2004 - JAMB Mathematics Past Questions and Answers - page 2

y = (2 + 3x)(i - x)

y = 2 - 2x + 3x - 3x²

y = 2 + x - 3x²

dy/dxb = 1 - 6x

V = 2/3π³

dv/dr = 2πr²

dv/dr = 2π(2)²

= 8π

y = 2x²(2x - 1)

y = 4x³ - 2x²

dy/dx = 12x² - 4x

at x = -1

dy/dx = 12(-1)² - 4(-1)

= 12 + 4

= 16

(\int_{1}^{3}(x^2 - 1)dx=\left[\frac{1}{3}x^2 - x\right] ^{3}_{1}=(9-3)-(\frac{1}{3}-1)=6-\left(-\frac{2}{3}\right)=6+\frac{2}{3}=6\frac{2}{3})

Number of white balls = x

Number of red balls = 12

Number of black balls = 16

Total number of balls = 28 + x

P(white balls) = 3/7

But P(white balls) (= \frac{x}{28+x}\

= \frac{3}{7} = \frac{x}{28+x}\

3(28 + x) = 7x\

84 + 3x = 7x\

7x - 3x = 84\

4x = 84\

x = 21)

X = 1, 2, 3, 4; ∑X = 10

x = ∑X/n = 10/4 = 2.5

X - x = -1.5, -0.5, 0.5, 1.5

lX - xl = 1.5, 0.5, 0.5, 1.5; ∑lX - xl = 4.0

mean deviation = (∑lX - xl)/n

= 4.0/4

= 1.0

Number of times 5 was obtained = 20

P(obtaining 5) = 20/100

= 1/5

Sectorial angle of beans = 360 - (150 + 90 + 60)

= 360 -300

= 60^o

if 150^o represents 3000 tonnes

1^o 3000/150

60^o will represents (3000/150) * (60/1)

= 1200 tonnes

(In\hspace{1mm} ^{5}C_{2}\hspace{1mm}ways\hspace{1mm}=\frac{5!}{(5-2)!2!}=\frac{5!}{3!2!}=\frac{5\times4\times3!}{3!\times2\times1}=10\hspace{1mm}ways)

Let the number of students = x

Total age of the students = 15x

Total age including age of the teacher = 15x + 45

Mean of their ages; (15x + 45)/(x + 1) = 18

= 15x + 45 = 18(x + 1)

= 15x + 45 = 18x + 18

= 18x - 15x = 45 - 18

= 3x = 27

x = 9