Technical Mathematics Grade 12 P2 NSC November Mock 2026 QP

Mock hlayiso mock exams Practice Series: Grade 12 NSC Preparation WWW.HLAYISO.COM NATIONAL SENIOR CERTIFICATE GRADE 12 TECHNICAL MATHEMATICS P2 MOCK EXAMINATION 2026 MARKS: 150 TIME: 3 hours This question paper consists of 12 pages and a 2-page information sheet. Copyright reserved Please turn over

Technical Mathematics/P2 2 Hlayiso/Mock 2026 NSC Mock INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. This question paper consists of 11 questions. 2. Answer ALL the questions in the SPECIAL ANSWER BOOK provided. 3. Clearly show ALL calculations, diagrams, graphs, etc. that you have used in determining your answers. 4. Answers only will NOT necessarily be awarded full marks. 5. If necessary, round off answers to TWO decimal places, unless stated otherwise. 6. Diagrams are NOT necessarily drawn to scale. 7. You may use an approved scientific calculator (non-programmable and non-graphical), unless stated otherwise. 8. An information sheet with formulae is included at the end of the question paper. 9. Write neatly and legibly. Copyright reserved Please turn over

Technical Mathematics/P2 3 Hlayiso/Mock 2026 NSC Mock QUESTION 1 The diagram below shows △ABC with vertices A(−3 ; 2), B(3 ; 4) and C(1 ; 0). M is the midpoint of AB. BC is produced to form the angle of inclination, θ , with the positive x-axis. y B(3 ; 4) M A(−3 ; 2) θ x O C(1 ; 0) 1.1 Determine the length of AB. Leave your answer in simplest surd form. (2) 1.2 Determine the coordinates of M. (2) 1.3 Determine the gradient of BC. (2) 1.4 Hence, determine the size of angle θ . (2) 1.5 Determine the equation of the line passing through A, parallel to BC, in the form y = ... (3) 1.6 Show that AĈB = 90◦ . (3) [14] Copyright reserved Please turn over

Technical Mathematics/P2 4 Hlayiso/Mock 2026 NSC Mock QUESTION 2 2.1 In the diagram below, O is the centre of the circle defined by x2 + y2 = r2 . • The straight line drawn is a tangent to the circle at T (−2 ; 4). • K is the y-intercept of the tangent and S is a positive x-intercept of the circle. y K T (−2 ; 4) x O S 2.1.1 Determine the equation of the circle. (2) 2.1.2 Write down the length of OS in simplest surd form. (1) 2.1.3 Determine the equation of the tangent at T in the form y = . . . (4) 2.1.4 Write down the coordinates of K. (1) 2.2 Given: 4x2 + y2 = 16 x2 y2 2.2.1 Express the equation in the form + = 1. (2) a2 b2 2.2.2 Hence, sketch the graph of the ellipse on the grid provided in the SPECIAL ANSWER BOOK. Clearly indicate ALL the intercepts with the axes. (2) [12] Copyright reserved Please turn over

Technical Mathematics/P2 5 Hlayiso/Mock 2026 NSC Mock QUESTION 3 3.1 Given: A = 2,3 rad and B = 47◦ Use a calculator to determine the following: 3.1.1 Convert 2,3 rad to degrees. (1) 3.1.2 sec B (2) 3.1.3 sin(A − B) (2) 3.2 Given: 5 tan θ = −12 and θ ∈ [90◦ ; 270◦ ] Use a sketch and, WITHOUT using a calculator, determine the value of: 3.2.1 cos θ (2) 3.2.2 1 − sec θ · sin θ (3) 3.3 Solve for x: 3 sin x = −1,8 for x ∈ [0◦ ; 360◦ ] (3) [13] QUESTION 4 4.1 Simplify the following: 4.1.1 cos(2π − x) (1) 4.1.2 sin(180◦ + x) (1) sin(180◦ + x) · cos(2π − x) 4.1.3 + sin2 (360◦ − x) (6) tan(π − x) 4.2 Answer the following questions: 4.2.1 Complete the identity: 1 − cos2 θ = . . . (1) sec θ − cos θ 4.2.2 Hence, prove that = sin θ (4) tan θ [13] Copyright reserved Please turn over

Technical Mathematics/P2 6 Hlayiso/Mock 2026 NSC Mock QUESTION 5 Given the functions defined by f (x) = 2 cos x and g(x) = sin 2x for x ∈ [0◦ ; 360◦ ]. 5.1 Sketch the graphs of f and g on the same set of axes on the grid provided in the SPECIAL ANSWER BOOK. Clearly indicate ALL the intercepts with the axes and the turning points. (6) 5.2 Use your graphs to write down the following: 5.2.1 The period of g (1) 5.2.2 The amplitude of f (1) 5.2.3 The values of x for which f (x) = g(x) (2) 5.2.4 The range of h if h(x) = f (x) + 1 (1) [11] QUESTION 6 In the diagram below, B, C and D are points in the same horizontal plane. AB is a vertical mast with AB ⊥ BC. • The angle of elevation of A from C is 28◦ . • BĈD = 35◦ and CB̂D = 115◦ • BC = 40 m A B 40 m 28◦ 115◦ D 35◦ C 6.1 Write down the size of D̂. (1) 6.2 Determine the length of BD. (3) 6.3 Determine the height of the mast, AB. (2) 6.4 Determine the length of CD. (3) 6.5 Determine the area of △BCD. (3) [12] Copyright reserved Please turn over

Technical Mathematics/P2 7 Hlayiso/Mock 2026 NSC Mock Give reasons for your statements in QUESTIONS 7, 8 and 9. QUESTION 7 7.1 Complete the following theorem: The angle subtended by an arc at the centre of a circle is … (1) 7.2 In the diagram below, O is the centre of circle ABCD. BÔD = 130◦ . A O B 130◦ D C Determine, stating reasons, the size of: 7.2.1 Â (2) 7.2.2 BĈD (2) 7.3 In the diagram below, O is the centre of the circle. OM ⊥ PQ and OM produced meets the circle at N. OP = 10 cm and PQ = 16 cm. O P Q M N 7.3.1 Determine, stating reasons, the length of OM. (3) 7.3.2 Hence, determine the length of MN. (2) [10] Copyright reserved Please turn over

Technical Mathematics/P2 8 Hlayiso/Mock 2026 NSC Mock QUESTION 8 8.1 In the diagram below, O is the centre of circle PQRS. T PV is a tangent to the circle at P. Chords PQ, QR, RS, SP and PR are drawn. T P̂Q = 62◦ and PQ̂R = 75◦ R Q 75◦ O S 62◦ V P T Determine, stating reasons, the size of EACH of the following angles: 8.1.1 PR̂Q (2) 8.1.2 PÔQ (2) 8.1.3 OP̂Q (2) 8.1.4 Ŝ (2) Copyright reserved Please turn over

Technical Mathematics/P2 9 Hlayiso/Mock 2026 NSC Mock 8.2 In the diagram below, A, B, C and D are points on the circle. AB = AD. Chord BC is produced to E. Chords AC and BD are drawn. BÂD = 80◦ and DB̂C = 35◦ A 80◦ B 35◦ D C E Determine, stating reasons, the size of: 8.2.1 DĈE (2) 8.2.2 AB̂D (2) 8.2.3 DÂC (2) 8.2.4 BD̂C (2) 8.2.5 Prove, stating reasons, that AC bisects BĈD. (4) [20] Copyright reserved Please turn over

Technical Mathematics/P2 10 Hlayiso/Mock 2026 NSC Mock QUESTION 9 In the diagram below, D and E are points on sides AB and AC of △ABC respectively, such that DE ∥ BC. AD = 6 units, DB = 4 units and AE = 9 units. A 6 9 D E 4 B C 9.1 Complete the following: AD . . . 9.1.1 = (Reason: …) (2) DB . . . 9.1.2 Hence, determine the length of EC. (2) 9.1.3 Write down the length of AC. (1) 9.2 Prove that △ADE ||| △ABC. (3) 9.3 Hence, determine the length of DE if BC = 15 units. (2) [10] Copyright reserved Please turn over

Technical Mathematics/P2 11 Hlayiso/Mock 2026 NSC Mock QUESTION 10 10.1 A workshop angle grinder uses a cutting disc with a diameter of 180 mm. The disc rotates at 6 000 revolutions per minute. 10.1.1 Convert 6 000 revolutions per minute to revolutions per second. (1) 10.1.2 Determine the angular velocity of the disc in radians per second. (2) 10.1.3 Determine the circumferential velocity of a point on the edge of the disc in metres per second. (3) 10.1.4 A smaller disc rotates at the same rotation frequency. A point on its edge has a circumferential velocity of 40 m/s. Determine the diameter of the smaller disc in millimetres. (2) 10.2 A garden sprinkler waters a sector-shaped region AOB of a lawn, as shown in the diagram below. The radius of the sector is 6 m and the central angle is 135◦ . A B 135◦ 6m O 10.2.1 Convert 135◦ to radians. (1) 10.2.2 Determine the length of arc AB. (2) 10.2.3 Determine the area of the watered sector. (2) 10.2.4 Determine the perimeter of the watered sector. (2) 10.3 The top of a window is in the shape of a segment of a circle, as shown below. The chord AB is 1,6 m long and the height of the segment, h, is 0,4 m. Determine the diameter of the circle. (3) h A B x = 1,6 m [18] Copyright reserved Please turn over

Technical Mathematics/P2 12 Hlayiso/Mock 2026 NSC Mock QUESTION 11 11.1 The diagram below shows the shape of a paved area at a taxi rank. The straight side of 12 m is divided into 6 equal parts. The ordinates dividing these parts are 3,0 m; 3,4 m; 3,8 m; 3,6 m; 3,2 m; 2,8 m and 2,4 m respectively. 3,0 m 3,4 m 3,8 m 3,6 m 3,2 m 2,8 m 2,4 m 12 m 11.1.1 Write down the width of each equal part. (1) 11.1.2 Use the mid-ordinate rule to determine the area of the paved region. (4) 11.1.3 Paving costs R185 per square metre. Determine the cost of paving the region. (2) 11.2 A grain silo consists of a cylinder with a cone on top, as shown in the diagram alongside. • The cylinder has a radius of 2 m and a height of 6 m. • The cone has the same radius as the cylinder and a perpendicular height of 1,5 m. The following formulae may be used: 1,5 m Volume of a cylinder = π r2 h 2m 1 Volume of a cone = π r2 h 3 Curved surface area of a cylinder = 2π rh 6m Curved surface area of a cone = π rs where s = slant height 11.2.1 Determine the volume of the cylindrical part of the silo. (2) 11.2.2 Determine the slant height, s, of the cone. (2) 11.2.3 Determine the total volume of the silo. (3) 11.2.4 The outside of the silo (excluding the base) must be painted. Determine the area to be painted. (3) [17] TOTAL: 150 Copyright reserved Please turn over

Technical Mathematics/P2 13 Hlayiso/Mock 2026 NSC Mock INFORMATION SHEET: TECHNICAL MATHEMATICS √ −b ± b2 − 4ac b 4ac − b2 x= x=− y= 2a 2a 4a a = b ⇔ x = loga b , a > 0 , a ̸= 1 and b > 0 x A = P(1 + ni) A = P(1 − ni) A = P(1 + i)n A = P(1 − i)n ( ) i m ieff = 1 + −1 m f (x + h) − f (x) f ′ (x) = lim h→0 h ∫ ∫ xn+1 k kxn dx = k · +C , n ̸= −1 dx = k · ln x +C , x>0 n+1 x ∫ anx k anx dx = k · +C , a>0 n ln a √ ( ) x2 + x1 y2 + y1 d = (x2 − x1 )2 + (y2 − y1 )2 M ; 2 2 y2 − y1 y = mx + c y − y1 = m(x − x1 ) m= tan θ = m x2 − x1 x2 y2 + =1 a2 b2 a b c In △ABC : = = a2 = b2 + c2 − 2bc · cos A sin A sin B sinC 1 Area of △ABC = ab · sinC 2 sin2 θ + cos2 θ = 1 1 + tan2 θ = sec2 θ 1 + cot2 θ = cosec2 θ Copyright reserved

Technical Mathematics/P2 14 Hlayiso/Mock 2026 NSC Mock π rad = 180◦ Angular velocity = ω = 2π n where n = rotation frequency Angular velocity = ω = 360◦ n where n = rotation frequency Circumferential velocity = v = π Dn where D = diameter and n = rotation frequency Circumferential velocity = v = ω r where ω = angular velocity and r = radius Arc length = s = rθ where r = radius and θ = central angle in radians rs Area of a sector = where r = radius and s = arc length 2 r2 θ Area of a sector = where r = radius and θ = central angle in radians 2 4h2 − 4dh + x2 = 0 where h = height of segment, d = diameter of circle and x = length of chord o1 + o2 AT = a (m1 + m2 + m3 + . . . + mn ) where a = width of equal parts, m1 = , 2 on = nth ordinate and n = number of ordinates OR ( ) o1 + on AT = a + o2 + o3 + . . . + on−1 where a = width of equal parts, on = nth ordinate and n 2 = number of ordinates Copyright reserved