Thus equation of normal is (y 2) = or 4y = 2x + 7 A1 – Eliminate between = + ln (2x) and normal equation to give an y x M1 equation in x Solve ln 2x = so = M1 x Subsitute to find a value for M1 y Point is A1 (8) Q (8 marks) 16. (a) Sets B1 Obtains and substitute into M1 P Uses with their substituted M1 y A1 (4) (b) > 0 and > 0 (distance) M1 x y As and are distances they are positive, so * A1 (2) x y Differentiates with negative index correct in ; M1 P A1 Sets and proceeds to = M1 x Substitutes their into to give perimeter = 59.8m. A1 (4) x (10 marks)
Thus equation of normal is (y – 2) = or 4y = 2x + 7 A1 Eliminate y between = x + ln (2x) and normal equation to give an equation in x M1 Solve ln 2x = so x = M1 Subsitute to find a value for y M1 Point Q is A1 (8) (8 marks) 16. (a) Sets B1 Obtains and substitute into P M1 Uses with their y substituted M1 A1 (4) (b) x > 0 and y > 0 (distance) M1 As x and y are distances they are positive, so * A1 (2) Differentiates P with negative index correct in ; M1 A1 Sets and proceeds to x = M1 Substitutes their x into to give perimeter = 59.8m. A1 (4) (10 marks)
01aa_as_paper_1_pure_mathematics_mark_scheme.docx