8 Variables x and y are such that y Axb = , where A and b are constants. When lgy is plotted against lgx, a straight line graph passing through the points (0.61, 0.57) and (5.36, 4.37) is obtained. (a) Find the value of A and of b. [5] Using your values of A and b, find (b) the value of y when x 3 = , [2] (c) the value of x when y 3 = . [2]

8(a) lg lg lg = + y b x A B1 May be implied by subsequent work 4.37 5.36 lg = + b A 0.57 0.61 lg = + b A M1 For at least one correct equation 0.8 = b A1 ( ) lg 0.082 = A k 10 = k A M1 Dep for substitution to obtain lg = A k and hence A 1.21 = A A1 Alternative 1 lg lg lg = + y b x A (B1) May be implied by subsequent work Gradient = 4.37 0.57 5.36 0.61 − − (M1) 0.8 = b (A1) ( ) lg 0.082 = A k 10 = k A (M1) Dep for substitution into a correct equation to obtain lg = A k and hence A 1.21 = A (A1) Alternative 2 4.37 5.36 10 10 = × b A or 0.57 0.61 10 10 = × b A (B1) 3.8 4.75 = b (M1) For eliminating A correctly Must have B1. 0.8 = b (A1) ( ) 4.37 5.36 ( ) 10 − × = theirb A oe (M1) For a correct attempt to find A. Must have B1 A = 1.21 (A1) 8(b) ( ) 0.8 1.21 3 or lg 0.8lg3 0.082 = = + y y B1 FT for substitution into their equation y = awrt 2.9 B1 8(c) 0.8 3 1.21 or lg3 0.8lg 0.082 = = + x x B1 FT for substitution into their equation x = awrt 3.1 B1

8 The curves y x x 1 2 = + - and y x x 2 6 2 2 = + - intersect at the points A and B. (a) Show that the mid-point of the line AB is (2, 9). [5] The line l is the perpendicular bisector of AB. (b) Show that the point C (12, 7) lies on the line l. [3] (c) The point D also lies on l, such that the distance of D from AB is two times the distance of C from AB. Find the coordinates of the two possible positions of D. [4]

8(a) 2 2 2 2 2 6 2 + − = + − x x x x 2 4 0 − = x x ( )4 0 − = x x 0, 4 = = x x M1 For obtaining an equation in one variable M1 Dep for a correct attempt to obtain at least one solution ( ) 0, 1 − A1 nfww ( ) 4, 19 A1 nfww Mid-point ( ) 2, 9 with sufficient detail B1 AG 8(b) Either Gradient of perpendicular 1 5 = − M1 ( ) 1 9 2 5 − = − − y x M1 Dep on previous M mark for perpendicular bisector using their mid- point and their perpendicular gradient ( ) 1 7 9 12 2 5 − = − − oe A1 For checking by substitution, must see evidence. Or Alternative 1 Gradient of perpendicular 1 5 = − (M1) ( ) 1 7 12 5 − = − − y x (M1) Dep on previous M mark for perpendicular bisector using ( ) 12, 7 and their perpendicular gradient ( ) 1 9 7 2 12 5 − = − − oe (A1) For checking by substitution, must see evidence Or Alternative 2 Gradient of perpendicular 1 5 = − (M1) Gradient of line joining their ( ) 2, 9 to ( ) 12, 7 = 1 5 = − (M1) ( ) 2, 9 is a common point and gradients of perpendicular bisector and l are the same so C lies on l. (A1) 8(c) ( ) 22,5 2 B1 for 22 B1 for 5 ( ) 18,13 − 2 B1 for 18 − B1 for 13

9 When e y 2 is plotted against x2, a straight line graph passing through the points (4, 7.96) and (2, 3.76) is obtained. (a) Find y in terms of x. [5] (b) Find y when x 1 = . [2] (c) Using your equation from part (a), find the positive values of x for which the straight line exists. [3]

9(a) 2 2 e = + y mx c B1 May be implied by later work Either 7.96 4 = + m c 3.76 2 = + m c M1 2.1 = m oe A1 0.44 =− c oe A1 ( ) 2 1 ln 2.1 0.44 2 = − y x oe A1 Do not isw Or gradient = 2.1 oe (B1) Use of either 7.96 4 = + m c or 3.76 2 = + m c (M1) For use with their m 0.44 =− c oe (A1) ( ) 2 1 ln 2.1 0.44 2 = − y x oe (A1) Must be bracketed correctly 9(b) ( ) 2 1 ln 2.1 0.44 2 = − y their x their oe M1 Must use the form ( ) 2 ln = ± y k px q 1 ≠ p and 0 ≠ q or 2 2 e = + y mx c 0.253 A1 9(c) 2 2.1 0.44 0 their x their − > or = 0 or ≥ 0 soi B1 Correct attempt to obtain the critical value using 2 2.1 0.44 0 their x − = M1 Must be from the form ( ) 2 ln = − y k px q , 1 ≠ p and 0 > q 0.458 > x or 22 105 x > oe A1 (a) You earn marks by correctly setting up the linear relationship between e^y and x^2, then using the given points to form simultaneous equations or calculate the gradient. You must correctly solve for the gradient (m) and y-intercept (c), and finally express y in terms of x by correctly applying the natural logarithm and algebraic manipulation. (b) To score, substitute x=1 into your derived equation from part (a) and correctly calculate the value of y. (c) You must set the argument of the natural logarithm from your part (a) equation to be greater than zero, then correctly solve this inequality for x to find the range of positive x-values for which the straight line exists.

2 Variables x and y are such that, when lny is plotted against lnx, a straight line graph passing through the points (6, 5) and (8, 9) is obtained. Show that y x e p q = where p and q are integers. [4]

2 9 5 8 6 − = − m oe M1 9 = their 2(8) + c oe or 5 = their 2(6) + c M1 lny = 2lnx – 7 A1 Correct completion to answer: 2 ln 7 7 2 e e − − = = x y x nfww A1 Alternative lny = p + qlnx soi (B1) 9 5 8 6 − = − m oe (M1) 9 = their 2(8) + c oe or 5 = their 2(6) + c (M1) 7 2 e− = y x (A1)

5 The curves y x2 = and y x 27 2 = intersect at O(0, 0) and at the point A. Find the equation of the perpendicular bisector of the line OA. [8]

5 ( ) 2 2 27 = x x or 2 2 27   =     y y oe M1 if M0 then, for first 4 marks, SC4 if (3, 9) only stated and verified in both equations, ignore (0, 0) or SC2 for (3, 9) only stated with no working, ignore (0, 0) If first M1 then (3, 9) with no additional working award M1SC1 4 27 0 − = x x or 4 729 0 − = y y oe nfww A1 x(x3 − 27) = 0 or y(y3 − 729) = 0 oe M1 A(3, 9) oe only nfww A1 Mid-point = ( ) 1.5,4.5 B1 9 3 = OA m oe B1 3 9 ⊥= − m oe M1 y – 4.5 = 3 ( 1.5) 9 − − x oe isw A1 FT their mid-point and 1 9 3 − their

Cambridge O Level Additional Mathematics · 4037

O-Level Additional Mathematics (4037) Straight-line graphs past paper questions

Revise O-Level Additional Mathematics (4037) — Straight-line graphs with real Cambridge past-paper questions grouped by topic, each shown with its marking scheme. Working one topic at a time targets exactly the skill each question tests.

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