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WAEC Further Mathematics 2018 Paper

Question : 26

Total: 40

y = 3x^{2} - 2x^{3}

Solution:

Concept:

Stationary points occur where the gradient of the curve is zero, i.e.

\frac{dy}{dx} = 0

Explanation:

Given

y = 3x^{2} - 2x^{3}

, differentiate with respect to

x

\frac{dy}{dx} = 6x - 6x^{2}

Set

\frac{dy}{dx} = 0

for a stationary point:

6x - 6x^{2} = 0

\Rightarrow 6x(1 - x) = 0

x = 0

x = 1

For

x = 1

y = 3(1)^2 - 2(1)^3 = 3 - 2 = 1

Hence,

(1,1)

is a stationary point.

Note:

x = 0

gives

(0,0)

, but that is not listed among the options.

Answer:

(1,1)

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