Match List-I with List-II.

Choose the correct answer from the options given below :

(A)-(IV), (B)-(III), (C)-(I), (D)-(II)

The correct answer is option (4) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

(A) $f(x)=\frac{1}{x}$ so $f(x_1)=f(x_2)⇒x_1=x_2$

for every $f(x)$ there exist one x in $R-\{0\}$ (IV)

(B) $f(x)=x^2$, injective as $f:N→N$

But not surjective as eq. for $f(x)=7$ x doesn't exist (III)

(C) $f(x)=x^2$, for $f(x)=1$

$x=-1$ as well as 1 (Not injective)

for $f(x)=-4$

x doesn't exist (Not injective) (I)

(D) Not injective

as for $y = 1,x=1,3$

yes surjective as for every $y∈\{1,2\}$ there exists atleast one $x∈\{1,2,3\}$ (II)