JAMB - Mathematics (1981 - No. 42)
What is the length of an arc of a circle that substends 2\(\frac{1}{2}\) radians at the centre when the raduis of the circle = \(\frac{k}{k + 1}\) + \(\frac{k + 1}{k}\) then
p< 0
p\(\geq\) 0
p \(\leq\) 0
p < 1
p > 0
Explanation
\(\frac{k}{k + 1}\) + \(\frac{k + 1}{k}\)
= \(\frac{k^2 + (k + 1)^2}{k(k + 10}\)
= \(\frac{2k^2 + 2k + 1}{k(k + 1}\)
let k = \(\frac{1}{2}\)
p = \(\frac{10}{3}\)
p > 0
= \(\frac{k^2 + (k + 1)^2}{k(k + 10}\)
= \(\frac{2k^2 + 2k + 1}{k(k + 1}\)
let k = \(\frac{1}{2}\)
p = \(\frac{10}{3}\)
p > 0
Comments (0)
