Class 12 Relations And Functions · Case Study4 Markstypes_of_functions
Class 12 Maths · Relations And Functions · Case Study #200
A parent discusses the functions f:R→R, f(x)=x, and g:R→R, g(x)=x².
Part (1) 2 Marks
Prove that f is bijective.
Answer: f is bijective.
- If f(u)=f(v), then u=v immediately, proving injectivity.
- For every y∈R, the input x=y gives f(x)=y, proving surjectivity.
- Thus the identity function f is bijective.
Part (2) 2 Marks
Is g bijective? Justify.
Answer: No; neither one-one nor onto R.
- The distinct inputs 1 and −1 both have image 1, so g is not injective.
- Since x²≥0 for every real x, the codomain element −1 has no preimage; g is not onto R.
- Therefore g is not bijective.