34 3 M1 2 H 4y = 2x2 M1 for rearranging and substituting given values of x y = gradient = = = = 3 A1 A1 cao 2 35 M1 2 M1 for appropriate substitution to enable comparison H = with Q35 = As above. A1 A1 for showing that the two functions give the same answer 2 36 a As working M1 2 M1 for clarity of proof H f(x) = + mx c A1 A1 for accuracy with manipulation of function The gradient = = = m f(x) = at = 2 x The gradient = = (2x + 0 h) h→ As working M1 M1 for clear reasoning = (2x) = x b A1 A1 for accuracy with manipulation of function to show a At = 2, gradient = 2. x gradient of 2 From the graph, points on the tangent are (1, 0) and (2, 2). The gradient = = 2
34 4y = 2x2 y = gradient = = = = 3 3 M1 A1 2 M1 for rearranging and substituting given values of x A1 cao H 2 35 = = As above. M1 A1 2 M1 for appropriate substitution to enable comparison with Q35 A1 for showing that the two functions give the same answer H 2 36 a b f(x) = mx + c The gradient = = = m f(x) = at x = 2 The gradient = = (2x + h) h→ 0 = (2x) = x At x = 2, gradient = 2. From the graph, points on the tangent are (1, 0) and (2, 2). The gradient = = 2 As working As working M1 A1 M1 A1 2 M1 for clarity of proof A1 for accuracy with manipulation of function M1 for clear reasoning A1 for accuracy with manipulation of function to show a gradient of 2 H