JAMB Mathematics Past Questions & Answers - Page 302

[$\frac{1}{0.03}$ + $\frac{1}{0.024}$]

= [$\frac{1}{0.03 \times 0.024}$]^-1

= [$\frac{0.024}{0.003}$]^-1

= $\frac{0.03}{0.024}$

= $\frac{30}{24}$ = 1.25

1011₂ + x₇ = 25₁₀ = 1011₂ = 1 x 2³ + 0 x 2² + 1 x 2¹ + 1 x 2^o

= 8 + 0 + 2 + 1

= 11₁₀

x₇ = 25₁₀ - 11₁₀

= 14₁₀

$\begin{array}{c|c} 7 & 14 \\ 7 & 2 R 0 \\ & 0 R 2 \end{array}$

X = 20₇

Equation: x$^3$ - 2x$^2$ - 5x + 6 = 0.

First, bring out a$_n$ which is the coefficient of x$^3$ = 1.

Then, a$_0$ which is the coefficient void of x = 6.

The factors of a$_n$ = 1; The factors of a$_0$ = 1, 2, 3 and 6.

The numbers to test for the roots are $\pm (\frac{a_0}{a_n})$.

= $\pm (1, 2, 3, 6)$.

Test for +1: 1$^3$ - 2(1$^2$) - 5(1) + 6 = 1 - 2 - 5 + 6 = 0.

Therefore x = 1 is a root of the equation.

Using long division method, $\frac{x^3 - 2x^2 - 5x + 6}{x - 1}$ = x$^2$ - x - 6.

x$^2$ - x - 6 = (x - 3)(x + 2).

x = -2, 3.

$\therefore$ The roots of the equation = 1, -2 and 3.

2(1) - (-1) = 3      2(2) - (0) = 4          2(-1) - (2) = -4

2(0) - (1) = -1      2(-1) - (-1) = -1      2(0) - (-1) = 1

2(1) - 2 = 0         2(0) - (-1) = 1         2(1) - (1) = 1

$\begin{pmatrix} 3 & -1 & 0 \\ 4 & -1 & 1 \\ -4 & 1 & 1\end{pmatrix}$

2p x 1 + 8 x 2 = 24

$\to$ 4p = 24 - 8 = 16,

p = 4

3 x 1 + -5q x 2 = -17

$\to$ -10q = -17 - 3

-10q = -20

q = 2