WAEC Further Mathematics Past Questions & Answers - Page 132

To find the maximum value, we can use the second derivative test where, given $f(x)$, the second derivative < 0, makes it a maximum value.

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

$\frac{\mathrm d y}{\mathrm d x} = 3x^{2} + 4x + 1 = 0$

Solving, we have $x = \frac{-1}{3}$ or $-1$.

$\frac{\mathrm d^{2} y}{\mathrm d x^{2}} = 6x + 4$

When $x = \frac{-1}{3}, \frac{\mathrm d^{2} y}{\mathrm d x^{2}} = 2 > 0$

When $x = -1, \frac{\mathrm d^{2} y}{\mathrm d x^{2}} = -2 < 0$

At maximum value of x being -1, $y = -1(-1 + 1)^{2} = 0$

We are given the equation $x^{2} + y^{2} - 4x - 2y = 0$

$y = x^{2} + y^{2} - 4x - 2y$

Using the method of implicit differentiation, 

$\frac{\mathrm d y}{\mathrm d x} = 2x + 2y\frac{\mathrm d y}{\mathrm d x} - 4 - 2\frac{\mathrm d y}{\mathrm d x}$

For the tangent, $\frac{\mathrm d y}{\mathrm d x} = 0$,

$\therefore 2x + 2y\frac{\mathrm d y}{\mathrm d x} - 4 - 2\frac{\mathrm d y}{\mathrm d x} = 0$

$(2y - 2)\frac{\mathrm d y}{\mathrm d x} = 4 - 2x \implies \frac{\mathrm d y}{\mathrm d x} = \frac{4 - 2x}{2y - 2}$

At (1, 3), $\frac{\mathrm d y}{\mathrm d x} = \frac{4 - 2(1)}{2(3) - 2} = \frac{2}{4} = \frac{1}{2}$

Equation: $\frac{y - 3}{x - 1} = \frac{1}{2} \implies 2y - 6 = x - 1$

= $2y - x - 6 + 1 = 2y - x - 5 = 0$

$180° = \pi radian$

$\frac{13}{4}\pi = \frac{13}{4} \times 180° = 585°$

$\begin{pmatrix} 2 & x \\ 3 & 5 \end{pmatrix} = 13$

$\begin{vmatrix} 2 & x \\ 3 & 5 \end{vmatrix} = (2 \times 5) - (3 \times x) = 13$

$10 - 3x = 13 \implies -3x = 3; x = -1$

The word has 8 letters with one letter repeated 3 times, therefore we have:

$\frac{8!}{3!} = 6720$ ways.