Mathematics Grade 11 Paper 2 KZN June 2024 Memo

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com NATIONAL SENIOR CERTIFICATE GRADE 11 MATHEMATICS P2 COMMON TEST JUNE 2024 MARKS: 100 TIME: 2 hours This question paper consists of 9 pages and 3 DIAGRAM SHEETS. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 2 Common Test June 2024 Downloaded f rom St anmorephysics. Grade 11-NSC com INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. This question paper consists of 7 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. which you have 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 correct to TWO decimal places, unless stated otherwise. 8. Diagrams are NOT necessarily drawn to scale. 9. THREE DIAGRAM SHEETS for QUESTION 1, QUESTION 2, QUESTION 5.1, QUESTION 5.2, QUESTION 6.1, QUESTION 6.2 and QUESTION 7 are attached at the end of this question paper. Detach the DIAGRAM SHEETS and hand them in together with your ANSWER BOOK. 10. Write neatly and legibly. Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 3 Common Test June 2024 Downloaded f rom St anmorephysics. Grade 11-NSC com QUESTION 1 In the diagram below, D(−10 ; 6) , E, F and G(1 ; 9) are the vertices of a quadrilateral. F is a point on the x-axis. The diagonals of the quadrilateral bisect each other at K. The equation of diagonal EG is 3x − y + 6 = 0 and β is the angle of inclination of EG. y K β O F x E 1.1 Calculate the size of β. (3) 1.2 Calculate the coordinates of F if the equation of DF is x + 3y = 8. (2) 1.3 Calculate the coordinates of E. (4) 1.4 Prove that DEFG is a rhombus. (3) [12] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 4 Common Test June 2024 Downloaded f rom St anmorephysics. Grade 11-NSC com QUESTION 2 In the diagram below, A(−2 ; −1), B(−3 ; 4) and C (1 ; 5) are the vertices of a triangle. BC is produced to D(5 ; y), AC cuts the y-axis at F. E is a point on the x-axis such that CE is perpendicular to the x-axis. y F O E x 2.1 Calculate the length of AC. Leave answer in simplified surd form. (2) 2.2 Calculate the gradient of BC. (2) 2.3 Calculate the value of y if B, C and D are collinear. (3) 2.4 If H is a point such that AH ⊥ BC, determine the equation of AH. (3) 2.5 Calculate the coordinates of G if CBAG, in that order, is a parallelogram. (3) 2.6 Calculate the area of CEOF. (4) [17] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 5 Common Test June 2024 Downloaded f rom St anmorephysics. Grade 11-NSC com QUESTION 3 If cos 23 = 1 − g 2 , express each of the following in terms of g: 3.1 sin 23 (3) 3.2 tan113 (2) 3.3 sin ( −653 ) (2) [7] QUESTION 4 4.1 Simplify the following without using a calculator: 4.1.1 cos 390 − tan ( 360 − x ) .cos (180 + x ) .cos ( x − 90 ) (8) cos ( −30 ) sin120.tan 330 4.1.2 cos 240 (4) cos x.tan 2 x 4.2 Prove the following identity: + cos x = 1 (5) 1 +1 cos x cos 2 x.tan x 4.3 Given the expression: . sin 2 x Determine the value(s) of x, in the interval x  0 ;180 , for which the expression will be undefined. (3) 4.4 Solve for α and  if: 2 cos ( +  ) = − for ( +  )   0 ;180 ; and 2 1 cos ( − 2 ) = for ( − 2 )  0 ;180 2 (4) [24] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 6 Common Test June 2024 Downloaded f rom St anmorephysics. Grade 11-NSC com GIVE REASONS FOR YOUR STATEMENTS IN QUESTION 5, QUESTION 6 AND QUESTION 7. QUESTION 5 5.1 In the diagram, O is the centre of the circle. AB and AC are chords, and BO and CO have been drawn. A O B C Use the diagram on the DIAGRAM SHEET to prove the theorem which states that the angle subtended by an arc or chord at the centre of the circle is equal to twice the angle subtended by the same arc or chord at the circumference, that is, prove that obtuse BOCˆ = 2BACˆ . (5) Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 7 Common Test June 2024 Downloaded f rom St anmorephysics. Grade 11-NSC com 5.2 In the diagram below, DF is a diameter of the circle with centre O. Chord EG intersects DF at H such that GH = HE. Chords EF, GF and DE and radius OE are drawn. EGF ˆ = 55 . E 1 2 3 D 3 H 2 1 11 2 O 2 G 11 F 5.2.1 Write down the size of D̂ . Provide a reason. (2) 5.2.2 Write down the size of Ô2 . Provide a reason. (2) 5.2.3 Calculate, giving reasons, the size of F̂2 (2) 5.2.4 Prove, with reasons, that DE is a tangent to the circle passing through E, F and H at E. (3) [14] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 8 Common Test June 2024 Downloaded f rom St anmorephysics. Grade 11-NSC com QUESTION 6 6.1 In the diagram below, CD is a diameter and PC is a tangent to the circle at C. Chord DK is produced to P. PT intersects KC at Q. CDP ˆ = 40 , DPTˆ = 25 and ˆ TQC = 65 . P K 1 D 1 2 Q 2 T 1 1 2 C Calculate, with reasons, the size of the following angles: 6.1.1 Ĉ2 (2) 6.1.2 K̂1 (2) 6.1.3 P̂1 (2) 6.1.4 Prove, with reasons, that TC = QC . (2) 6.2 In the diagram below, O is the centre of circle PTR. N is a point on chord PR such that ON ⊥ PR . RS and PS are tangents to the circle at R and P respectively. ˆ = 62 . RS = 15 units , TS = 9 units and RPS P N 9 O S T 15 R 6.2.1 ˆ . Calculate the size of NOR (5) 6.2.2 Calculate the length of the radius of the circle. (4) [17] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 9 Common Test June 2024 Downloaded f rom St anmorephysics. Grade 11-NSC com QUESTION 7 In the diagram, O is the centre of the circle and PVR is a tangent to the circle at V. T is a point on chord SU such that ST = TU. US extended meets PV at R. OV, OR, OS and OT are drawn. R V P S O T U 7.1 Prove that VOTR is a cyclic quadrilateral. (3) Prove that OR = OV + TR − ( RU − TR ) . 2 2 2 2 7.2 (6) [9] TOTAL: 100 MARKS Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 10 Common Test June 2024 Downloaded f rom St anmorephysics. Grade 11-NSC com NAME & SURNAME: DIAGRAM SHEET 1 QUESTION 1 y K β O F x E TEAR OFF QUESTION 2 y F O E Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 11 Common Test June 2024 Downloaded f rom St anmorephysics. Grade 11-NSC com NAME & SURNAME: DIAGRAM SHEET 2 QUESTION 5.1 A O B C TEAR OFF QUESTION 5.2 E 1 2 3 D 3 H 2 1 11 2 O 2 G 11 F Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 12 Common Test June 2024 Downloaded f rom St anmorephysics. Grade 11-NSC com NAME & SURNAME: DIAGRAM SHEET 3 QUESTION 6.1 P K 1 D 1 2 Q 2 T 1 1 2 C QUESTION 6.2 P TEAR OFF N 9 O S T 15 R QUESTION 7 R V P S O T U Copyright Reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com FINAL MATHEMATICS P2 COMMON TEST JUNE 2024 MARKING GUIDELINES NATIONAL SENIOR CERTIFICATE GRADE 11 MARKS: 100 These marking guidelines consist of 12 pages.

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Mathematics P2 2 com Common Test June 2024 GRADE 11 Marking Guideline QUESTION 1 1.1 Equation of EG: 3x – y  6  0 y  3x  6 A y  3x  6  mEG  3  tan   mEG  3 CA tan   mEG  3 β  71,57 CA answer (3) 1.2 x  30  8 A substitute y = 0 x 8  F 8 ; 0  Answer only: FULL MARKS A answer in coordinate form (2) 1.3 midptDF  midptGE [diagonals bisect]  10  8 6  0  K ;   2 2  K  1 ; 3 CACA co-ordinates of K x 1 y 9 1  and 3  2 2 x  3 and y  3  E  3 ;  3 CA x-value CAy-value OR (4) x  3y  8 x  3  3x  6   8 x  9 x  18  8 10 x  10 x  1 CA xK  y  3  1  6 y 3 CA yK  K  1 ; 3 x 1 y9  1 and 3 2 2 x  3 and y  3 CA only applies if both values are  E  3 ;  3 negative CA x-value CA y-value (4) 1.4 3y  x  8 1 8 y   x 3 3 1 1 A mDF   mDF  mGE    3 3 3 A mDF  mGE  1  1  DF  GE  A conclusion (3) Copyright Reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Mathematics P2 3 com Common Test June 2024 GRADE 11 Marking Guideline  DEFG is a rhombus (diagonals bisect at right angle) OR DEFG is a parallelogram (diagonals bisect each other) DG   10  1   6  9  2 2 = 130 A DG  130 DE   10   3   6   3  2 2  130 CA DE  130  DG  DE  DEFG is a rhombus (parm. with adjacent sides =) A conclusion (3) [12] QUESTION 2 AC  1  (2)    5  (1)  2 2 2.1 A substitution in correct formula  9  36 Penalise if answer 3 5 is not simplified CA answer (2) 54 A substitution in correct 2.2 mBC  1   3 formula 1  CA answer 4 (2) 2.3 mCD  mBC y 5 1 CA mCD  CA equating gradients 5 1 4 CA answer y6 (3) OR mBD  mBC y4 1 CA mBD  5   3 4 CA equating gradients y4 2  8 8 y6 CA answer (3) Copyright Reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Mathematics P2 4 com Common Test June 2024 GRADE 11 Marking Guideline 2.4 H (0 ; 3) I(1 ; 3) G(2 ; 0) mAH  4 (AH  BC) CA mAH  4 y   1  4  x   2   CA substitution of y  4x  9 A  2 ;  1 OR CA equation of AH mAH  4 (3) 1  4(2)  c CA mAH  4 CA substitution of c  9 A  2 ;  1  y  4 x  9 CA equation of AH (3) 2.5 By using transformation: B  C :  3 ; 4   1 ; 5   x ; y    x  4 ; y  1  A rule G  2  4 ;  1  1 G  2 ; 0 CA x-value CAy-value OR (3) B  A :  3 ; 4    2 ;  1  x ; y    x  1 ; y  5 A rule G 1  1 ; 5  5  G  2 ; 0 CA x-value CAy-value OR (3)  2  1 1  5  Midpoint of AC =  ;   2 2   1  =   ; 2 A Midpoint of AC  2  Midpoint of AC = Midpoint of BG x 3 1 y4  and 2 2 2 2 x2 and y0  G  2 ; 0 CA x-value CA y-value Answer only: FULL MARKS (3) Copyright Reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Mathematics P2 5 com Common Test June 2024 GRADE 11 Marking Guideline 2.6 F  0 ; 3 A coordinates of F & E or yF and xE E 1 ; 0  Area CEOF  Area OEIF + Area FIC CA area of OEIF 1  = (3 1)    2 1 CA area of FIC 2   4square units CA answer OR (4) F  0 ; 3 A coordinates of F & E or yF and xE E 1 ; 0  CEOF is a trapezium………………...[OF∥EC] 1 Area CEOF =  OF +EC   OE A formula for area of 2 trapezium 1 =  3  5  1 CA substitution into area of 2 trapezium  4square units CA answer (4) [17] QUESTION 3 1 g2 x 3.1 cos 23   1 r   2 y 2  1  1 g 2 A using x2  y 2  r 2 2 y2  11 g 2 y   g2 y g CA y = g sin 23  g CA answer (3) OR sin 2 23  cos 2 23  1 A square identity sin 23  1  cos 23 2 2   2  1 1 g2 CA substitution = 1  1  g 2  = g2  sin 23  g CA answer (3) Copyright Reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Mathematics P2 6 com Common Test June 2024 GRADE 11 Marking Guideline 3.2 tan113  tan 180  67    tan 67 A  tan 67 1 g2  CA answer g (2) 3.3 sin  653   sin 67 A sin  653   sin 67  1 g 2 CA answer (2) [7] QUESTION 4 cos390 4.1.1  tan  360  x  .cos 180  x  .cos  x  90  A cos390  cos30 cos  30 A cos(30)  cos30 cos 30 A tan(360  x)   tan x     tan x   cos x  sin x  cos 30 A cos(180  x)   cos x  sin x  A cos( x  90)  sin x  1      cos x  sin x   sin x   cos x  A  tan x      cos x   1  sin 2 x CA simplification  cos 2 x CA answer (8) sin120.tan 330 sin 60.   tan 30  A sin 60 and  tan 30 4.1.2  cos 240  cos 60 A  cos 60 3 1  A substitution of special 2 3  angles 1 2 1 CA answer (4) Copyright Reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Mathematics P2 7 com Common Test June 2024 GRADE 11 Marking Guideline cos x.tan 2 x 4.2 LHS   cos x 1 1 cos x  sin 2 x  cos x   A quotient identity  cos 2 x    cos x 1  cos x A denominator as a single cos x fraction  sin x  cos x 2    cos x  cos x  1  cos x sin 2 x   cos x 1  cos x 1  cos 2 x A square identity   cos x 1  cos x  1  cos x 1  cos x   cos x A factorisation 1  cos x  1  cos x  cos x A simplification 1 (5)  RHS OR 2 cos x.tan x LHS   cos x 1 1 cos x sin 2 x A quotient identity cos x .  cos 2 x  cos x 1  cos x A denominator as a single cos x fraction 2 sin x cos x    cos x cos x 1  cos x sin 2 x   cos x 1  cos x sin 2 x  cos x 1  cos x  A writing as a single  1  cos x fraction sin x  cos x  cos 2 x 2  A simplification 1  cos x 1  cos x  1  cos x A square identity 1  RHS (5) 4.3 x  0 A x  0 x  90 A x  90 x  180 A x  180 (3) Copyright Reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Mathematics P2 8 com Common Test June 2024 GRADE 11 Marking Guideline 2 4.4 cos       2  2      180  cos 1    2       180  45      135..............................(1) A      135 1 cos   2    2   2   cos 1   1 2   2   60............................(2) A   2   60 (1)  (2) : 3  75   25 A   25 =110 A   110 (4) [24] GEOMETRY  MEETKUNDE A mark for a correct statement (A statement mark is independent of a reason) S 'n Punt vir 'n korrekte bewering ('n Punt vir 'n bewering is onafhanklik van die rede) A mark for the correct reason (A reason mark may only be awarded if the statement is correct) R 'n Punt vir 'n korrekte rede ('n Punt word slegs vir die rede toegeken as die bewering korrek is) Award a mark if statement AND reason are both correct S/R Ken 'n punt toe as die bewering EN rede beide korrek is Copyright Reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Mathematics P2 9 com Common Test June 2024 GRADE 11 Marking Guideline QUESTION 5 5.1 A xy O x 2x 2y y B C D Construction: Draw AO extended to D A construction Proof: ˆ = ABO BAO ˆ x [∠’s opp. = sides] A S/R ˆ  2x BOD [ext. ∠ of ∆] A S/R Similarly: ˆ COD  2 y [ext. ∠ of ∆] ˆ  2x  2 y A S BOC  2 x  y A 2 (x + y) ˆ  2A (5) 5.2.1 ˆ  55 D̂  G [∠’s in the same segment] A S A R (2) 5.2.2 Ô 2  2(55)  110 [ at centre = 2 × at circumference] A S  A R (2) OR OR ˆ  DEO D ˆ  55 [∠’s opp = sides] A S/R  Ô2  110 [ext ∠ of ∆] A S/R (2) OR ˆ  DEO ˆ  55 OR D [∠’s opp = sides] A S/R [sum of s in ∆]  Ô1  70  A S/R  Ô2  180  70  110 [sum of ∠’s on a st line] (2) ˆ  180 5.2.3 ˆF2 + Eˆ 3 + O [sum of s in ∆] A S/R 2 Fˆ2 = Eˆ 3 [s opp equal sides]  F̂2  35 A S/R OR (2) ˆ  90 DEF [ in semi-circle] A S/R ˆ + DEF D ˆ + Fˆ  180 [sum of s in ∆] 2  F̂2  35 A S/R (2) Copyright Reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Mathematics P2 10 com Common Test June 2024 GRADE 11 Marking Guideline 5.2.4 ˆ = ˆ H H 2  90 [line from centre to midpt of chord] A S/R 3  Ê1  35 [sum of s in ∆ OR ext  of ] A S/R  Eˆ  Fˆ 1 2 ⸫ DE is a tangent [converse tan chord theorem] A R (3) [14] QUESTION 6 6.1.1 Ĉ 2  40 [tan-chord theorem] A S A R (2) 6.1.2 K̂1  90 [∠ in a semi-circle] A S A R (2) 6.1.3 P̂1  65  40 [exterior ∠ of a ∆] A S/R A answer  25 (2) 6.1.4 Tˆ1  D ˆ + DPT ˆ [exterior ∠ of a ∆] =40+25 =65 A S/R  TC = QC [ sides opp. =∠’s] A R (2) OR Ĉ1  50 [radius ⏊ to tangent] A S/R T̂1  65 [sum of ∠’s in a ∆] ˆ  65  T̂1  TQC  TC  QC [sides opp. = ∠’s] A R (2) 6.2.1 SP = SR [tans from same point] A S/R ˆ = SPR SRP ˆ  62 [∠’s opp. = sides] A S/R ˆ  90 ORS [tan ⏊ rad] A S/R ˆ  28  ORN A S ˆ  90  28 NOR [sum of ∠’s of ∆] ˆ  62 NOR A answer (5) Copyright Reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Mathematics P2 11 com Common Test June 2024 GRADE 11 Marking Guideline 6.2.2 In ∆ORS: let OT  r  OT  OR  r radii OS = OT  9  r  9 OS2  OR 2 + RS2 Pythagoras  r  9   r 2  152 2  A LHS A RHS r  18r  81  r  225 2 2 A simplification 18r  144 r  8 units  radius  8 units A answer (4) OR OR ˆ  NPR NRS ˆ  62 ∠’s opp = sides A S/R NR In NRS: cos 62  15  NR  15.cos 62 A NR ˆ  90  62  28 NRO radius ⏊ to tangent NR In ONR: cos 28  OR 15.cos 62 A OR  OR  cos 28  OR  7.98 units A Answer [17] Copyright Reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Mathematics P2 12 com Common Test June 2024 GRADE 11 Marking Guideline QUESTION 7 7.1 ˆ  90 PVO [rad ⏊ tan] A S/R ˆ  90 OTR [line from centre to midpoint A S/R of a chord] ∴VOTR is a cyclic quadrilateral [conv. ext. ∠of a cyclic quad.] A R (Any) [conv. opp. ∠’s suppl] (3) OR OR ˆ  90 OVR [rad ⏊ tan] ˆ  90 OTR [line from centre to midpoint A S/R ˆ ˆ  180 of a chord]  OVR+OTR A S/R  V OTR is a cyclic quadrilateral [conv. opp. ∠’s of cyclic A R quad.] (3) 7.2 In ROT: OR 2  TR 2 + OT 2 [Pythagoras] A S In SOT: OT 2  OS2  ST 2 [Pythagoras] A S  OR  TR  OS  ST 2 2 2 2 A substitution for but OS = OV [radii] OT2 [given] and ST = TU  OR 2  TR 2  OV 2  TU 2 also: TU = RU  TR A A substitution for OS and ST  OR 2  TR 2  OV 2   RU  TR  2  A TU = RU  TR (6) [9] TOTAL: 100 Copyright Reserved Please turn over