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− − + =