WAEC Further Mathematics Past Questions & Answers - Page 40

197.

If \(^nC_2\) = 15, find the value of n

A.

8

B.

7

C.

6

D.

5

Correct answer is C

\(^nC_2\) = 15

\(\frac{n!}{(n - 3)! 2!}\) = 15

\(\frac{n(n - 1)(n - 2)!}{(n - 2)!2}\) = 15 

n\(^2\) - n = 30

n\(^2\) - n - 30 = 0

(n - 6)(n + 5) = 0

n = 6 or n = -5 

198.

Rationalize; \(\frac{1}{\sqrt{2 + 1}}\)

A.

\(\sqrt{2}\) - 1

B.

1 - \(\sqrt{2}\)

C.

\(\frac{\sqrt{2} - 1}{2}\)

D.

\(\frac{1 - \sqrt{2}}{2}\)

Correct answer is A

\(\frac{1}{\sqrt{2} + 1}\) x \(\frac{\sqrt{2} - 1}{\sqrt{2} - 1}\)

= \(\frac{\sqrt{2} - 1}{2 - 1}\)

= \(\frac{\sqrt{2} - 1}{1} = \sqrt{2} - 1\)

199.

Evaluate tan 75\(^o\); leaving the answer in surd form (radicals) 

A.

\(\sqrt{3 + 2}\)

B.

\(\sqrt{3 + 1}\)

C.

\(\sqrt{3 - 1}\)

D.

\(\sqrt{3 - 2}\)

Correct answer is D

Tan 75\(^o\) = Tan (45\(^o\) + 30\(^o\))

= \(\frac{\tan 45^o + \tan 30^o}{1 - \tan 45^o \tan 30^o}\)

= \(\frac{\sqrt{3} + 1}{\sqrt{3}  - 1}\)

RATIONALIZE THE DENOMINATOR

= \(\frac{\sqrt{3} + 1}{\sqrt{3}  - 1}\) X \(\frac{\sqrt{3} + 1}{\sqrt{3}  +1}\)

= \(\frac{4 + 2\sqrt{3}}{3  - 1}\)

= \(\frac{2(2 + \sqrt{3})}{2}\)

= 2 + \(\sqrt{3}\)

200.

Solve; \(\frac{P}{2} + \frac{k}{3}\) = 5 and 2p = k = 6 simultaneously

A.

p = -6, k = -6

B.

p = -6, k = 6

C.

p = 6, k = 6

D.

p = 6, k = -6

Correct answer is C

\(\frac{P}{2} + \frac{k}{3}\) = 5

\(\frac{3p + 2k}{6}\) = 5

2p + 3k = 30

- 2p - k = 6

\(\overline{\frac{4k}{4} = \frac{24}{4}}\) 

k = 6

 

From 2p - k = 6

2p - 6 = 6

\(\frac{2p}{2} = \frac{12}{2}\)

p = 6