datavorous/entrance-exam-dataset huggingface dataset:
Let a ray of light passing through the point $(3,10)$ reflects on the line $2 x+y=6$ and the reflected ray passes through the point $(7,2)$. If the equation of the incident ray is $a x+b y+1=0$, then $a^2+b^2+3 a b$ is equal to_________

Options:

The correct answer is:
1

Solution:
For $B^{\prime} \quad \frac{x-7}{2}=\frac{y-2}{1}=-2\left(\frac{14+2-6}{5}\right)$
$\begin{aligned} & \frac{x-7}{2}=\frac{y-2}{1}=-4 \\ & x=-1 \quad y=-2 \quad B^{\prime}(-1,-2)\end{aligned}$
incident ray $\mathrm{AB}^{\prime}$
$\begin{aligned} & M_{\mathrm{AB}^{\prime}}=3 \\ & y+2=3(x+1) \\ & 3 x-y+1=0 \\ & a=3 b=-1 \\ & a^2+b^2+3 a b=9+1-9=1\end{aligned}$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2024 (08 Apr Shift 2)
Let $\mathrm{A}=\{\theta: \sin (\theta)=\tan (\theta)\}$ and $\mathrm{B}=(\theta: \cos (\theta)=$ 1\} be two sets. Then:

Options:
A
$\mathrm{A}=\mathrm{B}$

B
$A \not \subset B$

C
$B \not \subset A$

D
$A \subset B$ and $B-A \neq \phi$

The correct answer is:
$A \not \subset B$

Solution:
Let $\mathrm{A}=\{\theta: \sin \theta=\tan \theta\}$ and $\mathrm{B}=\{\theta: \cos \theta=1\}$
Now, $A=\left\{\theta: \sin \theta=\frac{\sin \theta}{\cos \theta}\right\}$ $=\{\theta: \sin \theta(\cos \theta-1)=0\}$ $=\{\theta=0, \pi, 2 \pi, 3 \pi, \ldots \ldots\}$
For $\mathrm{B}: \cos \theta=1 \Rightarrow \theta=\pi, 2 \pi, 4 \pi, \ldots \ldots$
This shows that $\mathrm{A}$ is not contained in B. i.e. $\mathrm{A} \not \subset \mathrm{B}$. but $\mathrm{B} \subset \mathrm{A}$.

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2013 (25 Apr Online)
Let a smooth curve y=fx be such that the slope of the tangent at any point x,y on it is directly proportional to -yx. If the curve passes through the points 1,2 and 8,1, then y18 is equal to

Options:
A 2loge2
B 4
C 1
D 4loge2

The correct answer is:
4

Solution:
Given, dydx∝-yx

⇒dydx=-Kyx (where K is proportionality constant)

Now integrating both side,

⇒∫dyy=-K∫dxx

⇒lny=-Klnx+C

If the above equation satisfy 1,2

⇒ln2=-K×0+C⇒C=ln2

So, lny=-Klnx+ln2

Now it also passes through 8,1

⇒ln1=-Kln8+ln2⇒K=13

So, equation becomes lny=-13lnx+ln2

So, at x=18

⇒lny=-13ln18+ln2=2ln2

⇒y=4

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2022 (25 Jul Shift 2)
Let a tangent be drawn to the ellipse x227+y2=1 at (33cosθ,sinθ) where θ∈0,π2. Then the value of θ such that the sum of intercepts on axes made by this tangent is minimum is equal to :

Options:
A π8
B π4
C π6
D π3

The correct answer is:
π6

Solution:
Equation of tangent be

xcosθ33+y·sinθ1=1, θ∈0,π2

intercept on x-axis

OA=33secθ

intercept on y-axis

OB=cosecθ

Now, sum of intercept

=33secθ+cosecθ=f(θ) let

f'(θ)=33secθtanθ-cosecθcotθ

=33sinθcos2θ-cosθsin2θ

=cosθsin2θ·33⏟⊕tan3θ-133=0⇒θ=π6

⇒ at θ=π6,fθ is minimum.

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Let a tangent to the curve 9x2+16y2=144 intersect the coordinate axes at the points A and B. Then, the minimum length of the line segment AB is ______

Options:

The correct answer is:
7

Solution:
Given:

9x2+16y2=144

⇒x216+y29=1

Equation of tangent at point P4cosθ, 3sinθ is

xcosθ4+ysinθ3=1

So, A is 4secθ,0 and point B is 0,3cosecθ.

Hence,

AB=16 sec2θ+9cosec2θ

⇒AB=25+16tan2 θ+9cot2 θ

⇒25+16tan2 θ+9cot2 θ≥7

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2023 (24 Jan Shift 1)
Let a total charge 2Q be distributed in a sphere of radius R, with the charge density given by ρ(r)=kr, where r is the distance from the centre. Two charges A and B , of -Q each, are placed on diametrically opposite points, at equal distance, a, from the centre. If A and B do not experience any force, then:

Options:
A a=3R214
B a=R3
C a=2-14R
D a=8-14R

The correct answer is:
a=8-14R

Solution:
As total charge on sphere is =2Q
2Q=∫0Rkr4πr2dr⇒kπr4=2Q
Charged enclosed in the sphere of radius a
q=∫0akr×4πr2dr=kπa4
As force on Q is zero
14πε0×Q22a2=kπa44πε0×a2×Q
⇒Q24=kπa4Q
=a42QR4Q
R4=8a4⇒R=814 a or a=8-14R

Topic: Physics, Electrostatics, JEE Main, JEE Main 2019 (12 Apr Shift 2)
Let a triangle ABC be inscribed in the circle x2-2x+y+y2=0 such that ∠BAC=π2. If the length of side AB is 2, then the area of the △ABC is equal to:

Options:
A 1
B 6+32
C 3+32
D 6+234

The correct answer is:
1

Solution:
Given,

circle x2-2x+y+y2=0

With centre 12,12 and radius 1.

Since ∠BAC=π2, so BC will act as diameter,

BC=diameter=2 (given)

So, AC=BC2-AB2=2 and AB=2

Now, △ABC=12AB·AC=1

Topic: Mathematics, Circle, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Let a triangle be bounded by the lines L1:2x+5y=10; L2:-4x+3y=12 and the line L3, which passes through the point P2,3, intersect L2 at A and L1 at B. If the point P divides the line-segment AB, internally in the ratio 1:3, then the area of the triangle is equal to

Options:
A 11013
B 13213
C 14213
D 15113

The correct answer is:
13213

Solution:
Given, point A lies on L2 : -4x+3y=12

Take x=α, so y=4+43α, Aα,4+43α

Points B lies on L1 : 2x+5y=10

Take x=β, so y=2-25β, Bβ,2-25β

Now point P divides AB internally in the ratio 1:3

⇒P2,3=P3α+β4,34+43α+12-25β4

⇒α=313,β=9513

We get, point A313,5613,B9513,-1213

Vertex C of triangle is the point of intersection of L1 and L2

⇒C-1513,3213

area ∆ABC=12313561319513-12131-151332131

=12×1333561395-1213-153213

area ∆ABC=13213sq. units

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2022 (28 Jun Shift 2)
Let a unit vector OP→ make angle α,β,γ with the positive directions of the co-ordinate axes OX,OY, OZ respectively, where β∈0,π2. OP→ is perpendicular to the plane through points 1,2,3, 2,3,4 and 1,5,7, then which one of the following is true ?

Options:
A α∈π2,π and γ∈π2,π
B α∈0,π2 and γ∈0,π2
C α∈π2,π and γ∈0,π2
D α∈0,π2 and γ∈π2,π

The correct answer is:
α∈π2,π and γ∈π2,π

Solution:
Equation of plane passing through the points 1,2,3, 2,3,4 and 1,5,7 is

x-1y-2z-3111034=0

⇒x-1-4y-2+3z-3=0

⇒x-4y+3z=2

Direction ratios of normal of the plane is <1,-4,3>

Direction cosines of normal of the plane is

±126,∓426,±326

Hence,

cosβ=426>0⇒β∈0,π2,

cosα=-126⇒α∈π2,π ,

cosγ=-326⇒γ∈π2,π

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (30 Jan Shift 1)
Let a unit vector u^=xi^+yj^+zk^ make angles π2,π3 and 2π3 with the vectors 12i^+12k^,12j^+12k^ and 12i^+12j^ respectively. If v→=12i^+12j^+12k^, then \(|\hat{u}-\vec{v}|^2\) is equal to

Options:
A 112
B 52
C 9
D 7

The correct answer is:
52

Solution:
Unit vector u^=xi^+yj^+zk^

p→1=12i^+12k^,p→2=12j^+12k^

p→3=12i^+12j^

Now angle between u^ and p→1=π2

u^·p→1=0⇒x2+z2=0

⇒x+z=0 …i

Angle between u^ and p→2=π3

u^·p→2=|u^|·p→2cosπ3

u^·p→2= y2+z2=12 …ii

Angle between u^ and p→3=2π3

u^·p→3=|u^|·p→3cos2π3

⇒x2+y2=-12⇒x+y=-12 …iii

from equation (i), (ii) and (iii) we get

x=-12, y=0, z=12

Thus u^-v→=-12i^+12k^-12i^-12j^-12k^

⇒u^-v→=-22i^-12j^

∴|u^-v→|2=42+122=52

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2024 (29 Jan Shift 2)
Let a unit vector which makes an angle of $60^{\circ}$ with $2 \hat{i}+2 \hat{j}-\hat{k}$ and angle $45^{\circ}$ with $\hat{i}-\hat{k}$ be $\overrightarrow{\mathrm{C}}$. Then $\overrightarrow{\mathrm{C}}+\left(-\frac{1}{2} \hat{i}+\frac{1}{3 \sqrt{2}} \hat{j}-\frac{\sqrt{2}}{3} \hat{k}\right)$ is :

Options:
A $\frac{\sqrt{2}}{3} \hat{i}-\frac{1}{2} \hat{k}$
B $\left(\frac{1}{\sqrt{3}}+\frac{1}{2}\right) \hat{i}+\left(\frac{1}{\sqrt{3}}-\frac{1}{3 \sqrt{2}}\right) \hat{j}+\left(\frac{1}{\sqrt{3}}+\frac{\sqrt{2}}{3}\right) \hat{k}$
C $\frac{\sqrt{2}}{3} \hat{i}+\frac{1}{3 \sqrt{2}} \hat{j}-\frac{1}{2} \hat{k}$
D $-\frac{\sqrt{2}}{3} \hat{i}+\frac{\sqrt{2}}{3} \hat{j}+\left(\frac{1}{2}+\frac{2 \sqrt{2}}{3}\right) \hat{k}$

The correct answer is:
$\frac{\sqrt{2}}{3} \hat{i}-\frac{1}{2} \hat{k}$

Solution:
$\overrightarrow{\mathrm{C}}=\mathrm{C}_1 \hat{\mathrm{i}}+\mathrm{C}_2 \hat{\mathrm{j}}+\mathrm{C}_3 \hat{\mathrm{k}}$
$\mathrm{C}_1{ }^2+\mathrm{C}_2{ }^2+\mathrm{C}_3{ }^2=1$
$\overrightarrow{\mathrm{C}} \cdot(2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})=|\mathrm{C}| \sqrt{9} \cos 60^{\circ}$
$2 \mathrm{C}_1+2 \mathrm{C}_2-\mathrm{C}_3=\frac{3}{2}$
$\mathrm{C}_1-\mathrm{C}_3=1$
$\mathrm{C}_1+2 \mathrm{C}_2=\frac{1}{2}$
$\mathrm{C}_1=\frac{\sqrt{2}}{3}+\frac{1}{2}$
$\mathrm{C}_2=\frac{-1}{3 \sqrt{2}}$
$\mathrm{C}_3=\frac{\sqrt{2}}{3}-\frac{1}{2}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2024 (04 Apr Shift 1)
Let a variable line of slope $m>0$ passing through the point $(4,-9)$ intersect the coordinate axes at the points $A$ and $B$. The minimum value of the sum of the distances of $A$ and $B$ from the origin is

Options:
A 30
B 25
C 15
D 10

The correct answer is:
25

Solution:
equation of line is
$\begin{aligned}
& y+9=m(x-4) \\
& \therefore \quad A=\left(\frac{9+4 m}{m}, 0\right) \\
& \quad B=(0,-9-4 m) \\
& \therefore \quad O A+O B=\frac{9+4 m}{m}+9+4 m
\end{aligned}$
$\begin{aligned} & \because \mathrm{m}>0 \\ & =13+\frac{9}{\mathrm{~m}}+4 \mathrm{~m} \\ & \because \frac{4 \mathrm{~m}+\frac{9}{\mathrm{~m}}}{2} \geq \sqrt{36} \Rightarrow 4 \mathrm{~m}+\frac{9}{\mathrm{~m}} \geq 12 \\ & \therefore \mathrm{OA}+\mathrm{OB} \geq 25\end{aligned}$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2024 (06 Apr Shift 1)
Let a variable line passing through a fixed point P in the first quadrant cuts the positive coordinate axes at points A and B respectively. If the area of ΔOAB is minimum, then OP is

Options:
A Altitude through vertex O of ΔAOB
B Median through vertex O of ΔAOB
C Internal angle bisector through vertex O of ΔAOB
D None of these

The correct answer is:
Median through vertex O of ΔAOB

Solution:
Let, the foot of perpendiculars from P on OA & OB are C & D respectively
Let, ∠PAC=θ=∠BPD
Now, CA=kcot⁡θ & BD=htan⁡θ
Area of ΔOAB= area of ΔPCA+ area of ΔBPD+ area of the rectangle PDOC
⇒ Area of ΔOAB=12k2cot⁡θ+12h2tan⁡θ+hk=12kcot⁡θ-htan⁡θ2+2hk
⇒ The minimum area of ΔOAB is 2hk when kcot⁡θ=htan⁡θ⇒tan⁡θ=kh
⇒hsin⁡θ=kcos⁡θ⇒hsec⁡θ=kcosecθ
⇒PB=PA⇒OP is the median

Topic: Mathematics, Straight Lines, JEE Main
Let a variable line passing through the centre of the circle x2+y2−16x−4y=0, meet the positive co-ordinate axes at the point A and B. Then the minimum value of OA+OB, where O is the origin, is equal to

Options:
A 12
B 18
C 20
D 24

The correct answer is:
18

Solution:
Given: A line passing through the centre of circle x2+y2-16x-4y=0 which is 8,2 intersects positive x & y axis at A & B respectively,

So, let the line be xa+yb=1

Now, given line passes through the centre,

So, 8a+2b=1

Now, using A.M≥H.M we get,

a2+a2+a2+a2+b1+b16≥62a+2a+2a+2a+1b+1b

⇒2a+2b6≥61

⇒2a+2b≥36

⇒a+b≥18

⇒OA+OB≥18

Hence, the minimum value of OA+OB=18

Topic: Mathematics, Circle, JEE Main, JEE Main 2024 (31 Jan Shift 2)
Let a vector a→ has a magnitude 9. Let a vector b→ be such that for every x,yR×R-0,0, the vector xa→+yb→ is perpendicular to the vector 6ya→-18xb→. Then the value of a→×b→ is equal to

Options:
A 93
B 273
C 9
D 81

The correct answer is:
273

Solution:
Given, a→=9

And xa→+yb→·6ya→-18xb→=0as they are perpendicular

⇒6xya→2-18x2a→·b→+6y2a→·b→-18xyb→2=0

⇒6xya→2-3b→2+a→·b→y2-3x2=0

This should hold ∀x,y∈R×R-0,0

So, a→2-3b→2=0 & a→·b→=0

∴ a→2=3b→2 & a→·b→=0⇒cosθ=0⇒θ=90°

Now a→×b→2=a→2b→2sin2θ=a→2b→2 as θ=90°

=a→2·a→23 as b→2=a→23

So, a→×b→=a→23=813=273

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2022 (28 Jul Shift 1)
Let a vector αi^+βj^ be obtained by rotating the vector 3i^+j^ by an angle 45° about the origin in counterclockwise direction in the first quadrant. Then the area (in sq. units) of triangle having vertices α,β,0,β and 0,0 is equal to

Options:
A 12
B 1
C 12
D 22

The correct answer is:
12

Solution:
Area of ΔOA'B=12OA'cos15°×OA'sin15°

=12OA'2sin30°2

Now, OA=OA', since rotation will not change the magnitude.

=3+1×18=12 sq. units

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (16 Mar Shift 1)
Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP=2AB. If ∠BPC=β, then tanβ is equal to:

Options:
A 67
B 14
C 29
D 49

The correct answer is:
29

Solution:
Let ∠APC=α
Given, AB=2AC=2BC
Also AP=2AB⇒AP=4AC=4BC
⇒tanα=ACAP=AC4AC=14
⇒tanα+β=ABAP=AB2AB=12
Let, tanβ=x
tanα+tanβ1-tanα ⋅tanβ=12
⇒ 14+x1-x4=12
⇒ 1+4x4-x=12
⇒ 2+8x-4+x=0
⇒ 9x-2=0
⇒ x=29

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2017 (02 Apr)
Let a vertical tower AB of height 2h stands on a horizontal ground. Let from a point P on the ground a man can see upto height h of the tower with an angle of elevation 2α. When from P, he moves a distance d in the direction of AP→, he can see the top B of the tower with an angle of elevation α. If d=7h, then tanα is equal to

Options:
A 5-2
B 3-1
C 7-2
D 7-3

The correct answer is:
7-2

Solution:
Given a vertical tower AB of height 2h stands on a horizontal ground and from a point P on the ground a man can see upto height h of the tower with an angle of elevation 2α. When from P, he moves a distance d in the direction of AP→, he can see the top B of the tower with an angle of elevation α. If d=7h, then tanα is equal to

Now on plotting the value in diagram we have,

âââââââ

Now in â³PAC

tan2α=hx .....1

And in â³HAB we have

tanα=2hx+7h

⇒tanα=2hhcot2α+7h from equation 1

⇒tanα=21-tan2α2tanα+7

Now put tanα=t and on simplifying we get,

⇒t=21-t22t+7

⇒t1-t22t+7=2

⇒1-t22+7t=2

⇒t2-27t+3=0

⇒t=7-2

⇒ tanα=7-2

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2022 (27 Jul Shift 1)
Let Ax=0x-2x-3x+20x-5x+3x+50, then the matrix A0A0T is a

Options:
A null matrix
B symmetric matrix
C skew symmetric matrix
D non singular matrix

The correct answer is:
symmetric matrix

Solution:
A0=0-2-320-5350

A0T=023-205-3-50=-A0
âA0 is a skew-symmetric matrix
A0A0TT=A0TTA0T=A0A0T
âA0A0T is a symmetric matrix
Also, A0A0T=A02=0

Topic: Mathematics, Matrices, JEE Main
Let A=x110, x ε R and A4=aij. If a11=109, then a22 is equal to_____________.

Options:

The correct answer is:
10

Solution:
A2=x110×x110=x2+1xx1

A4=x2+1xx1×x2+1xx1=x4+3x2+1x3+2xx3+2xx2+1

Given a11=109⇒x4+3x2+1=109

x2+12x2-9=0⇒x2=9

∴a22=x2+1=10

Topic: Mathematics, Matrices, JEE Main, JEE Main 2020 (03 Sep Shift 1)
Let A=x1,x2,…,x7 and B=y1,y2,y3 be two sets containing seven and three distinct elements respectively. Then the total number of functions f:A→B that are onto, if there exist exactly three elements x in A such that fx=y2, is equal to:

Options:
A 12·7C2
B 16·7C3
C 14· 7C3
D 14·7C2

The correct answer is:
14· 7C3

Solution:
A={x1, x2…..x7} and B= {y1, y2, y3}
Let us select 3 elements from A & connect it to y2

Number of ways =7C3
The number of ways would be equal to 7C3 and now we have 2 elements y1 & y3 in set B to be mapped from the remaining element of set A

Hence, total number of function 24=16 and out of which 2 would be "into" functions (when all four goes in y1 & when all four goes in y3) =24-2=14
∴So the total number of onto functions would be 14·7C3

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2015 (11 Apr Online)
Let Ax1,y1, Bx2,y2, Cx3,y3 and Dx4,y4 are four points which are at equal distance from the lines 3x-4y+1=0 and 8x+6y+1=0. The mean of the coordinates of the centroids of âABC, âBCD, âCDA and âDAB are

Options:
A -45,25
B -15,110
C -35,310
D -415,215

The correct answer is:
-15,110

Solution:
Point of intersection of the lines 3x-4y+1=0 and 8x+6y+1=0 is -15,110
Now, âi=14xi=4-15=-45
And, âi=18yi=4110=25
Mean of the centroids of âABC,âBCD,âCDA and âDAB is âi=14xi4,âi=14yi4=-15,110

Topic: Mathematics, Straight Lines, JEE Main
Let A=x2-3-13-22-11 be a matrix and adjadjA=124, then the sum of all the values of x is equal to

Options:
A -24
B 24
C -18
D 1

The correct answer is:
-18

Solution:
A=x2-3-13-22-11=x1-23-3-5
=x+9
adjadjA=A4 (By property)
⇒A4=124⇒A=±12
⇒x+9=±12
⇒x=3,-21

Topic: Mathematics, Determinants, JEE Main
Let a→=x2i^-3j^+x+3k^ and b→=i^+3j^-x-3k^ be two vectors such that a→=b→. If angle between 4a→+7b→ and 7a→-4b→ is equal to θ, then cos⁡2θ is equal to

Options:
A -12
B 32
C -1
D 15

The correct answer is:
-1

Solution:
∵a→·b→=x2-9-x2-9=0⇒a→⊥b→
4a→+7b→⋅7a→−4b→=28a→2−28b→2+33a→⋅b→=0 as a→=b→
⇒θ=90o⇒cos⁡2θ=-1

Topic: Mathematics, Vector Algebra, JEE Main
Let A=x4+4x3+2x2-4x+7 where x=cot⁡11π8 and B=1-cos⁡8θtan2⁡4θ+1+cos⁡8θcot2⁡4θ where θ=9o, then the value of A×B2 is equal to

Options:

The correct answer is:
6

Solution:
A=x4+4x3+2x2-4x+7
where x=cot⁡11π8=cot⁡π+3π8=cot⁡3π8
=cot⁡π2-π8=tan⁡π8=2-1
i.e. x=2-1⇒x2=2+1-22=1-22-1
⇒x2=1-2x⇒x2+2x-1=0
So, A=x2+2x-12+6=6
B=2sin2⁡4θtan2⁡4θ+2cos2⁡4θcot2⁡4θ=2cos2⁡4θ+sin2⁡4θ=2
⇒AB=12⇒AB2=6

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
Let A=x∈R:x+1<2 and B=x∈R:x-1≥2. Then which one the following statements is NOT true?

Options:
A A-B=-1,1
B B-A=R--3,1
C A∩B=(-3,-1]
D A∪B=R-[1,3)

The correct answer is:
B-A=R--3,1

Solution:
Given,

A=x∈R:x+1<2

⇒A=-3,1

Now, B=x∈R:x-1≥2

⇒B=(-∞,-1]∪[3,∞)

Now, B-A=(-∞,-3]∪[3,∞)

⇒B-A=R--3,3

So, option B is not true.

Topic: Mathematics, Basic of Mathematics, JEE Main, JEE Main 2022 (25 Jun Shift 2)
Let A= { x∈R:x is not a positive integer} . Define a function f:A→R as fx=2xx-1 , then f is:

Options:
A Injective but not surjective
B Not injective
C Surjective but not injective
D Neither injective nor surjective

The correct answer is:
Injective but not surjective

Solution:
Given, function fx=2xx-1 and domain of fx is -∞, 0
Now f'x=x-12-2x1-0x-12

f′x=-2x-12<0
⇒f is always decreasing in the given domain.
Clearly, given function is one-one or injective.
Since, codomain -∞, ∞≠ Range 0, 2 , hence f is into.

Topic: Mathematics, Functions, JEE Main, JEE Main 2019 (09 Jan Shift 2)
Let $A=\{x \in R / x$ is not a positive integer). Let a function $f$ be defined as $f: A \rightarrow R$ so that $f(x)=\frac{2 x}{x-1}$, then $f$ is

Options:
A Not injective.
B surjective but not injective.
C neither injective nor surjective.
D injective but not surjective.

The correct answer is:
injective but not surjective.

Solution:
$\begin{aligned} & f(x)=\frac{2 x}{x-1} \\ & \Rightarrow f^{\prime}(x)=\frac{(x-1) 2-2 x(1-0)}{(x-1)^2}=\frac{-2}{(x-1)^2}<0\end{aligned}$
$f(x)$ is strictly decreasing so $f(x)$ is injective
But $f(x)=4$
$\Rightarrow \frac{2 x}{x-1}=4$
$\Rightarrow x=2$ (which is a positive integer)
i.e., $4 \in R$ (co-domain of $f$ ) has no pre-image in $A$ (domain of $f$ )
So, ' $f$ ' is not surjective

Topic: Mathematics, Functions, MHT CET, MHT CET 2022 (10 Aug Shift 1)
Let A={x∈ℝ:x+3+x+4≤3}, B=x∈ℝ:3x∑r=1∞310rx-3<3-3x, where [t] denotes greatest integer function. Then,

Options:
A B⊂C, A≠B
B A∩B=ϕ
C A⊂B,A≠B
D A=B

The correct answer is:
A=B

Solution:
Given,

A={x∈ℝ:x+3+x+4≤3} & B=x∈ℝ:3x∑r=1∞310rx-3<3-3x,

Now solving A we get,

x+3+x+4≤3

⇒x+3+x+4≤3

⇒2x≤-4

⇒x≤-2

⇒x<-1⇒A=-∞,-1

Now solving B we get,

3x∑r=1∞310rx-3<3-3x

Now using infinite G.P formula S∞=a1-r we get,

⇒3x3·1101-110x-3<3-3x

⇒3x13x-3<3-3x

⇒3x-x+3+3x<1

⇒33x+1<1

⇒3x+1<0

⇒x<-1⇒B=-∞,-1

⇒A=B

Topic: Mathematics, Functions, JEE Main, JEE Main 2023 (06 Apr Shift 1)
Let A=X=x,y,zT:PX=0 and x2+y2+z2=1 where P=121-23-419-1 then the set A

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

The correct answer is:
Contains exactly two elements

Solution:
∵ P=0

∴ System of equations

x+2 y+z=0

2x-3y+4z=0

x+9y-z=0 has infinitely many solutions.

Let z=k∈R then x=-11k7, y=2k7, z=k

Also x2+y2+z2=1

∴k=±7174

∴ Two solutions are only possible.

Hence, set A contains exactly two elements.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2020 (02 Sep Shift 2)
Let $A=\{(x, y): 2 x+3 y=23, x, y \in \mathbb{N}\}$ and $B=\{x:(x, y) \in A\}$. Then the number of one-one functions from $A$ to $B$ is equal to _______

Options:

The correct answer is:
24

Solution:
$\begin{array}{ll}
2 x+3 y=23 \\
x=1 & y=7 \\
x=4 & y=5 \\
x=7 & y=3 \\
x=10 & y=1 \\
A & B \\
(1,7) & 1 \\
(4,5) & 4 \\
(7,3) & 7 \\
(10,1) & 10
\end{array}$
The number of one-one functions from $A$ to $B$ is equal to 4 !

Topic: Mathematics, Functions, JEE Main, JEE Main 2024 (09 Apr Shift 2)
Let A=x,y∈ℝ2:y≥0,2x≤y≤4-x-12 and B=x,y∈ℝ×ℝ:0≤y≤min2x,4-x-12. Then the ratio of the area of A to the area of B is

Options:
A π-1π+1
B ππ-1
C ππ+1
D π+1π-1

The correct answer is:
π-1π+1

Solution:
Given,

2x≤y≤4-x-12 :y≥0

⇒y2+x-12=4 and y=2x

Now plotting the diagram we get,

Area of A: area of shaded region as shown in the above figure,

Shaded portion = circular OABC

-Ar∆OAB=π44-1221

Hence, Area of A=π-1

Now, plotting the diagram of 0≤y≤min2x,4-x-12 we get,

Area of B: area of shaded region as shown in the above figure,

Area of B=Ar∆AOB+ Area of arc of circle ABC

=1212+π224=π+1

Hence, AB=area of Aarea of B=π-1π+1

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Let A=x,y:y2≤4x, y-2x≥-4. The area of the region A in square units is

Options:
A 10
B 8
C 9
D 11

The correct answer is:
9

Solution:
Given, y2=4x ...1

y=2x−4⇒2x=y+4 ...2

Solving equations 1 & 2, we get

⇒y2=2y+8

⇒ y2-2y-8=0

⇒y-4y+2=0

⇒y=4, -2

From equation 2,

If y=4 then x=4 and if y=-2 then x=1.

A=∫044x dx-12×2×4+∫014xdx+12×1×2

=2x3/23204-4+2x3/23201+1

=438-4+431+1

=363-3

=9

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2014 (09 Apr Online)
Let a, x, y, z be real numbers satisfying the equations
ax+ay=z
x+ay=z
x+ay=az, where
x, y, z are not all zero, then the number of the possible values of a is

Options:

The correct answer is:
2

Solution:
As x, y, z are not all zero, thus the system has a non-trivial solution.
Thus, D=0
⇒aa-11a-11a-a=0
⇒aa2-a-aa-1+1a-a=0 or a3-a2-a2+a=0
⇒aa2-2a+1=0
⇒a=0, 1

Topic: Mathematics, Determinants, JEE Main
Let A=xyzyzxzxy, where x, y and z are real numbers such that x+y+z>0 and xyz=2. If A2=I3, then the value of x3+y3+z3 is

Options:

The correct answer is:
7

Solution:
A2=I

⇒AA'=I (as A'=A)

⇒A is orthogonal

So, x2+y2+z2=1 and xy+yz+zx=0

⇒x+y+z2=1+2×0

⇒x+y+z=1

Thus,

x3+y3+z3=3×2+1×1-0

= 7

Topic: Mathematics, Matrices, JEE Main, JEE Main 2021 (25 Feb Shift 1)
Let Az1,Bz2 and C(z3) be complex numbers satisfying the equation z=1 and also satisfying the relation 3z1=2z2+2z3. Then z2-z32 is equal to

Options:

The correct answer is:
1.75

Solution:
A,B,C lie on the unit circle centered at the origin. Also,
3z1+1.04=z2+z32

i.e. the line segment joining z1 and originD bisects the line segment joining z2 & z3 at E

Also, DE:EA≡3:1
Let DE=3K and EA=K
Now, DA=4K=1⇒K=14

⇒DE=34
From △BED,

BE=1-916=74
⇒z2-z3=BC=72

Hence, z2-z32=74=1.75

Topic: Mathematics, Complex Number, JEE Main
Let A=z∈C:1⩽z-1+i⩽2 and B=z∈A:z-1-i=1. Then, B

Options:
A is an empty set
B contains exactly two elements
C contains exactly three elements
D is an infinite set

The correct answer is:
is an infinite set

Solution:
z-1+i=1 & z-1+i=2 represents two concentric circles with centre 1,1.

z-1-i=1 represents a circle with centre 1,-1 and radius 1

A represents the region between the concentric circles with centre 1,1

Set B contains all points on minor arc PQ of the circle z-1-i=1.

Hence, B has infinite elements.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2022 (24 Jun Shift 1)
Let \( A(2 \hat{i}+3 \hat{j}+5 \hat{k}), B(-\hat{i}+3 \hat{j}+2 \hat{k}) \) and \( C(\lambda \hat{i}+5 \hat{j}+\mu \hat{k}) \) are vertices of a triangle and its median through \( A \) is equally inclined to the positive directions of the axes. The value of \( 2 \lambda-\mu \) is equal to

Options:

The correct answer is:
4

Solution:
Given that A2i^+3j^+5k^, B–i^+3j^+2k^ and Cλi^+5j^+μk^ are vertices of a triangle

Since, D is the mid-point of B and C

So, position vector of D

=λ-12i^+4j^+μ+22k^

Direction ratio of AD is λ-52, 1, μ-82 or λ-523, 13, μ-823.

But direction cosine of AD should be

13, 13, 13

Therefore,

λ-52=1=μ-82

⇒λ=7, μ=10

⇒2λ-μ=4

Topic: Mathematics, Vector Algebra, JEE Main