Given, f(x)=tan(x+1+4π)For x+1 to be defined, we must have x+1≥0, i.e. x≥−1.For x≥−1, x+1≥0, hence x+1+4≥4.⇒0<x+1+41≤41⇒0<x+1+4π≤4πSo the argument of tan lies in (0,4π], an interval on which tan is strictly increasing.As x→∞, x+1→∞, so x+1+4π→0+ and tan→0+, but 0 is never attained.At x=−1, x+1=0, so x+1+4π=4π and f(−1)=tan4π=1.Therefore the range of the given function is (tan(0),tan(4π)], which is (0,1].∴ The range of f(x) is (0,1], i.e. option B is correct.