Edexcel IALMathematicsGeneral Applications9 Marks
A curve C has equation y = (2x+3)e^{-x}.
- (a)Find dy/dx.[3]
- (b)Find the exact coordinates of the stationary point of C.[3]
- (c)Determine the nature of this stationary point.[3]
Mark scheme
Marking points from the official mark scheme. The mark or mark type is shown in brackets.
- (a) M1: for attempting the product rule, u=(2x+3), v=e^{-x}, u'=2, v'=-e^{-x}.
- dy/dx = 2e⁻ˣ + (2x+3)(-e⁻ˣ)
- A1: for correctly differentiating both parts.
- A1: for simplified form dy/dx = e⁻ˣ(2 - (2x+3)) = e⁻ˣ(-2x-1).
- (b) M1: for setting dy/dx = 0.
- e⁻ˣ(-2x-1) = 0 ⇒ -2x-1 = 0 ⇒ x = -1/2.
- A1: for correct x-coordinate x = -1/2.
- A1: for exact y-coordinate y = (2(-1/2)+3)e^(-(-1/2)) = (-1+3)e^(1/2) = 2√e.
- Stationary point is (-1/2, 2√e).
- (c) M1: for attempting to find (d²y)/(dx²) using the product rule on dy/dx.
- Let u = (-2x-1), v = e^{-x}, then u'=-2, v'=-e^{-x}.
- (d²y)/(dx²) = -2e⁻ˣ + (-2x-1)(-e⁻ˣ) = e⁻ˣ(-2 + 2x+1) = e⁻ˣ(2x-1).
- A1: for correct second derivative (d²y)/(dx²) = e⁻ˣ(2x-1).
- C1: for substituting x = -1/2 into (d²y)/(dx²) and concluding the nature.
- At x = -1/2, (d²y)/(dx²) = e^(1/2)(2(-1/2)-1) = e^(1/2)(-1-1) = -2√e.
- Since (d²y)/(dx²) < 0, the stationary point is a maximum point.
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