datavorous/entrance-exam-dataset huggingface dataset:
If A=x∈R:x<2 and B=x∈R:x-2≥3; then

Options:
A A∩B=-2,-1
B B-A=R--2,5
C A∪B=R-2,5
D A-B=-1,2

The correct answer is:
B-A=R--2,5

Solution:
A=x:x∈-2,2

B=x:x∈(-∞,-1]∪[5,∞

A∩B=x:x∈(-2,-1]

A∪B=x:x∈-∞,2∪5,∞)

A-B=x:x∈-1,2

B-A=x:x∈-∞,-2∪5,∞

Topic: Mathematics, Basic of Mathematics, JEE Main, JEE Main 2020 (09 Jan Shift 2)
If A=x∈R:x<2 and B=x∈R:x-2≥3; then

Options:
A A∩B=-2,-1
B B-A=R--2,5
C A∪B=R-2,5
D A-B=-1,2

The correct answer is:
B-A=R--2,5

Solution:
A=x:x∈-2,2
B=x:x∈(-∞,-1]∪[5,∞
A∩B=x:x∈(-2,-1]
A∪B=x:x∈-∞,2∪5,∞)
A-B=x:x∈-1,2
B-A=x:x∈-∞,-2∪5,∞

Topic: Mathematics, Sets and Relations, JEE Main
If $A(x)=\left|\begin{array}{ccc}x+1 & 2 x+1 & 3 x+1 \\ 2 x+1 & 3 x+1 & x+1 \\ 3 x+1 & x+1 & 2 x+1\end{array}\right|$, then $\int_0^1 A(x) d x$ is equal to

Options:
A -15
B $-\frac{15}{2}$
C -30
D -5

The correct answer is:
$-\frac{15}{2}$

Solution:
Since,
$$
A(x)=\left|\begin{array}{ccc}
x+1 & 2 x+1 & 3 x+1 \\
2 x+1 & 3 x+1 & x+1 \\
3 x+1 & x+1 & 2 x+1
\end{array}\right|
$$
[applying $C_3 \rightarrow C_3-C_2$ and then $C_2 \rightarrow C_2-C_1$ ]
$\begin{aligned} & =\left|\begin{array}{ccc}x+1 & x & x \\ 2 x+1 & x & -2 x \\ 3 x+1 & -2 x & x\end{array}\right| \\ & \left.=x^2\left|\begin{array}{ccc}x+1 & 1 & 1 \\ 2 x+1 & 1 & -2 \\ 3 x+1 & -2 & 1\end{array}\right| \quad \text { [apply } C_2 \rightarrow C_2-C_3\right] \\ & =x^2\left|\begin{array}{ccc}x+1 & 0 & 1 \\ 2 x+1 & 3 & -2 \\ 3 x+1 & -3 & 1\end{array}\right| \quad \text { [apply } R_2 \rightarrow R_2+R_3 \text { ] }\end{aligned}$
$$
\begin{aligned}
& =x^2\left|\begin{array}{ccc}
x+1 & 0 & 1 \\
5 x+2 & 0 & -1 \\
3 x+1 & -3 & 1
\end{array}\right| \\
& =x^2(3)\{-x-1-5 x-2\} \\
& =3 x^2(-6 x-3)=-18 x^3-9 x^2 \\
& \therefore \quad \int_0^1 A(x) d x=\int_0^1\left(-18 x^3-9 x^2\right) d x \\
& =\left[\frac{-18 x^4}{4}-\frac{9 x^3}{3}\right]_0^1 \\
& =\left[\frac{-18}{4} x^4-\frac{9}{3} x^3\right]_0^1 \\
&
\end{aligned}
$$
[expand along $\mathrm{C}_2$ ]
$$
\begin{aligned}
& =-\frac{18}{4}-\frac{9}{3} \\
& =-\frac{18}{4}-3 \\
& =\frac{-18-12}{4} \\
& =-\frac{30}{4}=-\frac{15}{2}
\end{aligned}
$$

Topic: Mathematics, Determinants, JEE Main
If A=x:x2-5x+6=0, B=2,4, C=4,5, then AÃBâ©C is

Options:
A 2, 4, 3, 4
B 4, 2, 4, 3
C 2, 4, 3, 4, 4, 4
D 2,  2, 3,  3, 4,  4, 5,  5

The correct answer is:
2, 4, 3, 4

Solution:
Clearly, A = 2, 3, B= 2, 4, C= 4, 5
Bâ©C=4
â´ AÃBâ©C = 2, 4, 3, 4

Topic: Mathematics, Sets and Relations, JEE Main
If A=x:x=3n-2n-1,n∈N and B=x:x=4n-1,n∈N, then

Options:
A A⊂B
B B⊂A
C A∪B=A
D A∩B=B

The correct answer is:
A⊂B

Solution:
A=0,4,20,72……..
B=0,4,8,12……..

Clearly, A⊂B

Topic: Mathematics, Sets and Relations, JEE Main
If A=x:x=6n-5n-1,nâN and B=x:x=25(n-1),nâN, then

Options:
A A=B
B B⊂A
C AâB
D BâA

The correct answer is:
A⊆B

Solution:
A = {0, 25, 200, ......}
B = {0, 25, 50, 75 ......}
Clearly, AâB

Topic: Mathematics, Sets and Relations, JEE Main
If $A=\{x / x$ is a prime number, $0 \leq x \leq 9]$, then the number of elements of power
set of $A$ is

Options:
A 12
B 4
C 16
D 8

The correct answer is:
16

Solution:
$\begin{aligned} A=\{2,3,5,7\} & \\ \text { Power set of } A &=2^{4} \\ &=16 . \end{aligned}$

Topic: Mathematics, Sets and Relations, JEE Main
If $\mathrm{A}=\{x, y, z\}, \mathrm{B}=\{1,2\}$, then the total number of relations from set $\mathrm{A}$ to set $\mathrm{B}$ are

Options:
A 64
B 16
C 32
D 8

The correct answer is:
64

Solution:
It is given that $A=\{x, y, z\}$ and $B=\{1,2\} .$
$\therefore A \times B=\{(x, 1),(x, 2),(y, 1),(y, 2),(z, 1),(z, 2)\}$
Since $n(A \times B)=6$, the number of subsets of $A \times B$ is $2^{6} .$
Therefore, the number of relations from $A$ to $B$ is $2^{6} .$

Topic: Mathematics, Sets and Relations, JEE Main
If $A=\left\{x \in z^{+}: x < 10\right.$ and $x$ is a multiple of 3 or $4\}$, where $z^{+}$is the set of positive integers, then the total number of symmetric relations on $A$ is

Options:
A
$2^5$

B
$2^{15}$

C
$2^{10}$

D
$2^{20}$

The correct answer is:
$2^{15}$

Solution:
A relation on a set A is said to be symmetric iff $(a, b) \in A \Rightarrow(b, a) \in A, \forall a, b \in A$
Here $A=\{3,4,6,8,9\}$
Number of order pairs of $A \times A=5 \times 5=25$
Divide 25 order pairs of $A \times A$ in 3 parts as follows :
$$
\begin{aligned}
& \text { Part }-A:(3,3),(4,4),(6,6),(8,8),(9,9) \\
& \text { Part }-B:(3,4),(3,6),(3,8),(3,9),(4,6), \\
& (4,8),(4,9),(6,8),(6,9),(8,9) \\
& \text { Part-C:(4,3),(6,3),(8,3),(9,3),(6,4),(8,4), } \\
& (9,4),(8,6),(9,6),(9,8)
\end{aligned}
$$
In part $-A$, both components of each order pair are same.
In part $-B$, both components are different but not two such order pairs are present in which first component of one order pair is the second component of another order pair and vice-versa.
In part $-C$, only reverse of the order pairs of part $-B$ are present i.e., if $(a, b)$ is present in part $-B$, then (b, a) will be present in part $-C$ For example $(3,4)$ is present in part $-B$ and $(4,3)$ present in part $-C$.
Number of order pair in $A, B$ and $C$ are 5, 10 and 10 respectively.
In any symmetric relation on set $A$, if any order pair of part $-B$ is present then its reverse order pair of part $-C$ will must be also present.
Hence number of symmetric relation on set $A$ is equal to the number of all relations on a set $D$, which contains all the order pairs of part $-A$ and part $-B$.
Now $n(D)=n(A)+n(B)=5+10=15$
Hence number of all relations on set $D=(2)^{15}$
$\Rightarrow$ Number of symmetric relations on set $D=(2)^{15}$

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2012 (12 May Online)
If \( a_{1}, a_{2}, \ldots a_{n} \) are in arithmetic progression, where \( a_{1}>0 \) for all \( i \).
Then, \( \frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}} \) is equal to

Options:
A \( \frac{n^{2}(n+1)}{2} \)
B \( \frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}} \)
C \( \frac{n(n-1)}{2} \)
D None of these

The correct answer is:
\( \frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}} \)

Solution:
Since, a1,a2,…,an are in arithmetic progression.
Then, a2-a1=a3-a2=...=an-an-1=d
Where d is common difference
Now, 1a2+a1+1a3+a2+…+1an+an-1
=a2-a1d+a3-a2d+…+an-an-1d
=1d(an-a1)×an+a1an+a1
=1dan-a1an+a1=n-1an+a1

Topic: Mathematics, Sequences and Series, JEE Main
If \(A=\{1,2,3,4,5\}\), then the number of proper subsets of \(A\) is

Options:
A 31
B 38
C 48
D 54

The correct answer is:
31

Solution:
Number of proper subsets of \(A=2^n-1\) Given: \(A=\{1,2,3,4,5\}\)
Here \(n=5\)
\(\therefore\) no. of proper subsets \(=2^5-1\)

Topic: Mathematics, Sets and Relations, JEE Main
If AB4 molecule is a polar molecule, a possible geometry of AB4 is :

Options:
A Square pyramidal
B Tetrahedral
C Rectangular planar
D Square planar

The correct answer is:
Square pyramidal

Solution:
For AB4 compound possible geometry are

Square pyramidal can be polar due to lone pair moment as the bond pair moments will get cancelled out.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2020 (02 Sep Shift 1)
If \( a_{k} \) is the coefficient of \( x^{k} \) in the expansion of \( \left(1+x+x^{2}\right)^{n} \) for \( k=0,1,2, \ldots ., 2 n \) then value of
\( a_{1}+2 a_{2}+3 a_{3}+\ldots \ldots \ldots 2 n a_{2 n} \) is:

Options:
A \( -a_{0} \)
B \( 3^{n} \)
C \( n \cdot 3^{n+1} \)
D \( n \cdot 3^{n} \)

The correct answer is:
\( n \cdot 3^{n} \)

Solution:
We expand the equation by using binomial expansion,

1+x+x2n=a0+a1x+a2x3+....+a2nx2n

On differentiating both sides, we get
n(1+x+x2)n-11+2x=a1+2a2x+3a3x2
+...+2na2nx2n-1

Now, on putting x=1, we get
⇒n(3)n-1⋅3=a1+2a2+3a3+...+2na2n
⟹a1+2a2+3a3+....+2na2n=n⋅3n

Topic: Mathematics, Binomial Theorem, JEE Main
If \( a=\left(3 t^{2}+2 t+1\right) \mathrm{m} \mathrm{s}^{-2} \) is the expression according to which the acceleration of a particle varies. Find a displacement of the particle after \( 2 \mathrm{~s} \) of start.

Options:
A \( 27 \mathrm{~m} \)
B \( \frac{26}{3} \mathrm{~m} \)
C \( 30 / 7 \mathrm{~m} \)
D \( 26 / 2 \mathrm{~m} \)

The correct answer is:
\( \frac{26}{3} \mathrm{~m} \)

Solution:
According to the data given,

a=3t2+2t+1

We know that,

a=dvdt⇒dv=a·dt

Integrating both sides with given limits,

∫0tdv=∫0t(3t2+2t+1)dt

v0v =3t330t + 2t220t +t0t

Using the property of definite integration,

v=t3+t2+t

From the definition of instantaneous velocity,

v=dsdt⇒ds=vdt

Putting the values with proper limits,

∫0sds=∫02t3+t2+tdt⇒S0s=t44+t33+t22 02

After 2 s of start the total displacement is,S=263

Topic: Physics, Motion In One Dimension, JEE Main
If \( A=\left[a_{i j}\right]_{m \times n}, a_{i j}=\left(i^{2}+j^{2}-i j\right)(j-i) \) and \( n \) is odd, then which of the following is not the value of \( \operatorname{Tr}(A) \) ?

Options:
A \( 0 \)
B \( |A| \)
C \( 2|A| \)
D None of these

The correct answer is:
\( 0 \)

Solution:
Given,

A=aijm×n

aij=i2+j2-ijj-i

Now, Replace i→j, j→i

aji=i2+j2-iji-j⇒aji=-i2+j2-ijj-i⇒aji=-aij

It is a skew-symmetric matrix where aij+aji is equal to zero.

The determinant of Skew-Symmetric matrix of odd ordern=3 is always zero.

A=0

All diagonal elements of Skew-Symmetric matrix are zero.

So,
TrA=∑i=1naii=0

Topic: Mathematics, Matrices, JEE Main
If \(A=\left(\begin{array}{cc}3 & x-1 \\ 2 x+3 & x+2\end{array}\right)\) is a symmetric matrix, then the value of \(x\) is

Options:
A 4
B 3
C -4
D -3

The correct answer is:
-4

Solution:
Hints: \(A=A^T\)
\(\begin{aligned}
& \left(\begin{array}{cc}
3 & x-1 \\
2 x+3 & x+2
\end{array}\right)=\left(\begin{array}{cc}
3 & 2 x+3 \\
x-1 & x+2
\end{array}\right) \\
& \Rightarrow x-1=2 x+3 \text { or } x=-4
\end{aligned}\)

Topic: Mathematics, Matrices, JEE Main
If \( A=\left[\begin{array}{cc}-4 & -1 \\ 3 & 1\end{array}\right] \), then the determinant of the matrix \( \left(A^{2016}-2 A^{2015}-A^{2014}\right) \) is

Options:
A \( -175 \)
B \( 2014 \)
C \( 2016 \)
D \( -25 \)

The correct answer is:
\( -25 \)

Solution:
-4-131 ⇒ A2=-4-131 -4-131=133-9-2 and A=-1
Now, A2016-2A2015-A2014=A2014 A2-2A-I
⇒A2016-2A2015-A2014=A2014 A2-2A-I
A2-2A-I=205-15-5
=A2014 205-15-5
=-25

Topic: Mathematics, Determinants, JEE Main
If \( A=\left[\begin{array}{cc}5 a & -b \\ 3 & 2\end{array}\right] \) and \( A \cdot a d j A=A A^{T} \), then \( 5 a+b \) is equal to

Options:
A \( 4 \)
B \( 13 \)
C \( -1 \)
D \( 5 \)

The correct answer is:
\( 5 \)

Solution:
A=5a-b32 and AT=5a3-b2
AAT=25a2+b215a-2b15a-2b13
Now, A.adjA=AI2=10a+3b0010a+3b
Given AAT=A.adj A
15a-2b=0 ......(i)
10a+3b=13 .......(ii)
Solving we get
5a=2 and b= 3
∴ 5a+b=5

Topic: Mathematics, Matrices, JEE Main
If \( A=\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right] \) then the value of \( A^{2}-3 A+2 I \) would be

Options:
A \( \left.\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] \)
B \( \left.\begin{array}{ccc}1 & -1 & -1 \\ 3 & -3 & -4 \\ -3 & 2 & -4\end{array}\right] \)
C \( \left.\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right] \)
D \( \left[\begin{array}{ccc}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{array}\right] \)

The correct answer is:
\( \left.\begin{array}{ccc}1 & -1 & -1 \\ 3 & -3 & -4 \\ -3 & 2 & -4\end{array}\right] \)

Solution:
We have

A2=A.A=[2012131−10][2012131−10]

=[ 4+0+1 0+0−1 2+0+0 4+2+3 0+1−3 2+3+0 2−2+0 0−1+0 1−3+0 ] =[ 5 −1 2 9 −2 5 0 −1 −2 ]

∴ A2-3A+2I

=[ 5 −1 2 9 −2 5 0 −1 −2 ]−3[ 2 0 1 2 1 3 1 −1 0 ]+2[ 1 0 0 0 1 0 0 0 1 ] =[ 5 −1 2 9 −2 5 0 −1 −2 ]−[ 6 0 3 6 3 9 3 −3 0 ]+[ 2 0 0 0 2 0 0 0 2 ] =[ 1 −1 −1 3 −3 −4 −3 2 −4 ]

Topic: Mathematics, Matrices, JEE Main
If \(A=\left[\begin{array}{ll}3 & 2 \\ 0 & 1\end{array}\right]\), then \(\left(A^{-1}\right)^3\) is equal to

Options:
A \(\frac{1}{27}\left[\begin{array}{cc}1 & -26 \\ 0 & 27\end{array}\right]\)
B \(\frac{1}{27}\left[\begin{array}{cc}-1 & -26 \\ 0 & 27\end{array}\right]\)
C \(\frac{1}{27}\left[\begin{array}{ll}1 & -26 \\ 0 & -27\end{array}\right]\)
D \(\frac{1}{27}\left[\begin{array}{cc}-1 & -26 \\ 0 & -27\end{array}\right]\)

The correct answer is:
\(\frac{1}{27}\left[\begin{array}{cc}1 & -26 \\ 0 & 27\end{array}\right]\)

Solution:
\(|\mathrm{A}|=\left|\begin{array}{ll}
3 & 2 \\
0 & 1
\end{array}\right|=3-0=3 \neq 0 \quad \therefore \mathrm{A}^{-1} \text { exists }\)
Adj. \(A=\left[\begin{array}{cc}1 & 0 \\ -2 & 3\end{array}\right]^{\prime}=\left[\begin{array}{cc}1 & -2 \\ 0 & 3\end{array}\right]\)
\(\therefore \quad \mathrm{A}^{-1}=\frac{1}{3}\left[\begin{array}{cc}
1 & -2 \\
0 & 3
\end{array}\right]\)
\(\begin{aligned}
& \therefore\left(A^{-1}\right)^3=\frac{1}{27}\left[\begin{array}{cc}
1 & -2 \\
1 & 3
\end{array}\right]\left[\begin{array}{cc}
1 & -2 \\
0 & 3
\end{array}\right]\left[\begin{array}{cc}
1 & -2 \\
0 & 3
\end{array}\right] \\
& =\frac{1}{27}\left[\begin{array}{cc}
1 & -8 \\
0 & 9
\end{array}\right]\left[\begin{array}{cc}
1 & -2 \\
0 & 3
\end{array}\right]=\frac{1}{27}\left[\begin{array}{cc}
1 & -26 \\
0 & 27
\end{array}\right]
\end{aligned}\)

Topic: Mathematics, Matrices, JEE Main
If \(A=\left[\begin{array}{lll}2 & 0 & 0 \\ 2 & 2 & 0 \\ 2 & 2 & 2\end{array}\right]\), then \(\operatorname{adj}(\operatorname{adj} A)\) is equal to -

Options:
A \(8\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1\end{array}\right]\)
B \(16\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1\end{array}\right]\)
C \(64\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1\end{array}\right]\)
D None of these

The correct answer is:
\(16\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1\end{array}\right]\)

Solution:
\(|A|=\left[\begin{array}{lll}
2 & 0 & 0 \\
2 & 2 & 0 \\
2 & 2 & 2
\end{array}\right]=(2)(2)(2)=8\)
Now adj \((\operatorname{adj} A)=|\mathrm{A}|^{3-2} \mathrm{~A}\)
\(=8\left[\begin{array}{lll}
2 & 0 & 0 \\
2 & 2 & 0 \\
2 & 2 & 2
\end{array}\right]=16\left[\begin{array}{ccc}
1 & 0 & 0 \\
1 & 1 & 0 \\
1 & 1 & 1
\end{array}\right]\)

Topic: Mathematics, Matrices, JEE Main
If \( A=\left[\begin{array}{lll}6 & 8 & 5 \\ 4 & 2 & 3 \\ 9 & 7 & 1\end{array}\right] \) is the sum of a symmetric matrix \( B \) and skew-symmetric matrix, \( C \), then \( B \) is

Options:
A \( \left[\begin{array}{lll}6 & 6 & 7 \\ 6 & 2 & 5 \\ 7 & 5 & 1\end{array}\right] \)
B \( \left[\begin{array}{ccc}0 & 2 & -2 \\ -2 & 5 & -2 \\ 2 & 2 & 0\end{array}\right] \)
C \( \left[\begin{array}{ccc}6 & 6 & 7 \\ -6 & 2 & -5 \\ -7 & 5 & 1\end{array}\right] \)
D \( \left[\begin{array}{ccc}0 & 6 & -2 \\ 2 & 0 & -2 \\ -2 & -2 & 0\end{array}\right] \)

The correct answer is:
\( \left[\begin{array}{lll}6 & 6 & 7 \\ 6 & 2 & 5 \\ 7 & 5 & 1\end{array}\right] \)

Solution:
Given, A=685423971
amd symmetric matrix B=A+A′2
â´Â B=12685423971+649827531=667625751

Topic: Mathematics, Matrices, JEE Main
If \( A=\left\{x:-\frac{2}{5} \leq x \leq \frac{\pi-2}{5}\right\}, B=\{y:-1 \leq y \leq 1\} \) and \( f(x)=\cos (5 x+2) \), then the mapping \( f: A \rightarrow B \) is

Options:
A one-one but not onto
B onto but not one-one
C both one-one and onto
D neither one-one nor onto

The correct answer is:
both one-one and onto

Solution:
Let t=5x+2, then A=t:0≤t≤π

A bijective function is both one-one and onto.

∴ ft=cost, which is bijective in 0,π since cost is decreasing in 0,π and Range of cost is -1,1.

Hence, fx is bijective.

Topic: Mathematics, Functions, JEE Main
If all possible numbers are formed by using the digits 1, 2, 3, 5, 7 without repetition and they are arranged in descending order, then the rank of the number 327 is

Options:
A 31
B 175
C 149
D 271

The correct answer is:
271

Solution:
Given numbers are 1, 2, 3, 5, 7.

And also they are arranged in descending order.

i.e., the numbers with four and five digits are included.

The number of five-digit number=5!

The number of four digits number

=p45=5!(5-4)!=1201=120

The number of 3 digits starting with 7=

i.e., 4 digits have to be fit in 2 place in p24 ways,

p24=4!(4-2)!=4×3=12

The number of 3 digits numbers starting with 5=

p24=4!(4-2)!=4×3=12

The number of digits starting with 3

i.e., 3 numbers have to fit in 1 place in p13 way.

p13=3!(3-1)!=3×2×12×1=3=3

The number of digits starting with 3¯5¯ is,

p13=3!(3-1)!=3×2×12×1=3

And the next number is 327.

∴ The rank of 327 is

=120+120+12+12+3+3+1

=271

∴ Rank of 327=271

Topic: Mathematics, Permutation Combination, JEE Main
If all symbols have their usual meaning, then for a non-volatile non-electrolyte solute \( \operatorname{Lt}_{\mathrm{m} \rightarrow 1}\left(\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\mathrm{m}}\right) \) is equal to:

Options:
A Zero
B Infinity
C \( \Delta \mathrm{T}_{f} \)
D \( \mathrm{K}_{f} \)

The correct answer is:
\( \mathrm{K}_{f} \)

Solution:
For a non-volatile non-electrolyte solute, depression in freezing point
Δ T f⁡ = K f⁡ m or Δ T f⁡ m = K f⁡
Lt m → 1 Δ T f m = K f⁡

Topic: Chemistry, Solutions, JEE Main
If all the nucleophilic substitution reactions at saturated carbon atoms in the above sequence of reactions follow \(\mathrm{S}_{\mathrm{N}} 2\) mechanism, then \(\underline{E}\) and \(\underline{E}\) will be respectively,

Options:
A

B

C

D

The correct answer is:

Solution:
Hint :

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
If all the vertices of an equilateral triangle lie on the parabola $y^2=16 x$ and one of them coincides with the vertex of that parabola, then the length of the side of that triangle is

Options:
A $32 \sqrt{3}$
B $16 \sqrt{3}$
C $8 \sqrt{3}$
D 32

The correct answer is:
$32 \sqrt{3}$

Solution:
Given equation of parabola, $y^2=16 x$ and all the vertices are of an equilateral triangle with one of them coincides with the vertex of parabola
Since, given parabola is symmetrical about $X$-axis So, $\angle A O M=30^{\circ}$
In $\triangle A O M, \tan 30^{\circ}=\frac{A M}{O M}=\frac{8 t}{4 t^2} \Rightarrow \frac{1}{\sqrt{3}}=\frac{2}{t} \Rightarrow$
$t=2 \sqrt{3}$
$\therefore$ Coordinate of point $A$ lying on parabola is
$\left(4(2 \sqrt{3})^2, 8(2 \sqrt{3})\right) \equiv(48,16 \sqrt{3})$
$\therefore \text { Distance } O A=\sqrt{\left(x_A-x_0\right)^2+\left(y_A-y_0\right)^2}$
$=\sqrt{(48-0)^2+(16 \sqrt{3}-0)^2}$
$O A=32 \sqrt{3}$
$\therefore$ Length of the side of equilateral triangle $=32 \sqrt{3}$.

Topic: Mathematics, Parabola, JEE Main
If all the words, with or without meaning, are written using the letters of the word QUEEN and are arranged as in English dictionary, then the position of the word QUEEN is:

Options:
A 47th
B 45th
C 46th
D 44th

The correct answer is:
46th

Solution:
E, E, N, Q, U
(i) E……….= 4! = 24
(ii) N……….=4!2=12
(iii) Q E……..=3!=6
iv Q N……..=3!2!=3
v Q U E E N=1
Total =(i)+(ii)+(iii)+(iv)+(v)=46th

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2017 (08 Apr Online)
If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary; then the position of the word SMALL is

Options:
A 52nd
B 58th
C 46th
D 59th

The correct answer is:
58th

Solution:
SMALL
Total number of words formed would be 5!2!=60
When arranged as per dictionary. The words starting from A
A- - - -4!2!=12
The words starting from L
L- - - - = 4!=24
The words starting from M
M- - - - =4!2!=12
The words starting from S
S A- - - =3!2!=3
S L- - - 3!=6
S M A L L =1
∴ Rank would be 12 + 24 + 12 + 3 + 6 + 1 = 58

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2016 (03 Apr)
If \(\alpha\) and \(\beta\) are the roots of the equation \(x^2+x+1=0\), then the equation whose roots are \(\alpha^{19}\) and \(\beta^7\) is

Options:
A \(x^2-x-1=0\)
B \(x^2-x+1=0\)
C \(x^2+x-1=0\)
D \(x^2+x+1=0\)

The correct answer is:
\(x^2+x+1=0\)

Solution:
Hints: \(\alpha\) and \(\beta\) are the roots of \(x^2+x+1=0\)
\(\begin{array}{ll}
\alpha=\omega & \beta=\omega^2 \\
\alpha^{19}=\omega & \beta^7=\omega^2 \\
x^2-\left(\alpha^{19}+\beta^7\right) x+\alpha^{19} \beta^7=0
\end{array}\)
Thou,
\(\begin{aligned}
& x^2-\left(\omega+\omega^2\right) x+\omega \cdot \omega^2=0 \\
& x^2+x+1=0
\end{aligned}\)

Topic: Mathematics, Quadratic Equation, JEE Main
If amount of heat given to a system be \( 50 \mathrm{~J} \) and work done on the system be \( 15 \mathrm{~J} \), then change in internal energy of the
system is

Options:
A \( 35 \mathrm{~J} \)
B \( 50 \mathrm{~J} \)
C \( 65 \mathrm{~J} \)
D \( 15 \mathrm{~J} \)

The correct answer is:
\( 65 \mathrm{~J} \)

Solution:
Given that the amount of heat given to a system be dQ = 50 J and work done on the system be dW = -15 J

Here, dQ=50 J, dW=-15 J
from the first law of thermodynamics,dQ=dU+dW →1

the change in internal energy of the system is
dU=dQ-dW=50--15=65 J

Topic: Physics, Thermodynamics, JEE Main
If amplitude of waves at distance $r$ from a point source is $A$, the amplitude at a
distance $2 r$ will be

Options:
A $2 A$
B $A$
C $A / 2$
D $A / 4$

The correct answer is:
$A / 2$

Solution:
$f \propto a^2 \propto \frac{1}{d^2} \Rightarrow a \propto \frac{1}{d}$

Topic: Physics, Waves and Sound, JEE Main
If an ammeter is to be used in place of a voltmeter, then we must connect with the ammeter a

Options:
A
low resistance in parallel

B
high resistance in parallel

C
high resistance in series

D
low resistance in series

The correct answer is:
high resistance in series

Solution:
A voltmeter is a high resistance galvanometer and is connected in parallel to circuit and ammeter is a low resistance galvanometer so if we connect high resistance in series with ammeter its resistance will be much high.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2002
If an angle A of a ΔABC satisfies 5cosA+3=0, then the roots of the quadratic equation 9x2+27x+20=0 are

Options:
A secA, cotA
B secA, tanA
C tanA, cosA
D sinA, secA

The correct answer is:
secA, tanA

Solution:
We have,

5cosA+3=0⇒cosA=-35

Clearly, A∈90°,180°

Now, roots of equation

9x2+27x+20=0

⇒x=-27±729-72018=-27±318

⇒x=-53, -43

Then, -53=secA and -43=tanA.

Hence, the roots are secA and tanA.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2018 (16 Apr Online)
If an angle between the line, x+12=y-21=z-3-2 and the plane, x-2y-kz=3 is cos-1223, then a value of k is

Options:
A 53
B 35
C -35
D -53

The correct answer is:
53

Solution:
cos90-θ=n→1.v→n1→v→

⇒sinθ=i^-2j^-kk^.2i^+j^-2k^1+4+k24+1+4

∵θ=cos-1223⇒cosθ=223⇒sinθ=13

⇒13=2-2+2k35+k2

2k=5+k2⇒k=±53

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (12 Jan Shift 2)
If an atom has electronic configuration $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^3 4 s^2$, it will be placed in

Options:
A Second group
B Third group
C Fifth group
D Sixth group

The correct answer is:
Fifth group

Solution:
The outermost shell and one shell prior to it in d subshell, has total 5 electron so it will be placed in fifth group in transition element or 'd' block of periodic table.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
If an electron and a proton having same momenta enter perpendicular to a magnetic field, then

Options:
A
curved path of electron and proton will be same (ignoring the sense of revolution)

B
they will move undeflected

C
curved path of electron is more curved than that of the proton

D
path of proton is more curved

The correct answer is:
curved path of electron and proton will be same (ignoring the sense of revolution)

Solution:
r = mv / Bq is same for both

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2002
If an electron in a hydrogen atom is in its 2nd excited state, orbiting the nucleus in a stationary orbit of radius 4.65 A∘, then its de-Broglie wavelength is

Options:
A 12.9 A∘
B 3.5 Aâ
C 9.7 Aâ
D 6.6 A∘

The correct answer is:
9.7 Aâ

Solution:
2πr=nλ
⇒2π=3λ
λ=9.7A

Topic: Physics, Atomic Physics, JEE Main
If an electron is moving in the nth orbit of the hydrogen atom, then its velocity (vn) for the nth orbit is given as:

Options:
A vn∝n
B vn∝1n
C vn∝n2
D vn∝1n2

The correct answer is:
vn∝1n

Solution:
We know velocity of electron in nth shell of hydrogen atom is given by

v=2πkZe2nh

∴v∝1n

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2021 (17 Mar Shift 1)