datavorous/entrance-exam-dataset huggingface dataset:
Let $A$ and $B$ be non empty sets in $R$ and $f: A \rightarrow B$ is a bijective function.
Statement 1: $\mathrm{f}$ is an onto function.
Statement 2: There exists a function $g: B \rightarrow A$ such that fog $=I_B$.

Options:
A
Statement 1 is true, Statement 2 is false.

B
Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1.

C
Statement 1 is false, Statement 2 is true.

D
Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for Statement 1.

The correct answer is:
Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for Statement 1.

Solution:
Let $A$ and $B$ be non-empty sets in $R$.
Let $f: A \rightarrow B$ is bijective function.
Clearly statement - 1 is true which says that $f$ is an onto function.
Statement $-2$ is also true statement but it is not the correct explanation for statement-1

Topic: Mathematics, Functions, JEE Main, JEE Main 2012 (26 May Online)
Let a and b be real constants such that the function f defined by fx=x2+3x+a,x≤1bx+2,x>1 be differentiable on R. Then, the value of ∫-22fxdx equals

Options:
A 156
B 196
C 21
D 17

The correct answer is:
17

Solution:
Given: fx=x2+3x+a,x≤1bx+2,x>1

Now, for function to be continuous,

⇒12+3×1+a=b×1+2

⇒4+a=b+2

⇒a-b=-2 ...i

Now, differentiating and putting the value of x,

⇒2x+3=b

⇒21+3=b

⇒b=5

⇒a=3

⇒∫-22fxdx=∫-21x2+3x+adx+∫12bx+2dx

⇒∫-22fxdx=∫-21x2+3x+3dx+∫125x+2dx

⇒∫-22fxdx=x33+3x22+3x-21+5x22+2x12

⇒∫-22fxdx=13+32+3--83+6-6+5×42+2×2-52+2

⇒∫-22fxdx=13+32+3+83+14-92

⇒∫-22fxdx=152+14-92

⇒∫-22fxdx=17

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Let $A$ and $B$ be real matrices of the form $\left[\begin{array}{ll}\alpha & 0 \\ 0 & \beta\end{array}\right]$ and $\left[\begin{array}{ll}0 & \gamma \\ \delta & 0\end{array}\right]$, respectively.
Statement 1: $A B-B A$ is always an invertible matrix.
Statement $2: A B-B A$ is never an identity matrix.

Options:
A
Statement 1 is true, Statement 2 is false.

B
Statement 1 is false, Statement 2 is true.

C
Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation of Statement 1 .

D
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1.

The correct answer is:
Statement 1 is true, Statement 2 is false.

Solution:
Let $A$ and $B$ be real matrices such that $A=\left[\begin{array}{ll}\alpha & 0 \\ 0 & \beta\end{array}\right]$ and $B=\left[\begin{array}{ll}0 & \gamma \\ \delta & 0\end{array}\right]$ Now, $A B=\left[\begin{array}{cc}0 & \alpha \gamma \\ \beta \delta & 0\end{array}\right]$ and $B A=\left[\begin{array}{cc}0 & \gamma \beta \\ \delta \alpha & 0\end{array}\right]$ Statement-1:
$$
\begin{aligned}
& \left.A B-B A=\left[\begin{array}{cc}
0 & \gamma(\alpha-\beta \\
\delta(\beta-\alpha) & 0
\end{array}\right]\right) \\
& |A B-B A|=\left(\alpha-\beta^2\right\rangle \delta \neq 0
\end{aligned}
$$
$\therefore A B-B A$ is always an invertible matrix. Hence, statement $-1$ is true. But $A B-B A$ can be identity matrix if $\gamma=-\delta$ or $\delta=-\gamma$ So, statement $-2$ is false.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2012 (12 May Online)
Let $A$ and $B$ be sets. Show that $\mathrm{f}: \mathrm{A} \times \mathbf{B} \rightarrow \mathbf{B} \times \mathbf{A}$ such that $\mathbf{f}(\mathbf{a}, \mathbf{b})=(\mathrm{b}, \mathbf{a})$ is bijective function.

Options:

None

Solution:
We have $f:(A \times B) \rightarrow B \times A$ such that $f(a, b)=b, a$
(a) $\quad \therefore \mathrm{f}\left(\mathrm{a}_1, \mathrm{~b}_1\right)=\left(\mathrm{b}_1, \mathrm{a}_1\right) \quad \mathrm{f}\left(\mathrm{a}_2, \mathrm{~b}_2\right)=\left(\mathrm{b}_2, \mathrm{a}_2\right)$
$$
\mathrm{f}\left(\mathrm{a}_1, \mathrm{~b}_1\right)=\mathrm{f}\left(\mathrm{a}_2, \mathrm{~b}_2\right) \Rightarrow\left(\mathrm{b}_1, \mathrm{a}_1\right)=\left(\mathrm{b}_2, \mathrm{a}_2\right)
$$
$\Rightarrow \mathrm{b}_1=\mathrm{b}_2$ and $\mathrm{a}_1=\mathrm{a}_2 \quad \therefore \mathrm{f}$ is one-one
(b) Every member $(\mathrm{p}, \mathrm{q})$ belonging to its codomain has its pre-image in its domain as $(q, p) f$ is onto. Thus, $f$ is one-one and onto i.e. it is bijective.

Topic: Mathematics, Relations and Functions (Class 12)
Let A and B be two 3×3 matrices such that AB=I and A=18 then adjBadj2A is equal to

Options:
A 128
B 32
C 64
D 102

The correct answer is:
64

Solution:
Given, AB=I and A=18

Now by using property adjPQ=adjQadjP we get, adjB·adj2A=adjadj2AadjB ...i

Now adjadj2A=adj2A2 =2A4

=234A4=234A4=84A4 ...ii

{Property used adjA=An-1 and kA=knA where n is order of matrix}

Now adjB=B2 .............(iii)

Now putting the value of equation ii & iii in eq i

we get adjB·adj2A=84A4B2

Now usingA=1Bas AB=I, we get

adjB·adj2A=84A4×1A2 =84×A2=84×182=64

Topic: Mathematics, Matrices, JEE Main, JEE Main 2022 (27 Jun Shift 2)
Let A and B be two 3×3 non singular matrices, such that det ATBA=27 and det AB-1=8. Then det BTA-1B=

Options:
A 332
B 116
C 1
D 16

The correct answer is:
332

Solution:
Given, det.ATBA=27

⇒ATBA=27

⇒A2B=27 .....i

Also, detAB-1=8

⇒AB=8

⇒B=A8 .....ii

From i

⇒A2A8=27

⇒A=6

From ii,

B=A8=68=34

detBTA-1B

=BTA-1B

=B2A=3426

=9166=332

Topic: Mathematics, Determinants, JEE Main
Let A and B be two 3×3 non-zero real matrices such that AB is a zero matrix. Then

Options:
A The system of linear equations AX=0 has a unique solution
B The system of linear equations AX=0 has infinitely many solutions
C B is an invertible matrix
D adjA is an invertible matrix

The correct answer is:
The system of linear equations AX=0 has infinitely many solutions

Solution:
Given,

AB=0⇒AB=0

If A≠0, B=0 (not possible)

If B≠0, A=0 (not possible)

Hence A=B=0

So, if A=B=0 then A & B are non-invertible matrix,

So, AX=0 has infinitely many solutions

Topic: Mathematics, Matrices, JEE Main, JEE Main 2022 (29 Jul Shift 1)
Let A and B be two 3×3 real matrices such that A2-B2 is invertible matrix. If A5=B5 and A3 B2=A2 B3, then the value of the determinant of the matrix A3+B3 is equal to :

Options:
A 2
B 4
C 1
D 0

The correct answer is:
0

Solution:
Let C=A2-B2;C≠0

and A5=B5 ...1

A3 B2=A2 B3 ...2

Subtracting equation 2 from 1, we get A5-A3B2=B5-A2B3

⇒A3 A2-B2+B3 A2-B2=O

Post multiplying inverse of A2-B2:

A3+B3=O⇒A3+B3=0

Topic: Mathematics, Matrices, JEE Main, JEE Main 2021 (27 Jul Shift 2)
Let a and b be two distinct positive real numbers. Let 11th term of a GP, whose first term is a and third term is b, is equal to pth term of another GP, whose first term is a and fifth term is b. Then p is equal to

Options:
A 20
B 25
C 21
D 24

The correct answer is:
21

Solution:
Given,

1st GP⇒First term=a, a3=b

⇒ar2=b

⇒r2=ba

⇒a10=ar10=aba5 ...i

2nd GP⇒First term=a, a5=b

⇒aR4=b ...ii

⇒ap=aRp-1

⇒ap=aba14p-1 ...ii

Now, using the given condition of pth term we get,

⇒aba5=abap-14

⇒ba5=bap-14

⇒5=p-14

⇒p=21

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Let $a$ and $b$ be two distinct roots of a polynomial equation $f(x)=0$. Then there exists at least one root lying between $a$ and $b$ of the polynomial equation.

Options:
A $f(x)=0$
B $f^{\prime}(x)=0$
C $f^{\prime \prime}(x)=0$
D None of these

The correct answer is:
$f^{\prime}(x)=0$

Solution:
Rolle's theorem says between any two roots of a polynomial $f(x)$, there is always a root of its derivative $f^{\prime}(x)$
Therefore between $a$ and $b$. There exist at least one root of the polynomial equation $f^{\prime}(x)=0$

Topic: Mathematics, Application of Derivatives, JEE Main
Let $A$ and $B$ be two events such that $P(A)=0.3$ and $P(A \cup B)=0.8$. If $A$ and $B$ are independent events, then $P(B)=$

Options:
A $\frac{5}{6}$
B $\frac{5}{7}$
C $\frac{3}{5}$
D $\frac{2}{5}$

The correct answer is:
$\frac{5}{7}$

Solution:
$0.8=0.3+x-0.3 x \Rightarrow x=5 / 7$.

Topic: Mathematics, Probability, JEE Main
Let A and B be two events such that PA∪B¯=16, PA∩B=14 and PA¯=14, where A¯ stands for the complement of the event A. Then the events A and B are

Options:
A Independent but not equally likely.
B Independent and equally likely.
C Mutually exclusive and independent.
D Equally likely but not independent.

The correct answer is:
Independent but not equally likely.

Solution:
PA∪B¯=16

∴ PA∪B=56
PA∩B=14
PA¯=14

∴ PA=34
∴ PA∪B=PA+PB-PA∩B
∴ 56=34+PB-14
∴ PB=13
∴ PA·PB=13×34=14=PA∩B
Therefore, A and B are independent and since PA≠PB they are not equally likely.

Topic: Mathematics, Probability, JEE Main, JEE Main 2014 (06 Apr)
Let $A$ and $B$ be two events such that $P(\overline{A \cup B})=\frac{1}{6}, P(A \cap B)=\frac{1}{4}$ and $P(\bar{A})=\frac{1}{4}$, where $\bar{A}$ stands for complement of event $A$. Then events $A$ and $B$ are

Options:
A
equally likely and mutually exclusive

B
equally likely but not independent

C
independent but not equally likely

D
mutually exclusive and independent

The correct answer is:
independent but not equally likely

Solution:
$$
\begin{aligned}
& P(\overline{A \cup B})=\frac{1}{6}, P(A \cap B)=\frac{1}{4} \text { and } P(\bar{A})=\frac{1}{4} \\
& \Rightarrow P(A \cup B)=5 / 6 P(A)=3 / 4 \\
& \text { Also } P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
& \Rightarrow P(B)=5 / 6-3 / 4+1 / 4=1 / 3 \\
& P(A) P(B)=3 / 4-1 / 3=1 / 4=P(A \cap B)
\end{aligned}
$$
Hence $A$ and $B$ are independent but not equally likely.

Topic: Mathematics, Probability, JEE Main, JEE Main 2005
Let A and B be two events such that PB∣A=25, PA∣B=17 and PA∩B=19. Consider

S1PA'∪B=56,

S2PA'∩B'=118. Then

Options:
A Both S1 and S2 are true
B Both S1 and S2 are false
C Only S1 is true
D Only S2 is true

The correct answer is:
Both S1 and S2 are true

Solution:
Given,

A and B be two events such that PB∣A=25, PA∣B=17 and PA∩B=19,

Now using the formula of conditional probability we get,

PA∣B=17⇒PA∩BPB=17

⇒19PB=17

⇒PB=79

Now again using formula we get,

PB∣A=25⇒PA∩BPA=25

⇒19PA=25

⇒PA=518

Now we know that,

PA'∪B=1-PA∪B+PB

⇒PA'∪B=1-PA+PA∩B=56

And PA'∩B'=1-PA∪B

{also PA∪B=PA+PB-PA∩B}

⇒PA'∩B'=1-PA-PB+PA∩B=118

⇒ Both S1 and S2 are true.

Topic: Mathematics, Probability, JEE Main, JEE Main 2022 (28 Jul Shift 2)
Let A and B, be two events such that the probability that exactly one of them occurs is 25, and the probability that A or B, occurs is 12, then the probability of both of them occur together is.

Options:
A 0.02
B 0.20
C 0.01
D 0.10

The correct answer is:
0.10

Solution:
P (exactly one) =25
=PA+PB-2PA∩B=25 ...1
PA∪B=12
=PA+PB-PA∩B=12 ...2
∴PA∩B=12-25=5-410=110.

Topic: Mathematics, Probability, JEE Main, JEE Main 2020 (08 Jan Shift 2)
Let A and B be two finite sets with m and n elements respectively. The total number of subsets of the set A is 56 more than the total number of subsets of B. Then the distance of the point P(m,n) from the point Q(-2,-3) is

Options:
A 10
B 6
C 4
D 8

The correct answer is:
10

Solution:
We know that, the total number of subsets of a set having n elements is 2n.

⇒2m-2n=56

⇒2n2m-n-1=23×7

⇒2n=23 and 2m-n-1=7

⇒n=3 and 2m-n=8

⇒n=3 and m-n=3

⇒n=3 and m=6

Distance between P6, 3 and Q-2,-3 is given by,

PQ=6+22+3+32

⇒PQ=82+62

⇒PQ=100

⇒PQ=10 units

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2024 (27 Jan Shift 2)
Let A and B be two independent events such that PA=13 and PB=16. Then, which of the following is true?

Options:
A PAB=23
B PAB'=13
C PA'B'=13
D PAA∪B=14

The correct answer is:
PAB'=13

Solution:
A & B are independent events, so
PAB'=P(A∩B')P(B')=P(A)·P(B')P(B')=P(A)=13

Topic: Mathematics, Probability, JEE Main, JEE Main 2020 (08 Jan Shift 1)
Let $A$ and $B$ be two invertible matrices of order $3 \times 3$. If $\operatorname{det}\left(\mathrm{ABA}^{\mathrm{T}}\right)=8$ and det $\left(\mathrm{AB}^{-1}\right)=8$, then det $\left(\mathrm{BA}^{-1} \mathrm{~B}^{\mathrm{T}}\right)$ is

equal to

Options:
A $\frac{1}{4}$
B 1
C $\frac{1}{16}$
D 16

The correct answer is:
$\frac{1}{16}$

Solution:
Let $|A|=a,|B|=b$

$\left.\Rightarrow \quad \mid A^{\eta}\right\rceil=a\left|A^{-1}\right|=\frac{1}{a},\left|B^{\eta}\right|=b,\left|B^{-1}\right|=\frac{1}{b}$

$\because \quad A B A^{\eta}=8 \Rightarrow|A||B| \mid A^{\eta}-8 \ldots(1)$

$\Rightarrow \quad a \cdot b \cdot a=8 \Rightarrow a^{2} b=8$

$\because \quad A B^{-1}|=8 \Rightarrow H|\left|B^{-1}\right|=8 \Rightarrow a, \frac{1}{b}=8$

From (1) \& (2)

$a=4, b=\frac{1}{2}$

Then, $\left|B A^{-1} B^{\eta}\right|=|B|\left|A^{-1}\right|\left|B^{\eta}\right|=b \cdot \frac{1}{a} \cdot b=\frac{b^{2}}{a}=\frac{1}{16}$

Topic: Mathematics, Determinants, JEE Main, JEE Main 2019 (11 Jan Shift 2)
Let A and B be two invertible square matrices each of order $\mathrm{n}$. What is adj $(\mathrm{AB})$ equal to?

Options:
A $(\mathrm{adj} \mathrm{A})(\mathrm{adj} \mathrm{B})$
B $(\operatorname{adj} \mathrm{A})+(\operatorname{adj} \mathrm{B})$
C $(\operatorname{adj} \mathrm{A})-(\operatorname{adj} \mathrm{B})$
D $(\operatorname{adj} B)(\operatorname{adjA})$

The correct answer is:
$(\operatorname{adj} B)(\operatorname{adjA})$

Solution:
Since, $\mathrm{A}$ and $\mathrm{B}$ be two invertible square matrices each of order $\mathrm{n}$, then $(\mathrm{AB})^{-1}=\left(\mathrm{B}^{-1}\right)\left(\mathrm{A}^{-1}\right)$
$\Rightarrow \frac{\operatorname{adj}(\mathrm{AB})}{|\mathrm{AB}|}=\frac{\operatorname{adj} \mathrm{B}}{|\mathrm{B}|} \cdot \frac{\operatorname{adj} \mathrm{A}}{|\mathrm{A}|}$
Since $|\mathrm{AB}|=|\mathrm{B}||\mathrm{A}|$
$\operatorname{adj}(\mathrm{AB})=(\operatorname{adj} \mathrm{B})(\operatorname{adj} \mathrm{A})$

Topic: Mathematics, Matrices, JEE Main
Let A and B be two matrices such that the order of A is 5Ã7. If ATB and BAT are both defined, then (where AT is the transpose of matrix A)

Options:
A order of BT is 5×7
B order of BTA is 7×7
C order of BTA is 5×5
D BTA is undefined

The correct answer is:
order of BTA is 7×7

Solution:
Order of A is 5×7⇒ order of AT=7×5
If ATB is defined ⇒ order of B must be 5×n
If BAT is defined ⇒ order of B must be m×7
Hence, the order of B is 5×7
⇒Order of BT is 7×5
⇒Order of BTA is 7×7

Topic: Mathematics, Matrices, JEE Main
Let A and B be two non singular matrices of same order such that ABk=AkBk for consecutive positive integral values of k, then AB2A-1 is equal to

Options:
A A2
B B
C A
D B2

The correct answer is:
B2

Solution:
A≠0,B≠0 (given)
AB3=A3B3…i
Also, AB3=AB2AB=A2B2AB…ii
From i and ii
AB2=B2A
AB2A-1=B2AA-1=B2

Topic: Mathematics, Matrices, JEE Main
Let A and B be two non-null events such that A⊂B. Then, which of the following statements is always correct?

Options:
A PA|B≥P(A)
B PA|B=PB-PA
C PA|B≤ P(A)
D PA|B=1

The correct answer is:
PA|B≥P(A)

Solution:
Lets draw the venn diagram according to question,

PAB=P(A∩B)P(B)=P(A)P(B)      [A⊂B⇒A∩B=A]

Now, since P(B)≤1

⇒PAB=PAPB≥PA

⇒PA|B≥P(A)

Topic: Mathematics, Probability, JEE Main, JEE Main 2019 (08 Apr Shift 1)
Let a→ and b→ be two non-zero vectors perpendicular to each other and |a→|=|b→|, If |a→×b→|=|a→|, then the angle between the vectors
(a→+b→+(a→×b→)) and a→ is equal to :

Options:
A sin-113
B cos-113
C cos-112
D sin-116

The correct answer is:
cos-113

Solution:
|a→|=|b→|,|a→×b→|=|a→|,a→⊥b→

|a→×b→|=|a→|⇒|a→||b→|sin90°=|a→|⇒|b→|=1=|a→|

a→ and b→ are mutually perpendicular unit vectors.

Let a→=i^,b→=j^⇒a→×b→=k^

cosθ=(i^+j^+k^)i^31=13⇒θ=cos-113

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Let a→ and b→ be two orthogonal unit vectors and c→ is a vector such that c→×b→=a→-c→, then c→ is equal to

Options:
A 1
B 2
C 12
D 2

The correct answer is:
12

Solution:
c â Ã b â = a â â c â
Taking dot product with b â , we get,
0= a â â b â â c â â b â â c â â b â =0 As a â â b â =0 â¦â¦.. i
Taking dot product with c â , we get,
0= a â â c â â c â 2 â a â â c â = c â 2 â¦â¦.. ii
Taking dot product with a â , we get,
c â b â a â = a â â a â â a â â c â
â c â b â a â =1â c â 2 â¦â¦.. iii
Given, câÃbâ=aâ-câ
â c â Ã b â 2 = a â â c â 2 â
â c â 2 b â 2 = a â 2 + c â 2 â2 a â â c â
From i & ii, we get,
c â 2 =1+ c â 2 â2 c â 2 â c â = 1 2

Topic: Mathematics, Vector Algebra, JEE Main
Let A and B be two sets containing 2 elements and 4 elements respectively. The number of subsets of A×B having 3 or more elements is :

Options:
A 219
B 211
C 256
D 220

The correct answer is:
219

Solution:
As nA=2 and nB=4
Hence nA×B=8
Now subsets of A×B containing 3 or more elements will be
=Total - (no element - one element -2 elements)
=28-8C0-8C1-8C2=219

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2013 (07 Apr)
Let A and B be two sets containing four and two elements respectively. Then the number of subsets of the set A×B, each having at least three elements is

Options:
A 510
B 219
C 256
D 275

The correct answer is:
219

Solution:
Given nA=4

nB=2

Number of elements in nA × B=2×4=8

Number of subsets=28

Number of subsets having less than 3 elements =8C0+C81+C82

Number of subsets having at least 3 elements
=28-8C0-C81-C82

=256-1-8-28

=219

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2015 (04 Apr)
Let A and B be two sets. The set A has 2016 more subsets than B. If Aâ©B has 3 members, then the number of members in AâªB is

Options:
A 10
B 11
C 12
D 13

The correct answer is:
13

Solution:
If A has m members and B has n members, then
2m-2n=2016=25.63
â2n2m-n-1=2564-1=2526-1
ân=5 and m-n=6âm=11
The number of members in AâªB
=11+5-3=13

Topic: Mathematics, Sets and Relations, JEE Main
Let A and B be two square matrices of order 3 such that A=3 and B=2, then the value of A-1⋅adjB-1⋅adj2A-1 is equal to (where adj(M) represents the adjoint matrix of M)

Options:
A 72
B 6427
C 89
D 1627

The correct answer is:
1627

Solution:
A-1⋅adjB-1⋅adj2A-1=1A⋅B-122A-12
=1A×1B2⋅26A2=2633⋅22=1627

Topic: Mathematics, Determinants, JEE Main
Let $A$ and $B$ be two symmetric matrices of order 3 .
This question has Statement $-1$ and Statement $-2$. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement $-1$ : $\mathrm{A}(\mathrm{BA})$ and $(\mathrm{AB}) \mathrm{A}$ are symmetric matrices.
Statement - 2 : $\quad A B$ is symmetric matrix if matrix multiplication of $A$ and $B$ is commutative.

Options:
A
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is not a correct explanation for Statement $-1$

B
Statement $-1$ is true, Statement $-2$ is false.

C
Statement $-1$ is false, Statement- 2 is true.

D
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is a correct explanation for Statement $-1$

The correct answer is:
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is not a correct explanation for Statement $-1$

Solution:
$$
\begin{aligned}
& A^{\top}=A, B^{\top}=B \\
& (A(B A))^{\top}=(B A)^{\top} A^{\top}=\left(A^{\top} B^{\top}\right) A=(A B) A=A(B A) \\
& ((A B) A)^{\top}=A^{\top}(A B)^{\top}=A\left(B^{\top} A^{\top}\right)=A(B A)=(A B) A
\end{aligned}
$$
$\therefore$ Statement $-1$ is correct
Statement - 2
$$
(A B)^{\top}=B^{\top} A^{\top}=B A=A B
$$
( $\because \mathrm{AB}$ is commutative)
Statement $-2$ is also correct but it is not correct explanation of Statement $-1$

Topic: Mathematics, Matrices, JEE Main, JEE Main 2011
Let $\hat{a}$ and $\hat{b}$ be two unit vectors. If the vectors $\vec{c}=\hat{a}+2 \hat{b}$ and $\vec{d}=5 \hat{a}-4 \hat{b}$ are perpendicular to each other, then the angle between $\hat{\mathrm{a}}$ and $\hat{\mathrm{b}}$ is

Options:
A
$\frac{\pi}{6}$

B
$\frac{\pi}{2}$

C
$\frac{\pi}{3}$

D
$\frac{\pi}{4}$

The correct answer is:
$\frac{\pi}{3}$

Solution:
$\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}}=\overrightarrow{0} \quad \Rightarrow 5|\overrightarrow{\mathrm{a}}|^2+6 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}-8|\overrightarrow{\mathrm{b}}|^2=0$
$\Rightarrow 6 \vec{a} \cdot \vec{b}=3 \quad \Rightarrow \vec{a} \cdot \vec{b}=\frac{1}{2} \quad \Rightarrow(\vec{a} \cdot \vec{b})=\frac{\pi}{3}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2012 (Offline)
Let aâ and bâ be two unit vectors such that aâ.bâ=0. For some x,yâR, let câ=xaâ+ybâ+aâÃbâ. If câ=2 and the vector câ is inclined at the same angle Î± to both a â and b â , then the value of 8 cos 2 Î± is

Options:

The correct answer is:
3

Solution:
c→=xa→+yb→+a→×b→
c→.a→=x and x=2cos⁡α
c→.b→=y and y=2cos⁡α
Also, a→×b→=1
∴ c→=2cos α ⁡a→+b→+a→ ×b→
c→2=4cos2⁡αa→+b→2+a→×b→2+4cos⁡αa→+b→.a→×b→
4=8cos2⁡α+1
8cos2⁡α=3

Topic: Mathematics, Vector Algebra, JEE Main
Let a and b be two unit vectors such that a+b+2a×b=2. If θ∈0,π is the angle between a^ and b^, then among the statements:
S1:2a^×b^=a^-b^
S2 : The projection of a^ on a^+b^ is 12

Options:
A Only S1 is true.
B Only S2 is true.
C Both S1 and S2 are true.
D Both S1 and S2 are false.

The correct answer is:
Both S1 and S2 are true.

Solution:
Given,

a^+b^+2a^×b^=2,θ∈0,π

Squaring both side we get,

a^+b^+2a^×b^·a^+b^+2a^×b^=4

a^+b^2+4a^×b^2+0=4

Let the angle be θ between a^ and b^

⇒a^2+b^2+2a^b^cosθ+4a^b^sinθ=4

⇒2+2cosθ+4sin2θ=4

⇒2+2cosθ-4cos2θ=0

Let cosθ=t then

⇒2t2-t-1=0

⇒2t2-2t+t-1=0

⇒2tt-1+t-1=0

2t+1t-1=0

So, t=-12 or t=1

Then, cosθ=-12 or cosθ=1 which is not possible as 0∈0,π

So, θ=2π3

Now,

S1⇒ 2a×b=2sin2π3=2×32=3

And a^-b^=1+1-2cos2π3

=2-2×-12

=3

S1 is correct.

S2 projection of a^ on a^+b^

⇒a^.a^+b^∣a^+b^)=1+cos2π32+2cos2π3

⇒1-121=12

Which is true, So both are true.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2022 (24 Jun Shift 2)
Let a→and b→ be two unit vectors such that a→+b→=3. If c→=a→+2b→+(a→×b→) , then 2c→ is equal to:

Options:
A 51
B 37
C 43
D 55

The correct answer is:
55

Solution:
As aâ +bâ = 3
Squaring both the sides
aâ2+ bâ2+2aââbâ=3
1+1+2 â1â1âcosâ¡Î¸=3
2cosÎ¸=1
cosÎ¸=12
Î¸=60Â°
â´ Angle between aâ and bâ is 60o
Now,
| c â |â=â| a â â+â2 b â â+â3( a â âÃâ b â )|
Squaring both the sides
| c â | 2 â=â| a â â | 2 +â4| b â |â+â9| a â âÃâ b â | 2 +â4 a â .( b â )+3 a â .( a â âÃâ b â )+6 b â .( a â âÃâ b â )
| c â | 2 â=1+4+9 sin 2 Î¸+4cosÎ¸+0+0
| c â | 2 â=5+9. 3 4 +4. 1 2 = 55 4
â´ââ2â| c â |â= 55

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2015 (10 Apr Online)
Let a^ and b^ be two unit vectors such that the angle between them is π4. If θ is the angle between the vectors a^+b^ and a^+2b^+2a^×b^ then the value of 164cos2θ is equal to

Options:
A 90+272
B 45+182
C 90+32
D 54+902

The correct answer is:
90+272

Solution:
Let a^ and b^ be two unit vectors such that the angle between them is π4. If θ is the angle between the vectors a^+b^ and a^+2b^+2a^×b^ then the value of 164cos2θ is equal to

Now let angle between a^ & b^=π4=ϕ

So, a^·b^=a^b^cosϕ

⇒a^·b^=cosϕ=12

Now finding angle between a^+b^ & a^+2b^+2a^×b^ we get,

cosθ=a^+b^·a^+2b^+2a^×b^a^+b^a^+2b^+2a^×b^ ....1

Now finding the value of a^+b^2=a^+b^·a^+b^

⇒a^+b^2=2+2a^·b^=2+2

Also value of a^×b^=a^b^sinϕn^

⇒a^×b^=n^2 when n^ is vector ⊥to a^ and b^

Let c→=a^×b^

We know that, c→·a→=0 and c→·b→=0 as perpendicular vector dot product is zero.

Now value of a^+2b^+2c→2

=1+4+42+4a^·b^+8b^·c→+4c→·a^

=7+42=7+22

Now finding dot product we get, a^+b^·a^+2b^+2c→

=a^2+2a^·b^+0+b^·a^+2b^2+0

=1+22+12+2

=3+32

Now putting all the values in equation 1 we get,

cosθ=3+322+27+22

⇒cos2θ=92+1222+27+22

⇒cos2θ=9222+17+22

⇒164cos2θ=82922+17+227-227-22

⇒164cos2θ=822972-4+7-2241

⇒164cos2θ=9252+3

⇒164cos2θ=90+272

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2022 (29 Jul Shift 1)
Let a→ and b→ be two vector such that a→=14, b→=6 and a→×b→=48. Then a→·b→2 is equal to _____ .

Options:

The correct answer is:
36

Solution:
Given,

a→=14, b→=6 & a→×b→=48

Now we know that,

a→×b→2+a→·b→2=a→2b→2sin2θ+a→2b→2cos2θ

⇒a→×b→2+a→·b→2=a→2b→2

as cos2θ+sin2θ=1

⇒48+a→·b→2=14×6

⇒a→·b→2=84-48=36.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (31 Jan Shift 1)