WAEC 2014 · Paper 2 · Q11

(a)
If $f(x) = \dfrac{x - 3}{2x - 1}$ , $x \ne \frac12$ and $g(x) = \dfrac{x - 1}{x + 1}$ , $x \ne -1$ , find $g \circ f$ .
(b)(i)
Sketch the curve $y = 9x - x^3$ .
Model answer
Mark the crossings at $x = -3$ , $0$ and $3$ , the minimum $(-\sqrt3, -6\sqrt3)$ and the maximum $(\sqrt3, 6\sqrt3)$ , and join them with a smooth curve that rises from the bottom right to the top left overall.
(b)(ii)
Calculate the total area bounded by the $x$ -axis and the curve $y = 9x - x^3$ .

Worked solution (try it first)

$g \circ f(x) = g(f(x))$
$= \dfrac{f(x) - 1}{f(x) + 1}$ .
Top: $\dfrac{x - 3}{2x - 1} - 1$ has numerator $x - 3 - (2x - 1) = -x - 2$ over $2x - 1$ .
Bottom: $\dfrac{x - 3}{2x - 1} + 1$ has numerator $x - 3 + (2x - 1) = 3x - 4$ over $2x - 1$ .
Divide.
The $2x - 1$ cancels: $g \circ f(x) = \dfrac{-(x + 2)}{3x - 4}$ , $x \ne \frac43$ .

On the $x$ -axis $y = 0$ : $x(3 - x)(3 + x) = 0$ , so the curve crosses at $x = -3$ , $0$ and $3$ .
Turning points: $\dfrac{dy}{dx} = 9 - 3x^2 = 0$ gives $x = \pm\sqrt3$ .
At $x = \sqrt3$ : $y = 9\sqrt3 - 3\sqrt3$
$= 6\sqrt3$
$\approx 10.4$ , a maximum.
At $x = -\sqrt3$ : $y = -6\sqrt3 \approx -10.4$ , a minimum.
Sketch a smooth curve from the top left, down through $(-3, 0)$ to $(-\sqrt3, -6\sqrt3)$ , up through the origin to $(\sqrt3, 6\sqrt3)$ , and down through $(3, 0)$ .

The curve is below the axis for $-3 < x < 0$ and above it for $0 < x < 3$ , so find the two areas separately.
$\int_0^3 (9x - x^3)\,dx = \left[\frac{9x^2}{2} - \frac{x^4}{4}\right]_0^3$
$= \frac{81}{2} - \frac{81}{4}$
$= \frac{81}{4}$ .
$\int_{-3}^0 (9x - x^3)\,dx = -\frac{81}{4}$ , so that area is also $\frac{81}{4}$ .
Total area $= \frac{81}{4} + \frac{81}{4} = 40.5$ square units.