complete question is Let f:R→R is defined by f(x)= 1+∣x∣x. Then f(x) is option (a) Injective but not surjective(b)surjective but not Injective(b)Injective as well as surjective(b)Neither Injective nor surjective−−−−−−−−−−−−−−−−−−−−−−−−−−−−solution correct option is Aexplain:−for x<0f(x)=1−xxf(x1)=f(x2) implise 1−x1x1=1−x2x2x1−x1.x2=x2−x2.x1x1=x2...(i)and for x>0f(x)=1+xxf(x1)=f(x2) implise 1+x1x1=1+x2x2x1+x1.x2=x2+x2.x1x1=x2...(ii)now according to given f(x)so there not exist preimage for all real numbers∣f(x)∣<1...(iii)hence from i, ii and iii,f(x) is injective but not surjective.
Comments
Leave a comment