2011 - JAMB Mathematics Past Questions and Answers - page 4

tan(\theta) = (\frac{3}{4})

from Pythagoras tippet, the hypotenus is T

i.e. 3, 4, 5.

then sin (\theta) = (\frac{3}{5}) and cos(\theta) = (\frac{4}{3})

cos(\theta) - sin(\theta)

(\frac{4}{5}) - (\frac{3}{5}) = (\frac{1}{5})

tan(\theta) = (\frac{100}{100}) = 1

(\theta) = tan^-1(1) = 45^o

The bearing of x from z is N45^oE or 135^o

(2x + 1)(3x + 1) IS

2x + 1 (\frac{d(3x + 1)}{d}) + (3x + 1) (\frac{d(2x + 1)}{d})

2x + 1 (3) + (3x + 1) (2)

6x + 3 + 6x + 2 = 12x + 5

Mode = L₁ + ((\frac{D_1}{D_1 + D_2}))C

D₁ = frequency of modal class - frequency of the class before it

D₁ = 5 - 2 = 3

D₂ = frequency of modal class - frequency of the class that offers it

D₂ = 5 - 3 = 2

L₁ = lower class boundary of the modal class

L₁ = 5 - 5

C is the class width = 8 - 5.5 = 3

Mode = L₁ + ((\frac{D_1}{D_1 + D_2}))C

= 5.5 + (\frac{3}{2 + 3})C

= 5.5 + (\frac{3}{5}) x 3

= 5.5 + (\frac{9}{5})

= 5.5 + 1.8

= 7.3 (\approx) = 7

y = x³ + x² - x + 1

(\frac{dy}{dx}) = (\frac{d(x^3)}{dx}) + (\frac{d(x^2)}{dx}) - (\frac{d(x)}{dx}) + (\frac{d(1)}{dx})

(\frac{dy}{dx}) = 3x² + 2x - 1 = 0

(\frac{dy}{dx}) = 3x² + 2x - 1

At the maximum point (\frac{dy}{dx}) = 0

3x² + 2x - 1 = 0

(3x² + 3x) - (x - 1) = 0

3x(x + 1) -1(x + 1) = 0

(3x - 1)(x + 1) = 0

therefore x = (\frac{1}{3}) or -1

For the maximum point

(\frac{d^2y}{dx^2}) < 0

(\frac{d^2y}{dx^2}) 6x + 2

when x = (\frac{1}{3})

(\frac{dx^2}{dx^2}) = 6((\frac{1}{3})) + 2

= 2 + 2 = 4

(\frac{d^2y}{dx^2}) > o which is the minimum point

when x = -1

(\frac{d^2y}{dx^2}) = 6(-1) + 2

= -6 + 2 = -4

-4 < 0

therefore, (\frac{d^2y}{dx^2}) < 0

the maximum point is -1

(\int^{1}_{0})(3 - 2x)dx

[3x - x²]^o

[3(1) - (1)²] - [3(0) - (0)²]

(3 - 1) - (0 - 0) = 2 - 0

= 2

(\int^{1}_{0}) cos4 x dx

let u = 4x

(\frac{dy}{dx}) = 4

dx = (\frac{dy}{4})

(\int^{1}_{0})cos u. (\frac{dy}{4}) = (\frac{1}{4})(\int)cos u du

= (\frac{1}{4}) sin u + k

= (\frac{1}{4}) sin4x + k

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

a + a + 1 + a + 2 + a + 3 = 34

4a = 34 - 6

4a = 28

a = (\frac{28}{4})

= 7

The least of these numbers is a = 7

Median = (\frac{\sum fx}{\sum f})

(\begin{array}{c|c}

No & 0 & 1 & 2 & 3 & 4 & 5 \ F & 1 & 4 & 3 & 8 & 2 & 5 \ fx & 0 & 4 & 6 & 24 & 8 & 25 \end{array})

(\sum fx) = 0 + 4 + 6 + 24 + 8 + 25 = 67

(\sum f) = 23

Median = (\frac{\sum fx}{\sum f}) = (\frac{67}{23}) = 2.913

= (\approx) 3

Range = 5 - 0 = 5

(3, 5)

(\begin{array}{c|c}Class Interval & 3 - 3 & 6 - 8 & 9 - 11 \ x & 4 & 7 & 10 \ f & 2 & 2 & 2 \ f - x & 8 & 14 & 20 \ |x - \bar{x}|^2 & 9 & 0 & 9 \ |x - \bar{x}|^2 & 18 0 & 18 \end{array})

(\bar{x}) = (\frac {\sum fx}{\sum f})

= (\frac {8 + 14 + 20}{2 + 2 + 2})

= (\frac{42}{6})

(\bar{x}) = 7

S.D = (\sqrt\frac{\sum f(x - \bar{x})^2}{\sum f})

= (\sqrt\frac{18 + 0 + 18}{6})

= (\sqrt\frac{36}{6})

= (\sqrt {6})