1. f(x) = 2x – 3
f(x1) = f(x2)
⇒ 2x1 – 3 = 2x2 – 3 ⇒ x1 = x2
Therefore f is one-one.
Let f(x) = yeR, then
y = 2x – 3 ⇒ 2x = y + 3 ⇒ x = \(\frac{y+3}{2}\) ∈ R
Therefore f is onto. Then f-1(x) = \(\frac{x+3}{2}\).
2. Option (c)
[(b),(c),(d) are not one-one since for different values of x, we have same value of y. ie; horizontal line meets at more than one point, (b) and (d) are not onto since range and codomain are different].
3. f(x) = x2, f : R → [0, ∞) (Any other function).