JAMB Mathematics Past Questions & Answers - Page 43

$\begin{vmatrix} 2 & -5 & 3 \\ x & 1 & 4 \\ 0 & 3 & 2 \end{vmatrix} = 132$

$\implies 2 \begin{vmatrix} 1 & 4 \\ 3 & 2 \end{vmatrix} - (-5) \begin{vmatrix} x & 4 \\ 0 & 2 \end{vmatrix} + 3 \begin{vmatrix} x & 1 \\ 0 & 3 \end{vmatrix} = 132$

$2(2 - 12) + 5(2x) + 3(3x) = 132$

$-20 + 10x + 9x = 132$

$19x = 152$

$x = 8$

When you divide a polynomial p(x) by (x - a), the remainder = p(a)

i.e. In the case of 2x$^2$ + x - 3 $\div$ (x - 2), the remainder = p(2).

= 2(2)$^2$ + 2 - 3

= 8 + 2 - 3

= 7.

Given q(x) [quotient], d(x) [divisor] and r(x) [remainder], the polynomial is gotten by multiplying the quotient and the divisor and adding the remainder.

i.e In this case, the polynomial = (x$^2$ - x - 5)(3x - 1) + 7.

= (3x$^3$ - x$^2$ - 3x$^2$ + x - 15x + 5) + 7

= (3x$^3$ - 4x$^2$ - 14x + 5) + 7

= 3x$^3$ - 4x$^2$ - 14x + 12

Explanation

$x^2 - 7x + 10 \leq 0$

Solve for $x^2 - 7x + 10 = 0$

We have, (x - 5)(x - 2) $\leq$ 0.

Conditions:

Case 1: (x - 5) $\leq$ 0, (x - 2) $\geq$ 0.

$\implies$ x $\leq$ 5; x $\geq$ 2.

2 $\leq$ x $\leq$ 5.

Choosing x = 3,

3$^2$ - 7(3) + 10 = 9 - 21 + 10

= -2 $\leq$ 0.

$\therefore$ 2 $\leq$ x $\leq$ 5.

$\frac{2 + 2x}{3} - 2 \geq \frac{4x - 6}{5}$

$\frac{2 + 2x - 6}{3} \geq \frac{4x - 6}{5}$

$\frac{2x - 4}{3} \geq \frac{4x - 6}{5}$

$5(2x - 4) \geq 3(4x - 6)$

$10x - 20 \geq 12x - 18$

$10x - 12x \geq -18 + 20$

$-2x \geq 2$

$x \leq -1$