WAEC 2017 · Paper 2 · Q9

(a)
Simplify $\dfrac{1}{1 - \cos\theta} + \dfrac{1}{1 + \cos\theta}$ and leave the answer in terms of $\sin\theta$ .
(b)
Find the equation of the line joining the stationary points of $y = x^2(x - 3)$ and the distance between them.

Try it on a graph

Plot the curves, move them, and read values off the graph.

Worked solution (try it first)

(a)

The common denominator is $(1 - \cos\theta)(1 + \cos\theta) = 1 - \cos^2\theta$
$= \sin^2\theta$ .
The top is $(1 + \cos\theta) + (1 - \cos\theta) = 2$ .
So the sum is $\dfrac{2}{\sin^2\theta}$ .

(b)

$y = x^3 - 3x^2$ , so $\dfrac{dy}{dx} = 3x^2 - 6x = 3x(x - 2)$ .
It is zero at $x = 0$ and $x = 2$ .
The stationary points are $(0, 0)$ and $(2, 8 - 12) = (2, -4)$ .
The gradient of the line joining them is $\dfrac{-4 - 0}{2 - 0} = -2$ , so the line is $y = -2x$ , that is $y + 2x = 0$ .
The distance between them is $\sqrt{2^2 + 4^2} = \sqrt{20} \approx 4.47$ .

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