datavorous/entrance-exam-dataset huggingface dataset:
Let $M$ and $N$ be two $3 \times 3$ non-singular skew-symmetric matrices such that $M N=N M$. If $P^T$ denotes the transpose of $P$, then $M^2 N^2\left(M^T N\right)^{-1}\left(M N^{-1}\right)^T$ is equal to

Options:
A
$M^2$

B
$-N^2$

C
$-M^2$

D
$M N$

The correct answers are:
$-M^2$

Solution:
Given, $M^T=-M, N^T=-N$
and $\quad M N=N M$
$$
\begin{aligned}
\therefore & M^2 N^2\left(M^T N\right)^{-1}\left(M N^{-1}\right)^T \\
& =M^2 N^2 N^{-1}\left(M^T\right)^{-1}\left(N^{-1}\right)^T \cdot M^T \\
& =M^2 N\left(N N^{-1}\right)(-M)^{-1}\left(N^T\right)^{-1}(-M) \\
& =M^2 N\left(-M^{-1}\right)(-N)^{-1}(-M) \\
& =-M^2 N M^{-1} N^{-1} M \\
& =-M \cdot(M N) M^{-1} N^{-1} M \\
& =-M(N M) M^{-1} N^{-1} M \\
& =-M N\left(N M^{-1}\right) N^{-1} M \\
& =-M\left(N N^{-1}\right) M=-M^2
\end{aligned}
$$
Note Here, non-singular word should not be used, since there is no non-singular $3 \times 3$ skew-symmetric matrix.

Topic: Mathematics, Matrices, JEE Main
Let $M$ be $2 \times 2$ symmetric matrix with integer entries, then $M$ is invertible if

Options:
A the first column of $M$ is the transpose of second row of $M$.
B the second row of $M$ is the transpose of first column of $M$.
C $M$ is diagonal matrix with non-zero entries in the principal diagonal.
D The product of entries in the principal diagonal of $M$ is the product of entries in the other diagonal.

The correct answer is:
$M$ is diagonal matrix with non-zero entries in the principal diagonal.

Solution:
Let a symmetric matrix
$M=\left[\begin{array}{ll}
a & c \\
c & b
\end{array}\right]$
For matrix to be invertible, determinant must not be equal to zero.
$\begin{aligned}
& &|M| &=a b-c^{2} \neq 0 \\
\Rightarrow & & a b & \neq c^{2}
\end{aligned}$
Therefore, $M$ is a diagonal matrix with non-zero entries in the main diagonal and the product of entries in the main diagonal of $M$ is not the square of an integer.

Topic: Mathematics, Matrices, KCET, KCET 2021
Let M be a square matrix of order 3 whose elements are real numbers and
adjadjM=360-406 003 6, then the absolute value of TrM is
[Here, adjP denotes adjoint matrix of P and TrP denotes trace of matrix P i.e. sum of all principal diagonal elements of matrix P]

Options:

The correct answer is:
8

Solution:
adjadjM=Mn-2M
∴ order of M is 3
adjadjM=MM=360-4060066
Taking determinant on both the sides M4=360-4060036
M4=64
⇒ M=±6
hence, M=±60±230±100±12±1
Tr M=8

Topic: Mathematics, Matrices, JEE Main
Let M be any 3×3 matrix with entries from the set 0,1,2. The maximum number of such matrices, for which the sum of diagonal elements of MTM is seven, is______.

Options:

The correct answer is:
540

Solution:
abcdefghiadgbehcfi

a2+b2+c2+d2+e2+f2+g2+h2+i2=7

Case-I : Seven 1's and two 0's

C29=36

Case-II: One 2 and three 1's and five 0's

9!5!3!=504

∴ Total =540

Topic: Mathematics, Matrices, JEE Main, JEE Main 2021 (24 Feb Shift 1)
Let $M$ be the foot of the perpendicular from a point $P$ on the parabola $y^2=8(x-3)$ onto its directrix and let $S$ be the focus of the parabola. If $\triangle S P M$ is an equilateral triangle, then $P$ is equal to

Options:
A $(4 \sqrt{3}, 8)$
B $(8,4 \sqrt{3})$
C $(9,4 \sqrt{3})$
D $(4 \sqrt{3}, 9)$

The correct answer is:
$(9,4 \sqrt{3})$

Solution:
Given, that the $\triangle S P M$ (which is shown in figure) is equilateral.
Also, given parabola is $y^2=8(x-3)$ focus of this parabola is $S(5,0)$ and vertex $A(3,0)$.
Let coordinate of $P\left(h+a t^2, k+2 a t\right)$
$=P\left(3+2 t^2, 4 t\right)$
Then, coordinate of $M(-5,4 t)$.
We know that the side of this equilateral triangle is $4 a=4 \times 2=8$
Now, $P S=8$
$\sqrt{\left(3+2 t^2-5\right)^2+(4 t)^2}=8$
$\Rightarrow \quad \sqrt{\left(2 t^2-2\right)^2+(4 t)^2}=8$
$\Rightarrow \quad \sqrt{\left(2 t^2+2\right)^2}=8$
$\Rightarrow \quad 2 t^2+2=8$
$\Rightarrow \quad 2 t^2=6$
$\Rightarrow \quad t=\sqrt{3}$
$\therefore P(3+2 \times 3,4 \times \sqrt{3})=P(9,4 \sqrt{3})$

Topic: Mathematics, Parabola, JEE Main
Let M be the mass and L be the length of a thin uniform rod. In first case, axis of rotation is passing through center and perpendicular to the length of the rod. In second case axis of rotation is passing through one end and perpendicular to the length of the rod. The ratio of radius of gyration in first case to second case is

Options:
A 1
B 12
C 14
D 18

The correct answer is:
12

Solution:
Moment of inertia of rod whose axis of rotation is passing through center and perpendicular to the plane of rod is
I=ML212 and I=MK12 (where K 1 radius of gravitation)
âK1=L23 ..... (i)
When axis of rotation of rod is passing through one end of rod, then
I=MK22=ML23âK2=L3 ..... (ii)
Taking ratios of (i) and (ii) we get
K1K2=L23 Ã3L=12
â K1K2=12

Topic: Physics, Rotational Motion, JEE Main
Let M bea $3 \times 3 \mathrm{n} \times \mathrm{n}$ -singular matrix with $\operatorname{det}(\mathrm{M})$ $=\alpha$. If $\left[\mathrm{M}^{-1}\right.$ adj $(\mathrm{adj} (\mathrm{M})]=\mathrm{KI}$, then the value of $\mathrm{K}$ is

Options:
A 1
B $\alpha$
C $\alpha^{2}$
D $\alpha^{3}$

The correct answer is:
$\alpha$

Solution:
We know that, $M(\operatorname{adj} M)=|M|$ I

Replacing $\mathrm{M}$ by adj $\mathrm{M},$ we get $\operatorname{adj} M[\operatorname{adj}(\operatorname{adj} M)=\operatorname{det}(\operatorname{adj} M) I$

$=\operatorname{det}(M) M^{-1}\left[\operatorname{adj}(\operatorname{adj} M)=\alpha^{2} 1\right.$

$$

\left[\because \mathrm{M}^{-1}=\frac{1}{|\mathrm{M}|} \operatorname{adj}(\mathrm{M})\right]

$$

$\Rightarrow \alpha \mathrm{M}^{-1}[\operatorname{adj}(\operatorname{adj} M)]=\alpha^{2} \mathrm{I}$

$\Rightarrow \mathrm{M}^{-1}[\mathrm{adj}(\operatorname{adj} \mathrm{M})]=\alpha \mathrm{I}$

But $M^{-1}[\operatorname{adj}(\operatorname{adj} M)]=K I$

Hence, $\mathrm{K}=\alpha$

Topic: Mathematics, Matrices, JEE Main
Let M denote the median of the following frequency distribution.

Class 0-44-88-1212-1616-20 Frequency 391086

Then 20M is equal to :

Options:
A 416
B 104
C 52
D 208

The correct answer is:
208

Solution:
\(\begin{array}{|c|c|c|} \hline \text {Class } & \text {Frequency } & \text {Cumulative frequency } \\ \hline 0-4 & 3 & 3 \\ \hline 4-8 & 9 & 12 \\ \hline 8-12 & 10 & 22 \\ \hline 12-16 & 8 & 30 \\ \hline 16-20 & 6 & 36 \\ \hline \end{array}\)

M=l+N2-Cfh

Here, N=3+9+10+8+6=36

⇒N2=18

So, median class is 8-12

⇒l=8, C=22, f=10, h=4

⇒M=8+18-1210×4

⇒M=10.4

⇒20 M =208

Topic: Mathematics, Statistics, JEE Main, JEE Main 2024 (30 Jan Shift 1)
Let \( m, n \in N \) and \( g c d(2, n)=1 . \) If \( 30\left(\begin{array}{c}30 \\ 0\end{array}\right)+29\left(\begin{array}{c}30 \\ 1\end{array}\right)+\ldots \ldots+2\left(\begin{array}{c}30 \\ 28\end{array}\right)+1\left(\begin{array}{c}30 \\ 29\end{array}\right)=n .2^{m} \), then \( n+m \) is equal to
\( \left(\right. \) Here \( \left.\left(\begin{array}{l}n \\ k\end{array}\right)={ }^{n} C_{k}\right) \)

Options:

The correct answer is:
45

Solution:
30C030+29C130+…+2C2830+1C2930

=30C3030+29C2930+……+2C230+1C130

=∑r=130rCr30

=∑r=130r30rCr-129

=30∑r=130Cr-129

=30C029+C129+C229+…+C2929

=30229=15230=n2m

∴ n=15,m=30

n+m=45

Topic: Mathematics, Binomial Theorem, JEE Main
Let m (respectively, n ) be the number of 5 -digit integers obtained by using the digits 1,2,3,4,5 with repetitions (respectively, without repetitions) such that the sum of any two adjacent digits is odd. Then mn is equal to

Options:
A 9
B 12
C 15
D 18

The correct answer is:
15

Solution:
We have, m is 5-digits number using digits 1,2,3,4,5 with repetition such that sum of two adjacent digit is odd and n is 5 -digits number using digits 1,2 ,

3,4,5 without repetitions such that sum of any two adjacent digits is odd.

Sum of two digits are odd if one is even and other is odd.

Even =2,4
Odd =1,3,5

Case I Digit is repeated.
Two possibilities

(a) odd even odd even odd

=3×2×3×2×3=108

(b) even odd even odd even

=2×3×2×3×2=72

∴ m=108+72=180

Case II Digit is not repeated.
The possibility of arrangement is
odd even odd even odd

=3×2×2×1×1=12

n=12

∴ mn=18012=15

Topic: Mathematics, Permutation Combination, JEE Main
Let \(\mathbf{m}\) be a vector of magnitude \(\sqrt{3}\) and perpendicular to the vectors \(\hat{\mathbf{i}}+\hat{\mathbf{j}}\) and \(\hat{\mathbf{j}}-\hat{\mathbf{k}}\). Let \(\mathbf{n}\) be another vector of magnitude \(2 \sqrt{6}\) and perpendicular to the vectors \(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}\) and \(\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\). The area (in sq. units) of the triangle formed with \(\mathbf{m}\) and \(\mathbf{n}\) as sides is

Options:
A \(\sqrt{2}\)
B \(\sqrt{6}\)
C \(2 \sqrt{3}\)
D \(3 \sqrt{2}\)

The correct answer is:
\(3 \sqrt{2}\)

Solution:
Given,
\(\mathbf{m}=\sqrt{3} \times \text { unit vectors of }[(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \times(\hat{\mathbf{j}}-\hat{\mathbf{k}})]\)
\(\begin{aligned}
\because(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \times(\hat{\mathbf{j}}-\hat{\mathbf{k}}) & =\left[\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
1 & 1 & 0 \\
0 & 1 & -1
\end{array}\right] \\
& =\hat{\mathbf{i}}(-1-0)-\hat{\mathbf{j}}(-1-0)+\hat{\mathbf{k}}(\mathbf{l}) \\
& =-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\
\therefore \quad \mathbf{m} & =\sqrt{3} \times \frac{-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}}{\sqrt{1+1+1}} \\
& =\sqrt{3} \times \frac{-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}}{\sqrt{3}}=-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}
\end{aligned}\)
and \(\mathbf{n}=2 \sqrt{6} \times\) unit vectors of \([(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}) \times(\hat{\mathbf{j}}+2 \hat{\mathbf{k}})]\)
\(\begin{aligned}
& \therefore(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}) \times(\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\
&=\left[\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
2 & -1 & 0 \\
0 & 1 & 2
\end{array}\right] \\
&=\hat{\mathbf{i}}(-2-0)-\hat{\mathbf{j}}(4-0)+\hat{\mathbf{k}}(2+0)=-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\
& \mathbf{n}=2 \sqrt{6} \times \frac{-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}}{\sqrt{4+16+4}}=2 \sqrt{6} \times \frac{-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}}{2 \sqrt{6}}
\end{aligned}\)
\(=-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}\)
Now, required area
\(\begin{aligned}
& =\frac{1}{2}(\mathbf{m} \times \mathbf{n})=\frac{1}{2}|(-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \times(-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})| \\
& =\frac{1}{2}\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
-1 & 1 & 1 \\
-2 & -4 & 2
\end{array}\right| \\
& =\frac{1}{2} \times \mid \hat{\mathbf{i}}(2+4)-\hat{\mathbf{j}}(-2+2)+\hat{\mathbf{k}}(4+2|| \\
& =\frac{1}{2} \times|6 \hat{\mathbf{i}}+6 \hat{\mathbf{k}}|=\frac{1}{2} \sqrt{6^2+6^2}=\frac{1}{2} \times 6 \times \sqrt{2}=3 \sqrt{2}
\end{aligned}\)

Topic: Mathematics, Vector Algebra, JEE Main
Let \( \mathrm{AB} \) is the focal chord of a parabola and \( \mathrm{D} \) and \( \mathrm{C} \) be the foot of the perpendiculars from \( \mathrm{A} \) and \( \mathrm{B} \) on its directrix respectively. If \( \mathrm{CD}=6 \) units and area of trapezium \( \mathrm{ABCD} \) is \( 36 \) square units, then the length (in units) of the chord \( \mathrm{AB} \) is

Options:

The correct answer is:
12

Solution:
Let S be focus, AS=AD and BS=BC
Area of trapezium
=12AD+BC.6
=3(AS+BC)
=3AB
Hence, AB=12 units

Topic: Mathematics, Parabola, JEE Main
Let \( \mathrm{f}:[0,3] \rightarrow \mathrm{R} \) be defined by \( f(x)=\min \{x-[x], 1+[x]-x\} \) where \( [\mathrm{x}] \) is the greatest integer less than or equal to \( x \).
Let \( P \) denote the set containing all \( x \in[0,3] \) where \( \mathrm{f} \) is discontinuous, and \( \mathrm{Q} \) denote the set containing all \( x \in(0,3) \)
where \( f \) is not differentiable. Then the sum of number of elements in \( \mathrm{P} \) and \( \mathrm{Q} \) is equal to

Options:

The correct answer is:
5

Solution:
Given,

f(x)=min{x-[x],1+[x]-x}

So, the graph of the function is

1-{x}=1-x;0≤x<1

From the graph we can say that the function is continuous at all points in the interval 0,3.

So, P=0

Again, from the diagram, we can say that at x=12, 1, 32, 2, 52, there are sharp edges.

So, the function is non differentiable at x=12, 1, 32, 2, 52

Hence, Q=5

So, P+Q=5

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let \( \mathrm{f}:(-1,1) \rightarrow \mathrm{R} \) be a differentiable function with \( \mathrm{f}(0)=-1 \) and \( \mathrm{f}^{\prime}(0)=1, \mathrm{~g}(\mathrm{x})=\{\mathrm{f}(2 \mathrm{f}(\mathrm{x})+2)\}^{2} . \) Then \( \mathrm{g}^{\prime}(0)= \)

Options:
A \( 4 \)
B \( { }_{-4} \)
C \( 0 \)
D \( -2 \)

The correct answer is:
\( { }_{-4} \)

Solution:
g x = f⁡ ( 2 f⁡ x + 2 ) 2
We have on differentiation with respect to x,
g'(x) = 2f (2f (x) + 2) f '(2f (x) + 2) 2f ' (x)
g'(0) = 2f (2f (0) + 2) f ' (2f (0) + 2 f ' (0)
= - 4.

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let \( \mathrm{f}(x)=x+\frac{1}{2 x+\frac{1}{2 x+\frac{1}{2 x+\ldots \ldots . . \infty}}} . \) Then the value of \( f(100) . f^{\prime}(100) \) is

Options:
A \( 100 \)
B \( \frac{1}{100} \)
C \( -100 \)
D \( -\frac{1}{100} \)

The correct answer is:
\( 100 \)

Solution:
Let y= f x = x + 1 2 x + 1 2 x + 1 2 x + … … … ∞
⇒ y = x + 1 x + y = x 2 + x y + 1 x + y
⇒ x y + y 2 = x y + x 2 + 1
⇒ y 2 - x 2 = 1
Undefined y 2 = 1 + x 2
On differentiating
⇒ 2yy'=2x
⇒ yy'=x
⇒ f100.f'100=100

Topic: Mathematics, Differentiation, JEE Main
Let matrix A=xy-z12 311 2, where x, y, zâN. If adjadjadjadjA=48Â·516, then the number of such matrices A is equal to (where, M represents determinant of a matrix M)

Options:

The correct answer is:
36

Solution:
det.adjadjadjadjA=A3-14
=A16=48.516
âA=Â±10
A=xy-z123112=x+y+z=Â±10
âµx,y,zâNâx+y+z=-10 (not possible)
Hence, x+y+z=10
The number of such matrices =9C2
=36

Topic: Mathematics, Determinants, JEE Main
Let α=maxx∈R82sin3x·44cos3x and β=minx∈R82sin3x·44cos3x. If 8x2+bx+c=0 is a quadratic equation whose roots are α1/5 and β1/5, then the value of c-b is equal to :

Options:
A 42
B 47
C 43
D 50

The correct answer is:
42

Solution:
α=max82sin3x·44cos3x

=max26sin3x·28cos3x

=max26sin3x+8cos3x

and β=min82sin3x·44cos3x=min26sin3x+8cos3x

Now range of 6sin3x+8cos3x

=-62+82,+62+82=-10,10

α=210 & β=2-10

So, α1/5=22=4

⇒β1/5=2-2=14

The quadratic 8x2+bx+c=0,

sum of roots =-b8 and product of roots =c8

∴ c-b=8×[ (product of roots)+(sum of roots)]

=8×4×14+4+14=8×214=42

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2021 (27 Jul Shift 2)
Let $n=1 !+4 !+7 !+\ldots+400 !$. Then ten's digit of $n$ is

Options:
A 1
B 6
C 2
D 7

The correct answer is:
6

Solution:
Given, $n=1 !+4 !+7 !+\ldots+400 !$ $1 !=1,4 !=24,7 !=5040,10 !=3628800$ and further the terms has last two digits $=00$
So,
$1 !+4 !+7 !+\ldots+400 !=\ldots \ldots 65$
Here ten digit is 6 .
Hence, 6 is the answer.

Topic: Mathematics, Binomial Theorem, JEE Main
Let $n \geq 2$ be an integer. $A=\left[\begin{array}{ccc}\cos (2 \pi / n) & \sin (2 \pi / n) & 0 \\ -\sin (2 \pi / n) & \cos (2 \pi / n) & 0 \\ 0 & 0 & 1\end{array}\right]$ and $I$ is the identity matrix of order 3 . Then,

Options:
A $A^{n}=I$ and $A^{n-1} \neq I$
B $A^{m} \neq I$ for any positive integer $m$
C $A$ is not invertible
D $A^{m}=O$ for a positive integer $m$

The correct answer is:
$A^{n}=I$ and $A^{n-1} \neq I$

Solution:
Given,
$A=\left[\begin{array}{ccc}\cos \left(\frac{2 \pi}{n}\right) & \sin \left(\frac{2 \pi}{n}\right) & 0 \\ -\sin \left(\frac{2 \pi}{n}\right) & \cos \left(\frac{2 \pi}{n}\right) & 0 \\ 0 & 0 & 1\end{array}\right]$
Now, $A \times A=\left[\begin{array}{ccc}\cos \left(\frac{2 \pi}{n}\right) & \sin \left(\frac{2 \pi}{n}\right) & 0 \\ -\sin \left(\frac{2 \pi}{n}\right) & \cos \left(\frac{2 \pi}{n}\right) & 0 \\ 0 & 0 & 1\end{array}\right] \times$$\left[\begin{array}{ccc}\cos \left(\frac{2 \pi}{n}\right) & \sin \left(\frac{2 \pi}{n}\right) & 0 \\ -\sin \left(\frac{2 \pi}{n}\right) & \cos \left(\frac{2 \pi}{n}\right) & 0 \\ 0 & 0 & 1\end{array}\right]$
$\left[\begin{array}{ccc}\cos ^{2}\left(\frac{2 \pi}{n}\right)-\sin ^{2}\left(\frac{2 \pi}{n}\right) & \cos \left(\frac{2 \pi}{n}\right) \sin \left(\frac{2 \pi}{n}\right)+\sin \left(\frac{2 \pi}{n}\right) \cos \left(\frac{2 \pi}{n}\right) & 0\\ -\sin \left(\frac{2 \pi}{n}\right) \cos \left(\frac{2 \pi}{n}\right)-\sin \left(\frac{2 \pi}{n}\right) \cos \left(\frac{2 \pi}{n}\right) & -\sin ^{2}\left(\frac{2 \pi}{n}\right)+\cos ^{2}\left(\frac{2 \pi}{n}\right) & 0\\ 0 & 0 & 1\end{array}\right]$
$=\left[\begin{array}{cccc}\cos \left(2 \times \frac{2 \pi}{n}\right) & 2 \sin \left(\frac{2 \pi}{n}\right) \cos \left(\frac{2 \pi}{n}\right) & 0 \\ -2 \sin \left(\frac{2 \pi}{n}\right) \cos \left(\frac{2 \pi}{n}\right) & \cos \left(2 \times \frac{2 \pi}{n}\right) & 0 \\ 0 & 0 & 1\end{array}\right]$
$=\left[\begin{array}{ccc}
\cos \left(2 \times \frac{2 \pi}{n}\right) & \sin \left(2 \times \frac{2 \pi}{n}\right) & 0 \\
-\sin \left(2 \times \frac{2 \pi}{n}\right) & \cos \left(2 \times \frac{2 \pi}{n}\right) & 0 \\
0 & 0 & 1
\end{array}\right]$
Similarly,
$A^{n}=\left[\begin{array}{cccc}\cos \left(2^{n-1} \times \frac{2 \pi}{n}\right) & \sin \left(2^{n-1} \times \frac{2 \pi}{n}\right) & 0 \\ -\sin \left(2^{n-1} \times \frac{2 \pi}{n}\right) & \cos \left(2^{n-1} \times \frac{2 \pi}{n}\right) & 0 \\ 0 & 0 & 1\end{array}\right]$
$=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=I$
and
$\begin{aligned}
A^{n-1}=&\left[\begin{array}{cccc}
\cos \left(2^{n-2} \times \frac{2 \pi}{n}\right) & \sin \left(2^{n-2} \times \frac{2 \pi}{n}\right) & 0 \\
-\sin \left(2^{n-2} \times \frac{2 \pi}{n}\right) & \cos \left(2^{n-2} \times \frac{2 \pi}{n}\right) & 0 \\
0 & 0 & 1
\end{array}\right] \\
\neq I
\end{aligned}$
$\therefore A^{n-1} \neq I$ for any positive integer $m$

Topic: Mathematics, Matrices, JEE Main
Let $n \geq 3$. A list of numbers $0 < x_{1} < x_{2} < \ldots < x_{n}$ has mean $\mu$ and standard deviatiion $\sigma$. A new list of numbers is made as follows : $y_{1}=0, y_{2}=x_{2}, \ldots, y_{n-1}=x_{n-1}, y_{n}=x_{1}+x_{n}$. The mean and the standard deviation of the new list are $\hat{\mu}$ and $\hat{\sigma}$. Which of the following is necessarily true?

Options:
A $\mu=\hat{\mu}, \sigma \leq \hat{\sigma}$
B $\mu=\hat{\mu}, \sigma \geq \hat{\sigma}$
C $\sigma=\hat{\sigma}$
D $\mu$ may or may not be equal to $\hat{\mu}$

The correct answer is:
$\mu=\hat{\mu}, \sigma \leq \hat{\sigma}$

Solution:
$\hat{\mu}=\frac{0+x_{2}+x_{3} \ldots \ldots x_{n-1}+\left(x_{1}+x_{n}\right)}{n}=\mu$
$\hat{\sigma}^{2}=\frac{1}{n} \Sigma y_{i}^{2}-\hat{\mu}^{2}$
$=\frac{1}{n}\left[0^{2}+x_{2}^{2}+x_{3}^{2} \ldots\left(x_{n-1}\right)^{2}+\left(x_{1}+x_{n}\right)^{2}\right]-\mu^{2}$
$\hat{\sigma}^{2}-\sigma^{2}=\frac{2 x_{1} x_{n}}{n}>0$
so $\hat{\sigma}>\sigma$

Topic: Mathematics, Statistics, JEE Main
Let $n \geq 3$.A list of numbers $x_{1}, x_{2}, \ldots \ldots \ldots x_{n}$ has mean $\mu$ and standard deviation $\sigma .$ A new list of numbers $y_{1}, y_{2}, \ldots \ldots \ldots . y_{n}$ is made as follows: $y_{1}=\frac{x_{1}+x_{2}}{2}, y 2=\frac{x_{1}+x_{2}}{2}$ and $y_{j}=x$ for $j=3,4, \ldots \ldots . n$. The mean and the standard deviation of the list are $\hat{\mu}$ and $\hat{\sigma}$. Then which of the following is necessarily true?

Options:
A $\mu=\hat{\mu}$ and $\sigma \leq \hat{\sigma}$
B $\mu=\hat{\mu}$ and $\sigma \geq \hat{\sigma}$
C $\sigma=\hat{\sigma}$
D $\mu \neq \hat{\mu}$

The correct answer is:
$\mu=\hat{\mu}$ and $\sigma \geq \hat{\sigma}$

Solution:
$\mu=\frac{\sum x_{i}}{n} \quad n \geq 3$
$\sigma=\sqrt{\frac{\sum x_{i}^{2}}{n}-\left(\frac{\sum x_{i}}{n}\right)^{2}}$
$\hat{\mu}=\frac{y_{1}+y_{2}+y_{3}+\ldots \ldots+y_{n}}{n}$
$=\frac{\frac{x_{1}+x_{2}}{2}+\frac{x_{1}+x_{2}}{2}+x_{3}+x_{4}+\ldots \ldots \ldots+x_{n}}{n}$
$=\frac{\mathrm{x}_{1}+\mathrm{x}_{2}+\ldots \ldots . .+\mathrm{x}_{\mathrm{n}}}{\mathrm{n}}=\mu$
$\hat{\sigma}=\sqrt{\frac{\left(\frac{x_{1}+x_{2}}{2}\right)^{2}+\left(\frac{x_{1}+x_{2}}{2}\right)^{2}+x_{3}^{2}+\ldots \ldots+x_{n}^{2}}{n}-\left(\frac{\sum y_{1}}{n}\right)^{2}}$
$\begin{array}{l}=\sqrt{\frac{\frac{1}{4}\left(2 x_{1}^{2}+2 x_{2}^{2}+4 x_{1} x_{2}\right)+x_{3}^{2}+\ldots \ldots . .+x_{n}^{2}}{n}-\left(\frac{\sum y_{1}}{n}\right)^{2}} \\ =\sqrt{\frac{\frac{1}{2}\left(x_{1}^{2}+x_{2}^{2}\right)+x_{1} x_{2}+x_{3}^{2}+\ldots \ldots+x_{n}^{2}}{n}-\left(\frac{\sum x_{i}}{n}\right)^{2}} \\ \leq \sqrt{\frac{x_{1}^{2}+x_{2}^{2}+\ldots \ldots+x_{2}^{2}}{n}-\left(\frac{\sum x_{i}}{n}\right)^{2}} \\ \hat{\sigma} \leq \sigma & \ldots \ldots \ldots(2) \\ \text { by }(1) \text { and }(2)\end{array}$

Topic: Mathematics, Statistics, JEE Main
Let $n \geq 3$ and let $C_{1}, C_{2}, \ldots . C_{n}$, be circles with radii $r_{1}, r_{2}, \ldots, r_{n}$, respectively. Assume that $C_{i}$ and $C_{i+1}$ touch externally for $1 \leq i \leq n-1$. It is also given that the $x$-axis and the line $y=2 \sqrt{2} x+10$ are tangential to each of the circles. Then $r_{1}, r_{2}, \ldots \ldots, r_{n}$ are in

Options:
A An arithmetic progression with common difference $3+\sqrt{2}$
B A geometric progression with common ratio $3+\sqrt{2}$
C An arithetic progression with common diffrence $2+\sqrt{3}$
D Ageometric progression with common ratio $2+\sqrt{3}$

The correct answer is:
Ageometric progression with common ratio $2+\sqrt{3}$

Solution:
$\operatorname{Sin} \theta=\frac{1}{\sqrt{3}}=\frac{r_{1}}{\mathrm{AP}_{1}}=\frac{r_{2}}{\mathrm{AP}_{2}}$
$\mathrm{AP}_{2}=\mathrm{AP}_{1}+\mathrm{r}_{1}+\mathrm{r}_{2}$
$\mathrm{r}_{2} \sqrt{3}=\mathrm{r}_{1} \sqrt{3}+\mathrm{r}_{1}+\mathrm{r}_{2}$
$\mathrm{r}_{2}(\sqrt{3}-1)=\mathrm{r}_{1}(\sqrt{3}+1)$
$\frac{r_{2}}{r_{1}}=2+\sqrt{3}$

Topic: Mathematics, Circle, JEE Main
Let n≥5 be an integer. If 9n-8n-1=64α and 6n-5n-1=25 β, then α-β is equal to:

Options:
A 1+C2n8-5+C3n82-52+…+Cnn8n-1-5n-2
B 1+C3n8-5+C4n82-52+…+Cnn8n-2-5n-2
C C3n8-5+C4n82-52+…+Cnn8n-2-5n-2
D C4n8-5+C5n82-52+…+Cnn8n-3-5n-3

The correct answer is:
C3n8-5+C4n82-52+…+Cnn8n-2-5n-2

Solution:
Using the binomial expansion 1+xn=C0n+C1nx+C2nx2.......+Cnnxn

We get,

α=1+8n-8n-164=C2n+C3n8+C4n82+…Similarly β=C2n+C3n5+C4n52+….

So α-β=C3n8-5+C4n82-52+…+Cnn8n-2-5n-2

So, option (C) is the answer.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Let n(A) denotes the number of elements in the set A. If nA=4, nB=5 and nA∩B=3, then n[(A×B)∩(B×A)] is equal to

Options:
A 8
B 9
C 10
D 11

The correct answer is:
9

Solution:
Given, nA=4, nB=5 and nAâ©B=3
â´nAÃBâ©BÃA=nAâ©BÃBâ©A=32=9

Topic: Mathematics, Sets and Relations, JEE Main
Let n be a positive integer and a complex number with unit modulus is a solution of the equation Zn+Z+1=0, then the value of n can be

Options:
A 87
B 97
C 104
D 222

The correct answer is:
104

Solution:
Let, Z1 satisfies Zn+1=-Z
⇒Z1 also satisfies Zn+1=-Z
⇒Z1 also satisfies Zn+1=1
⇒Z1n+1=1 … 1
Now, Z1=1⇒Z1n=1 … 2
From 1 & 2,
we get, Z1n must be the point of intersection of
Z=1 & Zn+1=1
⇒Z1n=ω or ω2 {where, ω is non-real cube root of unity}
⇒Z1 can be ω or ω2

⇒n is of the form of 3k+2

Topic: Mathematics, Complex Number, JEE Main
Let $n$ be a positive integer. For a real number $x$, let $[x]$ denote the largest integer not exceeding $x$ and $\{x\}=$ $x-[x] .$ Then $\int_{1}^{n+1} \frac{(\{x\})^{[x]}}{[x]} d x$ is equal to

Options:
A $\log _{e}(n)$
B $\frac{1}{n+1}$
C $\frac{\mathrm{n}}{\mathrm{n}+1}$
D $1+\frac{1}{2}+\ldots+\frac{1}{n}$

The correct answer is:
$\frac{\mathrm{n}}{\mathrm{n}+1}$

Solution:
$\int_{1}^{n+1} \frac{\{x\}^{[x]}}{[x]} d x=\int_{1}^{2} \frac{\{x\}^{[x]}}{[x]} d x+\int_{2}^{3} \frac{\{x\}^{[x]}}{[x]} d x+\ldots .+\int_{1}^{n+1} \frac{\{x\}^{[x]}}{[x]} d x$
$=\sum_{r=1}^{n} \int_{r}^{r+1} \frac{\{x\}^{[x]}}{[x]} d x$
$=\sum_{r=1}^{n} \int_{r}^{r+1} \frac{(x-r)^{r}}{r} d x$
$=\sum_{r=1}^{n}\left[\frac{(x-r)^{r+1}}{r(r+1)}\right]_{r}^{r+1}$
$=\sum_{r=1}^{n} \frac{1}{r(r+1)}=\frac{n}{n+1}$

Topic: Mathematics, Definite Integration, JEE Main