JAMB Mathematics Past Questions & Answers - Page 148

$3x - (\frac{1}{4})^{-\frac{1}{2}} > \frac{1}{4} - x$

= $3x - 4^{\frac{1}{2}} > \frac{1}{4} - x$

= $3x - 2 > \frac{1}{4} - x$

= $3x + x > \frac{1}{4} + 2 \implies 4x > \frac{9}{4}$

$x > \frac{9}{16}$

Given the curve $y = x^{2} - 6x + 5$

At minimum or maximum point, $\frac{\mathrm d y}{\mathrm d x} = 0$

$\frac{\mathrm d y}{\mathrm d x} = 2x - 6$

$2x - 6 = 0 \implies x = 3$

Since 3 > 0, it is a minimum point.

When x = 3, $y = 3^{2} - 6(3) + 5 = -4$

Hence, the turning point has coordinates (3, -4).

Force to the east = 5 units

force to the North - east = 4 units.

Resultant of the two forces is the square root

52 + 42 = 41 and plus the sum of its resistance

5 x 4$\sqrt{2}$ = 20$\sqrt{2}$

= $\sqrt{41 + 20 \sqrt{2}}$ units

$\frac{a - b}{a + b}$ - $\frac{a + b}{a - b}$ = $\frac{(a - b)^2}{(a + b)}$ - $\frac{(a + b)^2}{(a - b0}$

applying the principle of difference of two sqrt. Numerator = (a - b) + (a + b) (a - b) - (a + b)

= (a = b + a = b)(a = b - a = b)

2a(-2b) = -4ab

= $\frac{-4ab}{(a + b)(a - b)}$

= $\frac{-4ab}{a^2 - b^2}$

Sum of interior angles of a polygon = (2n - 4) x 90

where n is the no. of sides

sum of interior angles of the quad. = ({2 x 4} -4) x 90

(8 - 4) x 90 = 4 x 90o

= 360o

(p + 10)o + (p - 30)o + (2p - 45)o + (p + 15)o = 360o

p + 10o + p - 30o + 2p - 45o + p + 15 = 360o

5p = 360o + 50o

= 82o