JAMB Mathematics Past Questions & Answers - Page 151

751.

If sec\(^2\) \(\theta\) + tan\(^2\) \(\theta\) = 3, then the angle \(\theta\) is equal to

A.

30o

B.

45o

C.

60o

D.

90o

E.

105o

Correct answer is B

sec\(^2\) \(\theta\) + tan\(^2\) \(\theta\) = 3

Where 1 + tan\(^2\) \(\theta\) = sec\(^2\) \(\theta\)

1 + tan\(^2\) \(\theta\) + tan\(^2\) \(\theta\) = 3

2 tan\(^2\) \(\theta\) = 2

tan\(^2\) \(\theta\) = 1

tan\(\theta\) = √1

where √1 = 1

tan\(\theta\) = 1

And  tan 45°  = 1 

∴ \(\theta\) = 45°

752.

In a soccer competition in one season, a club had scored the following goals: 2, 0, 3, 3, 2, 1, 4, 0, 0, 5, 1, 0, 2, 2, 1, 3, 1, 4, 1 and 1. The mean, median and mode are respectively

A.

1, 1.8 and 1.5

B.

1.8, 1.5 and 1

C.

1.8, 1 and 1.5

D.

1.51, 1 and 1.8

E.

1.5, 1.8 and 1

Correct answer is B

By re-arranging the goals in ascending order 0. 0. 0. 0. 0, 1. 1. 1. 1. 1. 1, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5.

Mean = \(\frac{36}{20}\) = 1.8

Median = \(\frac{1 + 2}{2}\)

= \(\frac{3}{2}\)

= 1.5

Mode = 1

= 1.8, 1.5 and 1

753.

A regular hexagon is constructed inside a circle of diameter 12cm. The area of the hexagon is

A.

36\(\pi\)cm2

B.

54\(\sqrt{3}\)cm2

C.

\(\sqrt{3}\)cm2

D.

\(\frac{1}{x - 1}\)

Correct answer is B

Sum of interior angle of hexagon = [2(6) - 4]90°

= 720°

sum of central angle = 360°

Each central angle = \(\frac{360}{6}\)

= 60°

Area of Hexagon = \(\frac{1}{2}\) x 6 x 6 sin 60°

\(\frac{36 \times 6\sqrt{3}}{2 \times 2}\)

= \(54 \sqrt{3}\)cm2

754.

A triangle has angles 30°, 15° and 135°. The side opposite to the angle 30° is length 6cm. The side opposite to the angle 135° is equal to

A.

12cm

B.

6cm

C.

6\(\sqrt{2}\)cm

D.

12\(\sqrt{2}\)cm

Correct answer is C

\(\frac{6}{\sin 30}\) = \(\frac{x}{\sin 135}\)

\(\frac{6}{\sin 30}\) = \(\frac{x}{\sin 45}\)

x = \(\frac{6 \times \sin 45}{\sin 30}\)

= \(6 \sqrt{2}\)cm

755.

An arithmetic progression has first term 11 and fourth term 32. The sum of the first nine terms is

A.

351

B.

531

C.

135

D.

153

Correct answer is A

1st term a = 11, 4th term = 32

nth term = a + (n - 1)d

4th term = 11 = (4 - 1)d

= 11 + 3d

= 32

3d = 21

d = 7

sn = n(2a + (n - 1)d)

sn = \(\frac{9}{2}\)(2 \times 11) + (9 - 1)7

\(\frac{9}{2}\)(22 + 56) = \(\frac{9}{2}\) x 78

= 351