NW-Maths-Grade-11-November-2025-P2-and-Memo.pdf

Downloaded from Stanmorephysics.com education Department: Education North West Provincial Government REPUBLIC OF SOUTH AFRICA MATHEMATICS P32 MARKS: 150 TIME: 3 hours This question paper consists of 13 pages and 1 information sheet. Copyright reserved Please turn over (Sweetness ii GG i Gl pl a

maDewmntoaded from Stanmorephysics.com NW/November 2025 Grade 11 INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. I, ae This question paper consists of 11 questions. Answer ALL the questions in the SPECIAL ANSWER BOOK provided. Clearly show ALL calculations, diagrams, graphs, etc. which you have used in determining your answers. Answers only will NOT necessarily be awarded full marks. You may use an approved scientific calculator (non-programmable and non- graphical), unless stated otherwise» ” If necessary, round of otherwise. TWO decimal places, unless stated Diagrams are NOT necessé An information sheet with fo uded at the end of the paper. Write neatly and legibly. Copyright reserved Please turn over

mapewstoaded from Stanmorep ics.com NW/November 2025 Grade 1 QUESTION 1 According to the SABS (South African Bureau of Standards), the acceptable standard deviation fora 340 ml can of Coke is equal to 2,74 ml. Out of a sampling of 20 cans the following: measured: Bal 342° | 338 336 340 340 345 338 339) 340 334 341 337 336 340 335 336 342 340 337 336 1.1 Calculate the mean volume of the 20 cans. (2) 1.2 Calculate the standard deviation of the 20 cans. qd) es between one standard deviation of 1.3. Calculate which percentage of the 0 can the mean. a (3) 1.4. What conclusion can the offi .BS make regarding the sampling? (1) 17] Copyright reserved Please turn over

Mawpanstoaded from Stanmorephysics.com NW/November 2025 rade 11 QUESTION 2 The fuel consumption of a car is expressed as the number of liters per 100 kilometers (//100-km) which the car use. The histogram below indicates the results of tests done on a different models of a specific car. 40 35 = 30 2 g 2 £ 20 § = 15 So Ss 10 E E 5 Zz 2.1 How many cars were tested? () 2.2. Write down the modal class. (1) 2.3. Use the histogram to complete the frequency table in the answer book. (3) 2.4 Draw an ogive to represent the data. (4) 2.5 Use the ogive to calculate the interquartile range. (3) [12] Copyright reserved Please turn over

mabewstoaded from Stanmore ics.com NW/November 2025 Grade 11 QUESTION 3 In the diagram below AABC has vertices A(-2 ; 4), B(-1;-3) and C(5;5). Point B lies on line DE. The x-intercepts of CB and AB are F and G respectively. B and 6 are the angles of inclination for lines BC and AB respectively. 3.1 Calculate the length of AC. (2) 3.2 Determine the gradient of AB. (2) 3.3. Prove that AABC is a right angle triangle. (3) 3.4 If BC1DE, calculate the size of ABE z (5) 3.5. Determine the equation of line DE. (3) 3.6 Calculate the area of AABC . (4) [19] Copyright reserved Please turn over

Mawpanstoaded from Stanmorep ics.com NW/November 2025 rade 11 QUESTION 4 In the diagram below, ABCD is a parallelogram with vertices A and D(0;2) lying on the y-axis, The side BC is produced to E such that BC = CE. B(3;9) and C(3;7) are given. The length of AD is 2 units. The line segment AE intersects DC at F. x —_ _ Oo ’ 4.1 Calculate the coordinates of A. qd) 4.2 Write down the equation of line BE. (1) 4.3. Why is F the midpoint of AE? (1) 4.4 Hence, find the coordinates of F. (2) 4.5 Determine whether points O, F and B are collinear. (4) 19] Copyright reserved Please turn over

mabewstoaded from Stanmore ics.com Grade 11 QUESTION 5 Answer question 5.1 to 5.4, without using a calculator: 5.1 If sina = —2 and 270° < A < 360°. Determine, without using a calculator, the value of: 5.1.1 cosA 5.1.2 tan(A —180°) 5.2 If tama@=p; sina <0 and p>0, determine sina in terms of p. = 5.3. Determine the value of: .tan330° x) + sin? (180° + x) 5.4 Prove the following identity: +2tan’ x) = 1+sin? x 5.5 Determine the general solution of the equation: 2sin@ + 3cos@ = 0 NW/November 2025 (3) (2) (4) (8) (6) (4) 5.6 Inthe diagram below, ABCD is a rectangle in a semicircle with center O, and COD = 8. For what value of @ (correct to 2 decimal places) will rectangle ABCD become a square? (Clearly show ALL calculations.) (4) Copyright reserved [31] Please turn over

mabewstoaded from Stanmore ics.com NW/November 2025 Grade 11 QUESTION 6 In the diagram below, the graphs of f(x) = acosx and g(x) = sinbx are drawn for the interval x €[0°;360°]. x (360°; 3) (180°; -3) 6.1 Determine the values of a and b. (2) 6.2 Write down the range of f- (2) 6.3. Determine the value of p rounded off to TWO decimal places. (2) 6.4 Give the coordinates of B. (2) 6.5 Calculate the length of CD, correct to TWO decimal places. Given CD is parallel to the y-axis and the y-coordinate of C is 2p. FI Copyright reserved Please turn over

Matemstoaded from Stanmorephysics.com NW/November 2025 ie 11 fidhimorephysics.com 7.1 Calculate the length of AC. (3) 7.2. Hence, prove that the area of AABD = 1200V3 mm?. (5) [8] Copyright reserved Please turn over

Matibewntoaded from Stanmorephysics. com NW/November 2025 The w emi-sphere with a radius of 200 mm, must be cast. The wall must be I shown below. 8.1 Calculate the volume of the mixture needed to cast the wall of the semi-sphere. (4) 8.2 Calculate the area of the FLAT surface of the wall of the cast. (3) 17] Copyright reserved Please turn over

matawntoaded from Stanmorephysics.com NW/November 2025 Grade 11 Provide reasons for your statements in QUESTIONS 9, 10 and 11. QUESTION 9 9.1 In the diagram, ABCD is a cyclic quadrilateral in a circle with center O. Prove the theorem that states: A + C = 180°. A (5) 9.2 In the diagram below, PQ! QA produced meets the circle M P eA u I 30° I 1 2 2 Q R 9.2.1 Calculate, with reasons, three angles in the diagram which are equal to 60°. (4) 9.2.2 Calculate QRS. (2) 9.2.3. Give a reason why PS|OR . (1) 9.2.4 Hence, prove that TR is a diameter of the circle. (3) [15] Copyright reserved Please turn over

matpawntpoaded from Stanmore ics.com NW/November 2025 Grade 11 QUESTION 10 In the diagram below, PA and PB are tangents to the circle with the center O. D lies on the circumference and C is a point on AD such that CD = CB. D = x. P 10.1 Give the reason why AP = BP. (1) 10.2 What is the size of PAO? Give a reason for your answer. (2) 10.3 Prove that AOBP is a cyclic quadrilateral. (2) 10.4 Calculate, with a reason, the size of AOB in terms of x. (2) 10.5 Calculate, with reasons, the size of ACB in terms of x. (3) 10.6 Prove that ACBP is a cyclic quadrilateral. (4) 10.7 Hence, prove with calculations that PC bisects ACB. Se [18] Copyright reserved Please turn over

matawntoaded from Stanmorephysics.com NW/November 2025 Grade 11 QUESTION 11 In the diagram below, KL is the diameter of a circle with center O. KL = 20 units. M is a point on the circle such that ML = 12 units. The bisector of PXM, line NX meets KM at X. Chord PXR cuts KL perpendicularly at S. Chord KQ = QM. 11.1 Prove that K, = 36,87°, using the necessary calculations and reasons. (3) 11.2 Calculate, with reasons, the size of X4. (3) 11.3 Prove that line NX is a tangent to a circle passing through L, M and X. (5) [11] TOTAL: 150 Copyright reserved

matbeswtoaded from Stanmore ics.com NW/November 2025 rade 11 INFORMATION SHEET: MATHEMATICS GRADE 11 _ ees b? —4ac 7 2a A=P(l+ni) A=P(l-ni) A=P(l-i)" A=P(1+i)" xX, +x. By +y. d=V(x,-%)? +0, -y,)" m2 122) y=mx+c mo22—1 m=tan0 Xg—™ In AABC: i sinA a2 =b2 area AABC =—ab.sin C « S, d(x, -x) yo oa oe == n n P(A) = a) P(A or B) = P(A) + P(B) — P(A and B) n Copyright reserved

Downloaded from Stanmorephysics.com NAME OF LEARNER: NAAM VAN LEERDER: CLASS: KLAS: PROVINCIAL ASSESSMENT/ PROVINSIALE ASSESSERING [ GRADE/GRAAD 11 | \y & MATHEMATICS P2/WISKUNDE V2 a 1 u a a 1 u ‘ NTWOORDEBOEK a were te... Ce QUESTION MARK MODERATION INITIAL VRAAG PUNT PARAAF MODERERING PARAAF 1 2 3 4 5 6 7 8 9 10 11 TOTAL TOTAAL (150) This answer book consist of 20 pages. Hierdie antwoordeboek bestaan uit 20 bladsye.

mabowaloaded. from Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Answer Book/Antwoordeboek QUESTION/VRAAG 1 342 338 336 340 340 345 338 339 340 334 341 337 336 340 335 336 342 340 337 336 Solution/Oplossing Marks Punte 1.1 (2) 12 Q) 1.3 QB) 1.4 Q) 17] QUESTION/VRAAG 2 Solution/Oplossing Marks Punte 2.1 Q) 22 () Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

mabewnloaded. from Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Answer Book/Antwoordeboek Solution/Oplossing Marks Punte 23 Interval FREQUENCY/ CUMULATIVE FREKWENSIE FREQUENCY / KUMULATIEWE (number of vehicles/ FREKWENSIE aantal karre) 5<x<6 62x<7 TS8 B8<x<9 9<x<10 (3) 2.4 2 zg E & 70 Fy ay = 0 E 3 50 S & 5 a z g 30 z = E 20 1S) 10 0 1 2 3 4 5 6 8 9 10 " 12 13 Fuel Consumption/ Brandstofverbruik (A00km) (4) 2S (3) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

mabowaloaded. from Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Answer Book/Antwoordeboek QUESTION/VRAAG 3 C(5:5) D Solution/Oplossing Marks Punte cat (2) 3.2 (2) 3.3 @) 3.4 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

mabeweloaded. from Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Answer Book/Antwoordeboek Solution/Oplossing Marks Punte (5) S| (3) (4) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

mabowaloaded. from Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Answer Book/Antwoordeboek QUESTION/VRAAG 4 Solution/Oplossing Marks Punte 4.1 qd) 4.2 dQ) 43 dd) 44 (2) 4.5 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

mabowaloaded. from Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Answer Book/Antwoordeboek Solution/Oplossing Marks Punte (4) 19] QUESTION/VRAAG 5 Solution/Oplossing Marks Punte Sl QB) 5.1.2 (2) 5.2 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

mabeweloaded. from Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Answer Book/Antwoordeboek Solution/Oplossing Marks Punte 5:3 orephYstes.com (8) 5.4 (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

mabowaloaded. from Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Answer Book/Antwoordeboek ution/Oplossing Marks Punte 55 —— (4) 5.6 ahmore A B (4) [31] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

mabowaloaded. from Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Answer Book/Antwoordeboek QUESTION/VRAAG 6 Marks Punte _——el 24 xs] AC1,6°; p) r ss 90" 180 270Y - me a CDN .CO _ poe Br a C 3) 6.1 (2) 6.2 Q) 6.3 (2) 6.4 Q) 6.5 (5) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

mabewnloaded. from Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Answer Book/Antwoordeboek QUESTION/VRAAG 7 Solution/Oplossing Marks Punte 7A (3) 72 (5) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

mabawaloaded. from Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Answer Book/Antwoordeboek Solution/Oplossing Marks Punte 8.1 h y a A (4) 8.2 GQ) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

mabeowaloaded. from Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Answer Book/Antwoordeboek QUESTION/VRAAG 9 Solution/Oplossing Marks Punte 9.1 G) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

mabowaloaded. from Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Answer Book/Antwoordeboek Solution/Oplossing Marks e Punte 9.2.1 (4) 9.2.2 (2) 9.2.3 ie) 9.2.4 (3) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

mabowaloaded. from Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Answer Book/Antwoordeboek QUESTION/VRAAG 10 P Solution/Oplossing Marks. Punte 10.1 Q) 10.2 (2) 10.3 (2) 10.4 (2) 10.5 GB) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

mabowaloaded. from Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Answer Book/Antwoordeboek Solution/Oplossing Marks Punte 10.6 (4) 10.7 (4) [18] ADDITIONAL SPACE/BYKOMENDE RUIMTE Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

mabeowaloaded. from Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Answer Book/Antwoordeboek QUESTION/VRAAG 11 Bai imorephyssocon: Solution/Oplossing Marks Punte 11.1 GB) 11.2 GB) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

mabewaloaded: from Stanmorephysics.com Grade/Graad 11 — Answer Book/Antwoordeboek NW/November 2025 Solution/Oplossing Marks Punte 11.3 (G) (11) TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

mabewaloaded. from Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Answer Book/Antwoordeboek ADDITIONAL SPACE/BYKOMENDE RUIMTE .com Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

mabewaloaded. from Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Answer Book/Antwoordeboek Copyright reserved/Kopiereg voorbehou

Downloaded from Stanmorephysics.com education Department: Education North West Provincial Government REPUBLIC OF SOUTH AFRICA PROVINCIAL ASSESSMENT/ PROVINSIALE ASSESSERING [ GRADE/GRAAD 11 J NOVEMBER 2 225) " ) ' Fj MARKING GUIDELINES/NASTENRIGE YNE " MARKS/PUNTE: 150 These marking guidelines consist of 17 pages. Hierdie nasienriglyne bestaan uit 17 bladsye.

maubawnoaded2from Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Marking Guidelines/Nasienriglyne NOTE: e Ifacandidate answers a question TWICE, only mark the FIRST attempt. e Ifacandidate has crossed out an attempt of a question and not redone the question, mark the crossed out version. e Consistent accuracy applies in ALL aspects of Marking Guidelines. Stop marking at the second Calculation error. e Assuming answers/values in order to solve a problem is NOT acceptable. LET WEL: e As ‘n kandidaat ‘n vraag TWEE KEER beantwoord, sien slegs die EERSTE poging na. e As ‘n kandidaat ‘n antwoord van ‘n vraag doodtrek en nie oordoen nie, sien die doodgetrekte poging na. Volgehoue akkuraatheid word in ALLE aspekte van Nasienriglyne toegepas. Hou op nasien by die tweede berekeningsfout. Aanvaar van antwoorde/waardes om ‘n probleem op te los, word NIE toegelaat NIE. GEOMETRY/MEETKUNDE A mark for correct statement (A statement mark is indepenent of a reason) ‘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) ‘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/Kopiereg voorbehou Please turn over/Blaai om asseblief

matpewnoadedfrom Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Marking Guidelines/Nasienriglyne QUESTION/VRAAG 1 342 338 336 340 340 345 338 339 340 334 341 337 336 340 335 336 342 340 337 336 1.1 _ | Mean/Gemiddeld == ¥ sum/som 20) Roy. SS Y answer/antwoord = 338.6 - 120 | o=2,71 ¥ wad? qd) 1.3 | Interval = ( 338, 6 — 2,71; 338,8 + 2,71 Y interval = (335,89 ; 341,31) CAfrom i and12 || 5 «5 x 100 = 75% lies between | standard deviation. ¥ answer/antwoord (3) 1.4. | The sample meets the requirements/Die steekproef voldoen Y answer/antwoord aan die vereistes. (1) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

mapawnoadedfrom Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Marking Guidelines/Nasienriglyne QUESTION/VRAAG 2 2.1 | 100 cars/karre v answer d) 2.2 | 7648 v answer a) 23 Interval FREQUENCY CUMULATIVE v5, 25, 40 (number of vehicles) FREQUENCY ¥20, & 10 5<x<6 5 5 ¥ Cumulative 6<x<7 25 30 colom 7<x<8 40 70 8<x<9 20 90 9<x<10 10 100 (3) 24 Y Grounding/ 100 grond (5;0) Y Using upper °0 intervals/gebruik 2 boonste intervalle s* Y End at/eindig by é .. (10;100) s —— ¥ shape/vorm = = é to 4 nore sez — ° la CoE a a 4 v . 3 7 & Ey 10 (4) nsumption/ Brandstofverbruik (/100km) 2.5 | IQR/IKV = Q3 — Qy %Q3 =8,5 =85-65 —2 (8-9) ¥%Q, = 6,5 Y answer GB) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

mattbawnMoadedfrom Stanmoreph ics.com NW/November 2025 Grade/Graad 11 — Marking Guidelines/Nasienriglyne QUESTION/VRAAG 3 A(-254 (13-3) 3.1) a. =f6-Cay +6-47 ¥ correct subst in distance formula/korrekte dic = /50 =5/2 =7.071 ee aan vervanging in afstandsformule. Yanswer/antwoord (2) 3.2 4—(-3) te eh Swap: Max 1/2 ¥ correct substitution into gradient formula korrekte vervanging in gradiént formule ¥my=-7 (2) Mag = —7 | 33 _ 5-4 Mac = 5 (=2) 1 Mmr=- ac VM 4c “> 1 aH Y Myc X My 7 v-l =-1 (3) “AB 1 AC and AABCis a right angle triangle Ba] 5= C3) BC en 4 4 Mgc = 3 ¥mMegc = 3 4 tan B = myo = = B BC 3 B =5313° ¥ B=53,13° Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

matpewnoadedfrom Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Marking Guidelines/Nasienriglyne tand=m,, tanO=—7 @=180°-8187° 0 =98,13° : ; ¥ 0 =98,13° ABC = 98,13°—53,13° [ext Zof A/ buite Z van A] ABC = 45° Vv ABC =45° ©. ABE = 45° +90° =135° V ABE =138 (5) OR/OF OR/OF AC = 5V2 BC=10 ¥ BC=10 on 52 sibaachalal 1. vv sinaBc = 2 Sv¥2 BC = sin-1(~— A : sin 70 ; ABC = 45° v¥ ABC =45° <. ABE = 45° +.90° = 135° Vv ABE =138° (5) 3.5 | mpp=-% BC1LDE Yor =—2 YY, =m(x—x,) alr ead te, 3 ¥ subst./vervang (-1 ; -3) 7==)a= gaia) -3=-2 +e 3 3 OR/OF yt3 =-—x-— 15 ys 3... 15 44 — ar’ 3 15 or’ yoo3x Bb 3) 4.4 3.6} dy = V(-1-(-2)° +(-3-4)" ¥ dy = 50 d yy = V50 V dye =10 dye =5X2=10 1 - « Area = gAB. BC. sinABE v5 x V50 x 10 x sin4s° 1 =3 x ¥50 x 10 x sin45° ¥ answer/antwoord =25 units” OR OR = 2 2 dy =V(-1--2 (3-4) ¥ dy = V50 dj, =V50 a a v Area = = AB.AC : Area ==AB.AC ‘ i 2 v5 x V50 x V50 => x ¥50 x ¥50 2 Yanswer/antwoord (4) =25 units* [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

mautbawnoaded2from Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Marking Guidelines/Nasienriglyne QUESTION/VRAAG 4 ie BG;9) C (G3; 7) E A F D/O; 2) x 4.1 |A(0;4) ¥ answer/antwoord (1) 4.2 |x=3 ¥x=3 (1) 4.3 | Line ae any to 2nd side. | rr v reason/rede OR (1) ACED is parm...AD=CE & AD|CE 44 r(0#3,24D) vx=s 2 2 Vy=2 2 FE;3) Q) 4.5 3 0 ¥ Mor = 3 Mor = 3— =3 “ 3 0 VY mp = 3 9 9-= v = Mar = - =3 Mor = Mgr Bie 2 Mor = Mpx and F is common/gemeenskaplik. Y conclusion/konklusie : O, F & B are collinear/is kollinieér. OR/ OF mgo = 3 (4) 19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

matpawnoadedfrom Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Marking Guidelines/Nasienriglyne QUESTION/VRAAG 5 5.1 5.1.1 | x? = (5)? = (43)? Y correct subst. Pm ate 8 V¥x=4 4 cosA == (3) 5.1.2 | tan(A — 180°) =tanA vtanA = 7 (2) 5.2 | sina [i+p? -1 a Y diagram P Yr=/1+p2 + p? vg v answer (4) 5.3 sin(—120°) . tan330° ¥(—sin 60°) cos(360° — x) sin(90° + x) + sin?(180° + x) v (—tan30°) __(~sin 60°).(—tan30°) a) (cosx.cosx)+sin2x° ¥ cos(360 x) = cOsx V3, 1 ¥sin(90° + x) = cosx _ pow? cos?x+sin2x vsin?x v3 1 Pj ae) oe (-2)(-2) ¥cos*x + sin?x =1 le ole ¥ answer/antwoord Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

matpawnoadedfrom Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Marking Guidelines/Nasienriglyne (8) 5.4 |(2—2sin?x)(1 + 2tan?x) _ 3 1+ sin?x ~ , (2-2sin?x)(14+2tan?x) Sint Ree 1+sin2x vtan®x = cos?x in? bat sin?x. - 2(1-sin x)(1420255) 1+sin?x in2 2 sin?x = 2(cos: ac142(22=2)) 1+sin?x __ 2(cos*x+2sin?x) 1+sin?x _ 2(1-sin?x+2sin?x) ¥ 2(1 — sin?x) Y¥1—-sin*x = cos?x ¥2(cos*x + 2sin?x 1+sin?x _ 2(1tsin?x) Ycos?x = 1-—sin?x 1+sin2x ¥1+sin?x =2 LH=RH (6) | 505 2sin@ + 3cos@ = 0 2sin@ = —3cos@ 2sin@ _ —3cos@ fx 7 2cos@ —_2cos@ ~ 608 3 =o? tan0d = -= ¥ tangs 2 2 RA = 56,31° I 6 = 180° — 56,31°+ k.180°.k EZ Y general solution. 6 = 123,69° + k.180°,k €Z V¥kEZ (4) 5.6 For ABCD a square, must/Vir ABCD ‘n vierkant, moet BC=CD v¥ BC=CD + 0C = cD cred (0C = 5BC) , Now/Nou voc =5CD tané = 0c _ CD 5 cD vtand =2=2 1 oc =7= 2 - ¥ 6 = 63,43° + = 63,43° (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

matpawnoadedfrom Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Marking Guidelines/Nasienriglyne | 131) | QUESTION/VRAAG 6 ad Y (360°: 3) ic = J A(71,6°: p) ® 90° 180) 270% cae D a é, 2] =a (180°; -3) 61 |a=3 Ya=3 b=1 VYb=1 (2) | 6.2 -3<y<3 ORye [-3 4 3] Y¥ notation/notasie Y Correct values cD = 1,9— (-0,77) =2,67 units/ eenh (2) 6.3 p = sin71,6° = 0,95 ¥ subst 71,6 in f(x) or OR/OF answer only full marks | g(x) p = 3cos71,6° = 0,95 Vp = 0,95 (2) 6.4 | B(251,6°; —0,95°) Vx = 251,6° Through symmetry/deur simmetrie Vy = -0,95° (2) 6.5 At/ By C: 2p = 3cosx « 1,9 = 3cosx “1,9 = 3cosx 1,9 v = cosx = — = 0,63 cosx = 0,63 2 x = 360° — 50,70° Ti = apo” ¥ x = 309,3° At/ By D: y = sin 309,3° Vy=- ay =-0,77 y 0,77 ¥ CD=2,67 units/eenh (5) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

mattpawnmoadedfrom Stanmoreph ics.com NW/November 2025 Grade/Graad 11 — Marking Guidelines/Nasienriglyne QUESTION/VRAAG 7 B 160 mm 7.1 | 4C? = AB? + BC? —2(AB)(BC) cos ABC AC? = 607 +1607 — 2(60)(160) cos60° ¥ subst. correctly into cos tule/ vervang korrek in cos-reél. i AD = pac = 70mm 1 Ps AreaAABD = =AB.AD.sinA AC’ =19600 ¥ AC? =19600 AC = 719600 AC =140mm Vv AC =140mm G3) 72 Sind sin dBC Y subst. correctly into sin BC AC tule/ vervang korrek in q sin-reél. sind _ sin60° 160 140 ~ 160sin60° v si = ——__—_. _ « 160sin 60° — sin A = ———— 140 . 1603) sinA= 140 ¥ sina = 43 4 43 7 sinA= 7 VAD =>AC = 70mm 2 1 4V3 _ E (60). 8 v= (60). (70). =1200V3 units?/eenh? (5) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

maibawroaded2from Stanmoreph Grade/Graad 11 — Marking QUESTION/VRAAG 8 ics .com NW/November 2025 uidelines/Nasienriglyne 81 | Wolmmestx cay — eect i me ==> xX -ar3 —-xX-ar olume => x57 3am 4 bs 3 tt 5 7(200)3 — 5 x 1 =5x $7(150)? =9686577,35 mm> ¥ r=150mm ¥ substitution into correct formula/ vervanging in korrekte formule. ¥ Volume ; — Volume 2 ¥ answer/antwoord (4) 8.2 | Area = nr? — nr? = n(200)? — 2(150)? = 54977,87 mm? v Area = mr? — mr? ¥ substitution into correct formula/ vervanging in korrekte formule. ¥ answer/antwoord (3) 7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

matbawnMoadedfrom Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Marking Guidelines/Nasienriglyne QUESTION/VRAAG 9 9.1 A D B c NOTE: If candidate fails to draw the construction or indicate the construction, it is an immediate BREAK DOWN. 0/5 marks. NOTA: Indien kandidaat nie die konstruksie teken of aandui dat ‘n konstruksie plaasvind nie, is dit ‘n onmiddellike “BREAK DOWN” en daar word nie verder gemerk nie. 0/5 punte. 9.1 Construction: Connect BO and OD Konstruksie: Verbind BO en OD 0, =2A [angle at centre = 2 x Z at circumference/ middelpunts 2 = 2 x omtreks Z | 02 = 2¢ {angle at centre =2 x Z at circumference/ middelpunts Z = 2 * omtreks Z| 0, + 0, = 360° [ Z around a point = 360°/ Z°om'n punt = 360°] «2A + 2€ = 360° A+ = 180° Y construction/ konstruksie YS YR YSIR VS 0, +0, = 360° (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

matpawnoadedfrom Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Marking Guidelines/Nasienriglyne 9.2 ala), 30° 1 1 2 2 Q R 9.21 | P=60° [ZsinaA/ Zein ‘n A] vS/R Ss, = P=60° [given/ gegee] a FR OTS =60° [Zs inthe same circle segment/ 4 e in dieselfde (4) sirkel segment] 9.2.2 | ORS =120° [opp Zs of cyclic quad =180°/t.0 Ze van YSYR KVH =180°] (2) 9.2.3 | $,+ORS =180° y * PS|OR [co-interior Zs =180°/ ko-binne Ze=180° | R () 9.2.4 | 4,=0,=90° [alt Zs/verw Ze; PS\OR] vs vR -. TR is diameter/ middellyn. [chord subtends 90° 7/ koord x @) onderspan 90° Z | [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

matpewnoadedfrom Stanmorephysics.com NW/November 2025 Grade/Graad 11 — Marking Guidelines/Nasienriglyne QUESTION/VRAAG 10 P B 10.1 | Tangents from the same point/raaklyne vanuit dieselfde punt. YR (1) 10.2 | P40=90° [radius 1 tangent/raaklyn] YSYR (2) 10.3 | PBO=90° [radius tangent/raaklyn] vSIR AOBP is cycl. quad/ KVH_ [COVERSE opp Zs of cycl quad supplementary/ OMGEKEERDE t.o. Ze | VR van KVH supplimentér] (2) OR [opp Zs of quard = 180°/ t.o. Z e = 180°] 104 | 40B=2x [angle at centre =2 x Zat circumference/ YSYR middelpunts Z = 2 * omtreks Z | (2) 10.5 | CBD=x [Zs opp=sides/ Ze teenoor = sye] vS/R ACB =2x [ext Zof A/buite Z van A] YSYR (3) 10.6 | 4PB=180°—2x [opp Zs of cycl quad supplementary/ t.0. Ze van VSYR KVH supplimentér] but/ maar ACB = 2x : APB + ACB= 180° — 2x + 2x = 180° -. ACBP is cycl. quad/ KVH vs [COVERSE opp 4 s of cycl quad supplementary/ OMGEKEERDE 1.0. 2 ¢ van KVH supplimentér| VR OR [opp Zs = 1807 t.o. Z e = 180°] (4) 10.7. | H=PAB=x [tan chord theorem/raaklyn-koord stelling] YSYR But/ maar PAB = PCB =x [4s in the same circel segment/ 7 e in YSVYR dieselfde sirkel segment] But/ maar ACB = 2x A oo « PCB ==ACB 2 4 (4) PC bisects/halveer ACB [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

matbawmMoaded2from Stanmorephysics.com NW/November 2024 Grade/Graad 11 — Marking Guidelines/Nasienriglyne QUESTION/VRAAG 11 11.1 | KML=90° [Z insemicircle/ 7 in halwe sirkel] YSYR sink, = ML _ 12 ¥ trig ratio/ _ 1 KL 20 verhouding Kis 36,87° GB) 11.2 | ¥,=5313° = [ZsinaA/ Zein ‘n A] vSIR X, =X, + Xs C20 vS/R But/ 4 vs (G3) 11.3 vs VSVY B SYR ¥8 YR RDE raaklyn/koord stelling.] OR/ OF 5 [ Z between line and chord/ Z tussen lyn en koord.] (5) [1] TOTAL/ TOTAAL: 150 Copyright reserved/Kopiereg voorbehou