datavorous/entrance-exam-dataset huggingface dataset:
Let $R=\{(1,3),(4,2),(2,4),(2,3),(3,1)\}$ be a relation on the set $A=\{1,2,3,4\}$. The relation $R$ is

Options:
A
a function

B
reflexive

C
not symmetric

D
transitive

The correct answer is:
not symmetric

Solution:
$(2,3) \in \mathrm{R}$ but $(3,2) \notin \mathrm{R}$
Hence R is not symmetric.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2004
Let R=1, 3, 4, 2, 2, 4, 2, 3, (3, 1) be a relation on the set A=1, 2, 3, 4. The relation R is

Options:
A Reflexive
B Transitive
C Not symmetric
D A function

The correct answer is:
Not symmetric

Solution:
Clearly, (1, 1), (2, 2), (3, 3), (4, 4) are not in R.
So, it is not reflexive
We observe that 2,3∈R but 3, 2 ∉R, so it is not symmetric.
Clearly, 1,3∈R and 3,1∈R, but 1,1∉R.
So, it is not transitive.
Since, 2,4 and 2,3 are in R. So, it is not a function.

Topic: Mathematics, Sets and Relations, JEE Main
Let r1 and r2 be the radii of the largest and smallest circles, respectively, which pass through the point (-4,1) and having their centres on the circumference of the circle x2+y2+2x+4y-4=0. If r1r2=a+b2, then a+b is equal to:

Options:
A 3
B 11
C 5
D 7

The correct answer is:
5

Solution:
We have,

x2+y2+2x+4y-4=0

⇒x2+2x+1+y2+4y+4=9

⇒x+12+y+22=32

Centre≡-1,-2

CP=-4+12+1+22=32

Centre of smallest circle is A

Centre of largest circle is B.

r2=|CP-CA|=32-3

r1=CP+CB=32+3

Hence,

r1r2=32+332-3=(32+3)29

⇒r1r2=(2+1)2=3+22

Hence,

a=3, b=2

a+b=5

Topic: Mathematics, Circle, JEE Main, JEE Main 2021 (20 Jul Shift 2)
Let R1 and R2 be two relations defined as follows :

R1=a, b∈R2:a2+b2∈Q and R2=a, b∈R2:a2+b2∉Q, where Q is the set of all rational numbers, then

Options:
A R1 is transitive but R2 is not transitive.
B R2 is transitive but R1 is not transitive.
C Neither R1 nor R2 is transitive.
D R1 and R2 are both transitive.

The correct answer is:
Neither R1 nor R2 is transitive.

Solution:
For R1 let a=1+2, b=1-2, c=814

aR1b⇒a2+b2=1+22+1-22=6∈Q

bR1c⇒b2+c2=1-22+8142=3∈Q

aR1c⇒a2+c2=1+22+8142=3+42∉Q

∴R1 is not transitive.

For R2 let a=1+2, b=2, c=1-2

aR2b⇒a2+b2=1+22+22=5+22∉Q

bR2c⇒b2+c2=22+1-22=5-22∉Q

aR2c⇒a2+c2=1+22+1-22=6∈Q

∴R2 is not transitive.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2020 (03 Sep Shift 2)
Let $R_1$ and $R_2$ respectively be the maximum ranges up and down an inclined plane and $R$ be the maximum range on the horizontal plane. Then $\mathrm{R}_1, \mathrm{R}, \mathrm{R}_2$ are in

Options:
A
Н.P

B
A.G.P

C
A.P

D
G.P.

The correct answer is:
Н.P

Solution:
Let $\beta$ be the inclination of the plane to the horizontal and $u$ be the velocity of projection of the projectile
$\mathrm{R}_1=\frac{\mathrm{u}^2}{\mathrm{~g}(1+\sin \beta)}$ and $\mathrm{R}_2=\frac{\mathrm{u}^2}{\mathrm{~g}(1-\sin \beta)}$
$\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}=\frac{2 \mathrm{~g}}{\mathrm{u}^2}$ or $\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}=\frac{2}{\mathrm{R}}\left[\because \mathrm{R}=\frac{\mathrm{u}^2}{\mathrm{~g}}\right]$
Therefore, $\mathrm{R}_1, \mathrm{R}, \mathrm{R}_2$ are in H.P.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2003
Let $R=\{(3,3)(5,5),(9,9),(12,12),(5,12),(3,9)$, $(3,12),(3,5)\}$ be a relation on the $\operatorname{set} A=\{3,5,9,12\}$. Then, $R$ is :

Options:
A
reflexive, symmetric but not transitive.

B
symmetric, transitive but not reflexive.

C
an equivalence relation.

D
reflexive, transitive but not symmetric.

The correct answer is:
reflexive, transitive but not symmetric.

Solution:
Let $\mathrm{R}=\{(3,3),(5,5),(9,9),(12,12),(5,12)$, $(3,9),(3,12),(3,5)\}$ be a relation on set $\mathrm{A}=\{3,5,9,12\}$
Clearly, every element of $A$ is related to itself. Therefore, it is a reflexive.
Now, $\mathrm{R}$ is not symmetry because 3 is related to 5 but 5 is not related to 3 .
Also $\mathrm{R}$ is transitive relation because it satisfies the property that if $a \mathrm{R} b$ and $b \mathrm{R} c$ then $a \mathrm{R} c$.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2013 (22 Apr Online)
Let $R=\{(3,3),(6,6),(9,9),(12,12),(6,12),(3,9),(3,12),(3,6)\}$ be a relation on the set $A=\{3,6,9,12\}$ . The relation is

Options:
A
reflexive and transitive only

B
reflexive only

C
an equivalence relation

D
reflexive and symmetric only

The correct answer is:
reflexive and transitive only

Solution:
Reflexive and transitive only.
e.g. $(3,3),(6,6),(9,9),(12,12)$ [Reflexive]
$(3,6),(6,12),(3,12)$
[Transitive].

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2005
Let $R=\{(3,3),(6,6),(9,9),(12,12),(6,12), (3,9),(3,12),(3,6)$ be a relation on the set $A=\{3,6,9,12\}$. Then, the relation is

Options:
A an equivalence relation
B reflexive and symmetric
C reflexive and transitive
D onlyreflexive

The correct answer is:
reflexive and transitive

Solution:
$(3,3),(6,6),(9,9),(12,12), \in R$.

$R$ isnot symmetricas (6,12)$\notin R$ but (12,6)$\in R$ $R$ is transitive as the only pair which needs verification is (3,6) and (6,12)$\in R$.

$$

\Rightarrow \quad(3,12) \in R

$$

Topic: Mathematics, Sets and Relations, JEE Main
Let R={a,b,c,d,e} and S={1,2,3,4}. Total number of onto functions f:R→S such that f(a)≠1, is equal to ________.

Options:

The correct answer is:
180

Solution:
Given,

R={a,b,c,d,e} and S={1,2,3,4}

Now taking, f(a)=1 we get, one of f(b), f(c), f(d), f(e)=1 then total such cases =4·3!=24
Now if, only f(a)=1, then we have distribute 2,3,4 amongst b,c,d,e,
So, total cases =34-C13·24+C23·1
=36
So, number of onto functions when f(a)=1 is 24+36=60

Now finding, total number of onto functions,
=45-C14·35+C24·25-C34·1
=1024-973+192-4
=240
Number of required functions when fa≠1 will be,

=240-60=180

Topic: Mathematics, Functions, JEE Main, JEE Main 2023 (08 Apr Shift 2)
Let $R$ and $S$ be two equivalence relations on a non-void set $A$. Then

Options:
A $\mathrm{R} \cup \mathrm{S}$ is equivalence relation
B $\mathrm{R} \cap \mathrm{S}$ is equivalence relation
C $\mathrm{R} \cap \mathrm{S}$ is not equivalence relation
D $\mathrm{R} \cup \mathrm{S}$ is not equivalence relation

The correct answers are:
$\mathrm{R} \cap \mathrm{S}$ is equivalence relation

Solution:
(B) $\mathrm{R} \cap \mathrm{S}$ is equivalence relation

Topic: Mathematics, Sets and Relations, JEE Main
Let α ∈R and the three vectors a→=αi^+j^+3k^,  b→=2i^+j^-αk^ and c→=αi^-2j^+3k^. Then the set S = {α:a→,b→ and c→ are coplanar}

Options:
A is singleton
B contains exactly two positive numbers
C is empty
D contains exactly two numbers only one of which is positive

The correct answer is:
is empty

Solution:
If a→, b→ and c→ are coplanar then,
α1321-αα-23=0
⇒3α2+18=0
⇒α∈ϕ

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2019 (12 Apr Shift 2)
Let R be a rectangle given by the lines x=0, x=2, y=0 and y=5. Let Aα,0 and B0,β, α∈0,2 and β∈0,5, be such that the line segment AB divides the area of the rectangle R in the ratio 4:1. Then, the mid-point of AB lies on a

Options:
A straight line
B parabola
C hyperbola
D circle

The correct answer is:
hyperbola

Solution:
Given,

R be a rectangle given by the lines x=0, x=2, y=0 and y=5

And point Aα,0 and B0,β, α∈0,2 and β∈0,5,

Now plotting the diagram we get,

Now given that line segment AB divides the ratio of area in 4:1, we get

10-12αβ12αβ=41

⇒20-αβ=4αβ

⇒αβ=4 ......1

Now using midpoint formula we get,

h=α2 & β=k2

Now using equation 1 we get,

⇒4hk=4

⇒xy=1 which is a equation of rectangular hyperbola.

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2023 (11 Apr Shift 1)
Let $R$ be a relation defined on the set $Z$ of all integers and $x R y,$ when $x+2 y$ is divisible by 3 , then

Options:
A $R$ is not transitive
B $A$ is symmetric only
C $R$ is an equivalence relation
D $A$ is not an equivalence relation

The correct answer is:
$R$ is an equivalence relation

Solution:
Reflexivity For reflexive $(x, x) \in R .$
$x+2 x=3 x$
which is divisible by 3.
$\Rightarrow$ xRx $\in R$ Hence, xRy is reflexive.
Symmetric Let $x+2 y=3 \lambda$
$\Rightarrow \quad x=3 \lambda-2 y$
$\begin{aligned}
\text {Now, } & \begin{aligned}
y+2 x &=y+2(3 \lambda-2 y) \\
&=y+6 \lambda-4 y \\
&=6 \lambda-3 y
\end{aligned}
\end{aligned}$
$\Rightarrow \quad y+2 x=3(2 \lambda-y)$
which is divisible by 3.
$\therefore$ $(x, y) \in R \Rightarrow(y, x) \in R$
Hence, xRy is symmetric.
$\begin{array}{ll}\text {Transifive } & x+2 y=3 \lambda \\ \text { and } & y+2 z=3 \mu\end{array}$
$x+3 y+2 z=3 x+3 \mu$
$\Rightarrow \quad x+2 z=3(\lambda+\mu-y)$
which is divisible by $3 .$
$\because (x, y) \in$ R and $(y, z) \in R$
$\Rightarrow \quad(x, z) \in R$
Hence, xRy is transitive
$\because$ R is reflexive, symmetric and transitive.
$\therefore$ R is an equivalence relation.

Topic: Mathematics, Sets and Relations, JEE Main
Let R be a relation from the set 1,2,3………,60 to itself such that R={a,b:b=pq, where p,q≥3 are prime numbers}. Then, the number of elements in R is

Options:
A 600
B 660
C 540
D 720

The correct answer is:
660

Solution:
Given b=pq where p & q are prime numbers ≥3 and 1≤pq≤60

If p=3,q=3,5,7,11,13,17,19

And if p=5 q=5,7,11 and so on,

So, b will be 3×3,3×5,3×7,3×11,3×13,3×17,3×19,5×5,5×7,5×11,7×7

i.e. total 11 possibilities

Now a can be 1,2,⋯,60

i.e. total 60 possibilities

Hence, the number of relations =60×11=660

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2022 (29 Jul Shift 1)
Let R be a relation on ℝ, given by R={a,b:3a-3b+7 is an irrational number }. Then R is

Options:
A Reflexive but neither symmetric nor transitive
B Reflexive and transitive but not symmetric
C Reflexive and symmetric but not transitive
D An equivalence relation

The correct answer is:
Reflexive but neither symmetric nor transitive

Solution:
Given:

R=a,b:3a-3b+7 is irrational

Reflexive:

3a-3a+7=7 is irrational.

So, a,a∈R, hence R is reflexive.

Given:

R=a,b:3a-3b+7 is irrational

Symmetric:

Let a,b∈R, then

3a-3b+7 is irrational.

So, 3b-3a+7 is not necessarily irrational, since if 3a=7 and 3b∈Z, then 3a-3b+7 is irrational., but 3b-3a+7 is not irrational.

So, a,b∈R⇒b,a∈R, hence R is not symmetric

Transitive:

Take a,b as 73,1 and b,c as 1,273

So, a,b∈R and b,c∈R but a,c∉R which means relation is not transitive.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2023 (01 Feb Shift 1)
Let R be a relation on Z×Z defined by (a, b)R(c, d) if and only if ad-bc is divisible by 5 . Then R is

Options:
A Reflexive and symmetric but not transitive
B Reflexive but neither symmetric not transitive
C Reflexive, symmetric and transitive
D Reflexive and transitive but not symmetric

The correct answer is:
Reflexive and symmetric but not transitive

Solution:
Given relation R : a,b R c,d

And ad-bc=5k where k∈Z

Now, for reflexive R : a,b R a,b

⇒ab-ba=0 which is divisible by 5

So, the relation is reflexive,

Now, for symmetric R : c,d R a,b

⇒ bc-ad=-5k which is divisible by 5

So, the relation is symmetric,

Now for transitive relation, R : a,b R c,d R e,f

Taking example a,b≡1,2, c,d≡10,10, e,f≡4,4

Now, ad-bc=10-20=-10

cf-ed=40-40=0

But af-be=8-4=4 which is not divisible by 5,

Hence, the relation is reflexive, symmetric but not transitive.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2024 (29 Jan Shift 1)
Let $r$ be a root of the equation $x^{2}+2 x+6=0$. The value of $(r+2)(r+3)(r+4)(r+5)$ is equal to .

Options:
A 51
B $-51$
C $-126$
D 126

The correct answer is:
$-126$

Solution:
$r$ be a root $\Rightarrow r^{2}+2 r+6=0\quad ......\text {(1)}$
$\begin{array}{l}\text { now }(r+2)(r+3)(r+4)(r+5) \\ =\left(r^{2}+5 r+6\right)(r 2+9 r+20) \\ =(3 r)(7 r+14) \quad \text { using (i) } \\ =21\left(r^{2}+2 r\right) & \\ =-126 \quad \text { using (i) } \\\end{array}$

Topic: Mathematics, Quadratic Equation, JEE Main
Let $R$ be an equivalence relation defined on a set containing 6 elements. The minimum number or ordered pairs that $R$ should contain is

Options:
A 12
B 6
C 64
D 36

The correct answer is:
6

Solution:
Given that,
$\mathrm{R} \rightarrow$ equivalence relation on a set $\mathrm{A}$
Let $\mathrm{A}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}, \mathrm{f}\}$
Since, $R$ is an equivalence relation on set $A$, then it must be satisfies reflexive property, for this.
$$
\mathrm{aRa}, \forall \mathrm{a} \in \mathrm{A}
$$
That mean minimum number of elements in ' $R$ ' should be six
$e g, \quad \mathrm{R}=\{(\mathrm{a}, \mathrm{a}),(\mathrm{b}, \mathrm{b}),(\mathrm{c}, \mathrm{c}),(\mathrm{d}, \mathrm{d})$
$(e, e),(f, f)\}$

Topic: Mathematics, Sets and Relations, JEE Main
Let $\mathrm{R}$ be relation defined on the set of natural number $\mathrm{N}$ as follows, $R=\{(x, y): x \in N, y \in N, 2 x+y=41\}$. Find the domain and range of the relation $R$. Also verify whether $R$ is reflexive, symmetric and transitive.

Options:

None

Solution:
We have,
$\mathrm{R}=\{(\mathrm{x}, \mathrm{y}): \mathrm{x} \in \mathrm{N}, \mathrm{y} \in \mathrm{N}, 2 \mathrm{x}+\mathrm{y}=41\}$
Domain $=\{1,2,3, \ldots, 20\}$
Range $=\{1,3,5,7, \ldots \ldots ., 39\}$
$\mathrm{R}=\{(1,39),(2,37),(3,35), \ldots .,(19,3),(20,1)\}$
$\mathrm{R}$ is not reflexive as $(2,2) \notin \mathrm{R}[\because \quad 2 \times 2+2 \neq 41]$
Also, $R$ is not symmetric.
As $\quad(1,39) \in \operatorname{R}$ but $(39,1) \notin \mathrm{R}$
$R$ is not transitive.
As $(11,19) \in \mathrm{R},(19,3) \in \mathrm{R}$
But $\quad(11,3) \notin \mathrm{R}$
Hence, $\mathrm{R}$ is neither reflexive, nor symmetric and nor transitive.

Topic: Mathematics, Relations and Functions (Class 11)
Let α∈R be such that the function fx=cos-11-x2sin-11-xx-x3,x≠0α,x=0 is continuous at x=0, where x=x-x,x is the greatest integer less than or equal to x. Then :

Options:
A α=π2
B α=0
C no such α exists
D α=π4

The correct answer is:
no such α exists

Solution:
limx→0+fx=f0=Limx→0-x

=limx→0+cos-11-x2·sin-11-xx1-x1+x

=limx→0+cos-11-x2x·1·1·π2

Let 1-x2=cosθ

=π2limθ→0+θ1-cosθ

=π2limθ→0+θ2sinθ2=π2

Now, limx→0-cos-11-1+x2sin-1-x1+x-1+x3

limx→0-π2-sin-1x1+x2+x-x

limx→0-π21·2·sin-1xx=π4

⇒RHL≠LHL

Function can't be continuous

⇒ No value of α exist

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let α∈R be such that the function fx=cos-11-x2sin-11-xx-x3,x≠0α,x=0 is continuous at x=0, where x=x-x,x is the greatest integer less than or equal to x. Then :

Options:
A α=π2
B α=0
C no such α exists
D α=π4

The correct answer is:
no such α exists

Solution:
limx→0+fx=f0=Limx→0-x

=limx→0+cos-11-x2·sin-11-xx1-x1+x

=limx→0+cos-11-x2x·1·1·π2

Let 1-x2=cosθ

=π2limθ→0+θ1-cosθ

=π2limθ→0+θ2sinθ2=π2

Now, limx→0-cos-11-1+x2sin-1-x1+x-1+x3

limx→0-π2-sin-1x1+x2+x-x

limx→0-π21·2·sin-1xx=π4

⇒RHL≠LHL

Function can't be continuous

⇒ No value of α exist

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2021 (16 Mar Shift 2)
Let R be the focus of the parabola y2=20x and the line y=mx+c intersect the parabola at two points P and Q. Let the points G10, 10 be the centroid of the triangle PQR. If c-m=6, then PQ2 is

Options:
A 296
B 325
C 317
D 346

The correct answer is:
325

Solution:
Given,

R be the focus of the parabola y2=20x and the line y=mx+c intersect the parabola at two points P and Q,

And the points G10, 10 be the centroid of the triangle PQR,

Now focus of the parabola y2=20x will be, R5,0

And parametric points of PQ be 5t2,10t,

Now plotting the diagram we get,

Now finding the centroid of the triangle we get,

For x- coordinate we get,

⇒5t12+5t22+53=10

⇒t12+t22=5 ...(i)

Now for y-coordinate we get,

10t1+t23=10

⇒t1+t2=3 ...(ii)

Now solving both equations we get, t1=1, t2=2

So, points will be, P≡(5, 10) & Q≡20,20

Hence, equation of PQ=y-10=1015x-5

⇒3y-30=2x-10

⇒y=23x+203, so on comparing with y=mx+c, we get c-m=6

Hence, PQ2=225+100=325

Topic: Mathematics, Parabola, JEE Main
Let R be the focus of the parabola y2=20x and the line y=mx+c intersect the parabola at two points P and Q. Let the points G10, 10 be the centroid of the triangle PQR. If c-m=6, then PQ2 is

Options:
A 296
B 325
C 317
D 346

The correct answer is:
325

Solution:
Given,

R be the focus of the parabola y2=20x and the line y=mx+c intersect the parabola at two points P and Q,

And the points G10, 10 be the centroid of the triangle PQR,

Now focus of the parabola y2=20x will be, R5,0

And parametric points of PQ be 5t2,10t,

Now plotting the diagram we get,

Now finding the centroid of the triangle we get,

For x- coordinate we get,

⇒5t12+5t22+53=10

⇒t12+t22=5 ...(i)

Now for y-coordinate we get,

10t1+t23=10

⇒t1+t2=3 ...(ii)

Now solving both equations we get, t1=1, t2=2

So, points will be, P≡(5, 10) & Q≡20,20

Hence, equation of PQ=y-10=1015x-5

⇒3y-30=2x-10

⇒y=23x+203, so on comparing with y=mx+c, we get c-m=6

Hence, PQ2=225+100=325

Topic: Mathematics, Parabola, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Let R be the interior region between the lines 3x-y+1=0 and x+2y-5=0 containing the origin. The set of all values of a, for which the points a2,a+1 lie in R, is :

Options:
A (-3,-1)∪-13,1
B (-3,0)∪13,1
C (-3,0)∪23,1
D (-3,-1)∪13,1

The correct answer is:
(-3,0)∪13,1

Solution:
Let, P≡a2,a+1, L1:3x-y+1=0 and L2:x+2y-5=0.

Now, origin and P lies on the same side with respect to line L1.

⇒L1(0)·L1(P)>0

⇒3a2-(a+1)+1>0

⇒3a2-a>0

⇒a3a-1>0

⇒a∈(-∞,0)∪13,∞ ...i

Now, L2:x+2y-5=0

Origin and P lies on same side with respect to L2

⇒L2(0)·L2(P)>0

⇒a2+2(a+1)-5<0

⇒a2+2a-3<0

⇒(a+3)(a-1)<0

∴a∈(-3,1) ...ii

Using intersection of i and ii,

⇒a∈(-3,0)∪13,1

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2024 (27 Jan Shift 2)
Let R be the point 3,7 and let P and Q be two points on the line x+y=5 such that PQR is an equilateral triangle. Then the area of ∆PQR is

Options:
A 2543
B 2532
C 253
D 2523

The correct answer is:
2523

Solution:
Given PQR is an equilateral triangle.

Let the side of the triangle be a

Perpendicular distance from R3,7 to the line x+y=5 is

3+7-51+1=52

So sin60°=52a

a=523

Area of ∆PQR=34a2=2523

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2022 (26 Jun Shift 1)
Let $R$ be the real line. Consider the following subsets of the plane $R \times R$.
$S=\{(x, y): y=x+1$ and $0 < x < 2\}, T=\{(x, y): x-y$ is an integer $\}$. Which one of the following is true?

Options:
A
neither $S$ nor $T$ is an equivalence relation on $R$

B
both $S$ and $T$ are equivalence relations on $R$

C
$S$ is an equivalence relation on $R$ but $T$ is not

D
$\mathrm{T}$ is an equivalence relation on $\mathrm{R}$ but $S$ is not

The correct answer is:
$\mathrm{T}$ is an equivalence relation on $\mathrm{R}$ but $S$ is not

Solution:
$$
T=\{(x, y): x-y \in l\}
$$
as $0 \in \mathrm{I} \mathrm{T}$ is a reflexive relation.
If $x-y \in 1 \Rightarrow y-x \in I$
$\therefore \mathrm{T}$ is symmetrical also
If $x-y=l_1$ and $y-z=l_2$
Then $x-z=(x-y)+(y-z)=l_1+I_2 \in I$
$\therefore \mathrm{T}$ is also transitive.
Hence $T$ is an equivalence relation.
Clearly $x \neq x+1 \Rightarrow(x, x) \notin S$
$\therefore S$ is not reflexive.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2008
Let $R$ be the real line. Let the relations $S$ and $T$ on $R$ be defined by $S=\{(x, y): y=x+1,0 < x < 2\}, T=\{(x, y):(x-y)$ is an integer $\}$. Then

Options:
A both $S$ and $T$ are equivalence relations on $R$
B $\mathrm{T}$ is an equivalence on $\mathrm{R}$ but $\mathrm{S}$ is not
C neither $S$ nor $T$ is an equivalence relation on $R$
D $\mathrm{S}$ is an equivalence relation on $\mathrm{R}$ but $\mathrm{T}$ is not

The correct answer is:
$\mathrm{T}$ is an equivalence on $\mathrm{R}$ but $\mathrm{S}$ is not

Solution:
$\mathrm{T}$ is an equivalence but $\mathrm{S}$ is not

Topic: Mathematics, Sets and Relations, JEE Main
Let $R$ be the region of the disc $x^{2}+y^{2} \leq 1$ in the first quadrant. Then the area of the largest possible circle contained in $\mathrm{R}$ is

Options:
A $\pi(3-2 \sqrt{2})$
B $\pi(4-3 \sqrt{2})$
C $\frac{\pi}{6}$
D $\pi(2 \sqrt{2}-2)$

The correct answer is:
$\pi(3-2 \sqrt{2})$

Solution:
Required equation of circle $(x-h)^{2}+(y-h)^{2}=h^{2}$
Both circle touch internally
$$
\begin{array}{l}
\mathrm{C}_{1} \mathrm{C}_{2}=\left|\mathrm{r}_{1}-\mathrm{r}_{2}\right| \\
\sqrt{\mathrm{h}^{2}+\mathrm{h}^{2}}=|\mathrm{h}-1|
\end{array}
$$
Solve this $\mathrm{h}=\sqrt{2}-1$
Area $\pi(\sqrt{2}-1)^{2}=\pi(3-2 \sqrt{2})$

Topic: Mathematics, Area Under Curves, JEE Main
Let $R$ be the relation on the set $R$ of all real numbers, defined by $a R b$ If $|a-b| \leq 1$. Then, $R$ is

Options:
A reflexive and symmetric only
B reflexive and transitive only
C equivalence
D None of the above

The correct answer is:
reflexive and symmetric only

Solution:
Since, $|a-a|=0 \leq 1,$ so $a R a, \forall a \in R$ $\therefore R$ is reflexive. Now, $a R b \Rightarrow|a-b| \leq 1 \Rightarrow|b-a| \leq 1 \Rightarrow b R a$

$\therefore R$ is symmetric. But $R$ is not transitive as

$1 R 2,2 R 3$ but $1 \not R^{\prime} 3$

$$

[\because|1-3|=2>1]

$$

Topic: Mathematics, Sets and Relations, JEE Main
Let $\mathrm{R}$ be the relation on $Z$ defined by $R=\{(a, b): a \in Z$, $b \in Z, a-b$ is an even integer $\}$. Find the domain and range of $R$.

Options:

None

Solution:
Here, $R=\{(a, b): a \in Z, b \in Z, a-b$ is an even integer $\}$
Now $(a-b)$ is an even integer, provided $a$ and $b$ are both even or both odd.
$\therefore$ Domain $=Z, \because a$ is any integer
Range $=Z, \because b$ is any integer.

Topic: Mathematics, Relations and Functions (Class 11)
Let r be the remainder when 20212020 is divided by 20202. Then r lies between

Options:
A 0 and 5
B 10 and 15
C 20 and 100
D 107 and 120

The correct answer is:
0 and 5

Solution:
(2021)2020=(1+2020)2020

C02020+C12020·2020+C22020·20202+…+C20202020·20202020

1+20202+C22020·20202+…+C20202020·20202020

1+202021+C22020+…+20202018

1+(2020)2·λ

Hence (2021)2020=λ(2020)2+1

Hence remainder =1

Topic: Mathematics, Binomial Theorem, JEE Main
Let r be the remainder when 20212020 is divided by 20202. Then r lies between

Options:
A 0 and 5
B 10 and 15
C 20 and 100
D 107 and 120

The correct answer is:
0 and 5

Solution:
(2021)2020=(1+2020)2020

C02020+C12020·2020+C22020·20202+…+C20202020·20202020

1+20202+C22020·20202+…+C20202020·20202020

1+202021+C22020+…+20202018

1+(2020)2·λ

Hence (2021)2020=λ(2020)2+1

Hence remainder =1

Topic: Mathematics, Binomial Theorem, KVPY, KVPY 2020 (SA)
Let R be the set of all real numbers and f:R→R be a continuous function. Suppose |f(x)-f(y)|≥|x-y| for all real numbers x and y. Then

Options:
A f is one-one, but need not be onto
B f is onto, but need not be one-one
C f need not be either one-one or onto
D f is one-one and onto

The correct answer is:
f is one-one and onto

Solution:
Let f(x)=f(y)
So, |f(x)-f(y)|≥|x-y|
⇒0≥|x-y|⇒x-y=0⇒x=y
⇒f is one-one
Since, f is continuous
So f(0) is finite Now, |f(x)-f(0)|≥|x-0|
⇒limx→∞|f(x)-f(0)|≥limx→∞|x|
⇒limx→∞f(x)=∞
⇒f is unbounded
⇒f is surjective

Topic: Mathematics, Functions, JEE Main
Let $R$ be the set of all real numbers and let $f$ be a fucntion $R$ to $R$ such that $f(x)+\left(x+\frac{1}{2}\right) f(1-x)=1$, for all $x \in R .$ Then $2 f(0)+3 f(1)$ is equal to.

Options:
A 2
B 0
C $-2$
D $-4$

The correct answer is:
$-2$

Solution:
Given $f(x)+\left(x+\frac{1}{2}\right) f(1-x)=1 \quad ......\text {(1)}$
but $x=0$
$\begin{array}{l}
f(0)+\frac{1}{2} f(1)=1 \\
\Rightarrow 2 f(0)+f(1)=2 \quad ......\text {(2)}\end{array}$
$\begin{array}{l}
\text { put } x=1 \text { in }(1) \\
\Rightarrow f(1)+\frac{3}{2} f(0)=1 \\
\Rightarrow 2 f(1)+3 f(0)=2 \quad ......\text {(3)}
\end{array}$
Solving $(2) \&(3)$ we have
$\begin{array}{l}
F(0)=2 \& f(1)=-2 \\
\therefore 2 f(0)+f(1)=4-6=-2
\end{array}$

Topic: Mathematics, Functions, JEE Main
Let $R$ be the set of real numbers and the functions $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined by $f(x)=x^{2}+2 x-3$ and $g(x)=x+1 .$ Then, the value of $x$ for which $f(g(x))=g(f(x))$ is

Options:
A -1
B 0
C 1
D 2

The correct answer is:
-1

Solution:
According to question,
$$
f(g(x))=g(f(x))
$$
$\Rightarrow \quad f(x+1)=g\left(x^{2}+2 x-3\right)$
$\Rightarrow \quad(x+1)^{2}+2(x+1)-3=x^{2}+2 x-3+1$
$\Rightarrow \quad x^{2}+1+2 x+2 x+2-3=x^{2}+2 x-2$
$\Rightarrow \quad x^{2}+4 x=x^{2}+2 x-2$
$\Rightarrow \quad x^{2}+4 x-x^{2}-2 x+2=0$
$\Rightarrow \quad 2 x+2=0$
$\Rightarrow \quad 2 x=-2$
$$
\Rightarrow \quad x=-1
$$

Topic: Mathematics, Functions, JEE Main