Mathematics and Mathematical Literacy - Thutong

Mathematics(NSC) /Grade 11/P1 85 ExemplarMEMORANDUMThe roots are x = 3 or x = -1 4.2.23+( −1)Axis of symmetry: x =2= 1 24.2.3 x = 1 ∴ f ( 1) = −1+ 2 + 3 = 4 Thus TP (1 ; 4) (must write in co-ordinate form) 24.3.1 a4 = ∴ a = 4 114.3.2 Everywhere except at x = 0 ORx ∈ R but x ≠ 024.3.3 4= x − 1x2∴ x − x − 4 = 0 1 ± 1 + 16∴ x =2∴ x = 2,56 because x > 0 Thus P (2,56 ; 1,56) 54.3.4 2f (2,56) = −(2,56)+ 2(2,56) + 3 = 1, 56 Thus P also lies on the parabola 24.4 h ( x)> g(x)1 − 17x–coordinate of T = = −1,56 2∴ x < −1,56 or 0 < x < 2, 56 45.1 A(-1,5; -3) and B (0,5; -5) (substitute into formula) 25.2−5− ( −3)Average gradient = = −10,5 − ( −1,5)25.3 Axis of symmetry is the average of the rootsthus x = -0,25 Thus f (x) is increasing on x > 0,25 35.4 For average gradient to be 0, C must havesame y-coordinate as A, Thus C(1; -3) 25.5 By inspection: D(1,5; 0) 2y6.1 4500Population of fish in damFish population40003500300025002000150010005007.17.20,97 = 0,48Thus during the 7 th day. 3 x + 5−= 3 x + 2 ( x − 2)( x + 2)( − 2) − ( x + 5) = 3( x2 − 4)3 x 3x− 6 − x − 5 = 3x3x2 − 2x− 1 = 0( 3x+ 1)( x − 1) = 02 −121x = − or x = 1 3x + y + 7 = 0 thus x = -y - 7 2 2( −y − 7) + y = 25 y22+ 14y+ 49 + yy + 7 y + 12 = 0 2= 25( y + 3)( y + 4) = 0 ∴ y = −3 or y = −4If y = −3 then x = −4If y = −4 then x = −378.1 a) 1 ≤ x ≤ 3 b) y ≤ −2 x + 10 c) y ≥ 0, 5x 68.2 A (1; 8) then P = 9 B (3; 4) then P = 7 Thus P = 9 is a maximum at point P 38.3 If gradient of T < gradient of AB then B isthe point that would give a maximum. Thus k < -2 36-0.5 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5-500Days after disease identified36.2 4500 = A (1 − 0,1)1 ∴ A = 5000 36.3 x5000 (1 − 0,1) = 2500 ∴0 ,9= 0,5 By trial and error: 0,96 = 0,53 5Copyright reserved