Mathematics Grade 11 Paper 1 Gauteng June 2016 Memo

Downloaded from hlayiso.com GAUTENG DEPARTMENT OF EDUCATION PROVINCIAL EXAMINATION JUNE 2016 GRADE 11 MATHEMATICS P1 MEMORANDUM 11 pages

Downloaded from hlayiso.com MATHEMATICS Grade 11 MEMORANDUM (Paper 1) GAUTENG DEPARTMENT OF EDUCATION− PROVINCIAL EXAMINATION MATHEMATICS (Paper 1) MEMORANDUM QUESTION 1 1.1  4  OR x   3 (2) 1.2 1.2.1 = x+4  Squaring ( both sides  Standard form 2–x = + 8x +16 2  Factors x + 8x +16 + x – 2 = 0  Answers x2 + 9x +14 = 0  x= –2 only (x + 2)(x + 7) =0 x = –2 OR x = –7 NA (5) 1.2.2 2x( x - 3) = 1  Standard form 2x2 - 6x – 1 =0  Substitution a = 2; b = -6; c = -1  Answer  answer x = x= x= x = 3,2 OR x = -0,2 (4) 2

Downloaded from hlayiso.com MATHEMATICS Grade 11 MEMORANDUM (Paper 1) 1.2.3 >0  x 1  x3  Critical values in >0 context of an inequality >0  correct notation Only focus on -3 < < -1 or > 1 numerator OR (-3 ; -1)  (1 ; ) (4) 1.3  Divide by 2 73  2 16  Finding square root 2 ( )  x = 2,89 2 =  x = -1,39 Use of quadratic x = formula max 2/5 for the two correct answers. x= OR x= x = 2,89 OR x = -1,39 (5) [20] 3

Downloaded from hlayiso.com MATHEMATICS Grade 11 MEMORANDUM (Paper 1) QUESTION 2 2.1  Expansion  33 3  4.9 =  3x 1  3  Answer = = If k-method is used exactly the same mark = allocation = 16 OR (4) 2.2 2.2.1   5  10 5  2    5  10 5  2  5 2 5 52 5  simplification  5  10 5  20   5 5 2 5  52 5 5  2 55 25  10 5  52 5  answer   552 5  5  2 55 5 OR  + Rationalizing the denominator = +   Rationalizing the denominator =5 - 2  Simplification  Answer =5 - 2 =5 (4) 4

Downloaded from hlayiso.com MATHEMATICS Grade 11 MEMORANDUM (Paper 1) 2.2.2 (  7 2008  7 3  71 =(  (7 =( +  Answer =(7 - =(7 =343 OR 2009 2.2.2 (  7 2 cont.  (7–1)  (7 =( + )2  Answer = + )2 = (7 = (49)(7) = 343 (4) 2.3 from: 2 . 4y = 1 x  2 2y 2x . 22 y = 20  x + 2y = 0 x + 2y = 0 OR x = –2y................. x= –2y subst  into (4y)x =  substitution (4y)–2x =  same bases on either 2 4 y  2 4 side 2 –4y2 = –4  y = 1 y = 1  x = 2 subst y =  1 into  y= 1 x = –2 y = –1 x =2 OR (6) 5

Downloaded from hlayiso.com MATHEMATICS Grade 11 MEMORANDUM (Paper 1) 2 x .4 y  1  22 y 2 x .2 2 y  2 0  x + 2y = 0 x  2 y  0.......... .1 OR   4y  x 1 16 x= –2y  22xy  24 2  2 4 2 xy  2 xy  4 2 xy  4 2  y = 1 y   .......... ...... 2 x  x = 2  2 x  2    0  x x2  4  0 x  2x  2  0 x  2 or x  2 2 y x 2 2  or  2 2 y  1 y 1 [18] 6

Downloaded from hlayiso.com MATHEMATICS Grade 11 MEMORANDUM (Paper 1) QUESTION 3 3.1 3.1.1 Roots are non-real (imaginary) Non-real roots therefore  < 0. OR   < 0 (2) 3.1.2 Roots are real and equal,  Roots are real = 0  Roots equal (2) 3.2  2 = (2k -1) – 4(k)(k-1)  Substitution = 4k2 – 4k +1 -4k2 +4k  Simplification =1  Value of 1 1 is a perfect square, the coefficients  Perfect square are rational, so the roots are rational. (4) [8] QUESTION 4 4.1 4.1.1 -1; 8; 23; .........  –1 8 23 (3) 4.1.2 2 3k – 4 = 71 3k2 – 4 = 71 k2 = 25  k2 = 25 or k = 5 k  5k  5 k = 5  k=5 No marks for k =  5 (3) 4.2 4.2.1 Quadratic number pattern  Quadratic  second difference is 0 ; 5 ; 12 ; 21; .... constant or illustration 5 7 9 2 2 If the learner only Quadratic number pattern show the pattern without justification OR/OF only 1 mark Quadratic number pattern The first difference is not constant but the second difference is constant. (2) 7

Downloaded from hlayiso.com MATHEMATICS Grade 11 MEMORANDUM (Paper 1) 4.2.2 2a = 2 a = 1 a = 1 3a + b = 5 3(1) + b = 5 b = 2 b = 2 T1 = a + b + c 0 = 1 + 2 +c c = –3 c = –3 Tn = an2 + bn + c Tn = n2 + 2n -3 Tn = n2 + 2n -3 (4) 4.3 4.3.1 Row 4  Row 4 = 22 7  6  5  4  22 2 2 2 2 Row 20 Row 20 = 86 23  22  21  20  86 2 2 2 2 (2) 4.3.2 (n  3)  (n  2)  (n  1)  n  4n  6  a  n  3 2 2 2 2  b  n  2; c  n  1; d  n  Tn  4n  6 If only the general term was given 1/3 (3) [17] QUESTION 5 5.1 x = –2 and y = 1  x = –2  y=1 Both has to be in equation form. If not 0/2 If p  2 and q  1 0/2 (2) 5.2 Sub B = (0; -2) in y = +1  Substitution of (0 ; –2) and q = 1 -2 = +1 -2 = +1 –3 = then k = -6  y= +1  k value  Answer (3) 8

Downloaded from hlayiso.com MATHEMATICS Grade 11 MEMORANDUM (Paper 1) 0 = +1  y=0 5.3  x26 –1 =  x=4 (x +2) = 6  Writing Point D in x = 4 coordinate form. D(4 ; 0) (4) 5.4 C(–2 ; 0) and B(4 ; 0) CA from 5.3 y = a(x+ 2)(x – 4)  x2 –2 = a(0+2)(0 – 4)   x  4 –2 = a (–8)  Sub. B (0; –2) =a  a = y = (x + 2)(x – 4)  answer in any form y = -2x – 8) = – x–2 (5) 5.5  Shape  Coordinates of (0 ; 4) y  Graph not crossing the x- axis (0; 4) x (3) 5.6 y=  y = (2) 5.7 y is real ,  y is real, (y R) both condition (1) [20] 9

Downloaded from hlayiso.com MATHEMATICS Grade 11 MEMORANDUM (Paper 1) QUESTION 6 6.1 [0 ; 4]  0 OR 4 0 x4 (2) 6.2 h( x)  ( x2  4x  4  4)  a = –1.   x  2   4 2  p = –2.  q = –4. (3) 6.3 CA from 6.1 y y (2 ;4)  (2;4) h(x)  Shape (neg graph)  Turning point  y-intercept. 2 4 x  positive y-values 0 only. 0 4 x (4) 6.4 h(x) = - x + 4x2  Sub x with x  5 = –( x – 5 )2 + 4( x – 5)  Simplification = – (x2 – 10x +25) + 4x -20  Answer = – x2 +10x –25 + 4x -20 h(x – 5) = – x2 + 14x -45 OR OR y = – (x – 2)2 + 4  (x – 2 – 5)2 = – (x – 2 – 5)2 + 4 = – (x – 7)2 + 4 = – (x2 – 14x + 49) + 4  Simplification = – x2 + 14x – 49 + 4 = – x2 + 14x – 45  Answer (3) 10

Downloaded from hlayiso.com MATHEMATICS Grade 11 MEMORANDUM (Paper 1) 6.5 k(x) = x2 – 4x  x2  –4x (2) 6.6 9 9 p(–3) = 2  p(–3) = 2 1 1 p(–1) = 2  p(–1) = 2 Average gradient  Answer = = Average gradient of p = –2 (3) [17] 11