WAEC 2022 · Paper 2 · Q10

Worked solution (try it first)

The top has degree 3 and the bottom degree 2, so divide first.
$x(x^2 - 1) = x^3 - x$ .
Take it away: $-x^2 + x - 4$ is left.
$-1(x^2 - 1) = -x^2 + 1$ .
Take it away: $x - 5$ is left.
So the fraction is $x - 1 + \dfrac{x - 5}{x^2 - 1}$ .
Split: $x - 5 = A(x - 1) + B(x + 1)$ for $\dfrac{A}{x + 1} + \dfrac{B}{x - 1}$ .
Put $x = 1$ : $-4 = 2B$ , so $B = -2$ .
Put $x = -1$ : $-6 = -2A$ , so $A = 3$ .
So the answer is $x - 1 + \dfrac{3}{x + 1} - \dfrac{2}{x - 1}$ .

Use the quotient rule with $u = \sin x$ and $v = 1 + \cos x$ : $u' = \cos x$ and $v' = -\sin x$ .
$\dfrac{dy}{dx} = \dfrac{\cos x(1 + \cos x) - \sin x(-\sin x)}{(1 + \cos x)^2}$
$= \dfrac{\cos x + \cos^2x + \sin^2x}{(1 + \cos x)^2}$ .
Use $\cos^2x + \sin^2x = 1$ : the top is $1 + \cos x$ .
Cancel one factor: $\dfrac{dy}{dx} = \dfrac{1}{1 + \cos x}$ .