Standard +0.3 This is a straightforward exponential equation requiring standard algebraic manipulation (multiplying through, substituting y = e^x to get a quadratic, then taking logarithms). While it involves multiple steps for 5 marks, the technique is routine for P3 students with no conceptual difficulty or novel insight required—slightly easier than average.
Substitute and obtain 3-term quadratic 3u2 +4u−1=0, or
equivalent
B1
( )
e.g.3 ex 2 +4ex −1=0
Answer
Marks
Guidance
Solve a 3 term quadratic for u
M1
Must be an equation with real roots
( )
Answer
Marks
Guidance
Obtain root 7 −2 /3, or decimal in [0.21, 0.22]
A1
Or equivalent. Ignore second root (even if incorrect)
Use correct method for finding x from a positive value of ex
M1
Must see some indication of method: use of x=lnu
Obtain answer x = – 1.536 only
A1
CAO. Must be 3 dp
5
Answer
Marks
Guidance
Question
Answer
Marks
Question 4:
4 | Substitute and obtain 3-term quadratic 3u2 +4u−1=0, or
equivalent | B1 | ( )
e.g.3 ex 2 +4ex −1=0
Solve a 3 term quadratic for u | M1 | Must be an equation with real roots
( )
Obtain root 7 −2 /3, or decimal in [0.21, 0.22] | A1 | Or equivalent. Ignore second root (even if incorrect)
Use correct method for finding x from a positive value of ex | M1 | Must see some indication of method: use of x=lnu
Obtain answer x = – 1.536 only | A1 | CAO. Must be 3 dp
5
Question | Answer | Marks | Guidance