WAEC 2010 · Paper 2 · Q12

(a)
Given that $\displaystyle\int_1^m (x^2 - 2x + 1)\,dx = \dfrac13$ , $m > 0$ , determine the value of $m$ .
(b)
If $(x - y)^2 = 3xy + 1$ , find the gradient at the point $(1, 0)$ .

Worked solution (try it first)

(a)

Integrate: $\left[\dfrac{x^3}{3} - x^2 + x\right]_1^m = \dfrac{m^3}{3} - m^2 + m - \dfrac13$ .
Set this equal to $\frac13$ : $\dfrac{m^3}{3} - m^2 + m - \dfrac23 = 0$ .
Multiply by 3: $m^3 - 3m^2 + 3m - 2 = 0$ .
Factorise ( $m = 2$ makes it zero): $(m - 2)(m^2 - m + 1) = 0$ .
$m^2 - m + 1 = 0$ has no real roots, so $m = 2$ .

(b)

Differentiate each side with respect to $x$ : $2(x - y)\left(1 - \dfrac{dy}{dx}\right) = 3y + 3x\dfrac{dy}{dx}$ .
Put $x = 1$ , $y = 0$ : $2\left(1 - \dfrac{dy}{dx}\right) = 3\dfrac{dy}{dx}$ .
So $2 = 5\dfrac{dy}{dx}$ and the gradient is $\dfrac25 = 0.4$ .

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