datavorous/entrance-exam-dataset huggingface dataset:
If A and B are matrices with 24 and 40 elements respectively, then the number of possible orders of A and B such that AB is defined is

Options:
A 2
B 3
C 4
D 8

The correct answer is:
4

Solution:
If AB is defined then order of A must be m×n and order of B is n×p
mn=24=23×3
np=40=23×5
n can be 20,21,22,23
There are only 4 possibilities of n, so the number of possible orders of A,B is 4

Topic: Mathematics, Matrices, JEE Main
If $\vec{a}$ and $\vec{b}$ are non-collinear vectors, then the value of $\alpha$ for which the vectors $\vec{u}=(\alpha-2) \vec{a}+\vec{b}$ and $\vec{v}=(2+3 \alpha) \vec{a}-3 \vec{b}$ are collinear is :

Options:
A
$\frac{3}{2}$

B
$\frac{2}{3}$

C
$-\frac{3}{2}$

D
$-\frac{2}{3}$

The correct answer is:
$\frac{2}{3}$

Solution:
Since, $\vec{u}$ and $\vec{v}$ are collinear, therefore
$$
\begin{aligned}
& \vec{u}+\vec{v}=0 \\
\Rightarrow & {[k(\alpha-2)+2+3 \alpha] \vec{a}+(k-3) \vec{b}=0 }
\end{aligned}
$$
Since $\vec{a}$ and $\vec{b}$ are non-collinear, then for some constant $m$ and $n$,
$$
m \vec{a}+n \vec{b}=0 \Rightarrow m=0, n=0
$$
Hence from equation (i)
$$
\begin{aligned}
& k-3=0 \Rightarrow k=3 \\
& \text { And } k(\alpha-2)+2+3 \alpha=0 \\
& \Rightarrow 3(\alpha-2)+2+3 \alpha=0 \Rightarrow \alpha=\frac{2}{3}
\end{aligned}
$$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2013 (23 Apr Online)
If A and B are non-singular matrices of order 3Ã3, such that A=adjB and B=adjA , then detA+detB is equal to (where det(M) represents the determinant of matrix M and adjM represents the adjoint matrix of matrix M)

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
A=adjB=adjadjA=AA
⇒A=A4⇒A=1⇒B=1
Hence, A+B=2

Topic: Mathematics, Determinants, JEE Main
If A and B are non-singular matrices of order three such that adjAB=1111α111α and B2adjA=α2+3α-8, then the value of α is equal to

Options:
A 95
B 85
C 3
D 2

The correct answer is:
95

Solution:
adjAB=1111α111α
=1α2-1-1α-1+11-α
=α2-2α+1=α-12
⇒AB2=α-12 and B2adjA=B2A2
=AB2=α-12=α2+3α-8
⇒α2-2α+1=α2+3α-8
⇒5α=9⇒α=95

Topic: Mathematics, Determinants, JEE Main
If $A$ and $B$ are non-singular square matrices of same order, then $\operatorname{adj}(A B)$ is equal to

Options:
A $(\operatorname{adj} A)(\operatorname{adj} B)$
B $(\operatorname{adj} B)(\operatorname{adj} A)$
C $\left(\operatorname{adj} B^{-1}\right)\left(\operatorname{adj} A^{-1}\right)$
D $\left(\operatorname{adj} A^{-1}\right)\left(\operatorname{adj} B^{-1}\right)$

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

Solution:
\(A\) and \(B\) are non-singular, so \(A B\) is non-singular.
\(\begin{aligned}
& \text {Hence } \mathrm{AB} \operatorname{adj}(\mathrm{AB})=|\mathrm{AB}| \mathrm{I}(1) \\
& \mathrm{AB}(\operatorname{adj} \mathrm{B} \operatorname{adj} \mathrm{A})=\mathrm{A}(\mathrm{B} \operatorname{adj} \mathrm{B}) \operatorname{adj} \mathrm{A} \\
& =\mathrm{A}(|\mathrm{B}| \mathrm{I}) \operatorname{adj} \mathrm{A} \\
& =|\mathrm{B}|(\mathrm{A}(\operatorname{adj} \mathrm{A})) \\
& =|\mathrm{B}||\mathrm{A}| \mathrm{I}(2) \\
& \operatorname{adj}(\mathrm{AB})=(\operatorname{adj} \mathrm{B})(\operatorname{adj} \mathrm{A})
\end{aligned}\)

Topic: Mathematics, Matrices, JEE Main
If A and B are non-singular square matrix of same order 3×3, then which of the following options is correct?

Options:
A adj(AB)=AB
B (adjAB)-1=adj(AB)
C adj(AB)-1=(adjAB)-1
D adj(AB)T=AB-2

The correct answer is:
adj(AB)-1=(adjAB)-1

Solution:
adj(AB)=AB2
(adj AB)-1=adj AB-1
adj(AB)-1=(AB)-12=AB-2
adj(AB)-1=adj AB-1=AB2-1=AB-2
adj(AB)T=(AB)T2=AB2

Topic: Mathematics, Determinants, JEE Main
If a→ and b→ are perpendicular, then a→×a→×a→×a→×b→ is equal to:

Options:
A a→4b→
B 0→
C a→×b→
D 12a→4b→

The correct answer is:
a→4b→

Solution:
a→·b→=0

a→×a→×b→=a→·b→a→-a→·a→b→=-a→2b→

Now a→×a→×-a→2b→

=-a→2a→×a→×b→

=-a→2-a→2b→=a→4b→

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (26 Feb Shift 1)
If $A$ and $B$ are positive acute angles satisfying $3 \cos ^{2} A+2 \cos ^{2} B=4$ and $\frac{3 \sin A}{\sin B}=\frac{2 \cos B}{\cos A}$

Then the value of $\mathrm{A}+2 \mathrm{~B}$ is equal to :

Options:
A $\frac{\pi}{6}$
B $\frac{\pi}{2}$
C $\frac{\pi}{3}$
D $\frac{\pi}{4}$

The correct answer is:
$\frac{\pi}{2}$

Solution:
Given, $3 \cos ^{2} \mathrm{~A}+2 \cos ^{2} \mathrm{~B}=4$

$\Rightarrow 2 \cos ^{2} \mathrm{~B}-1=4-3 \cos ^{2} \mathrm{~A}-1$

$\Rightarrow \cos 2 \mathrm{~B}=3\left(1-\cos ^{2} \mathrm{~A}\right)=3 \sin ^{2} \mathrm{~A} \ldots(1)$

and $2 \cos \mathrm{B} \sin \mathrm{B}=3 \sin \mathrm{A} \cos \mathrm{A}$

$\sin 2 \mathrm{~B}=3 \sin \mathrm{A} \cos \mathrm{A}$

Now, $\cos (\mathrm{A}+2 \mathrm{~B})=\cos \mathrm{A} \cos 2 \mathrm{~B}-\sin \mathrm{A}$

$\sin 2 \mathrm{~B}$

$=\cos A\left(3 \sin ^{2} A\right)-\sin A(3 \sin A \cos A)=0$

[using eqs. (1) and (2)] $\Rightarrow A+2 B=\frac{\pi}{2}$

Topic: Mathematics, Trigonometric Equations, JEE Main
If a and b are positive integers such that N=a+ib3-107i (where N is a natural number), then the value of a is equal to (where i2=-1)

Options:
A 4
B 5
C 6
D 9

The correct answer is:
6

Solution:
a+ib3-107i= Natural number
⇒a3-3b2a+i3a2b-b3-107i=Natural number
⇒3a2b-b3-107=0
⇒a2=b3+1073b=b23+1073b
⇒b3+107 is a multiple of 3 and b is a factor of 107
⇒b=1 and a2=1083=36⇒a=6
{b cannot be equal to 107, because 1073+107 is not a multiple of 3}

Topic: Mathematics, Complex Number, JEE Main
If a and b are real numbers such that 2+α4=a+bα, where α=-1+i32, then a+b is equal to:

Options:
A 9
B 24
C 33
D 57

The correct answer is:
9

Solution:
2+α4=a+bα
4+α2+4α2=a+bα

Now using, α=-1+i32⇒2α+12=i32

⇒α2=-1+α

So, a+bα=4-α-1+4α2

a+bα=3+3α2=91+2α+α2
a+bα=91+2α+-1-α=9α

⇒a=0, b=9⇒a+b=9

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2020 (04 Sep Shift 2)
If A and B are square matrices of order 3 such that A=3 and B=2, then the value of A-1adjB-1adj3A-1 is equal to

Options:
A 27
B 274
C 1108
D 14

The correct answer is:
274

Solution:
A-1adjB-1adj3A-1=A-1adjB-1adj3A-1
= 1AB-12×3A-12
= 1A×1B2×36A2
= 3633×22=274

Topic: Mathematics, Determinants, JEE Main
If A and B are square matrices of order 3 such that AAT=3B and 2AB-1=3A-1B, then the value of B216 is equal to

Options:

The correct answer is:
4

Solution:
Given, AAT=3B
Taking determinant on both sides, we get,
AAT=33B⇒A2=27B … 1
Also given, 2AB-1=3A-1B
Taking determinant on both sides, we get,
23A1B=331A⋅B
23A2=33B2 … 2
From 1 and 2, we get,
2327B=33B2⇒B2=8B
⇒B=8

Topic: Mathematics, Determinants, JEE Main
If \( A \) and \( B \) are square matrices of order \( 3 \) such that \( |A|=3 \) and \( |B|=2 \), then the value of \( \left|A^{-1} a d j B^{-1} a d j\left(3 A^{-1}\right)\right| \) is equal
to

Options:
A \( 27 \)
B \( \frac{27}{4} \)
C \( \frac{1}{108} \)
D \( \frac{1}{4} \)

The correct answer is:
\( \frac{27}{4} \)

Solution:
A-1adjB-1adj3A-1=A-1adjB-1adj3A-1

= 1AB-12×3A-12

= 1A×1B2×36A2

= 3633×22=274

Topic: Mathematics, Matrices, JEE Main
If $\mathrm{A}$ and $\mathrm{B}$ are square matrices of second order such that $|\mathrm{A}|=-1,|\mathrm{~B}|=3$, then what is $|3 \mathrm{AB}|$ equal to?

Options:
A 3
B $-9$
C $-27$
D None of these

The correct answer is:
$-27$

Solution:
We know that,$|\mathrm{kA}|=\mathrm{k}^{\mathrm{n}}|\mathrm{A}|$, where $\mathrm{n}$ is order of matrix $\mathrm{A}$.
$\therefore|3 \mathrm{AB}|=3^{2}|\mathrm{~A}||\mathrm{B}| \quad(\because|\mathrm{AB}|=|\mathrm{A}||\mathrm{B}|)$
$=9(-1)(3)$
$=-27 \quad(\because|\mathrm{A}|=-1,|\mathrm{~B}|=3)$

Topic: Mathematics, Determinants, JEE Main
If $A$ and $B$ are square matrices of size $n \times n$ such that $A^2-B^2=(A-B)(A+B)$, then which of the following will be always true?

Options:
A
A = B

B
AB = BA

C
either of A or B is a zero matrix

D
either of A or B is an identity matrix

The correct answer is:
AB = BA

Solution:
$A^2-B^2=(A-B)(A+B)$
$A^2-B^2=A^2+A B-B A-B^2$
$\Rightarrow A B=B A$

Topic: Mathematics, Matrices, JEE Main, JEE Main 2006
If A and B are square matrices of the same order such that A=-B-1AB, then A+3B2 is equal to

Options:
A A+3B
B A2+9B2
C A2+6AB+AB2
D O

The correct answer is:
A2+9B2

Solution:
A=-B-1AB⇒BA=-AB⇒BA+AB=0
A+3B2=A2+9B2+3AB+3BA=A2+9B2

Topic: Mathematics, Matrices, JEE Main
If A and B are square matrices such that A2020=O and AB=A+B, then B is equal to (where, O is a null matrix)

Options:
A 0
B 1
C -1
D 4

The correct answer is:
0

Solution:
A2020=O⇒A=0
⇒B=AB-I
Taking determinant on both sides, we get,
B=AB-I⇒B=0 (as A=0)

Topic: Mathematics, Determinants, JEE Main
$\begin{array}{l}
\text { If } \mathrm{A} \text { and } \mathrm{B} \text { are subsets of universal set } \mathrm{X} \text { such that } \mathrm{n}(\mathrm{X})=200, \mathrm{n}(\mathrm{A})=90, \mathrm{n}(\mathrm{B})=80 \text { , }\\
n\left(A^{\prime} \cap B^{\prime}\right)=40, \text { then } n\left(A \cap B^{\prime}\right)=
\end{array}$

Options:
A 70
B 80
C 20
D 10

The correct answer is:
80

Solution:
$$
\begin{aligned}
n\left(A \cap B^{\prime}\right) &=n(A)-n(A \cap B) \\
n(x) &=200 \\
& \sim(A \cap B)=n(A)+n(B)+2\left(A^{\prime} \cap B^{\prime}\right) \\
&-n(X) \\
n\left(A \cap B^{\prime}\right) &=90-10 \\
&=80
\end{aligned}
$$

Topic: Mathematics, Sets and Relations, JEE Main
If $A$ and $B$ are symmetric matrices of the same order, then which one of the following is not true?

Options:
A $A+B$ is symmetric
B $A-B$ is symmetric
C $A B+B A$ is symmetric
D $A B-B A$ is symmetric

The correct answer is:
$A B-B A$ is symmetric

Solution:
Given, $A$ and $B$ are symmetric matrix of same order.
$$
\Rightarrow \quad A=A^{\prime}, B=B^{\prime}
$$
(i) $(A+B)^{\prime}=A^{\prime}+B^{\prime}=A+B \quad$ [from Eq. (i)] $\Rightarrow(A+B)$ is symmetric.
(ii) $(A-B)^{\prime}-\Lambda^{\prime}-B^{\prime}-\Lambda-B$ [from Eq. (i)] $\Rightarrow(A-B)$ is symmetric.
$$
\text { (iii) } \begin{aligned}
(A B+B A)^{\prime} &=(A B)^{\prime}+(B A)^{\prime} \\
&=B^{\prime} A^{\prime}+A^{\prime} B^{\prime} \\
&=B A+A B \quad\lfloor\text { from Eq. (i) }] \\
&=(A B+B A)
\end{aligned}
$$
$\Rightarrow(A B+B A)$ is symmetric.
$$
\text { (iv) } \begin{aligned}
(A B-B A)^{\prime} &=(A B)^{\prime}-(B A)^{\prime} \\
&=B^{\prime} A^{\prime}-A^{\prime} B^{\prime} \\
&=B A-A B \quad \text { [from Eq. (i)] } \\
&=-(A B-B A)
\end{aligned}
$$
$\Rightarrow(A B-B A)$ is skew-symmetric matrix.

Topic: Mathematics, Matrices, JEE Main
If a and b are the roots of the equation x2-7x-1=0, then the value of a21+b21+a17+b17a19+b19 is equal to

Options:

The correct answer is:
51

Solution:
Since, a and b are roots of x2-7x-1=0, so

a2-7a-1=0

⇒a2-1=7a

⇒a4+1-2a2=49a2

⇒a4+1=51a2 ...1

Similarly,

b4+1=51b2 ...2

Now,

a21+b21+a17+b17a19+b19

=a17a4+1+b17b4+1a19+b19

=51a19+51b19a19+b19

=51

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2023 (11 Apr Shift 1)
If $a$ and $b$ are two different positive real numbers, then which of the following relations is true

Options:
A $2 \sqrt{a b} \gt (a+b)$
B $2 \sqrt{a b} \lt (a+b)$
C $2 \sqrt{a b}=(a+b)$
D None of these

The correct answer is:
$2 \sqrt{a b} \lt (a+b)$

Solution:
We know that $A \gt G \gt H$
Where $A$ is arithmetic mean, $G$ is geometric mean and $H$ is harmonic mean, then $A \gt G$
$\Rightarrow \frac{a+b}{2} \gt \sqrt{a b}$ or $(a+b) \gt 2 \sqrt{a b}$

Topic: Mathematics, Sequences and Series, JEE Main
If A and B are two events defined on a sample space with the probabilities PA=0.5, PB=0.69 and PAB=0.5, then the value of PAAc∪Bc is equal to

Options:
A 25
B 313
C 31131
D 100131

The correct answer is:
31131

Solution:
PAB=PA∩BPB=0.5⇒PA∩B=0.5×0.69=0.345
PAAc∪Bc=PA∩Ac∪BcPAc∪Bc=PA∩BcPA∩Bc
=PA-PA∩B1-PA∩B=0.5-0.3451-0.345=0.1550.655=31131

Topic: Mathematics, Probability, JEE Main
If A and B are two events such that PA=13,PB=15 and PA∪B=12, then PAB'+PBA' is equal to

Options:
A 34
B 58
C 54
D 78

The correct answer is:
58

Solution:
Given,

PA=13,PB=15 and PA∪B=12

We know that,

PA∪B=PA+PB-PA∩B

⇒PA∩B=13+15-12=130

Now PA∩B'=Only A=13-130=930

And PB∩A'=Only B=15-130=530

Now, PA∣B'+PB∣A'=PA∩B'PB'+PB∩A'PA'

=93045+53023=58

Topic: Mathematics, Probability, JEE Main, JEE Main 2022 (25 Jul Shift 2)
If A and B are two events such that PA=47, PAâ©B=328 and the conditional probability PAAcâªBc (where Ac denotes the compliment of the event A) is equal to Î», then the value of 26Î» is equal to

Options:

The correct answer is:
50

Solution:
PAAc∪Bc=PA∩A∩BcPA∩Bc=PA∩Bc1-PA∩B
=PA-PA∩B1-PA∩B=47-3282528=1325=λ

Topic: Mathematics, Probability, JEE Main
If $A$ and $B$ are two events such that $P(A \cup B)+P(A \cap B)=\frac{7}{8}$ and $P(A)=2 P(B)$, then $P(A)=$

Options:
A $\frac{7}{12}$
B $\frac{7}{24}$
C $\frac{5}{12}$
D $\frac{17}{24}$

The correct answer is:
$\frac{7}{12}$

Solution:
$P(A \cup B)+P(A \cap B)=P(A)+P(B)$ $=P(A)+\frac{P(A)}{2}$
$\Rightarrow \frac{7}{8}=\frac{3 P(A)}{2} \Rightarrow P(A)=\frac{7}{12}$

Topic: Mathematics, Probability, JEE Main
If Aand B are two events such that PA∪B=PA∩B, then the incorrect statement amongst the following statements is :

Options:
A PA+PB=1
B PA∩B'=0
C A & B are equally likely
D PA'∩B=0

The correct answer is:
PA+PB=1

Solution:
PA∪B=PA∩B

Using the addition theorem of probability,

PA∪B=PA+PB-PA∩B

⇒PA=PB

PA∩B'=0

PA'∩B=0

Here, A and B are equally likely.

But we can't say that PA=PB=12

Therefore, PA+PB≠1

Topic: Mathematics, Probability, JEE Main, JEE Main 2014 (09 Apr Online)
If A and B are two independent events such that PA>12, PAâ©BC=325 and PACâ©B=825, then PA is equal to (where, Ac and Bc represent the complement of events A and B respectively)

Options:
A 15
B 35
C 34
D 45

The correct answer is:
35

Solution:
A,B are independent events
PA∩BC=PAPBC=PA1-PB=325 … 1
PAC∩B=PAC⋅PB=1-PAPB=825 ... 2
From 1 and 2, we get,
PB-PA=15 ... 3
From 1 and 3
PB-151-PB=325
Let, PB=t
⇒65t-15-t2=325⇒30t-5-25t2=3
⇒25t2-30t+8=0⇒t=25,45
PB=25⇒PA=15
PB=45⇒PA=35
∵PA>12⇒PA=35

Topic: Mathematics, Probability, JEE Main
If A and B are two independent events such that PA=23 and PBC=14, then which of the following is false?

Options:
A PAA∪B=711
B PA∩BAc∩Bc=0
C PAB=23
D P(AC∩BC)=112

The correct answer is:
PAA∪B=711

Solution:
PA=23,PB=34
∵A,B are independent events
PA∩B=PA·PB=12
PA∪B=PA+PB-PA∩B
=23+34-12=8+9-612=1112
PAA∪B=PA∩A∪BPA∪B=PAPAUB=231112=811
PA∩BA∪Bc=PA∩B∩A∪BCPA∪BC=0

(As A∩B & A∪BCare disjoint)
PAB=PA∩BPB=1234=23
PAc∩Bc=PA∪BC=1-PA∪B
=1-1112=112

Topic: Mathematics, Probability, JEE Main
If $A$ and $B$ are two independent events such that $P\left(A \cap B^{\prime}\right)=\frac{3}{25}$ and $P\left(A^{\prime} \cap B\right)=\frac{8}{25}$, then $P(A)=$

Options:
A $\frac{1}{5}$
B $\frac{3}{8}$
C $\frac{2}{5}$
D $\frac{4}{5}$

The correct answer is:
$\frac{1}{5}$

Solution:
Since events are independent.
So, $\quad P\left(A \cap B^{\prime}\right)=P(A) \times P\left(B^{\prime}\right)=\frac{3}{25}$ ldots...(1)
$\Rightarrow P(A) \times\{1-2 P(B)\}=\frac{3}{25}$ $\ldots.$ (i)
Similarly,
$P(B) \times\{1-P(A)\}=\frac{8}{25}$ $\ldots.$ (ii)
On solving (i) and (ii), we get $P(A)=\frac{1}{5}$ and $\frac{3}{5}$.

Topic: Mathematics, Probability, JEE Main
If A and B are two matrices of order 3×3 satisfying AB=A and BA=B, then A+B5 is equal to

Options:
A 5A+B
B 5I
C 16A+B
D 32I

The correct answer is:
16A+B

Solution:
AB=A,BA=B (Given)
∴A2=A⋅A=ABA=ABA=AB=A
⇒A5=A4=A3=A
Similarly,
B2=B⋅B=BA⋅B=BAB=BA=B
⇒B5=B4=B3=B
So, A+B2=A2+B2+AB+BA=2A+B
A+B3=A+B2A+B=2A+B2=4A+B
Similarly,
A+B4=8A+B
A+B5=16A+B

Topic: Mathematics, Matrices, JEE Main
If $A$ and $B$ are two matrices such that $A B=B$ and $B A=A$, then $A^2+B^2=$

Options:
A $2 A B$
B $2 B A$
C $A+B$
D $A B$

The correct answer is:
$A+B$

Solution:
We have $A B=B$ and $B A=A$.
Therefore
$A^2+B^2=A A+B B$ $=A(B A)+B(A B)$
$=(A B) A+(B A) B=B A+A B=A+B$
( $A B=B$ and $B A=A$ )

Topic: Mathematics, Matrices, JEE Main
If $A$ and $B$ are two mutually exclusive events, then what is P(A B) equal to?

Options:
A 0
B $P(A)+P(B)$
C $P(A) P(B)$
D $\mathrm{P}(\mathrm{A}) P\left(\frac{B}{A}\right)$

The correct answer is:
0

Solution:
Since, A and B are mutually exclusive events. $\therefore \quad \mathrm{P}(\mathrm{AB})=\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0$

Topic: Mathematics, Probability, JEE Main
If $\mathbf{a}$ and $\mathbf{b}$ are two non-zero and non-collinear vectors, then $\mathbf{a}+\mathbf{b}$ and $\mathbf{a}-\mathbf{b}$ are

Options:
A Linearly dependent vectors
B Linearly independent vectors
C Linearly dependent and independent vectors
D None of these

The correct answer is:
Linearly independent vectors

Solution:
Since $\mathbf{a}$ and $\mathbf{b}$ are non-collinear, so $\mathbf{a}+\mathbf{b}$ and $\mathbf{a}-\mathbf{b}$ will also be non-collinear. Hence, $\mathbf{a}+\mathbf{b}$ and $\mathbf{a}-\mathbf{b}$ are linearly independent vectors.

Topic: Mathematics, Vector Algebra, JEE Main
If A and B are two non-zero n×n matrices such that A2+B=A2B, then

Options:
A AB=I
B A2B=I
C A2=I or B=I
D A2B=BA2

The correct answer is:
A2B=BA2

Solution:
Given:

A2B=A2+B

⇒A2B-A2-B+I=I

⇒A2B-I-B-I=I

⇒A2-IB-I=I

⇒A2-IB-IB-I-1=IB-I-1

⇒A2-I=B-I-1

Also,

A2+B=A2B

⇒A2B-I=B

⇒A2=BB-I-1

⇒A2=BA2-I

⇒A2=BA2-B

⇒A2+B=BA2

⇒A2B=BA2

Topic: Mathematics, Matrices, JEE Main, JEE Main 2023 (24 Jan Shift 1)
If a and b are two real numbers lying between 0 and 1, such that Z1=a+i, Z2=1+bi and Z3=0 form an equilateral triangle, then

Options:
A a=2+3
B b=4-3
C a=b
D a=2,b=3

The correct answer is:
a=b

Solution:
If Z1, Z2, Z3 form an equilateral triangle, then we know that,
Z12+Z22+Z32=Z1Z2+Z2Z3+Z3Z1
âa2-1+2ai+1-b2+2bi-a+b-i-abi=0
âa-ba+b-1+2a+2b-ab-1i=0
Case-I:
a=b & 2a+2b-ab-1=0
âa=b & a2-4a+1=0 âa=b=2-3
Case-II:
a+b-1=0 & 2a+2b-ab-1=0
âab=1 (not possible because a,bâ0,1 )
âa=b=2-3 is the only solution

Topic: Mathematics, Complex Number, JEE Main
If A and B are two sets such that nA=2 and nB=4, then the total number of subsets of AÃB not having more than 5 elements are

Options:
A 219
B 247
C 239
D 248

The correct answer is:
219

Solution:
nA×B=nA×nB=2×4=8
∴ total number of subsets of A×B are 28=256
For the number of subsets not having more than 5 number of elements.
The required subsets =total number of subsets-(sets having 6 elements) - (sets having 7 elements) - (sets having 8 elements)
=256-8×72-8-1=219

Topic: Mathematics, Sets and Relations, JEE Main
If A→ and B→ are two vectors satisfying the relation A→·B→=|A→×B→|. Then the value of |A→-B→| will be:

Options:
A A2+B2
B A2+B2+2AB
C A2+B2+2AB
D A2+B2-2AB

The correct answer is:
A2+B2-2AB

Solution:
A→·B→=|A→×B→|
ABcosθ=ABsinθ⇒θ=45°
|A→-B→|=A2+B2-2ABcos45°
=A2+B2-2AB
Hence option (4).

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2021 (20 Jul Shift 1)
If a→ and b→ are unit vectors and a→+3b→ is perpendicular to 7a→-5b→ and a→-4b→ is perpendicular to 7a→-2b→, then the angle between a→ and b→ (in degrees) is _________.

Options:

The correct answer is:
60

Solution:
We have, a→+3b→⊥7a→-5b→

Therefore, a→+3b→·7a→-5b→=0

⇒7a→2-15b→2+16a→·b→=0 …1

and a→-4b→·7a→-2b→=0

⇒7a→2+8b→2-30a→·b→=0 …2

From 1 and 2, we get

2a→cosθ=b→

∴ cosθ=b→2a→

⇒θ=60°.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (25 Jul Shift 2)
If aâ and bâ are unit vectors making an angle Î± with each other, such that Î±â0,Ï and aâ+2bâ<5, then Î± lies in the interval

Options:
A 0,π
B 0,π2
C π2,2π3
D π6,5π6

The correct answer is:
0,π

Solution:
a→+2b→<5
⇒a→+2b→2<25
⇒a→2+4b→2+4a→⋅b→<25
⇒1+4+4cos⁡α<25
⇒cos⁡α<5⇒α∈0,π

Topic: Mathematics, Vector Algebra, JEE Main
If a→ and b→ are unit vectors, then the greatest value of 3|a→+b→|+|a→-b→| is

Options:

The correct answer is:
4

Solution:
Let angle between a→ and b→=α.
So, 3|a→+b→|+|a→-b→|=32+2cosα+2-2cosα
=32×2cos2α2+2×2sin2α2
=23cosα2+sinα2

As α∈0,π⇒α2∈0,π2

So, given expression is

=23cosα2+sinα2
Thus, maximum value is=232+12=2×2=4

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2020 (06 Sep Shift 1)
If $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ are vectors in space given by $\overrightarrow{\mathbf{a}}=\frac{\hat{\mathbf{i}}-2 \hat{\mathbf{j}}}{\sqrt{5}}$ and $\overrightarrow{\mathbf{b}}=\frac{2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}}{\sqrt{14}}$, then the value of $(2 \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot[(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times(\overrightarrow{\mathbf{a}}-2 \overrightarrow{\mathbf{b}})]$ is

Options:

The correct answer is:
5

Solution:
From the given information, it is clear that
$$
\begin{gathered}
\overrightarrow{\mathbf{a}}=\frac{\hat{\mathbf{i}}-2 \hat{\mathbf{j}}}{\sqrt{5}} \\
\Rightarrow \quad|\overrightarrow{\mathbf{a}}|=1,|\overrightarrow{\mathbf{b}}|=1, \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=0 \\
\text { Now, }(2 \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot[(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times(\overrightarrow{\mathbf{a}}-2 \overrightarrow{\mathbf{b}})]
\end{gathered}
$$
$$
\begin{aligned}
& =(2 \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot\left[a^2 \overrightarrow{\mathbf{b}}-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}) \cdot \overrightarrow{\mathbf{a}}+2 b^2 \cdot \overrightarrow{\mathbf{a}}\right. \\
& =[2 \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}] \cdot[\overrightarrow{\mathbf{b}}+2 \overrightarrow{\mathbf{a}}]=4 \overrightarrow{\mathbf{a}}^2+\overrightarrow{\mathbf{b}}^2 \\
& =4 \cdot 1+1=5 \quad \text { [as } a \cdot b=0]
\end{aligned}
$$

Topic: Mathematics, Vector Algebra, JEE Main
If a→ and b→ are vectors perpendicular to each other and a→=2, b→=3 and c→×a→=b→, then the least value of c→-a→ is

Options:

The correct answer is:
1.5

Solution:
c→×a→=b→
⇒c→a→sin⁡θ=3
⇒c→sin⁡θ=32
c→-a→2=c→2+a→2-2a→⋅c→
=c→2+4-2a→c→cos⁡θ
=c→2+4-4c→cos⁡θ
=94cosec2θ+4-6cot⁡θ
=94+94cot2θ-6cot⁡θ+4
⇒c-a2=94+32cot⁡θ-22≥94
Hence, c→-a→≥32

Topic: Mathematics, Vector Algebra, JEE Main
If $\mathbf{a}$ and $\mathbf{b}$ be parallel vectors, then [$\mathbf{a} \mathbf{c} \mathbf{b}]$ =

Options:
A 0
B 1
C 2
D None of these

The correct answer is:
0

Solution:
$\quad \mathbf{a} \cdot(\mathbf{c} \times \mathbf{b})=\mathbf{c} .(\mathbf{b} \times \mathbf{a})=0$, (Since $\mathbf{a}$ and $\mathbf{b}$ are parallel)

Topic: Mathematics, Vector Algebra, JEE Main
If $A$ and $B$ have $n$ elements in common, then the number of elements common to $A \times B$ and $B \times A$ is

Options:
A $\mathrm{n}$
B $2 \mathrm{n}$
C $\mathrm{n}^{2}$
D 0

The correct answer is:
$\mathrm{n}^{2}$

Solution:
Let $\quad \mathrm{C}=\mathrm{A} \cap \mathrm{B}$, then
$C \times C=(A \cap B) \times(A \cap B)$
$=(A \cap B) \times(B \cap A)$
$$
\begin{gathered}
=(A \times B) \cap(B \times A) \\
{[\because(A \cap C) \times(B \cap D)=(A \times B) \cap(C \times D)]}
\end{gathered}
$$
Since, $A \cap B$ has $\mathrm{n}$ elements, so $C$ has n elements. Hence, $\mathrm{C} \times \mathrm{C}$ has $\mathrm{n}^{2}$ elements. $\therefore \quad(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{B} \times \mathrm{A})$ has $\mathrm{n}^{2}$ elements.
Hence, $\mathrm{A} \times \mathrm{B}$ and $\mathrm{B} \times \mathrm{A}$ has $\mathrm{n}^{2}$ elements in common.

Topic: Mathematics, Sets and Relations, JEE Main
If $\vec{a}$ and $\vec{b}$ makes an angle $\cos ^{-1}\left(\frac{5}{9}\right)$ with each other, then $|\vec{a}+\vec{b}|=\sqrt{2}|\vec{a}-\vec{b}|$ for $|\vec{a}|=n|\vec{b}|$ The integer value of $\mathrm{n}$ is ____

Options:

The correct answer is:
3

Solution:
$\begin{aligned} & \cos \theta=\frac{5}{9} \\ & \frac{\vec{a} \cdot \vec{b}}{a b}=\frac{5}{9} \\ & a^2+b^2+2 \vec{a} \cdot \vec{b}=2 a^2+2 b^2-4 \vec{a} \cdot \vec{b} \\ & 6 \vec{a} \cdot \vec{b}=a^2+b^2\end{aligned}$
$\begin{aligned} & 6 \times \frac{5}{9} a b=a^2+b^2 \\ & \frac{10}{3} a b=a^2+b^2 \quad \& \quad a=n b \\ & \frac{10}{3} n b^2=n^2 b^2+b^2 \\ & 3 n^2-10 n+3=0 \\ & n=\frac{1}{3} \text { and } n=3\end{aligned}$
integer value $n=3$

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2024 (09 Apr Shift 1)