ANSWERING TECHNIQUES: SPM MATHEMATICS

Paper 2 Section A

Simultaneous Linear equation (4 m) Simultaneous linear equations with two unknowns can be solved by (a) substitution or (b) elimination. Example: (SPM07-P2) Calculate the values of p and q that satisfy the simultaneous : g + 2h = 1 4g  3h = 18 g + 2h = 1  4g  3h = 18   : g = 1  2h   into : 4(1  2h)  3h = 18 4  8h 3h = 18  11h = 22  h = 2 When h = 2, from : g = 1  2(2) g = 1  4  g = 3 1 1 Hence, h = 2 and g = 3 2

Simultaneous Linear equation Simultaneous linear equations with two unknowns can be solved by (a) elimination or (b) substitution. Example: (SPM04-P2) Calculate the values of p and q that satisfy the simultaneous : ½p – 2q =13 3p + 4q = 2 When p = 6, from : ½ (6) – 2q = 13 2q = 3 – 13  2q = - 10 q = - 5 ½p – 2q =13  3p + 4q = 2    2: p – 4q = 26   + : 4p = 24  p = 6 1 1 Hence, p = 6 and q = - 5 2

Solis geometry (4 marks) Include solid geometry of cuboid, prism, cylinder, pyramid, cone and sphere. Example : (SPM04-P2) The diagram shows a solid formed by joining a cone and a cylinder. The diameter of the cylinder and the base of the cone is 7 cm. The volume of the solid is 231 cm3. Using  = 22/7, calculate the height , in cm of the cone. Let the height of the cone be t cm. Radius of cylinder = radius of cone= 7/2 cm (r) Volume of cylinder = pj2t = 154 cm3 Hence volume of cone = 231 – 154 = 77 cm3 = 77 t = t = 6 cm 4 cm t cm 7/2 cm 1 Rujuk rumus yang diberi dalam kertas soalan. 2 1

Perimeters & Areas of circles (6 m) Usually involve the calculation of both the arc and area of part of a circle. Example : (SPM04-P2) In the diagram, PQ and RS are the arc of two circles with centre O. RQ = ST = 7 cm and PO = 14 cm. Using  = 22/7 , calculate (a) area, in cm2, of the shaded region, (b) perimeter, in cm, of the whole diagram. 60 O T R S P Q

Perimeters & Areas of circles Formula given in exam paper. (a) Area of shaded region = Area sector ORS –Area of DOQT = = 346½ – 98 = 248½ cm2 2 60 O T R S P Q 1 (b) Perimeter of the whole diagram = OP + arc PQ + QR + arc RS + SO = 14 + + 7 + + 21 = 346½ – 98 = 248½ cm2 2 Formula given . 1

Mathematical Reasoning (5 marks) (a) State whether the following compound statement is true or false Ans: False 1

Mathematical Reasoning (b) Write down two implications based on the following compound statement. Ans: Implication I : If x3 = -64, then x = -4 Implication II : If x = -4, then x3 = -64 2 (c) It is given that the interior angle of a regular polygon of n sides is Make one conclusion by deduction on the size of the size of the interior angle of a regular hexagon. Ans: 2

The Straight Line ( 5 or 6 marks) Diagram shows a trapezium PQRS drawn on a Cartesian plane. SR is parallel to PQ. Find (a) The equation of the straight line SR. Ans: 1 1 1 1

The Straight Line Diagram shows a trapezium PQRS drawn on a Cartesian plane. SR is parallel to PQ. Find (b) The y-intercept of the straight line SR Ans: The y-intercept of SR is 13. 1

Graphs of Functions (6 marks) Diagram shows the speed-time graph for the movement of a particle for a period of t seconds.

Graphs of Functions (a) State the uniform speed, in m s-1, of the particle. Ans: 20 m s-1 1 (b) Calculate the rate of change of speed, in m s-1, of the particle in the first 4 seconds. 1 Ans: 1 (c) The total distance travelled in t seconds is 184 metres. Calculate the value of t. Ans: 2 1

Probability (5 or 6 marks) Diagram shows three numbered cards in box P and two cards labelled with letters in box Q. 2 3 6 Y R P Q A card is picked at random from box P and then a card is picked at random from box Q.

Probability (5 or 6 marks) By listing the sample of all the possible outcomes of the event, find the probability that (a) A card with even number and the card labeled Y are picked, 1 1 1 (b) A card with a number which is multiple of 3 or the card labeled R is picked. 1 1

Lines and planes in 3-Dimensions(3m) Diagram shows a cuboid. M is the midpoint of the side EH and AM = 15 cm. (a) Name the angle between the line AM and the plane ADEF. Ans: 1 M D H G C B F 8 cm E A (b) Calculate the angle between the line AM and the plane ADEF. A M E 15 cm 4 cm θ Ans: 1 1

Matrices This topic is questioned both in Paper 1 & Paper 2 Paper 1: Usually on addition, subtraction and multiplication of matrices. Paper 2: Usually on Inverse Matrix and the use of inverse matrix to solve simultaneous equations.

Matrices (objective question) Example 1: (SPM03-P1) 5(-2) + 14 3(-2) + 44

Matrices (6 or 7 marks) Example 2: (SPM04-P2) (a) Inverse Matrix for is Inverse matrix formula is given in the exam paper. 1 Hence, m = ½ , p = 4. 2

Matrices Change the simultaneous equation into matrix equation: Example 2: (SPM04-P2) (cont’d) (b) Using the matrix method , find the value of x and y that satisfy the following matrix equation: 3x – 4y =  1 5x – 6y = 2 Change the simultaneous equation into matrix equation: Solve the matrix equation: 1 1 1 Maka, x = 7, y = 5½ 2

Paper 2 Section B

Graphs of functions (12 marks) This question usually begins with the calculation of two to three values of the function.( Allocated 2-3 marks) Example: (SPM04-P2) y = 2x2 – 4x – 3 Using calculator, find the values of k and m: When x = - 2, y = k. hence, k = 2(-2)2 – 4(-2) – 3 = 13 When x = 3, y = m. hence, m = 2(3)2 – 4(3) – 2 = 3 Usage of calculator: Press 2 ( - 2 ) x2 - 4 ( - 2 ) - 2 = . Answer 13 shown on screen. To calculate the next value, change – 2 to 3. 2

Graphs of functions To draw graph (i) Must use graph paper. (ii) Must follow scale given in the question. (iii) Scale need to be uniform. (iv) Graph needs to be smooth with regular shape. Example: (SPM04-P2) y = 2x2 – 4x – 3 

Graphs of functions Example: (SPM04-P2) Draw y = 2x2 – 4x – 3 To solve equation 2x2 + x – 23 = 0, 2x2 + x + 4x – 4x – 3 -20 = 0 2x2 – 4x – 3 = - 5x + 20 y = - 5x + 20 Hence, draw straight line From graph find values of x  4 1 1 2

Plans & Elevations (12 marks) NOT ALLOW to sketch. Labelling not important. The plans & elevations can be drawn from any angle. (except when it becomes a reflection) Points to avoid: Inaccurate drawing e.g. of the length or angle. Solid line is drawn as dashed line and vice versa. The line is too long. Failure to draw plan/elevation according to given scale. Double lines. Failure to draw projection lines parallel to guiding line and to show hidden edges.

Plans & Elevations (3/4/5 marks) X H G D E L K M J F N 4 cm 6 cm 3 cm

Statistics (12 marks) Use the correct method to draw ogive, histogram and frequency polygon. Follow the scale given in the question. Scale needs to be uniform. Mark the points accurately. The ogive graph has to be a smooth curve. Example (SPM03-P2) The data given below shows the amount of money in RM, donated by 40 families for a welfare fund of their children school.

Statistics To draw an ogive, Show the Upper boundary column, 40 24 17 30 22 26 35 19 23 28 33 33 39 34 39 28 27 35 45 21 38 22 27 35 30 34 31 37 40 32 14 28 20 32 29 26 32 22 38 44 Amount (RM) Frequency Cumulative Upper boundary 10.5 To draw an ogive, Show the Upper boundary column, An extra row to indicate the beginning point. 11 - 15 16 - 20 21 - 25 26 - 30 31 - 35 36 - 40 41 - 45 1 3 6 10 11 7 2 1 4 10 20 31 38 40 15.5 20.5 25.5 30.5 35.5 40.5 45.5 3

Statistics d) To use value from graph to solve question given (2 m) The ogive drawn is a smooth curve. Q3 4 d) To use value from graph to solve question given (2 m)

Combined Transformation (SPM03-P2) (a) R – Reflection in the line y = 3, T – translasion Image of H under (i) RT (ii) TR 10 x -6 -4 -2 2 4 6 8 y O P N M L 2  2 C B A D G H J K  E F  

Combined Transformation (12 marks) (SPM03-P2) (b) V maps ABCD to ABEF V is a reflection in the line AB. W maps ABEF to GHJK. W is a reflection in the line x = 6. 2 10 x -6 -4 -2 2 4 6 8 y O P N M L C B A D G H J K E F 2

Combined Transformation (SPM03-P2) (b) (ii) To find a transformation that is equivalent to two successive transformations WV. Rotation of 90 anti clockwise about point (6, 5). 3 10 x -6 -4 -2 2 4 6 8 y O C B A D G H J K

Combined Transformation (SPM03-P2) (c) Enlargement which maps ABCD to LMNP. Enlargement centered at point (6, 2) with a scale factor of 3. Area LMNP = 325.8 unit2 Hence, Area ABCD = 36.2 unit2 3 10 x -6 -4 -2 2 4 6 8 y O P N M L C B A D 1 1

THE END GOD BLESS & Enjoy teaching