2006 - JAMB Mathematics Past Questions and Answers - page 1

The question will be answered in

(^5C_4 = \frac{6!}{(6-4)!4!}=\frac{6!}{2!4!}=\frac{6\times5\times4!}{2\times1\times4!}=15\hspace{1mm}ways)

Let the consecutive numbers be a, a+1, a+2, a+3, a+4

Mean = (\frac{(a+a+1+a+2+a+3+a+4)}{5}\

30 = \frac{5a+10}{5}\

30 \times 5 = 5a + 10\

5a = 150 - 10\

5a = 140\

a = 28 \

∴ a + 4 = 28 + 4\

= 32)

Black = 5

White = 4

Red = x

Total = 9+x

P(red) = ²/₃

x / (9+x) = ²/₅

5x = 2(9+x)

5x = 18+2x

5x-2 = 18

3x = 18

x = 6

(\sum x = 6x\

\sum(x-\bar{x})^2 = 2\

\bar{x} = \frac{\sum x}{n}\

= \frac{6x}{3}\

= 2x\

Variance = \frac{\sum(x-\bar{x})^2}{n}\

= \frac{2}{3})

P(GTN) = 5/20

P(Qtel) = 3/20

∴P(GTN or Qtel) = (5/20) + (3/20)

= 8/20

= 2/5

Expenditure on housing = x

Expenditure on food = 2x

Expenditure on school fees = 90^o

Expenditure on transport = 45^o

x + 2x + 90 + 45 = 360^o

3x + 135 = 360^o

3x = 360 - 135

3x = 225

x = 225/3 = 75^o

∴∠ for food = 2x = 2 * 75

= 150

360^o = N30,000

1^o = ?

1^o = 30,000/360

150^o = (30000/360) * (150/1)

= N12500

(^{n+1}C_3 = 4(^nC_3)\frac{(n+1)!}{(n+1-3)!3!} = 4\left(\frac{n!}{(n-3)!3!}\right)\frac{(n+1)n!}{(n-2)(n-3)!}=4\left(\frac{n!}{n-3!}\right)=\frac{n+1}{n-2}=\frac{4}{1}\n+1 = 4(n-2)\n+1 = 4n-8-3n = -9\frac{-9}{-3}\n=3)

dy/dx = 2x + 7

y = ∫2x + 7

y = x² + 7x + C at (2,0)

0 = 2² + 7(2) + C

0 = 4 + 14 + C

0 = 18 + C

C = -18

∴ The equation is y = x² + 7x - 18

y = (x² - ¹/_x)²

y = (x² - ¹/_x)(x² - ¹/_x)

y = x⁴ - x - x + ¹/_x²

y = x⁴ - 2x + ¹/_x²

y= x⁴ - 2x + x^-2

dy/dx = 4x² - 2 - 2x^-3

= 4x² - 2 - ²/_x³