datavorous/entrance-exam-dataset huggingface dataset:
Let 3,a,b,c be in A.P. and 3,a−1,b+1,c+9 be in G.P. Then, the arithmetic mean of a, b and c is:

Options:
A -4
B -1
C 13
D 11

The correct answer is:
11

Solution:
Given: 3, a, b, c are in A.P

So, the common difference will be d=a-3

Now, b=2a-3 and c=3a-6

Also, 3, a-1, b+1, c+9 are in G.P

So, 3c+9=b+1a-1

⇒33a-6+9=2a-3+1a-1

⇒33a+3=2a-2a-1

⇒9a+9=2a2-4a+2

⇒2a2-13a-7=0

⇒a=7, -12rejected

So, b=2×7-3=11 & c=3×7-6=15

Now, the A.M of a, b, c is given by,

⇒A=a+b+c3

⇒A=7+11+153

⇒A=333

⇒A=11

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2024 (01 Feb Shift 1)
Let α→=3i^+j^ and β→=2i^-j^+3k^. If β→=β1→-β2→, where β1→ is parallel to α→ and β2→ is perpendicular to α→, then β1→×β2→ is equal to:

Options:
A 12(-3i^+9j^+5k^)
B 3i^-9j^-5k^
C -3i^+9j^+5k^
D 123i^-9j^+5k^

The correct answer is:
12(-3i^+9j^+5k^)

Solution:
If two vectors are parallel then they are proportional.

So, let β1→=kα→

Given β→=β1→-β2→

⇒α→⋅β→=α→⋅β1→-α→⋅β2→ ...i

Now, α→·β→=3i^+j^·2i^-j^+3k^

=6-1=5

And, α→ is perpendicular to β1→, hence α→⋅β1→=0

And, α→⋅β2→=α→⋅kα→=kα→2=kα→2

Also, α→=32+12=10

Put, all these values in the equation i, to get 5=k×10

∴ k=12

∴ β1→=12(3i^+j^)

=32i^+12j^

Now β2→=β1→-β→=-12i^+32j^-3k^

∴ β1→×β2→=i^j^k^32120-1232-3

=i^-32-0-j^-92-0+k^94+14

=-32i^+92j^+52k^

=12(-3i^+9j^+5k^).

Topic: Mathematics, Vector Algebra, JEE Main
Let α→=3i^+j^ and β→=2i^-j^+3k^. If β→=β1→-β2→, where β1→ is parallel to α→ and β2→ is perpendicular to α→, then β1→×β2→ is equal to:

Options:
A 12(-3i^+9j^+5k^)
B 3i^-9j^-5k^
C -3i^+9j^+5k^
D 123i^-9j^+5k^

The correct answer is:
12(-3i^+9j^+5k^)

Solution:
If two vectors are parallel then they are proportional.

So, let β1→=kα→

Given β→=β1→-β2→

⇒α→⋅β→=α→⋅β1→-α→⋅β2→ ...i

Now, α→·β→=3i^+j^·2i^-j^+3k^

=6-1=5

And, α→ is perpendicular to β1→, hence α→⋅β1→=0

And, α→⋅β2→=α→⋅kα→=kα→2=kα→2

Also, α→=32+12=10

Put, all these values in the equation i, to get 5=k×10

∴ k=12

∴ β1→=12(3i^+j^)

=32i^+12j^

Now β2→=β1→-β→=-12i^+32j^-3k^

∴ β1→×β2→=i^j^k^32120-1232-3

=i^-32-0-j^-92-0+k^94+14

=-32i^+92j^+52k^

=12(-3i^+9j^+5k^).

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2019 (09 Apr Shift 1)
Let $\sqrt{3} \hat{i}+\hat{j}, \hat{i}+\sqrt{3} \hat{j}$ and $\beta \hat{i}+(1-\beta) \hat{j}$ respectively be the

position vectors of the points $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ with respect to the origin $\mathrm{O}$. If the distance of $\mathrm{C}$ from the bisector of the

acute angle between $\mathrm{OA}$ and $\mathrm{OB}$ is $\frac{3}{\sqrt{2}},$ then the sum of

all possible values of $\beta$ is:

Options:
A 4
B 3
C 2
D 1

The correct answer is:
1

Solution:
Since, the angle bisector of acute angle between $O A$

and $O B$ would be $y=x$

Since, the distance of $C$ from bisector $=\frac{3}{\sqrt{2}}$

$\Rightarrow\left|\frac{\beta-(1-\beta)}{\sqrt{2}}\right|=\frac{3}{\sqrt{2}}=2 \beta=\pm 3+1$

$\beta=2$ or $\beta=-1$

Hence, the sum of all possible value of $\beta=2+(-1)=1$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2019 (11 Jan Shift 2)
Let $\int \frac{2-\tan x}{3+\tan x} \mathrm{~d} x=\frac{1}{2}\left(\alpha x+\log _{\mathrm{e}}|\beta \sin x+\gamma \cos x|\right)+C$, where $C$ is the constant of integration. Then $\alpha+\frac{\gamma}{\beta}$ is equal to :

Options:
A 7
B 4
C 1
D 3

The correct answer is:
4

Solution:
$\begin{aligned} & \int \frac{2-\tan x}{3+\tan x} d x=\int \frac{2 \cos x-\sin x}{3 \cos x+\sin x} d x \\ & 2 \cos x-\sin x=A(3 \cos x+\sin x)+B(\cos x-3 \sin x) \\ & 3 A+B=2 \\ & A-3 B=-1\end{aligned}$
$\begin{aligned} & \Rightarrow A=\frac{1}{2}, B=\frac{1}{2} \\ & \therefore \int \frac{2 \cos x-\sin x}{3 \cos x+\sin x} d x \\ & =\frac{x}{2}+\frac{1}{2} \ln |3 \cos x+\sin x|+C \\ & =\frac{1}{2}(x+\ln |3 \cos x+\sin x|)+C \\ & =\frac{1}{2}(\alpha x+\ln |\beta \sin x+\gamma \cos x|)+C \\ & \alpha=1, \beta=1, \gamma=3 \\ & \therefore \alpha+\frac{\gamma}{\beta}=1+\frac{3}{1}=4\end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2024 (09 Apr Shift 1)
Let α=(4!)!(4!)3! and β=(5!)!(5!)4!. Then :

Options:
A α∈N and β∉N
B α∉N and β∈N
C α∈N and β∈N
D α∉N and β∉N

The correct answer is:
α∈N and β∈N

Solution:
Given: α=(4!)!(4!)3!, β=(5!)!(5!)4!

⇒α=(24)!(4!)6, β=(120)!(5!)24

If 24 distinct objects are divided into 6 groups of 4 objects in each group, then the number of ways of formation of group is=24!(4!)6.6!∈N

Similarly, let 120 distinct objects are divided into 24 groups of 5 objects in each group.

So, the number of ways of formation of groups

=(120)!(5!)24·24!∈N

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2024 (27 Jan Shift 2)
Let α→=4i^+3j^+5k^ and β→=i^+2j^-4k^. Let β→1 be parallel to α→ and β→2 be perpendicular to α→. If β→=β→1+β→2, then the value of 5β→2·i^+j^+k^ is

Options:
A 6
B 11
C 7
D 9

The correct answer is:
7

Solution:
Given that α→=4i^+3j^+5k^ and β→=i^+2j^-4k^ and β→=β1→+β2→

⇒β2→=β→-β1→ ..........1

since β→1 is parallel to α→, β1→=tα→

⇒β1→=t4i^+3j^+5k^=4ti^+3tj^+5tk^ .........2

Substituting the values of β1→ and α→ in (1), we get

β2→=i^+2j^-4k^-4ti^+3tj^+5tk^=1-4ti^+2-3tj^+-4-5tk^ ............3

since β2→ is perpendicular to α→, so β2→·α→=0

⇒1-4ti^+2-3tj^+-4-5tk^ ·4i^+3j^+5k^=0

⇒41-4t+32-3t+5-4-5t=0

⇒4-16t+6-9t-20-25t=0

⇒-50t=10⇒t=-15

From 2 and 3, we get β1→=-154i^+3j^+5k^

β2→=95i^+135j^-3k^=159i^+13j^-15k^

So, the value of 5β→2·i^+j^+k^ will be,

=5×15(9i^+13j^-15k^)·i^+j^+k^=9+13-15=7

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (24 Jan Shift 2)
Let 5,a4, be the circumcenter of a triangle with vertices A(a,-2), B(a, 6) and Ca4,-2. Let α denote the circumradius, β denote the area and γ denote the perimeter of the triangle. Then α+β+γ is

Options:
A 60
B 53
C 62
D 30

The correct answer is:
53

Solution:
Given: Aa,-2, Ba,6 and Ca4,-2

⇒AB2=a-a2+6+22

⇒AB2=64 ...i

⇒BC2=a-a42+6+22

⇒BC2=9a216+64 ...ii

⇒AC2=a-a42+-2+22

⇒AC2=9a216 ...iii

Using i, ii and iii,

⇒AC2+AB2=BC2

So, △ABC is right angled at A.

So, circumcentre will be mid-point of BC.

⇒S≡a+a42,6-22

⇒S≡5a8,2=5,a4

⇒a=8

⇒BC=9×6416+64

⇒BC=100=10

⇒AB=8

⇒AC=9×6416

⇒AC=6

Now, circumradius is given by,

R=BC2=5⇒α=5

Now, area of triangle will be β=12×6×8=24 and perimeter will be γ=6+8+10=24

⇒α=5, β=24, γ=24

⇒α+β+γ=53

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2024 (29 Jan Shift 1)
Let θ=π5 and A=cosθsinθ-sinθcosθ. If B=A+A4, then det B :

Options:
A is one
B lies in 2, 3
C is zero
D lies in 1, 2

The correct answer is:
lies in 1, 2

Solution:
A2=cosθsinθ-sinθcosθcosθsinθ-sinθcosθ

A2=cos2θsin2θ-sin2θcos2θ

⇒A4=cos4θsin4θ-sin4θcos4θ

B=cos4θsin4θ-sin4θcos4θ+cosθsinθ-sinθcosθ

=cos4θ+cosθsin4θ+sinθ-sin4θ+sinθcos4θ+cosθ

B=cos4θ+cosθ2+sin4θ+sinθ2

=2+2cos4θ·cosθ+sin4θ·sinθ

=2+2cos4θ-θ

=2+2cos3θ

B=2+2cos3π5

=2-5-12=5-52∈1,2

Topic: Mathematics, Matrices, JEE Main, JEE Main 2020 (06 Sep Shift 2)
Let 5 digit numbers be constructed using the digits 0,2,3,4,7,9 with repetition allowed, and are arranged in ascending order with serial numbers. Then the serial number of the number 42923 is _____ .

Options:

The correct answer is:
2997

Solution:
Number of five-digit numbers formed using the digits 0, 2, 3, 4, 7, 9 with 2 at the first place is

=6×6×6×6=1296

Since, 2+6+6+6+6=1296

Number of five-digit numbers formed using the given digits with 3 at the first place is

=1296

Since, 3+6+6+6+6=1296

Number of five-digit numbers formed using the given digits with 40 at the first place is

=216

Since, 40+6+6+6=216

And,

420+6+6=36

4 2 2+6+6=36

423+6+6=36

424+6+6=36

427+6+6=36

429 0+6=6

429 2 0=1

429 2 2=1

429 2 3=1

Required serial number is

=1296×2+216+5×36+6+1+1+1=2997

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (31 Jan Shift 1)
Let 5fx+4f1x=1x+3, x>0. Then 18∫12fxdx is equal to

Options:
A 5 loge2+3
B 10 loge2+6
C 10 loge2-6
D 5loge2-3

The correct answer is:
10 loge2-6

Solution:
Given:

5fx+4f1x=1x+3 ....1

Replace x→1x

5f1x+4fx=x+3 ....2

Eliminating f1x from 1 & 2, we get

9fx=5x+15-4x-12

⇒9fx=5x-4x+3

⇒18fx=10x-8x+6

⇒18∫12fxdx=∫1210x-8x+6dx

⇒18∫12fxdx=10logx-4x2+6x12

⇒18∫12fxdx=10log2-4-10log1+2

⇒18∫12fxdx=10loge2-6

Hence this is the correct option.

Topic: Mathematics, Functions, JEE Main, JEE Main 2023 (06 Apr Shift 1)
Let 9 distinct balls be distributed among 4 boxes, B1, B2, B3 and B4. If the probability that B3 contains exactly 3 balls is k349 then k lies in the set :

Options:
A {x∈R:|x-3|<1}
B {x∈R:|x-2|≤1}
C {x∈R:|x-1|<1}
D {x∈R:|x-5|≤1}

The correct answer is:
{x∈R:|x-3|<1}

Solution:
The number of ways in which 9 distinct balls can be distributed among 4 boxes is 49

The number of ways in which B3 contains 3 balls is C39·36

Using the classical definition of probability we can say the required probability =C39·3649

=C3927·349

=289·349⇒k=289≈3.11

Now, consider |x-3|<1

⇒-1<x-3<1

⇒-1+3<x<1+3

⇒2<x<4

∴k satisfies |x-3|<1

Topic: Mathematics, Probability, JEE Main, JEE Main 2021 (25 Jul Shift 1)
Let 10 vertical poles standing at equal distances on a straight line, subtend the same angle of elevation α at a point O on this line and all the poles are on the same side of O. If the height of the longest pole is h and the distance of the foot of the smallest pole from O is a; then the distance between two consecutive poles, is

Options:
A hsinα+acosα9 cosα
B hcosα-a sinα9sinα
C hsinα+acosα9sinα
D hcosα-asinα9cosα

The correct answer is:
hcosα-a sinα9sinα

Solution:
from diagram

tanα=ha+9b

a+9b=hcosαsinα

9b= hcosα-asinαsinα

b=hcosα-asinα9sinα

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2015 (11 Apr Online)
Let 10 vertical poles standing at equal distances on a straight line, subtend the same angle of elevation α at a point O on this line and all the poles are on the same side of O. If the height of the longest pole is h and the distance of the foot of the smallest pole from O is a; then the distance between two consecutive poles, is

Options:
A hsinÎ±+a cosÎ±9 cosÎ±
B hcosÎ±-a sinÎ±9sinÎ±
C hsinÎ±+a cosÎ±9sinÎ±
D hcosÎ±-a sinÎ±9cosÎ±

The correct answer is:
hcosÎ±-a sinÎ±9sinÎ±

Solution:
from diagram
tanâ¡Î±=ha+9b
a+9b=hcosâ¡Î±sinâ¡Î±
9b= hcosâ¡Î±-asinâ¡Î±sinâ¡Î±
b=hcosâ¡Î±-asinâ¡Î±9sinâ¡Î±

Topic: Mathematics, Heights and Distances, JEE Main
Let $\alpha \beta \gamma=45 ; \alpha, \beta, \gamma \in \mathbb{R}$. If $x(\alpha, 1,2)+y(1, \beta, 2)+z(2,3, \gamma)=(0,0,0)$ for some $x, y, z \in \mathbb{R}, x y z \neq 0$, then $6 \alpha+4 \beta+\gamma$ is equal to _______

Options:

The correct answer is:
55

Solution:
$\begin{aligned} & \alpha \beta \gamma=45, \alpha \beta \gamma \in \mathrm{R} \\ & \mathrm{x}(\alpha, 1,2)+\mathrm{y}(1, \beta, 2)+\mathrm{z}(2,3, \gamma)=(0,0,0) \\ & \mathrm{x}, \mathrm{y}, \mathrm{z} \in \mathrm{R}, \mathrm{xyz} \neq 0 \\ & \alpha \mathrm{x}+\mathrm{y}+2 \mathrm{z}=0 \\ & \mathrm{x}+\beta \mathrm{y}+3 \mathrm{z}=0 \\ & 2 \mathrm{x}+2 \mathrm{y}+\gamma \mathrm{z}=0 \\ & \mathrm{xyz} \neq 0 \Rightarrow \text { non-trivial } \\ & \left|\begin{array}{lll}\alpha & 1 & 2 \\ 1 & \beta & 3 \\ 2 & 2 & \gamma\end{array}\right|=0\end{aligned}$
$\begin{aligned} & \Rightarrow \alpha(\beta \gamma-6)-1(\gamma-6)+2(2-2 \beta)=0 \\ & \Rightarrow \alpha \beta \gamma-6 \alpha-\gamma+6+4-4 \beta=0 \\ & \Rightarrow 6 \alpha+4 \beta+\gamma=55\end{aligned}$

Topic: Mathematics, Determinants, JEE Main, JEE Main 2024 (06 Apr Shift 1)
Let $A=\left[\begin{array}{ll}0 & \alpha \\ 0 & 0\end{array}\right]$ and $(A+I)^{50}-50 A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$,
find $a b c+a b d+b c d+a c d$

Options:
A 0
B -1
C 1
D None of these

The correct answer is:
0

Solution:
As $\mathrm{A}^{2}=0, \mathrm{~A}^{k}=0 \forall k \geq 2$
Thus, $\quad(\mathrm{A}+\mathrm{I})^{50}=\mathrm{I}+50 \mathrm{~A} \Rightarrow(\mathrm{A}+\mathrm{I})^{50}-50 \mathrm{~A}=\mathrm{I}$
$\therefore \quad \mathrm{a}=1, \mathrm{~b}=0, \mathrm{c}=0, \mathrm{~d}=1$
$\mathrm{abc}+\mathrm{abd}+\mathrm{bed}+\mathrm{acd}=0$

Topic: Mathematics, Matrices, JEE Main
Let $A=\left(\begin{array}{ccc}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right)$
The only correct statement about the matrix $A$ is

Options:
A
$A$ is a zero matrix

B
$A^2=I$

C
$\mathrm{A}^{-1}$ does not exist

D
$A=(-1) I$, where $I$ is a unit matrix

The correct answer is:
$A^2=I$

Solution:
$A . A=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=I$

Topic: Mathematics, Matrices, JEE Main
Let $A=\left(\begin{array}{ccc}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right)$
The only correct statement about the matrix $A$ is

Options:
A
$A$ is a zero matrix

B
$A^2=I$

C
$\mathrm{A}^{-1}$ does not exist

D
$A=(-1) I$, where $I$ is a unit matrix

The correct answer is:
$A^2=I$

Solution:
$A . A=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=I$

Topic: Mathematics, Matrices, JEE Main, JEE Main 2004
Let $a_{0}=0$ and $a_{x}=3 a_{n-1}+1$ for $n \geq 1$. Then the remainder obtained dividing aco ob by 11 is-

Options:
A 0
B 7
C 3
D 4

The correct answer is:
0

Solution:
$$
\begin{array}{l}
a_{n}=3 a_{n-1}+1 \\
a_{2010}=3 a_{2009}+1 \\
=3\left(3 a_{2008}+1\right)+1=3^{2} a_{2008}+3+1 \\
\quad=3^{3} a_{2007}+3+3+1
\end{array}
$$
$$
\begin{array}{l}
3^{2010} \mathrm{a}_{0}+\underbrace{(3+3+\ldots 3)}_{2009 \text { times }}+1 \\
=0+6027+1=6028
\end{array}
$$
Remainder $\left(\frac{6028}{11}\right)=0$

Topic: Mathematics, Binomial Theorem, JEE Main
Let $a_{0}=0$ and $a_{x}=3 a_{n-1}+1$ for $n \geq 1$. Then the remainder obtained dividing aco ob by 11 is-

Options:
A 0
B 7
C 3
D 4

The correct answer is:
0

Solution:
$$
\begin{array}{l}
a_{n}=3 a_{n-1}+1 \\
a_{2010}=3 a_{2009}+1 \\
=3\left(3 a_{2008}+1\right)+1=3^{2} a_{2008}+3+1 \\
\quad=3^{3} a_{2007}+3+3+1
\end{array}
$$
$$
\begin{array}{l}
3^{2010} \mathrm{a}_{0}+\underbrace{(3+3+\ldots 3)}_{2009 \text { times }}+1 \\
=0+6027+1=6028
\end{array}
$$
Remainder $\left(\frac{6028}{11}\right)=0$

Topic: Mathematics, Binomial Theorem, KVPY, KVPY 2010 (SB/SX)
Let A={0, 1, 2, 3, 4, 5, 6, 7}. Then the number of bijective functions f:A→A such that f(1)+f(2)=3-f(3) is equal to

Options:

The correct answer is:
720

Solution:
Given the set A={0, 1, 2, 3, 4, 5, 6, 7}.

And, also a function f:A→A satisfying f(1)+f(2)=3-f(3)

⇒ f(1)+f(2)+f(3)=3

Since, the range of the function is also A, hence the only possibility satisfying the given condition is: 0+1+2=3

We know that, the number of arrangements of n objects at n places is n!.

Since, the given function is bijective i.e. one-one and onto, hence, the elements 1, 2, 3 in the domain can be mapped with only 0, 1, 2 in the co-domain in 3! ways and the remaining 5 elements 0, 4, 5, 6, 7 in the domain can be mapped with any of the remaining 5 elements 3, 4, 5, 6, 7 in 5! ways.

So, the number of bijective functions are =3!×5!=6×120=720.

Topic: Mathematics, Functions, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Let A=012a031c0, where a,c∈R. If A3=A and the positive value of a belongs to the interval (n-1, n], where n∈ℕ, then n is equal to ____.

Options:

The correct answer is:
2

Solution:
Given,

A=012a031c0

⇒A2=012a031c0012a031c0

⇒A2=a+22c33a+3c2aca12+3c

Now finding,

A3=a+22c33a+3c2aca12+3c012a031c0

⇒A3=2ca+3α+2+3c2a+4+6ca2+3ca+2a3+2ac6+3a+9ca+2+3cca+2c+3c22ca+3

Now equating A3=A and comparing both side we get,

2ca+3=0⇒c=-32a

And a+2+3c=1

⇒a+2+3-32a=1

⇒a+1-92a=0

⇒2a2+2a-9=0

⇒a=-2+4+4×9×24=-2+764≈6.74≈1.4

Hence, a∈(1,2]

So, on comparing with a∈(n-1,n] we get n=2

Topic: Mathematics, Matrices, JEE Main, JEE Main 2023 (11 Apr Shift 1)
Let A=θ∈0, 2π:1+2isinθ1-isinθ is purely imaginary Then the sum of the elements is in A is

Options:
A 4π
B 3π
C π
D 2π

The correct answer is:
4π

Solution:
Let

z=1+2isinθ1-isinθ

⇒z=1+2isinθ1-isinθ×1+isinθ1+isinθ

⇒z=1-2sin2θ+3isinθ1+sin2θ

⇒z=1-2sin2θ1+sin2θ+i3sinθ1+sin2θ

Since, z is a purely imaginary number, so real part must be zero, hence

1-2sin2θ1+sin2θ=0

⇒1-2sin2θ=0

⇒cos2θ=0

⇒θ=π4,3π4,5π4,7π4, for θ∈0,2π

∴Sum of all values =16π4=4π

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2023 (08 Apr Shift 2)
Let A=0-220. If M and N are two matrices given by M=∑k=110 A2k and N=∑k=110 A2k-1 then MN2 is

Options:
A a non-identity symmetric matrix
B a skew-symmetric matrix
C neither symmetric nor skew-symmetric matrix
D an identity matrix

The correct answer is:
a non-identity symmetric matrix

Solution:
Since A=0-220

So A2=0-2200-220=-400-4=-4I

and A3=-4A

Similarly A4=-4I-4I=-42I,

A5=-42A, A6=-43I

Now M=∑k=110A2k=A2+A4+….+A20

=-4+-42+-43+….+-420I

=-k1I

So M is symmetric matrix

N=∑k=110 A2k-1=A+A3+……+A19

=A1+-4+-42+……+-49

=k2A

So N is skew symmetric

⇒N2 is symmetric matrix

Hence, MN2 is non-identity symmetric matrix

Topic: Mathematics, Matrices, JEE Main, JEE Main 2022 (25 Jun Shift 1)
Let A=0 2y zx y-zx-y z such that ATA=I, then the value of x2+y2+z2 is

Options:

The correct answer is:
1

Solution:
AT=0xx2yy-yz-zz
ATA=0xx2yy-yz-zz02yzxy-zx-yz
=2x20006y20003z2=100010001
ATA=I

2x2=1⇒x2=12
3z2=1⇒z2=13
6y2=1⇒y2=16
x2+y2+z2=12+13+16=12+12=1

Topic: Mathematics, Matrices, JEE Main
Let A(α,0) and B(0,β) be the points on the line 5x+7y=50. Let the point P divide the line segment AB internally in the ratio 7: 3. Let 3x-25=0 be a directrix of the ellipse E:x2a2+y2b2=1 and the corresponding focus be S. If from S, the perpendicular on the x-axis passes through P, then the length of the latus rectum of E is equal to

Options:
A 253
B 329
C 259
D 325

The correct answer is:
325

Solution:
Given: A(α,0) and B(0,β) lies on the line 5x+7y=50.

⇒5α+70=50, 50+7β=50

⇒α=10, β=507

Also, P divide the line segment AB internally in the ratio 7: 3.

⇒P≡7×0+3×107+3,7×507+3×07+3

⇒P≡3,5

⇒ae=3 as perpendicular from S passes through P

Now, 3x-25=0 is a directrix of the ellipse E:x2a2+y2b2=1

⇒x=253

We know that, directrix of an ellipse is given by x=±ae

⇒ae=253

Also, ae=3

⇒a=5, b=4

Length of LR =2b2a=325

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Let $a \neq 0$ and $p(x)$ be a polynomial of degree greater than 2. If $p(x)$ leaves remainders a and $-a$ when divided respectively by $x+a$ and $x-a$, then the remainder when $p(x)$ is divided by $x^2-a^2$ is :

Options:
A $x$
B $-x$
C $-2 x$
D $2 x$

The correct answer is:
$-x$

Solution:
Let the remainder be $R(x)$, then
$R(x)=P(x)+q$
Given, $\quad R(-a)=a \Rightarrow-P a+q=a$ $\ldots$ (i)
and $R(a)=-a \Rightarrow P a+q=-a$$\ldots$ (ii)
On solving Eqs. (i) and (ii), we get
$P=1, q=0$
$\therefore \quad R(x)=-x$

Topic: Mathematics, Quadratic Equation, AP EAMCET, AP EAMCET 2003
Let $a \neq 0$ and $p(x)$ be a polynomial of degree greater than 2. If $p(x)$ leaves remainders a and $-a$ when divided respectively by $x+a$ and $x-a$, then the remainder when $p(x)$ is divided by $x^2-a^2$ is :

Options:
A $x$
B $-x$
C $-2 x$
D $2 x$

The correct answer is:
$-x$

Solution:
Let the remainder be $R(x)$, then
$R(x)=P(x)+q$
Given, $\quad R(-a)=a \Rightarrow-P a+q=a$ $\ldots$ (i)
and $R(a)=-a \Rightarrow P a+q=-a$$\ldots$ (ii)
On solving Eqs. (i) and (ii), we get
$P=1, q=0$
$\therefore \quad R(x)=-x$

Topic: Mathematics, Quadratic Equation, JEE Main
Let $a \neq 0$ and $p(x)$ be a polynomial of degree greater than 2. If $p(x)$ leaves remainders a and $-a$ when divided respectively by $x+a$ and $x-a$, then the remainder when $p(x)$ is divided by $x^2-a^2$ is :

Options:
A $x$
B $-x$
C $-2 x$
D $2 x$

The correct answer is:
$-x$

Solution:
Let the remainder be $R(x)$, then
$R(x)=P(x)+q$
Given, $\quad R(-a)=a \Rightarrow-P a+q=a$ $\ldots$ (i)
and $R(a)=-a \Rightarrow P a+q=-a$$\ldots$ (ii)
On solving Eqs. (i) and (ii), we get
$P=1, q=0$
$\therefore \quad R(x)=-x$

Topic: Mathematics, Quadratic Equation, TS EAMCET, TS EAMCET 2003
Let $a>0$ be a real number. Then the limit $\lim _{x \rightarrow 2} \frac{a^{x}+a^{3-x}-\left(a^{2}+a\right)}{a^{3-x}-a^{x / 2}}$
is

Options:
A $2 \log a$
B $-\frac{4}{3} a$
C $\frac{a^{2}+a}{2}$
D $\frac{2}{3}(1-a)$

The correct answer is:
$\frac{2}{3}(1-a)$

Solution:
$\lim _{x \rightarrow 2} \frac{a^{x}+a^{3-x}-\left(a^{2}+a\right)}{a^{3-x}-a^{x / 2}}$
$\lim _{x \rightarrow 2} \frac{\left(a^{x}-a\right)\left(a^{x / 2}-a\right)\left(a^{x / 2}+a\right)}{\left(a-a^{x / 2}\right)\left(a^{2}+a^{x}+a^{x / 2} \cdot a\right)}$
$\lim _{x \rightarrow 2}-\frac{\left(a^{x}-a\right)\left(a^{x / 2}+a\right)}{a^{2}+a^{x}+a^{x / 2} \cdot a}$
$=\frac{2}{3}(1-a)$

Topic: Mathematics, Limits, JEE Main
Let $\mathrm{a}>0$ be a root of the equation $2 x^2+x-2=0$. If $\lim _{x \rightarrow \frac{1}{\mathrm{a}}} \frac{16\left(1-\cos \left(2+x-2 x^2\right)\right)}{(1-\mathrm{a} x)^2}=\alpha+\beta \sqrt{17}$, where $\alpha, \beta \in Z$, then $\alpha+\beta$ is equal to_______

Options:

The correct answer is:
170

Solution:
$\begin{aligned} & \lim _{x \rightarrow \frac{1}{a}} 16 \cdot \frac{\left(1-\cos 2\left(x-\frac{1}{a}\right)\left(x-\frac{1}{b}\right)\right)}{4\left(x-\frac{1}{b}\right)^2} \times \frac{4\left(x-\frac{1}{b}\right)^2}{a^2\left(x-\frac{1}{a}\right)^2} \\ & =16 \times \frac{2}{a^2}\left(\frac{1}{a}-\frac{1}{b}\right)^2 \\ & =\frac{32}{a^2}\left(\frac{17}{4}\right)=\frac{17.8}{a^2}=\frac{17 \times 8 \times 16}{(-1+\sqrt{117})^2} \\ & =\frac{136.16}{18.2 \sqrt{7}} \times \frac{18+2 \sqrt{7}}{18+2 \sqrt{7}} \\ & =\frac{136}{256}(18+2 \sqrt{7}) \cdot 16 \\ & =153+17 \sqrt{17}=\alpha+\beta \sqrt{17} \\ & \alpha+\beta=153+17=170\end{aligned}$

Topic: Mathematics, Limits, JEE Main, JEE Main 2024 (05 Apr Shift 2)
Let A=0ii0, where i2=-1. Let I denotes the identity matrix of order 2, then I+A+A2+A3…….A110 is equal to

Options:
A 0ii0
B 0000
C 1001
D -1000

The correct answer is:
0ii0

Solution:
A2=0ii00ii0=-100-1=-I
A 3 = A 2 A=−I⋅A=−A
A4=A22=I
A5=A,A6=-I,A7=-A,A8=I…. So on
I+A+A2+A3=I+A-I-A=0
A4+A5+A6+A7=0…………A104+A105+A106+A107=0
A108+A109+A110=I+A-I=A
⇒I+A+A2+.…….+A110=A

Topic: Mathematics, Matrices, JEE Main
Let $A=\left(\begin{array}{ccc}0 & 2 q & r \\ p & q & -r \\ p & -q & r\end{array}\right)$. If $\mathrm{AA}^{\mathrm{T}}=\mathrm{I}_{3},$ then $|\mathrm{p}|$ is:

Options:
A $\frac{1}{\sqrt{5}}$
B $\frac{1}{\sqrt{3}}$
C $\frac{1}{\sqrt{2}}$
D $\frac{1}{\sqrt{6}}$

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

Solution:
$A=\left[\begin{array}{ccc}0 & 2 q & r \\ p & q & -r \\ p & -q & r\end{array}\right]$

$\begin{aligned} \therefore & A \cdot A^{T}=\left[\begin{array}{ccc}0 & 2 q & r \\ p & q & -r \\ p & -q & r\end{array}\right] \times\left[\begin{array}{ccc}0 & p & p \\ 2 q & q & -q \\ r & -r & r\end{array}\right] \\ &=\left[\begin{array}{ccc}4 q^{2}+r^{2} & 2 q^{2}-r^{2} & -2 q^{2}+r^{2} \\ 2 q^{2}-r^{2} & p^{2}+q^{2}+r^{2} & p^{2}-q^{2}-r^{2} \\ -2 q^{2}+r^{2} & p^{2}-q^{2}-r^{2} & p^{2}+q^{2}+r^{2}\end{array}\right] \end{aligned}$

Given, $A A^{T}=I$

$\therefore \quad 4 q^{2}+r^{2}=p^{2}+q^{2}+r^{2}=1$

$\Rightarrow \quad p^{2}-3 q^{2}=0$ and $r^{2}=1-4 q^{2}$

and $2 q^{2}-r^{2}=0 \Rightarrow r^{2}=2 q^{2}$

$\begin{array}{ll}\therefore \quad & p^{2}=\frac{1}{2}, q^{2}=\frac{1}{6} \text { and } r^{2}=\frac{1}{3} \\ \therefore \quad & |p|=\frac{1}{\sqrt{2}}\end{array}$

Topic: Mathematics, Matrices, JEE Main, JEE Main 2019 (11 Jan Shift 1)
Let A=1000αβ0βα and |2A|3=221 where α,β∈Z, Then a value of α is

Options:
A 3
B 5
C 17
D 9

The correct answer is:
5

Solution:
Given: A=1000αβ0βα and 2A3=221

Now, 2A3=221

⇒2A=27

⇒23A=27

⇒A=16

⇒α2-β2=16

Now, taking value from the option we get,

α=5 as α,β ∈Z, so for other value of α, β is not a integer.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2024 (29 Jan Shift 1)
Let A=100011100. Then A2025-A2020 is equal to

Options:
A A6-A
B A6
C A5
D A5-A

The correct answer is:
A6-A

Solution:
A=100011100

A2=100011100100011100=100111100A3=100111100100011100=100211100

A4=100311100

An=100n-111100

A2025-A2020=100202411100-100201911100=000500000=A6-A

Topic: Mathematics, Matrices, JEE Main, JEE Main 2021 (26 Aug Shift 2)
Let A=10004-1012-3. Then the sum of the diagonal elements of the matrix A+I11 is equal to:

Options:
A 6144
B 4094
C 4097
D 2050

The correct answer is:
4097

Solution:
Given,

A=10004-1012-3

A2=10004-1012-310004-1012-3

⇒A2=10004-1012-3=A

⇒A2=A

⇒A3=A·A2=A2=A

⇒A4=A2·A2=A2=A

⇒A3=A4=……=A

A+I11=C011 A11+C111 A10+….C1011 A+C1111I

=C011+C111+….C1011A+I

=211-1A+I=2047 A+I

⇒A+I11=2047A+I

⇒A+I11=2047+10004×2047+1-1×2047012×2047-3×2047+1

∴ Sum of diagonal elements

=20471+4-3+3

=4094+3=4097

Topic: Mathematics, Matrices, JEE Main, JEE Main 2023 (31 Jan Shift 1)