LImp-Maths-Gr11-June-2022-P1-and-Memo_hlayiso.com_.pdf

Downloaded from hlayiso.com LI.M.POPu PROVl�CIAL GO\IERNMEHT REPUBUC OF SOUTH "-�CA DEPARTMENT OF EDUCATION [ VHEMBE EAST DISTRICT ] Downloaded from testpapers.co.za I GRADE 11 I This question paper consists of 6 pages

Downloaded from hlayiso.com TERM 2: 2022 INSRUCTIONS AND INFORMATION READ THE FOLLOWING INSTRUCTIONS CAREFULLY BEFORE ANSWERING THE QUESTIONS. 1. This question paper consists of 5 questions. Answer ALL the questions. 2. Clearly show ALL calculations, diagrams, graphs, et cetera that you have used in determining the answers. 3. An approved scientific calculator (non-programming and non-graphical) may be used, unless stated otherwise. 4. If necessary, answers should be rounded off to TWO decimal places, unle�ated otherwise. 5. Diagrams are NOT necessary drawn to scale. ;.f'�G 6. Number the answers correctly according to the numbering system a,,lin this question paper. ♦ 7. It is in your own interest to write legibly and to present neatly. Copyright reserved 2 Please turn over

Downloaded from hlayiso.com TERM 2: 2022 QUESTION 1 1.1. Solve for x in each of the following: 1.1.1. x 2 + X - 12 = 0 (3) 1.1.2. ✓zx+l=x-1 (5) 1.1.3. 2x..Jx = 227 (4) 1.1.4. x 2 - 2x - 8 < 0 (3) 1.2. Given: f(x) = 5x 2 + 6x - 7 1.2.1. Solve for x if f(x) = 0 (correct to TWO decimal places). c,O (4) � 1.2.2. Hence, or otherwise, calculate the value of d for whi�G'3 � 6x - d = 0 has equal roots (3) 1.3. Solve for x and y simultaneously: x- 2y = -3 and xy = 20 (6) [28) QUESTION2 2.1. The solution to a quad �uation is x = - where PEQ. �� 3+� 4 Determine the v �6f p such that: 2.1.1. � 0 �ots of the equation are equal (2) 2.1.2. *�he roots of the equation are non-real (2) 2.2. Gi0.9 ✓s - x = x + 1 2.2.1. Without solving the equation, show that the solution to the above equation lies in the interval -1 � x � 5. (3) 2.2.2. Solve the equation. (5) 2.2.3. Without any further calculations, solve the equation -✓s - x = x + 1. (1) [13) Copyright reserved 3 Please turn over

Downloaded from hlayiso.com TERM 2: 2022 QUESTION 3 3.1. Consider the following number pattern: 9; 14; ..... . } 3.1.1. Write down the next two terms of the pattern. (2) 3.1.2. Determine the expression for the n th term of the'pattern. (2) 3.1.3. Determine if 1099 is a term of the number pa em. (3) 3.2. Consider the following quadratic number pattern: 6; 10; 18; ..... 3.2.1. Write down the following two terms of the pattern. (2) 3.2.2. Determine the equation of the general term in the form: Tn=an2 +bn+c (4) 3.2.3. Calculate the value of T12 (2) 3.2.4. What term of the pattern will have a value of 766? (4) 3.3. A certain number pattern has the following properties: Determine the value of k. (5) [24] Copyright reserved 4 Please turn over

Downloaded from hlayiso.com TERM 2: 2022 QUESTION 4 The diagram represents the functions f(x) = ax 2 +bx+c and g(x) = mx + k 4.1. Calculate the values of a, b and c. (5) 4.2. Find the equation of g(x). (5) 4.3. Calculate the co-ordinates of A, a point of · (6) [16) QUESTION 5 3 Given: f(x) = -+2 + 1 and g(x) = 2-x - 4 x 5.1. Determine f(-3) (2) 5.2. Determine x if g(x) = 4 (3) 5.3. Write down the asymptotes off (x) (2) Copyright reserved 5 Please turn over

Downloaded from hlayiso.com TERM 2: 2022 5.4. Write the range of g (1) 5.5. Determine the coordinates of the x and y- intercepts off (4) 5.6. Sketch the graphs off and g on the same system of axes. Clearly show ALL the intercepts with the axes and any asymptotes. (4) 5.7. If it is given that f(-1) = g(-1), determine the values of x for which g(x) � f(x) (3) [19] TOTAL: 100 Copyright reserved 6 Please turn over

Downloaded from hlayiso.com LI.M.POPO PROVl�CIAL GO\IERNMEHT REPUBUC OF SOUTH "-�CA DEPARTMENT OF EDUCATION [ VHEMBE EAST DISTRICT ] I GRADE 11 I MARKS: 100 TIME: 2 hours

Downloaded from hlayiso.com TERM 2: 2022 NOTE: • If a candidate answered a question TWICE, mark only the FIRST attempt. • If a candidate crossed out an answer and did not redo it, mark the crossed-out answer. • Consistent accuracy applies to ALL aspects of the marking memorandum. • Assuming values/answers in order to solve a problem is unacceptable. QUESTION 1 1.1.1 x 2 + X -12 = 0 (x +4)(x - 3) = 0 ✓ factors x = -4 or x = 3 ✓ answer ✓ answer 3 1.1.2 ✓2x + 1 = X -1 ✓ squaring both sides 2x + 1 = (x - 1)2 ✓ standard form 2x + 1 = x 2 - 2x + 1 ✓ factors x 2 - 4x = 0 ✓ answer x(x - 4) = 0 ✓ x = 4 (correct selection) x=0 or x = 4 n/a (5) 2xrx = 227 3 1.1.3 ✓ 2x2 3 2xz = 2 27 ✓ x2 = 27 X2 = 27 ✓ Raise both sides to � 2 ✓ Answer (4) X = 273 x=9 1.1.4 x 2 - 2x - 8::::; 0 ✓ (x - 4)(x + 2) < 0 (x - 4)( X + 2) < 0 -2 < x < 4 OR/OF x E (-2;4) ✓ Critical values ✓ Inequalities (3) 1.2.1 Sx 2 + 6x - 7 = 0 ✓ Formula ✓ Substitution ✓ -b ± bL4ac x= 2a X= -6 ± ✓6 -4(5)(-7) 2 2(5) ✓ Answers = 0, 73 or - 1, 93 ✓ (4) 2

Downloaded from hlayiso.com TERM 2: 2022 1.2.2 Sx 2 + 6x - d = 0 x= ✓ Substitution -6 ± ✓6 2 -4(5)(-d) x=----'----- ✓ 36 + 20d = 0 2(5) 36 + 20d = 0 ✓ Answer (3) d = - -9 5 OR for equal roots: t1 = 0 t1 = b 2 - 4ac ✓ Substitution = (6)2 - 4(5)(- d) ✓ 36 + 20d = 0 36 + 20d = 0 d=- � ✓ answer (3) OR Sx 2 + 6x - d = 0 x2 + 6x = � 5 5 ✓ completing the square 5d+9 25 ✓ 5d + 9 = O ✓ Answer For equal roots Sd+9 =Q (3) 25 -9 3

Downloaded from hlayiso.com TERM 2: 2022 1.3 X = 2y- 3 ... . (1) ✓ Making x the subject xy = 20 ... . (2) Substitute (1) into (2): ✓ Substitution (2y - 3)y = 20 ✓ Standard form 2y 2 - 3y - 20 = 0 ✓ Factors (2y + 5)(y - 4) = 0 ✓ y - values 5 y= --2 or y=4 x = -8 or x = 5 ✓ x - values (6) OR X + 3 = 2y ✓ Making y the subject ..... (1) ✓ Subst xy = 20 ✓ Standard form e;3 ✓ Factors ✓ x - values X ) = 20 x 2 + 3x = 40 ✓ y - values x 2 + 3x- 40 = 0 (6) (x + 8)(x - 5) = 0 ✓ Making y the subject x = -8 or x = 5 ✓ Substitution y= 5 or y = 4 ✓ Standard form 2 ✓ Factors OR X - 2y = -3 ...... (1) 20 ✓ x - values y= - ...... (2) ✓ y - values Substitute (2) into (1) (6) 20 X - 2( ) = -3 x 2 - 40 = -3x x 2 + 3x - 40 = 0 (x + 8)(x - 5) = 0 x = -8 or x = 5 y = - -2 or y = 4 OR 4

Downloaded from hlayiso.com TERM 2: 2022 X - 2y = -3 ..... ( 1) 20 x= - ...... (2) y ✓ Making x the subject ✓ Subst Substitute (2) into (1): -20 -2y =-3 y ✓ Standard form 20- 2y 2 = -3y ✓ Factors 0 = 2y 2 - 3y- 20 O= (2y + S)(y- 4) ✓ y values y= - -2 5 or y = 4 ✓ x values (6) X = -8 or x=5 [28] QUESTION 2 2.1.1 4-BP = 0 ✓ 4- BP= 0 - 8P = -4 ✓ Answer P= !2 (2) 2.1.2 4-Bp <0 ✓ 4-Bp <0 1 ✓ Answer p > -2 (2) 2.2.1 ✓s - x = x + 1 ✓ 5-x � 0 5-x � 0 and x+ 1 � 0 ✓ X+1 �0 X :::; 5 and X > -1 ✓ And Hence -1 :::; X :::; 5 (3) 2.2.2 5 - x = x 2 + 2x + 1 ✓ Square both sides x + 3x- 4 = 0 2 ✓ Standard form (x + 4)(x- 1) = 0 ✓ Factors X = -4 or x=l ✓ Answers ✓ Selection of 1 Since -1 :::; x :::; 5, x = l only (5) 2.2.3 X = -4 ✓ Answer (1 ) [13] 5

Downloaded from hlayiso.com TERM 2: 2022 QUESTION 3 3.1.1 19; 24 ✓ 19 ✓ 24 (2) 3.1.2 Tn = Sn-1 ✓ Sn-1 (2) 3.1.3 Tn =1099 ✓ Equating Sn- 1 = 1099 ✓ Simplification Sn= 1100 Sn= 1100 n = 220 ✓ Answer (3) :. T220 = 1099 :. it is in the sequence 3.2.1 30; 46 ✓ 30 ✓ 46 (2) 3.2.2 Tn = an +bn+c 2 ✓ Second difference = 4 2a = 4 ✓ a= 2 :. a= 2 3a + b = 4 :. 3(2) + b = 4 ✓ b = -2 :. b = -2 a+b+c=6 2- 2+c=6 ✓ C =6 c=6 (4) :. T, = 2n2 - 2n + 6 3.2.3 T12 = 2(12)2 - 2(12)+ 6 ✓ Correct substitution in Tn = 270 ✓ Answer (2) 3.2.4 Tn = 766 ✓ Equating 2n2 - 2n+ 6 = 766 ✓ Standard form = 0 2n2 - 2n+ 760 = 0 ✓ Factors n2 - n-380 = 0 (n- 20)(n+ 19) = 0 ✓ n = 20 (4) n = 20 :. it is term number 20 3.3 T1 ; T2 ; T3 ; T4 ✓ First difference k , 14 , 24 ; 7k ✓ Second differences -4 + k = 7k - 24 ✓ Equating 6k = 30 ✓ 6k = 30 K= 5 ✓ Answer (5) [24] 6

Downloaded from hlayiso.com TERM 2: 2022 QUESTION 4 4.1 y = a(x -x1)(x-x2) y = a(x -(-2))(x- 4) y = a(x + 2)(x- 4) Substituting (0; -16), we have: -16 = a(O + 2)(0 - 4) ✓ Substitution -16 = -8a ✓ Value of a 2=a Substituting 2 for a: y = 2(x + 2)(x- 4) ✓ Substitution & y = 2(x2 - 2x - 8) simplification y = 2x2 - 4x- 16 ✓ Value of b :. a = 2, b = -4, c = -16 ✓ Value of c ( 5) 4.2 The points (-2;0)(0;-8) lie on g - Y1 m= Y2 ✓ Substitution Xz - X1 -8 -0 ✓ Gradient m= 0 -(-2) = -4 ✓ Value of c :. y = -4X + C ✓ Equation of g Substituting (0; -8) , we have: -8 = -4(0) + C (4) :. -8 = C :. g(x) = -4x- 8 4.3 At A' 2x2 - 4x- 16 = -4x- 8 ✓ Equating equations 2x2 - 8 = 0 x2 - 4 = 0 (x - 2)(x + 2) = 0 ✓ Factorization x = 2 or x = -2 ✓ Values of x y = 4( 2) - 8 y = -4(-2) - 8 ✓ Substitution y = -8 - 8 or y = 8 - 8 y= -16 or y=O :.y= -16 ✓ y value :.A( 2;-16) ✓ Coordinate of A (6) [15) 7

Downloaded from hlayiso.com TERM 2: 2022 QUESTION 5 5.1 -3 f(-3) = +1 -3 + 2 =4 (2) 5.2 4 = 2-x - 4 ✓ Substitution ✓ Raise to exponent 8 = 2-x ✓ x - value 23 = 2-x (3) :. X = -3 5. 3 X = -2 ✓ x values & ✓ y value =1 (2) 5.4 y � -4 ✓ Answer 1) 5.5 -3 f(x) = -+ 1 x+2 ✓ Substitution -3 0= -+1 x+2 ✓ (1; 0) -1= - -3 Multiplication x+2 -l(x + 2) = -3 ✓ x - value -x- 2 = -3 x=l y= -+ 1 -3 y - value 0+2 1 y= 2 (4) 8

Downloaded from hlayiso.com TERM 2: 2022 5.6\ ✓ Asymptotes I I n ✓ I I I I ✓ Correct I ✓ f - ··-·········· r···�································· -·····-······- I .,,,..,.....- l I 11 rww X -13 -12 11 10 9 8 , 1 5 � 3 5 6 9 10 11 11 13 (4) 5.7 x :s; -3 or -2 :s; x :s; -1 ✓ -2 :s; X :s; -1 ✓ ✓ X :s; -3 (3) [19] 9