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
Correct Answer: Option E
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