The curve y = f(x) passes through the point (1/2, 7/2) and is such that f'(x) = e^(2x-1).
- (i)Find the equation of the curve.[4]
- (ii)Find the value of x for which f''(x) = 4, giving your answer in the form a + b ln √2, where a and b are constants.[4]
Mark scheme
Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.
- [M1] me^(2x-1) where m is numeric constant
- [A1] f(x) = (1/2)e^(2x-1) (+c)
- [DM1] correct attempt to find arbitrary constant
- [A1] must be an equation f(x) = (1/2)e^(2x-1) + 3
- [M1] ke^(2x-1) where k is a numeric constant
- [A1] f''(x) = 2e^(2x-1)
- [DM1] attempt to equate to 4 and use logarithms
- [A1] x = 1/2 + ln√2
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