WAEC 2014 · Paper 2 · Q9

(a)
Resolve $\dfrac{x^2 + 1}{(x + 2)^3}$ into partial fractions.
(b)
The gradient of a curve at the point $(x, y)$ is $(2x - 3)$ . If the minimum value on the curve is 3, find the equation of the curve.
Show the answer
$4y = 4x^2 - 12x + 21$

Worked solution (try it first)

(a)

A cubed bracket needs three fractions: $\dfrac{A}{x + 2} + \dfrac{B}{(x + 2)^2} + \dfrac{C}{(x + 2)^3}$ .
Multiply through by $(x + 2)^3$ : $x^2 + 1 = A(x + 2)^2 + B(x + 2) + C$ .
Put $x = -2$ : $5 = C$ .
Expand: $A(x^2 + 4x + 4) + Bx + 2B + C$ .
The $x^2$ terms give $A = 1$ .
The $x$ terms give $0 = 4A + B$ , so $B = -4$ .
Check the constants: $4 - 8 + 5 = 1$ ✓.
So the answer is $\dfrac{1}{x + 2} - \dfrac{4}{(x + 2)^2} + \dfrac{5}{(x + 2)^3}$ .

(b)

Integrate the gradient: $y = x^2 - 3x + c$ .
The minimum is where the gradient is zero: $2x - 3 = 0$ , so $x = \frac32$ .
There $y = 3$ : $\frac94 - \frac92 + c = 3$ , so $c = 3 + \frac94 = \frac{21}{4}$ .
So $y = x^2 - 3x + \dfrac{21}{4}$ , that is $4y = 4x^2 - 12x + 21$ .

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