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