WAEC 2017 · Paper 2 · Q10✱

(a)
If $y = 3xy + 16$ , find $\dfrac{dy}{dx}$ at $(-1, 4)$ .
(b)
If $\alpha$ and $\beta$ are the roots of $3x^2 - 5x + 1 = 0$ , find the equation whose roots are $\left(\alpha - \frac1\beta\right)$ and $\left(\beta - \frac1\alpha\right)$ .
Show the answer
$3x^2 + 10x + 4 = 0$

Worked solution (try it first)

Differentiate both sides with respect to $x$ , using the product rule on $3xy$ : $\dfrac{dy}{dx} = 3y + 3x\dfrac{dy}{dx}$ .
Collect the $\dfrac{dy}{dx}$ terms: $\dfrac{dy}{dx}(1 - 3x) = 3y$ , so $\dfrac{dy}{dx} = \dfrac{3y}{1 - 3x}$ .
At $(-1, 4)$ , substitute $x = -1$ and $y = 4$ : $\dfrac{dy}{dx} = \dfrac{3 \times 4}{1 - 3(-1)}$ .
Work out the top and the bottom: $\dfrac{12}{1 + 3} = \dfrac{12}{4}$ , so $\dfrac{dy}{dx} = 3$ at $(-1, 4)$ .

For $3x^2 - 5x + 1 = 0$ : $\alpha + \beta = \frac53$ and $\alpha\beta = \frac13$ .
New sum: $\left(\alpha - \frac1\beta\right) + \left(\beta - \frac1\alpha\right) = (\alpha + \beta) - \dfrac{\alpha + \beta}{\alpha\beta}$ .
Substitute: $\frac53 - \dfrac{\frac53}{\frac13} = \frac53 - 5$
$= -\frac{10}{3}$ .
New product: $\left(\alpha - \frac1\beta\right)\left(\beta - \frac1\alpha\right) = \alpha\beta - 1 - 1 + \dfrac{1}{\alpha\beta}$ .
Substitute: $\frac13 - 2 + 3 = \frac43$ .
The equation: $x^2 + \frac{10}{3}x + \frac43 = 0$ .
Multiply through by 3: $3x^2 + 10x + 4 = 0$ .