Mathematics and Mathematical Literacy - Thutong

Mathematics(NSC)/Grade 12/P1 90 ExemplarMEMORANDUM3.24.1.14.1.20.1i = 12⎛ ⎞⎜(1 + i)n −120000 = 300 ⎟⎝ i ⎠n 14∴( 1+i ) = 9log14n ∴ ( i + i)=9Thus n = 53,2 monthsThus need 54 months 50f (0) = a × b = 2∴ a = 21f (1) = 2b= 6∴ b = 3x∴ f ( x)= 2. 3 3g(0)= a × b∴ a = 2g(2)= 2b∴b22= 8,5∴ b = 2,920= 2= 17x∴ g ( x)= 2.(2,92)34.1.3 f(2,3) = 25,03f(6) = 1458g(2,3) = 23,52g(6) = 1239,72 24.1.4 f(x) is the closer approximation as thevalues of f(2,3) and f(6) are closer to thecollected data than those of g(2,3) and g(6). 24.2.1 f ( x)= ( x + 3)( x = 1)∴ x = −3 or x = −1 34.2.2 E = 3 – 6 = – 3 E (0 ; -3) 14.2.3 g (x) = ( x 2 + 4x+ 3)(2x−1)32∴ g ( x)= 2x+ 7x+ 2x− 3 34.2.4 22( x + 4x+ 3) = ( x + 4x+ 3)(2x−1)2∴ (2x−1−1)(x + 4x+ 3) = 0 ∴ x = 1 ∴ K (1; 8) 44.2.5 2g′ ( x)= 6x+ 14x+ 2 = 0 − 7 ± 37∴ x = 6∴ x = −2,18 or x = −0, 15 Axis of symmetry of f(x) is x = -2 Thus F does not lie on the axis of symmetryof f (x). 54.2.6 f ′( x)= g′( x)2∴2x + 4 = 6x+ 14x+ 2 ∴6x 2 + 12x− 2 = 0 ∴3x 2 + 6x−1= 06− 6 ± 36 − 4 × 3×( −1)∴ x =6∴ x = 0.15 or x = −2.15 5.1 Line 5 Can’t ÷ by ( a – b) because a – b = 0 35.2.1 2f ( x)= ( x − 2)(2x+ 5x+ 3 ) By factor theorem or inspection∴ f ( x)= ( x − 2)(2x+ 3)( x + 1) ∴ x = 2 or x = -1,5 or x = -1 55.2.2 x − 2 = 2 x = 4Thus x − 2 = −1,5thus x = 0,5 x − 2 = −1x = 136.1f ( x + h)− f ( x)f ′(x)= limh→0h1 1−∴ f ′(x)= limx + h − 2 x − 2h→0h( x − 2) − ( x + h − 2)∴ f ′(x)= limh →0 ( x − 2)( x + h − 2)h−1∴ f ′(x)= limh→0 ( x − 2)( x + h −2)−1∴ f ′(x)= 2( x − 2)56.2 2 −0.5−1y = x + 2x− 3xdy−1.5−2∴ = 2x− x + 3xdx56.3.1 f ′( 0) = 2 thus gradient of tangent is 2 16.3.2 Increasing implies that f ′( x)> 0 2∴ −x− x + 2 > 0∴(x = 2)( x −1)< 0∴ −2 < x < 1 46.4.1 3 = 2m + c ∴ c = 3 − 2m 26.4.2 y = mx + ( 3 − 2m)Thus x – intercept is when y = 0 3 − 2m∴ x = m26.4.3 Area = 0,5 (x-intercept)(y-intercept) =1 ⎛ 3 − 2m⎞(3 − 2m)2⎜ ⎟ ⎝ m ⎠26.4.4 Multiply out above expression to get9 −1Area = m − 6 + 2m2dA 9Thus = − + 2 = 0 dx22m9m 2 = 43m = ± 2−3But m < 0 thus m = 25Copyright reserved