Match List-I with List-II.
| List-I | List-II | ||
| (A) | $f(x)=\frac{1}{x}, R-\begin{Bmatrix}0 \end{Bmatrix}→R-\begin{Bmatrix}0 \end{Bmatrix}$ | (I) | neither injective nor subjective |
| (B) | $f(x)=x^2, f: N →N$ | (II) | subjective but not injective |
| (C) | $f(x)=x^2, f: R →R$ | (III) | injective but not surjective |
| (D) | $f: \begin{Bmatrix}1, 2, 3 \end{Bmatrix}→\begin{Bmatrix}1, 2\end{Bmatrix}$defined as $f:\begin{Bmatrix} (1, 1), (2, 2), (3, 1)\end{Bmatrix}$ | (IV) | injective and surjective |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 4
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)