Technical Mathematics Grade 12 P2 2026 QP (Eng)

Prediction hlayiso mock exams Practice Series: Grade 12 NSC Preparation WWW.HLAYISO.COM NATIONAL SENIOR CERTIFICATE GRADE 12 TECHNICAL MATHEMATICS P2 PREDICTION EXAMINATION 2026 MARKS: 150 TIME: 3 hours This question paper consists of 15 pages and a 2-page information sheet. Copyright reserved Please turn over

Technical Mathematics/P2 2 Hlayiso/Prediction 2026 NSC Prediction 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/Prediction 2026 NSC Prediction QUESTION 1 In the diagram below, A(−4 ; 5), B(2 ; 1) and C(1 ; t) are the vertices of △ABC. M is the midpoint of AB. AB is produced to form the angle of inclination, θ , with the positive x-axis. y C(1 ; t) A(−4 ; 5) M B(2 ; 1) θ x O 1.1 Determine the gradient of AB. (2) 1.2 Hence, determine the size of θ . (2) 1.3 Determine the length of AB. Leave your answer in simplest surd form. (2) 1.4 Determine the coordinates of M. (2) 1.5 Determine the equation of the perpendicular bisector of AB in the form y = . . . (4) 1.6 C lies on the perpendicular bisector of AB. Determine the value of t. (1) [13] Copyright reserved Please turn over

Technical Mathematics/P2 4 Hlayiso/Prediction 2026 NSC Prediction QUESTION 2 2.1 In the diagram below, O is the centre of the circle defined by x2 + y2 = r2 . • T (−3 ; 6) is a point on the circle. • The straight line drawn is a tangent to the circle at T . • K is the y-intercept of the tangent. y K T (−3 ; 6) x O 2.1.1 Determine the equation of the circle. (2) 2.1.2 Write down the length of the radius of the circle 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: 25x2 + 4y2 = 100 x2 y2 2.2.1 Express the equation in the form + = 1. (2) a2 b2 2.2.2 Hence, sketch the graph defined by 25x2 + 4y2 = 100 on the grid provided in the SPECIAL ANSWER BOOK. Clearly show ALL the intercepts with the axes. (2) [12] Copyright reserved Please turn over

Technical Mathematics/P2 5 Hlayiso/Prediction 2026 NSC Prediction QUESTION 3 3.1 Given: A = 2,5 rad and B = 41,3◦ Use a calculator to determine the following: 3.1.1 Convert 2,5 rad to degrees. (1) 3.1.2 cot(A − B) (2) sin2 B 3.1.3 (2) cos A 3.2 In the diagram below, R(k ; −5) is a point in the Cartesian plane such that OR = 6 units and k > 0. θ is the reflex angle formed by OR and the positive x-axis. y θ x O 6 R(k ; −5) WITHOUT using a calculator, determine: 3.2.1 The value of k (2) √ 3.2.2 The numerical value of 6 sin θ + 11 tan θ (3) 3.3 Determine the values of x if 5 cos x + 2 = 0 and x ∈ [0◦ ; 360◦ ]. (4) [14] Copyright reserved Please turn over

Technical Mathematics/P2 6 Hlayiso/Prediction 2026 NSC Prediction QUESTION 4 4.1 Simplify the following: 4.1.1 cos(180◦ + x) (1) 4.1.2 tan(360◦ − x) (1) cos(180◦ + x) · tan(360◦ − x) 4.1.3 + cos2 (−x) · tan2 (360◦ + x) (6) sin(180◦ + x) 4.2 Answer the following questions: 4.2.1 Complete the identity: 1 + tan2 θ = . . . (1) sin θ + sin θ · tan2 θ 4.2.2 Hence, prove that = tan θ (4) sec θ [13] QUESTION 5 Given the functions defined by f (x) = cos 2x and g(x) = tan x for x ∈ [0◦ ; 180◦ ]. 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, turning points, end points and asymptotes. (6) 5.2 Write down: 5.2.1 The period of f (1) 5.2.2 The equation of the asymptote of g (1) 5.3 Use your graphs to determine the values of x for which f (x) · g(x) < 0. (2) 5.4 Write down the range of h if h(x) = 2 f (x) − 1. (2) [12] Copyright reserved Please turn over

Technical Mathematics/P2 7 Hlayiso/Prediction 2026 NSC Prediction QUESTION 6 In the diagram below, AB represents a vertical cellphone tower. B, C and D are points in the same horizontal plane. • BC = 52 m, CD = 38 m and BD = 70 m • The angle of elevation of A from D is 24◦ . A 70 m 24◦ B D 52 m 38 m C 6.1 Determine the size of BĈD. (3) 6.2 Determine the area of △BCD. (3) 6.3 Determine the height of the tower, AB. (2) 6.4 Determine the angle of elevation of A from C. (3) [11] Copyright reserved Please turn over

Technical Mathematics/P2 8 Hlayiso/Prediction 2026 NSC Prediction Give reasons for your statements in QUESTIONS 7, 8 and 9. QUESTION 7 7.1 Complete the following theorem: The line drawn from the centre of a circle perpendicular to a chord … (1) 7.2 In the diagram below, O is the centre of the circle. AB is a chord with AB = 24 cm. • OM ⊥ AB with M on AB. OM produced meets the circle at C, and CO produced meets the circle at D. • MC = 8 cm • Chords AC and AD are drawn. D O A B M C 7.2.1 Write down, with a reason, the length of AM. (2) 7.2.2 Determine, with reasons, the length of the radius of the circle. (4) 7.2.3 Give a reason why CÂD = 90◦ . (1) 7.2.4 Hence, determine the length of AD. (2) [10] Copyright reserved Please turn over

Technical Mathematics/P2 9 Hlayiso/Prediction 2026 NSC Prediction QUESTION 8 8.1 Complete the following theorem: The angle between a tangent to a circle and a chord drawn from the point of contact is equal to … (1) 8.2 In the diagram below, A, B, C and D are points on the circle. TAV is a tangent to the circle at A. Chords AB, BC, CD, DA, AC and BD are drawn. T ÂB = 40◦ and BD̂C = 58◦ C 58◦ D B 40◦ V A T Determine, stating reasons, the size of EACH of the following angles: 8.2.1 AD̂B (2) 8.2.2 AĈB (2) 8.2.3 AB̂C (3) 8.2.4 It is further given that AB = AD. Prove, stating reasons, that TV ∥ BD. (4) Copyright reserved Please turn over

Technical Mathematics/P2 10 Hlayiso/Prediction 2026 NSC Prediction 8.3 In the diagram below, O is the centre of circle ABC. T B and TC are tangents to the circle at B and C respectively. Radii OB and OC and chords AB and AC are drawn. BÂC = 64◦ A 64◦ O B C T 8.3.1 Determine, stating a reason, the size of BÔC. (2) 8.3.2 Prove, stating reasons, that OBTC is a cyclic quadrilateral. (3) [17] Copyright reserved Please turn over

Technical Mathematics/P2 11 Hlayiso/Prediction 2026 NSC Prediction QUESTION 9 9.1 Complete the following theorem: A line drawn parallel to one side of a triangle divides the other two sides … (1) 9.2 In △PQR below, S is a point on PQ and T is a point on PR such that ST ∥ QR. U is a point on QR such that TU ∥ PQ. PS = 6 units, SQ = 9 units, PT = 8 units and QR = 30 units P 6 8 S > T 9 >> >> > Q U R 9.2.1 Determine, stating reasons, the length of T R. (3) 9.2.2 Determine the length of QU. (2) 9.2.3 Prove, stating reasons, that △PST ||| △PQR. (3) 9.2.4 Hence, determine the length of ST . (2) [11] Copyright reserved Please turn over

Technical Mathematics/P2 12 Hlayiso/Prediction 2026 NSC Prediction QUESTION 10 10.1 A table saw in a school woodwork workshop has a circular blade with a diameter of 250 mm. The blade rotates at 3 000 revolutions per minute. The diagram below models the blade. • O is the centre of the blade. • PQ is the chord formed where the blade meets the top of the table. • The blade protrudes h = 60 mm above the table. rotation h P Q table x O 10.1.1 Convert 3 000 revolutions per minute to revolutions per second. (1) 10.1.2 Determine the angular velocity of the blade in radians per second. (2) 10.1.3 Determine the circumferential velocity of a point on the edge of the blade in metres per second. (3) 10.1.4 Determine the length of chord PQ, that is x. (4) Copyright reserved Please turn over

Technical Mathematics/P2 13 Hlayiso/Prediction 2026 NSC Prediction 10.2 The diagram below shows an open belt drive. • The larger pulley (centre O1 ) has a radius of 15 cm and the smaller pulley (centre O2 ) has a radius of 6 cm. • The belt touches the larger pulley along major arc AB, which subtends a reflex angle of 220◦ at O1 . • The belt touches the smaller pulley along minor arc CD, which subtends an angle of 140◦ at O2 . • AC and BD are the straight sections of the belt. Each straight section is 24,73 cm long. A 15 C cm 6 cm 220◦ O1 O2 140◦ D B 10.2.1 Convert 220◦ to radians. (1) 10.2.2 Determine the length of the belt that is in contact with the larger pulley (major arc AB). (2) 10.2.3 Determine the length of arc CD. (2) 10.2.4 Hence, determine the total length of the belt. (2) 10.2.5 Determine the area of the shaded minor sector AO1 B. (3) [20] Copyright reserved Please turn over

Technical Mathematics/P2 14 Hlayiso/Prediction 2026 NSC Prediction QUESTION 11 11.1 The diagram below shows an irregular section of a school sports field that will be covered with artificial turf. The straight side of 18 m is divided into 6 equal parts. The 7 ordinates, including the two end ordinates, are 4,2 m; 5,1 m; 5,8 m; 6,0 m; 5,4 m; 4,6 m and 3,9 m respectively. 4,2 m 5,1 m 5,8 m 6,0 m 5,4 m 4,6 m 3,9 m 18 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 irregular section. (4) 11.1.3 Artificial turf costs R245 per square metre. An extra 10% of turf must be bought to allow for cutting and waste. The school has budgeted R24 000 for the turf. Determine whether the budget is sufficient. (4) Copyright reserved Please turn over

Technical Mathematics/P2 15 Hlayiso/Prediction 2026 NSC Prediction 11.2 A concrete road bollard is made in the shape of a right cylinder with a hemisphere on top, as shown in the diagram alongside. • The radius of the cylinder and of the hemisphere is 15 cm. • The height of the cylindrical part is 80 cm. The following formulae may be used: 15 cm Volume of a cylinder = π r2 h 4 Volume of a sphere = π r3 3 Curved surface area of a cylinder = 2π rh 80 cm Surface area of a sphere = 4π r2 1 m3 = 1 000 000 cm3 11.2.1 Determine the volume of concrete needed to make ONE bollard. (3) 11.2.2 The outside of the bollard, excluding the base, must be painted. Determine the area to be painted. (3) 11.2.3 A contractor has 2 m3 of concrete. Determine whether this is enough concrete to make 30 bollards. (2) [17] TOTAL: 150 Copyright reserved Please turn over

Technical Mathematics/P2 16 Hlayiso/Prediction 2026 NSC Prediction 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 17 Hlayiso/Prediction 2026 NSC Prediction 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