datavorous/entrance-exam-dataset huggingface dataset:
Let Î» Î± , Î» Î² and Î» Î± â² denote the wavelengths of the X-rays of the K Î± , K Î² and L Î± lines in the characteristic X-rays for a metal. Then,

Options:
A 1 Î» Î² = 1 Î» Î± + 1 Î» Î± â²
B 1λ'α=1λβ+1λα
C 1λα=1λβ+1λ'α
D Î»Î±=Î»Î²+Î»'Î±

The correct answer is:
1 λ β = 1 λ α + 1 λ α ′

Solution:
E L â E K = hc Î» Î± ...(i)
E M â E K = hc Î» Î² ...(ii)
E M â E L = hc Î» Î± ' ...(iii)
( ii )â( i )â E M â E L = hc Î» Î± ' = hc Î» Î² â hc Î» Î±
1 Î» Î² = 1 Î» Î± + 1 Î» Î± '

Topic: Physics, Atomic Physics, JEE Main
Let λ∈ℝ and let the equation E be |x|2-2|x|+|λ-3|=0. Then the largest element in the set S= {x+λ:x is an integer solution of E} is ______

Options:

The correct answer is:
5

Solution:
Given:

|x|2-2|x|+|λ-3|=0

⇒|λ-3|=x2-x

At λ=3, x=0 and 2,

At λ=4 or 2, then,

x=1 or -1

So, maximum value of x+λ=5 when x=1 and λ=4.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2023 (24 Jan Shift 1)
Let \( A=(x+4 y) \vec{a}+(2 x+y+1) \vec{b} \) and \( B=(y-2 x+2) \vec{a}+(2 x-3 y-1) \vec{b} \) where \( \vec{a} \) and \( \vec{b} \) are non-collinear vectors, if \( 3 A=2 B \) then

Options:
A \( x=1, y=2 \)
B \( x=2, y=1 \)
C \( x=2, y=-1 \)
D \( x=-1, y=2 \)

The correct answer is:
\( x=2, y=-1 \)

Solution:
Given:

A=x+4ya→+2x+y+1b→ and B=y-2x+2a→+2x-3y-1b→

3A=3x+4ya→+32x+y+1b→

2B=2y-2x+2a→+22x-3y-1b→

In question it is given,

3A=2B

⇒3x+4y=2y-2x+2....1

and 32x+y+1=22x-3y-1....2

From the equation1

⇒7x+10y=4

⇒x=4-10y7

From the equation 2

2x+9y=-5

⇒24-10y7+9y=-5

⇒8-20y+63y=-35

⇒y=-1

⇒x=4-10-17=2

Hence, x=2, y=-1.

Topic: Mathematics, Vector Algebra, JEE Main
Let B=13α123αα4,α>2 be the adjoint of a matrix A and A=2. Then α-2ααBα-2αα is equal to

Options:
A 0
B 16
C -16
D 32

The correct answer is:
-16

Solution:
We have been given that B=13α123αα4,α>2

And adj(A)=B,|A|=2

⇒|adj(A)|=|B|

We know that adjA=An-1

⇒22=(8-3α)-3(4-3α)+α(-α)

⇒α2-6α+8=0

⇒(α-4)(α-2)=0

α=4,2 but α>2 so α=4

Now α-2ααBα-2αα= 4-841341234444-84

=121284-84=48-96+32=-16

Hence this is the correct option.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2023 (13 Apr Shift 1)
Let B and C are points of intersection of the parabola y=x2 and the circle x2+y-22=8 . The area of the triangle OBC, where O is the origin, is

Options:
A 2
B 4
C 6
D 8

The correct answer is:
8

Solution:
Solving y=x2 and x2+y-22=8
we get, y+y-22=8
⇒y2-3y-4=0⇒y=-1,4
⇒y=4 { y≠-1 otherwise x is not defined}
⇒x2=4⇒x=±2
Let, B=-2,4, C=2,4
Now, the area of ΔOBC=12×4×4=8 units

Topic: Mathematics, Straight Lines, JEE Main
Let B and C are the points of intersection of the parabola x=y2 and the circle y2+x-22=8. The perimeter (in units) of the triangle OBC, where O is the origin, is

Options:
A 8
B 45
C 45+2
D 45+1

The correct answer is:
45+1

Solution:
On solving x=y2 and y2+x-22=8
we get, x+x-22=8
⇒x2-3x-4=0⇒x=-1,4
⇒x=4 (as x≠-1 is rejected because it makes y not defined)
⇒y2=4⇒y=±2
Let, B=4,-2, C=4,2
Now, perimeter of ∆OBC=4+45 units

Topic: Mathematics, Parabola, JEE Main
Let B and C be the two points on the line y+x=0 such that B and C are symmetric with respect to the origin. Suppose A is a point on y-2x=2 such that ∆ABC is an equilateral triangle. Then, the area of the ∆ABC is

Options:
A 33
B 23
C 83
D 103

The correct answer is:
83

Solution:
On plotting the diagram of the given data we get,

Now given, x=y and y-2x=2

On solving we get, A(-2,-2)

Now height h of ∆ABC = Distance of point A to origin,

h=-22+(-2)2

⇒h=22

Given ∆ABC is an equilateral triangle,

So, sinB=hAB

⇒sin60°=hAB (∵∠A=∠B=∠C=60°)

Area of Δ=34AB2

⇒Area of Δ=34h2sin260∘=34×323

So, Area of ∆ABC=83

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Let B be the centre of the circle x2+y2-2x+4y+1=0. Let the tangents at two points P and Q on the circle intersect at the point A(3, 1). Then 8area ΔAPQarea ΔBPQ is equal to

.

Options:

The correct answer is:
18

Solution:
Given, centre of circle x2+y2-2x+4y+1=0 .....i

Points B 1,-2 & A3,1

For point P, y=-2

Putting value of y in equation i

x2+-22-2x+4×-2+1=0

x2-2x-3=0

x-3x+1=0

x=-1 (not possible)

x=3

Point P3,-2

AP=3-32+-2-12

AP=3=AQ

r=12+-22-1

r=1+4-1

r=2
tanθ=32

Area(ΔAPQ)Area(ΔBPQ)=ARRB

=3sinθ2cosθ=94

8Area(ΔAPQ)Area(ΔBPQ)=18

Topic: Mathematics, Circle, JEE Main, JEE Main 2021 (31 Aug Shift 2)
Let $B, C$ be $n \times n$ matrices such that $A=B+C, B C=C B$ and $C^2$ is a null matrix. Then, $B^{2020}[B+(2021) C]=$

Options:
A $A^{2020}$
B null zero matrix of order $n \times n$
C $A^{2021}$
D $B^{2021}$

The correct answer is:
$A^{2021}$

Solution:
Given that,
$A=B+C$
$B C=C B$ and $C^2=0$
From Eq. (i)
$\begin{aligned} & A^{n+1}=(B+C)^{n+1} \\ & \Rightarrow \quad A^{n+1}={ }^{n+1} C_0 \cdot B^{n+1}+{ }^{n+1} C_1 C \cdot B^n+\ldots \\ & \text { But, } \quad C^2=0, C^3=C^4 \ldots=C^r=0 \\ & \therefore \quad A^{n+1}={ }^{n+1} C_0 B^{n+1}+{ }^{n+1} C_1 B^n \cdot C \\ & =B^{n+1}+(n+1) B^n C \\ & A^{n+1}=B^n[B+(n+1) C] \\ & \therefore B^{2020}[B+(2020+1) C]=A^{2020+1} \\ & \Rightarrow \quad B^{2020}[B+(2021) C]=A^{2021} \\ & \end{aligned}$

Topic: Mathematics, Matrices, JEE Main
Let b→=i^+j^+λk^,λ∈ℝ. If a→ is a vector such that a→×b→=13i^-j^-4k^ and a→·b→+21=0, then b→-a→·k^-j^+b→+a→·i^-k^ is equal to

Options:

The correct answer is:
14

Solution:
Let a→=xi^+yj^+zk^

So,i^j^k^xyz11λ=i^λy-z+j^z-λx+k^x-y

On comparing with 13i^-j^-4k^ we get,

⇒λy-z=13,z-λx=-1,x-y=-4

and x+y+λz=-21

⇒ Clearly, λ=3,x=-2,y=2 and z=-7

So, b→-a→=3i^-j^+10k^ and b→+a→=-i^+3j^-4k^

⇒b→-a→·k^-j^+b→+a→·i^-k^=11+3

=14

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2022 (25 Jun Shift 2)
Let $\mathrm{BC}$ be a fixed line segment in the plane. The locus of a point $A$ such that the triangle $\mathrm{ABC}$ is isosceles, is (with finitely many possible exceptional points)

Options:
A a line
B a circle
C the union of a circle and a line
D the union of two circles and a line

The correct answer is:
the union of two circles and a line

Solution:
Case (i):
If $\angle \mathrm{B}=\angle \mathrm{C}$
locus of $\mathrm{A}$ is $\perp$ bisector of $\mathrm{BC}$
So it is straight line
Case (ii):
If $\angle \mathrm{A}=\angle \mathrm{C}$
BC fixed $\mathrm{B}(\mathrm{a}, 0), \mathrm{C}(0, \mathrm{a})$
$\mathrm{BC}=\mathrm{AB}$
So, $(x-a)^{2}+y^{2}=2 a^{2}$
Circle
Case (iii):
$\begin{array}{l}
\angle \mathrm{A}=\angle \mathrm{B} \\
\mathrm{AC}=\mathrm{BC} \\
\sqrt{\mathrm{h}^{2}+(\mathrm{k}-\mathrm{a})^{2}}=\sqrt{2 \mathrm{a}^{2}} \\
\mathrm{x}^{2}+(\mathrm{y}-\mathrm{a})^{2}=2 \mathrm{a}^{2}
\end{array}$
also a circle
So union of two circle and a line.

Topic: Mathematics, Straight Lines, JEE Main
Let ω be a complex number such that 2ω+1=z where z=-3 . If
1111-ω2-1ω21ω2ω7=3k,
Then k can be equal to:

Options:
A –z
B 1z
C -1
D 1

The correct answer is:
–z

Solution:
ω= -12+32 i (cube root of unity)
So, 1+ω+ω2=0 and ω3=1
Now 1111ωω21ω2ω=3ω2-ω
So k=ω2-ω

⇒k=-12-32i--12+32i
⇒k=-3i

⇒k=-z

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2017 (02 Apr)
Let $\omega$ be a cube root of unity not equal to 1 . Then the maximum possible value of $\left|a+b w+c w^{2}\right|$ where $a$, b, $c \in\{+1,-1\}$ is

Options:
A 0
B 2
C $\sqrt{3}$
D $1+\sqrt{3}$

The correct answer is:
2

Solution:
$\left|a+b w+c w^{2}\right|$
$\quad|a-c+(b-c) w| \quad$, for maximum value taking $a=1, c=-1, b=1$
$\left|a+b w+c w^{2}\right|=|2+2 w|=2\left|w^{2}\right|=2$

Topic: Mathematics, Complex Number, JEE Main
Let α be a non-zero real number. Suppose f:R→R is a differentiable function such that f0=1 and limx→−∞fx=1. If f'x=αfx+3, for all x∈R, then f−loge2 is equal to ________.

Options:
A 1
B 5
C 9
D 7

The correct answer is:
1

Solution:
Given: f0=1,limx→−∞fx=1

And f'x−α·fx=3

⇒dydx−α·y=3

Which is a linear differential equation, so

⇒I.F.=e−αx

Solution of the differential equation is given by,

ye−αx=∫3·e−αxdx

⇒fx⋅e−αx=3e−αx−α+c

⇒fx=−3α+c·eαx

Now, taking case I: when α>0 and limx→−∞fx=1 we get,

⇒1=−3α+c0

α=−3 (rejected)

Case-II: α<0

as limx→−∞fx=1

⇒c=0 & −3α=1 as e--∞→∞, so c=0

⇒α=−3

⇒fx=1 and also f0=1

Hence, fx=1 is constant function, so f-loge2=1

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2024 (01 Feb Shift 2)
Let λ be a real number for which the system of linear equations

x+y+z=6,

4x+λy-λz=λ-2 and

3x+2y-4z=-5

has infinitely many solutions. Then λ is a root of the quadratic equation:

Options:
A λ2+3λ-4=0
B λ2-λ-6=0
C λ2-3λ-4=0
D λ2+λ-6=0

The correct answer is:
λ2-λ-6=0

Solution:
The system of equations a1x+b1y+c1z+d1=0, a2x+b2y+c2z+d2=0 and a3x+b3y+c3z+d3=0 have a infinitely many solution, if

D=a1b1c1a2b2c2a3b3c3=0, D1=d1b1c1d2b2c2d3b3c3=0, D2=a1d1c1a2d2c2a3d3c3=0 and D3=a1b1d1a2b2d2a3b3d3=0.

Thus, for infinitely many solutions, we have D=0

⇒1114λ-λ32-4=0

⇒1-4λ+2λ-1-16+3λ+18-3λ=0

⇒-2λ+16-3λ+8-3λ=0

⇒24-3λ=0

⇒λ=3

And, λ=3 satisfies the equation λ2-λ-6=0.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2019 (10 Apr Shift 2)
Let $\rho$ be a relation defined on $N$, the set of natural numbers, as $\rho=\{(x, y) \in N \times N: 2 x+y=41\} .$ Then

Options:
A $\rho$ is an equivalence relation
B $\rho$ is only reflexive relation
C $\rho$ is only symmetric relation
D $\rho$ is not transitive

The correct answer is:
$\rho$ is not transitive

Solution:
We have, $\rho=\{(x, y) \in N \times N: 2 x+y=41\}$
For reflexive,
$\begin{array}{l}
x p x \Rightarrow 2 x+x=41 \\
\Rightarrow 3 x=41
\end{array}$
$\Rightarrow x=\frac{41}{3} \notin N$
So, $\rho$ is not reflexive.
For symmetric,
$x \rho y \Rightarrow 2 x+y=41$
and $y \rho x \Rightarrow 2 y+x=41$
$\Rightarrow x p y \neq y \rho x$
So, $\rho$ is not symmetric.
For transitive,
$x p y \Rightarrow 2 x+y=41$
and $y \rho z \Rightarrow 2 y+z=41$
$\Rightarrow x p z$
$\Rightarrow \rho$ is not transitive.

Topic: Mathematics, Sets and Relations, JEE Main
Let α be a root of the equation 1+x2+x4=0. Then the value of α1011+α2022-α3033 is equal to:

Options:
A 1
B α
C 1+α
D 1+2α

The correct answer is:
1

Solution:
Given, x4+x2+1=0

Using the formula a4+a2+1=a2+a+1a2-a+1

We get, x2+x+1x2-x+1=0

⇒x=±ω, ±ω2 where ω=11/3 and imaginary.

So α1011+α2022-α3033=1+1-1=1

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Let α be a root of the equation a-cx2+b-ax+c-b=0 where a, b, c are distinct real numbers such that the matrix α2α1111abc is singular. Then the value of a-c2b-ac-b+b-a2a-cc-b+c-b2a-cb-a is

Options:
A 6
B 3
C 9
D 12

The correct answer is:
3

Solution:
It is given that the matrix α2α1111abc is singular, so

α2α1111abc=0

⇒α2c-b-αc-a+b-a=0

Above equation holds true if α=1. Now,

a-c2b-ac-b+b-a2a-cc-b+c-b2a-cb-a

=a-b3+b-c3+c-a3a-bb-cc-a

=3a-bb-cc-aa-bb-cc-a=3

∵x3+y3+z3=3xyz if x+y+z=0

Topic: Mathematics, Determinants, JEE Main, JEE Main 2023 (24 Jan Shift 1)
Let α be a root of the equation x2+x+1=0 and the matrix A=131111αα21α2α4, then the matrix A31 is equal to

Options:
A A3
B I3
C A2
D A

The correct answer is:
A3

Solution:
Here, α=ω,ω2 where ω is complex cube root of unity.

A2=131111ωω21ω2ω1111ωω21ω2ω=100001010

⇒A4=100001010100001010=100010001

⇒A31=A28×A3=I×A3=A3

Topic: Mathematics, Matrices, JEE Main, JEE Main 2020 (07 Jan Shift 1)
Let λ be an integer. If the shortest distance between the lines x-λ=2y-1=-2z and x=y+2λ=z-λ is 722, then the value of λ is _______.

Options:

The correct answer is:
1

Solution:
x-λ1=y-1212=z-0-12

x-01=y+2λ1=z-λ1

Shortest distance=a2→-a1→·b1→×b2→b1→×b2→

b1→×b2→=i^j^k^112-12111

=i^12+12-j^1+12+k^1-12

=i^-32j^+k^2=2i^-3j^+k^2

b1→×b2→b1→×b2→=2i^-3j^+k^14

a2→-a1→·b1→×b2→b1→×b2→=

-λi^+-2λ+12j^+λk^·2i^-3j^+k^14

=-2λ+6λ-32+λ14=722

5λ-32=72

5λ=32±72

5λ=5,-2

λ=1,-25

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (24 Feb Shift 2)
Let α,β,γ be mirror image of the point 2,3,5 in the line x−12=y−23=z−34. Then 2α+3β+4γ is equal to

Options:
A 32
B 33
C 31
D 34

The correct answer is:
33

Solution:
Given: x-12=y-23=z-34

Let, P≡2,3,5 and R≡α,β,γ.

Now, direction ratio of PR=α-2, β-3, γ-5

And the line x-12=y-23=z-34 will be perpendicular,

So, using perpendicular condition we get,

⇒α-2×2+β-3×3+γ-5×4=0

⇒2α-4+3β-9+4γ-20=0

⇒2α+3β+4γ=33

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (31 Jan Shift 2)
Let $\alpha, \beta$ be real and $z$ be a complex number. If $z^2+\alpha z+\beta=0$ has two distinct roots on the line $\operatorname{Re} z=1$, then it is necessary that

Options:
A
$\beta \in(-1,0)$

B
$|\beta|=1$

C
$\beta \in(1, \infty)$

D
$\beta \in(0,1)$

The correct answer is:
$\beta \in(1, \infty)$

Solution:
Suppose roots are $1+\mathrm{pi}, 1+\mathrm{qi}$
Sum of roots $1+p i+1+q i=-\alpha$ which is real $\Rightarrow$ roots of $1+\mathrm{pi}, 1-\mathrm{pi}$
Product of roots $=\beta=1+p^2 \in(1, \infty)$
$\mathrm{p} \neq 0$ since roots are distinct.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2011
Let $\alpha, \beta, \gamma$ be real numbers. If
$\left(\begin{array}{ccc}7 & 5 & \alpha \\ \beta & 2 & 11 \\ 3 & \gamma & 1\end{array}\right)\left(\begin{array}{l}1 \\ 3 \\ 2\end{array}\right)=\left(\begin{array}{c}\alpha+\beta \\ -2 \alpha+\beta-2 \gamma \\ \alpha+2 \beta+3 \gamma\end{array}\right)$ then $100+\frac{2 \alpha+11 \beta}{\gamma}=$

Options:
A 27
B -25
C 225
D -227

The correct answer is:
27

Solution:
We have
$\left[\begin{array}{ccc}7 & 5 & \lambda \\ \beta & 2 & 11 \\ 3 & \gamma & 1\end{array}\right]\left[\begin{array}{l}1 \\ 3 \\ 2\end{array}\right]=\left[\begin{array}{c}\alpha+\beta \\ -2 \alpha+\beta-2 \gamma \\ \alpha+2 \beta+3 \gamma\end{array}\right]$
On solving we get,
$$
\begin{aligned}
& \alpha=-13, \beta=9, \gamma=-1, \text { then, } 100+\frac{2 \alpha+11 \beta}{\gamma} \\
& \Rightarrow 100-73=27
\end{aligned}
$$
On solving we get,
$$
\begin{aligned}
& \alpha=-13, \beta=9, \gamma=-1, \text { then, } 100+\frac{2 \alpha+11 \beta}{\gamma} \\
& \Rightarrow 100-73=27
\end{aligned}
$$

Topic: Mathematics, Matrices, JEE Main
Let $\alpha, \beta$ be roots of $x^2+\sqrt{2} x-8=0$. If $\mathrm{U}_{\mathrm{n}}=\alpha^{\mathrm{n}}+\beta^n$, then $\frac{\mathrm{U}_{10}+\sqrt{2} \mathrm{U}_9}{2 \mathrm{U}_8}$ is equal to______

Options:

The correct answer is:
4

Solution:
$\begin{aligned} & \frac{\alpha^{10}+\beta^{10}+\sqrt{2}\left(\alpha^9+\beta^9\right)}{2\left(\alpha^8+\beta^8\right)} \\ & \frac{\alpha^8\left(\alpha^2+\sqrt{2} \alpha\right)+\beta^8\left(\beta^2+\sqrt{2} \beta\right)}{2\left(\alpha^8+\beta^8\right)} \\ & \frac{8 \alpha^8+8 \beta^8}{2\left(\alpha^8+\beta^8\right)}=4\end{aligned}$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2024 (06 Apr Shift 2)
Let $\alpha, \beta$ be such that $\pi < \alpha-\beta < 3 \pi$. If $\sin \alpha+\sin \beta=-\frac{21}{65}$ and $\cos \alpha+\cos \beta=-\frac{27}{65}$, then the value of $\cos \frac{\alpha-\beta}{2}$ is

Options:
A
$-\frac{3}{\sqrt{130}}$

B
$\frac{3}{\sqrt{130}}$

C
$\frac{6}{65}$

D
$-\frac{6}{65}$

The correct answer is:
$-\frac{3}{\sqrt{130}}$

Solution:
$\sin \alpha+\sin \beta=-\frac{21}{65}$ and $\cos \alpha+\cos \beta=-\frac{27}{65}$.
Squaring and adding, we get
$$
\begin{aligned}
& 2+2 \cos (\alpha-\beta)=\frac{1170}{(65)^2} \\
& \Rightarrow \cos ^2\left(\frac{\alpha-\beta}{2}\right)=\frac{9}{130} \Rightarrow \cos \left(\frac{\alpha-\beta}{2}\right)=\frac{-3}{\sqrt{130}} \quad\left(\because \frac{\pi}{2} < \frac{\alpha-\beta}{2} < \frac{3 \pi}{2}\right) .
\end{aligned}
$$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2004
Let △,∇∈{∧,∨} be such that (p→q)△(p∇q) is a tautology. Then

Options:
A △=∧,∇=∨
B △=∨,∇=∧
C △=∨,∇=∨
D △=∧,∇=∧

The correct answer is:
△=∨,∇=∨

Solution:
Given,

△,∇∈{∧,∨} be such that (p→q)△(p∇q) is a tautology

Now checking all options by making truth table we get,

Option A △=∧,∇=∨

p	q	(p→q)	(p∨q)	(p→q)∧(p∨q)

T	T	T	T	T

T	F	F	T	F

F	T	T	T	T

F	F	T	F	F

Option B △=∨,∇=∧

p	q	(p→q)	(p∧q)	(p→q)∨(p∧q)

T	T	T	T	T

T	F	F	F	F

F	T	T	F	T

F	F	T	F	T

Option C △=∨,∇=∨

p	q	(p→q)	(p∨q)	(p→q)∨(p∧q)

T	T	T	T	T

T	F	F	T	T

F	T	T	T	T

F	F	T	F	T

Hence, it is a tautology.

Option D △=∧,∇=∧

p	q	(p→q)	(p∧q)	(p→q)∧(p∧q)

T	T	T	T	T

T	F	F	F	F

F	T	T	F	F

F	F	T	F	F

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Let Δ,∇∈∧,∨ be such that p∇q→pΔq∇r is a tautology. Then p∇q∆r is logically equivalent to

Options:
A pΔr∨q
B pΔr∧q
C p∧rΔq
D p∇r∧q

The correct answer is:
pΔr∨q

Solution:
Case-I

If Δ≡∇≡∧

p∧q→p∧q∧r

then it is false if r is false.

So not a tautology

Case-II

If Δ≡∇≡∨

p∨q→p∨q∨r≡ tautology

then p∨q∨r≡pΔr∨q

Case-III

If Δ=∨,∇=∧

then p∧q→p∨q∧r is not a tautology

when p→T,q→T,r→F

Case-IV

If Δ=∧,∇=∨

p∧q→p∧q∨r is not a tautology

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2022 (26 Jun Shift 1)
Let *,□∈{∧,∨} be such that the Boolean expression (p*~q)⇒(p□q) is a tautology. Then :

Options:
A *=∨, □=∧
B *=∨, □=∨
C *=∧, □=∨
D *=∧, □=∧

The correct answer is:
*=∧, □=∨

Solution:
Let (p∧~q)→(p∨q) is a tautology.

p	q	~q	p∨q	p∧~q	p∧~q⇒p∨q

T	T	F	T	F	T

T	F	T	T	T	T

F	T	F	T	F	T

F	F	T	F	F	T

All are true so it's a tautology.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2021 (31 Aug Shift 1)
Let θ be the acute angle between the tangents to the ellipse x29+y21=1 and the circle x2+y2=3 at their point of intersection in the first quadrant. Then tanθ is equal to :

Options:
A 523
B 43
C 23
D 2

The correct answer is:
23

Solution:
Given, x29+y21=1 ...1

x2+y2=3 ...2

On solving equations 1 & 2, we get

x29+3-x21=1

⇒x2+27-9x2=9

⇒8x2=18

⇒x2=94⇒x=±32

⇒x=32

Since, point of intersection in the first quadrant.

Put, x=32 in equation 2, we get y=32.

Hence, the point of intersection of the curves x29+y21=1 and x2+y2=3 in the first quadrant is 32, 32

Now, differentiate the equation 1 on both sides with respect to x, we get

y'=-x9y

∴Slope of tangent to the ellipse x29+y21=1 at 32, 32 is m1=-133

Now, differentiate the equation 2 on both sides with respect to x, we get

y'=-xy

∴Slope of tangent to the circle x2+y2=3 at 32, 32 is m2=-3

So, if acute angle between both curves is θ then tanθ=m1-m21+m1m2=-133+31+-133(-3)=23

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2021 (01 Sep Shift 2)
Let α be the angle between the lines whose direction cosines satisfy the equations l+m-n=0 and l2+m2-n2=0. Then the value of sin4α+cos4α is :

Options:
A 58
B 12
C 38
D 34

The correct answer is:
58

Solution:
n=ℓ+m

Now, ℓ2+m2=n2=ℓ+m2

⇒ 2ℓm=0

l2+m2+n2=1

If ℓ=0⇒ 2n2=1⇒n=±12

m=n=±12

And, If m=0⇒n=ℓ=±12

∴ l2+m2=12& l+m=12

⇒12+2 lm=12

∴ l=0, m=12 or l=12, m=0

So, direction cosines of two lines are

0,12,12 and 12,0,12

Thus,

∴ cos α=0+0+12=12

⇒α=π3

∴ sin4α+cos4α=1-12sin2(2α)=1-12·34=58

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (25 Feb Shift 1)
Let θ be the angle between the vectors a→ and b→, where a→=4,b→=3 and θ∈π4,π3. Then a→-b→×a→+b→2+4a→·b→2 is equal to ______

Options:

The correct answer is:
576

Solution:
a→-b→×a→+b→2+4a→·b→2

=a→×a→+a→×b→−b→×a→−b→×b→2+4a→⋅b→2

=4a→×b→2+a→·b→2=4a→2b→2=4×9×16=576

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2022 (25 Jun Shift 1)
Let ω be the angular velocity of the earth's rotation about its axis. Assume that the acceleration due to gravity on the earth's surface has the same value at the equator and the poles. An object weighed at the equator gives the same reading as a reading taken at a depth d below earth's surface at a pole d<<R. The value of d is

Options:
A ω 2 R 2 g
B ω 2 R 2 2 g
C 2 ω 2 R 2 g
D ω2R24g

The correct answer is:
ω 2 R 2 g

Solution:
g eff due to rotation = g eff due to depth d
g - R ω 2 = g 1 - d R ⇒ R ω 2 = gd R ⇒ d = R 2 ω 2 g

Topic: Physics, Gravitation, JEE Main
Let α be the area of the larger region bounded by the curve y2=8x and the lines y=x and x=2, which lies in the first quadrant. Then the value of 3α is equal to

Options:

The correct answer is:
22

Solution:
We have,

y2=8x

y=x

Point of intersection of curve and a line is A8,8.

Required area

α=∫288x-xdx

⇒α=42x323-x2228

⇒α=422233-642-42233-42

⇒α=1123-30

⇒α=223

So, 3α=22

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2023 (30 Jan Shift 1)
Let Δ be the area of the region x,y∈ℝ2:x2+y2≤21,y2≤4x,x≥1. Then 12Δ-21sin-127 is equal to

Options:
A 23-13
B 3-23
C 23-23
D 3-43

The correct answer is:
3-43

Solution:
Plotting the diagram of x2+y2≤21 & y2≤4x for x≥1 we get,

Now from above diagram required Area Δ=2∫132xdx+2∫32121-x2dx

⇒Δ=8333-1+21sin-127-63

So, 12Δ-21sin-127

=128333-1+21sin-127-63-21sin-127

=128333-1-63

=23-832

=3-43

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Let (α,β) be the centroid of the triangle formed by the lines 15x-y=82, 6x-5y=-4 and 9x+4y=17. Then α+2β and 2α-β are the roots of the equation

Options:
A x2-7x+12=0
B x2-14x+48=0
C x2-13x+42=0
D x2-10x+25=0

The correct answer is:
x2-13x+42=0

Solution:
The given set of equations are:

15x-y=82

6x-5y=-4

9x+4y=17

Now on solving the above equation by taking two at a time and plotting the diagram we get,

(α,β)≡1+5+63,2-7+83≡(4,1)

Hence, α+2β=6 and 2α-β=7

The required equation is x-6x-7=0

Equation x2-13x+42=0

Hence this is the required option.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2023 (13 Apr Shift 2)
Let α be the constant term in the binomial expansion of x-6x32n,n≤15. If the sum of the coefficients of the remaining terms in the expansion is 649 and the coefficient of x-n is λα, then λ is equal to ________.

Options:

The correct answer is:
36

Solution:
Given,

x-6x32n,n≤15

Now rth term is given by

Tr+1=Crnx12n-r-6x-32r

⇒Tr+1=-6r·Crn·xn-r2·x-3r2

⇒Tr+1=-6r·Crn·xn-4r2

Now we have to take, n=4r for constant term so n is divisible by 4.

So, α=Cn4n-6n4

And sum of all coefficients will be 1-6132n=-5n

Also given, -5n-Cn4n-6n4=649

Now by hit and trial we get, n=4

Hence, α=C14-644=-24

Now for coefficient of x-4

n-4r2=-4

⇒n=4r-8⇒r=3

So, λα=-63·C34

⇒λ-24=-63·C34

⇒λ=36

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (13 Apr Shift 1)
Let (α,β,γ) be the foot of perpendicular from the point (1,2,3) on the line x+35=y-12=z+43. then 19(α+β+γ) is equal to :

Options:
A 102
B 101
C 99
D 100

The correct answer is:
101

Solution:
Given: Pα,β,γ be the foot of the perpendicular from the point Q1,2,3 on the line x+35=y-12=z+43

Now, let x+35=y-12=z+43=λ

So, any point on the line is given by 5λ-3, 2λ+1, 3λ-4, so let the point be P5λ-3, 2λ+1, 3λ-4

Now, direction ratio of PQ=5λ-4, 2λ-1, 3λ-7

Now, the line and PQ will be perpendicular,

So, 55λ-4+22λ-1+33λ-7=0

⇒38λ-43=0

⇒λ=4338

So, the point Pα,β,γ=P5×4338-3, 2×4338+1, 3×4338-4

Then 19α+β+γ=195×4338-3+2×4338+1+ 3×4338-4

⇒19α+β+γ=19101+124-2338

⇒19α+β+γ=101

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (30 Jan Shift 1)
Let (α,β,γ) be the image of point P(2,3,5) in the plane 2x+y-3z=6. Then α+β+γ is equal to

Options:
A 5
B 10
C 12
D 9

The correct answer is:
10

Solution:
Given plane is 2x+y-3z=6.

Since, α,β,γ is the image point of 2,3,5 in the given plane, so

α-22=β-31=γ-5-3=-24+3-15-64+1+9

⇒α-22=β-31=γ-5-3=2

⇒α-22=2; β-31=2; γ-5-3=2

∴α=6, β=5, γ=-1

Hence, α+β+γ=6+5-1=10

Hence this is the required option.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (11 Apr Shift 1)
Let $(\alpha, \beta, \gamma)$ be the image of the point $(8,5,7)$ in the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{5}$. Then $\alpha+\beta+\gamma$ is equal to :

Options:
A 16
B 20
C 14
D 18

The correct answer is:
14

Solution:
$\begin{aligned} & \overrightarrow{\mathrm{AM}} \cdot(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})=0 \\ & (2 \lambda-7)(2)+(3 \lambda-6)(3)+(5 \lambda-5)(5)=0 \\ & 38 \lambda=57 \\ & \lambda=\frac{3}{2} \\ & \mathrm{M}\left(4, \frac{7}{2}, \frac{19}{2}\right) \\ & \mathrm{A}^{\prime}(0,2,12)\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (05 Apr Shift 2)
Let λ* be the largest value of λ for which the function fλx=4λx3-36λx2+36x+48 is increasing for all x∈ℝ. Then fλ*1+fλ,*-1 is equal to:

Options:
A 36
B 48
C 64
D 72

The correct answer is:
72

Solution:
Given,

fλx=4λx3-36λx2+36x+48

Differentiating w.r.t x both side we get,

fλ'x=12λx2-72λx+36

fλx'=12λx2-6λx+3≥0

∴ λ>0 and D≤0

36λ2-4×λ×3≤0

9λ2-3λ≤0

3λ3λ-1≤0

λ∈0,13

So, λlargest =13

fx=43x3-12x2+36x+48

So, f1+f-1=72

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2022 (24 Jun Shift 2)