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