JAMB Mathematics Past Questions & Answers - Page 266

2x + 3 \(\neq\) 2x - 3 for any value of x

∴ for the \(\bigtriangleup\) to be isosceles, either

2x - 3 = x + 3 or 2x + 3 = x + 3

solve the two equations we arrive at

x = 6 or x = 0

When x = 6, the sides are 9, 15, 9

When x = 0, the sides are 3, 4, -3 since lengths of a \(\bigtriangleup\)can never be negative then the value of x = 6

Diameter = 8cm

∴ Radius = 4cm

Length of arc = \(\frac{\theta}{360}\) x 2 \(\pi\)r but Q = 22\(\frac{1}{2}\)

∴ Length \(\frac{22\frac{1}{2}}{360}\) x 2 x \(\pi\) x 4

= \(\frac{22\frac{1}{2} \times 8\pi}{360}\)

= \(\frac{180}{360}\)

= \(\frac{\pi}{2}\)

A rectangular polygon has each interior angle to be 150^o

let the polygon has n-sides

therefore, Total interior angle 150 x n = 150n

hence 150n = (2n - 4)90

150n = 180n - 360

360 = (180 - 150)n

30n = 360

n = 12

Initial salary = N540 increment = N36 (every 6 months) Period of increment = 2 yrs and 6 months amount(increment) = N36 x 5 = N180 The man's new salary = N540 = N180 = N720.00

\(\frac{2k - 1}{k + 1} = \frac{3k + 1}{2k - 1}\)

\((k + 1)(3k + 1) = (2k - 1)(2k - 1)\)

\(3k^{2} + 4k + 1 = 4k^{2} - 4k + 1\)

\(4k^{2} - 3k^{2} - 4k - 4k + 1 - 1 = 0\)

\(k^{2} - 8k = 0\)

\(k(k - 8) = 0\)

\(\therefore \text{k = 0 or 8}\)

The terms of the sequence given k = 0: (1, -1, 1)

\(\implies \text{The common ratio r = -1}\)

The terms of the sequence given k = 8: (9, 15, 25)

\(\implies \text{The common ratio r = } \frac{5}{3}\)

The possible values of the common ratio are -1 and \(\frac{5}{3}\).