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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.

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456.

Given that $^{n}P_{r} = 90$ and $^{n}C_{r} = 15$ , find the value of r.

View answer & explanation

Correct answer is B

$\frac{^{n}P_{r}}{^{n}C_{r}} = \frac{\frac{n!}{(n - r)!}}{\frac{n!}{(n - r)! r!}} = \frac{90}{15} = 6$

$\frac{n!}{(n - r)!} \times \frac{(n - r)! r!}{n!} = r! = 6$

$r = 3$

457.

In computing the mean of 8 numbers, a boy mistakenly used 17 instead of 25 as one of the numbers and obtained 20 as the mean. Find the correct mean

View answer & explanation

Correct answer is B

$Mean = \frac{sum of items}{total number of items}$

$20 = \frac{x}{8} \implies x = 160$

The sum of the 7 nos = 160 - 17 = 143

Correct mean = $\frac{143 + 25}{8} = \frac{168}{8} = 21$

458.

Find the unit vector in the direction of the vector $-12i + 5j$

$\frac{-12i}{13} - \frac{5j}{13}$

$\frac{-1i}{13} + \frac{5j}{13}$

$\frac{-12i}{13} + \frac{5j}{13}$

$\frac{-5i}{13} + \frac{12j}{13}$

View answer & explanation

Correct answer is C

$\hat{n} = \frac{\overrightarrow{n}}{|n|}$

$\hat{n} = \frac{-12i + 5j}{\sqrt{(-12)^{2} + (5)^{2}}}$

= $\frac{-12i}{13} + \frac{5j}{13}$

459.

Two forces 10N and 6N act in the directions 060° and 330° respectively. Find the x- component of their resultant.

$5\sqrt{3} - 3$

$3 - 5\sqrt{3}$

$5 - 3\sqrt{3}$

$3\sqrt{3} - 5$

View answer & explanation

Correct answer is A

$F = F \cos \theta i + F \sin \theta j$

$10N = 10 \cos 60 i + 10 \sin 60 j$

$6N = -6 \cos 330 i - 6 \sin 330 j$

$R_{x} = 10 \cos 60 - 6 \cos 330$

= $10 \times \frac{1}{2} - 6 \times \frac{\sqrt{3}}{2}$

= $5 - 3\sqrt{3}$

460.

Differentiate $\frac{x}{x + 1}$ with respect to x

$\frac{1}{x + 1}$

$\frac{1}{(x + 1)^{2}}$

$\frac{1 - x}{x + 1}$

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

View answer & explanation

Correct answer is B

$y = \frac{x}{x + 1}$

Using quotient rule because the function is of the form $\frac{u(x)}{v(x)}$

$\frac{\mathrm d y}{\mathrm d x} = \frac{v\frac{\mathrm d u}{\mathrm d x} - u\frac{\mathrm d v}{\mathrm d x}}{v^{2}}$

$\frac{\mathrm d y}{\mathrm d x} = \frac{(x + 1) . 1 - x . 1}{(x + 1)^{2}}$

= $\frac{1}{(x + 1)^{2}}$

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