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

Correct answer is E

\(\frac{k}{k + 1}\) + \(\frac{k + 1}{k}\)

= \(\frac{k^2 + (k + 1)^2}{k(k + 10}\)

= \(\frac{2k^2 + 2k + 1}{k(k + 1}\)

let k = \(\frac{1}{2}\)

p = \(\frac{10}{3}\)

p > 0

See all Mathematics questions & answers

See more JAMB questions & answers

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

Aptitude Tests