2024-kcse-maths-p2-ms-teacher_co_ke_1
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DATE 07 Dec 2025
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About This Document
Document Type: This is a Study Notes, designed for Reviewing core curriculum material.
Context: Standard material from the 2024 academic period.
Key Content: Likely covers essential definitions, problem solving, and theoretical concepts necessary for mastery of the subject.
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Recommendation: comprehensive resource for students aiming to deepen their understanding of General Studies.
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SECTION I (50 marks) Answer all the questions in this section in the spaces provided. An arithmetic progression (AP) is given as 600 + 650 + 700 + 750 +. Determine: (a) the 30th term of the AP; Tn = a + ( n - 1) d T30 = 600 + 50 ( 30 - 1) a = 600, n = 30 = 60 + ( 29× 50 ) d = 700 - 650 = 50 = 2050 (b) the sum of the first 30 terms of the AP. 14/11/2024 (2 marks) (2 marks) n n Sn = 2a + ( n − 1) d a + l ; l = last term 30th term 2 2 30 30 ( 600 2 ) + 50 ( 30 − 1) S 30 = 600 + 2050 Alternatively, S 30 = 2 2 = 15 (1200 + 1450 ) = 15 2650 Sn = = 39750 2. = 39750 The quadratic equation 5x2 + kx + 20 = 0 has only one root. Determine the possible values of k. (2 marks) For a repeated root the discriminant, b 2 − 4ac = 0 hence b 2 = 4ac. Without using mathematical tables or a calculator, evaluate log125 + log 64. log 6 5 + log 3 2 1 (3 marks) 1 125 = 53 , 64 = 26 , 6 5 = 5 6 , 3 2 = 2 3 log ( 5 2 ) 3 6 log 5 2 1 6 1 3 log ( 5 2 ) 1 = 2 3 = 1 2 6 log ( 51 2 ) 3log ( 51 22 ) 1 1 log ( 51 22 ) 6 1 3 6 = 1 = 3 = 18 1 6 Alternatively log ( 53 26 ) log 5 2 1 6 = log 5 + log 2 6 1 6 1 3 log 5 + log 2 Make x the subject of the formula y = 3 ( log 5 + 2 log 2 ) 6 a. bx (3 marks) yb x = a b x = yb x = a b x = a y Taking logs on both sides : a x log b = log b hence x= 1 3log 5 + 6 log 2 3 6 = 1 = = 1 = 3 = 18 1 1 1 1 6 log 5 + 3 log 2 ( log 5 + 2 log 2 ) 6 log ( ba ) Alternatively 4.
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SECTION I (50 marks) Answer all the questions in this section in the spaces provided. 1. An arithmetic progression (AP) is given as 600 + 650 + 700 + 750 + ... Determine: (a) the 30th term of the AP; Tn = a + ( n - 1) d T30 = 600 + 50 ( 30 - 1) a = 600, n = 30 = 60 + ( 29× 50 ) d = 700 - 650 = 50 = 2050 (b) the sum of the first 30 terms of the AP. 14/11/2024 (2 marks) (2 marks) n n Sn = 2a + ( n − 1) d a + l ; l = last term 30th term 2 2 30 30 ( 600 2 ) + 50 ( 30 − 1) S 30 = 6...
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