Further Mathematics questions and answers

Further Mathematics Questions and Answers

Test your knowledge of advanced level mathematics with this aptitude test. This test comprises Further Maths questions and answers from past JAMB and WAEC examinations.

141.

If the binomial expansion of (1 + 3x)\(^6\) is used to evaluate (0.97)\(^6\), find the value of x. 

A.

0.03

B.

0.01

C.

-0.01

D.

-0.03

Correct answer is C

(1 + 3x )\(^6\) = (0.97)\(^6\) 

1 + 3x = 0.97 

3x = 0.97.1

\(\frac{3x}{3} = \frac{0.03}{3}\)

x = -0.01

142.

In which of the following series can be the formula S = \(\frac{a}{1 - r}\) where a is the first term and r is the common ratio, be used to find the sum of all the terms? 

A.

4 + 8 + 16 + 32 + ...

B.

\(\frac{1}{2}\) + 2 \(\frac{1}{2}\) + 12\(\frac{1}{2}\) + 62 \(\frac{1}{2}\) + ..

C.

\(\frac{4}{81}\) + \(\frac{2}{27}\) + \(\frac{1}{9}\) + \(\frac{1}{6}\) + ...

D.

128 + 64 + 32 + 16 + ...

Correct answer is D

No explanation has been provided for this answer.

143.

A stone was dropped from the top of a building 40m high. Find, correct to one decimal place, the time it took the stone to reach the ground. [Take g = 9.8ms\(^{-2}\)]

A.

2.9.seconds

B.

2.8 seconds

C.

2.6 seconds

D.

1.4 seconds

Correct answer is A

s = \(\frac{1}{2}gt^2\) 

40 = \(\frac{1}{2} \times 9.8 \times t^2\)

\(\frac{80}{9.8} = t^2\)

t = \(\sqrt{8.16326}\)

t  2.9 secs

144.

Given that F = 3i - 12j, R = 7i + 5j and N = pi + qj are forces acting on a body, if the body is in equilibrium. find the values of p and q.

A.

p=-10, q=7

B.

p=-10, q=-7

C.

p=10, q=- 7

D.

p-10, q=7

Correct answer is C

If the body is in equilibrium then 

F + R = N 

3i - 12j + 7i + 5j = pi + qj

10i - 7j = pi + qj

p = 10, q = -7

145.

Find the angle between i + 5j and 5i - J

A.

0\(^o\)

B.

45\(^o\)

C.

60\(^o\)

D.

90\(^o\)

Correct answer is D

a = i + 5j and b = 5i - j

cos\(\theta\) = \(\frac{a.b}{|a|.|b|}\)

= \(\frac{(1 \times 5) + (5x - 1)}{(\sqrt{1^2 + 5^2}) (5^2 + (-1))^2}\)

= \(\frac{5 - 5}{\sqrt{26}\times \sqrt{26}}\) = 0

x = cos\(^{-1}\)(0), x = 90\(^o\)