datavorous/entrance-exam-dataset huggingface dataset:
Let A=100110111 and B=A20. Then the sum of the elements of the first column of B is

Options:
A 210
B 211
C 251
D 231

The correct answer is:
231

Solution:
Given, A=100110111
⇒A2=100210321; A3=100310631
A4=1004101041; ,…; A20=1002010210201 ∵a31=20th term in 1+3+6+10+....

Hence, sum of the elements of first column=231.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2018 (16 Apr Online)
Let $A=\left(\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right)$ Then, for positive integer
$n, A^{n}$ is

Options:
A $\left(\begin{array}{lll}1 & n & n^{2} \\ 0 & n^{2} & n \\ 0 & 0 & n\end{array}\right)$
B $\left(\begin{array}{lll}1 & n & n\left(\frac{n+1}{2}\right) \\ 0 & 1 & n \\ 0 & 0 & 1\end{array}\right)$
C $\left(\begin{array}{lll}1 & n^{2} & n \\ 0 & n & n^{2} \\ 0 & 0 & n^{2}\end{array}\right)$
D $\left(\begin{array}{ccc}1 & n & 2 n-1 \\ 0 & \frac{n+1}{2} & n^{2} \\ 0 & 0 & \frac{n+1}{2}\end{array}\right)$

The correct answer is:
$\left(\begin{array}{lll}1 & n & n\left(\frac{n+1}{2}\right) \\ 0 & 1 & n \\ 0 & 0 & 1\end{array}\right)$

Solution:
We have, $A=\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]$
$\therefore \quad A^{2}=A \cdot A$
$=\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]$
$=\left[\begin{array}{lll}1+0+0 & 1+1+0 & 1+1+1 \\ 0+0+0 & 0+1+0 & 0+1+1 \\ 0+0+0 & 0+0+0 & 0+0+1\end{array}\right]$
$=\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{lll}1 & 2 & \frac{2(2+1)}{2} \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{array}\right]$
Again, $A^{3}=A^{2} \cdot A$
$=\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]$
$\begin{aligned} &=\left[\begin{array}{lll}1+0+0 & 1+2+0 & 1+2+3 \\ 0+0+0 & 0+1+0 & 0+1+2 \\ 0+0+0 & 0+0+0 & 0+0+1\end{array}\right] \\=\left[\begin{array}{lll}1 & 3 & 6 \\ 0 & 1 & 3 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{lll}1 & 3 & \frac{3(3+1)}{2} \\ 0 & 1 & 3 \\ 0 & 0 & 1\end{array}\right] \\\therefore A^{n} =\left[\begin{array}{lll}1 & n & \frac{n(n+1)}{2} \\ 0 & 1 & n \\ 0 & 0 & 1\end{array}\right] \end{aligned}$

Topic: Mathematics, Matrices, JEE Main
Let $\mathrm{A}=\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{ll}1 & \mathrm{x} \\ 0 & 1\end{array}\right]$
If $\mathrm{AB}=\mathrm{BA}$, then what is the value of $\mathrm{x}$?

Options:
A $-1$
B 0
C 1
D Any real number

The correct answer is:
0

Solution:
As given $\mathrm{A}=\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right]$ and $\mathrm{B}=\left|\begin{array}{ll}1 & \mathrm{x} \\ 0 & 1\end{array}\right|$
$\mathrm{AB}=\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right]\left[\begin{array}{ll}1 & \mathrm{x} \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1 & \mathrm{x} \\ 0 & -1\end{array}\right]$
and $\mathrm{BA}=\left[\begin{array}{ll}1 & \mathrm{x} \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right]=\left[\begin{array}{cc}1 & -\mathrm{x} \\ 0 & -1\end{array}\right]$
Given that $\mathrm{AB}=\mathrm{BA}$
We have, $\left[\begin{array}{cc}1 & \mathrm{x} \\ 0 & -1\end{array}\right]=\left[\begin{array}{cc}1 & -\mathrm{x} \\ 0 & -1\end{array}\right]$
$\Rightarrow \mathrm{x}=-\mathrm{x} \Rightarrow 2 \mathrm{x}=0 \Rightarrow \mathrm{x}=0$

Topic: Mathematics, Matrices, JEE Main
Let A=1-1001-1001 and B=7A20-20A7+2I,
where I is an identity matrix of order 3×3. If B=bij, then b13 is equal to

Options:

The correct answer is:
910

Solution:
Let A=1-1001-1001=I+C

where I=100010001, C=0-1000-1000

Then,

C2=001000000,

C3=000000000=C4=C5=…….

Now,

B=7 A20-20 A7+2I

=7(I+C)20-20(I+C)7+2I

=7I+20C+C220C2+...-20I+7C+C27C2+..+2I

=–11I+910 C2

=-1109100-11000-11

So,

b13=910

Topic: Mathematics, Matrices, JEE Main, JEE Main 2021 (20 Jul Shift 1)
Let A=1111-1001-1, A1 be a matrix formed by the cofactors of the elements of the matrix A and A2 be a matrix formed by the cofactors of the elements of matrix A1. Similarly, if A10 be a matrix formed by the cofactors of the elements of matrix A9, then the value of A10 is

Options:
A 310
B 320
C 9
D 31024

The correct answer is:
31024

Solution:
A=11-0-1-1-1=3
A1=A2 by property
A2=A12=A4
A3=A22=A8
Similarly, A10=A210=3210=31024

Topic: Mathematics, Determinants, JEE Main
Let $A^{-1}=\left[\begin{array}{lll}1 & 2017 & 2 \\ 1 & 2017 & 4 \\ 1 & 2018 & 8\end{array}\right] .$ Then $|2 \mathrm{~A}|-\left|2 \mathrm{~A}^{-1}\right|$ is equal to

Options:
A 3
B $-3$
C 12
D $-12$

The correct answer is:
12

Solution:
$\begin{array}{l}
2^{3}|\mathrm{~A}|-2^{3} \frac{1}{|\mathrm{~A}|} \\
\left|\mathrm{A}^{-1}\right|=\left|\begin{array}{lll}
1 & 2017 & 2 \\
1 & 2017 & 4 \\
1 & 2018 & 8
\end{array}\right| \\
\frac{1}{|\mathrm{~A}|}=-2 \Rightarrow|\mathrm{A}|=\frac{-1}{2}
\end{array}$
Put the value answer is $=12$

Topic: Mathematics, Determinants, JEE Main
Let A=111 and B=92-102112122132-142-152162172, then the value of A'BA is;

Options:
A 1224
B 1042
C 540
D 539

The correct answer is:
539

Solution:
Given A=111 and B=92-102112122132-142-152162172

then A'=111

So A'B=11192-102112122132-142-152162172

=92+122-152-102+132+162112-142+172

=0325214

Now A'BA=0325214111

=325+214

=539

Topic: Mathematics, Matrices, JEE Main, JEE Main 2022 (26 Jul Shift 2)
Let A=1-12-1 and B=a 1b-1 are two matrices. If A+B2=A2+B2, then the value of 3a+4b is equal to

Options:
A 15
B 17
C 19
D 21

The correct answer is:
19

Solution:
Given, A+B2=A2+B2
A2+B2+AB+BA=A2+B2⇒AB+BA=O
AB=1-12-1a 1b-1=a-b22a-b3
BA=a 1b-11-12-1=a+2-a-1b-2-b+1
AB+BA=2a-b+2-a+12a-2-b+4=O
2a-b+2=0
-a+1=0⇒a=1
2a-2=0⇒a=1
-b+4=0⇒b=4

Topic: Mathematics, Matrices, JEE Main
Let A=1-12α and B=β110,α,β∈R. Let α1 be the value of α which satisfies A+B2=A2+2222 and α2 be the value of α which satisfies A+B2=B2. Then α1-α2 is equal to

Options:

The correct answer is:
2

Solution:
Given A=1-12α and B=β110,α,β∈R,

So, A+B=β+103α

Now A+B2=β+103αβ+103α

=β+1203β+1+3αα2

Also, A2=1-12α1-12α=-1-1-α2+2αα2-2

Now solving A+B2=A2+2222

⇒β+1203α+β+1α2=1-α+12α+4α2

Now on comparing both side we get, α=1=α1

And B2=β110β110=β2+1ββ1

Now using A+B2=B2

⇒β2+1ββ1=β+1203β+1+3αα2

Again on comparing both side we get, β=0,α=-1=α2

So, α1-α2=1--1=2

Topic: Mathematics, Matrices, JEE Main, JEE Main 2022 (28 Jul Shift 1)
Let A=1133 and B=A+A2+A3+A4. If B=λA, ∀λ∈R, then the value of λ is equal to

Options:

The correct answer is:
85

Solution:
A2=11331133=441212=4A
So, A3=4A2=16A, A4=64A
Hence, B=1+4+16+64A=85A

Topic: Mathematics, Matrices, JEE Main
Let a1=1, a2, a3, a4,….. be consecutive natural numbers. Then tan-111+a1a2+tan-111+a2a3 +…..+tan-111+a2021a2022 is equal to

Options:
A π4-cot-1(2022)
B cot-1(2022)-π4
C tan-1(2022)-π4
D π4-tan-1(2022)

The correct answer is:
π4-cot-1(2022)

Solution:
Given,

tan-111+a1a2+tan-111+a2a3................tan-111+a2021a2022

=tan-1a2-a11+a1a2+tan-1a3-a21+a2a3................tan-1a2022-a20211+a2021a2022

=tan-1a2-tan-1a1+tan-1a3-tan-1a2..........+tan-1a2022-tan-1a2021

using the formula tan-1x-tan-1y=tan-1x-y1+xy

=tan-1a2022-tan-1a1

=tan-1a2022-tan-11 as given a1=1

=tan-1a2022-π4

=π2-cot-1a2022-π4

=π4-cot-1a2022

=π4-cot-12022

As a1=1, a2=2 .... and so on a2022=2022

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2023 (30 Jan Shift 2)
Let A1,1, B-4,3,C-2,-5 be vertices of a triangle ABC, P be a point on side BC, and Δ1 and Δ2 be the areas of triangle APB and ABC.

Respectively.

If Δ1:Δ2=4:7, then the area enclosed by the lines AP,AC and the x-axis is

Options:
A 14
B 34
C 12
D 1

The correct answer is:
12

Solution:
Plotting the given value in diagram we have,

Now let Px,y be a point on BC, so area formed by △ABP will be,

Δ1=12xy1111-431=12-2x-5y+7

And area of triangle formed by △ABC will be, Δ2=12111-431-2-51=362

Given Δ1Δ2=47⇒-2x-5y+736=47

⇒ 14x+35y=-95 ⋯1

Equation of BC by two point form will be 4x+y=-13 ⋯2

Solving equation 1 and 2 we get,

Point P-207,-117

Now equation of AP will be y-1=1+1171+207x-1

⇒3y-2x=1,

So x-intercept will be point Q-12,0 and similarly from AC we get R12,0

So Area of triangle AQR=12×1×1=12

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2022 (27 Jul Shift 1)
Let $A=\left(\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right)(10) B=\left(\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right)$. If $B$

Options:
A
$-2$

B
5

C
2

D
$-1$

The correct answer is:
5

Solution:
$A B=I \Rightarrow A(10 B)=10 \mathrm{I}$
$\Rightarrow\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right]=\left[\begin{array}{ccc}10 & 0 & 5-\alpha \\ 0 & 10 & \alpha-5 \\ 0 & 0 & 5+\alpha\end{array}\right]=10\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$ if $\alpha=5$.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2004
LetA=12-14. If A-1=αI+βA, α,β∈R, I is a 2×2 identity matrix, then 4α-β is equal to :

Options:
A 5
B 83
C 2
D 4

The correct answer is:
4

Solution:
Given,

A=12-14⇒A=6

So, A-1=adjAA=164-211=23-131616

Here, 23-131616=α00α+β2β-β4β

α+β=23β=-16⇒α=23+16=56

Hence, 4α-β=41=4

Topic: Mathematics, Matrices, JEE Main, JEE Main 2021 (27 Jul Shift 1)
Let A=12-2-5. Let α,β∈ℝ be such that αA2+βA=2I. Then α+β is equal to

Options:
A -10
B -6
C 6
D 10

The correct answer is:
10

Solution:
We know that characteristic equation of matrix A is given by,

A-λI=0

⇒1-λ2-2-5-λ=0

⇒λ2+4λ=1

⇒A2+4A=I

Multiplying by 2 both side we get,

⇒2A2+8A=2I ⋯1

Given that αA2+βA=2I ⋯2

Now on Comparing equation 1 & 2 we get

α=2, β=8

∴ α+β=10

Topic: Mathematics, Matrices, JEE Main, JEE Main 2022 (27 Jul Shift 1)
Let $A=\left(\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1\end{array}\right)$. If $u_1$ and $u_2$ are column matrices such that $A u_1=\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right)$ and $A u_2=\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right)$, then $u_1+u_2$ is equal to

Options:
A
$\left(\begin{array}{c}-1 \\ 1 \\ 0\end{array}\right)$

B
$\left(\begin{array}{c}-1 \\ 1 \\ -1\end{array}\right)$

C
$\left(\begin{array}{c}-1 \\ -1 \\ 0\end{array}\right)$

D
$\left(\begin{array}{c}1 \\ -1 \\ -1\end{array}\right)$

The correct answer is:
$\left(\begin{array}{c}1 \\ -1 \\ -1\end{array}\right)$

Solution:
$A=\left(\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1\end{array}\right)$
Let $u_1=\left[\begin{array}{l}a \\ b \\ c\end{array}\right] ; u_2=\left[\begin{array}{l}d \\ e \\ f\end{array}\right]$
$\mathrm{Au}_1=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right] \quad \Rightarrow u_1=\left[\begin{array}{c}1 \\ -2 \\ 1\end{array}\right]$
$\mathrm{Au}_2=\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right] \quad \Rightarrow \mathrm{u}_2=\left[\begin{array}{c}0 \\ 1 \\ -2\end{array}\right] \quad \Rightarrow \mathrm{u}_1+\mathrm{u}_2=\left[\begin{array}{c}1 \\ -1 \\ -1\end{array}\right]$

Topic: Mathematics, Matrices, JEE Main, JEE Main 2012 (Offline)
Let A=-1 2-3-2 0 3 3-3 1 be a matrix, then Aadj A-1 is equal to

Options:
A O3Ã3
B -1 2-3-2 0 3 3-3 1
C I3
D -3-3&nbsp&nbsp&nbsp1&nbsp&nbsp&nbsp3&nbsp&nbsp&nbsp0-2-1&nbsp&nbsp&nbsp2-3

The correct answer is:
-1 2-3-2 0 3 3-3 1

Solution:
We know that,
A-1adj A-1= A-1I3
⇒A. A-1adj A-1= A-1AI3
⇒ adj A-1= A-1A
⇒ Aadj A-1= A

Topic: Mathematics, Determinants, JEE Main
Let A=-1 2-3-2 0 3 3-3 1 be a matrix, then Aadj A-1 is equal to

Options:
A O3×3
B -1&nbsp&nbsp&nbsp2-3-2&nbsp&nbsp&nbsp0&nbsp&nbsp&nbsp3&nbsp&nbsp&nbsp3-3&nbsp&nbsp&nbsp1
C I3
D -3-3&nbsp&nbsp&nbsp1&nbsp&nbsp&nbsp3&nbsp&nbsp&nbsp0-2-1&nbsp&nbsp&nbsp2-3

The correct answer is:
-1&nbsp&nbsp&nbsp2-3-2&nbsp&nbsp&nbsp0&nbsp&nbsp&nbsp3&nbsp&nbsp&nbsp3-3&nbsp&nbsp&nbsp1

Solution:
Aadj A-1=?
A-1adj A-1= A-1I3
âA. A-1adj A-1= A-1AI3
â adj A-1= A-1A
â Aadj A-1= A

Topic: Mathematics, Determinants, JEE Main
Let A=12341-12, then the value of sum of all the elements of A100 is

Options:

The correct answer is:
2

Solution:
A2=12341-1212341-12=1001
⇒A2=Î⇒A100=Î⇒sum=2

Topic: Mathematics, Matrices, JEE Main
Let $A=\left[\begin{array}{ccc}1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & -6\end{array}\right]$ and $B=\left[b_{i j}\right]_{3 \times 3}$ with
$b_{11}=2, b_{13}=-2, b_{12}=0$ is such that $A B=\left[\begin{array}{ccc}2 & 14 & -4 \\ 4 & 1 & -8 \\ -6 & 15 & 12\end{array}\right]$, then $|B|+\operatorname{trace}(B)=$

Options:
A -2
B 10
C -8
D 6

The correct answer is:
-2

Solution:
We have, $A=\left[\begin{array}{ccc}1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & -6\end{array}\right]$
and $B=\left[\begin{array}{lll}b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33}\end{array}\right]=\left[\begin{array}{ccc}2 & 0 & -2 \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33}\end{array}\right]$
$\therefore A B=\left[\begin{array}{ccc}1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & -6\end{array}\right]\left[\begin{array}{ccc}2 & 0 & -2 \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33}\end{array}\right]$
$=\left[\begin{array}{ccc}2+4 b_{21}+2 b_{31} & 0+4 b_{22}+2 b_{32} & -2+4 b_{23}+2 b_{33} \\ 4-b_{21}+4 b_{31} & 0+b_{22}+4 b_{32} & -4-b_{23}+4 b_{33} \\ -6+7 b_{21}-6 b_{31} & 0+7 b_{22}-6 b_{32} & 6+7 b_{23}-6 b_{33}\end{array}\right]$
$=\left[\begin{array}{ccc}2 & 14 & -4 \\ 4 & 1 & -8 \\ -6 & 15 & 12\end{array}\right]$
On solving above equal matrices with corresponding elements, we get
$b_{21}=b_{31}=0, b_{22}=3, b_{32}=1, b_{23}=0 \text { and } b_{33}=-1$
$\therefore \quad B=\left[\begin{array}{ccc}2 & 0 & -2 \\ 0 & 3 & 0 \\ 0 & 1 & -1\end{array}\right]$
$\therefore|B|=2(-3-0)+(-2)(0-0)=-6$
and Trace $(B)=2+3-1=4$
$\therefore|B|+\operatorname{trace}(B)=-6+4=-2$

Topic: Mathematics, Matrices, JEE Main
Let A={1,2,3,4,5,6,7}. Then the relation R={(x,y)∈A×A:x+y=7} is

Options:
A an equivalence relation
B symmetric but neither reflexive nor transitive
C transitive but neither symmetric nor reflexive
D reflexive but neither symmetric nor transitive

The correct answer is:
symmetric but neither reflexive nor transitive

Solution:
We have, A=1,2,3,4,5,6,7

Reflexive: A relation R on a set A is said to be reflexive if every element of A is related to itself.

Thus, R is reflexive⇔(a,a)∈R for all a∈A

∵(1,1),2,2,(3,3),......(7,7) does not satisfy x+y=7

Hence R is not reflexive.

Symmetric: A relation R is symmetric on a set A iff

(a,b)∈R⇒(b,a)∈R for all a,b∈A

⇒x+y=7

Now on interchanging y and x we get, y+x=7 which is always true for given set,

Hence R is symmetric.

Transitive: A relation R on A is said to be transitive relation iff

a,b∈R and (b,c)∈R

⇒a,c∈R for all a,b,c∈A

Now taking a,b≡3,4 and b,c≡4,3 so a,c≡3,3 does not satisfy x+y=7,

Hence, R is not transitive and not equivalence.

Therefore, R is only Symmetric.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2023 (08 Apr Shift 2)
Let $A=\{1,2,3,4,5\}$. Let $\mathrm{R}$ be a relation on $\mathrm{A}$ defined by $x \mathrm{R} y$ if and only if $4 x \leq 5 \mathrm{y}$. Let $\mathrm{m}$ be the number of elements in $\mathrm{R}$ and $\mathrm{n}$ be the minimum number of elements from $\mathrm{A} \times \mathrm{A}$ that are required to be added to $\mathrm{R}$ to make it a symmetric relation. Then $\mathrm{m}+\mathrm{n}$ is equal to :

Options:
A 25
B 24
C 26
D 23

The correct answer is:
25

Solution:
Given : $4 x \leq 5 y$
then
$\begin{aligned}
\mathrm{R}=\{ & (1,1),(1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4) \\
& (2,5),(3,3),(3,4),(3,5),(4,4),(4,5),(5,4),(5,5)\}
\end{aligned}$
i.e. 16 elements.
i.e. $\mathrm{m}=16$
Now to make R a symmetric relation add
$\{(2,1)(3,2)(4,3)(3,1)(4,2)(5,3)(4,1)(5,2)(5,1)\}$
i.e. $n=9$
So $m+n=25$

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2024 (06 Apr Shift 2)
Let $A=\{1,2,3,4,6\}$. Let $R$ be the relations on $A$ defined by $\{(a, b)\}: a, b \in A, a$ divides $b\}$.
(i) Write $R$ in roster form
(ii) Find the domain of $R$
(iii) Find the range of $R$

Options:

None

Solution:
(i) Here $A=\{1,2,3,4,6\}$ and $R$ is a relation on A defined by $\{(a, b): a \in A, b \in A$, a divides
Clearly, $R=\{(1,1),(1,2),(1,3),(1,4),(1,6),(2,2),(2,4)$, $(3,3),(3,6),(4,4),(6,6)\}$.
(ii) Domain $=\{1,2,3,4,6\}$.
(iii) Range $=\{1,2,3,4,6\}$.

Topic: Mathematics, Relations and Functions (Class 11)
Let A=1,2,3,4,..........10 and B=0,1,2,3,4. The number of elements in the relation R=(a,b)∈A×A:2a-b2+3a-b∈B is __________.

Options:

The correct answer is:
18

Solution:
Given,

A={1,2,3,…,10} B={0,1,2,4}

(a,b)∈A×A such that

2a-b2+3a-b-k=0

where k∈{0,1,2,3,4}

Now finding discriminant D=9-4×2-k

And 9-4×2-k a perfect square for any possible a,b as a-b will be a integer,

So, 9+8k is a perfect square

⇒k=0 or k=2

Now for k=0,

2a-b2+3a-b=0

⇒a-b2a-b2+3=0

⇒a-b=0⇒(a,b)∈{(1,1),(2,2)….(10,10)}

⇒Total 10 elements belonging to R.

And, a-b=32 is not possible

Now for k=2,

2a-b2+3a-b-2=0

⇒a-b=-2 or a-b=12 (not possible)

Now for a-b=-2 possible pair will be,

⇒(a,b)∈{(1,3),(2,4),….(8,10)}

⇒8 elements belonging to R

Total number of elements will be=18

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2023 (06 Apr Shift 1)
Let A=1234 and B=pqrs are two matrices such that AB=BA and r≠0, then the value of 3p-3s5q-4r is equal to

Options:
A 32
B 4
C 92
D 5

The correct answer is:
92

Solution:
AB=1234pqrs=p+2rq+2s3p+4r3q+4s
BA=pqrs1234=p+3q2p+4qr+3s2r+4s
∵AB=BA⇒p+2r=p+3q⇒2r=3q
⇒q+2s=2p+4q⇒2s=2p+3q
⇒3p+4r=r+3s⇒p+r=s⇒p-s=-r
⇒3q+4s=2r+4s⇒2r=3q
3p-s5q-4r=3-r52r3-4r=-3103-4=92=4.5

Topic: Mathematics, Matrices, JEE Main
Let A=1, 2, 3, 4 and R=(1, 2), (2, 3), (1, 4) be a relation on A. Let S be the equivalence relation on A such that R⊂S and the number of elements in S is n. Then, the minimum value of n is _______

Options:

The correct answer is:
16

Solution:
Given,

A=1, 2, 3, 4 and R=(1, 2), (2, 3), (1, 4) be a relation on A.

Now, given S be the equivalence relation on A and R⊂S

So, for S to be equivalence it should be Reflexive, Symmetric and Transitive on given relation,

Now, for reflexive Relation we need, 1,1, 2,2, 3,3, 4,4→4 elements,

For Symmetric relation we need, 2,1, 3,2, 4,1→3 elements,

And for Transitive relation we need, 1,3, 3,1, 3,4, 4,2, 2,4, 4,3→6 elements

So, minimum 13 elements should be added in R

Hence , nS=16 elements.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2024 (31 Jan Shift 1)
Let $A=\{1,2,3,4\}$ and $R: A \rightarrow A$ be the relation defined by $R=\{(1,1),(2,3),(3,4),(4,2)\}$. The correct statement is :

Options:
A
$R$ does not have an inverse.

B
$R$ is not a one to one function.

C
$R$ is an onto function.

D
$R$ is not a function.

The correct answer is:
$R$ is an onto function.

Solution:
$\begin{aligned} & \text { Domain }=\{1,2,3,4\} \\ & \text { Range }=\{1,2,3,4\} \\ & \therefore \quad \text { Domain }=\text { Range }\end{aligned}$

Topic: Mathematics, Functions, JEE Main, JEE Main 2013 (09 Apr Online)
Let $A=\{1,2,3, \ldots .14\}$. Define a relation $R$ from $A$ to $A$ by $R=\{(x, y): 3 x-y=0$ where $x, y \in A\}$. Write down the domain, co-domain and range.

Options:

None

Solution:
(i) By the definition of the relation, $R=\{(1,3)(2,6)$, $(3,9),(4,12)\}$. The corresponding arrow diagram is:
(ii) We can see that the set of first element i.e., the domain $=\{1,2,3,4\}$.
Similarly, the set of second elements (images) i.e., the range $=\{3,6,9,12\}$ and the co-domain $=\{1,2,3, \ldots .14\}$.

Topic: Mathematics, Relations and Functions (Class 11)
Let A=1,2,3,...20. Let R1 and R2 two relation on A such that

R1={a,b:b is divisible by a}

R2={a,b:a is an integral multiple of b}

Then, number of elements in R1−R2 is equal to __________.

Options:

The correct answer is:
46

Solution:
Given: A=1,2,3,...20, R1={a,b:b is divisible by a} and R2={a,b:a is an integral multiple of b}.

⇒R1=1,1,1,2,...,1,20,2,2,2,4,...,2,20,3,3,3,6,...,3,18,4,4,4,8,...,4,20,5,5,5,10,...,5,20,6,6,6,18,...,6,18,7,7,7,14,8,8,8,16,9,9,9,18,10,10,10,20,11,11,...,20,20

⇒nR1=20+10+6+5+4+3+2+2+2+2+10=66

⇒R1∩R2=1,1,2,2,3,3,4,4,...,20,20

⇒nR1∩R2=20

⇒nR1-R2=nR1-nR1∩R2

⇒nR1-R2=66-20

⇒nR1-R2=46

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2024 (01 Feb Shift 1)
Let A=1,2,3,....100. Let R be a relation on A defined by x, y∈R if and only if 2x=3y. Let R1 be a symmetric relation on A such that R⊂R1 and the number of elements in R1 is n. Then the minimum value of n is _______.

Options:

The correct answer is:
66

Solution:
Given,

2x=3y

⇒x3=y2

Now, let x3=y2=k

⇒x=3k & y=2k

Now, given set A=1, 2, 3, .......,100

So, for k≤33 we get,

⇒R=3,2,6,4,9,6,12,8,....99,66

⇒nR=33

Now, R1 is symmetric relation on A and R⊂R1 so, we need elements 3,2,6,4,9,6,12,8,....99,66, 2,3,4,6,6,9, 8,12,....66,99 in R1

Hence, the minimum number of elements in R1 is 33+33=66

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2024 (31 Jan Shift 2)
Let $A=\left(\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right)$ and $B=\left(\begin{array}{ll}a & 0 \\ 0 & b\end{array}\right), a, b \in N$. Then

Options:
A
there cannot exist any B such that AB = BA

B
there exist more than one but finite number of B’s such that AB = BA

C
there exists exactly one B such that AB = BA

D
there exist infinitely many B’s such that AB = BA

The correct answer is:
there exist infinitely many B’s such that AB = BA

Solution:
$A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right] \quad B=\left[\begin{array}{ll}a & 0 \\ 0 & b\end{array}\right]$
$A B=\left[\begin{array}{ll}a & 2 b \\ 3 a & 4 b\end{array}\right]$
$B A=\left[\begin{array}{ll}a & 0 \\ 0 & b\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]=\left[\begin{array}{cc}a & 2 a \\ 3 b & 4 b\end{array}\right]$
$A B=B A$ only when $a=b$

Topic: Mathematics, Matrices, JEE Main, JEE Main 2006
Let A=1,3,4,6,9 and B=2,4,5,8,10. Let R be a relation defined on A×B such that R=a1,b1,a2,b2:a1≤b2 and b1≤a2. Then the number of elements in the set R is

Options:
A 160
B 52
C 26
D 180

The correct answer is:
160

Solution:
Given,

A=1,3,4,6,9 & B=2,4,5,8,10

And relation is given by,

R=a1,b1,a2,b2:a1≤b2 and b1≤a2

Now taking cases for a1≤b2 we get,

a1=1, b2∈2,4,5,8,10→5 cases

a1=3, b2∈4,5,8,10→4 cases

a1=4, b2∈4,5,8,10→4 cases

a1=6, b2∈8,10→2 cases

a1=9, b2∈10→1 cases

So, total 16 cases will be there

Now finding cases of b1≤a2 we get,

b1=2, a2∈3,4,6,9→4 cases

b1=4, a2∈4,6,9→3 cases

b1=5, a2∈6,9→2 cases

b1=8, a2∈9→1 cases

So, here total 10 cases,

Hence, total elements in relation =16×10=160

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2023 (11 Apr Shift 2)
Let A1→=3, A2→=5 and A1→+A2→=5. The value of 2A1→+3A2→∙3A1→-2A2→ is:

Options:
A -112.5
B -118.5
C -106.5
D -99.5

The correct answer is:
-118.5

Solution:
A→1+A→2=5
A→1+A→22=25
A→12+A→22+2A→1A→2=25
9+25+2A→1∙A→2=25
A→1.A→2=-92
Now, 2A→1+3A→2∙3A→1-2A→2=6A→12-4A→1A→2+9A→1A→2-6A→22
=6.A→12+5A→1A→2-6A→22
=6×9+5-92-6×25
=54-452-150
= -118.5

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2019 (08 Apr Shift 2)
Let A=13cos⁡2θ1sin⁡2θ13cos⁡2θ1sin⁡2θ1. If the maximum value of A is equal to k, then k-32 is equal to

Options:

The correct answer is:
49

Solution:
A=11-3sin⁡2θcos⁡2θ-3cos⁡2θsin⁡2θ-3cos⁡2θ+1sin2⁡2θ-1
=1-3sin⁡4θ+9cos2⁡2θ+sin2⁡2θ-1
=1-3sin⁡4θ+1+8cos2⁡2θ-1
=1-3sin⁡4θ+4+4cos⁡4θ
A=5-3sin⁡4θ+4cos⁡4θ
Amax=10

Topic: Mathematics, Determinants, JEE Main
Let A1,4 and B1,-5 be two points. Let P be a point on the circle x-12+y-12=1, such that PA2+PB2 have maximum value, then the points, P, A and B lie on

Options:
A a hyperbola
B a straight line
C an ellipse
D a parabola

The correct answer is:
a straight line

Solution:
P be a point on x-12+y-12=1, so P1+cosθ,1+sinθ.

Given A1,4 & B1,-5

PA2+PB2

=cosθ2+sinθ-32+cosθ2+sinθ+62

=47+6sinθ is maximum if sinθ=1

⇒sinθ=1,cosθ=0

P1,2, A1,4, B1,-5

The points P, A and B lie on the line x=1.

Topic: Mathematics, Circle, JEE Main, JEE Main 2021 (26 Feb Shift 2)