JAMB Mathematics Past Questions & Answers - Page 101

The coordinate of the midpoint is $(\frac{-3 + 4}{2}, \frac{5 + (-1)}{2})$

= $(\frac{1}{2}, 2)$

Let the extension be E and the tension be T.

Then $E \propto T$

$E = kT$

when T = 8N, E = 2cm

$2 = k \times 8$

$k = \frac{2}{8} = 0.25$

$\therefore E = 0.25T$

when T = 12N, $E = 0.25 \times 12 = 3cm$

log₅20 = x

5x = 20(Take log₁₀ of both sides)

log_5x = log₂₀

xlog₅ = log₂₀

x= [log₂₀ ÷ log₅]

[1.30103 ÷ 0.69897]

x = 1.861

nth term of a linear sequence (AP) = a+(n − 1)d

first term = 6, last term = 10 sum − 40

i.e. a = 6, l = 10, S = 40

S$_{n}$= n/2(2a + (n − 1)d or Sn = ÷2 (a + l)

S$_{n}$ = n/2(a + l)

40 = n/2(6 + 10)

40 = 8n

8n = 40

8n = 40

n = 40/8

= 5

The number of terms = 5

Using the quadratic formula, we have

$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$

From the equation $3x^{2} - 4x - 5 = 0$, a = 3, b = -4 and c = -5.

$\therefore x = \frac{-(-4) \pm \sqrt{(-4)^{2} - 4(3)(-5)}}{2(3)}$

= $\frac{4 \pm \sqrt{16 + 60}}{6}$

= $\frac{4 \pm \sqrt{76}}{6}$

= $\frac{4 \pm 8.72}{6}$

= $\frac{4 + 8.72}{6} \text{ or } \frac{4 - 8.72}{6}$

= $\frac{12.72}{6} \text{ or } \frac{-4.72}{6}$

$x = \text{2.12 or -0.79}$