Match List-I with List-II
Let $f: A → B$ be a function given by $f(x) = x^2$
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List-I Domain and Co-domain |
List-II Kind |
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(A) $A= R$ and $B = R$ |
(I) $f$ is both one-one and onto |
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(B) $A= R$ and $B = [0, ∞]$ |
(II) $f$ is one-one but not onto |
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(C) $A= B = [0, ∞]$ |
(III) $f$ is not one-one but onto |
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(D) $A= [0, ∞]$ and $B = R$ |
(IV) $f$ is neither one-one nor onto |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
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List-I Domain and Co-domain |
List-II Kind |
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(A) $A= R$ and $B = R$ |
(IV) $f$ is neither one-one nor onto |
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(B) $A= R$ and $B = [0, ∞]$ |
(III) $f$ is not one-one but onto |
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(C) $A= B = [0, ∞]$ |
(I) $f$ is both one-one and onto |
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(D) $A= [0, ∞]$ and $B = R$ |
(II) $f$ is one-one but not onto |
Given $f(x)=x^{2}$.
(A) $A=\mathbb{R},\;B=\mathbb{R}$: $f$ is not one-one since $f(1)=f(-1)$ and not onto since negative values are not in the range.
So (A) → (IV).
(B) $A=\mathbb{R},\;B=[0,\infty)$: not one-one but onto because every non-negative real has a square root.
So (B) → (III).
(C) $A=B=[0,\infty)$: one-one and onto since $x^{2}$ is strictly increasing on $[0,\infty)$ and covers all $[0,\infty)$.
So (C) → (I).
(D) $A=[0,\infty),\;B=\mathbb{R}$: one-one but not onto since negative values are not attained.
So (D) → (II).
Final answer: (A)–(IV), (B)–(III), (C)–(I), (D)–(II)