WAEC Further Mathematics Past Questions & Answers - Page 134

The terms of the sequence can be written as : $u_{r} = ar + b$ in this case, being that they have a regular common difference for each of the r terms.

We can rewrite the sequence as $a + b, 2a + b, 3a + b,...$ where a is the common difference of the sequence and b is a given constant gotten by solving

$a + b = 9$ or $2a + b = 4$ or any other one. 

The common difference here is 4 - 9 = -1 - 4 = -5.

$-5 + b = 9 \implies b = 9 + 5 = 14$

$\therefore$ The equation can be written as $u_{r} = -5r + 14$.

$x - \frac{3}{x^{2}} = x - 3x^{-2}$

Let the power on x be t, so that the power on $x^{-2}$ = 9 - t

$(x)^{t}(x^{-2})^{9 - t} = x^{3}  \implies t - 18 + 2t = 3$

$3t = 3 + 18 = 21 \therefore t = 7$

To obtain the coefficient of $x^{3}$, we have

$^{9}C_{7}(x)^{7}(3x^{-2))^{2} = \frac{9!}{(9 - 7)! 7!}(x)^{7}(9x^{-4})$

= $\frac{9 \times 8 \times 7!}{7! 2!} \times 9(x^{3}) = 324x^{3}$

$\log_{2}(12x - 10) = 1 + \log_{2}(4x + 3)$

Recall, $1 = \log_{2}2$, so

$\log_{2}(12x - 10) = \log_{2}2 + \log_{2}(4x + 3)$

= $\log_{2}(12x - 10) = \log_{2}(2(4x + 3))$

$\implies (12x - 10) = 2(4x + 3) \therefore 12x - 10 = 8x + 6$

$12x - 8x = 4x = 6 + 10 = 16 \implies x = 4$

An equation can be written as $x^{2} - (\alpha + \beta)x + (\alpha\beta) = 0$

Making the coefficient of $x^{2}$ = 1 in the given equation, we have

$x^{2} + \frac{5}{2}x + \frac{n}{2} = 0$

Comparing, we have $\alpha\beta = \frac{n}{2} = 2 \implies n = 4$

$\frac{3x - 1}{(x - 2)^{2}} = \frac{A}{(x - 2)} + \frac{Bx}{(x - 2)^{2}}$

$\frac{3x - 1}{(x - 2)^{2}} = \frac{A(x - 2) + Bx}{(x - 2)^{2}}$

Comparing, we have

$3x - 1 = Ax - 2A + Bx  \implies -2A = -1;  A + B = 3$

$\therefore A = \frac{1}{2}; B = \frac{5}{2}$

= $\frac{1}{2(x - 2)} + \frac{5x}{2(x - 2)^{2}}$