datavorous/entrance-exam-dataset huggingface dataset:
Let $R$ be the set of real numbers and the mapping $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined by $f(x)=5-x^2$ and $g(x)=3 x-4$, then the value of $(f \circ g)(-1)$ is

Options:
A -44
B -54
C -32
D -64

The correct answer is:
-44

Solution:
Hints: $\mathrm{f}(\mathrm{g}(-1))=\mathrm{f}(-3-4)=\mathrm{f}(-7)=5-49=-44$

Topic: Mathematics, Functions, JEE Main
Let $R$ be the set of real numbers
This question has Statement $-1$ and Statement $-2$. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement-1 : $A=\{(x, y) \in R \times R: y-x$ is an integer $\}$ is an equivalence relation on $R$.
Statement-2 : $B=\{(x, y) \in R \times R: x=\alpha y$ for some rational number $\alpha\}$ is an equivalence relation on $\mathrm{R}$.

Options:
A
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is not a correct explanation for Statement $-1$

B
Statement $-1$ is true, Statement- 2 is false.

C
Statement $-1$ is false, Statement $-2$ is true.

D
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is a correct explanation for Statement $-1$

The correct answer is:
Statement $-1$ is true, Statement- 2 is false.

Solution:
$\mathrm{x}-\mathrm{y}$ is an integer
$x-x=0$ is an integer $\Rightarrow A$ is Reflexive
$x-y$ is an integer $\Rightarrow y-x$ is an integer $\Rightarrow A$ is symmetric
$x-y, y-z$ are integers
As sum of two integers is an integer.
$\Rightarrow(x-y)+(y-z)=x-z$ is an integer
$\Rightarrow \mathrm{A}$ is transitive. Hence statement $-1$ is true.
Also $\frac{\mathrm{x}}{\mathrm{x}}=1$ is a rational number $\Rightarrow B$ is reflexive
$\frac{\mathrm{x}}{\mathrm{y}}=\alpha$ is rational $\Rightarrow \frac{\mathrm{y}}{\mathrm{x}}$ need not be rational
i.e., $\frac{0}{1}$ is rational $\Rightarrow \frac{1}{0}$ is not rational
Hence B is not symmetric
$\Rightarrow B$ is not an equivalence relation.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2011
Let $R$ denote the set of all real numbers and $R^{+}$denote the set of all positive real numbers. For the subsets $A$ and $B$ of $R$ define $f: A \rightarrow B$ by $f(x)=x^2$ for $x \in A$. Observe the two lists given below

Options:
A

B

C

D

The correct answer is:

Solution:
Given, $f(x)=-x^2, f: A \rightarrow B$
where, $A$ and $B \in R$
$R \rightarrow$ Real No's $x \in A$
Here, the domain of the function is positive real number only
ie, Domain $=A \in R^{+}$
Example $\{1,2,3, \ldots\}$ and $B=\{1,4,9, \ldots\}$
Both set have $f: A \rightarrow B$ unique image.
But, $A=\{-1,1,2, \ldots\}$ and $B=\{1,1,4, \ldots\}$
In $f: A \rightarrow B$ have not unique image and for Range $x^2=y, x= \pm \sqrt{y}$
ie, (Range $\in R^{+}$)
In onto function the (Range $=$ Co-domain $=B$ ) ie, $\left(B \in R^{+}\right)$
So,
(A) $f$ is one-one and onto, if $A=B=R^{+}$
(B $f$ is one-one but not onto, if $A=R^{+}, B=R$
(C) $f$ is onto but not one-one, if $A=R, B=R^{+}$
(D) $f$ is neither one-one nor onto, if $A=B=R$
Hence, the answer is
$(\mathrm{A}) \rightarrow 4,(\mathrm{~B}) \rightarrow 1,(\mathrm{C}) \rightarrow 3,(\mathrm{D}) \rightarrow 2$

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2010
Let $R$ denote the set of all real numbers and $R^{+}$denote the set of all positive real numbers. For the subsets $A$ and $B$ of $R$ define $f: A \rightarrow B$ by $f(x)=x^2$ for $x \in A$. Observe the two lists given below

Options:
A

B

C

D

The correct answer is:

Solution:
Given, $f(x)=-x^2, f: A \rightarrow B$
where, $A$ and $B \in R$
$R \rightarrow$ Real No's $x \in A$
Here, the domain of the function is positive real number only
ie, Domain $=A \in R^{+}$
Example $\{1,2,3, \ldots\}$ and $B=\{1,4,9, \ldots\}$
Both set have $f: A \rightarrow B$ unique image.
But, $A=\{-1,1,2, \ldots\}$ and $B=\{1,1,4, \ldots\}$
In $f: A \rightarrow B$ have not unique image and for Range $x^2=y, x= \pm \sqrt{y}$
ie, (Range $\in R^{+}$)
In onto function the (Range $=$ Co-domain $=B$ ) ie, $\left(B \in R^{+}\right)$
So,
(A) $f$ is one-one and onto, if $A=B=R^{+}$
(B $f$ is one-one but not onto, if $A=R^{+}, B=R$
(C) $f$ is onto but not one-one, if $A=R, B=R^{+}$
(D) $f$ is neither one-one nor onto, if $A=B=R$
Hence, the answer is
$(\mathrm{A}) \rightarrow 4,(\mathrm{~B}) \rightarrow 1,(\mathrm{C}) \rightarrow 3,(\mathrm{D}) \rightarrow 2$

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2010
Let $\lambda, \mu \in \mathbf{R}$. If the system of equations
$\begin{aligned}
& 3 x+5 y+\lambda z=3 \\
& 7 x+11 y-9 z=2 \\
& 97 x+155 y-189 z=\mu
\end{aligned}$
has infinitely many solutions, then $\mu+2 \lambda$ is equal to :

Options:
A 24
B 25
C 22
D 27

The correct answer is:
25

Solution:
$\begin{array}{l}
3 x+5 y+\lambda z=3 \\
7 x+11 y-9 z=2 \\
97 x+155 y-189 z=\mu \\
93 x+155 y+31 \lambda z=93 \\
97 x+155 y-189 z=\mu \\
-\quad+\quad+\quad- \\
\hline-4 x+(31 \lambda+189) z=93-\mu \\
1085 x+1705 y-1395 z=310 \\
1067 x+1705 y-2079 z=11 \mu \\
-\quad+\quad-\quad \\
\hline 18 x+684 z=310-11 \mu
\end{array}$
$\begin{aligned}
& -36 x+9(31 \lambda+189) z=9(93-\mu) \\
& 36 x+1368 z=2(310-11 \mu) \\
& (279 \lambda+3069) z=1457-31 \mu
\end{aligned}$
for infinite solutions -
$\begin{aligned}
& \lambda=\frac{-3069}{279}=\frac{-341}{31} \\
& \mu=\frac{1457}{31}
\end{aligned}$
$\mu+2 \lambda=\frac{1457-682}{31}=\frac{775}{31}=25$

Topic: Mathematics, Determinants, JEE Main, JEE Main 2024 (09 Apr Shift 1)
Let $\alpha, \beta \in \mathbf{R}$. Let the mean and the variance of 6 observations $-3,4,7,-6, \alpha, \beta$ be 2 and 23 , respectively. The mean deviation about the mean of these 6 observations is :

Options:
A $\frac{13}{3}$
B $\frac{16}{3}$
C $\frac{11}{3}$
D $\frac{14}{3}$

The correct answer is:
$\frac{13}{3}$

Solution:
$\begin{aligned}
& \frac{\sum x_i}{6}=2 \text { and } \frac{\sum x_i^2}{N}-\mu^2=23 \\
& \alpha+\beta=10 \\
& \alpha^2+\beta^2=52
\end{aligned}$
solving we get $\alpha=4, \beta=6$
$\frac{\sum\left|x_i-\bar{x}\right|}{6}=\frac{5+2+5+8+2+4}{6}=\frac{13}{3}$

Topic: Mathematics, Statistics, JEE Main, JEE Main 2024 (04 Apr Shift 1)
Let R={P,Q|P and Q are at the same distance from the origin} be a relation, then the equivalence class of 1,-1 is the set

Options:
A S=x,y|x2+y2=1
B S=x,y|x2+y2=2
C S=x,y|x2+y2=2
D S=x,y|x2+y2=4

The correct answer is:
S=x,y|x2+y2=2

Solution:
Equivalence class of 1,-1 is a circle with centre at 0,0and radius =2

⇒x2+y2=2

S=x,y|x2+y2=2

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2021 (26 Feb Shift 1)
Let r∈P,q,~p,~q be such that the logical statement r∨~p⇒p∧q∨r is a tautology. Then r is equal to

Options:
A p
B q
C ~p
D ~q

The correct answer is:
~p

Solution:
Solving all option one by one we get following truth table,

(1)

p=r	q	~p	r∨~p	p∧q	p∧q∨r	r∨~p⇒p∧q∨r

T	F	F	T	F	T	T

F	T	T	T	F	F	F

T	T	F	T	T	T	T

F	F	T	T	F	F	T

(2)

p	~p	r∨~p	q=r	p∧q	p∧q∨r	r∨~p⇒p∧q∨r

T	F	T	T	T	T	T

F	T	T	T	F	T	T

T	F	F	F	F	F	T

F	T	T	F	F	F	F

(3)

p	q	r=~p	r∨~p	p∧q	p∧q∨r	r∨~p⇒p∧q∨r

T	T	F	F	T	T	T

F	T	T	T	F	T	T

T	F	F	F	F	F	T

F	F	T	T	F	T	T

(4)

~p	p	q	r∨~p	r=~q	p∧q	p∧q∨r	r∨~p⇒p∧q∨r

F	T	T	F	F	T	T	T

F	T	F	T	T	F	T	T

T	F	T	T	F	F	F	F

T	F	F	T	T	F	T	T

Now final answer is option C.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2022 (26 Jun Shift 2)
Let λ∈R. The system of linear equations
2x1-4x2+λx3=1
x1-6x2+x3=2
λx1-10x2+4x3=3
is inconsistent for :

Options:
A exactly one positive value of λ
B exactly one negative value of λ
C every value of λ
D exactly two values of λ

The correct answer is:
exactly one negative value of λ

Solution:
Here, we will use Cramer's Rule according to which, D=0 and, D1≠0 if the system of linear equations is inconsistent. D=2-4λ1-61λ-104=2-24+10+44-λ+λ-10+6λ

=23λ2-7λ-6=2λ-33λ+2

D=0⇒λ=3,-23

D1=1-4λ2-613-104=1-24+10+48-3+λ-20+18=23-λ

For λ=-23,D1≠0

Topic: Mathematics, Determinants, JEE Main, JEE Main 2020 (05 Sep Shift 1)
Let $r(x)$ be the remainder when the polynomial $x^{135}+x^{125}-x^{185}+x^{5}+1$ is divided by $x^{3}-x .$ Then

Options:
A $r(x)$ is the zero polynomial
B $r(x)$ is a nonzero constant
C degree of $r(x)$ is one
D degree of $r(x)$ is two

The correct answer is:
degree of $r(x)$ is one

Solution:
$\mathrm{x}^{135}+\mathrm{x}^{125}-\mathrm{x}^{115}+\mathrm{x}^{5}+1=\mathrm{k}\left(\mathrm{x}^{3}-\mathrm{x}\right)+\mathrm{A} \mathrm{x}^{2}+\mathrm{Bx}+\mathrm{C}$
put $\mathrm{x}=0 \quad \mathrm{C}=1$
Put $\mathrm{x}=1$
$3=\mathrm{A}+\mathrm{B}+1$
$A+B=2$ .........(i)
put $x=-1$
$-1=A-B+1$
$A-B=-2$ .........(ii)
(i) $+(i i)$
$A=0$
$B=2$

Topic: Mathematics, Quadratic Equation, KVPY, KVPY 2017 (5 Nov SA)
Let $\mathrm{R}=\left\{(x, y): x, y \in N\right.$ and $\left.x^2-4 x y+3 y^2=0\right\}$, where $N$ is the set of all natural numbers. Then the relation $R$ is :

Options:
A
reflexive but neither symmetric nor transitive.

B
symmetric and transitive.

C
reflexive and symmetric,

D
reflexive and transitive.

The correct answer is:
reflexive and transitive.

Solution:
$\mathrm{R}=\left\{(x, y): x, y \in \mathrm{N}\right.$ and $\left.x^2-4 x y+3 y^2=0\right\}$
Now, $x^2-4 x y+3 y^2=0$
$\Rightarrow(x-y)(x-3 y)=0$
$\therefore x=y$ or $x=3 y$
$\therefore \quad \mathrm{R}=\{(1,1),(3,1),(2,2),(6,2),(3,3)$, $(9,3), \ldots \ldots\}$
Since $(1,1),(2,2),(3,3), \ldots \ldots$ are present in the relation, therefore $\mathrm{R}$ is reflexive.
Since $(3,1)$ is an element of $R$ but $(1,3)$ is not the element of $\mathrm{R}$, therefore $\mathrm{R}$ is not symmetric
Here $(3,1) \in R$ and $(1,1) \in R \Rightarrow(3,1) \in R$ $(6,2) \in \mathrm{R}$ and $(2,2) \in \mathrm{R} \Rightarrow(6,2) \in \mathrm{R}$
For all such $(a, b) \in \mathrm{R}$ and $(b, c) \in \mathrm{R}$ $\Rightarrow(a, c) \in \mathrm{R}$
Hence $\mathrm{R}$ is transitive.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2013 (23 Apr Online)
Let \(\rho\) be a relation defined on set of natural numbers \(\mathbb{N}\), as \(\rho=\{(x, y) \in \mathbb{N} \times \mathbb{N}: 2 x+y=41\}\). Then domain \(A\) and range B are

Options:
A \(\mathrm{A} \subset\{\mathrm{x} \in \mathbb{N}: 1 \leq \mathrm{x} \leq 20\}\) and \(B \subset\{\mathrm{y} \in \mathbb{N}: 1 \leq \mathrm{y} \leq 39\}\)
B \(A=\{x \in \mathbb{N}: 1 \leq x \leq 15\}\) and \(B=\{y \in \mathbb{N}: 2 \leq \mathrm{y} \leq 30\}\)
C \(\mathrm{A} \equiv \mathbb{N}, \mathrm{B} \equiv \mathbb{Q}\)
D \(A \equiv \mathbb{Q}, B \equiv \mathbb{Q}\)

The correct answer is:
\(\mathrm{A} \subset\{\mathrm{x} \in \mathbb{N}: 1 \leq \mathrm{x} \leq 20\}\) and \(B \subset\{\mathrm{y} \in \mathbb{N}: 1 \leq \mathrm{y} \leq 39\}\)

Solution:
\(\begin{aligned} \text {Hint: } &
2 x +y=41 \\
& y=41-2 x \\
\therefore & x=\{1,2,3,4, \ldots .20\} \\
& y=\{1,3,5, \ldots .39\}
\end{aligned}\)

Topic: Mathematics, Functions, WBJEE, WBJEE 2023
Let S=0,2π-π2,3π4,3π2,7π4. Let y=yx, x∈S, be the solution curve of the differential equation dydx=11+sin2x, yπ4=12. If the sum of abscissas of all the points of intersection of the curve y=yx with the curve y=2sinx is kπ12, then k is equal to _____.

Options:

The correct answer is:
42

Solution:
Given differential equation dydx=11+sin2x

∫dy=∫dxsinx+cosx2

∫dy=∫sec2x1+tanx2

y=-11+tanx+C

So yπ4=12

So 12=-12+C⇒C=1

So y=-11+tanx+1

y=-1+1+tanx1+tanx

y=tanx1+tanx

Solving this equation with y=2sinx

tanx1+tanx=2sinx

i.e. sinx=0, 12=sinx+cosx

x=π

or 12=sinx+π4

⇒sinπ6=sinx+π4

x+π4=π-π6,2π+π6

x=5π6-π4, x=13π6-π4

x=7π12, x=23π12

Hence sum of abscissas of all the points of intersection

=π+7π12+23π12

=12π+7π+2312=42π12

kπ12=42π12⇒k=42

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2022 (26 Jun Shift 1)
Let S={1, 2, 3, 4, 5, 6}. Then the probability that a randomly chosen onto function g from S to S satisfies g3=2 g1 is :

Options:
A 115
B 15
C 130
D 110

The correct answer is:
110

Solution:
Given:S=1,2,3,4,5,6

Total number of onto functions from S tto S=6!

Now, for g3=2g1:

g3g1214263

∴g3=2g1can be defined in 3 ways.

g2,g4,g5 & g6 can be anything and can be defined in 4! ways.

∴Number of onto functions for which g3=2 g1=3.4!

Now required probability=3.4!6!=3×4!30×4!=110

Topic: Mathematics, Probability, JEE Main, JEE Main 2021 (31 Aug Shift 2)
Let S=1,2,3,5,7,10,11. The number of non-empty subsets of S that have the sum of all elements a multiple of 3, is _____ .

Options:

The correct answer is:
43

Solution:
Given set is

S=1,2,3,5,7,10,11

Elements of the type 3k is 3.

Elements of the type 3k+1 are 1,7,9.

Elements of the type 3k+2 are 2,5,11

Subset containing one element S1=1

Subsets containing two elements S2

=C13×C13=9

Subsets containing three elements S3

=C13×C13+1+1=11

Subsets containing four elements S4

=C33+C33+C23×C23=11

Subsets containing five elements S5

=C23×C23×1=9

Subsets containing six elements S6=1

Subsets containing seven elements S7=1

Required number

=1+9+11+11+9+1+1

=43

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (25 Jan Shift 1)
Let S={1,2,3,…,10}. Suppose M is the set of all the subsets of S, then the relation R={(A,B):A∩B≠ϕ;A,B∈M} is :

Options:
A symmetric and reflexive only
B reflexive only
C symmetric and transitive only
D symmetric only

The correct answer is:
symmetric only

Solution:
Given: S={1,2,3,…,10} and

R={(A,B):A∩B≠ϕ;A,B∈M}

For Reflexive, M is subset of 'S'

⇒ϕ∈M for ϕ∩ϕ=ϕ

But it is given that relation is A∩B≠ϕ.

So, it is not reflexive.

For symmetric,

If RBA⇒A∩B≠ϕ, then RAB⇒B∩A≠ϕ,

So, it is symmetric.

For transitive,

If A={(1,2),(2,3)}, B={(2,3),(3,4)} then C={(3,4),(5,6)}.

⇒RBA, RCB is applicable but RCA is not followed.

So, it not transitive

Hence, the given relation is symmetric only.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2024 (27 Jan Shift 1)
Let S=1, 2, 3, ….,100, then number of non-empty subsets A of S such that the product of elements in A is even is :

Options:
A 2100-1
B 250+1
C 250250-1
D 250-1

The correct answer is:
250250-1

Solution:
As we know that the product of numbers is even when at least one of the number must be even.

Hence total subsets of A in which product of numbers is even = Total subsets – total subsets in which all the elements are odd

=2100-250=250250-1.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2019 (12 Jan Shift 1)
Let $\mathrm{S}=\{1,2, \ldots . ., 20\}$. A subset $\mathrm{B}$ of $\mathrm{S}$ is said to be "nice", if the sum of the elements of $\mathrm{B}$ is 203 . Than the probability that a randomly chosen subset of $S$ is "nice" is :

Options:
A $\frac{7}{2^{20}}$
B $\frac{5}{2^{20}}$
C $\frac{4}{2^{20}}$
D None of the above

The correct answer is:
$\frac{5}{2^{20}}$

Solution:
Since total number of subsets of the set $S=2^{20}$

Now, the sum of all number from 1 to $20=\frac{20 \times 21}{2}=210$ Then, find the sets which has sum 7 .

(1) $\{7\}$

(2) $\{1,6\}$

(3) $\{2,5\}$

(4) $\{3,4\}$

(5) $\{1,2,4\}$

Then, there is only 5 sets which has sum 203

Hence required probability $=\frac{5}{2^{20}}$

Topic: Mathematics, Probability, JEE Main, JEE Main 2019 (11 Jan Shift 2)
Let S=−1, ∞ and f:S→ℝ be defined as fx=∫−1xet−1112t−15t−27t−3122t−1061dt. Let p= Sum of square of the values of x, where fx attains local maxima on S. and q=Sum of the values of x, where fx attains local minima on S. Then, the value of p2+2q is ________

Options:

The correct answer is:
27

Solution:
Given,

fx=∫−1xet−1112t−15t−27t−3122t−1061dt

Now, differentiating the above integral using Newton Leibnitz's theorem, we get,

f'x=ex−1112x−15x−27x−3122x−1061

Now, from the above diagram using first derivative test we get,

Local minima at x=12, x=5

Local maxima at x=0, x=2

So, p=0+4=4, q=12+5=112

Hence, p2+2q=16+11=27

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2024 (31 Jan Shift 1)
Let S1 and S2 be respectively the sets of all a∈R-0 for which the system of linear equations
ax+2ay-3az=1

2a+1 x+2a+3 y+a+1z=2

3a+5 x+a+5 y+a+2 z=3

has unique solution and infinitely many solutions. Then

Options:
A nS1=2 and S2 is an infinite set
B S1 is an infinite set an nS2=2
C S1=ϕ and S2=ℝ-0
D S1=ℝ-0 and S2=ϕ

The correct answer is:
S1=ℝ-0 and S2=ϕ

Solution:
Given,

S1 and S2 be respectively the sets of all a∈R-0 for which the system of linear equations

ax+2ay-3az=1 .........1

2a+1 x+2a+3 y+a+1z=2 ....2

3a+5 x+a+5 y+a+2 z=3 .....3

Now from above equations finding Δ=a2a-3a2a+12a+3a+13a+5a+5a+2

=a15a2+31a+36=0⇒a=0

As 15a2+31a+36 cannot be zero as 312-4×15×36<0

So, Δ≠0 for all a∈R-0

Hence, S1=R-0 & S2=ϕ

Topic: Mathematics, Determinants, JEE Main, JEE Main 2023 (25 Jan Shift 1)
Let S1 be the sum of first 2n terms of an arithmetic progression. Let S2 be the sum of first 4n terms of the same arithmetic progression. If S2-S1 is 1000, then the sum of the first 6n terms of the arithmetic progression is equal to:

Options:
A 1000
B 7000
C 5000
D 3000

The correct answer is:
3000

Solution:
S2n=2n22a+2n-1d, S4n=4n22a+4n-1d

⇒S2-S1=4n2[2a+(4n-1)d]-2n22a+2n-1d

=4an+(4n-1)2nd-2na-(2n-1)dn

=2na+nd[8n-2-2n+1]

⇒2na+nd[6n-1]=1000

2a+6n-1d=1000n

Now, S6n=6n22a+6n-1d

=3n·1000n=3000

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Let S1 is the complete solution set of the inequality cos-1x>cos-1x2 and S2 is the complete solution set of the inequality cot-1x2-5cot-1x+6>0, then S1∩S2 is

Options:
A -1,0
B cot3,0
C cot2,0
D -1,cot2

The correct answer is:
cot2,0

Solution:
cos-1x>cos-1x2 ⇒x<x2 & x∈-1,1
{ ∵ cos-1x is a decreasing function}
⇒ xx–1>0 & x∈-1,1
⇒ x∈-∞,0∪1,∞ & x∈-1,1
⇒x∈-1,0 … 1
Now cot-1x2-5cot-1x+6>0
⇒cot-1x-2cot-1x-3>0
⇒cot-1x<2 or x<cot3 … 2
{ ∵ cot-1x is a decreasing function}
1∩2⇒cot2,0

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
Let $\mathrm{S}_1=\sum_{\mathrm{j}=1}^{10} \mathrm{j}(\mathrm{j}-1)^{10} \mathrm{C}_{\mathrm{j}}, \mathrm{S}_2=\sum_{\mathrm{j}=1}^{10} \mathrm{j}^{10} \mathrm{C}_{\mathrm{j}}$ and $\mathrm{S}_3=\sum_{\mathrm{j}=1}^{10} \mathrm{j}^2{ }^{10} \mathrm{C}_{\mathrm{j}}$.
Statement-1: $\mathrm{S}_3=55 \times 2^9$
Statement-2: $S_1=90 \times 2^8$ and $S_2=10 \times 2^8$.

Options:
A
Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1

B
Statement-1 is true, Statement-2 is false

C
Statement-1 is false, Statement-2 is true

D
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1

The correct answer is:
Statement-1 is true, Statement-2 is false

Solution:
$S_1=\sum_{j=1}^{10} j(j-1) \frac{10 !}{j(j-1)(j-2) !(10-j) !}=90 \sum_{j=2}^{10} \frac{8 !}{(j-2) !(8-(j-2)) !}=90 \cdot 2^8$.
$S_2=\sum_{j=1}^{10} j \frac{10 !}{j(j-1) !(9-(j-1)) !}=10 \sum_{j=1}^{10} \frac{9 !}{(j-1) !(9-(j-1)) !}=10 \cdot 2^9$.
$S_3=\sum_{j=1}^{10}[j(j-1)+j] \frac{10 !}{j !(10-j) !}=\sum_{j=1}^{10} j(j-1)^{10} C_j=\sum_{j=1}^{10} j^{10} C_j=90 \cdot 2^8+10 \cdot 2^9$
$=90 \cdot 2^8+20 \cdot 2^8=110 \cdot 2^8=55 \cdot 2^9$.

Topic: Mathematics, Binomial Theorem, JEE Main
Let $\mathrm{S}_1=\sum_{\mathrm{j}=1}^{10} \mathrm{j}(\mathrm{j}-1)^{10} \mathrm{C}_{\mathrm{j}}, \mathrm{S}_2=\sum_{\mathrm{j}=1}^{10} \mathrm{j}^{10} \mathrm{C}_{\mathrm{j}}$ and $\mathrm{S}_3=\sum_{\mathrm{j}=1}^{10} \mathrm{j}^2{ }^{10} \mathrm{C}_{\mathrm{j}}$.
Statement-1: $\mathrm{S}_3=55 \times 2^9$
Statement-2: $S_1=90 \times 2^8$ and $S_2=10 \times 2^8$.

Options:
A
Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1

B
Statement-1 is true, Statement-2 is false

C
Statement-1 is false, Statement-2 is true

D
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1

The correct answer is:
Statement-1 is true, Statement-2 is false

Solution:
$S_1=\sum_{j=1}^{10} j(j-1) \frac{10 !}{j(j-1)(j-2) !(10-j) !}=90 \sum_{j=2}^{10} \frac{8 !}{(j-2) !(8-(j-2)) !}=90 \cdot 2^8$.
$S_2=\sum_{j=1}^{10} j \frac{10 !}{j(j-1) !(9-(j-1)) !}=10 \sum_{j=1}^{10} \frac{9 !}{(j-1) !(9-(j-1)) !}=10 \cdot 2^9$.
$S_3=\sum_{j=1}^{10}[j(j-1)+j] \frac{10 !}{j !(10-j) !}=\sum_{j=1}^{10} j(j-1)^{10} C_j=\sum_{j=1}^{10} j^{10} C_j=90 \cdot 2^8+10 \cdot 2^9$
$=90 \cdot 2^8+20 \cdot 2^8=110 \cdot 2^8=55 \cdot 2^9$.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2010
Let S1, S2 and S3 be three sets defined as

S1={z∈ℂ:|z-1|≤2},

S2={z∈ℂ:Re((1-i)z)≥1} and

S3={z∈ℂ:Im(z)≤1}.

Then, the set S1∩S2∩S3

Options:
A is a singleton
B has exactly two elements
C has infinitely many elements
D has exactly three elements

The correct answer is:
has infinitely many elements

Solution:
We know that the complex number z satisfying z-z0=r represents a circle with centre z0 and radius r units.

Hence, for S1=|z-1|≤2, z lies on and inside the circle
of radius 2 units and centre (1, 0).

For S2, let z=x+iy

Now, (1-i)(z)=(1-i)(x+i y)

⇒(1-i)(z)=x+iy-ix-i2y

⇒(1-i)(z)=x+iy-ix+y

⇒Re((1-i)z)=x+y

⇒x+y≥1.

And, for S3, again let z=x+iy,

⇒y≤1.

Plotting all the inequalities on the graph, we get

Now, the common part is shown in the shaded part, hence, we get infinite points in the shaded region.

⇒S1∩S2∩S3 has infinitely many elements.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2021 (17 Mar Shift 2)
Let s1,s2,s3....,s10 respectively be the sum of 12 terms of 10 A.Ps whose first terms are 1, 2, 3,....,10 and the common differences are 1, 3, 5,...,19 respectively. Then ∑i=110si is equal to

Options:
A 7220
B 7360
C 7260
D 7380

The correct answer is:
7260

Solution:
Given,

First term of A.P are 1,2,3....10, so general term of the first term will be i

And common difference are 1,3,5,...., so general term of common difference is given by 2i-1

Now sum of the A.P is given by,

Si=1222×i+12-12i-1

⇒Si=62×i+112i-1

⇒Si=144i-66

So, ∑i=110Si=∑i=110144i-66∑i=1101

⇒∑i=110Si=14410×112-66×10

⇒∑i=110Si=79210-66×10

⇒∑i=110Si=7260

Hence this is the correct option.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (13 Apr Shift 1)
Let S1 : x2+y2=9 and S2 : (x-2)2+y2=1.

Then the locus of center of a variable circle S which touches S1 internally and S2 externally always passes through the points :

Options:
A (0,±3)
B 12,±52
C 2,±32
D (1,±2)

The correct answer is:
2,±32

Solution:
∵c1c2=r1-r2

∴ Given circles are touching internally

Let a variable circle with centre P and radius r

⇒PA=r1-r and PB=r2+r

⇒PA+PB=r1+r2

⇒PA+PB=4(>AB)

⇒ Locus of P is an ellipse with foci at A(0,0) and B(2,0) and length of major axis is 2a=4, e=12

⇒ Centre is at (1,0) and b2=a21-e2=3

if x-axis is the axis of ellipse then

⇒E : (x-1)24+y23=1

which is satisfied by 2,±32

Topic: Mathematics, Circle, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Let S=θ∈-2π,2π:2cos2θ+3sinθ=0. Then the sum of the elements of S is:

Options:
A π
B 13π6
C 5π3
D 2π

The correct answer is:
2π

Solution:
The given equation 2cos2θ+3sinθ=0 can be written as

21-sin2θ+3sinθ=0

⇒2sin2θ-3sinθ-2=0

⇒2sinθ+1sinθ-2=0

⇒sinθ=2, -12, sinθ≠2

⇒sinθ=-12

Hence, solutions in -2π, 2π are
-π+π6, -π6, π+π6, 2π-π6

Sum of the elements is =2π.

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2019 (09 Apr Shift 1)
Let S=2+67+1272+2073+3074+….. then 4S is equal to

Options:
A 722
B 733
C 73
D 734

The correct answer is:
733

Solution:
S=2+67+1272+2073+⋯ ...i

S7=0+27+672+1273+2074+⋯ ...ii

On subtracting the equation ii from equation i, we get

6S7=2+47+672+873+⋯ ...iii

6S49=0+27+472+673+⋯ ...iv

On subtracting the equation iv from equation iii , we get

67-649S=2+27+272+273+⋯

3649S=21+17+172+⋯∞

⇒ S=2·11-17·4936

⇒ 4S=733

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2022 (27 Jun Shift 2)
Let S=θ∈[-π,π]-±π2:sinθtanθ+tanθ=sin2θ. If T=∑θ∈Scos2θ, then T+nS is equal to

Options:
A 7+3
B 5
C 8+3
D 9

The correct answer is:
9

Solution:
Given sin θ tan θ+tan θ=sin 2θ; θ≠±π2

⇒sin2 θcos θ+sin θcos θ=2 sin θ cos θ

⇒sin θ=0 or sin θcos θ+1cos θ=2 cos θ

⇒θ=0, π, -π or sin θ+1=21-sin2 θ

⇒θ=0, π, -π or 2 sin2 θ +sin θ-1=0

⇒θ=0, π, -π or sin θ=12, sin θ=-1

⇒θ=0, π, -π or θ=π6, 5π6 or θ=-π2 (rejected)

Hence, θ=0, π, -π, π6, 5π6

⇒nS=5

Now, T=Σcos 2θ

=cos 0+cos 2π+cos-2π+cosπ3+cos5π3

=1+1+1+12+12=4

So, T+nS=9

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2022 (24 Jun Shift 1)