BITSAT Mathematics Question (2024) — Solution

Step-by-step solution

Given that, f(x)= x 1+x^2 For injective: Let x_1, x_2 R such that f (x_1 )=f (x_2 ) x_1 1+x_1^2 = x_2 1+x_2^2 x_1^2 1+x_1^2 = x_2^2 1+x_2^2 x_1^2+x_1^2 x_2^2=x_2^2+x_1^2 x_2^2 x_1^2=x_2^2 x_1=x_2 So, f(x) is injective. For surjective: Let y= x 1+x^2 y^2 (1+x^2 )=x^2 y^2+y^2 x^2=x^2 x^2 (1-y^2 )=y^2 x= y^2 1-y^2 y^2 1-y^2 0 y (-1,1) So, f(x) is not surjective.