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.

206.

A lift moving upwards with a uniform acceleration of 5\(ms^{-2}\) carries a body of mass p kg. If the reaction on the floor is 480 N, find the value of p. [Take g = \(10 ms^{-2}\)].

A.

32

B.

36

C.

48

D.

64

Correct answer is A

Reacting force - Weight = Net force

R = W + ma = mg + ma

480 = 10p + 5p

480 = 15p

p = 32 kg

207.

If \(a = \begin{pmatrix} 3 \\ 2 \end{pmatrix}\) and \(b = \begin{pmatrix} -3 \\ 5 \end{pmatrix}\), find a vector c such that \(4a + 3c = b\).

A.

\(\begin{pmatrix} 3 \\ -1 \end{pmatrix}\)

B.

\(\begin{pmatrix} -5 \\ -1 \end{pmatrix}\)

C.

\(\begin{pmatrix} -5 \\ 1 \end{pmatrix}\)

D.

\(\begin{pmatrix} -5 \\ -9 \end{pmatrix}\)

Correct answer is B

\(4 \begin{pmatrix} 3 \\ 2 \end{pmatrix} + 3 \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -3 \\ 5 \end{pmatrix}\)

\(\begin{pmatrix} 12 \\ 8 \end{pmatrix} + \begin{pmatrix} 3x \\ 3y \end{pmatrix} = \begin{pmatrix} -3 \\ 5 \end{pmatrix}\)

\(\begin{pmatrix} 3x \\ 3y \end{pmatrix} = \begin{pmatrix} -3 - 12 \\ 5 - 8 \end{pmatrix}\)

\(\begin{pmatrix} 3x \\ 3y \end{pmatrix} = \begin{pmatrix} -15 \\ -3 \end{pmatrix}\)

\(\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -5 \\ -1 \end{pmatrix}\)

208.

The mean and median of integers x, y, z and t are 5 and z respectively. If x < y < z < t and y = 4, find (x + t).

A.

12

B.

11

C.

10

D.

8

Correct answer is A

\(Mean : \frac{x + 4 + z + t}{4} = 5 \implies x + 4 + z + t = 20\)

\(\implies x + z + t = 16 ... (1)\)

\(Median : \frac{4 + z}{2} = z \implies 4 + z = 2z\)

\(4 = z\)

From 1, 

\(\implies x + 4 + t = 16 \)

\(x + t = 12\)

209.

Find, in surd form, the value of \(\cos 165\).

A.

\(\frac{1}{4}(\sqrt{6} + \sqrt{2})\)

B.

\(\frac{1}{4}(\sqrt{6} - \sqrt{2})\)

C.

\(-\frac{1}{4}(\sqrt{6} - \sqrt{2})\)

D.

\(-\frac{1}{4}(\sqrt{6} + \sqrt{2})\)

Correct answer is D

\(\cos 165 = -\cos (180 - 165) = -\cos 15\)

\(\cos 15 = \cos (45 - 30)\)

\(\cos (x - y) = \cos x \cos y + \sin x \sin y\)

\(\cos (45 - 30) = \cos 45 \cos 30 + \sin 45 \sin 30\)

= \((\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) + (\frac{\sqrt{2}}{2})(\frac{1}{2})\)

= \(\frac{1}{4}(\sqrt{6} + \sqrt{2})\)

\(\therefore \cos 165 = -\frac{1}{4}(\sqrt{6} + \sqrt{2})\)

210.

The parallelogram PQRS has vertices P(-2, 3), Q(1, 4), R(2, 6) and S(-1,5). Find the coordinates of the point of intersection of the diagonals.

A.

\((-1, 5)\)

B.

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

C.

\((0, 4\frac{1}{2})\)

D.

\((1\frac{1}{2}, 5)\)

Correct answer is C

No explanation has been provided for this answer.