datavorous/entrance-exam-dataset huggingface dataset:
Let S=109+1085+10752+….+25107+15108. Then the value of 16S-(25)-54 is equal to

Options:

The correct answer is:
2175

Solution:
Given,

S=109+1085+10752+…+25107+15108 ....1

S5=1095+10852+…+…+25108+15109 ....2

Now subtracting above equations we get,
-4S5=-109+15+152+…+15108+15109

⇒-4S5=-109+151-151091-15

⇒4S5=-109+141-15109

⇒16S5=-436+1-15109

⇒16S=-436×5+5-5-108

⇒16S-(25)-54=2175

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (11 Apr Shift 1)
Let $S=\left\{\left(\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right): a_{i j} \in\{0,1,2\}, a_{11}=a_{22}\right\}$
Then the number of non-singular matrices in the set $S$ is :

Options:
A
27

B
24

C
10

D
20

The correct answer is:
20

Solution:
The matrices in the form
$$
\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right], a_{i j} \in\{0,1,2\}, a_{11}=a_{12} \text { are }
$$
$$
\left[\begin{array}{cc}
0 & 0 / 1 / 2 \\
0 / 1 / 2 & 0
\end{array}\right],\left[\begin{array}{cc}
1 & 0 / 1 / 2 \\
0 / 1 / 2 & 1
\end{array}\right],\left[\begin{array}{cc}
2 & 0 / 1 / 2 \\
0 / 1 / 2 & 2
\end{array}\right]
$$
At any place, $0 / 1 / 2$ means 0,1 or 2 will be the element at that place.
Hence there are total $27=3 \times 3+3 \times 3+3 \times 3$ ) matrices of the above form. Out of which the matrices which are singular are
$$
\left[\begin{array}{cc}
0 & 0 / 1 / 2 \\
0 & 0
\end{array}\right],\left[\begin{array}{cc}
0 & 0 \\
1 / 2 & 0
\end{array}\right],\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right],\left[\begin{array}{ll}
2 & 2 \\
2 & 2
\end{array}\right]
$$
Hence there are total $7(=3+2+1+1)$ singular matrices.
Therefore number of all non-singular matrices in the given form $=27-7=20$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2013 (25 Apr Online)
Let $S=\{(a, b, c) \in N \times N \times N: a+b+c=21$, $a \leq b \leq c\}$ and $T=\{(a, b, c) \in N \times N \times N: a, b, c$ are in $A P \}$ where $N$ is the set of all natural numbers. Then, the number of elements in the set $\mathrm{S} \cap T$

Options:
A 6
B 7
C 13
D 14

The correct answer is:
7

Solution:
We have, $a+b+c=21$ and $\frac{a+c}{2}=b$ $\Rightarrow \quad a+c+\frac{a+c}{2}=21$
$\begin{array}{ll}\Rightarrow & a+c=14 \Rightarrow \frac{a+c}{2}=7 \\ \Rightarrow & b=7\end{array}$
So, a can take values from 1 to 6 . C can have values from 8 to 13
or a=b=c=7
So, there are 7 such triplets.

Topic: Mathematics, Sets and Relations, JEE Main
Let Sa denote the sum of first n terms an arithmetic progression. If S20=790 and S10=145, then S15- S5 is :

Options:
A 395
B 390
C 405
D 410

The correct answer is:
395

Solution:
Given,

In an arithmetic progression sum of first 20 terms is 790

So, let a be the first term and d be common difference,

So, S20=2022a+19d=790 .....1

And sum of the first 10 terms is 145

So, S10=1022a+9d=145 .....2

By solving equation 1 & 2 we get,

a=-8 & d=5

Then, S15-S5=1522×-8+14×5-522×-8+4×5

⇒S15-S5=405-10=395

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2024 (30 Jan Shift 1)
Let S and S â² are the focii of the ellipse x=3+5cosâ¡Î¸, y=-2+4sinâ¡Î¸. If B is one of the ends of one of the latus rectum, then the area (in sq. units) of the triangle BS S â² is equal to

Options:
A 245
B 485
C 125
D 645

The correct answer is:
485

Solution:
Assume the base of ΔS S ′ B is S S ′ , then
base =S S ′ =2ae
height =122b2a=length of latus rectum2
⇒area of ΔBS S ′ = 1 2 2ae b 2 a =b2e
Now, equation of the ellipse is x-3225+y+2216=1
⇒a2=25,b2=16,e2=925
⇒b2e=16×35=485

Hence, area of ΔBS S ′ =485 sq. units

Topic: Mathematics, Ellipse, JEE Main
Let S and S' be the foci of an ellipse and B be any one of the extremities of its minor axis. If ΔS'BS is a right angled triangle with right angle at B and area ΔS'BS=8 sq.units, then the length of a latus rectum of the ellipse is :

Options:
A 2 2
B 2
C 4
D 42

The correct answer is:
4

Solution:
Given that,

∠SBS'=π2

⇒∠OBS=π4

⇒b=ae ...i

Also, area ∆SBS'=8

⇒12.2ae.b=8

⇒aeb=8 ...ii

From equations i & ii,

b2=8 and a2=b2+a2e2=8+8=16

⇒b=22 and a=4

Thus, the length of the latus rectum is 2b2a=164=4

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2019 (12 Jan Shift 2)
Let $S$ be a non-empty subset of R. Consider the following statement:
$\mathrm{P}$ : There is a rational number $\mathrm{x} \in \mathrm{S}$ such that $\mathrm{x}>0$.
Which of the following statements is the negation of the statement $P$ ?

Options:
A
There is no rational number $x \in S$ such that $x \leq 0$

B
Every rational number $x \in S$ satisfies $x \leq 0$

C
$x \in S$ and $x \leq 0 \Rightarrow x$ is not rational

D
There is a rational number $x \in S$ such that $x \leq 0$

The correct answer is:
Every rational number $x \in S$ satisfies $x \leq 0$

Solution:
$P$ : there is a rational number $x \in S$ such that $x>0$
$\sim P$ : Every rational number $\mathrm{x} \in \mathrm{S}$ satisfies $\mathrm{x} \leq 0$

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2010
Let $S$ be the circle in $x y$-plane which touches the $x$-axis at point $A$, the $y$-axis at point $B$ and the unit circle $x^{2}+y^{2}=1$ at point $C$ externally. If $O$ denotes the origin, then the angle OCA equals

Options:
A $\frac{5 \pi}{8}$
B $\frac{\pi}{2}$
C $\frac{3 \pi}{4}$
D $\frac{3 \pi}{5}$

The correct answer is:
$\frac{5 \pi}{8}$

Solution:
$\therefore$ Slope of $O M=1$ $\therefore \angle \mathrm{COA}=45^{\circ}$ $\mathrm{C}\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
Now OM $=O C+C M$
$\sqrt{2} \mathrm{~h}=1+\mathrm{h}$
Squaring
$\begin{array}{l}
2 h^{2}=h^{2}+1+2 h \\
h^{2}-2 h-1=0
\end{array}$
$\mathrm{h}=\frac{2 \pm \sqrt{8}}{2} \Rightarrow \mathrm{h}=1 \pm \sqrt{2}$
$\mathrm{~A}(1+\sqrt{2}, 0)$
Slope of $A C=\frac{0-\frac{1}{\sqrt{2}}}{1+\sqrt{2}-\frac{1}{\sqrt{2}}}=-\frac{1}{\sqrt{2}+1}=-(\sqrt{2}-1)$
$=-\tan 22 \frac{1}{2}^{\circ}=\tan \left(180-22 \frac{1}{2}\right)^{\circ}=\tan 157 \frac{1}{2}^{\circ}$
$\therefore \angle \mathrm{CAX}=157 \frac{1}{2}^{\circ}$
$\therefore \angle \mathrm{OCA}=157 \frac{1}{2}^{\circ}-45^{\circ}=112 \frac{1}{2}^{\circ}=\frac{5 \pi}{8}$

Topic: Mathematics, Circle, JEE Main
Let $\mathrm{S}$ be the focus of the hyperbola $\frac{x^2}{3}-\frac{y^2}{5}=1$, on the positive $x$-axis. Let $\mathrm{C}$ be the circle with its centre at $A(\sqrt{6}, \sqrt{5})$ and passing through the point $S$. If $O$ is the origin and $S A B$ is a diameter of $C$, then the square of the area of the triangle OSB is equal to___________

Options:

The correct answer is:
40

Solution:
Area $=\frac{1}{2}(O S) h=\frac{1}{2} \sqrt{8} 2 \sqrt{5}=\sqrt{40}$

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2024 (08 Apr Shift 2)
Let S be the focus of the hyperbola x216-y29=1 lying on the positive X- axis and P5,y1 be point on the hyperbola. Then SP=

Options:
A 14
B 34
C 94
D 54

The correct answer is:
94

Solution:
Since, point P5,y1 lies on the hyperbola x216-y29=1, therefore

5216-y129=1

⇒y129=2516-1

⇒y12=8116

⇒y1=±94

So, P≡5,±94.

Now, eccentricity of hyperbola is

e=1+916=54

Therefore, S≡5,0

Hence,

SP=5-52+0-942=94

Topic: Mathematics, Hyperbola, JEE Main
Let \( S \) be the focus of the parabola \( y^{2}=4 a x, X \) be the foot of the directrix, \( P P \) ' be a double ordinate of the curve and \( P X \) meets the curve again at \( Q \). Then the locus of point of intersection of tangents at \( P^{\prime} \) and \( Q \) is a

Options:
A line
B circle
C parabola
D None of these

The correct answer is:
line

Solution:
Consider the foot of directrix, X is -a, 0.

Now, equation of PX is y=2at-0at2+ax+a

⇒1+t2y=2tx+a

It intersects parabola again at Q.

Therefore, on solving this with parabola y2=4ax, we get

4t2x+a2 1+t22=4ax

⇒t2x+a2=ax1+t22

⇒t2x2+a2t2-ax-axt4=0

⇒xt2x-a-at2t2x-a=0

⇒x=at2, x=at2( coincide with P)

∴Qat2,2at ∵x=at2⇒y=2at

But if we observe, P'at2,-2at and Qat2,2at are extremities of the focal chord.P', Q, S are collinear

So, tangents at P' and Q intersect on the directrix of the parabola.

∴The locus of point of intersection of tangents is a straight line.

Topic: Mathematics, Parabola, JEE Main
Let $\mathrm{S}$ be the infinite sum given by $\mathrm{S}=\sum_{\mathrm{n}=0}^{\infty} \frac{\mathrm{a}_{\mathrm{n}}}{10^{2 \mathrm{n}}}$
where $\left\{a_{n}\right\}_{n \geq 0}$ is a sequence defined by $a_{0}=a_{1}=1$ and $a_{j}=20 a_{j-1}-108 a_{j-2}$ for $j \geq 2$.
If $S$ is expressed in the form $\frac{a}{b}$, where $a, b$ are coprime positive integers, then a equals

Options:
A 2017
B 2020
C 2023
D 2025

The correct answer is:
2025

Solution:
$\begin{array}{l}
a_{n}=20 a_{n-1}-108 a_{n-2} \\
\frac{a_{n}}{10^{2 n}}=\frac{20 a_{n-1}}{10^{2 n}}-\frac{108 a_{n-2}}{10^{2 n}} \\
\frac{a_{n}}{10^{2 n}}=\frac{20}{100} \frac{a_{n-1}}{10^{2(n-1)}}-\frac{108}{10000} \frac{a_{n-2}}{10^{2(n-2)}}
\end{array}$
apply summation
$\begin{array}{l}
\sum_{\mathrm{n}=2}^{\infty} \frac{\mathrm{a}_{\mathrm{n}}}{10^{2 \mathrm{n}}}=\frac{1}{5} \sum_{\mathrm{n}=2}^{\infty} \frac{\mathrm{a}_{\mathrm{n}-1}}{10^{2(\mathrm{n}-1)}}-\frac{27}{2500} \sum_{\mathrm{n}=2}^{\infty} \frac{\mathrm{a}_{\mathrm{n}-2}}{10^{2(\mathrm{n}-2)}} \\
\mathrm{S}-1-\frac{1}{100}=\frac{1}{5}(\mathrm{~S}-1)-\frac{27}{2500} \mathrm{~S} \\
\mathrm{~S}-1-\frac{1}{100}=\frac{1}{5} \mathrm{~S}-\frac{1}{5}-\frac{27}{2500} \mathrm{~S} \\
\mathrm{~S}\left(1-\frac{1}{5}+\frac{27}{2500}\right)=-\frac{1}{5}+1+\frac{1}{100}
\end{array}$
$\mathrm{~S}\left(\frac{2500-500+27}{2500}\right)=\frac{81}{100}$
$\mathrm{~S}=\frac{81 \times 25}{2027}$
$\mathrm{~S}=\frac{2025}{2027}$

Topic: Mathematics, Sequences and Series, JEE Main
Let S be the mirror image of the point Q(1,3,4) with respect to the plane 2x-y+z+3=0 and let R(3,5,γ) be a point of this plane. Then the square of the length of the line segment SR is

Options:

The correct answer is:
72

Solution:
Let point S(α,β,γ) be an image of Q(1,3,4)thenα-12=β-3-1=γ-41=-2(2-3+4+3)4+1+1=-2 ⇒S(α,β,γ)=(-3,5,2) and

point R(3,5,γ) lies on the plane 2x-y+z+3=0

put the points of R in equation of plane anf find the value
⇒6−5+γ+3=0⇒γ=−4

⇒R(3,5,−4)&S(−3,5,2)

using distance formula find

distance of SR=36+0+36⇒(SR)2=72

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (27 Aug Shift 2)
Let S be the sample space of all five digit numbers. If p is the probability that a randomly selected number from S, is a multiple of 7 but not divisible by 5, then 9p is equal to

Options:
A 1.0146
B 1.2085
C 1.0285
D 1.1521

The correct answer is:
1.0285

Solution:
Given, five-digit number,

So nS=total 5 digit number will be=9×104

Now let A: number is multiple of 7 but not divisible by 5

Smallest 5 digit divisible by 7 is 10003

Largest 5 digit divisible by 7 is 99995

Now by using nth term of A.P formula we have,

∴ 99995=10003+n-17

⇒n=12857

Now finding number which is divisible by 35

Again using A.P formula we have,

99995=10010+P-135

⇒P=2572

Now number divisible by 7 but not by 35 are

12857-2572=10285

So, probability will be P=1028590000

And 9P=1.0285

Topic: Mathematics, Probability, JEE Main, JEE Main 2022 (27 Jul Shift 1)
Let S be the set of all α,β,π<α,β<2π, for which the complex number 1-isinα1+2isinα is purely imaginary and 1+icosβ1-2icosβ is purely real. Let Zαβ=sin2α+icos2β,α,β∈S.
Then ∑α,β∈SiZαβ+1iZ¯αβ is equal to

Options:
A 3
B 3i
C 1
D 2-i

The correct answer is:
1

Solution:
Given π<α,β<2π

1-isinα1+i2sinα is purely imaginary

⇒1-isinα1-i2sinα1+4sin2α is purely imaginary

⇒1-2sin2α1+4sin2α=0

⇒sin2α=12

⇒α=5π4,7π4

Also 1+icosβ1+i-2cosβ is purely real

⇒1+icosβ1+2icosβ1+4cos2β is purely real

⇒3cosβ=0

⇒β=3π2

Now Zαβ=sin2α+icos2β

⇒Zαβ=sin5π2+icos3π=1-i

or Zαβ=sin7π2+icos3π=-1-i

∑α,β∈SiZαβ+1iZ¯αβ=i1-i+1i1+i+i-1-i+1i-1+i

=i+1-i1-i2+-i+1+i1+i2

=2-1=1

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2022 (27 Jul Shift 2)
Let S be the set of all a∈R for which the angle between the vectors u→=alogebi^-6j^+3k^ and v→=logebi^+2j^+2alogebk^,b>1 is acute. Then S is equal to

Options:
A -∞,-43
B Φ
C -43,0
D 127,∞

The correct answer is:
Φ

Solution:
Given,

u→=alogebi^-6j^+3k^ and v→=logebi^+2j^+2alogebk^

For angle to be acute u→·v→>0

⇒alogebi^-6j^+3k^.logebi^+2j^+2alogebk^>0

⇒alogeb2-12+6alogeb>0 ∀b>1

Let y=alogeb2-12+6alogeb

Now let logeb=t⇒t>0 as b>1

Now y=at2+6at-12>0,∀ t>0

Since y>0, so a>0 & D<0

⇒D=36a2+48a<0

⇒a∈-43,0 but a>0,

So, a∈ϕ

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2022 (28 Jul Shift 2)
Let S be the set of all λ, μ for which the vectors λi^-j^+k^, j^+2j^+μk^ and 3i^-4j^+5k^, where λ-μ=5, are coplanar, then ∑λ, μ∈S80λ2+μ2 is equal to

Options:
A 2210
B 2130
C 2290
D 2370

The correct answer is:
2290

Solution:
The given equations are coplanar.

⇒ λ-1112μ3-45=0

⇒λ10+4μ+15-3μ+1-10=0

⇒4λμ+10λ-3μ=5 .......i

And λ-μ=5

So, by 1

⇒45+μμ+105+μ-3μ=5

⇒4μ2+27μ+45=0

⇒4μ2+15μ+12μ+45=0

∴ μ=-3, -154

⇒λ=2, 54

∴ ∑λμ∈S80λ2+μ2=809+4+22516+2516

=8013+1258=10×229

=2290

Hence this is the correct option.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (15 Apr Shift 1)
Let S, be the set of all functions f:0,1→R, which are continuous on 0,1, and differentiable on 0,1. Then for every f in S, there exists c∈0,1, depending on f, such that.

Options:
A fc-f1<1-cf'c
B f1-fc1-c=f'c
C fc+f1<1+cf'c
D fc-f1<f'c

The correct answer is:
f1-fc1-c=f'c

Solution:
Let us consider fx, a constant function, as constant functions are continuous and differentiable everywhere.

Then, f1=fc & f'c=0

So, f1-fc1-c=f'c=0.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2020 (08 Jan Shift 2)
Let S be the set of all integer solutions x, y, z of the system of equations

x-2 y+5 z=0

-2 x+4 y+z=0

-7 x+14 y+9 z=0

such that 15≤x2+y2+z2≤150. Then, the number of elements in the set S is equal to ..........

Options:

The correct answer is:
8

Solution:
x-2y+5z=0 ...........i

-2x+4y+z=0 ..........ii

-7x+14y+9z=0 ..........iii

Applying 2× equation i+equation ii, we get

z=0

Putting z=0 in equation i, we get

x=2y

Given 15≤x2+y2+z2≤150

⇒15≤4y2+y2≤150

⇒3≤y2≤30

⇒y=±2, ±3, ±4, ±5

⇒8 solutions.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2020 (03 Sep Shift 2)
Let \( S \) be the set of all \( \lambda \in R \) for which the system of linear equations
\( 2 \mathrm{x}-\mathrm{y}+2 \mathrm{z}=2 \)
\( \mathrm{x}-2 \mathrm{y}+\lambda \mathrm{z}=-4 \)
\( \mathrm{x}+\lambda \mathrm{y}+\mathrm{z}=4 \)
has no solution. Then the set \( \mathrm{S} \)

Options:
A Contains more than two elements
B Is an empty set
C Is a singleton
D Contains exactly two elements

The correct answer is:
Contains exactly two elements

Solution:
For no. solution Δ=0 and at least one of Δ1,Δ2,Δ3 is non-zero.
Δ=2-121-2λ1λ1=-λ-12λ+1
Δ1=2-12-4-2λ4λ1=-2λ2+6λ-4
Δ=0⇒ λ=1,-12

And for λ=1,-12, ∆1≠0
Hence, S=1,-12

Topic: Mathematics, Determinants, JEE Main
Let S be the set of all passwords which are six to eight characters long, where each character is either an alphabet from A,B,C,D,E or a number from 1,2,3,4,5 with the repetition of characters allowed. If the number of passwords in S whose at least one character is a number from 1,2,3,4,5 is α×56, then α is equal to

Options:

The correct answer is:
7073

Solution:
Given total available characters is 10 and total digits are 5,

So, for 6-digit password, the number of ways =106-56

For 7-digit password, the number of ways =107-57

For 8-digit password, the number of ways =108-58

∴ Total ways =106+107+108-56+57+58

=2656+2757+2858-56-57-58

=5626+5·27+25·28-1-5-25=7073·56

Now on comparing with α×56 we get α=7073

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2022 (28 Jul Shift 1)
Let S be the set of all points in -π,π at which the function, fx=minsinx, cosx is not differentiable. Then S is a subset of which of the following?

Options:
A -3π4,-π2,π2,3π4
B -3π4,-π4,3π4,π4
C -π4,0,π4
D -π2,-π4, π4, π2

The correct answer is:
-3π4,-π4,3π4,π4

Solution:
Sharp points are at -3π4 and π4.

∴ Set of non-differentiable points of function are -3π4,π4 which is a subset of -3π4,-π4, 3π4, π4.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2019 (12 Jan Shift 1)
Let S be the set of all λ∈R for which the system of linear equations

2x-y+2z=2

x-2y+λz=-4

x+λy+z=4

has no solution. Then the set S

Options:
A Contains more than two elements
B Is an empty set
C Is a singleton
D Contains exactly two elements

The correct answer is:
Contains exactly two elements

Solution:
For no solution Δ=0 and at least one of Δ1,Δ2,Δ3 is non-zero.
Δ=2-121-2λ1λ1=-λ-12λ+1
Δ1=2-12-4-2λ4λ1=-2λ2+6λ-4
Δ=0⇒ λ=1,-12

And for λ=1,-12, ∆1≠0
Hence, S=1,-12

Topic: Mathematics, Determinants, JEE Main, JEE Main 2020 (02 Sep Shift 1)
Let S be the set of all real numbers. Then, the relation R={a, b:1+ab>0} on S is

Options:
A reflexive and symmetric but not transitive
B reflexive and transitive but not symmetric
C reflexive, transitive and symmetric
D None of the above

The correct answer is:
reflexive and symmetric but not transitive

Solution:
R={a, b:1+ab>0}
(i) (1,1)∈R

(ii) (1,2)∈R ⇒(2,1)∈R

(iii) (1,2)∈R and (2,3)∈R⇒(1,3)∈R
It is clear that the given relation on S is reflexive, symmetric but not transitive.

Topic: Mathematics, Sets and Relations, JEE Main
Let $S$ be the set of all real values of $k$ for which the system of linear equations
$$
\begin{aligned}
&x+y+z=2 \\
&2 x+y-z=3 \\
&3 x+2 y+k z=4
\end{aligned}
$$
has a unique solution. Then $S$ is

Options:
A
an empty set

B
equal to $\mathrm{R}-\{0\}$

C
equal to $\{0\}$

D
equal to $\mathrm{R}$

The correct answer is:
equal to $\mathrm{R}-\{0\}$

Solution:
The system of linear equations is:
$$
\begin{aligned}
&x+y+z=2 \\
&2 x+y-z=3 \\
&3 x+2 y+k z=4
\end{aligned}
$$
As, system has unique solution.
$$
\begin{aligned}
&\text { So, }\left|\begin{array}{ccc}
1 & 1 & 1 \\
2 & 1 & -1 \\
3 & 2 & k
\end{array}\right| \neq 0 \\
&\Rightarrow k+2-(2 k+3)+1 \neq 0 \\
&\Rightarrow k \neq 0
\end{aligned}
$$
Hence, $k \in R-\{0\} \equiv S$

Topic: Mathematics, Determinants, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
Let S be the set of all real values of k for which the system of linear equations
x+y+z=2
2x+y-z=3
3x+2y+kz=4
has a unique solution. Then, S is :

Options:
A equal to R â{0}
B an empty set
C equal to R
D equal to 0

The correct answer is:
equal to R â{0}

Solution:
For unique solution, ∆≠0

⇒11121-132k≠0
⇒1k+2-12k+3+14-3≠0

⇒-k≠0⇒k≠0

Thus, S=R-0

Topic: Mathematics, Determinants, JEE Main, JEE Main 2018 (15 Apr)
Let S be the set of all real values of λ such that a plane passing through the points -λ2, 1, 1, 1, -λ2, 1 and 1, 1, -λ2 also passes through the point -1, -1, 1. Then S is equal to :

Options:
A {3}
B 3, -3
C 1, -1
D 3, -3

The correct answer is:
3, -3

Solution:
As we know that if four points P, A, B, C are coplanar, the vectors PA→, PB→, PC→ must be coplanar.

⇒PA→.PB→×PC→=0

⇒-λ 2+1202-λ2+1022-λ2-1=0

⇒-λ2+1-λ2+1-λ2-1-0-22-λ2-1-0+04-2-λ2+1=0

⇒-λ2+1λ4-1+4λ2+1=0

⇒-λ2+1λ2+1λ2-1+4λ2+1=0

⇒λ2+1-λ2+1λ2-1+4=0

⇒-λ2+1λ4-2λ2+3=0

⇒-λ2+1λ2-3λ2+1=0

⇒λ=±3

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let S be the set of all real values of λ such that a plane passing through the points -λ2, 1, 1, 1, -λ2, 1 and 1, 1, -λ2 also passes through the point -1, -1, 1. Then S is equal to :

Options:
A {3}
B 3, -3
C 1, -1
D 3, -3

The correct answer is:
3, -3

Solution:
As we know that if four points P, A, B, C are coplanar, the vectors PA→, PB→, PC→ must be coplanar.

⇒PA→.PB→×PC→=0

⇒-λ 2+1202-λ2+1022-λ2-1=0

⇒-λ2+1-λ2+1-λ2-1-0-22-λ2-1-0+04-2-λ2+1=0

⇒-λ2+1λ4-1+4λ2+1=0

⇒-λ2+1λ2+1λ2-1+4λ2+1=0

⇒λ2+1-λ2+1λ2-1+4=0

⇒-λ2+1λ4-2λ2+3=0

⇒-λ2+1λ2-3λ2+1=0

⇒λ=±3

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (12 Jan Shift 2)
Let S be the set of all solutions of the equation cos-12x-2cos-11-x2=π,x∈-12,12. Then ∑x∈S2sin-1x2-1 is equal to

Options:
A 0
B -2π3
C π-sin-134
D π-2sin-134

The correct answer is:
0

Solution:
Given equation is:

cos-12x-2cos-11-x2=π,x∈-12,12

It is possible if

cos-12x=π ....(1)

and 2cos-11-x2=0 ....(2)

From equation (1), x=-12 which is not satisfying equation (2).

So, no such x exists.

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2023 (01 Feb Shift 1)
Let S be the set of all the natural numbers, for which the line xa+yb=2 is a tangent to the curve xan+ybn=2 at the point a,b, ab≠0. Then

Options:
A S=ϕ
B nS=1
C S=2k:k∈N
D S=N

The correct answer is:
S=N

Solution:
xan+ybn=2

Differentiating, we get,

nxan−1⋅1a+n⋅ybn−1⋅1b⋅dydx=0

dydx=-1axan−11bybn−1=−baxbyan−1⇒dydxa,b=−baabban−1 =−ba

So, equation of the tangent to the curve at a,b will be y−b=−bax−a ⇒yb−1=−xa+1

⇒xa+yb=2

i.e. S∈N

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2022 (26 Jun Shift 1)
Let S be the set of all triangles in the xy -plane, each having one vertex at the origin and the other two vertices lie on coordinate axes with integral coordinates. If each triangle in S has area 50 sq. units, then the number of elements in the set S is:

Options:
A 36
B 32
C 9
D 18

The correct answer is:
36

Solution:
Since the vertex of given triangle is O (origin) and other two vertices A and B lies on the coordinate axes, hence triangle must be a right-angled triangle.
Area =12x.y ( x, y∈I)
As per given condition
12x.y=50⇒x. y=100
Possible ordered pairs of x, y when ∆ formed in 1st quadrant are 9 i. e. 1,100, 2,50, 4,25, 5,20, 10,10, 25, 4, 50, 2, 100,1
∆ can be formed in any of the quadrants, hence total number of cases 9×4=36.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2019 (09 Jan Shift 2)
Let S be the set of all values of a1 for which the mean deviation about the mean of 100 consecutive positive integers a1,a2,a3,….,a100 is 25 . Then S is

Options:
A ϕ
B {99}
C ℕ
D {9}

The correct answer is:
ℕ

Solution:
Given,

a1 be any natural number

So, a1,a1+1,a1+2,….,a1+99 are values of ai or observation.

Now mean of the observation is given by,

x¯=a1+a1+1+a1+2+….+a1+99100

⇒x¯=100a1+(1+2+…..+99)100

⇒x¯=a1+99×1002×100

⇒x¯=a1+992

Mean deviation about mean=∑i=1100xi-x¯100

=2992+972+952+….+12100

=1+3+….+99100

=502[1+99]100

=25

So, it is true for every natural no. ' a1 ' as there is no term of a1 in mean deviation.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2023 (30 Jan Shift 2)
Let S be the set of all values of λ, for which the shortest distance between the lines x-λ0=y-34=z+61and x+λ3=y-4=z-60 is 13. Then 8∑λ∈Sλ is equal to

Options:
A 306
B 304
C 308
D 302

The correct answer is:
306

Solution:
The given lines arex-λ0=y-34=z+61and x+λ3=y-4=z-60

We know that the shortest distance between two lines is d=a→2-a→1·n→1×n→2n→1×n→2 & d=13 given

⇒n→1×n→2=i^j^k^0413-40=4i^+3j^-12k^

Now using the above formula of distance we get,

⇒(2λi^+3j^-12k^)·(4i^+3j^-12k^)16+9+144=13

⇒(2λi^+3j^-12k^)·(4i^+3j^-12k^)13=13

⇒8λ+9+144=169

⇒8λ+153=169

⇒8λ=±169-153

⇒λ=168, -3228

⇒8∑λ∈sλ=8168-3228=306

Hence this is the correct option.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (15 Apr Shift 1)
Let S be the set of all values of x for which the tangent to the curve y=fx=x3-x2-2x at (x, y) is parallel to the line segment joining the points (1, f(1)) and -1, f-1, then S is equal to

Options:
A -13,-1
B -13,1
C 13,1
D 13,-1

The correct answer is:
-13,1

Solution:
Given y=fx=x3-x2-2x

⇒f1=1-1-2=-2

⇒f-1=-1-1+2=0

The slope of a line joining the points x1, y1 & x2, y2 is y2-y1x2-x1

Thus, the slope of the line segment joining the points 1, f1 & -1, f-1 is m=f(1)-f-11+1=-2-02=-1

Given, this line segment is parallel to the tangent of the curve y=fx, at x, y and we know that the slope of the tangent to y=fx is dydx, also the slope of two parallel lines is same, hence

m=dydx=3x2-2x-2

⇒3x2-2x-2=-1

⇒3x2-2x-1=0

⇒x-13x+1=0

⇒x=1, -13.

Hence, the set S is -13, 1.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2019 (09 Apr Shift 1)
Let S be the set of points where the function ,fx=|2-|x-3|,x∈R, is not differentiable. Then ∑x∈Sffx is equal to

Options:

The correct answer is:
3

Solution:
Check non-differentiability of function, we get,

∵fx is non differentiable at x=1,3,5
∑ffx=ff1+ff3+ff5
=1+1+1
=3

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2020 (07 Jan Shift 1)