A function f :

Solution:

(a) Given:

f (x) = x2     

Range of f = R+     (Set of all real numbers greater than or equal to zero)

(b) Given:

f (x) = x2   

⇒ x2 = 4

⇒ x = ± 2

∴ {x : f (x) = 4 } = { -">−- 2, 2}.

(c) { y : f (y) = -">−-1}

⇒ f (y) = -">−1   

It is clear that $x^{2}=-1$ but $x^{2} \geq 0$.

$\Rightarrow f(y) \neq-1$

 

$\therefore\{y: f(y)=-1\}=\Phi$