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

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