Question

If `f : R -> R^+  U {0}` be defined by `f(x) = x^2, x ∈ R`. The mapping is

विकल्प

• Injective but ont surjective

• Surjective but not injective

• Both injective and surjective

• Neither injective nor surjective

Answer 1

Surjective but not injective

Explanation:

Given `f(x) = x^2`

Where `f : R -> R^(-1) ∪ {0}`

Let `x_1, x_2 ∈ R`

Now let `f(x_1) = f(x_2)`

⇒ `x_1^2 = x_2^2` ⇒ `x_1 = +- x_2`

`f(x)` is many one

For onto

Let `f(x) = x^2` ⇒ `y = x^2` 

or `x = +- sqrt(y)`

∴ `y ∈ R^+ ∪ {0}`

Hence `x ∈ R`

So `f(x)` is onto.