Variables x and y are such that y = Ax^b, 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]
- (b)the value of y when x = 3,[2]
- (c)the value of x when y = 3.[2]
Mark scheme
Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.
- [1] B1 lg y = b lg x + lg A May be implied by subsequent work
- [1] M1 For at least one correct equation
- [1] A1 b = 0.8
- [1] M1 Dep for substitution to obtain lg A = k and hence A
- [1] A1 A = 1.21
- [1] (B1) lg y = b lg x + lg A May be implied by subsequent work
- [1] (M1) Gradient = (4.37-0.57)/(5.36-0.61)
- [1] (A1) b = 0.8
- [1] (M1) Dep for substitution into a correct equation to obtain lg A = k and hence A
- [1] (A1) A = 1.21
- [1] (B1) 10^4.37 = A×10^5.36b or 10^0.57 = A×10^0.61b
- [1] (M1) For eliminating A correctly
- [1] (A1) b = 0.8
- [1] (M1) For a correct attempt to find A. Must have B1
- [1] (A1) A = 1.21
- [1] B1 FT for substitution into their equation
- [1] B1 y = awrt 2.9
- [1] B1 FT for substitution into their equation
- [1] B1 x = awrt 3.1
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