WAEC 2008 · Paper 2 · Q11

(a)(i)
Find $\dfrac{dy}{dx}$ if $x^2 + 4xy - y^2 = 7$ .
(a)(ii)
If $x = -2$ and $y = 4$ , evaluate $\dfrac{dy}{dx}$ in (a)(i).
(b)
Express $\dfrac{x + 3}{x^2 - 9x + 18}$ in partial fractions.

Worked solution (try it first)

(a)(i)

Differentiate each term with respect to $x$ , using the product rule on $4xy$ : $2x + 4y + 4x\dfrac{dy}{dx} - 2y\dfrac{dy}{dx} = 0$ .
Collect the $\dfrac{dy}{dx}$ terms: $(4x - 2y)\dfrac{dy}{dx} = -2x - 4y$ .
Divide and simplify by $-2$ : $\dfrac{dy}{dx} = \dfrac{x + 2y}{y - 2x}$ .

(ii)

Put $x = -2$ , $y = 4$ : $\dfrac{-2 + 8}{4 + 4} = \dfrac68$
$= \dfrac34$ .

(b)

Factorise: $x^2 - 9x + 18 = (x - 6)(x - 3)$ .
Write $\dfrac{x + 3}{(x - 6)(x - 3)} = \dfrac{A}{x - 6} + \dfrac{B}{x - 3}$ .
Multiply up: $x + 3 = A(x - 3) + B(x - 6)$ .
Put $x = 6$ : $9 = 3A$ , so $A = 3$ .
Put $x = 3$ : $6 = -3B$ , so $B = -2$ .
So $\dfrac{x + 3}{x^2 - 9x + 18} = \dfrac{3}{x - 6} - \dfrac{2}{x - 3}$ .

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