WAEC Further Mathematics Past Questions & Answers - Page 95

471.

A polynomial is defined by \(f(x + 1) = x^{3} + px^{2} - 4x + 2\), find f(2)

A.

-8

B.

-2

C.

2

D.

8

Correct answer is C

Given: \(f(x + 1) = x^{3} + 3x^{2} - 4x + 2\). 

\(f(2) = f(x + 1) \implies x + 1 = 2; x = 1\)

\(f(2) = 1^{3} + 3(1)^{2} - 4(1) + 2 = 1 + 3 - 4 + 2 = 2\)

472.

If (x + 1) is a factor of the polynomial \(x^{3} + px^{2} + x + 6\). Find the value of p.

A.

-8

B.

-4

C.

4

D.

8

Correct answer is B

If (x + 1) is a factor, then f(-1) = 0.

\((-1)^{3} + p(-1)^{2} + (-1) + 6 = 0\)

\(-1 + p - 1 + 6 = 0 \implies p + 4 = 0\)

\(p = -4\)

473.

QRS is a triangle such that \(\overrightarrow{QR} = (3i + 2j)\) and \(\overrightarrow{SR} = (-5i + 3j)\), find \(\overrightarrow{SQ}\).

A.

8i + j

B.

2i - j

C.

-2i - 3j

D.

-8i - j

Correct answer is A

\(\overrightarrow{SQ} = \overrightarrow{SR} + \overrightarrow{RQ}\)

\(\overrightarrow{RQ} = -\overrightarrow{QR} = - (3i + 2j) = -3i - 2j\)

\(\overrightarrow{SQ} = (-5i + 3j) - 3i - 2j = -8i + j\)

474.

Evaluate \(\log_{10}(\frac{1}{3} + \frac{1}{4}) + 2\log_{10} 2 + \log_{10} (\frac{3}{7})\)

A.

-3

B.

0

C.

\(\frac{5}{6}\)

D.

1

Correct answer is B

\(\log_{10} (\frac{1}{3} + \frac{1}{4}) + 2\log_{10} 2 + \log_{10} (\frac{3}{7})\)

\(\frac{1}{3} + \frac{1}{4} = \frac{7}{12}\)

= \(\log_{10} (\frac{7}{12} \times 2^{2} \times \frac{3}{7})\)

= \(\log_{10} 1 = 0\)

475.

Given that \(\sin x = \frac{-\sqrt{3}}{2}\) and \(\cos x > 0\), find x

A.

300°

B.

240°

C.

120°

D.

60°

Correct answer is A

In order to do this, simply find the option in the range where only the cos is +ve. This occurs in the range \(270° \leq x \leq 360°\).

Check: \(\sin 300 = - \sin 60 = \frac{-\sqrt{3}}{2}\)

\(\cos 300 = \cos 60 = \frac{1}{2}\)