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May/June 2023 Pure Mathematics 1 (9709/12)

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CAMBRIDGE ASSESSMENT INTERNATIONAL EDUCATION (CAIE)9709/12
May/June 2023 Pure Mathematics 1 (9709/12) • AS & A Level Mathematics
base
[4 Marks]
1
Stationary Points & Second Derivative
Worth 4 Marks • Challenging Level
The curve y=2x3−9x2+12x+5y = 2x^3 - 9x^2 + 12x + 5 has two stationary points where dydx=0\frac{dy}{dx} = 0. What is the coordinates and nature of the stationary point where x>1.5x > 1.5?
Formula Guide:
dydx=6x2−18x+12=6(x−1)(x−2)=0  ⟹  x=1,x=2\frac{dy}{dx} = 6x^2 - 18x + 12 = 6(x-1)(x-2) = 0 \implies x=1, x=2. At x=2x=2, d2ydx2=12x−18=6>0\frac{d^2y}{dx^2} = 12x - 18 = 6 > 0 (Local Minimum), and y=2(2)3−9(2)2+12(2)+5=9y = 2(2)^3 - 9(2)^2 + 12(2) + 5 = 9.

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May/June 2023 Pure Mathematics 1 (9709/12) – AS & A Level Mathematics (Cambridge) Digital CBT | TutorVerse