datavorous/entrance-exam-dataset huggingface dataset:
If 400 gm radioactive Iodine is taken initially. Then calculate the amount of 53 I 128 t 1 2  = 25 min left after 75 minutes

Options:

The correct answer is:
50

Solution:
N N 0 â=â 1 2 T t 1 2 â N N 0 â=â 1 2 75 25
N 0 â=â400âg
Nâ=â 400/8
N = 50 gm

Topic: Chemistry, Chemical Kinetics, JEE Main
If $5.85 \mathrm{gms}$ of $\mathrm{NaCl}$ are dissolved in $90 \mathrm{gms}$ of water, the mole fraction of $\mathrm{NaCl}$ is

Options:
A 0.1
B 0.2
C 0.3
D 0.0196

The correct answer is:
0.0196

Solution:
$\begin{aligned} & 5.85 \mathrm{~g} \mathrm{NaCl}=\frac{5.85}{58.5} \text { mole }=0.1 \mathrm{~mol} \\ & 90 \mathrm{~g}\end{aligned}$
mole fraction of $\mathrm{NaCl}=$$\frac{0.1}{5+0.1} \approx 0.0196$

Topic: Chemistry, Solutions, JEE Main
If 917 Å be the lowest wavelength of Lyman series then the lowest wavelength of Balmer series will be Å.

Options:

The correct answer is:
3668

Solution:
Using Rydberg's formula,

1λ=RZ21n21-1n22

For the lowest wavelength in Lyman series, n1=1, n2=∞.

1λ=RZ2112-1∞2=RZ2

For Balmer series, n1=2, n2=∞

1λ'=RZ2122-1∞2=RZ24

So,

1λ'=14λ⇒λ'=4×917 Å=3668 Å

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2023 (10 Apr Shift 2)
If 1000 droplets of water of surface tension 0.07 N m-1. having same radius 1 mm each, combine to from a single drop. In the process the released surface energy is-

Take π=227

Options:
A 7.92×10-6 J
B 7.92×10-4 J
C 9.68×10-4 J
D 8.8×10-5J

The correct answer is:
7.92×10-4 J

Solution:
Let the radius of bigger droplet is R, then by volume conservation:

43πR3=100043π13

⇒R=10 mm

Final potential energy:

Uf=10004πR2T=1000×4π10×10-32T

Initial potential energy:

Ui=4π×r2T=4π×10-32T

Therefore, change in potential energy,

∆E=T×1000×4π10-32-T×4π10×10-32

∆E=4×π×7×10-21000-100×10-6

∆E=7.92×10-4J

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2023 (31 Jan Shift 1)
If (2021)3762 is divided by 17, then the remainder is _______.

Options:

The correct answer is:
4

Solution:
(2023-2)3762=2023k1+23762

=17k2+23762 (as2023=17×17×9)

=17k2+4×16940

=17k2+4×(17-1)940

=17k2+417k3+1

=17k+4⇒remainder=4

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (17 Mar Shift 1)
If 64,27,36 are the Pth $Q$ th and $R$ th terms of a GP, then $P+2 Q$ is equal to

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

The correct answer is:
$3 R$

Solution:
Let a be the first term and \(r\) be the common ration of a GP. \(\therefore P^{\text {th }}, Q^{\text {th }}\) and \(R^{\text {th }}\) terms of a GP are respectively ar \({ }^{\mathrm{P}-1}\), ar \(\mathrm{Q}^{-1}\) and \({a r}^{\mathrm{R}-1}\).
According to question, \(a r^{P-1}=64 \ldots . .\) (i)
\(a r^{Q-1}=27 \ldots . .\) (ii)
\(a r^{R-1}=36 \ldots . .\) (iii)
Dividing Eq. (i) by Eq. (ii), we get
\(\mathrm{r}^{\mathrm{P}-\mathrm{Q}}=\left(\frac{4}{3}\right)^{3} \ldots \ldots\) (iv)
Dividing Eq. (ii) by Eq. (iii), we get
\(\begin{array}{l}
\mathrm{r}^{\mathrm{Q}-\mathrm{R}}=\frac{3}{4} \\
\Rightarrow \mathrm{r}^{3 \mathrm{Q}-3 \mathrm{R}}=\left(\frac{3}{4}\right)^{3}
\end{array}\)
Multiplying Eq. (iv) and Eq. (v), we get
\(\begin{array}{l}
\mathrm{r}^{\mathrm{P}-\mathrm{Q}} \times \mathrm{r}^{3 \mathrm{Q}-3 \mathrm{R}}=1 \\
\Rightarrow \mathrm{r}^{\mathrm{P}-\mathrm{Q}+3 \mathrm{Q}-3 \mathrm{R}}=1 \\
\Rightarrow \mathrm{r}^{\mathrm{P}+2 \mathrm{Q}-3 \mathrm{R}}=\mathrm{r}^{0} \\
\Rightarrow \mathrm{P}+2 \mathrm{Q}-3 \mathrm{R}=0 \\
\Rightarrow \mathrm{P}+2 \mathrm{Q}=3 \mathrm{R}
\end{array}\)

Topic: Mathematics, Sequences and Series, JEE Main
If \( 53^{53}-33^{3} \) is divided by \( 10 \), then the remainder obtained is

Options:

The correct answer is:
6

Solution:
We know that,
xn-yn=x-yxn-1+xn-2y+....+xyn-2+yn-1
i.e. xn-yn is divisible by x-y
5353-333=5353-353+353-333+33-33
=5353-353+353-33-333-33
Clearly, 1st bracket is divisible by 50 and the last one is divisible by 30. So, both are divisible by 10.
353-33=33â350-33=33350-1
=3310-125-1=2710-125-27=2710k-1-27
=270k-54=270k-60+6=10Î¼+6.
Hence, the remainder obtained is 6.

Topic: Mathematics, Binomial Theorem, JEE Main
If $A=\left[\begin{array}{ccc}0 & 2 & 0 \\ 0 & 0 & 3 \\ -2 & 2 & 0\end{array}\right] \text {and } B=\left[\begin{array}{ccc}1 & 2 & 3 \\ 3 & 4 & 5 \\ 5 & -4 & 0\end{array}\right]$, then the element of $3^{\text {rd }}$ row and third column in $A B$ will be

Options:
A $-18$
B $4$
C $-12$
D None of these

The correct answer is:
$4$

Solution:
In the product $A B$, the required element
$C_{33}=(-2) 3+2.5+0.0=-6+10=4$

Topic: Mathematics, Matrices, JEE Main
If A0,0, BÎ¸,cosâ¡Î¸ and Csin3â¡Î¸,0 are the vertices of a triangle ABC, then the value of Î¸ for which the triangle has the maximum area is where Î¸â0,Ï2

Options:
A π6
B π4
C π3
D π2

The correct answer is:
π3

Solution:
Area of triangle ABC, △=12001θcos⁡θ1sin3⁡θ01
⇒△=12sin3⁡θcos⁡θ
d△dθ=3sin2⁡θcos2⁡θ-sin4⁡θ2=0
⇒sin2⁡θ3cos2⁡θ-sin2⁡θ=0
⇒sin⁡θ=0 or tan2⁡θ=3
θ=0 or θ=π3
d2△dθ2<0 for θ=π3

Topic: Mathematics, Application of Derivatives, JEE Main
If $A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right],B=\left[\begin{array}{cc}0 & -i \\ i & 0\end{array}\right]$
then $(A+B)^2$ equals

Options:
A $A^2+B^2$
B $A^2+B^2+2 A B$
C $A^2+B^2+A B-B A$
D None of these

The correct answer is:
$A^2+B^2$

Solution:
$A B=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{cc}0 & -i \\ i & 0\end{array}\right]$ $=\left[\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right]$ and
$B A=\left[\begin{array}{cc}0 & -i \\ i & 0\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$$=\left[\begin{array}{cc}-i & 0 \\ 0 & i\end{array}\right]=-A B$
$\therefore A B+B A=0$
Hence, $(A+B)^2=A^2+B^2$

Topic: Mathematics, Matrices, JEE Main
If A=01 -10 , then which one of the following statements is not correct?

Options:
A A3+I=AA3- I
B A4-I=A2+I
C A2+I=AA2-I
D A3-I=A(A-I)

The correct answer is:
A2+I=AA2-I

Solution:
A =0-110
A2= 0-1100-110= -100-1
A3=-100-10-110=01-10
A4=01-100-110=1001
Option 1 A3+I= A A3-I
01 -10 -1-1 1-1= 1-1 11
1-1 11 = 1-1 11 .....Correct
Option 2 A4-I=A2+I
0000= 0000 ....Correct
Option 3 0000= 0-110× -200-2 = 02-20 .....Incorrect
Option 4
A3-I=A(A-I)
-11-1-1= 0-110 × -1-11-1 = -11-1-1
......Correct.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2015 (10 Apr Online)
If $A=\left[\begin{array}{ccc}0 & 1 & -2 \\ -1 & 0 & 5 \\ 2 & -5 & 0\end{array}\right]$, then

Options:
A $A^{\prime}=A$
B $A^{\prime}=-A$
C $A^{\prime}=2 A$
D None of these

The correct answer is:
$A^{\prime}=-A$

Solution:
$A^{\prime}=\left[\begin{array}{ccc}0 & -1 & 2 \\ 1 & 0 & -5 \\ -2 & 5 & 0\end{array}\right]=-A$

Topic: Mathematics, Matrices, JEE Main
If a=∫01cossin⁡xsec⁡xdx, then the value of a2+cos2sin⁡1 is equal to

Options:
A 0
B 1
C sin1
D sinsin⁡1

The correct answer is:
1

Solution:
a=∫01cos⁡sin⁡xcos⁡xdx

Let sin⁡x=t
⇒cos⁡xdx=dt
∴a=∫0sin⁡1cos⁡tdt=(sin⁡t)0sin⁡1
=sinsin⁡1-sin⁡0
=sinsin⁡1
∴a2+cos2sin⁡1=sin2⁡θ+cos2⁡θ=1; θ=sin⁡1

Topic: Mathematics, Definite Integration, JEE Main
If A=â«0Ï2sin3â¡x1+cos2â¡xdx and B=â«0Ï2cos3â¡x1+sin2â¡xdx, then 2AB is equal to

Options:
A -2
B 4
C 2
D -4

The correct answer is:
2

Solution:
In A, let cos⁡x=t
⇒-sin⁡xdx=dt
∴A=-∫101-t21+t2dt
=∫011-t21+t2dt
In B, let sin⁡x=u
cos⁡xdx=du
∴B=∫011-u21+u2du=A
∴2AB=2

Topic: Mathematics, Definite Integration, JEE Main
If \(a > 0, a \in R, z=a+2 i\) and \(z|z|-a z+1=0\) then

Options:
A \(z\) is always a positive real number
B \(z\) is always a negative real number
C \(z\) is purely imaginary number
D such a complex z does not exist

The correct answer is:
such a complex z does not exist

Solution:
Putting \(z=a+2 i\) in the given equation and comparing imaginary parts, we get \(\mathrm{a}^2+4=\mathrm{a}^2\), which is not possible.

Topic: Mathematics, Complex Number, JEE Main
If $a \neq 0$ and the line $2 b x+3 c y+4 d=0$ passes through the points of intersection of the parabolas $y^2=4 a x$ and $x^2=4 a y$, then

Options:
A
$d^2+(2 b+3 c)^2=0$

B
$d^2+(3 b+2 c)^2=0$

C
$d^2+(2 b-3 c)^2=0$

D
$d^2+(3 b-2 c)^2=0$

The correct answer is:
$d^2+(2 b+3 c)^2=0$

Solution:
Points of intersection of given parabolas are $(0,0)$ and $(4 a, 4 a)$ $\Rightarrow$ equation of line passing through these points is $y=x$ On comparing this line with the given line $2 b x+3 c y+4 d=0$, we get $d=0$ and $2 b+3 c=0 \Rightarrow(2 b+3 c)^2+d^2=0$

Topic: Mathematics, Parabola, JEE Main, JEE Main 2004
If a>0 and z=(1+i)2a-i , has magnitude 25 , then z- is equal to:

Options:
A -15-35i
B -35-15i
C 15-35i
D -15+35i

The correct answer is:
-15-35i

Solution:
Z=1+i2a-i=2ia-i
z=25⇒2ia-i=2a2+1=25
⇒a=3 ∵a>0
So, z-=2i3-i-= -2i3+i=-2i3+i×3-i3-i
=-6i-232+1=-15-35i

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2019 (10 Apr Shift 1)
If a>0, b>0 and a2+b=2, then the maximum value of the term independent of x in the expansion of ax16+bx-139 is

Options:
A 48
B 84
C 42
D 168

The correct answer is:
84

Solution:
Let k+1th term be independent of x
Tk+1=9Ckax169-kbx-13k
=9Ck a9-k bk x9-k6-k3
For this to be independent of x,
9-k6-k3=0⇒96=k6+k3=k2⇒k=3.
⇒ Tk+1=9C3a6b3=84a2b6
Using A.M.≥G.M. we get a2+b2≥a2b
⇒ a2b≤1
⇒ Tk+1≤84

Topic: Mathematics, Binomial Theorem, JEE Main
If A>0, B>0 and A+B=π6, then the minimum positive value of tan⁡A+tan⁡B is :

Options:
A 3-2
B 4-23
C 23
D 2-3

The correct answer is:
4-23

Solution:
tan⁡A+B=tan⁡A+tan⁡B1-tan⁡Atan⁡B
⇒ 13=y1-tan⁡A tan⁡B where y=tan⁡A+tan⁡B
⇒ tan⁡A.tan⁡B=1-3y
and, also AM≥GM
⇒ tan⁡A+tan⁡B2 ≥ tan⁡Atan⁡B
⇒ y ≥21- 3y
⇒ y2 ≥4-43y≥0
⇒ y ≤ -23-4 or y ≥ -23+4
y ≤ -23-4, is not possible as tan⁡A+tan⁡B>0

Therefore, y ≥ -23+4

Hence, minimum positive value is 4-23.

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2016 (10 Apr Online)
If $a>0, b>0, c>0$ and $a, b, c$ are distinct, then $(a+$b) $(b+c)(c+a)$ is greater than

Options:
A $2(a+b+c)$
B $3(a+b+c)$
C $6 \mathrm{abc}$
D $8 \mathrm{abc}$

The correct answer is:
$8 \mathrm{abc}$

Solution:
$\mathrm{AM}>\mathrm{GM}$

$\Rightarrow \frac{\mathrm{a}+\mathrm{b}}{2}>\sqrt{\mathrm{ab}}, \frac{\mathrm{b}+\mathrm{c}}{2}>\sqrt{\mathrm{bc}}, \frac{\mathrm{c}+\mathrm{a}}{2}>\sqrt{\mathrm{ca}}$

So, $(a+b)(b+c)(c+a)>8 a b c$

Topic: Mathematics, Sequences and Series, JEE Main
If $a>0$ discriminant of $a x^2+2 b x+c$ is -ve, then
$\left|\begin{array}{ccc}a & b & a x+b \\ b & c & b x+c \\ a x+b & b x+c & 0\end{array}\right|$ is

Options:
A
+ve

B
$\left(a c-b^2\right)\left(a x^2+2 b x+c\right)$

C
-ve

D
0

The correct answer is:
-ve

Solution:

Topic: Mathematics, Determinants, JEE Main, JEE Main 2002
If $a>0, \lim _{x \rightarrow a} \frac{a^x-x^a}{x^x-a^a}=-1$, then $a$ is equal to

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

The correct answer is:
1

Solution:
We have,
$\lim _{x \rightarrow a} \frac{a^x-x^a}{x^x-a^a}=-1$
Using $L^{\prime}$ hospital's rule
$\Rightarrow \quad \lim _{x \rightarrow a} \frac{a^x \log a-a x^{a-1}}{x^x(1+\log x)}=-1$
$\Rightarrow \quad \frac{a^a \log a-a \cdot a^{a-1}}{a^a(1+\log a)}=-1$
$\Rightarrow \quad \log a-1=-1-\log a$
$\therefore \quad 2 \log _e a=0 \Rightarrow a=e^0=1$

Topic: Mathematics, Limits, JEE Main
If A=0sinαsinα0 and detA2-12I=0, then a possible value of α is

Options:
A π2
B π3
C π4
D π6

The correct answer is:
π4

Solution:
A2=sin2αI

So, A2-I2=sin2α-122=0

⇒sinα=12

⇒α=π4

Topic: Mathematics, Matrices, JEE Main, JEE Main 2021 (17 Mar Shift 1)
If A=0-tanθ2tanθ20 and I2+AI2-A-1=a-bba, then 13a2+b2 is equal to _____ .

Options:

The correct answer is:
13

Solution:
A=0-tanθ2tanθ20

⇒I+A=1-tanθ2tanθ21

⇒I-A=1tanθ2-tanθ21 ∴ |I-A|=sec2θ/2

⇒I-A-1=1sec2θ21-tanθ2tanθ21

⇒I+AI-A-1=1sec2θ21-tanθ2tanθ2 1 1-tanθ2tanθ2 1

=1sec2θ21-tan2θ2-2 tanθ22 tanθ21-tan2θ2

a=1-tan2θ2sec2θ2

b=2 tanθ2sec2θ2

∴ a2+b2=1

∴ 13a2+b2=13

Topic: Mathematics, Matrices, JEE Main
If A=0-tanθ2tanθ20 and I2+AI2-A-1=a-bba, then 13a2+b2 is equal to _____ .

Options:

The correct answer is:
13

Solution:
A=0-tanθ2tanθ20

⇒I+A=1-tanθ2tanθ21

⇒I-A=1tanθ2-tanθ21 ∴ |I-A|=sec2θ/2

⇒I-A-1=1sec2θ21-tanθ2tanθ21

⇒I+AI-A-1=1sec2θ21-tanθ2tanθ2 1 1-tanθ2tanθ2 1

=1sec2θ21-tan2θ2-2 tanθ22 tanθ21-tan2θ2

a=1-tan2θ2sec2θ2

b=2 tanθ2sec2θ2

∴ a2+b2=1

∴ 13a2+b2=13

Topic: Mathematics, Matrices, JEE Main, JEE Main 2021 (25 Feb Shift 1)
If A=1000110-24, I=100010001 and A-1=16A2+cA+d then the sum of values of c and d is

Options:

The correct answer is:
5

Solution:
Method I
We evaluate A2 and A3 and write the given equation as AAâ1=I=16A3+cA2+dA .
Comparing the corresponding elements on both the sides, we get c =-6, d=11.
Method II
Characteristic equation for A is A-xI=0=1-x0001-x10-24-x=0
â1-x1-x4-x+2=0
Hence characteristic equation is x3-6x2+11x-6=0
Then by Caley Hamilton theorem
A3-6A2+11A-6I=0
multiply by A-1 both the sides,
we get 16A2-6A+11I=A-1 ...(i)
given A-1=16A2+cA+d ...(ii)
then from equation (i) and (ii)
we get c=-6d=11
then c+d=5

Topic: Mathematics, Matrices, JEE Main
If $A=\left(\begin{array}{c}\alpha-1 \\ 0 \\ 0\end{array}\right), B=\left(\begin{array}{c}\alpha+1 \\ 0 \\ 0\end{array}\right)$ be two matrices, then $A B^T$ is a non-zero matrix for $|\alpha|$ not equal to

Options:
A
2

B
0

C
1

D
3

The correct answer is:
1

Solution:
Let $A=\left(\begin{array}{c}\alpha-1 \\ 0 \\ 0\end{array}\right), B=\left(\begin{array}{c}\alpha+1 \\ 0 \\ 0\end{array}\right)$ be two matrices.
$$
\begin{aligned}
& A B^T=\left(\begin{array}{c}
\alpha-1 \\
0 \\
0
\end{array}\right)(\alpha+1 \quad 0 \quad 0) \\
& =\left(\begin{array}{ccc}
\alpha^2-1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right) \\
&
\end{aligned}
$$

Topic: Mathematics, Matrices, JEE Main, JEE Main 2012 (07 May Online)
If a1(>0),a2,a3,a4,a5 are in a G.P. , a2+a4=2a3+1 and 3a2+a3=2a4, then a2+a4+2a5 is equal to _____.

Options:

The correct answer is:
40

Solution:
Given, a1>0,a2,a3,a4,a5→G.P.

3a2+a3=2a4

Using the formula an=arn-1 and assuming a1=a we get, 3ar+ar2=2ar3

⇒3+r=2r2

⇒2r2-r-3=0

⇒r=-1 & r=32

Also, a2+a4=2a3+1

⇒ar+ar3=2ar2+1

⇒ar+r3-2r2=1

⇒a32+278-184=1

⇒a=83

When r=-1,a=-14 (rejected, a1>0)

So, at r=23, a=83

Now a2+a4+2a5

=83×32+83×278+2×83×8116

=4+9+27=40

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2022 (26 Jun Shift 2)
If $A=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]$ and $I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$, then which one of the following holds for all $n \geq 1$, by the principle of mathematical indunction

Options:
A
$A^n=n A-(n-1) I$

B
$A^n=2^{n-1} A-(n-1) I$

C
$A^n=n A+(n-1) I$

D
$A^n=2^{n-1} A+(n-1) I$

The correct answer is:
$A^n=n A-(n-1) I$

Solution:
By the principle of mathematical induction (A) is true.

Topic: Mathematics, Mathematical Induction, JEE Main, JEE Main 2005
If A=111011001 and M=A+A2+A3+…+A20, then the sum of all the elements of the matrix M is equal to _______.

Options:

The correct answer is:
2020

Solution:
We have, A=111011001 and M=A+A2+A3+…+A20

Now, A2=A.A=111011001111011001=123012001

⇒A3=A2.A=111011001123012001=136013001

.

.

.

.

An=1nnn+1201n001

So, required sum

=20×3+2∑rr=120+∑r=120rr+12

=20×3+2∑rr=120+12∑r=120r2+12∑r=120r

=20×3+52∑r=120r+12∑r=120r2

=20×3+5220×212+1220×21×416

∵∑r=1nr=nn+12, ∑r=1nr2=nn+12n+16

=60+55×21+5×7×41

=60+525+1435

=2020

Topic: Mathematics, Matrices, JEE Main, JEE Main 2021 (27 Jul Shift 2)
If $A=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$ then $A^{100}=$

Options:
A $2^{100} \mathrm{~A}$
B $2^{99} \mathrm{~A}$
C $2^{101} \mathrm{~A}$
D None of these

The correct answer is:
$2^{99} \mathrm{~A}$

Solution:
$A=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$
$A^2=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$$=\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]=$ 2 $\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$
$A^3=2\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$ $\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$ $=2^2\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$
$A^n=2^{n-1}\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$ $\Rightarrow A^{100}=2^{99}\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$

Topic: Mathematics, Matrices, JEE Main
If $\mathrm{A}=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$ then $\mathbf{A}^{100}$ :

Options:
A $2^{100} \mathrm{~A}$
B $2^{99} \mathrm{~A}$
C $2^{101} \mathrm{~A}$
D None of the above

The correct answer is:
$2^{99} \mathrm{~A}$

Solution:
Let $A=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$

$$

\begin{array}{l}

\mathrm{A}^{2}=\left[\begin{array}{ll}

1 & 1 \\

1 & 1

\end{array}\right]\left[\begin{array}{ll}

1 & 1 \\

1 & 1

\end{array}\right]=2\left[\begin{array}{ll}

1 & 1 \\

1 & 1

\end{array}\right]=2 \mathrm{~A} \\

\mathrm{~A}^{3}=2^{2}\left[\begin{array}{ll}

1 & 1 \\

1 & 1

\end{array}\right], \mathrm{A}^{4}=2^{3}\left[\begin{array}{ll}

1 & 1 \\

1 & 1

\end{array}\right] \\

\mathrm{A}^{3}=2^{2} \mathrm{~A}, \quad \mathrm{~A}^{4}=2^{3} \mathrm{~A} \\

\therefore \mathrm{A}^{\mathrm{n}}=2^{\mathrm{n}-1}\left[\begin{array}{ll}

1 & 1 \\

1 & 1

\end{array}\right] \\

\Rightarrow \mathrm{A}^{100}=2^{100-1} \mathrm{~A} \therefore \mathrm{A}^{100}=2^{99} \mathrm{~A}

\end{array}

$$

Topic: Mathematics, Matrices, JEE Main
If $\mathbf{a}=(1,-1,1)$ and $\mathbf{c}=(-1,-1,0)$, then the vector $\mathbf{b}$ satisfying $\mathbf{a} \times \mathbf{b}=\mathbf{c}$ and $\mathbf{a} \cdot \mathbf{b}=1$ is

Options:
A $(1,0,0)$
B $(0,0,1)$
C $(0,-1,0)$
D None of these

The correct answer is:
$(0,0,1)$

Solution:
$\begin{aligned}
& \text {Let } \mathbf{b}=b_1 \mathbf{i}+b_2 \mathbf{j}+b_3 \mathbf{k} \\
& \text { But }(\mathbf{i}-\mathbf{j}+\mathbf{k}) .\left(b_1 \mathbf{i}+b_2 \mathbf{j}+b_3 \mathbf{k}\right)=1 \Rightarrow b_1-b_2+b_3=1 \\
& \qquad \mathbf{a} \times \mathbf{b}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & -1 & 1 \\
b_1 & b_2 & b_3
\end{array}\right| \\
& \text { and } \quad \begin{aligned}
& \\
&=-\mathbf{i}\left(b_2+b_3\right)+\mathbf{j}\left(b_1-b_3\right)+\mathbf{k}\left(b_2+b_1\right) \\
& \Rightarrow \mathbf{a} \times \mathbf{b}=\mathbf{c}
\end{aligned}
\end{aligned}$
Comparing the coefficients of $\mathbf{i}, \mathbf{j}$ and $\mathbf{k}$ respectively,
we get $b_2+b_3=1$
$\begin{aligned}
& b_1-b_3=-1 \\
& b_2+b_1=0
\end{aligned}$
..(iv)
By solving the equations (i), (ii), (iii) and (iv), we get
$b_1=0, \quad b_2=0 \text { and } b_3=1$

Topic: Mathematics, Vector Algebra, JEE Main
If A=1121341-13, B=adjA and C=3A, then adjBC is equal to

Options:
A 8
B 16
C 72
D 2

The correct answer is:
8

Solution:
A=1121341-13=9+4-13-4+2-1-3=13+1-8=6
adjB=adjadjA=An-12=A4=362
C=3A=33×6
adjBC=36×3633×6=8

Topic: Mathematics, Matrices, JEE Main, JEE Main 2020 (09 Jan Shift 1)
If A=1121341-13, B=adjA and C=3A, then adjBC is equal to

Options:
A 8
B 16
C 72
D 2

The correct answer is:
8

Solution:
A=1121341-13=9+4-13-4+2-1-3=13+1-8=6
adjB=adjadjA=An-12=A4=362
C=3A=33×6
adjBC=36×3633×6=8

Topic: Mathematics, Determinants, JEE Main
If $\mathbf{a}=(1,-1,2), \mathbf{b}=(-2,3,5), \mathbf{c}=(2,-2,4)$ and $\mathbf{i}$ is the unit vector in the $x$-direction, then $(\mathbf{a}-2 \mathbf{b}+3 \mathbf{c}) . \mathbf{i}=$

Options:
A 11
B 15
C 18
D 36

The correct answer is:
11

Solution:
$\begin{aligned}
& \mathbf{a}=(1,-1,2), \mathbf{b}=(-2,3,5), \mathbf{c}=(2,-2,4) \\
& \text { So, } \mathbf{a}=(1,-1,2)=\mathbf{i}-\mathbf{j}+2 \mathbf{k} ; \mathbf{b}=(-2,3,5)=-2 \mathbf{i}+3 \mathbf{j}+5 \mathbf{k} \\
& \text { and } \mathbf{c}=(2,-2,4)=2 \mathbf{i}-2 \mathbf{j}+4 \mathbf{k} \\
& \Rightarrow \mathbf{a}-2 \mathbf{b}+3 \mathbf{c}=(\mathbf{i}-\mathbf{j}+2 \mathbf{k})-2(-2 \mathbf{i}+3 \mathbf{j}+5 \mathbf{k})
\end{aligned}$
$+3(2 \mathbf{i}-2 \mathbf{j}+4 \mathbf{k})$
$=11 \mathbf{i}-13 \mathbf{j}+4 \mathbf{k}$ and $(\mathbf{a}-2 \mathbf{b}+3 \mathbf{c}) \cdot \mathbf{i}=11$.

Topic: Mathematics, Vector Algebra, JEE Main
If $\mathrm{A}(1,-1,2), \mathrm{B}(5,7,-6), \mathrm{C}(3,4,-10)$ and $\mathrm{D}(-1,-4,-2)$ are the vertices of a quadrilateral $A B C D$, then its area is :

Options:
A $48 \sqrt{7}$
B $12 \sqrt{29}$
C $24 \sqrt{7}$
D $24 \sqrt{29}$

The correct answer is:
$12 \sqrt{29}$

Solution:
$\begin{aligned} & \mathrm{A}(1,-1,2) \\ & \mathrm{B}(5,7,-6) \\ & \mathrm{C}(3,4,-10) \\ & \mathrm{D}(-1,-4,-2) \\ & \text { Area }=\frac{1}{2}|\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}}|=\frac{1}{2}|(2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-12 \hat{\mathrm{k}}) \times(6 \hat{\mathrm{i}}+11 \hat{\mathrm{j}}-4 \hat{\mathrm{k}})| \\ & =\frac{1}{2}|112 \hat{\mathrm{i}}-64 \hat{\mathrm{j}}-8 \hat{\mathrm{k}}| \\ & =4|14 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}-\hat{\mathrm{k}}| \\ & =4 \sqrt{196+64+1} \\ & =4 \sqrt{261} \\ & =12 \sqrt{29}\end{aligned}$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2024 (05 Apr Shift 1)
If A=113526-2-1-3, where Ax=O (where, O is a null matrix and x<15, xâN), then which of the following is true?

Options:
A Greatest value of x is 13
B Sum of the values of x is 102
C Difference between the largest and the smallest value of x is 10
D Number of values of x is 7

The correct answer is:
Sum of the values of x is 102

Solution:
A2=113526-2-1-3 113526-2-1-3

=000339-1-1-3
A3=A2A=113526-2-1-3000339-1-1-3

=0000000-10=O
A4=A5=A6=……A14=O
x∈3,4,5,6,……14

Topic: Mathematics, Matrices, JEE Main
If A=1-2 12 λ-21 3-3 be the adjoint matrix of matrix B such that B=9, then the value of λ is equal to

Options:
A 1
B -774
C 232
D -392

The correct answer is:
-774

Solution:
A=adjB
A=B2
1-212λ-213-3=81
⇒1-3λ+6+2-6+2+16-λ=81
⇒-4λ+4=81
⇒λ=-774

Topic: Mathematics, Determinants, JEE Main
If A=1213-31 then,

Options:
A A30-A25=2I
B A30+A25+A=I
C A30+A25-A=I
D A30=A25

The correct answer is:
A30+A25-A=I

Solution:
Given A=1213-31

A2=1413-3113-31

=141-3 23-23-3+1

A2=12-1 3-3-1

A3=14-1 3-3-1-1 3-3-1

⇒A3=14-4 00-4=-1 001

⇒A3=-I .....(1)

Now from equation 1,

A30=(-I)10=I .....(2)

Again from equation 1,

A25=A38.A=-I8.A

⇒A25=A

⇒A25-A=0 ......(3)

On adding equations 2 and 3 we get,

A30+A25-A=I

Topic: Mathematics, Matrices, JEE Main, JEE Main 2023 (01 Feb Shift 2)
If A=12221-2a2b is a matrix satisfying the equation AAT=9I , where I is 3×3 identity matrix, then the ordered pair a, b is equal to

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

The correct answer is:
-2, -1

Solution:
Given, A A T =9I
12221−2a2b12a2122-2b=9I
⇒90a+4+2b092a+2−2ba+4+2b2a+2−2ba2+4+b2=900090009
Equation a+4+2b=0 ⇒a+2b=−4 ...1
2a+2−2b=0 ⇒2a−2b=−2 ...2
&  a2+4+b2=9⇒a2+b2=5 ...3
On solving, we get a=−2, b=−1

Topic: Mathematics, Matrices, JEE Main, JEE Main 2015 (04 Apr)
If $A=\left[\begin{array}{ccc}1 & 0 & 0 \\ 2 & 1 & 0 \\ -3 & 2 & 1\end{array}\right]$ and $B=\left[\begin{array}{ccc}1 & 0 & 0 \\ -2 & 1 & 0 \\ 7 & -2 & 1\end{array}\right]$ then $A B$ equals

Options:
A
$I$

B
$A$

C
$B$

D
0

The correct answer is:
$I$

Solution:
$$
\begin{aligned}
& A=\left[\begin{array}{ccc}
1 & 0 & 0 \\
2 & 1 & 0 \\
-3 & 2 & 1
\end{array}\right], B=\left[\begin{array}{ccc}
1 & 0 & 0 \\
-2 & 1 & 0 \\
7 & -2 & 1
\end{array}\right] \\
& A B=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=I
\end{aligned}
$$

Topic: Mathematics, Matrices, JEE Main, JEE Main 2012 (26 May Online)
If A=123456 and B=142536, then the determinant value of BA is

Options:
A 8
B 0
C -8
D 24

The correct answer is:
0

Solution:
BA=142536123456=172227222936273645
BA=175522772799 (Using C3âC3-C2 and C2âC2-C1 )
Hence, BA=0

Topic: Mathematics, Determinants, JEE Main
If $A=\{1,2,3,4\}$, then which one of the following is reflexive?

Options:
A $\{(1,1),(2,3),(3,3)\}$
B $\{(1,1),(2,2),(3,3),(4,4)\}$
C $\{(1,2),(2,1)(3,2),(2,3)\}$
D $\{(1,2),(1,3),(1,4)\}$

The correct answer is:
$\{(1,1),(2,2),(3,3),(4,4)\}$

Solution:
Given, $A=[1,2,3,4]$
Let $R$ be a reflexive relation on $A$, then for each $x \in A,(a, a) \in R$
$\therefore$ Option (b) is true.

Topic: Mathematics, Sets and Relations, JEE Main
If $A=\{1,2,3, \ldots, 10\}$, then number of subsets of $A$ containing only odd numbers is

Options:
A 31
B 27
C 32
D 30

The correct answer is:
31

Solution:
Given, $A=\{1,2,3, \ldots, 10\}$
Set $A$ containing only odd number $=\{1,3,5,7,9\}$ Number of subsets $=2^5$
[it contains empty set also]
The required number of subsets of $A$ containing only odd number $=2^5-1$
$$
=32-1=31
$$

Topic: Mathematics, Sets and Relations, JEE Main