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Correct answer: (C) f(x) >= 2
By AM-GM, (e^x + e^(-x))/2 >= sqrt(e^x * e^(-x)) = sqrt(1) = 1, so e^x + e^(-x) >= 2, with equality at x = 0. As x -> +/- infinity the function grows without bound. Hence the range is [2, infinity), i.e. f(x) >= 2.