Letf: (0,1)Rbe a function and leta(0,1).Match eachstatement in Group A with a statement from Group B which meansthe same thing.Group A:(i) >0, >0 such that|xa|< implies|f(x)f(a)|< .(ii) >0, >0,|xa|< implies|f(x)f(a)|< .(iii) >0 such that >0,|xa|< implies|f(x)f(a)|< .(iv) >0 and >0 such that|xa|< implies|f(x)f(a)|< .(v) >0, >0 such that|xa|< implies|f(x)f(a)|< .(vi) >0 such that >0,|xa|< implies|f(x)f(a)|< .Group B:(a)fis continuous ata.(b)fis bounded on (0,1).(c)fis constant on (0,1)(d) There is some neighbourhood ofaon whichfis bounded.(e) There is some neighbourhood ofaon whichfis constant.Solution:(i) (a)(ii) (c)(iii) (b)(iv) (d)(v) (b)(vi) (e).