Technical Mathematics Grade 12 P2 Gauteng Mock 2018 QP

Downloaded from hlayiso.com GAUTENG DEPARTMENT OF EDUCATION PREPARATORY EXAMINATION 2018 11092 TECHNICAL MATHEMATICS PAPER 2 TIME: 3 hours MARKS: 150 13 pages and 2 information sheets P.T.O.

Downloaded from hlayiso.com TECHNICAL MATHEMATICS 2 (Paper 2) 11092/18 GAUTENG DEPARTMENT OF EDUCATION PREPARATORY EXAMINATION – 2018 TECHNICAL MATHEMATICS (Paper 2) TIME: 3 hours MARKS: 150 INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. This question paper consists of 12 questions. 2. Answer ALL the questions. 3. Number the answers correctly according to the numbering system used in this question paper. 4. Clearly show ALL calculations, diagrams, graphs, etc. that you used in determining your answers. 5. Answers only will not necessarily be awarded full marks. 6. You may use an approved scientific calculator (non-programmable and non-graphical), unless stated otherwise. 7. If necessary, round-off answers to TWO decimal places, unless stated otherwise. 8. Diagrams are NOT necessarily drawn to scale. 9. Write neatly and legibly. P.T.O.

Downloaded from hlayiso.com TECHNICAL MATHEMATICS 3 (Paper 2) QUESTION 1 In the given diagram A(3 ; 7) , B(1 ; – 6) and C(– 3 ; 4) are the vertices of ∆ABC. D is the x – intercept of the straight line AD. K(e ; f ) is a point in the first quadrant of the Cartesian Plane with e > 0 and f > 0. 𝛼𝛼 and 𝜃𝜃 are the angles of inclination of lines BC and AD respectively, and DC� B = 𝛽𝛽. y A (3 ; 7) C (– 3 ; 4) K (e ; f ) 𝛽𝛽 • 𝜃𝜃 𝛼𝛼 x D O B (1 ; – 6) 1.1 Determine the coordinates of the midpoint of BC. (2) 1.2 Determine the gradient of AD. (2) 1.3 Determine the equation of AD. (3) −5 1.4 Given: The gradient of line BC is equal to 2 . 1.4.1 Show that 𝛼𝛼 = 111,80°. (2) 1.4.2 Determine the value of 𝛽𝛽. (2) 1.5 If f = 2e + 1 and KC = √17, determine the possible value(s) of f . (5) [16] P.T.O.

Downloaded from hlayiso.com TECHNICAL MATHEMATICS 4 (Paper 2) QUESTION 2 2.1 In the given diagram O(0 ; 0) is the centre of the circle with equation x2 + y2 = 25. CA and CB are tangents to the circle at A(d ; 0) and B(–3 ; – 4) respectively. y A(d ; 0) x O B(– 3 ; – 4) C 2.1.1 Write down the value of d. (1) 2.1.2 Why is OA = OB? (1) 2.1.3 Show that the equation of tangent BC is 3x + 4y + 25 = 0. (4) 2.1.4 Determine the coordinates of the point C. (3) 2.2 Sketch the graph of the ellipse defined by 4x2 + 9y2 = 36. Clearly indicate ALL intercepts with the axes. (3) [12] P.T.O.

Downloaded from hlayiso.com TECHNICAL MATHEMATICS 5 (Paper 2) QUESTION 3 3.1 Given: √3 tan 𝜃𝜃 = – 1 and 900 ≤ 𝜃𝜃 ≤ 2700 . With the aid of a diagram and WITHOUT using a calculator, determine the value of: 3.1.1 sin 𝜃𝜃 (4) 3.1.2 cos2 𝜃𝜃 (2) 3.1.3 𝜃𝜃 (2) 3𝜋 𝜋 3.2 Determine the value of cot(x –y), if x = radians and y = radians. (3) 4 3 [11] QUESTION 4 sin(180°−𝑥).cos(360°−𝑥).tan x 4.1 Show that: 1+cos 𝑥.cos(180°−𝑥) =1 (6) 1 1+ 𝑡𝑎𝑛2 𝑥 4.2 Prove the identity: = (5) 1–2𝑠𝑖𝑛2 𝑥 1− 𝑡𝑎𝑛2 𝑥 4.3 Solve for x: 2 cos 2x = √3 , if 2x ∈ [0° ; 360°] (4) [15] P.T.O.

Downloaded from hlayiso.com TECHNICAL MATHEMATICS 6 (Paper 2) QUESTION 5 The diagram below shows the boundaries of a sports field QVWR, with QV ∥ RW and VW ⊥ RW. PQ = 25 m is a vertical pole for a floodlight. The angle of elevation of P � V = 75° and RV from R is 30°. QR �W = 30°. P 25 m Q V 2 1 30° 30° 75° R W 5.1 � W = 90o ? Why is QV (1) 5.2 Determine the size of � Q1 . (2) 5.3 Show that VR = 25√2 m. (4) 5.4 The groundsman needs to determine the area of the sports field QVWR and requests you to assist him. Calculate the area of QVWR. (7) [14] P.T.O.

Downloaded from hlayiso.com TECHNICAL MATHEMATICS 7 (Paper 2) QUESTION 6 Given: f (x) = 2tan x and g(x) = sin x – 2 for x ∈ [0o ; 360o] 6.1 Draw the graphs of f and g on the same set of axes. Clearly show the intercepts with the axes as well as the asymptotes and turning points of the graphs. (6) 6.2 Write down the amplitude of g. (1) 6.3 Write down the period of f. (1) 6.4 For which value(s) of x is 𝑔′ (𝑥) < 0? (2) [10] QUESTION 7 7.1 Complete the following theorem statement: The line drawn from the centre of a circle to the midpoint of a chord, ….. (2) 7.2 In the given diagram, O is the centre of the circle with a radius of 10 cm. The length of chord BC is 16 cm and OD ⊥ BC. M is the midpoint of chord CE. E M O B C D 7.2.1 Write down, with reason, the length of BD. (2) 7.2.2 Calculate the length of OD if OB = 10 cm and BD = 8 cm. (3) 7.2.3 Show that MODC is a cyclic quadrilateral. (4) [11] P.T.O.

Downloaded from hlayiso.com TECHNICAL MATHEMATICS 8 (Paper 2) QUESTION 8 In the given diagram RS is a diameter of the circle with centre O. Chord ST is produced to a point W not on the circle. Chord SP produced meets the tangent RW at V. VR� T = 50o. V R W 1 2 50° 1 2 P 1 2 3 •O T 1 2 1 S 8.1 � S = 90 ? Why is VR o (1) 8.2 � S. Calculate the size of TR (1) 8.3 Name, with reason, one other angle equal to 90°. (2) 8.4 � R = 40°. Show that TW (2) 8.5 Determine, with reason, the size of � P1. (2) 8.6 �1 = PT Show that V �S. (4) [12] P.T.O.

Downloaded from hlayiso.com TECHNICAL MATHEMATICS 9 (Paper 2) QUESTION 9 9.1 In ∆PQR given below, MS ∥ QR. PM = x cm, MQ = 2 cm, PS = (x + 2) cm and SR = 3 cm. P (x + 2) cm x cm M S 3 cm 2 cm Q R Determine, with reasons, the value of x. (4) 9.2 In ∆AEM given below, BF ∥ CG ∥ DM and DG ∥ EM. Furthermore, AB = 1 cm = CD and BC = 3 cm. A B F C D G E M 9.2.1 Write down, with reasons, the values of the following ratios: (a) AG (2) GM (b) FG (2) GM 9.2.2 Hence, or otherwise, determine the length of DE. (2) [10] P.T.O.

Downloaded from hlayiso.com TECHNICAL MATHEMATICS 10 (Paper 2) QUESTION 10 In the given figure, KLMN is a quadrilateral with KL = 10 units, LM = 15 units, MN = 40 units and NK = 30 units. The diagonal KM = 20 units. 10 L K 15 20 30 M N 40 10.1 Use calculations to show that ∆LKM ||| ∆KMN. (3) 10.2 Hence, deduce that LK ∥ MN. (2) 10.3 If LN intersects KM at O such that ∆KLO ||| ∆MNO, show that KO = 4 units. (4) [9] P.T.O.

Downloaded from hlayiso.com TECHNICAL MATHEMATICS 11 (Paper 2) QUESTION 11 11.1 In the given diagram, O is the centre of the circle. Points A, C, B and E are on the circle. Diameter COE and chord AB intersect at right angles at D. AD = DB = 15 cm and CD = 6 cm. C 6 cm A 15 cm 15 cm B D • O E 11.1.1 Calculate the height of DOE. (3) � B. 11.1.2 Calculate the size of AO (3) 11.1.3 Determine the arc length ACB, correct to TWO decimal places, if � B = 87,21o. the radius of the circle is 21,75 cm and AO (3) 11.2 A pulley rotates at 420 r/min. Calculate: 11.2.1 The angular velocity of the pulley in radians per second (2) 11.2.2 The circumferential velocity of the pulley in metres per second, if the diameter of the pulley is 240 mm (3) [14] P.T.O.

Downloaded from hlayiso.com TECHNICAL MATHEMATICS 12 (Paper 2) QUESTION 12 12.1 A cylinder and a cone have the same curved surface area. Both objects have the same radius r cm and height h cm. h cm l cm r cm r cm The formulae given below can be used to answer the questions that follow. Volume of cylinder = π𝑟 2 ℎ Surface area of cylinder = 2𝜋𝑟 2 + 2𝜋𝑟ℎ Curved area of cone = 𝜋𝑟𝑙 1 Volume of cone = 𝜋𝑟 2 ℎ 3 Surface area of cone = 𝜋𝑟 2 + 𝜋𝑟𝑙 12.1.1 Express the slant height, l of a cone in terms of r and h. (2) 12.1.2 If 𝜋𝑟𝑙 = 2𝜋𝑟ℎ , show that r2 = 3h2. (3) 12.1.3 Express the volume of the cone in terms of h only. (2) 12.1.4 If the two volumes differ by 54𝜋, (a) show that h = 3. (2) (b) find the value of r. (2) P.T.O.

Downloaded from hlayiso.com TECHNICAL MATHEMATICS 13 (Paper 2) 12.2 The map below represents a part of Southern Africa. Ordinates are drawn on the map such that the distance between the ordinates is 110 km. All measurements are in kilometres. y1 y2 y3 y4 y5 y6 y7 y8 y9 y10 y11 430 793 1167 1475 1409 1431 1365 1205 870 529 347 y1 y2 y3 y4 y5 y6 y7 y8 y9 y10 y11 Part of Southern Africa Determine: 12.2.1 The area of the region represented by the map, using the mid- ordinate rule (3) 12.2.2 The area of the shaded region (2) [16] TOTAL: 150 END

Downloaded from hlayiso.com TECHNICAL MATHEMATICS 14 (Paper 2) INFORMATION SHEET: TECHNICAL MATHEMATICS − b ± b 2 − 4ac b 4ac − b 2 x= x=− y= 2a 2a 4a a x = b ⇔ x = log a b , a > 0 , a ≠ 1 and b > 0 A = P(1 + ni ) A = P(1 − ni ) A = P(1 − i ) n A = P(1 + i ) n m  im  ieff = 1 +  − 1  m   f ( x + h) − f ( x) f / ( x ) = lim h →0 h  x + x2 y1 + y 2  d = ( x2 − x1 ) 2 + ( y 2 − y1 ) 2 M  1 ;   2 2  y 2 − y1 y = mx + c y − y1 = m( x − x1 ) m= m = tan θ x 2 − x1 x2 y2 + =1 a2 b2 a b c In ∆ABC: = = a 2 = b 2 + c 2 − 2bc. cos A sin A sin B sin C area of ∆ ABC = 1 ab ⋅ sin C 2 sin 2 θ + cos 2 θ = 1 1 + tan 2 θ = sec 2 θ cot 2 θ + 1 = cos ec 2θ f ( x + h) − f ( x) x n +1 f / ( x ) = lim ∫ x= n dx +C , n ≠ −1 h →0 h n +1 1 ax ∫ x dx = ∫ a dx =+ x ln( x) + C , x>0 C , a>0 ln a

Downloaded from hlayiso.com TECHNICAL MATHEMATICS 15 (Paper 2) π rad = 180° Angular velocity= ω= 2π n= 3600 n where n = rotation frequency Circumferencial velocity= ν= π Dn where D = diameter and n = rotation frequency s = rθ where r = radius and θ = central angle in radians rs r 2θ = = Area of a sector where r = radius, s = arc length and θ = central 2 2 angle in radians 4h 2 − 4dh + x 2 = 0 where h = height of segment, d = diameter of circle and x = length of chord o +o  = A T a  1 n + o2 + o3 + o4 + . . . + on −1  where a = equal parts, oi = ith ordinate and  2  n = number of ordinates OR o +o A T= a ( m1 + m2 + m3 + . . . + mn ) where a = equal parts, m1 = 1 2 2 and n = number of ordinates