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

  • A p
  • B p\(\geq\) 0
  • C p \(\leq\) 0
  • D p
  • E p > 0
jamb 1981mathematics

The correct answer is E. p > 0

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

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

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

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

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

p > 0

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