JAMB Mathematics Past Questions & Answers

901.

If sin $\theta$ = $\frac{m - n}{m + n}$ ; Find the value of 1 + tan2 $\theta$

$\frac{(m^2 + n^2)}{m + n}$

$\frac{(m^2 + n^2 + 2mn)}{4mn}$

$\frac{2(m^2 + n^2 + mn)}{m + n}$

$\frac{(m^2 + n^2 + mn)}{m + n}$

View answer & explanation

Correct answer is B

$(m + n)^{2} = (m - n)^{2} + x^{2}$

$m^{2} + 2mn + n^{2} = m^{2} - 2mn + n^{2} + x^{2}$

$x^{2} = 4mn$

$x = \sqrt{4mn} = 2\sqrt{mn}$

1 + tan2 $\theta$ = sec2 $\theta$

= $\frac{1}{cos^2\theta}$

$\cos \theta = \frac{2\sqrt{mn}}{(m + n)}$

$\frac{1}{\cos \theta} = \frac{(m + n)}{2\sqrt{mn}}$

$\sec^{2} \theta = \frac{(m + n)^{2}}{4mn}$

= $\frac{(m^2 + n^2 + 2mn)}{4mn}$

902.

Given that p:q = $\frac{1}{3}$ : $\frac{1}{2}$ and q:r = $\frac{2}{5}$ , find p:r

4:105

7:15

20:21

2:35

3:20

View answer & explanation

Correct answer is B

$p : q = \frac{1}{3} : \frac{1}{2}$

$\frac{p}{q} = \frac{2}{3}$

$2q = 3p ... (1)$

$q : r = \frac{2}{5} : \frac{4}{7}$

$\frac{q}{r} = \frac{2}{5} \times \frac{7}{4} = \frac{7}{10}$

$10q = 7r ... (2)$

Eliminating q, we have

(1) : $2q = 3p$

$10q = 15p$

$\implies 15p = 7r$

$\therefore \frac{p}{r} = \frac{7}{15}$

$p : r = 7 : 15$

903.

A sector of a circle is bounded by two radii 7cm long and an arc length 6cm. Find the area of the sector.

42cm²

3cm²

21cm²

24cm²

12cm²

View answer & explanation

Correct answer is C

Length of arc = $\frac{\theta}{360}$ x 2 $\pi$ r = 6

$\theta$ x 2 $\pi$ r = 360 x 6

$\theta$ = $\frac{360 \times 6}{2\pi r}$

Area of the sector = $\frac{\theta}{360}$ x $\pi$ r²

$\frac{360 \times 6}{2\pi r}$ x $\frac{1}{360}$ x $\pi$ r² = r

= 3 x 7

= 21cm²