datavorous/entrance-exam-dataset huggingface dataset:
If $A=\left[\begin{array}{cccc}1 & 0 & 0 & 0 \\ 2 & 3 & 0 & 0 \\ 4 & 5 & 6 & 0 \\ 7 & 8 & 9 & 10\end{array}\right]$, then $A$ is

Options:
A An upper triangular matrix
B A null matrix
C A lower triangular matrix
D None of these

The correct answer is:
A lower triangular matrix

Solution:
Since a square matrix $A$ whose elements $a_{i j}=0$ for $i \lt j$. Then $A$ is the lower triangular matrix.

Topic: Mathematics, Matrices, JEE Main
If a12+a22+a32=1, b12+b22+b32=4, c12+c22+c32=9, a1b1+a2b2+a3b3=0, a1c1+a2c2+a3c3=0, b1c1+b2c2+b3c3=0 and A=a1a2a3b1b2b3c1c2c3, then A4 is equal to

Options:
A 36
B 49
C 1296
D 216

The correct answer is:
1296

Solution:
A2=a1a2a3b1b2b3c1c2c3a1a2a3b1b2b3c1c2c3
=a12+a22+a32a1b1+a2b2+a3b3a1c1+a2c2+a3c3a1b1+a2b2+a3b3b12+b22+b32b1c1+b2c2+b3c3a1c1+a2c2+a3c3b1c1+b2c2+b3c3c12+c22+c32
=100040009=36⇒A4=1296

Topic: Mathematics, Determinants, JEE Main
If $A=\left[\begin{array}{ccc}1 & 2 & x \\ 3 & -1 & 2\end{array}\right]$ and $B=\left[\begin{array}{c}y \\ x \\ 1\end{array}\right]$ be such that $\mathrm{AB}=\left[\begin{array}{l}6 \\ 8\end{array}\right]$, then:

Options:
A
$y=2 x$

B
$y=-2 x$

C
$y=x$

D
$y=-x$

The correct answer is:
$y=2 x$

Solution:
Let $\mathrm{A}=\left[\begin{array}{ccc}1 & 2 & x \\ 3 & -1 & 2\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{l}y \\ x \\ 1\end{array}\right]$
$$
\begin{aligned}
&\mathrm{AB}=\left[\begin{array}{ccc}
1 & 2 & x \\
3 & -1 & 2
\end{array}\right]\left[\begin{array}{l}
y \\
x \\
1
\end{array}\right] \\
&\Rightarrow\left[\begin{array}{l}
6 \\
8
\end{array}\right]=\left[\begin{array}{l}
y+2 x+x \\
3 y-x+2
\end{array}\right] \\
&\Rightarrow\left[\begin{array}{l}
6 \\
8
\end{array}\right]=\left[\begin{array}{l}
y+3 x \\
3 y-x+2
\end{array}\right] \\
&\Rightarrow y+3 x=6 \text { and } 3 y-x=6
\end{aligned}
$$
On solving, we get
$$
\begin{aligned}
&x=\frac{6}{5} \text { and } y=\frac{12}{5} \\
&\Rightarrow y=2 x
\end{aligned}
$$

Topic: Mathematics, Matrices, JEE Main, JEE Main 2014 (12 Apr Online)
If $A=\left(\begin{array}{lll}1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1\end{array}\right)$ and $B=\left(\begin{array}{ccc}-5 & 7 & 1 \\ 1 & -5 & 7 \\ 7 & 1 & -5\end{array}\right)$ then $A B$ is equal to

Options:
A $I_3$
B $2 I_3$
C $4 I_3$
D $18 I_3$

The correct answer is:
$18 I_3$

Solution:
We have $A=\left[\begin{array}{lll}1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1\end{array}\right]$ and $B=\left[\begin{array}{ccc}-5 & 7 & 1 \\ 1 & -5 & 7 \\ 7 & 1 & -5\end{array}\right]$
$\therefore A B=\left[\begin{array}{lll}1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1\end{array}\right]\left[\begin{array}{ccc}-5 & 7 & 1 \\ 1 & -5 & 7 \\ 7 & 1 & -5\end{array}\right]$
$A B=\left[\begin{array}{ccc}18 & 0 & 0 \\ 0 & 18 & 0 \\ 0 & 0 & 18\end{array}\right]=18\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$A B=18 I_3$

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

Options:
A $\frac{-1}{2}\left[\begin{array}{cc}4 & -2 \\ -3 & 1\end{array}\right]$
B $\frac{1}{2}\left[\begin{array}{cc}4 & -2 \\ -3 & 1\end{array}\right]$
C $\left[\begin{array}{cc}-2 & 4 \\ 1 & 3\end{array}\right]$
D $\left[\begin{array}{ll}2 & 4 \\ 1 & 3\end{array}\right]$

The correct answer is:
$\frac{-1}{2}\left[\begin{array}{cc}4 & -2 \\ -3 & 1\end{array}\right]$

Solution:
If $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$, then
$$
A^{-1}=\frac{1}{|A|}\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right]
$$
So, if $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$, then
$$
\begin{array}{r}
A^{-1}=\frac{1}{4-6}\left[\begin{array}{cc}
4 & -2 \\
-3 & 1
\end{array}\right] \\
=-\frac{1}{2}\left[\begin{array}{cc}
4 & -2 \\
-3 & 1
\end{array}\right]
\end{array}
$$

Topic: Mathematics, Matrices, JEE Main
If A=1-32 k and A2-4A+10I=A, then k is equal to

Options:
A 0
B -4
C 4
D 1 or 4

The correct answer is:
4

Solution:
A2-4A+10I=A
â1 -32 k1 -32 k-41 -32 k+101 00 1=1 -32 k
â â5 â3â3k 2+2k â6+ k 2 âââ 4 â12 8 4k + 10 0 0 10 = 1 â3 2 k
â1 9-3k-6+2k 4+k2-4k=1 -32 k
â9-3k=-3,-6+2k=2 ...(i)
and 4+k2-4k=k
â k 2 â5k+4=0âk=4,â1
But, k=1 does not satisfy the Eq (i).

Topic: Mathematics, Matrices, JEE Main
If A=1,3,5,7,9,11,13,15, B=2,4,â¦,16 and N is the universal set, then A â² âª AâªB â© B â² is (where, N is the set of natural numbers)

Options:
A A
B N
C B
D None of these

The correct answer is:
N

Solution:
We know that,
A∪B ∩ B ′ =A (as, A∩B=ϕ)
So, A∪B ∩ B ′ ∪ A ′ =A∪ A ′ =N

Topic: Mathematics, Sets and Relations, JEE Main
If $\mathrm{A}=\left[\begin{array}{rr}1 & 2 \\ -4 & -1\end{array}\right]$ then $\mathrm{A}^{-1}$ is

Options:
A $\frac{1}{7}\left[\begin{array}{rr}-1 & -2 \\ 4 & 1\end{array}\right]$
B $\frac{1}{7}\left[\begin{array}{rr}1 & 2 \\ -4 & -1\end{array}\right]$
C $\frac{1}{7}\left[\begin{array}{rr}-1 & -2 \\ 4 & -1\end{array}\right]$
D Does not exist

The correct answer is:
$\frac{1}{7}\left[\begin{array}{rr}-1 & -2 \\ 4 & 1\end{array}\right]$

Solution:
Hints: $|\mathrm{A}|=-1+8=7$
$$
\begin{aligned}
& \operatorname{adj}(\mathrm{A})=\left[\begin{array}{rr}
+(-1) & -(2) \\
-(-4) & +(1)
\end{array}\right]=\left[\begin{array}{rr}
-1 & -2 \\
4 & 1
\end{array}\right] \\
& \mathrm{A}^{-1}=\frac{1}{7}\left[\begin{array}{rr}
-1 & -2 \\
4 & 1
\end{array}\right] \quad \operatorname{Both}(\mathrm{A} \text { and } C)
\end{aligned}
$$

Topic: Mathematics, Matrices, JEE Main
If A=15!6!7!5!6!7!6!7!8!7!8!9!, then adjadj2A is equal to

Options:
A 220
B 28
C 212
D 216

The correct answer is:
216

Solution:
Given,

A=15!6!7!5!6!7!6!7!8!7!8!9!

⇒A=15!6!7!5!6!7!6!7!8!7!8!9!

⇒A=15!6!7!5!×6!×7!164217561872

⇒|A|=2

Now using the property of adjoint of matrix we get,

adjadjA=A(n-1)2 & kA=knA, where n is order of square matrix,

Now using the above two formula we get,

adjadj2A=2A4=212A4

⇒adjadj2A=212·24

⇒adjadj2A=216

Topic: Mathematics, Matrices, JEE Main, JEE Main 2023 (10 Apr Shift 2)
If A=15λ10, A-1=αA+βI and α+β=-2, then 4α2+β2+λ2 is equal to :

Options:
A 12
B 19
C 14
D 10

The correct answer is:
14

Solution:
Given,

A=15λ10, A-1=αA+βI and α+β=-2,

Now solving by using characteristic equation, we get
|A-k|=0⇒1-k5λ10-k=0

⇒k2-11k+10-5λ=0

⇒A2-11A+(10-5λ)I=0 {by putting k=A}

⇒ A-1=110-5λ(-A+11I)

Now on comparing with A-1=αA+βI we get,

⇒α=-110-5λ, β=1110-5λ

And given,

α+β=-2⇒λ=3, α=15, β=-115

Hence, the value of 4α2+β2+λ2=14

Topic: Mathematics, Matrices, JEE Main, JEE Main 2023 (08 Apr Shift 2)
If a1,a2,a3,a4,a5 are consecutive terms of an arithmetic progression with common difference 3, then the value of a32a2a1a42a3a2a52a4a3 is

Options:
A 0
B 27
C 81
D 162

The correct answer is:
162

Solution:
Apply R3↔R3-R2, R2↔R2-R1
a32a2a1a42-a32a3-a2a2-a1a52-a42a4-a3a3-a2
=a32a2a13a3+a4333a4+a533
Applying C3↔C3-C2
a32a2-33a3+a4303a4+a530
=-39a3+a4-9a4+a5
=-27a3-a5=-27×-6=162

Topic: Mathematics, Determinants, JEE Main
If A1,A2,A3â¦â¦.A20 are 20 skew-symmetric matrices of same order and B=âr=1202rAr2r+1, then the sum of the principal diagonal elements of matrix B is equal to

Options:

The correct answer is:
0

Solution:
A1T=-A1, A2T=-A2, ….., A20T=-A20

B=∑r=1202rAr2r+1
BT=∑r=1202rAr2r+1T=∑r=1202rArT2r+1
=∑r=1202r-Ar2r+1=-∑r=1202rAr2r+1

=-B
⇒B is skew-symmetric

Hence, the sum of principal diagonal elements of B=0

Topic: Mathematics, Matrices, JEE Main
If a1,a2,a3.........,an are in A.P. and a1+ a4+a7.........+a16=114 , then a1+a6+a11+a16 is equal to :

Options:
A 64
B 98
C 38
D 76

The correct answer is:
76

Solution:
∴ We know that sum of equidistant term from end & beginning is same in any A.P.⇒ a1+a16=a2+a15=a3+a14=a4+a13=a5+a12=…=a7+a10

∵ a1+a4+a7+a10+a13+a16=114

⇒3a1+a16=114⇒a1+a16=38

Then,

a1+a6+a11+a16

=2a1+a16=76

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2019 (10 Apr Shift 1)
If $a_1, a_2, a_3, \ldots, a_n, \ldots$. are in A.P. such that $a_4-a_7$ $+a_{10}=m$, then the sum of first 13 terms of this A.P., is :

Options:
A
$10 \mathrm{~m}$

B
$12 \mathrm{~m}$

C
$13 \mathrm{~m}$

D
$15 \mathrm{~m}$

The correct answer is:
$13 \mathrm{~m}$

Solution:
If $d$ be the common difference, then
$$
\begin{aligned}
m & =a_4-a_7+a_{10}=a_4-a_7+a_7+3 \mathrm{~d}=a_7 \\
\mathrm{~S}_{13} & =\frac{13}{2}\left[a_1+a_{13}\right]=\frac{13}{2}\left[a_1+a_7+6 d\right] \\
& =\frac{13}{2}\left[2 a_7\right]=13 a_7=13 \mathrm{~m}
\end{aligned}
$$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2013 (23 Apr Online)
If $a_1, a_2, a_3, \ldots, a_n, \ldots$ are in G.P., then the determinant $\Delta=\left|\begin{array}{lll}\log a_n & \log a_{n+1} & \log a_{n+2} \\ \log a_{n+3} & \log a_{n+4} & \log a_{n+5} \\ \log a_{n+6} & \log a_{n+7} & \log a_{n+8}\end{array}\right|$ is equal to

Options:
A
1

B
0

C
4

D
2

The correct answer is:
0

Solution:
$\mathrm{C}_1-\mathrm{C}_2, \mathrm{C}_2-\mathrm{C}_3$
two rows becomes identical
Answer: 0.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2005
If $a_1, a_2, a_3, \ldots, a_n, \ldots$. are in G.P., then the value of the determinant
$$
\left|\begin{array}{ccc}
\log a_n & \log a_{n+1} & \log a_{n+2} \\
\log a_{n+3} & \log a_{n+4} & \log a_{n+5} \\
\log a_{n+6} & \log a_{n+7} & \log a_{n+8}
\end{array}\right| \text {, is }
$$

Options:
A
0

B
$-2$

C
2

D
1

The correct answer is:
0

Solution:
$$
\begin{aligned}
& \left|\begin{array}{lll}
\log a_n & \log a_{n+1} & \log a_{n+2} \\
\log a_{n+3} & \log a_{n+4} & \log a_{n+5} \\
\log a_{n+6} & \log a_{n+7} & \log a_{n+8}
\end{array}\right| \\
& C_3 \rightarrow C_3-C_2, C_2 \rightarrow C_3-C_1 \\
& =\left|\begin{array}{lll}
\log a_n & \log r & \log r \\
\log a_{n+3} & \log r & \log r \\
\log a_{n+6} & \log r & \log r
\end{array}\right|=0
\end{aligned}
$$
(where $r$ is a common ratio).

Topic: Mathematics, Determinants, JEE Main, JEE Main 2004
If a1,a2,a3… and b1,b2,b3…. are A.P. and a1=2,a10=3,a1b1=1=a10b10 then a4b4 is equal to

Options:
A 2827
B 2824
C 2326
D 2223

The correct answer is:
2827

Solution:
Given, a1,a2,a3......& b1,b2,b3.....are in A.P

Also given, a1=2, a10=3

and a1b1=1=a10b10

So b1=1a1=12 and b10=1a10=13

Now using nth term formula we get, a10=a1+9da⇒3=2+9da

⇒da=19

Using same formula we get,

b10=b1+9db ⇒13=12+9db ⇒db=-154

Now calculating a4b4=a1+3dab1+3db

=2+3×1912+3×-154

= 73×818=5654=2827

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2022 (27 Jun Shift 2)
If a1â,a2â,a3â and b1â,b2â,b3â be two sets of non-coplanar vectors, such that apââbqâ=0if pâ q4if p=q for p=1,2,3 and q=1,2,3, then the value of a1â2a2â3a3âb1â+b2âb2â+b3âb3â+b1â is equal to

Options:

The correct answer is:
768

Solution:
a1→2a2→3a3→b1→+b2→b2→+b3→b3→+b1→
=6a1→a2→a3→2b1→b2→b3→
=12a1→⋅b1→a1→⋅b2→a1→⋅b3→a2→⋅b1→a2→⋅b2→a2→⋅b3→a3→⋅b1→a3→⋅b2→a3→⋅b3→
=12400040004=12×64=768

Topic: Mathematics, Vector Algebra, JEE Main
If a1,a2,a3,.... are in A.P. such that a1+a7+a16=40, then the sum of the first 15 terms of this A.P is:

Options:
A 280
B 120
C 150
D 200

The correct answer is:
200

Solution:
Let the 1st terms of A.P. be 'a' and common difference 'd' .
Given  a1+a7+a16=40
⇒a+a+6d+a+15d=40
⇒a+7d=403 ….(1)
Sum of first 15 term
S15=152[2a+14d]
=152×2×403=200

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2019 (12 Apr Shift 2)
If a1, a2, a3 are in arithmetic progression and d is the common difference, then tan-1â¡ d1+a1a2+tan-1â¡ d1+a2a3=

Options:
A tan-1â¡ 2d1+a1a3
B tan-1â¡Â d1+a1a3
C tan-1â¡Â 2d1+a2a3
D tan-1â¡Â 2d1-a1a3

The correct answer is:
tan-1â¡ 2d1+a1a3

Solution:
Since, a1, a2, a3 are in A.P.
⇒ a2-a1=d=a3-a2
Now, tan-1⁡d1+a1a2+tan-1⁡d1+a2a3
=tan-1⁡a2-a11+a1a2+tan-1⁡a3-a21+a2a3
∵tan−1x−tan−1y=tan−1x−y1+xy
=tan-1⁡a2-tan-1⁡a1+tan-1⁡a3-tan-1⁡a2
=tan-1⁡a3-tan-1⁡a1
=tan-1⁡a3-a11+a1a3
=tan-1⁡a3-a2+a2-a11+a1a3
=tan-1⁡2d1+a1a3

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $a_1, a_2, \ldots, a_n$ are in H.P., then the expression $a_1 a_2+a_2 a_3+\ldots+a_{n-1} a_n$ is equal to

Options:
A
$n\left(a_1-a_n\right)$

B
$(n-1)\left(a_1-a_n\right)$

C
$n a_1 a_n$

D
$(n-1) a_1 a_n$

The correct answer is:
$(n-1) a_1 a_n$

Solution:

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2006
If $a_1, a_2, \ldots . a_n$ are positive real numbers whose product is a fixed number $c$, then the minimum value of $a_1+a_2+\ldots+a_{n-1}+2 a_n$ is

Options:
A $n(2 c)^{1 / n}$
B $(n+1) c^{1 / n}$
C $2 n c^{1 / n}$
D $(n+1)(2 c)^{1 / n}$

The correct answer is:
$n(2 c)^{1 / n}$

Solution:
Given that $\mathrm{a}_{1,} \mathrm{a}_2 \ldots, \mathrm{a}_n$ be positive real numbers such that $\mathrm{a}_1 \cdot \mathbf{a}_2 \cdot \ldots \mathrm{a}_{\mathrm{n}}=\mathrm{c} .$. (1)
We know that, A.M z G.M
$\Rightarrow \frac{a_1+a_2+\ldots+a_{n-1}+2 a_n}{n} \geq\left(a_1 a_2 \ldots a_{n-1}\left(2 a_n\right)\right)^{\frac{1}{n}}$
$\Rightarrow a_1+a_2+\ldots+a_{n-1}+2 a_n \geq n(2 c)^{\frac{1}{n}} \ldots($ from(1) $)$
So, the minimum value of ${ }_{a_1+a_2+\ldots+a_{n-1}+2 a_n}$ is
$\mathrm{n}(2 \mathrm{c})^{\frac{1}{\mathrm{n}}}$.

Topic: Mathematics, Sequences and Series, JEE Main
If a1, a2, b1 and b2 take values in the set 1,-1,0, then the probability that the equation a1a2=b1b2 is satisfied is pq, ( p & q are co-prime) then, q-2p is

Options:

The correct answer is:
5

Solution:
Total ways is 34
Now, a1a2=b1b2
If,
Case 1a1a2=b1b2=0
If at least one of a1 or a2 and b1 or b2 is “0” i.e. 32-22 ways
Case 2 a1a2=b1b2=+1
Both a1,a2 is +1
or –1
i.e. 22 ways
Case 3 a1a2=b1b2=-1
22 ways
Hence, required probability
=32-22+4+434 =3381=1127

Topic: Mathematics, Probability, JEE Main
If $A_1, A_2$ be two arithmetic means between $\frac{1}{3}$ and $\frac{1}{24}$, then their values are

Options:
A $\frac{7}{72}, \frac{5}{36}$
B $\frac{17}{72}, \frac{5}{36}$
C $\frac{7}{36}, \frac{5}{72}$
D $\frac{5}{72}, \frac{17}{72}$

The correct answer is:
$\frac{17}{72}, \frac{5}{36}$

Solution:
Here $\frac{1}{3}, A_1, A_2, \frac{1}{24}$ will be in A.P.,
then $A_1-\frac{1}{3}=\frac{1}{24}-A_2 \Rightarrow A_1+A_2=\frac{3}{8}$\ldots(i)
Now, $A_1$ is a arithmetic mean of $\frac{1}{3}$ and $A_2$, we have
$2 A_1=\frac{1}{3}+A_2 \Rightarrow 2 A_1-A_2=\frac{1}{3}$\ldots(ii)
From (i) and (ii), we get,$A_1=\frac{17}{72}$ and $A_2=\frac{5}{36}$
Aliter: As we have formula $A_m=a+\frac{m(b-a)}{n+1}$
where, $n=2, a=\frac{1}{3}, b=\frac{1}{24}$
$\therefore \quad A_1=\frac{1}{3}+\frac{-7 / 24}{3}=\frac{17}{72}$
$A_2=\frac{1}{3}+\frac{-14 / 24}{3}=\frac{10}{72}=\frac{5}{36}$

Topic: Mathematics, Sequences and Series, JEE Main
If $A_1, A_2 ; G_1, G_2$ and $H_1, H_2$ be $A M^{\prime} s, G M^{\prime} s$ and $H M^{\prime} s$ between two quantities, then the value of $\frac{G_1 G_2}{H_1 H_2}$

Options:
A $\frac{A_1+A_2}{H_1+H_2}$
B $\frac{A_1-A_2}{H_1+H_2}$
C $\frac{A_1+A_2}{H_1-H_2}$
D $\frac{A_1-A_2}{H_1-H_2}$

The correct answer is:
$\frac{A_1+A_2}{H_1+H_2}$

Solution:
The correct option is $\mathbf{1} \frac{A_1+A_2}{H_1+H_2}$
Let the two quantities be $a$ and $b$. Then $a, A_1, A_2, b$ are in A.P.
$\therefore A_1-a=b-A_2 \Rightarrow A_1+A_2=a+b \ldots$ (i)
Again $a, \mathrm{G}_1, \mathrm{G}_2, \mathrm{~b}$ are in G.P.
$\therefore \frac{\mathrm{G}_1}{\mathrm{a}}=\frac{\mathrm{b}}{\mathrm{G}_2} \Rightarrow \mathrm{G}_1 \mathrm{G}_2=\mathrm{ab} \ldots$ (ii)
Also $\mathrm{a}, \mathrm{H}_1, \mathrm{H}_2, \mathrm{~b}$ are in H.P.
$\therefore \frac{1}{\mathrm{H}_1}-\frac{1}{\mathrm{a}}=\frac{1}{\mathrm{~b}}-\frac{1}{\mathrm{H}_2} \Rightarrow$ $\frac{1}{\mathrm{H}_1}+\frac{1}{\mathrm{H}_2}=\frac{1}{\mathrm{a}}+\frac{1}{b}$
$\Rightarrow \frac{H_1+H_2}{H_1 H_2}=\frac{a+b}{a b}= \frac{A_1+A_2}{G_1 G_2}$ [By (i) and (ii)]
$\therefore \frac{\mathrm{G}_1 \mathrm{G}_2}{\mathrm{H}_1 \mathrm{H}_2}=\frac{\mathrm{A}_1+\mathrm{A}_2}{\mathrm{H}_1+\mathrm{H}_2}$.

Topic: Mathematics, Sequences and Series, JEE Main
If a1+a5+a10+a15+a20+a24=225, then the sum of the first 24 terms of the arithmetic progression a1, a2, a3..... is equal to

Options:
A 450
B 675
C 900
D 1200

The correct answer is:
900

Solution:
S=a1+a2+a3+....+a23+a24
S=242a1+a24=12a1+a24
Given that, a1+a5+a10+a15+a20+a24=225
⇒a1+a1+4d+a1+9d+a24-9d+a24-4d+a24=225
⇒3a1+a24=225
⇒a1+a24=75
⇒S=12a1+a24=12×75=900

Topic: Mathematics, Sequences and Series, JEE Main
If a<1 and b<1, then the sum of the series aa+b+a2a2+b2+a3a3+b3+… is

Options:
A a1-a+ab1-ab
B a21-a2+ab1-ab
C b1-b+a1-a
D b21-b2+ab1-ab

The correct answer is:
a21-a2+ab1-ab

Solution:
We have, a<1, b<1
∴ ab=ab<1
Now, aa+b+a2a2+b2+a3a3+b3+…
= a2+a4+a6+…+ab+ab2+ab3+…
= a21-a2+ab1-ab

Topic: Mathematics, Sequences and Series, JEE Main
If $a^{-1}+b^{-1}+c^{-1}=0$ such that $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=\lambda$, then the value of $\lambda$ is

Options:
A 0
B $a b c$
C $- a b c$
D None of these

The correct answer is:
$a b c$

Solution:
$\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=\lambda$
Applying $C_2 \rightarrow C_2-C_1$ and $C_3 \rightarrow C_3-C_1$,
$\left|\begin{array}{ccc}1+a & -a & -a \\ 1 & b & 0 \\ 1 & 0 & c\end{array}\right|$
On expanding w.r.t. $R_3$
Given, $a b+b c+c a+a b c=\lambda$ ....(i)
$\begin{aligned} & \text { Given, } a^{-1}+b^{-1}+c^{-1}=0 \\ & \Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0 \Rightarrow a b+b c+c a=0\end{aligned}$
$\Rightarrow \lambda=a b c, \quad$ (From equation (i)).

Topic: Mathematics, Determinants, JEE Main
If a1b1c1a2b2c2a3b3c3=5, then the value of b2c3-b3c2a3c2-a2c3a2b3-a3b2b3c1-b1c3a1c3-a3c1a3b1-a1b3b1c2-b2c1a2c1-a1c2a1b2-a2b1 is

Options:
A 5
B 25
C 125
D 0

The correct answer is:
25

Solution:
Let ∆=a1b1c1a2b2c2a3b3c3=5 and ∆'=b2c3-b3c2a3c2-a2c3a2b3-a3b2b3c1-b1c3a1c3-a3c1a3b1-a1b3b1c2-b2c1a2c1-a1c2a1b2-a2b1
here ∆' is co-factor determinant of ∆
hence Δ'=Δn-1⇒ Δ3-1= Δ2
= 52= 25

Topic: Mathematics, Determinants, JEE Main
If $a \neq 1, b \neq-1, c \neq-1$ and the system of equations, $x=a(y+z), y=b(z+x), z=c(x+y)$ has a non-trivial solution, then.

Options:
A $\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}=0$
B $\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}=1$
C $\frac{a b c}{(a+1)(b+1)(c+1)}=1$
D $\frac{a+b+c}{(a+1)(b+1)(c+1)}=2$

The correct answer is:
$\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}=1$

Solution:
For system of homogeneous equation, if it has non-trivial solution, then $\Delta=0$, so
$$
\left|\begin{array}{ccc}
1 & -a & -a \\
-b & 1 & -b \\
-c & -c & 1
\end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc}
-\frac{1}{a} & 1 & 1 \\
1 & -\frac{1}{b} & 1 \\
1 & 1 & -\frac{1}{c}
\end{array}\right|=0
$$
$$
\begin{aligned}
& R_2 \rightarrow R_2-R_1 \text { and } R_3 \rightarrow R_3-R_1 \\
& \Rightarrow\left|\begin{array}{ccc}
-\frac{1}{a} & 1 & 1 \\
1+\frac{1}{a} & -\frac{1}{b}-1 & 0 \\
1+\frac{1}{a} & 0 & -\frac{1}{c}-1
\end{array}\right|=0 \\
& \Rightarrow-\frac{1}{a}\left(\frac{1}{b}+1\right)\left(\frac{1}{c}+1\right)+\left(1+\frac{1}{a}\right)\left(1+\frac{1}{c}\right) \\
& \left.\Rightarrow \frac{(1+a)(1+c)}{a c}+\frac{(1+a)(1+b)}{a}+1\right)\left(1+\frac{1}{a}\right)=0 \\
& \Rightarrow \quad \frac{(1+b)(1+c)}{a b c} \\
& \Rightarrow \quad \frac{a}{1+b}+\frac{c}{1+c}=\frac{1}{1+a}-1+1 \\
& \Rightarrow \frac{b}{1+a}+\frac{c}{1+c}=1
\end{aligned}
$$

Topic: Mathematics, Determinants, JEE Main
If a+α=1,b+β=2 and afx+αf1x=bx+βx,x≠0, then the value of the expression fx+f1xx+1x is ___________.

Options:

The correct answer is:
2

Solution:
afx+αf1x=bx+βx …1

replace x by 1x

af1x+αfx=bx+βx …2

1+2

a+αfx+a+αf1x=xb+β+b+β1x

fx+f1xx+1x=b+βa+α=21=2

Topic: Mathematics, Functions, JEE Main, JEE Main 2021 (24 Feb Shift 2)
If A=1sinθ1-sinθ1sinθ-1-sinθ1, then for all θ∈3π4,5π4, detA lies in the interval :

Options:
A 1, 52
B 52, 4
C 32, 3
D 0, 32

The correct answer is:
32, 3

Solution:
$A=\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right]$
$\Rightarrow \operatorname{det}(A)=\left(11+\sin ^2 \theta\right)-\sin \theta(-\sin \theta+\sin \theta)+1\left(\sin ^2 \theta+1\right)$
$=2+2 \sin ^2 \theta$
$\because \theta \in\left(\frac{3 \pi}{4}, \frac{5 \pi}{4}\right)$
$\therefore \sin 2 \theta \in\left[0, \frac{1}{2}\right]$
$\therefore \operatorname{det}(A) \in[2,3]$ which is a subset of $\left(\frac{3}{2}, 3\right]$

Topic: Mathematics, Determinants, JEE Main, JEE Main 2019 (12 Jan Shift 2)
If $A=\left[\begin{array}{cc}1 & \tan \alpha / 2 \\ -\tan \alpha / 2 & 1\end{array}\right]$ and $A B=I$, then $B$ is equal to

Options:
A $\cos ^2 \alpha / 2 \cdot A$
B $\cos ^2 \alpha / 2 \cdot I$
C $\sin ^2 \alpha / 2 \cdot A$
D $\cos ^2 \alpha / 2 \cdot A^T$

The correct answer is:
$\cos ^2 \alpha / 2 \cdot A^T$

Solution:
G̣iven,
$A=\left[\begin{array}{cc}1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1\end{array}\right]$ and $A B=I$
$\Rightarrow B=A^{-1}$
$\begin{aligned} & \text { Now } A=\left[\begin{array}{cc}1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1\end{array}\right] \\ & |A|=\left[\begin{array}{cc}1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1\end{array}\right]=1+\operatorname{an}^2 \frac{\alpha}{2}=\sec ^2 \frac{\alpha}{2} \\ & \operatorname{adj} A=\left[\begin{array}{cc}1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1\end{array}\right] \\ & A^{-1}=\frac{1}{\sec ^2 \frac{\alpha}{2}}\left[\begin{array}{cc}1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1\end{array}\right] \\ & =\cos ^2 \frac{\alpha}{2}\left[\begin{array}{cc}1 & \left.-\tan \frac{\alpha}{2}\right] \\ \tan \frac{\alpha}{2} & 1\end{array} \cos ^2 \frac{\alpha}{2} A T\right. \\ & \end{aligned}$
$B=\left[\cos ^2 \frac{\alpha}{2}\right] A T$

Topic: Mathematics, Matrices, JEE Main
If $a_1 x+b_1 y+c_1 z=0, a_2 x+b_2 y+c_2 z=0 \quad a_3 x+b_3 y+c_3 z=0$ and $\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|=0$, then the given system has

Options:
A One trivial and one non-trivial solution
B No solution
C One solution
D Infinite solution

The correct answer is:
Infinite solution

Solution:
It is based on fundamental concept.

Topic: Mathematics, Determinants, JEE Main
If $a^{1 / x}=b^{1 / y}=c^{1 / z}$ and ${ }^{a, b, c}$ are in G.P., then $x, y, z$ will be in

Options:
A A.P.
B G.P.
C H.P.
D None of these

The correct answer is:
A.P.

Solution:
Let $a^{1 / x}=b^{1 / y}=c^{1 / z}=k \Rightarrow a=k^x, b=k^y, c=k^z$
Now, $a, b, c$ are in G.P.
$\Rightarrow b^2=a c \Rightarrow k^{2 y}=k^x \cdot k^z=k^{x+z} \Rightarrow 2 y=x+z$
$\Rightarrow \quad x, y, z$ are in A.P.

Topic: Mathematics, Sequences and Series, JEE Main
If $A=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]$ then $A^5=$

Options:
A $5 A$
B $10 \mathrm{~A}$
C $16 \mathrm{~A}$
D $32 A$

The correct answer is:
$16 \mathrm{~A}$

Solution:
$A=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]$ $\Rightarrow A^5=\left[\begin{array}{ccc}2^5 & 0 & 0 \\ 0 & 2^5 & 0 \\ 0 & 0 & 2^5\end{array}\right]=$ $=2^4\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]$
$=16 \mathrm{~A}$

Topic: Mathematics, Matrices, JEE Main
If A=21−12,B1011,C=ABAT and X=ATC2A, then det X is equal to:

Options:
A 243
B 729
C 27
D 891

The correct answer is:
729

Solution:
Given: A=21-12

⇒A=2+1=3

B=1011⇒B=1

And C=ABAT

⇒C=ABAT

⇒C=A2B as A=AT

⇒C=9

Now, solving X=AC2AT

.⇒X=AC2AT

⇒X=3×92×3

⇒X=729

Topic: Mathematics, Matrices, JEE Main, JEE Main 2024 (01 Feb Shift 1)
If A=21-1352161, then trAadjadjA is equal to (where, trP denotes the trace of the matrix P i.e. the sum of all the diagonal elements of the matrix P and adjP denotes the adjoint of matrix P)

Options:
A 7
B 18
C -58
D -1624

The correct answer is:
-1624

Solution:
∴adjadjA=AA (by property)
AadjadjA=AA2…..i
A=21-1352161=25-12-13-2-118-5
=-14-1-13=-28
A2=21-135216121-1352161=61-123409213712
trAadjadjA=trAA2=-286+40+12
=-28×58=-1624

Topic: Mathematics, Matrices, JEE Main
If A=2-2-4-1341-2-3 and B=-4-3-3101443 are two matrices, then the value of the determinantA+A2B2+A3+A4B4+.........20 terms is

Options:
A 203
B 2203
C -203
D 0

The correct answer is:
0

Solution:
A2=2-2-4-1341-2-32-2-4-1341-2-3=2-2-4-1341-2-3=A
B2=-4-3-3101443-4-3-3101443=100010001=Ι
B2=B4=B6=B8=.......=B20=Ι
A2=A3=A4=.......A20=A
So, A+A2B2+A3.......+A20B20=A+A+A.......+A=20A=203A ….(1)
A=2-2-4-1341-2-3=2-9+8+23-4-42-3
=-2-2+4=0
Hence, from equation (1),
A+A2B2+A3.......+A20B20=0

Topic: Mathematics, Determinants, JEE Main