9 (a) Solve the following simultaneous equations. ex+y × e3x-2y = 1 x²y = 256
- (a)Solve the following simultaneous equations. ex+y × e3x-2y = 1 x²y = 256[5]
- (b)Solve the equation 10e(2x-1) - 11 = 6e(1-2x), giving your answer in exact form.[4]
Mark scheme
Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.
- [5] B1 (for 9a) For use of eº = 1 or ln1 = 0
- [5] B1 For simplifying powers of e leading to a linear equation in x and y
- [5] M1 For attempt to obtain a cubic equation in one variable using their linear 4x - y = 0 with an attempt to solve
- [5] A1 for each (x=4, y=16) A0 for y = ±16
- [4] B1 (for 9b) Correct use of the reciprocal
- [4] M1 For a correct attempt to form a quadratic equation in e2x-1 or in e1-2x
- [4] M1 Dep on attempt to solve their quadratic equation
- [4] A1 x=1/2 + 1/2ln(3/2) or exact equivalent only. Allow x=1+ln(3/2)/2 and x=1-ln(2/5)/2 for A1 if negative root not discounted. A0 if negative root not discounted
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