JAMB Mathematics Past Questions & Answers - Page 187

0 \(\geq\) x \(\geq\) 2 \(\to\) 0, 1, 2

If x = 0, \(\frac{9}{1 + 2x^2}\)

\(\frac{9}{1 + 2(0)^2}\) = \(\frac{9}{1}\)

= 3

If x = 2, \(\frac{9}{1 + 2(1)^2}\)

= \(\frac{9}{3}\)

= 3

If x = 2, \(\frac{9}{1 + 2(2)^2}\)

= \(\frac{9}{9}\)

= 1

The least value of \(\frac{9}{1 + 2x^2}\) is 1 when x = 2

Two lines are said to be parallel if the slope of the two lines are equal.

The equation : \(3y + 2x + 7 = 0\)

\(3y = -2x - 7\)

\(y = \frac{-2}{3} x - \frac{7}{3}\)

\(\frac{\mathrm d y}{\mathrm d x} = - \frac{2}{3}\)

All the options have the same slope except \(3y - 2x + 7 = 0\).

x\(\frac{3}{8}\) x y\(\frac{-6}{7}\) x (\(\frac{y^{\frac{9}{7}}}{x^{\frac{45}{8}}}\))\(\frac{1}{9}\) = \(\frac{y^{\alpha}}{y^{\beta}}\)

x\(\frac{3}{8}\) x y\(\frac{-6}{7}\) x y\(\frac{1}{7}\) = x\(\alpha\)

= x\(\frac{3}{8}\) + \(\frac{5}{8}\) + y\(\frac{6}{7}\) + \(\frac{1}{7}\)

= x\(\alpha\)y\(\beta\)

x1y\(\frac{-5}{7}\) = x\(\alpha\)y\(\beta\)

\(\alpha\) = 1, \(\beta\) = \(\frac{5}{7}\)

2log2 x - x log2 (1 + y) = 3

log2 \(\frac{x^2}{(1 + y)^x}\) = 3

= \(\frac{x^2}{(1 + y)^x}\)

= 23

= 8

The graphical method of solving the equation x3 + 3x2 + 4x - 28 = 0 is by drawing the graphs of the curves

y = x2 + 3x + 4 and y = \(\frac{28}{x}\)`.