JEE MAIN Sets And Relations Previous Year Questions (PYQs) – Page 1 of 8
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Consider function f : A $\to$ B and g : B $\to$ C (A, B, C $ \subseteq $ R) such that (gof)$-$1 exists, then :
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Solution
Let A={-3,-2,-1,0,1,2,3} and R be a relation on A defined by xRy iff 2x-y $\in\{0,1\}$. Let $l$ be the number of elements in R. Let m and n be the minimum number of elements required to be added in R to make it reflexive and symmetric relations, respectively. Then l+m+n is equal to
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Solution
Given $A=\{-3,-2,-1,0,1,2,3\}$ and $xRy \iff 2x-y\in\{0,1\}$
So $y=2x$ or $y=2x-1$
All valid pairs in $A$:
$(-1,-2),\ (-1,-3),\ (0,0),\ (0,-1),\ (1,2),\ (1,1),\ (2,3)$
So $l=7$
Reflexive pairs needed: $(x,x)$ for all 7 elements
Present: $(0,0),(1,1)$ ⇒ missing $5$ ⇒ $m=5$
For symmetry, add reverse of non-symmetric pairs:
Missing: $(-2,-1),\ (-3,-1),\ (-1,0),\ (2,1),\ (3,2)$ ⇒ $n=5$
$l+m+n=7+5+5=17$
$\boxed{17}$
So $y=2x$ or $y=2x-1$
All valid pairs in $A$:
$(-1,-2),\ (-1,-3),\ (0,0),\ (0,-1),\ (1,2),\ (1,1),\ (2,3)$
So $l=7$
Reflexive pairs needed: $(x,x)$ for all 7 elements
Present: $(0,0),(1,1)$ ⇒ missing $5$ ⇒ $m=5$
For symmetry, add reverse of non-symmetric pairs:
Missing: $(-2,-1),\ (-3,-1),\ (-1,0),\ (2,1),\ (3,2)$ ⇒ $n=5$
$l+m+n=7+5+5=17$
$\boxed{17}$
Let $A, B$ and $C$ be sets such that $\varnothing \ne A\cap B \subseteq C$. Which of the following statements is not true?
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Solution
Let $R=\{(1,2),(2,3),(3,3)\}$ be a relation on the set $\{1,2,3,4\}$. The minimum number of ordered pairs that must be added to $R$ so that it becomes an equivalence relation is:
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Solution
Let $A = {2,3,5,7,9}$. Let $R$ be the relation on $A$ defined by $xRy$ if and only if $2x \le 3y$. Let $\ell$ be the number of elements in $R$, and $m$ be the minimum number of elements required to be added in $R$ to make it a symmetric relation. Then $\ell + m$ is equal to:
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Solution
$A = {2,3,5,7,9}$
$y \ge \frac{2x}{3}$
$x = 2 \Rightarrow y = 2,3,5,7,9$
$x = 3 \Rightarrow y = 2,3,5,7,9$
$x = 5 \Rightarrow y = 5,7,9$
$x = 7 \Rightarrow y = 5,7,9$
$x = 9 \Rightarrow y = 7,9$
$\ell = 18$
To make symmetric, elements to be added:
${(5,2), (7,2), (9,2), (5,3), (7,3), (9,3), (9,5)}$
$m = 7$
$\ell + m = 25$
If {p} denotes the fractional part of the number p, then $\left\{ {{{{3^{200}}} \over 8}} \right\}$, is equal to :
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Solution
Let $P(S)$ denote the power set of $S=\{1,2,3,\ldots,10\}$. Define the relations $R_{1}$ and $R_{2}$ on $P(S)$ as
$A\,R_{1}\,B \iff (A\cap B^{c})\cup(B\cap A^{c})=\varnothing$ and
$A\,R_{2}\,B \iff A\cup B^{c}=B\cup A^{c}$, for all $A,B\in P(S)$. Then:
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Solution
Let X = ℝ × ℝ. Define a relation R on X by
(a₁,b₁) R (a₂,b₂) ⇔ b₁ = b₂.
Statement I: R is an equivalence relation.
Statement II: For some (a,b) ∈ X, the set S = { (x,y) ∈ X : (x,y) R (a,b) } represents a line parallel to y = x.
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Solution
Let the relation $R$ on the set $M = {1,2,3,\ldots,16}$ be given by
$R = {(x,y) : 4y = 5x - 3,; x,y \in M}$.
Then the minimum number of elements required to be added in $R$, in order to make the relation symmetric, is equal to:
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Solution
$R = {(3,3), (7,8), (11,13)}$
To make it symmetric $(8,7), (13,11)$ must be added
$ \text{Let } R_1 \text{ and } R_2 \text{ be two relations defined on } \mathbb{R} \text{ by } a R_1 b \Leftrightarrow ab \ge 0 \text{ and } aR_2b \Leftrightarrow a \ge b. \text{ Then,}$
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Solution
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