2007 - JAMB Mathematics Past Questions and Answers - page 5

K + 5K + 3K + 75 + 105 = 360 (at a point)
6K + 180 = 360
6K = 360 - 180
6K = 180
k = 180/6 = 30^o

P(At east 11 yrs) = P(11yrs) + P(12yrs)
= ²⁷/₄₀ + ⁷/₄₀
= ³⁴/₄₀
= ¹⁷/₂₀

Mean \(= \bar{x} = \frac{60}{6} = 10\)
mean deviation \(= \frac{\sum|x-\bar{x}|}{n} = \frac{28}{6}\)
= 4.7

Total mark scored = 200
∴200 = 15 + 4y - 4 + 5y + 54 + 28 + 8
200 = 9y + 101
200 - 101 = 9y
99 = 9y
∴y = 11

\(^{10}C_6 = \frac{10!}{(10-6)!6!}=\frac{10!}{4!6!}\\ =\frac{(10\times 9\times 8\times 7 \times 6!)}{4\times 3\times 2\times 1\times 6!}\\ =210\)

\(\int \frac{x^2 -\sqrt{x}}{x} = \int \frac{x^2}{x} - \frac{x^{\frac{1}{2}}}{x}\\ \int x - x^{\frac{-1}{2}}\\ =\left(\frac{1}{2}\right)x^2 - \frac{x^{\frac{1}{2}}}{\frac{1}{2}}+K\\ =\frac{x^2}{2}-2x^{\frac{1}{2}}+K\\ =\frac{x^2}{2}-2\sqrt{x}+K\)

y = x cos x
dy/dx = 1. cos x + x (-sin x)
= cos x - x sin x

f(x) = 2x³ - x² - 4x – 4
f’(x) = 6x² - 2x – 4
As f’(x) = 0
Implies 6x² - 2x – 4 = 0
3x – x – 2 = 0 (By dividing by 2)
(3x – 2)(x + 1) = 0
3x – 2 = 0 implies x = -²/₃
Or x + 1 = 0 implies x = -1
f’(x) = 6x² - 2x – 4
f’’(x) = 12x – 2
At max point f’’(x) < 0
∴f’’(x) = 12x – 2 at x = -1
= 12(-1) – 2
= -12 – 2 = -14
∴Max at x = 1

\(\int_0 ^{\frac{\pi}{2}}(-2cos x)dx = [-2sin x + c]_0 ^{\frac{\pi}{2}}\\ =(-2sin\frac{\pi}{2}+c+2sin0-c)\\ =-2sin90+c+2sin0-c\\ =-2(1)+2(0)\\ =-2\)

x⊕y = xy + x + y
= -³/₄ (6) + (-³/₄) + 6
= -⁹/₂ - ³/₄ + 6
= (-18-3+3+24) / 4
= ³/₄