Match the LIST-I with LIST-II (Consider N and R as set of natural numbers and set of real numbers respectively)
| LIST-I | LIST-II | ||
| Function | |||
| A. | \( f:N\rightarrow N,f(x)=x^2 \) is | I. | Neither one-one nor onto |
| B. | \( f:R\rightarrow R,f(x)=3-2x \) is | II. | one-one but not onto |
| C. | \( f:R\rightarrow R,f(x)=x^2 \) is | III. | onto but not one-one |
| D. | \( f:\{-1,1\}\rightarrow\{1\},f(x)=|x| \) is | IV. | one-one and onto |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A - II, B - IV, C - I, D - III
(A) \( f:N\rightarrow N,f(x)=x^2 \) is
This matches (II) one-one but not onto.
For natural numbers, \( x_1^2=x_2^2 \) gives \( x_1=x_2 \), so f is one-one. The value 2 is not a perfect square, so it has no pre-image and f is not onto.
(B) \( f:R\rightarrow R,f(x)=3-2x \) is
This matches (IV) one-one and onto.
\( 3-2x_1=3-2x_2 \) gives \( x_1=x_2 \). For any real y, \( x=\frac{3-y}{2} \) gives f(x) = y, so f is onto.
(C) \( f:R\rightarrow R,f(x)=x^2 \) is
This matches (I) Neither one-one nor onto.
f(−1) = f(1) = 1, so f is not one-one; −1 has no pre-image, since \( x^2\ge 0 \), so f is not onto.
(D) \( f:\{-1,1\}\rightarrow\{1\},f(x)=|x| \) is
This matches (III) onto but not one-one.
f(−1) = f(1) = 1: two elements share one image, so f is not one-one; the only element 1 of the co-domain is an image, so f is onto.
So the matching is A - II, B - IV, C - I, D - III.
Tip: One-one and onto depend on the domain and co-domain, not only on the formula: \( x^2 \) is one-one on N but not on R.