complete \space question \space is \space \\ Let \space f:R→R \space is \space defined \space by \space f(x)= \space \frac{x}{1+∣x∣}.\\ \space Then \space f(x) \space is \space \\ option \space \\(a) \space Injective \space but \space not \space surjective\\ (b)surjective \space but \space not \space Injective\\ (b)Injective \space as \space well \space as \space surjective\\ (b)Neither \space Injective \space nor \space \space surjective\\ ----------------------------\\ solution \space \\ correct \space option \space is \space A\\ explain:-\\ for \space x<0\\ f(x)=\frac{x}{1-x}\\ f(x_1)=f(x_2) \space implise \space \\ \frac{x_1}{1-x_1}=\frac{x_2}{1-x_2}\\ x_1-x_1.x_2=x_2-x_2.x_1\\ x_1=x_2...(i)\\ and \space for \space x>0\\ f(x)=\frac{x}{1+x}\\ f(x_1)=f(x_2) \space implise \space \\ \frac{x_1}{1+x_1}=\frac{x_2}{1+x_2}\\ x_1+x_1.x_2=x_2+x_2.x_1\\ x_1=x_2...(ii)\\ now \space according \space to \space given \space f(x)\\ so \space there \space not \space exist \space preimage \space for \space all \space real \space numbers\\ |f(x)|<1...(iii) hence \space from \space i, \space ii \space and \space iii,\\ f(x) \space is \space injective \space but \space not \space surjective.\\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 A explain:− for x<0 f(x)= 1−x x ​ f(x 1 ​ )=f(x 2 ​ ) implise 1−x 1 ​ x 1 ​ ​ = 1−x 2 ​ x 2 ​ ​ x 1 ​ −x 1 ​ .x 2 ​ =x 2 ​ −x 2 ​ .x 1 ​ x 1 ​ =x 2 ​ ...(i) and for x>0 f(x)= 1+x x ​ f(x 1 ​ )=f(x 2 ​ ) implise 1+x 1 ​ x 1 ​ ​ = 1+x 2 ​ x 2 ​ ​ x 1 ​ +x 1 ​ .x 2 ​ =x 2 ​ +x 2 ​ .x 1 ​ x 1 ​ =x 2 ​ ...(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.