Given that a super additive function satisfies the following property:f(x1 + x2) ? f(x1) + f(x2)Checking for each option, i.e.Option 1:Let us take x1 = 2, and x2 = 4ex1+x2 = e 2+4 = e6 = 403.42\({e^{{x_1}}} + {e^{{x_2}}} = {e^2} + {e^4} = 61.98\) Since \({e^{{x_1} + {x_2}}} > {e^{{x_1}}} + {e^{{x_2}}}\)ex satisfies the superadditive property.Option 2:If \(\sqrt x \) is to be a super-additive function, then it must also satisfy:\(\sqrt {{x_1} + {x_2}} \ge \sqrt {{x_1}} + \sqrt {{x_2}} \)Again for x1 = 2 and x2 = 4, we get:\(\sqrt {2 + 4} = \sqrt 6 = 2.44\)\(\sqrt 2 + \sqrt 4 = 1.414 + 2 = 3.414\)\(\sqrt {{x_1} + {x_2}} < \sqrt {{x_1}} + \sqrt {{x_2}} \)? ?x does not follow the super-additive property.Option 3: 1/x For x1 = 2 and x2 = 4\(\frac{1}{{{x_1} + {x_2}}} = \frac{1}{6} = 0.166\)\(\frac{1}{{{x_1}}} + \frac{1}{{{x_2}}} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4} = 0.75\)Since, \(\frac{1}{{{x_1} + {x_2}}} < \frac{1}{{{x_1}}} + \frac{1}{{{x_2}}}\)1/x does not satisfy the superadditive property.Option 4:Checking for e-xFor x1 = 2, x2 = 4\({e^{ - \left( {{x_1} + {x_2}} \right)}} = {e^{ - 6}} = 0.0024\)\({e^{ - {x_1}}} = 0.135\)\({e^{ - {x_2}}} = 0.028\)Since, \({e^{ - \left( {{x_1} + {x_2}} \right)}} < {e^{ - {x_1}}} + {e^{ - {x_2}}}\)e-x does not satisfy the superadditive property.