Mathematics Grade 11 KwaZulu Natal June 2022 Paper & Memo

Downloaded from Stanmorephysics.com @@aRHRES.. KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE GRADE 11 : MATHEMATICS : : COMMON TEST . ; JUNE 2022 : MARKS: _ 100 TIME: 2 hours This question paper consists of 6 pages and 2 DIAGRAM SHEETS. Copyright Reserved Please Turn Over SS

Downloaded from hlayiso.com — Downloaded from Stanmagephysi¢s-com INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. This question paper consists of 6 questions. 2. Answer ALL the questions. 3. Number the answers correctly according to the numbering system used in this question paper. OS 4. Clearly show ALL calculations, diagrams, graphs, etc. which you have used) determining your answers. o: 5. Answers only will NOT necessarily be awarded full marks. x 6. You may use an approved scientific calculator (non- pg and non-graphical), unless stated otherwise. 0 7. If necessary, round off answers correct to TWO oy places, unless stated otherwise. 8. Diagrams are NOT necessarily drawn to scale. 9. TWO DIAGRAM SHEETS for UBSTION 3.2, QUESTION 4, and QUESTION 5 are attached at the end of this question paper. Detach the DIAGRAM SHEETS and hand in together with your ANSWER BOOK. @ 10. Write neatly and ne ww ee Copyright Reserved Please Turn Over

Matherpatics Downloaded from hlayiso.com wnloaded from Stanmoxephysic¢s.com —_‘1#"* 2022 Common Test QUESTION 1 1.1 Consider the following linear number pattern: 11; 17; 23; 293 wow. ; 299, peees| Determine 7, , the general term of this pattern, in the form 7, =an+b. ell) Calculate the value of the 20" term in this pattern. 1.1.3 Calculate the number of terms in this pattern. 1.1.4 The terms of this linear number pattern is the sequence of 1“ differences of a quadratic number pattern, i.e. the 1“ differences of the quadratic number pattern are 11;17;23;... If the fifth term of this quadratic number pattern is 100, what will value of the second term? S 12 Consider the following quadratic number pattem: -16; -—12; e 1.2.1 Write down the next two terms in the pattern. Ss 1.2.2 Determine T,,, the general term of this rth form T, =an’ +bn+c. 1.2.3 Show that all the terms of this number are even numbers. 1.2.4 A new pattern with general rade formed such that P, = 7-128. How many negative terms me e in this new pattern? 0 K\ QUESTION 2 S oe Given: f(x)=—x? +3x+ 10e 2.1 Calculate the cota of the turning point of f 2.2 Write range of f- 2.3 Calculate the x-intercepts of /: 24 Sketch the graph of f, clearly indicating all the intercepts with the axes and the coordinates of the turning point. 2.5 For which values of k will ~x° +3x+10=k have two positive, unequal real roots? 2.6 The graph of f is translated 2 units to the right and 3 units down to obtain the graph of g. Write down the equation of g in the form g(x) = a(x + py +q. Copyright Reserved (2) (2) (2) (2) (2) (4) (2) (4) [20] (3) (1) (3) (4) (2) (3) [16] Please Turn Over

Downloaded from hlayiso.com Downloaded from Stanmoagephysics-com = QUESTION 3 1 x 3.1 Given: ats)=(5) +4 3.1.1 Write down the equation of the asymptote of h. (1) 3.1.2 Draw a sketch graph of A, clearly indicating the asymptote and any intercepts with the axes. (3) 3.1.3 Two transformations are applied to / to obtain the graph of mx)= -3* -4, Write down the two transformations. (2) 3.2 The diagram below shows the graphs of (x) = 74 q and g(x) =mx-8. X+p e The asymptotes of / intersect at (1 od 2). © Graphs fand g intersect the x-axis and y-axis at C and D respectively. y 3.2.1 Write down the values of p and q. (2) 3.2.2 Write down the domain of f qy 3.2.3 Calculate the value of a. (3) 3.2.4 Calculate the value of m. (4) 3.2.5 Determine the equation of the axis of symmetry of g that has a positive gradient. (2) 3.2.6 For which values of x will f (x)s g(x)? @) [21] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Meherswnloaded from Stanmorephysi¢s.com QUESTION 4 June 2022 Common Test In the diagram below, A(2 ; 11), BC-10 ; -5), C and D(6 ; 9) are the vertices of a quadrilateral in the Cartesian plane. BC is a horizontal line and CD a vertical line. E is the midpoint of AB. F is a point on CD such that EF is parallel to AD. *) ast D6 ; 9) E oO x F BC-10; 5) c 41 Write down the coordinates of C. (2) 4.2 Calculate the coordinates Of E. ses (2) . 43 Determin ‘ (4) 4.4 Calculate the C00 (2) [10] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com .---Downloaded from Stanmorephysics.com NSC- Grade 11 QUESTION 5 In the diagram below, straight line GH cuts the x-axis at J(-6 ; 0). K is the point (1 5 4). KJ is produced to M and GK is drawn. L(-3 i 2) is a point on GH and KL is drawn. Gik =a. yd 5.1 Calculate the gradient of GH. 5.2 Calculate the angle of inclination of MK. 5.3 Determine the size of angle a. 5.4 Prove that KL 1 JL. 5.5 Calculate the area of AKJL. QUESTION 6 6.1 Prove that (1—sin? @)(1+ tan? 6) =1. 6.2 Solve for x if sin(2x + 30°) = —0,4 and x €[—90° ;90°] 6.3 Determine the general solution of 3sinxcosx—4cos* x=0. G 7 Kd; 4) Copyright Reserved (2) (4) (3) (3) (5) {17] (3) ©) (6) [16] TOTAL: 100

TEAR OFF Mathematics ownloaded from Stanmorephysies.com NAME & SURNAME: Downloaded from hlayiso.com June 2022 Common Test DIAGRAM SHEET 1 Copyright Reserved QUESTION 3.2 QUESTION 4 y A(2; 11) D(6 ; 9) E oO x F B(-10;-5) [e Please Turn Over

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Downloaded from hlayiso.com Downloaded from Stanmorephysics.com |} KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA GRADE 11 | MARKS: — 100 These marking guideline consist of 8 pages. Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathefaownloaded from Stanmorephysics.com Common Test June 2022 GRADE 11 Marking Guideline QUESTION 1 11d T, =an+b T, =6n+b T, =6(1)+b=11 ¥ b=11-6=5 1 T, =6n+5 vv 6n4+5 (2) 1.1.2 T, =6n+5 Tyy = 6(20)+5 Y substitution =125 vY answer (2) 1.1.3 299 =6n+5 Y substitution 6n = 294 n=49 Y answer There are 49 terms in the pattern. (2) 1.1.4 Quadratic number pattern: tee PR PR vo a i 17 23 29 Term 2 =100—(17+ 23+29) ¥ 100-(17+23429) =31 Y answer (2) 1.2.1 24; 44 v 24 v 44 Q) 1.2.2 — 16 -12 -8 —4 \, 4 \ : 4OoO™N » a N\ SN 4 4 4 Second difference = 4 a= 2nd difference 2 =2 ¥ value of a first Ist difference = 3a+b=4 3(2)+b=4 b=-2 v value of b Term 1=a+b+c=-16 2-2+c=-16 c=-16 Y value of c Therefore: 7, =2n>—2n-16 v answer (4) Copyright Reserved Please turn over

Mathefagwnloaded from Stanmorephysics.com Downloaded from hlayiso.com Common Test June 2022 GRADE 11 Marking Guideline 1.23 T, =2n* —2n-16 = An? -n-8) Y¥ An? -n-8) Because n is a natural number, n? —2—8 will also be a natural number. And: 2(a natural number) will always be even. Y reasoning: multiples of 2 (2) OR ; OR The given terms in the number pattern are even. The 1“ | / given terms are even differences between terms are also even. An even number plus an even number will always give an even number. Therefore all terms are even numbers. ¥ reasoning Q) 124 P,=T, -128=2n* —2n—-144 Y expression for P, Let P, <0 .2n? -2n-144 <0 Y 2n? -2n-144<0 nv —n-72<0 (n—-9Xn+8)<0 \-8 yl 9 n o —8<n<9 Y¥ solution for inequality But neN, son>d: «. O<n<9 .. there are 8 negative terms. v answer (4) Answer only: | mark only [20] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathefaownloaded from Stanmorephysics.com Common Test June 2022 GRADE 11 Marking Guideline QUESTION 2 2A yee 2a -3 = ~2E0 Y substitution 3 => v x-value of TP 2 ; - 2 3 3 3 =J=4 =] +3)—]+10 a)G) “G) =P! 4 Turning point: 3, vi g point: 3° °°4 @) 2.2 ys vi OR y €(—00 ; 12,25] Y answer 4 ) 23 -x° +3x+410=0 ¥ f(x)=0 x —3x-10=0 y (x+2\(x-5)=0 factors y=? or x=5 v x-values (3) 24 y (1,5 ; 12,25) 10 ¥ shape ¥ x-intercepts oy ¥ y-intercept = 5\ x ¥ turning point (4) 1 1 2.5 10<k<127 OR ke 10127 Y VY answer (2) 2 ¥ —(value of a=-1) 2.6 es)=-{x-2) vot 7 7 2 4 v 5 (value of p=-—) v +90 (value of ¢=95) (3) [16] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathefagwnloaded from Stanmorephysics.com Common Test June 2022 GRADE 11 Marking Guideline QUESTION 3 3.1.1 y=4 Y answer dQ) 3.1.2 y ¥ shape h ¥ asymptote 5 i) ¥ y-intercept < (3) 3.1.3 reflection in x-axis ¥ reflection in x-axis reflection in y-axis ¥ reflection in y-axis (2) 3.2.1 p=-l v¥ p=-l q=? vq=2 2) 3.2.2 xeR,but x #1 ¥xeER,but x41 qd) OR OR xe(-0 ; 1) or (1; «) ¥ xe(-o ; 1) or (1; «) qd) 3.2.3 Coordinates of D: (0 - 8) v D0 ; -8) Equation of fi y= 1-2 x-1 Substitute (0 7 8): -8= ra -2 Y substitution a=8-2 a=6 Y answer (3) 3.2.4 Cis the x-intercept of f Substitute y=0: 0= a Y substitution x-1 y= 8% x-l Ix-2=6 ; x=4 Y x-coordinate of C Va -8-0 ma 222i 8 ¥ substitution x,-x, 0-4 =2 ¥ answer (4) 3.2.5 yHxte Substitute (1;-2): —2=1+e c=-3 pex-3 vv x-3 (2) Copyright Reserved Please turn over

Mathefaownloaded from Stanmorephysics.com Downloaded from hlayiso.com Common Test June 2022 GRADE 11 Marking Guideline 3.2.6 O<x<l or x24 vv 0sx<l VY x>4 (3) OR OR x €[0; 1) or [4;00) ¥¥xe[0;1) v [450) (3) [21] QUESTION 4 41 C(6 ;-5) v6 v -5 (2) 4.2 E Xi +X) V+. 2° 2 _ “10+2 =5+1 Y substitution 2 2 = (-4 ; 3) ¥ answer (2) Answer only: full marks 43 Myy = v2 71 x, ~% aoc Y substitution in gradient 6-2 formula 1 2 Mop = May =~ Y Mey Substitute m,, an y=mx+e 3=(— 1 (- 4) m3 v substitution in formula for 2 equation of line 3=2+¢ c=1 1 ya-pxtl ¥ answer (4) 44 AtF: x=6 y= 61 Y substitution of x=6 in 2 equation for EF. ya-2 Y yp=2 F(6 ; —2) (2) [10] Copyright Reserved Please turn over

Mathefagwnloaded from Stanmorephysics.com Downloaded from hlayiso.com Common Test June 2022 GRADE 11 Marking Guideline QUESTION 5 0-(-2) 5.1 Mon = ~6-(3) Y substitution 2 “3 v answer (2) 5.2 Tyg = 22! x2 —% = 4-0 Y substitution 1-(-6) 4 , =a answer * 4 2 4 tan KJO = my =5 v tankJO=— KJO = 29,74° v answer (4) 53 tan GJO = mg, --3 GJO = 180° — 33,69” - 146,31 . Y size of GIO a= GJO-KIO Y subtracting = 146,31 —29,74 =116,57° v answer (3) 5.4 m., = 4- (- 2) ¥ substitution in gradient formula for KL 1-(-3) KL _3 _) Y gradient of KL Mg, X My = 2 x 3 -l ee 302 Y multiplying gradients and getting a Therefore KL 1 JL. product of — 1 (3) 5.5 JL= (x, -m) +0. -H) = \(2-0F +E 3-COF = +13 units KL=,\(-2-4) +(-3-1) =V52 OR 2v13 units 1 . Area of AKJL= 5 x base x height ¥ substitution in distance formula for JL Y length of JL Y length of KL =] ox KL 2 = 1 x VIB x 2713 ¥ substitution in formula for area of 2 triangle =13 square units v answer (5) {17] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathefagwnioaded from Stanmorephysics.com GRADE 11 Marking Guideline QUESTION 6 Common Test June 2022 6.1 | LHS=(1-sin? 4) + tan? 4) = cos’? 6+sin’ 6 =1 = RHS Y 1-sin* @=cos’@ a) Y tan? 9 = 5 o cos’ 0 Y cos? @+sin’ 9 (3) 62 sin(2x +30°)=-0,4 reference 7 = 23,58° 2x+30° =180° + 23,58° +k.360°,k EZ 2x =173,58° +k.360° x = 86,79" +k.180° or 2x +30° = 360° — 23,58° + k.360°, ke Z 2x = 306,42° + k.360° x =153,21° +k.180° v reference 7 = 23,58° Y 2x+30° =180° + 23,58° +4.360° VY x =86,79° +k.180° ¥ 2x+30° =360° —23,58° +k.360° Y x=153,21 +k.180° In the interval [-90°;90°|: x = 86,79° or — 26,79° Y 86,79" ¥ —26,79° (7) 6.3 3sin x cos x—4cos” x =0 cos x(3sin.x—4cos x)= 0 Y factorisation cosx=0 or 3sinx—4cosx=0 Y both equations x=90°+4.180°>,keEZ or 3sinx=4cosx Y x¥=90° +k.180° sinx 4 cosx 3 tanx= 4 Y tanx= 4 3a 3 oe 4 Y x=53,13° x =5313°+k.180, kez Y +k.180°sk EZ \ (6) ! [16] Copyright Reserved TOTAL: 100