WAEC Mathematics Past Questions & Answers - Page 50

246.

Given that t = \(2 ^{-x}\), find \(2 ^{x + 1}\) in terms of t. 

A.

\(\frac{2}{t}\)

B.

\(\frac{t}{2}\)

C.

\(\frac{1}{2t}\)

D.

t

Correct answer is A

t = \(2^{-x} = \frac{1}{2^{x}}\)

\(\implies 2^{x} =\frac{1}{t}\)

\(2^{x+1} = 2^{x} \times 2^{1}\)

= \(\frac{1}{t} \times 2 = \frac{2}{t}\)

247.

Consider the statements: p it is hot, q: it is raining

Which of the following symbols correctly represents the statement "It is raining if and only if it it is cold"?

A.

p \(\iff\) \(\sim\)q

B.

p \(\iff\) q

C.

\(\sim\)p \(\iff\) \(\sim\)q

D.

q \(\iff\) \(\sim\)p

Correct answer is D

No explanation has been provided for this answer.

248.

Make x the subject of the relation d = \(\sqrt{\frac{6}{x} - \frac{y}{2}}\)

A.

x = \(\frac{6 + 12}{d^2 + y}\)

B.

x = \(\frac{12}{d^2 - y}\)

C.

x = \(\frac{12}{y} - 2d^2\)

D.

x = \(\frac{12}{2d^2 + y}\)

Correct answer is D

d = \(\sqrt{\frac{6}{x} - \frac{y}{2}}\)

\(d^2 = \frac{6}{x} - \frac{y}{2}\)

\(2xd^2 = 12 - xy\)

\(2xd^2 + xy = 12\)

x = \(\frac{12}{2d^2 + y}\)

249.

The roots of a quadratic equation are \(\frac{-1}{2}\) and \(\frac{2}{3}\). Find the equation.

A.

\(6x^2 - x + 2 = 0\)

B.

\(6x^2 - x - 2 = 0\)

C.

\(6x^2 + x - 2 = 0\)

D.

\(6x^2 + x + 2 = 0\)

Correct answer is B

(x + \(\frac{1}{2}\)) (n - \(\frac{2}{3}\))

\(x^2 - \frac{2}{3^x} + \frac{x}{2} - \frac{1}{3}\)

\(6x^2 - 4n + 3n - 2 = 0\)

\(6x^2 - x - 2 = 0\)

250.

Find the 6th term of the sequence \(\frac{2}{3} \frac{7}{15} \frac{4}{15}\),...

A.

-\(\frac{1}{3}\)

B.

-\(\frac{1}{5}\)

C.

-\(\frac{1}{15}\)

D.

\(\frac{1}{9}\)

Correct answer is A

a = \(\frac{2}{3}\), d = \(\frac{7}{15}\) - \(\frac{2}{3}\)

= 7 - 10

= \(\frac{-3}{15}\)

d = - \(\frac{-1}{5}\)

T6 = a + 5d

= \(\frac{2}{3}\) + 5(\(\frac{-1}{5}\)

= \(\frac{2}{3}\) - 1

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

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