in series W are odd numbers. Alternative Method 2 (induction argument) 22 1 3 5 7 3( 1) 5( 1) 7 6 3 5 68 2(3 4) nnw w n n n n n n n −− = − + − − + − − = − − =− =−  The difference of any two consecutive terms in W is even. 2 1 3(1) 5(1) 7 5w = − + = , which is odd. Since an odd and even number add up to produce an odd number, (by induction) all the terms of W are odd. (ii) Hence explain why series U and series W do not have any terms in common. [1] Solutions 50 204 2(25 102) 2 n n un n M =− =− = where 25 102nMn=− is an integer. Hence all terms in series U are even numbers. Since all terms in series W are odd numbers, and a number cannot be both even and odd, there are no common terms in series U and series W.

2024 GCE A-Level H2 Mathematics 9758 Paper 2 Solutions Page 8 of 23 5 (a) The graph of f ( )yx= intersects the x-axis at the point ( , 0)a and the y-axis at the point (0, )b . (i) The graph of f ( )yx= is shown in Fig. 1. The scales on the x- and y-axes are the same. Sketch the graph of 1f ( )yx −= on Fig. 1, labelling the intersection with the axes. [1] Solutions (ii) The graph of f ( )yx= is also shown in Fig. 2. The scales on the x- and y-axes are the same. Sketch the graph of 12f ( 1)yx −=− on Fig. 2, labelling the intersection with the x- axis. [2] Solutions (b, 0) (0, a)

2024 GCE A-Level H2 Mathematics 9758 Paper 2 Solutions Page 9 of 23 5 (b) The graph of g( )yx= intersects the x-axis at the points ( , 0)s , ( , 0)t and ( , 0)u and the y-axis at the point (0, )v . (i) The graph of g( )yx= is shown in Fig. 3.1. Sketch t