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

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

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

= \(\sqrt {6}\)