states value of T = 56 t ●3 ●3 7(b) ans: 64 (3 marks) applies given information ●1 ●1 simplifies ●2 ●2 solves for ●3 x ●3 8(a) ans: (4 marks) finds value of k ●1 ●1 equates coefficients ●2 ●2 finds tan ●3 ●3 α finds = 31o ●4 ●4 α α 8(b) ans: 101o, 321o (4 marks) substitutes for original equation = 2 ●1 ●1 rearranges ●2 ●2 1st value of 31 = 70o; = 101o x x x ●3 ●3 – finds 2nd value of 31 = 290o; = 321o x x x ●4 ●4 – 9 ans: (5 integrates 1st term ●1 ●1 integrates 2nd term ●2 ●2 substitutes limits ●3 ●3 substitutes exact values ●4 ●4 evaluates to answer ●5 ●5 10 ans: 0∙486 (4 marks) substitutes for and finds t k substitutes given values or takes logs finds t
●3 states value of T ●3 t = 56 7(b) ans: 64 (3 marks) ●1 applies given information ●1 ●2 simplifies ●2 ●3 solves for x ●3 8(a) ans: (4 marks) ●1 finds value of k ●1 ●2 equates coefficients ●2 ●3 finds tan α ●3 ●4 finds α ●4 α = 31o 8(b) ans: 101o, 321o (4 marks) ●1 substitutes for original equation ●1 = 2 ●2 rearranges ●2 ●3 1st value of x ●3 x – 31 = 70o; x = 101o ●4 finds 2nd value of x ●4 x – 31 = 290o; x = 321o 9 ans: (5 ●1 integrates 1st term ●1 ●2 integrates 2nd term ●2 ●3 substitutes limits ●3 ●4 substitutes exact values ●4 ●5 evaluates to answer ●5 10 ans: 0∙486 (4 marks) substitutes for t and finds k substitutes given values or takes logs finds t
Unit-3-Extended-Test-2012.docx