WAEC 2013 · Paper 2 · Q9

(a)
Differentiate $\dfrac{x^2 + 1}{(x + 1)^2}$ with respect to $x$ .
(b)
(i) Evaluate $\begin{vmatrix} 1 & 2 & -1 \\ 2 & 3 & -1 \\ -1 & 1 & 3 \end{vmatrix}$ . (ii) Using the answer in (b)(i), solve the system of equations $x + 2y - z = 4$ , $2x + 3y - z = 2$ , $-x + y + 3z = -1$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Quotient rule with $u = x^2 + 1$ and $v = (x + 1)^2$ : $u' = 2x$ and $v' = 2(x + 1)$ .
$\dfrac{dy}{dx} = \dfrac{(x + 1)^2 \cdot 2x - (x^2 + 1) \cdot 2(x + 1)}{(x + 1)^4}$ .
Take out $2(x + 1)$ from the top: $2(x + 1)[x(x + 1) - (x^2 + 1)] = 2(x + 1)(x - 1)$ .
Cancel one $(x + 1)$ : $\dfrac{dy}{dx} = \dfrac{2(x - 1)}{(x + 1)^3}$ .

(b)(i)

Expand along the top row: $1(9 + 1) - 2(6 - 1) + (-1)(2 + 3) = 10 - 10 - 5$
$= -5$ .

(ii)

By Cramer's rule, replace each column by $(4, 2, -1)$ in turn: $\Delta_x = 25$ , $\Delta_y = -15$ and $\Delta_z = 15$ .
So $x = \dfrac{25}{-5} = -5$ , $y = \dfrac{-15}{-5} = 3$ and $z = \dfrac{15}{-5} = -3$ .
Check in the first equation: $-5 + 6 + 3 = 4$ ✓.

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