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Relations and Functions — JEE MCQs with answers

NCERT Class 11 · Mathematics · Ch 210 free questions518 in the full bank

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Question 1 · 2.4 Functions · Easy
The real valued function $f:\mathbf{R}\to\mathbf{R}$ defined by $y=f(x)=x$ for each $x\in\mathbf{R}$ is called the
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Answer: B — (B) The text names $y=f(x)=x$ on $\mathbf{R}$ the identity function. The constant function is $f(x)=c$, not $f(x)=x$; the signum function is the three-line rule giving $1,0,-1$; the modulus function is $f(x)=|x|$.

Class 11 · Mathematics · Ch 2 · §2.4.1 · p.32 · ("Define the real valued function f : R → R by y = f(x) = x for each x ∈ R")

Question 2 · 2.2 Cartesian Product of Sets · Easy
According to Definition 1, if either $P$ or $Q$ is the null set, then $P \times Q$ is:
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Answer: A — A is correct: the definition adds that if either $P$ or $Q$ is the null set, then $P \times Q$ will also be the empty set, $P \times Q = \phi$. B is wrong because the surviving non-empty set contributes nothing on its own — no ordered pair can be completed. C is wrong because $\{\phi\}$ is a set containing one element, not the empty set. D is wrong because the product is empty, not a set with a single ordered pair.

Class 11 · Mathematics · Ch 2 · §2.2 · p.25 · Definition 1 ("If either P or Q is the null set, then P x Q will also be empty set, i.e., P x Q = phi")

Question 3 · 2.3 Relations · Medium
For $A = \{1,\, 2,\, 3,\, 4,\, 5,\, 6,\, 7\}$ and the relation $R = \{(x, y) : y = x + 3,\ x, y \in A\}$ on $A$, the domain and the range of $R$ are respectively:
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Answer: B — B is correct: $R = \{(1,4), (2,5), (3,6), (4,7)\}$. The domain is the set of all first elements, $\{1,2,3,4\}$, and the range is the set of all second elements, $\{4,5,6,7\}$. A is wrong because it swaps the two: $\{4,5,6,7\}$ consists of second elements, so it is the range, not the domain. C is wrong because the whole of $A$ is the codomain of $R$, and NCERT distinguishes the codomain from the range. D is wrong because $5$ and $6$ are never first elements of a pair of $R$, since $5 + 3 = 8 \notin A$.

Class 11 · Mathematics · Ch 2 · §2.3 · p.28 · Definitions 3 and 4 ("The set of all first elements of the ordered pairs in a relation R ... is called the domain"; "The set of all second elements in a relation R ... is called the range ... The whole set B is called the codomain")

Question 4 · 2.1 Introduction · Hard
Which sequence correctly reproduces the programme of work announced in the Introduction?
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Answer: B — B is correct: the announced order is to link pairs of objects from two sets, then to introduce relations between the two objects of the pair, and finally to study those special relations that qualify to be functions. A is wrong because it puts functions before relations. C is wrong because relations cannot be introduced before the pairs they are built from. D is wrong because it makes the linking of pairs the last step, whereas the text makes it the first.

Class 11 · Mathematics · Ch 2 · §2.1 · p.24 · ¶1 ("we will learn how to link pairs of objects from two sets and then introduce relations ... Finally, we will learn about special relations which will qualify to be functions.")

Question 5 · 2.4 Functions · Hard
Let $A = \{1,2,3,4,6\}$ and let $R$ be the relation on $A$ defined by $\{(a,b) : a,b \in A,\ b \text{ is exactly divisible by } a\}$. Which statement is correct?
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Answer: B — B is correct: every element of $A$ is divisible by $1$, so $R$ contains $(1,1),(1,2),(1,3),(1,4),(1,6)$ and the single element $1$ carries five different images; Definition 5 is violated at once. A is wrong because $R$ is not a function, so speaking of its domain and range as those of a function is beside the point. C is wrong because $4$ does have an image, since $4$ is exactly divisible by $4$. D is wrong because $6$ is not divisible by $4$, so the claim it makes is simply false.

Class 11 · Mathematics · Ch 2 · Exercise 2.2 · p.30 · Q5 ("Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by {(a, b): a, b ∈ A, b is exactly divisible by a}")

Question 6 · 2.2 Cartesian Product of Sets · Medium
Let $\mathbf{R}$ be the set of all real numbers and $A = \{0, 1\}$. Which statement about $A \times \mathbf{R}$ is correct?
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Answer: C — C is correct: $A$ is non-empty and $\mathbf{R}$ is infinite, so by Remark (iii) the product is an infinite set. A is wrong because the product is certainly not finite with two elements — each element of $A$ is paired with every real number. B is wrong because both factors are non-empty, so the product is non-empty. D is wrong because the product consists of ordered pairs, not of real numbers, so it is not the same as $\mathbf{R}$.

Class 11 · Mathematics · Ch 2 · §2.2 · p.26 · Remarks (iii) ("If A and B are non-empty sets and either A or B is an infinite set, then so is A x B.")

Question 7 · 2.3 Relations · Medium
NCERT notes that in Example 8 the element $1$ is related to no element. Consequently, $1$:
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Answer: A — A is correct: an element that is paired with nothing never occurs as a first element, and the domain is exactly the set of first elements. B is wrong because $1$ is not a second element of any pair either. C is wrong because it is still an element of its set; it simply has no partner. D is wrong because the codomain is the whole target set and is not trimmed.

Class 11 · Mathematics · Ch 2 · §2.3 · p.29 · Example 8 (Fig 2.6; R is "x is the square of y"; roster form {(9,3),(9,-3),(4,2),(4,-2),(25,5),(25,-5)}; domain {4,9,25}, range {-2,2,-3,3,-5,5}, Q is the codomain)

Question 8 · 2.1 Introduction · Medium
In the Introduction, the statement "set $A$ is a subset of set $B$" is quoted as an illustration of:
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Answer: D — D is correct: it appears in the list of relations that occur inside mathematics, alongside $m$ less than $n$ and $l$ parallel to $m$. A is wrong because the Introduction cites it as a relation, not as a definition of equality of sets. B is wrong because it is not offered as an everyday, non-mathematical relation like brother and sister. C is wrong because the Introduction does not present it as an example of a function.

Class 11 · Mathematics · Ch 2 · §2.1 · p.24 · ¶1 ("In mathematics also, we come across many relations such as ... set A is a subset of set B")

Question 9 · 2.4 Functions · Medium
The relation $R = \{(9,3),(9,-3),(4,2),(4,-2),(25,5),(25,-5)\}$ is not a function because:
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Answer: D — D is correct: §2.4 rejects this relation of Example 8 because the elements in its domain are connected to more than one image, and indeed each of $9$, $4$ and $25$ carries two images. A is wrong because nothing in Definition 5 forbids negative images. B is wrong because the count of ordered pairs is irrelevant. C is wrong because $3$ and $-3$ are second elements here, they are images and not preimages, so they have no images of their own in this relation.

Class 11 · Mathematics · Ch 2 · §2.4 · p.31 · ¶2 ("the relation in Example 8 is not a function because the elements in the domain are connected to more than one images")

Question 10 · 2.3 Relations · Medium
For $R = \{(x, y): y = x + 5,\ x$ a natural number less than $4;\ x, y \in \mathbf{N}\}$, the domain and range are respectively:
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Answer: B — B is correct: $R = \{(1, 6), (2, 7), (3, 8)\}$, so the first elements are $1, 2, 3$ and the second elements are $6, 7, 8$. A is wrong because $\mathbf{N}$ is the codomain here; the range is only $\{6, 7, 8\}$. C is wrong because domain and range have been interchanged. D is wrong because $x = 4$ is not less than $4$, so this pair is not in $R$.

Class 11 · Mathematics · Ch 2 · Exercise 2.2 · p.30 · Q2 (R = {(x, y): y = x + 5, x is a natural number less than 4})

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