WAEC Further Mathematics Past Questions & Answers - Page 127

The remainder theorem states that if f(x) is divided by (x - a), the remainder is f(a). 

$f(x) = x^{3} - 2x + m$ divided by (x - 1), so that a = 1.

Remainder = $f(1) = 1^3 - 2(1) + m = -1 + m$

$f(x) = 2x^{3} + x - m$ divided by (2x + 1), so that a = $\frac{-1}{2}$

$f(\frac{-1}{2}) = 2(\frac{-1}{2}^{3}) + (\frac{-1}{2}) - m = \frac{-3}{4} - m$

$\implies m - 1 = \frac{-3}{4} - m$, collecting like terms,

$2m = \frac{1}{4} \therefore m = \frac{1}{8}$

Given x = (-1, 5) for the equation $x^{2} + kx - 5 = 0$

$x = -1 \implies x + 1 = 0$; $x = 5 \implies x - 5 = 0$

$(x + 1)(x - 5) = 0$, expanding,

$x^{2} - 5x + x - 5 = 0   \therefore  x^{2} - 4x -  5 = 0$

$\therefore$ k = -4.

$x^{2} + x^{2}y + 3x - 10y + 3xy -10$

= $x^{2} + x^{2}y + 3x + 3xy - 10y - 10  = x^{2}(1 + y) + 3x(1 + y) - 10(y + 1)$

= $(x^{2} + 3x - 10)(y + 1)$

= $(x^{2} + 3x - 10) = x^{2} - 2x + 5x - 10$

= $x(x - 2) + 5(x - 2) = (x - 2)(x +5)$

$\therefore x^{2} + x^{2}y + 3x - 10y + 3xy -10 = (x - 2)(x + 5)(y + 1)$.

The elements of x are {-2, -1, 0, 1, 2}

$y = 4x - 1$ = 4(-2) - 1 = -9; 4(-1) - 1 = -5; 4(0) - 1 = -1; 4(1) - 1 = 3; 4(2) - 1 = 7.

The range of x is {-9, -5, -1, 3 7}.

$f(x) = \frac{4}{x} - 1$. Let y = f(x)

$y = \frac{4 - x}{x}  \implies xy + x = 4$

$x(y + 1) = 4  \therefore  x = \frac{4}{y + 1}$

$f^{-1}(7) = \frac{4}{7 + 1} = \frac{1}{2}$