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.

371.

Two forces (2i - 5j)N and (-3i + 4j)N act on a body of mass 5kg. Find in \(ms^{-2}\), the magnitude of the acceleration of the body.

A.

\(\frac{\sqrt{2}}{5}\)

B.

\(5\sqrt{2}\)

C.

\(2\sqrt{5}\)

D.

\(\frac{5\sqrt{2}}{2}\)

Correct answer is A

\(F = F_{1} + F_{2}\)

\((2i - 5j) + (-3i + 4j) = (-i - j)\)

\(F = ma \implies (-1, -1) = 5a\)

\(a = (-\frac{1}{5}, -\frac{1}{5})\)

\(|a| = \sqrt{(\frac{-1}{5})^2 + (\frac{-1}{5})^2} = \sqrt{2}{25}\)

\(|a| = \frac{\sqrt{2}}{5} ms^{-2}\)

372.

Yomi was asked to label four seats S, R, P, Q. What is the probability he labelled them in alphabetical order?

A.

\(\frac{1}{24}\)

B.

\(\frac{1}{6}\)

C.

\(\frac{2}{13}\)

D.

\(\frac{1}{4}\)

Correct answer is A

The number of arrangements for the 4 letters = \(^{4}P_{4} = \frac{4!}{(4 - 4)!}\)

\(4! = 24\)

Alphabetical order is just 1 of the arrangements for the letters 

= \(\frac{1}{24}\)

373.

Find the direction cosines of the vector \(4i - 3j\).

A.

\(\frac{9}{10}, \frac{27}{10}\)

B.

\(\frac{17}{27}, -\frac{17}{27}\)

C.

\(\frac{4}{5}, -\frac{3}{5}\)

D.

\(\frac{4}{7}, \frac{-3}{7}\)

Correct answer is C

Given \(V = xi +yj\), the direction cosines are \(\frac{x}{|V|}, \frac{y}{|V|}\).

\(|4i - 3j| = \sqrt{4^{2} + (-3)^{2}} = \sqrt{25} = 5\)

Direction cosines = \(\frac{4}{5}, \frac{-3}{5}\).

374.

If \(\overrightarrow{OA} = 3i + 4j\) and \(\overrightarrow{OB} = 5i - 6j \) where O is the origin and M is the midpoint of AB, find OM

A.

-2i - 10j

B.

-2i + 2j

C.

4i - j

D.

4i + j

Correct answer is C

\(\overrightarrow{OA} = (3, 4)\)

\(\overrightarrow{OB} = (5, -6)\)

\(\overrightarrow{OM} = (\frac{3 + 5}{2}, \frac{4 + (-6)}{2})\)

= \((4, -1) = 4i - j\)

375.

Find the least value of n for which \(^{3n}C_{2} > 0, n \in R\)

A.

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

B.

\(\frac{1}{6}\)

C.

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

D.

1

Correct answer is C

\(^{3n}C_{2} > 0 \implies \frac{3n!}{(3n - 2)! 2!} > 0\)

\(\frac{3n(3n - 1)(3n - 2)!}{(3n - 2)! 2} > 0\)

\(\frac{3n(3n - 1)}{2} > 0\)

\(3n(3n - 1) > 0 \implies n > 0; n > \frac{1}{3}\)

The least number in the option that satisfies \(n > 0; n > \frac{1}{3} = \frac{2}{3}\)