Variables x and y are such that y = e^(2x) + e^(-2x).
- (i)Find dy/dx.[2]
- (ii)By using the substitution u = e^(2x), find the value of y when dy/dx = 3.[4]
- (iii)Given that x is decreasing at the rate of 0.5 units s⁻¹, find the corresponding rate of change of y when x = 1.[3]
Mark scheme
Marking points from the official mark scheme (May/June 2012). The mark or mark type is shown in brackets.
- [B1] (i) dy/dx = 2e^(2x)
- [B1] (i) -2e^(-2x)
- [M1] (ii) for attempt to obtain in 'quadratic' form (3 = 2e^(2x) - 2e^(-2x) => 2(e^(2x))² - 3(e^(2x)) - 2 = 0)
- [DM1] (ii) for attempt to solve for e^(2x) ((2e^(2x) + 1)(e^(2x) - 2) = 0)
- [M1] (ii) for using e^(2x) = 2 to find y (y = 2 + 1/2)
- [A1] (ii) y = 5/2 (or 2.5)
- [B1] (iii) for dx/dt = -0.5
- [M1] (iii) for substitution of x = 1 into dy/dx (dy/dx = 2e² - 2e⁻²)
- [M1] (iii) for correct application of chain rule (dy/dt = dy/dx * dx/dt)
- [A1] (iii) -7.25 (3 s.f.)
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