datavorous/entrance-exam-dataset huggingface dataset:
If $A=\left[\begin{array}{cc}2-k & 2 \\ 1 & 3-k\end{array}\right]$ is a singular matrix, then the value of $5 k-k^2$ is

Options:
A 0
B 6
C -6
D 4

The correct answer is:
4

Solution:
Given, $A=\left[\begin{array}{cc}2-k & 2 \\ 1 & 3-k\end{array}\right]$
Since, the Matrix $A$ is singular
$$
\begin{aligned}
& \therefore \quad|A|=0 \\
& \Rightarrow \quad\left|\begin{array}{cc}
2-k & 2 \\
1 & 3-k
\end{array}\right|=0 \\
& \Rightarrow \quad(2-k)(3-k)-2=0 \\
& \Rightarrow \quad 6-5 k+k^2-2=0 \\
& \Rightarrow \quad 4-5 k+k^2=0 \\
& \Rightarrow \quad 5 k-k^2=4 \\
&
\end{aligned}
$$

Topic: Mathematics, Determinants, JEE Main
If $\int \frac{1}{\mathrm{a}^2 \sin ^2 x+\mathrm{b}^2 \cos ^2 x} \mathrm{~d} x=\frac{1}{12} \tan ^{-1}(3 \tan x)+$ constant, then the maximum value of $\mathrm{a} \sin x+\mathrm{b} \cos x$, is :

Options:
A $\sqrt{40}$
B $\sqrt{41}$
C $\sqrt{39}$
D $\sqrt{42}$

The correct answer is:
$\sqrt{40}$

Solution:
$\begin{aligned}
& \int \frac{\sec ^2 x d x}{a^2 \tan ^2 x+b^2} \\
& \text { let } \tan x=t \\
& \sec ^2 d x=d t \\
& \int \frac{d t}{a^2 t^2+b^2} \\
& \frac{1}{a^2} \int \frac{d t}{t^2+\left(\frac{b}{a}\right)^2} \\
& \frac{1}{a^2} \frac{1}{\frac{b}{a}} \tan ^{-1}\left(\frac{t}{b} a\right)+c \\
& \frac{1}{a b} \tan ^{-1}\left(\frac{\alpha}{b} \tan x\right)+c
\end{aligned}$
on comparing $\frac{\mathrm{a}}{\mathrm{b}}=3$
$\begin{aligned}
& a b=12 \\
& a=6, b=2
\end{aligned}$
maximum value of
$6 \sin x+2 \cos x \text { is } \sqrt{40}$

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2024 (06 Apr Shift 2)
If a>2, then the roots of the equation 2-ax2+3ax-1=0 are

Options:
A one positive and one negative
B both negative
C both positive
D both imaginary

The correct answer is:
both positive

Solution:
Let, fx=x2-3aa-2x+1a-2=0
D=9a2a-22-41a-2=9a2-4a+8a-22
=8a2+a-22+4a-22>0
f0=1a-2>0 and
-B2A=3a2a-2>0
Since, D>0,f0>0,-B2A>0
Hence, both roots of the given equation are positive.

Topic: Mathematics, Quadratic Equation, JEE Main
If $A(3,1,-1), B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right), C(2,2,1)$ and $D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right)$ are the vertices of a quadrilateral $A B C D$, then its area is

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

The correct answer is:
$\frac{4 \sqrt{2}}{3}$

Solution:
Area $=\frac{1}{2}|\overline{\mathrm{BD}} \times \overline{\mathrm{AC}}|$
$\begin{aligned} & \overline{\mathrm{BD}}=\frac{5}{3} \hat{\mathrm{i}}-\frac{5}{3} \hat{\mathrm{j}}-\frac{2}{3} \hat{\mathrm{k}} \\ & \overline{\mathrm{AC}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}}\end{aligned}$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2024 (06 Apr Shift 1)
If $A\left[\begin{array}{lll}3 & 2 & 4 \\ 1 & 2 & 1 \\ 3 & 2 & 6\end{array}\right]$ and $\mathrm{A}_{\mathrm{ij}}$ are cofactors of the elements $\mathrm{a}_{\mathrm{ij}}$ of $\mathrm{A}$, then $a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13}$ is equal to

Options:
A 8
B 6
C 4
D 0

The correct answer is:
8

Solution:
$\begin{aligned} & a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13}=|A| \\ & \therefore|A|=\left|\begin{array}{lll}3 & 2 & 4 \\ 1 & 2 & 1 \\ 3 & 2 & 6\end{array}\right| \\ & =3(12-2)-2(6-3)+4(2-6)=30-6-16=8\end{aligned}$

Topic: Mathematics, Determinants, JEE Main
If $A=\left[\begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right]$ then $A^{-1}=$

Options:
A $A$
B $A^2$
C $A^3$
D $A^4$

The correct answer is:
$A^3$

Solution:
Here, $C_{11}=1, C_{12}=-2, C_{13}=-2$
$C_{21}=-1, C_{22}=3, C_{23}=3$
$C_{31}=0, C_{32}=-4, C_{33}=-3$
$\left.\Rightarrow \operatorname{det} A \neq A|=| \begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array} \right\rvert\,=1$
$\Rightarrow A^{-1}=\frac{1}{|A|} \cdot(A d j A)=\frac{1}{1} \cdot\left[\begin{array}{lll}C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33}\end{array}\right]$
$=\left[\begin{array}{ccc}1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3\end{array}\right]$
Now, $A^2=\left[\begin{array}{ccc}3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3\end{array}\right]$
and $A^3=A^2 \cdot A=\left[\begin{array}{ccc}3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3\end{array}\right] \times\left[\begin{array}{ccc}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right]$
$=\left[\begin{array}{ccc}1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3\end{array}\right]=A^{-1}$.

Topic: Mathematics, Matrices, JEE Main
If A=3-27-5, then the value of -3A2019+A2020 is equal to

Options:
A -14
B 28
C 14
D 22019⋅14

The correct answer is:
-14

Solution:
A=-15+14=-1
A2019-3Ι+A=A2019A-3Ι
=-10-27-8=-114=-14

Topic: Mathematics, Determinants, JEE Main
If $A=\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]$ then $A^2-5 A=$

Options:
A $I$
B $14I$
C $0$
D None of these

The correct answer is:
$14I$

Solution:
$A^2=A \cdot A=\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]$
$\Rightarrow A^2=\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right]$ and $5 A=\left[\begin{array}{cc}15 & -25 \\ -20 & 10\end{array}\right]$
$\therefore A^2-5 A=14\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=14 I$.

Topic: Mathematics, Matrices, JEE Main
If $A=\left[\begin{array}{ll}3 & 4 \\ 5 & 7\end{array}\right]$, then $\mathrm{A}$. ( $\operatorname{adj} A$ ) is equal to

Options:
A $\mathrm{A}$
B $|A|$
C $|A|$
D None of these

The correct answer is:
$|A|$

Solution:
$\mathrm{A}=\left[\begin{array}{ll}3 & 4 \\ 5 & 7\end{array}\right]$
$|\mathrm{A}|=21-20=1$
$\begin{array}{l}
\therefore \quad \mathrm{A}(\operatorname{adj} \mathrm{A})=\left[\begin{array}{ll}
3 & 4 \\
5 & 7
\end{array}\right]\left[\begin{array}{cc}
7 & -4 \\
-5 & 3
\end{array}\right] \\
=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=1 .\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=|\mathrm{A}| . \mathrm{I}
\end{array}$

Topic: Mathematics, Matrices, JEE Main
If $A=35^{\circ}, B=15^{\circ}$ and $C=40^{\circ}$, then $\tan A \cdot \tan B+\tan B \cdot \tan C+\tan C \cdot \tan A$ is equal to

Options:
A $0$
B $1$
C $2$
D $3$

The correct answer is:
$1$

Solution:
Given that, $A=35^{\circ}, B=15^{\circ}$ and $C=40^{\circ}$
$$
\begin{aligned}
& \therefore \tan (A+B+C) \\
& =\frac{\left[\begin{array}{c}
\tan A+\tan B+\tan C \\
-\tan A \tan B \tan C
\end{array}\right]}{\left[\begin{array}{c}
1-\tan A \tan B-\tan B \tan C \\
-\tan C \tan A
\end{array}\right]} \\
& \Rightarrow \tan \left(90^{\circ}\right) \\
& =\frac{\tan A+\tan B+\tan C-\tan A \tan B \tan C}{1-\tan A \tan B-\tan B \tan C-\tan C \tan A} \\
& \Rightarrow \quad \tan A \tan B+\tan B \tan C \\
& +\tan C \tan A=1 \\
&
\end{aligned}
$$

Topic: Mathematics, Trigonometric Equations, JEE Main
If $A=3 \hat{i}+4 \hat{j}$ and $B=7 \hat{i}+24 \hat{j}$, the vector having the same magnitude as $B$ and parallel to $A$ is

Options:
A $5 \hat{i}+20 \hat{j}$
B $15 \hat{i}+10 \hat{j}$
C $20 \hat{i}+15 \hat{j}$
D $15 \hat{i}+20 \hat{j}$

The correct answer is:
$15 \hat{i}+20 \hat{j}$

Solution:
$|B|=\sqrt{7^2+(24)^2}=\sqrt{625}=25$
Unit vector in the direction of $A$ will be
$\hat{A}=\frac{3 \hat{i}+4 \hat{j}}{5}$
So required vector $=25\left(\frac{3 \hat{i}+4 \hat{j}}{5}\right)=15 \hat{i}+20 \hat{j}$

Topic: Physics, Mathematics in Physics, JEE Main
If $\overline{\mathrm{a}}=3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}, \overline{\mathrm{b}}=6 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}$ are two vectors and $\overline{\mathrm{c}}$ is vector such that $\bar{c}=\bar{a} \times \bar{b}$, then $a: b: c$ is

Options:
A $\sqrt{34}: \sqrt{45}: \sqrt{39}$
B $\sqrt{34}: \sqrt{45}: 39$
C $34: 39: 45$
D $39: 35: 34$

The correct answer is:
$\sqrt{34}: \sqrt{45}: 39$

Solution:
$$
\begin{aligned}
& \overline{\mathrm{c}}=\overline{\mathrm{a}} \times \overline{\mathrm{b}}=\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
3 & -5 & 0 \\
6 & 3 & 0
\end{array}\right|=\hat{\mathrm{i}}(0)-\hat{\mathrm{j}}(0)+\hat{\mathrm{k}}(9+30)=39 \hat{\mathrm{k}} \\
& |\overline{\mathrm{a}}|=\sqrt{(3)^2+(-5)^2}=\sqrt{34} \text { and }|\overline{\mathrm{b}}|=\sqrt{(6)^2+(3)^2}=\sqrt{45} \\
& |\overline{\mathrm{c}}|=\sqrt{39^2}=39 \\
& \therefore \mathrm{a}: \mathrm{b}: \mathrm{c}=\sqrt{34}: \sqrt{45}: 39
\end{aligned}
$$

Topic: Mathematics, Vector Algebra, JEE Main
If $\mathbf{a}=3 \mathbf{i}-\mathbf{j}+2 \mathbf{k}, \mathbf{b}=2 \mathbf{i}+\mathbf{j}-\mathbf{k}$, then $\mathbf{a} \times(\mathbf{a} \cdot \mathbf{b})=$

Options:
A $3 a$
B $3 \sqrt{14}$
C 0
D None of these

The correct answer is:
None of these

Solution:
$\quad \mathbf{a} \times(\mathbf{a} \cdot \mathbf{b})$ is not meaningful.

Topic: Mathematics, Vector Algebra, JEE Main
If a→=3i^-j^10 and b→=i^+3j^+k^11, then the value of 2a→+b→⋅a→×b→×a→-3b→ is

Options:
A 5
B 7
C 8
D 9

The correct answer is:
7

Solution:
Given, a→=3i^-j^10 and b→=i^+3j^+k^11

So, |a→|=1, |b→|=1, a→⋅b→=0
2a→+b→⋅a→×b→×a→-3b→

=2a→+b→·a→·a→-3b→b→-b→·a→-3b→a→
=2a→+b→·b→+3a→
=7

Topic: Mathematics, Vector Algebra, JEE Main
If A=-4-131 , then the determinant of the matrix A2016-2A2015-A2014 is :

Options:
A -175
B 2014
C 2016
D -25

The correct answer is:
-25

Solution:
A=-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, Matrices, JEE Main, JEE Main 2016 (10 Apr Online)
If $|\vec{a}|=4,|\vec{b}|=2$ and the angle between $\vec{a}$ and $\vec{b}$ is $\pi / 6$ then $(\vec{a} \times \vec{b})^2=2$ is equal to

Options:
A
48

B
16

C
$\vec{a}$

D
none of these

The correct answer is:
16

Solution:
We have, $\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \frac{\pi}{6}=4 \times 2 \times \frac{\sqrt{3}}{2}=4 \sqrt{3}$
Now, $(\vec{a} \times \vec{b})^2+(\vec{a} \cdot \vec{b})^2=a^2 b^2 \Rightarrow(\vec{a} \times \vec{b})^2+48=16 \times 4 \Rightarrow(\vec{a} \times \vec{b})^2=16$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2002
If $\vec{A}=4 \hat{i}-3 \hat{j}$ and $\vec{B}=6 \hat{i}+8 \hat{j}$ then magnitude and direction of $\dot{A}+\dot{B}$ will be

Options:
A $5, \tan ^{-1}(3 / 4)$
B $5 \sqrt{5}, \tan ^{-1}(1 / 2)$
C $10, \tan ^{-1}(5)$
D $25, \tan ^{-1}(3 / 4)$

The correct answer is:
$5 \sqrt{5}, \tan ^{-1}(1 / 2)$

Solution:
$\vec{A}+\vec{B}=4 \hat{i}-3 \hat{j}+6 \hat{i}+8 \hat{j}=10 \hat{i}+5 \hat{j}$
$|\vec{A}+\vec{B}|=\sqrt{(10)^2+(5)^2}=5 \sqrt{5}$
$\tan \theta=\frac{5}{10}=\frac{1}{2} \Rightarrow \theta=\tan ^{-1}\left(\frac{1}{2}\right)$

Topic: Physics, Mathematics in Physics, JEE Main
If a $4 \mathrm{~kg}$ of ice is inside a closed cubical thermocol box of side length $20 \mathrm{~cm}$ and wall thickness $4 \mathrm{~cm}$ then the mass of the ice remaining after 10 hours is nearly
(The out side temperature $=50^{\circ} \mathrm{C}$
co-efficient of thermal conductivity of thermocol $=0.01 \mathrm{Js}^{-1} \mathrm{~m}^{-1{ }^{\circ}} \mathrm{C}^{-1}$
Latent heat of fusion of ice $=335 \times 10^3 \mathrm{~J} \mathrm{~kg}^{-1}$)

Options:
A $3.678 \mathrm{~kg}$
B $6.378 \mathrm{~kg}$
C $2.87 \mathrm{~kg}$
D $1.87 \mathrm{~kg}$

The correct answer is:
$3.678 \mathrm{~kg}$

Solution:
No solution. Refer to answer key.

Topic: Physics, Thermal Properties of Matter, AP EAMCET, AP EAMCET 2017 (25 Apr Shift 2)
If a 0.004 M solution of Na2SO4 is isotonic with a 0.010 M solution of glucose at same temperature. Then what is the apparent degree of disociation of Na2SO4 is

Options:
A 25%
B 50%
C 75%
D 85%

The correct answer is:
75%

Solution:
Na2SO4(0.004-x)⇌2Na+2x+SO42-x
Since both the solution are isotonic 0.004+2x=0.01
x=0.003
HNO2 Percent dissociation. =3×10-30.004×100=75% .

Topic: Chemistry, Solutions, JEE Main
If $A=\left(\begin{array}{ccc}5 & 5 x & x \\ 0 & x & 5 x \\ 0 & 0 & 5\end{array}\right)$ and $\left|A^{2}\right|=25,$ then $|x|$ is equal to

Options:
A $\frac{1}{5}$
B 5
C $5^{2}$
D 1

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

Solution:
Given
$$
\begin{aligned}
A &=\left[\begin{array}{ccc}
5 & 5 x & x \\
0 & x & 5 x \\
0 & 0 & 5
\end{array}\right] \\
\therefore \quad A^{2}=A \cdot A
&=\left[\begin{array}{ccc}
5 & 5 x & x \\
0 & x & 5 x \\
0 & 0 & 5
\end{array}\right]\left[\begin{array}{ccc}
5 & 5 x & x \\
0 & x & 5 x \\
0 & 0 & 5
\end{array}\right] \\
&\left[\begin{array}{ccc}
25 & 25 x+5 x^{2} & 5 x+25 x^{2}+5 x \\
0 & x^{2} & 5 x^{2}+25 \\
0 & 0 & 25
\end{array}\right] \\
&=25\left(25 x^{2}-0\right) \\
&=25\left(25 x^{2}\right)
\end{aligned}
$$
since, given that $|A|^{2}=25$
Now, $25\left(25 x^{2}\right)=25$
$\begin{array}{lr}\Rightarrow & 25 x^{2}=1 \\ \Rightarrow & x^{2}=\frac{1}{25} \\ \Rightarrow & x=\pm \frac{1}{5}\end{array}$
$\Rightarrow \quad |x|=\frac{1}{5}$

Topic: Mathematics, Matrices, JEE Main
If A=5a-b32 and A.adjA=A AT , then 5a+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, JEE Main 2016 (03 Apr)
If $|a|=5,|b|=4,|c|=3$ thus what will be the value of $|a \cdot b+b . c+c . a|$, given that $\vec{a}+\vec{b}+\vec{c}=0$

Options:
A
25

B
50

C
-25

D
-50

The correct answer is:
25

Solution:
We have, $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0} \Rightarrow(\vec{a}+\vec{b}+\vec{c})^2=0$
$\Rightarrow|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0 \Rightarrow 25+16+9+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0$
$\Rightarrow(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=-25 \quad \therefore|\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}|=25$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2002
If $\bar{a}=\hat{\imath}+5 \hat{k}, \bar{b}=2 \hat{\imath}+3 \hat{k}, \bar{c}=4 \hat{\imath}-\hat{\jmath}+2 \hat{k}$ and $\bar{d}=\hat{\imath}-\hat{\jmath}$,
then $(\bar{c}-\bar{a}) \cdot(\bar{b} \times \bar{d})=$

Options:
A $12$
B $20$
C $30$
D $10$

The correct answer is:
$12$

Solution:
$\bar{b} \times \bar{d}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 3 \\ 1 & -1 & 0\end{array}\right|=\hat{i}(3)-\hat{j}(0-3)+\hat{k}(-2)$
$\bar{b} \times \bar{d}=3 \hat{i}+3 \hat{j}-2 \hat{k}$
and $\bar{c}-\bar{a}=3 \hat{i}-\hat{j}-3 \hat{k}$
then $(\bar{c}-\bar{a}) \cdot(\bar{b} \times \bar{d})=3(3)+(-1)(3)+(-3)(-2)$
$=9-3+6=12$

Topic: Mathematics, Vector Algebra, JEE Main
If $\mathrm{A}=75^{\circ}, \mathrm{B}=45^{\circ}$, then $\mathrm{b}+\mathrm{c} \sqrt{2}$ equals

Options:
A $\mathrm{a}$
B $a+b+c$
C $2 a$
D $\frac{1}{2}(a+b+c)$

The correct answer is:
$2 a$

Solution:
$C=180^{\circ}-120^{\circ}=60^{\circ}$
Use sine rule $\frac{a}{\sin 75^{\circ}}=\frac{b}{\sin 45^{\circ}}=\frac{c}{\sin 60^{\circ}}=k$
$\Rightarrow \quad(b+c \sqrt{2})=k\left(\sin 45^{\circ}+\sqrt{2} \sin 60^{\circ}\right)$
$=\mathrm{k} \frac{\sqrt{3}+1}{\sqrt{2}}=2 \mathrm{k} \frac{\sqrt{3}+1}{2 \sqrt{2}}$
$=2 \mathrm{k} \sin 75^{\circ}=2 \mathrm{k} \sin \mathrm{A}=2 \mathrm{a}$

Topic: Mathematics, Properties of Triangles, JEE Main
If $\overrightarrow{\mathrm{a}}=\frac{1}{\sqrt{10}}(3 \hat{\mathrm{i}}+\hat{\mathrm{k}})$ and $\overrightarrow{\mathrm{b}}=\frac{1}{7}(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}})$, then the value of $(2 \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}) \cdot[(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times(\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}})]$ is

Options:
A
$-3$

B
$5$

C
$3$

D
$-5$

The correct answer is:
$-5$

Solution:
$(2 \overline{\mathrm{a}}-\overline{\mathrm{b}}) \cdot\{(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times(\overline{\mathrm{a}}+2 \overline{\mathrm{b}})\}=(2 \overline{\mathrm{a}}-\overline{\mathrm{b}}) \cdot\{[\overline{\mathrm{a}} \cdot(\overline{\mathrm{a}}+2 \overline{\mathrm{b}})] \overline{\mathrm{b}}-[\overline{\mathrm{b}} \cdot(\overline{\mathrm{a}}+2 \overline{\mathrm{b}}) \overline{\mathrm{a}}]\}$
$=-5(\overline{\mathrm{a}})^2(\overline{\mathrm{b}})^2+5(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}})^2=-5$

Topic: Mathematics, Vector Algebra, JEE Main
If $\overrightarrow{\mathrm{a}}=\frac{1}{\sqrt{10}}(3 \hat{\mathrm{i}}+\hat{\mathrm{k}})$ and $\overrightarrow{\mathrm{b}}=\frac{1}{7}(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}})$, then the value of $(2 \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}) \cdot[(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times(\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}})]$ is

Options:
A
$-3$

B
$5$

C
$3$

D
$-5$

The correct answer is:
$-5$

Solution:
$(2 \overline{\mathrm{a}}-\overline{\mathrm{b}}) \cdot\{(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times(\overline{\mathrm{a}}+2 \overline{\mathrm{b}})\}=(2 \overline{\mathrm{a}}-\overline{\mathrm{b}}) \cdot\{[\overline{\mathrm{a}} \cdot(\overline{\mathrm{a}}+2 \overline{\mathrm{b}})] \overline{\mathrm{b}}-[\overline{\mathrm{b}} \cdot(\overline{\mathrm{a}}+2 \overline{\mathrm{b}}) \overline{\mathrm{a}}]\}$
$=-5(\overline{\mathrm{a}})^2(\overline{\mathrm{b}})^2+5(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}})^2=-5$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2011
If a=111â¦..1â55 times , b=1+10+102+103+104 and c=1+105+1010+â¦.+1050, then

Options:
A b,a2,c are in arithmetic progression
B b,a,c are in geometric progression
C a,b,c are in geometric progression
D a,b,c are in arithmetic progression

The correct answer is:
b,a,c are in geometric progression

Solution:
a=1+10+102+……..+1054=1055-19
b=105-19
c=10511-1105-1=1055-1105-1
⇒ab=c
⇒a=bc
⇒b,a,c are in G.P.

Topic: Mathematics, Sequences and Series, JEE Main
If $\left(a, a^2\right)$ falls inside the angle made by the lines $y=\frac{x}{2}, x>0$ and $y=3 x, x>0$, then a belongs to

Options:
A
$\left(0, \frac{1}{2}\right)$

B
$(3, \infty)$

C
$\left(\frac{1}{2}, 3\right)$

D
$\left(-3,-\frac{1}{2}\right)$

The correct answer is:
$\left(\frac{1}{2}, 3\right)$

Solution:
$a^2-3 a < 0$ and $a^2-\frac{a}{2}>0 \Rightarrow \frac{1}{2} < a < 3$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2006
If $A=\{(a, b): 4 a=5 b ; a, b \in\{1,2,3, \ldots 30\}\}$, then the number of such ordered pairs $(a, b)$ is

Options:
A 4
B 6
C 8
D 10

The correct answer is:
6

Solution:
(b) The only possible pairs are:
$(5,4),(10,8),(15,12),(20,16),(25,20),(30,24)$

Topic: Mathematics, Sets and Relations, JEE Main
If $A=\left[\begin{array}{ll}a & b \\ b & a\end{array}\right]$ and $A_2=\left[\begin{array}{ll}\alpha & \beta \\ \beta & \alpha\end{array}\right]$, then

Options:
A
$\alpha=2 a b, \beta=a^2+b^2$

B
$\alpha=\mathrm{a}_2+\mathrm{b}_2, \beta=\mathrm{ab}$

C
$\alpha=a^2+b^2, \beta=2 a b$

D
$\alpha=a^2+b^2, \beta=a^2-b^2$

The correct answer is:
$\alpha=a^2+b^2, \beta=2 a b$

Solution:
$A^2=\left[\begin{array}{ll}\alpha & \beta \\ \beta & \alpha\end{array}\right]=\left[\begin{array}{ll}a & b \\ b & a\end{array}\right]\left[\begin{array}{ll}a & b \\ b & a\end{array}\right]$
$\alpha=a^2+b^2 ; \beta=2 a b$

Topic: Mathematics, Matrices, JEE Main, JEE Main 2003
If A=abcbcacab, abc=1 and ATA=I, then the value of a3+b3+c3 (where a,b,c>0 and AT is transpose of matrix A) is

Options:
A 2
B 0
C 1
D 4

The correct answer is:
4

Solution:
ATA=I
As AT=A
⇒A2=I
⇒a2+b2+c2=1 and ab+bc+ca=0
As a+b+c2=a2+b2+c2+2ab+bc+ca
⇒a+b+c=1
Now, a3+b3+c3-3abc=a+b+ca2+b2+c2-ab-bc-ca
⇒a3+b3+c3-3=a+b+c
⇒a3+b3+c3=3+1
⇒a3+b3+c3=4

Topic: Mathematics, Matrices, JEE Main
If a ∆ABC has vertices A–1,7, B–7,1 and C5,–5, then its orthocentre has coordinates:

Options:
A – 3,3
B 3,-3
C -35,35
D 35,-35

The correct answer is:
– 3,3

Solution:
Slope of line BC is mBC=6-12=-12

Slope of line AD=1mBC=2

∴ Equation of line AD is y-7=2x+1

⇒y=2x+9 ...1

Slope of line AC is mAC=12-6=-2

Slope of line BE=1mAC=12

∴ Equation of BE is y-1=12x+7

⇒y=x2+92 ...2

Comparing equations 1 & 2, we get

2x+9=x+92

⇒4x+18=x+9

⇒3x=-9⇒x=-3

∴ y=2x+9=3

Therefore, orthocentre has co-ordinates -3,3.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2020 (03 Sep Shift 2)
If ∫-aax+x-2dx=22, a>2 and x denotes the greatest integer ≤x, then ∫a-ax+xdx is equal to

Options:

The correct answer is:
3

Solution:
∫-a0-2x+2dx+∫02x+2-xdx+∫2a2x-2dx=22

⇒-x2+2x|-a0+2x|02+x2-2x|0a=22

⇒a2+2a+4+a2-2a=22

⇒2a2=18⇒a=3 (a≠-3)

∫3-3x+xdx=∫3-3xdx+∫3-3xdx=0+--3-2-1+1+2=3

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (24 Feb Shift 1)
If $A$ and $B$ an two events such that $P(A \cup B)=\frac{5}{6} \quad P(A \cap B)=\frac{1}{3}$ and $P(\bar{B})=\frac{1}{3}$, then $P(A)=$

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

The correct answer is:
$\frac{1}{2}$

Solution:
$P(A)=P(A \cap B)+P(A \cup B)-P(B)$
$=\frac{1}{3}+\frac{5}{6}-\frac{2}{3}=\frac{3}{6}=\frac{1}{2}$.

Topic: Mathematics, Probability, JEE Main
If $A$ and $B$ are any two events of a sample space, then set-theoretic description for the event: "Exactly one of the events A, B to occur" is
(Here $\mathrm{E}^{\mathrm{c}}$ denotes the compliment of the event $\mathrm{E}$ )

Options:
A $A \cap B^c$
B $(A-B) \cup(A \cup B)$
C $\left(A \cap B^c\right) \cup\left(A^c \cap B\right)$
D $(A \cap B)^c \cup\left(A^c \cap B^c\right)$

The correct answer is:
$\left(A \cap B^c\right) \cup\left(A^c \cap B\right)$

Solution:
When only event $A$ occurs, then it can be written as $\left(A \cap B^C\right)$
And when only event B occurs, then it can be written as $\left(A^C \cap B\right)$
Then, 'exactly one of the events $A, B$ to occur' can be written as : $\left(A \cap B^C\right) \cup\left(A^C \cap B\right)$

Topic: Mathematics, Sets and Relations, JEE Main
If A and B are any two events such that PA=25 and PA∩B=320, then the conditional probability, PAA′∪B′, where A' denotes the complement of A, is equal to :

Options:
A 1120
B 517
C 817
D 14

The correct answer is:
517

Solution:
Given PA=25=820 & PA∩B=320
∴ PA'∪B'=PA∩B'
=1-PA∩B
=1-320=1720
PA∩A'∪B'=PA∩B'
=PA-PA∩B
=820-320=520
∴ PAA'∪B'=PA∩A'∪B'PA'∪B'
=5201720=517

Topic: Mathematics, Probability, JEE Main, JEE Main 2016 (09 Apr Online)
If a and b are arbitrary constants, then the order and degree of the differential equation of the family of curves ax2+by2=2 respectively are

Options:
A 2,2
B 1,2
C 1,1
D 2,1

The correct answer is:
2,1

Solution:
ax2+by2=1
On differentiating w.r.t. x
ax+bydydx=0 ... 1
Again differentiating w.r.t. x
a+byd2ydx2+bdydx2=0 ... 2
From 1 & 2, we get,
x-byd2ydx2-bdydx2+bydydx=0
-xyd2ydx2-xdydx2+ydydx=0
Hence, order =2, degree =1

Topic: Mathematics, Differential Equations, JEE Main
If $a$ and $b$ are arbitrary positive real numbers, then the least possible value of $\frac{6 a}{5 b}+\frac{10 b}{3 a}$ is

Options:
A 4
B $\frac{6}{5}$
C $\frac{10}{3}$
D $\frac{68}{15}$

The correct answer is:
4

Solution:
Hint:
$\frac{6 a}{5 b}+\frac{10 b}{3 a} \geq 2 \sqrt{\frac{6 a}{5 b} \times \frac{10 b}{3 a}}, \quad \frac{6 a}{5 b}+\frac{10 b}{3 a} \geq 2 \times 2 \geq 4$

Topic: Mathematics, Sequences and Series, JEE Main
If $A$ and $B$ are events such that $P(A \cup B)=0.5, P(\bar{B})=0.8$
and $P(A / B)=0.4$, then what is $P(A \cap B)$ equal to?

Options:
A $0.08$
B 0.02
C $0.8$
D $0.2$

The correct answer is:
$0.08$

Solution:
Let $P(A \cup B)=0.5, P(\bar{B})=0.8, P\left(\frac{A}{B}\right)=0.4$
$\quad P(\bar{B})=1-P(B)$
$\Rightarrow 0.8=1-P(B)$
$\Rightarrow P(B)=1-0.8=0.2$
Now, $P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}$
$\Rightarrow P(B) \times P\left(\frac{A}{B}\right)=P(A \cap B)$
$\Rightarrow 0.4 \times 0.2=P(A \cap B)$
$\Rightarrow 0.08=P(A \cap B)$

Topic: Mathematics, Probability, JEE Main
If $A$ and $B$ are matrices and $B=A B A^{-1}$ then the value of $(A+B)(A-B)$ is

Options:
A $\mathrm{A}^{2}+\mathrm{B}^{2}$
B $\mathrm{A}^{2}-\mathrm{B}^{2}$
C $\mathrm{A}+\mathrm{B}$
D $\mathrm{A}-\mathrm{B}$

The correct answer is:
$\mathrm{A}^{2}-\mathrm{B}^{2}$

Solution:
$\mathrm{B}=\mathrm{ABA}^{-1}$ (Given)
But $B=B A A^{-A}$
$\therefore \mathrm{ABA}^{-1}=\mathrm{BAA}^{-1} \Rightarrow \mathrm{AB}=\mathrm{BA}$ $\mathrm{Now}(\mathrm{A}+\mathrm{B})(\mathrm{A}-\mathrm{B})=\mathrm{A}^{2}-\mathrm{AB}+\mathrm{BA}-\mathrm{B}^{2}$ $=\mathrm{A}^{2}-\mathrm{AB}+\mathrm{AB}-\mathrm{B}^{2} \quad[\because \mathrm{AB}=\mathrm{BA}]$ $=\mathrm{A}^{2}-\mathrm{B}^{2}$

Topic: Mathematics, Matrices, JEE Main