ACSI 2021 Y3 Core Mathematics Paper 1 Solution

ACS(Independent)/Y3IPCoreMathP1/2021/FinalExamination 1 Anglo - Chinese School (Independent) FINAL EXAMINATION 2021 YEAR 3 INTEGRATED PROGRAMME CORE MATHEMATICS PAPER 1 Friday 1st October 2021 1 hour 30 minutes Candidates answer on the Question Paper. No additional materials are required. INSTRUCTIONS TO CANDIDATES • Write your index number in the boxes above. • Do not open this examination paper until instructed to do so. • You are not permitted access to any calculator for this paper. • Answer all questions in the spaces provided. • Unless otherwise stated in the question, all numerical answers must be given exactly or correct to three significant figures. • The maximum mark for this paper is 80. Paper 1 containts more straight-forward questions, but many students made careless mistakes in expansion, simplification and factorisation. They do not check the workings and verify the answers with the knowledge or the facts they learned. Many students still could not give the form of equation properly and they wrote expression instead. _______________________________________________________________________________________ This paper consists of 15 printed pages and 1 blank page. [Turn over For Examiner’s Use Candidate Index Number

ACS(Independent)/Y3IPCoreMathP1/2021/FinalExamination 2 Full marks are not necessarily awarded for a correct answer with no working. Answers must be supported by working and/or explanations. Where an answer is incorrect, some marks may be given for a correct method, provided this is shown by written working. You are therefore advised to show all working. Answer all the questions in the spaces provided. 1. [Maximum mark: 9] (a) Evaluate 34 33 33 4 − − − . [3 marks] (b) Make q the subject of the formula, 32 .5 qpp q −= + [3 marks] (c) Factorise 3 2 2 2 12 4 3x x y xz yz− − + completely. [3 marks] 1 (a) 3344 3333 393 44 3 4 43 3 3 4 5 3 9 4 5 11 5 − = − −− − =− − =− =− = (b) ( ) 2 22 22 2 2 32 5 32 5 5 3 2 3 2 5 25 3 qpp q qpp q p q p q p q p p p ppq p −= + −= + + = − − =− − −−= − © ( ) ( ) ( )( ) ( )( )( ) 3 2 2 2 22 22 12 4 3 4 3 3 43 2 2 3 x x y xz yz x x y z x y x z x y x z x z x y − − + = − − − = − − = + − − Students made careless mistake in simplifying the fraction in fraction. Students did not factorise completely the last step. Since the question wants q to be subject, q must not appear on the right hand side of the equation.

ACS(Independent)/Y3IPCoreMathP1/2021/FinalExamination 3 2. [Maximum mark: 8] (a) Sketch the graph of 2 9yx= − clearly labelling the coordinates of the axes -intercepts and turning point. [2 marks] (b) The curve 2 9yx= − meets the line 26yx=+ at points P and Q. Find the coordinates of P and Q. [4 marks] (c) Hence, find the area of POQ where O is the origin. [2 marks] (a) (b) ( )( ) ( ) ( ) 2 2 2 When 9, 9 2 6 2 15 0 5 3 0 5 0 or 3 0 5 3 2 5 +6 2 3 6 16 0 yx xx xx xx xx xx yy =− − = + − − = − + = − = + = = =− = = − + == The coordinates of P and Q are (5, 16) and (-3, 0). (c) Area of POQ ( ) 2 5 3 0 51 16 0 0 162 1 0 0 0 48 0 02 24 units = − = + + − − − − = y O x 2 9,yx= − 2 9yx= − The two x-intercepts and turning point must be clearly labelled. Since the question asked for a sketch, students SHOULD NOT plot the graph to scale! This question is meant to be a simultaneous equation. Solution to simultaneous equation = point(s) of intersection between the two equations. However, there are some students who tried to solve them by plotting the two graphs to scale. This is time consuming and not making use of the right concept. Alternate solution: Area = 1 2 × 3 × 16 = 24

ACS(Independent)/Y3IPCoreMathP1/2021/FinalExamination 4 3. [Maximum mark: 8] (a) Solve 2 65.yy + [2 marks] (b) Solve ( )21 4 2 5 192 x xx+ − +  + and hence state the integer values of x that satisfies the inequality. [6 marks] 3 (a) ( )( ) 2 2 60 65 5 3 2 0 23 y y y yy y y + + − − −   (b) ( ) ( ) ( ) 21 4 2 5 192 21 4 2 4 2 5 192 2 1 8 16 4 8 5 19 10 17 27 9 31.7 x xx x x x x x x x x xx x x + − +  + + − + − +  + + − − − −  + − −  − − 3 1.7x−  − The integer values are -3 and -2. 4. [Maximum mark: 5] A cuboid has a square base of length 12+ units and the volume is 7 5 2+ units3. Express the height of the cuboid, H, in the form 2,ab+ where a and b are constants. What do you notice about this cuboid? [5 marks] 4 ( )( ) ( ) ( ) ( ) ( ) 1 2 1 2 7 5 2 7 5 2 3 2 2 3 2 2 3 2 2 21 14 2 15 2 10 2 9 4 2 21 20 2 15 14 1 2 units H H + + = + +−= +− − + −= − = − + − =+ The height of the cuboid is equal to the length of its square base. Hence it is a cube. Students must state the integer values which satisfies the final inequality. It is preferred that students state the cuboid is a cube. However, students MUST compare the lengths and not merely state the lengths of the sides.

ACS(Independent)/Y3IPCoreMathP1/2021/FinalExamination 5 5. [Maximum mark: 5] Given that A is an acute angle and 1cos 3 ,A= (a) find the value of sin ,A [2 marks] (b) hence, show that 2 tan 1 3sin 22. 4 A A − =− [3 marks] 5 (a) 2 sin 31 3 8 2 2 / 33 A= − = (b) 2 tan 1 3sin 4 2 1 223 3 12 2 or 2 (shown) 442 A A − −=    = − − Angle A is acute and it is in Quadrant 1. Hence all the trigo ratios are positive. Some students gave sin A as a negative value which don’t make sense. When asked to “show”, students SHOULD not cross multiply with the right-hand side of the equation. Students are expected to manipulate the left- hand side of the equation until it looks like the expression on the right hand side.

ACS(Independent)/Y3IPCoreMathP1/2021/FinalExamination 6 6. [Maximum mark: 9] Given that the roots of the quadratic equation 23 2 3 0xx− − = are  and . (a) State the value of + and of  . [2 marks] (b) Show that 22 22 .9+= [2 marks] (c) Find the quadratic equation with roots 2+ and 2.+ . [5 marks] (a) 2 2 3 2 3 0 2 103 2 , 1.3 xx xx    − − = − − =  + = =− (b) ( ) ( ) 222 2 2 2 2 1 3 22 (shown)9     + = + − = − − = (c) ( ) ( )2 2 3 3     + + + = + 23 3 2 =  = ( )( ) 222 2 2 4 2       + + = + + + ( ) ( ) ( ) 2225 222 5 19 1 9   = + + = + − =− Quadratic equation, 22 12 0 / 9 18 1 09x x x x− − = − − = NOTE: It is a MUST for students to equate the expression to 0!

ACS(Independent)/Y3IPCoreMathP1/2021/FinalExamination 7 7. [Maximum mark: 8] The points ( ) ( )2,0 , 1, 1AB− and ( )3, 3D −− are the three vertices of a rhombus ABCD. E is a point at the foot of the perpendicular from A to BD. (a) Find the equation of CD. [3 marks] (b) Find the length of AE, leaving your answer in surd form. [3 marks] (c) Find the area of the rhombus. [2 marks] (a) Coordinates of C are (0, -2) Equation of CD, ( ) ( )0133 21 1 23 y x yx −= −− − −  −− =− (b) Method 1 Method 2 Coordinates of E ( ) 3 1 3 1, 22 1, 1 − + − += = − − Distance of AE ( ) ( ) 2 2 1 2 1 0 2 units = − − − + − − = (c) Area of rhombus ( ) ( ) ( ) 2 2 1 2 3 0 11 2 Are