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.

306.

Two statements are represented by p and q as follows:

p : He is brilliant; q : He is regular in class

Which of the following symbols represent "He is regular in class but dull"?

A.

\(q \vee \sim p\)

B.

\(q \edge \sim p\)

C.

\(\sim q \edge \sim p\)

D.

\(\sim q \vee \sim p\)

Correct answer is B

No explanation has been provided for this answer.

307.

If \(P = \begin{pmatrix} 1 & -2 \\ 3 & 4 \end{pmatrix}\) and \(Q = \begin{pmatrix} -2 & 3 \\ 1 & 0 \end{pmatrix}\), find PQ.

A.

\(\begin{pmatrix} 4 & 1 \\ -2 & 9 \end{pmatrix}\)

B.

\(\begin{pmatrix} -4 & 1 \\ 2 & 9 \end{pmatrix}\)

C.

\(\begin{pmatrix} -4 & 3 \\ -2 & 13 \end{pmatrix}\)

D.

\(\begin{pmatrix} -4 & 3 \\ -2 & 9 \end{pmatrix}\)

Correct answer is D

\(\begin{pmatrix} 1 & -2 \\ 3 & 4 \end{pmatrix} \(\begin{pmatrix} -2 & 3 \\ 1 & 0 \end{pmatrix}\)

\(\begin{pmatrix} (1 \times -2) + (-2 \times 1) & (1 \times 3) + (-2 \times 0) \\ (3 \times -2) + (4 \times 1) & (3 \times 3) + (4 \times 0) \end{pmatrix}\)

= \(\begin{pmatrix} -4 & 3 \\ -2 & 9 \end{pmatrix}\)

308.

Express \(\frac{7\pi}{6}\) radians in degrees.

A.

315°

B.

210°

C.

105°

D.

75°

Correct answer is B

\(\pi = 180°\)

\(\frac{7\pi}{6} = \frac{7 \times 180}{6} \)

= \(210°\)

309.

A rectangle has a perimeter of 24m. If its area is to be maximum, find its dimension.

A.

12, 12

B.

6, 6

C.

4, 8

D.

9, 3

Correct answer is B

\(Perimeter = 2(l + b) = 24\)

\(l + b = 12 \implies l = 12 - b\)

\(Area = (12 - b) \times b = 12b - b^{2}\)

\(\frac{\mathrm d A}{\mathrm d b} = 12 - 2b = 0\) (at maximum)

\(2b = 12 \implies b = 6\)

\(l = 12 - 6 = 6m\)

310.

Evaluate \(\int_{1}^{2} (2 + 2x - 3x^{2}) \mathrm {d} x\).

A.

-2

B.

2

C.

8

D.

10

Correct answer is A

\(\int_{1}^{2} (2 + 2x - 3x^{2}) \mathrm {d} x\)

= \((2x + x^{2} - x^{3})|_{1}^{2}\)

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

= \(0 - 2 = -2\)