Correct answer is (a) True
For explanation I would say: The endurance of cruise method 2 is identical to that of cruise method 1. E=\(\Big[\frac{E_{max}}{C}\Big]\Big\{\frac{2u^2}{u^4+1}\Big\}l_n\omega\) where V is true airspeed, C is specific fuel consumption, L is lift, D is drag, Emax is endurance, u is relative airspeed and ω is fuel ratio.