WAEC 2011 · Paper 2 · Q9

(a)
Find, from first principles, the derivative of $f(x) = (2x + 3)^2$ .
(b)
Evaluate $\displaystyle\int_1^2 \frac{(x + 1)(x^2 - 2x + 2)}{x^2}\,dx$ .

Worked solution (try it first)

(a)

Expand: $f(x) = (2x + 3)^2 = 4x^2 + 12x + 9$ .
$f(x + h) = 4(x + h)^2 + 12(x + h) + 9$
$= 4x^2 + 8xh + 4h^2 + 12x + 12h + 9$ .
Take away $f(x)$ : $f(x + h) - f(x) = 8xh + 4h^2 + 12h$ .
Divide by $h$ : $8x + 4h + 12$ .
Let $h \to 0$ : $f'(x) = 8x + 12$ .

(b)

Multiply out the top: $(x + 1)(x^2 - 2x + 2) = x^3 - x^2 + 2$ .
Divide each term by $x^2$ : the integrand is $x - 1 + 2x^{-2}$ .
Integrate: $\left[\dfrac{x^2}{2} - x - \dfrac2x\right]_1^2$ .
At $x = 2$ : $2 - 2 - 1 = -1$ .
At $x = 1$ : $\frac12 - 1 - 2 = -\frac52$ .
Subtract: $-1 - \left(-\frac52\right) = \dfrac32$ .

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