datavorous/entrance-exam-dataset huggingface dataset:
Consider the infinite ladder circuit shown below.
For which angular frequency $\omega$ will the circuit behave like a pure inductance?

Options:
A $\frac{\mathrm{LC}}{\sqrt{2}}$
B $\frac{1}{\sqrt{\mathrm{LC}}}$
C $\frac{2}{\sqrt{L C}}$
D $\frac{2 \mathrm{~L}}{\sqrt{\mathrm{C}}}$

The correct answer is:
$\frac{2}{\sqrt{L C}}$

Solution:
Let the equivalent impedance of the circuit be $Z$
So $Z=\omega L+Z^{\prime}$
now $Z^{\prime}=\frac{Z X_{C}}{Z+X_{C}}$
$$
\therefore Z=\omega L+\left(\frac{Z \times \frac{1}{\omega C}}{Z+\frac{1}{\omega C}}\right)
$$
on ving we get $Z=\frac{\omega L C \pm \sqrt{(\omega L C)^{2}-4 L C}}{2 C}$
for $Z$ to be purely inductive
$$
\omega^{2} L^{2} C^{2}-4 L C=0 \text { or } \omega=\frac{2}{\sqrt{L C}}
$$

Topic: Physics, Alternating Current, JEE Main
Consider the integral A=∫01ex-1xdx and B=∫01xex-1dx. Then, which of the following is incorrect?

Options:
A B<1
B A>1
C B>A
D A>12

The correct answer is:
B>A

Solution:
As, ex-1>x ∀x∈0,∞
∴ex-1x>1
Hence, A=∫01ex-1xdx>∫011dx
⇒A>1
And, B=∫01xex-1dx<∫011dx
⇒B<1
∴ B<1<A

Topic: Mathematics, Definite Integration, JEE Main
Consider the integral I=∫010xexex-1dx where x denotes the greatest integer less than or equal to x. Then the value of I is equal to :

Options:
A 9e-1
B 45e+1
C 45e-1
D 9e+1

The correct answer is:
45e-1

Solution:
I=∫010x·ex-x+1

I=∫010dx+∫121·e2-x+∫232·e3-x+…..+∫9109·e10-xdx

⇒I=∑n=09∫nn+1n·en+1-xdx

=-∑n=09nen+1-xnn+1

=-∑n=09n·e0-e1

=e-1∑n=09n

=e-1·9·102

=45e-1

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (16 Mar Shift 2)
Consider the integral In=∫0π4sin2n−1xsin xdx, then the value of I20-I19 is

Options:
A 120
B -119
C -125
D 119

The correct answer is:
-119

Solution:
I 20 − I 19 = ∫ 0 π 4 sin 39x −sin 37x sin x dx
⇒ I 20 − I 19 = ∫ 0 π 4 2sin x cos 38x sin x dx
=sin38x19π40
= sin 38π 4 19
=-119

Topic: Mathematics, Definite Integration, JEE Main
Consider the integrals I1=â«ex2cosâ¡xdx and I2=â«xex2sinâ¡xdx. Then, I1+2I2 simplifies to (where, c is the constant of integration)

Options:
A exsin⁡x+c
B ex2cos⁡x+c
C 2ex2sin⁡x+c
D ex2sin⁡x+c

The correct answer is:
ex2sin⁡x+c

Solution:
I1+2I2=∫ex2cos⁡x+2xex2sin⁡xdx
=∫dex2sin⁡x
=ex2sin⁡x+c

Topic: Mathematics, Indefinite Integration, JEE Main
Consider the integrals I=∫sin⁡x3cos⁡x+4sin⁡xdx and J=∫cos⁡x3cos⁡x+4sin⁡xdx. Then, the integral 4J-3I is equal to (where, c is the constant of integration)

Options:
A ln⁡3cos⁡x-4sin⁡x+c
B ln⁡3cos⁡x+4sin⁡x+c
C ln⁡3sin⁡x+4cos⁡x+c
D ln⁡3sin⁡x-4cos⁡x+c

The correct answer is:
ln⁡3cos⁡x+4sin⁡x+c

Solution:
4J-3I=∫4cos⁡x-3sin⁡x3cos⁡x+4sin⁡xdx
Let, 3cos⁡x+4sin⁡x=t
⇒4cos⁡x-3sin⁡xdx=dt
⇒4J-3I=∫dtt=ln⁡t+c
⇒4J-3I=ln⁡3cos⁡x+4sin⁡x+c

Topic: Mathematics, Indefinite Integration, JEE Main
Consider the interference at P between waves emanating from three coherent sources in the same phase located at S1, S2 and S3. If the intensity due to each source is I0=12 W m-2 at P and d22D=Î»3 then what will be the resultant intensity (in W m-2) at P?

Options:

The correct answer is:
36

Solution:
S2PâS1P=dsinâ¡Î¸1=d.d/2D=d22D=Î»3
S3PâS2P=dsinâ¡Î¸2=d3d/2D=3d22D=Î»
IResultant = I0+4I0+2I04I0cos1200
=3I0

Topic: Physics, Wave Optics, JEE Main
Consider the ions/molecule O2+,O2,O2-,O22-. For increasing bond order the correct option is

Options:
A O2-<O22-<O2<O2+
B O2-<O22-<O2+<O2
C O2-<O2+<O22-<O2
D O22-<O2-<O2<O2+

The correct answer is:
O22-<O2-<O2<O2+

Solution:
Bond order =number of bonding e--number of antibonding e-2

Electronic configuration of O2, O2+, O2- and O2-2 are:

O2=σ1s2σ1s2*σ2s2σ2s2*σ2px2π2py2=π2pz2π2py1*=π2pz1* (Bond order=2)O2+=σ1s2σ1s2*σ2s2σ2s2*σ2px2π2py2=π2pz2π2py1*=π2pz0* (Bond order=2.5)O2-=σ1s2σ1s2*σ2s2σ2s2*σ2px2π2py2=π2pz2π2py2*=π2pz1* (Bond order=1.5)O2-2=σ1s2σ1s2*σ2s2σ2s2*σ2px2π2py2=π2pz2π2py2*=π2pz2* (Bond order=1)

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2022 (26 Jun Shift 1)
Consider the line L given by the equation x-32=y-11=z-21. Let Q be the mirror image of the point (2,3,-1) with respect to L. Let a plane P be such that it passes through Q, and the line L is perpendicular to P. Then which of the following points is on the plane P?

Options:
A -1, 1, 2
B 1, 1, 1
C 1, 1, 2
D 1, 2, 2

The correct answer is:
1, 2, 2

Solution:
We have,

L:x-32=y-11=z-21=k

Let R≡2,3,-1 and Q≡a,b,c

And, M≡2k+3,k+1,k+2.

Direction ratios of RM is 2k+1,k-2,k+3,

Then, RM⊥L, therefore

22k+1+k-2+k+3=0

⇒6k=-3

⇒k=-12.

M≡2,12,32

Let Q≡a,b,c

Also, mid-point of RQ is M,

a+22=2⇒a=2

b+32=12⇒b=-2

c-12=32⇒c=4

Hence, Q≡2,-2,4.

Plane P is perpendicular to line

x-32=y-11=z-21

and passes through Q2,-2,4, therefore equation of plane P is

2(x-2)+1(y+2)+1(z-4)=0

⇒2x+y+z-6=0

Point 1, 2, 2 satisfies above equation.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (20 Jul Shift 2)
Consider the line $L$ passing through the points $(1,2,3)$ and $(2,3,5)$. The distance of the point $\left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right)$ from the line $\mathrm{L}$ along the line $\frac{3 x-11}{2}=\frac{3 y-11}{1}=\frac{3 z-19}{2}$ is equal to

Options:
A 6
B 5
C 4
D 3

The correct answer is:
3

Solution:
$\begin{aligned} & \frac{x-1}{2-1}=\frac{y-2}{3-2}=\frac{z-3}{5-3} \\ & \Rightarrow \frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{2}=\lambda\end{aligned}$
$\begin{aligned} & \mathrm{B}(1+\lambda, 2+\lambda, 3+2 \lambda) \\ & \text { D.R. of } \mathrm{AB}= < \frac{3 \lambda-8}{3}, \frac{3 \lambda-5}{3}, \frac{6 \lambda-10}{3}> \\ & \mathrm{B}\left(\frac{5}{3}, \frac{8}{3}, \frac{13}{3}\right) \frac{3 \lambda-8}{3 \lambda-5}=\frac{2}{1} \Rightarrow 3 \lambda-8=6 \lambda-10 \\ & 3 \lambda=2 \\ & \lambda=\frac{2}{3} \\ & \mathrm{AB}=\frac{\sqrt{36+9+36}}{3}=\frac{9}{3}=3\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (09 Apr Shift 2)
Consider the line L:x-12=y+1-3=z+108 and a family of planes P containing the line L. The member of the family of planes P which is situated at a maximum distance from A1,0,0 will be

Options:
A x-2y-z=13
B x+2y-z=7
C 2x+y-z=7
D x+2y-2z=13

The correct answer is:
x-2y-z=13

Solution:
Let the foot of the perpendicular from A1,0,0 to given line L=0 is B, then the required plane must be perpendicular to AB and contain the given line.
Any general point on the given line is B=2Î»+1,-3Î»-1,8Î»-10
ABâ=<2Î»,-3Î»-1,8Î»-10>
â´ABâ is perpendicular to the given line
â22Î»+-3-3Î»-1+88Î»-10=0
â4Î»+9Î»+3+64Î»-80=0
âÎ»=1
So, the coordinates of point B are 3,-4,-2
Equation of the required plane is
2x-1-4y+1-2z+10=0
â2x-2-4y-4-2z-20=0
â2x-4y-2z-26=0
âx-2y-z=13

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Consider the line L≡x-12=y+23=z-76. Point P2,-5,0 and Q are such that PQ is perpendicular to the line L and the midpoint of PQ lies on line L, then coordinates of Q are

Options:
A -4,-5,2
B -3,0,1
C 1,6,2
D 1,5,7

The correct answer is:
-4,-5,2

Solution:
Q is reflection of P about line L
Coordinate of point R are 2λ+1,3λ-2,6λ+7
PR→=2λ-1,3λ+3,6λ+7
From diagram
PR→⋅2i^+3j^+6k^=0
4λ-2+9λ+9+36λ+42=0
49λ+49=0⇒λ=-1
So, coordinates of R are -1,-5,1
⇒ coordinates of Q are -4,-5,2

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Consider the lines L1 and L2 given by

L1:x-12=y-31=z-22

L2:x-21=y-22=z-33

A line L3 having direction ratios 1,-1,-2, intersects L1 and L2 at the points P and Q respectively. Then the length of line segment PQ is

Options:
A 26
B 32
C 43
D 4

The correct answer is:
26

Solution:
Given,

The lines L1 and L2 given by

L1:x-12=y-31=z-22

L2:x-21=y-22=z-33

Now any point on line L1 is given by,

P=2λ+1,λ+3,2λ+2

And any point on line L2 is given by,

Q=μ+2,2μ+2,3μ+3

Now equation of line L3 which passes through Q is given by,

x-μ+11=y-2μ+2-1=z-3μ+3-2

Now it also passes through P, so we get,

⇒ 2λ-μ-11=λ-2μ+1-1=2λ-3μ-1-2

Now on solving above equations we get, λ=μ=3

Hence, P7,6,8 and Q5,8,12

So, the distance PQ=26

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (25 Jan Shift 1)
Consider the LR circuit shown in the figure. If the switch S is closed at t=0 then the amount of charge that passes through the battery between t=0 and t=LR is:

Options:
A 7.3 ELR2
B EL7.3R2
C 2.7 ELR2
D  EL2.7R2

The correct answer is:
EL2.7R2

Solution:
The current in the LR circuit at any time t is
I=ER1-e-RtL

Where E, R, L, I are the EMF, resistance, inductance and the current respectively.
∴ Charge flown through the circuit from time 0 to LR is
(charge) ∆q=∫0LRI.dt=∫0LRER1-e-RtLdt
⇒∆q=ERt+LR e-RtL0LR
⇒∆q=ERLR+LRe-1-LR
⇒∆q=ER×LeR
⇒∆q=ELeR2=EL2.7R2

Topic: Physics, Alternating Current, JEE Main, JEE Main 2019 (12 Apr Shift 2)
Consider the matrices A, B, C and D with order 2Ã4, 4Ã3, 3Ã3 and 3Ã2 respectively. Let M=2ABC2020D6 and M=Î»ABC2020D6, then Î» is equal to

Options:
A 1024
B 4096
C 64
D 4

The correct answer is:
4096

Solution:
M=2ABC2020D6 (Given)
M=2ABC2020D6
âµABC2020D is a 2Ã2 matrix
Hence, 2ABC2020D=22ABC2020D
Therefore,
M=46ABC2020D6
âÎ»=4096

Topic: Mathematics, Determinants, JEE Main
Consider the matrix A=31-6-2, then Ι+A40 is equal to

Options:
A Ι+238A
B Ι+239A
C Ι+240+1A
D Ι+240-1A

The correct answer is:
Ι+240-1A

Solution:
A2=31-6-231-6-2=31-6-2=A
A3=A4=A5.........=A40=A
Ι+A40=Ι+40C1A+40C2A2.......+40C40A40
=Ι+A40C1+40C2.......40C40
=Ι+A240-1

Topic: Mathematics, Matrices, JEE Main
Consider the matrix A=3-201 and B be a square matrix of order 2 such that BBT=BTB=I. Let C=BABT and D=dij2×2=BTC6B, then d11 is equal to

Options:
A 25
B 27
C 30
D 81

The correct answer is:
27

Solution:
D=BTBABTBABTBABTBABTBABTBABTB
D=A6
Now, A2=3-2013-201=3-23-201
A3=A2⋅A=3-23-2013−201=33−23−801
A6=A3⋅A3=33−23−80133−23−801
=27-26-26301=D

⇒d11=27

Topic: Mathematics, Matrices, JEE Main
Consider the matrix A=x2yz2yzxzx2y and AAT=9I. If TrA>0 and xyz=16, then the value of x3+8y3+z3 is equal to (where, TrA, I and AT denote the trace of matrix A i.e. the sum of all the principal diagonal elements, the identity matrix of the same order of matrix A and the transpose of matrix A respectively)

Options:
A 20
B 22
C 26
D 28

The correct answer is:
28

Solution:
TrA>0⇒x+z+2y>0
AAT=x2yz2yzxzx2yx2yz2yzxzx2y
=x2+4y2+z22yx+2yz+xz2yx+2yz+zx2yx+2yz+xz4y2+x2+z22yz+2xy+zxzx+2xy+2yz2xy+2yz+zxx2+4y2+z2
∵AAT=9I⇒x2+4y2+z2=9, 2yx+2yz+xz=0
x+2y+z2=x2+4y2+z2+22xy+2yz+xz
x+2y+z2=9⇒x+2y+z=3 As, x+2y+z>0
x3+8y3+z3-6xyz=x+2y+zx2+4y2+z2-xz-2xy-2yz
⇒x3+8y3+z3-1=3×9
⇒x3+8y3+z3=28

Topic: Mathematics, Matrices, JEE Main
Consider the matrix fx=cosx-sinx0sinxcosx0001. Given below are two statements :
Statement I:f(-x) is the inverse of the matrix f(x).
Statement II: f(x) f(y)=f(x+y).
In the light of the above statements, choose the correct answer from the options given below

Options:
A Statement I is false but Statement II is true
B Both Statement I and Statement II are false
C Statement I is true but Statement II is false
D Both Statement I and Statement II are true

The correct answer is:
Both Statement I and Statement II are true

Solution:
Given: fx=cosx-sinx0sinxcosx0001

⇒f-x=cosxsinx0-sinxcosx0001

⇒fx×f-x=cosx-sinx0sinxcosx0001cosxsinx0-sinxcosx0001

⇒fx×f-x=100010001

⇒fx×f-x=I

⇒f-x=f-1x

Hence statement-I is correct

Now, checking statement II

⇒fy=cosy-siny0sinycosy0001

⇒fxfy=cosx-sinx0sinxcosx0001cosy-siny0sinycosy0001

⇒fxfy=cosxcosy-sinxsiny-cosxsiny-sinxcosy0sinxcosy+cosxsiny-sinxsiny+cosxcosy0001

⇒fxfy=cos(x+y)-sin(x+y)0sin(x+y)cos(x+y)0001

⇒f(x)·f(y)=f(x+y)

Hence statement-II is also correct.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2024 (27 Jan Shift 1)
Consider the methyl substituted benzoic acids.
1. PhCOOH
2. o-CH3C6H4COOH
3. p-CH3C6H4COOH
4. m-CH3C6H4COOH
The correct sequence of acidity is

Options:
A 1 < 2 < 3 < 4
B 2 < 3 < 4 < 1
C 3 < 4 < 1 < 2
D 3 < 4 < 2 < 1

The correct answer is:
3 < 4 < 1 < 2

Solution:
Both p- and m-methyl groups are electron-donating and acid weakening. Hyperconjugation of p-CH3 produces a little extra δ- on the ring carbon, bearing the COO- . This results in more effective electron-donation and acid-weakening. This makes p-CH3C6H4COOH less acidic than m-CH3C6H4COOH .
o-CH3C6H4COOH is a stronger acid than benzoic acid, due to ortho-effect.
p-CH3C6H4COOH < m-CH3C6H4COOH < PhCOOH < o-CH3C6H4COOH

Topic: Chemistry, General Organic Chemistry, JEE Main
Consider the nuclear fission, Ne20→2He4+C12. Given that the binding energy/nucleon of Ne20, He4 and C12 are 8.03 MeV, 7.86 MeV, respectively. Identify the correct statement:

Options:
A Energy of 12.4 MeV will be supplied.
B Energy of 9.72 MeV has to be supplied.
C Energy of 3.6 MeV will be released.
D 8.3 MeV energy will be released.

The correct answer is:
Energy of 9.72 MeV has to be supplied.

Solution:
Q=Ef-Ei
=12×7.86+4×7.07+4×7.07-(20×8.03)
Q=-9.72 MeV

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2019 (10 Jan Shift 2)
Consider the number \( N=10 ! \). If the number of positive divisors of \( N \) is \( \lambda \), then \( \frac{\lambda}{30} \) equals

Options:

The correct answer is:
9

Solution:
10!=28·34·52·7

Number of divisors =9×5×3×2=270=λ

Topic: Mathematics, Permutation Combination, JEE Main
Consider the oxides of group 14 elements SiO2, GeO2, SnO2, PbO2, CO and GeO. The amphoteric oxides are

Options:
A GeO, GeO2
B SiO2, GeO2
C SnO2, PbO2
D SnO2, CO

The correct answer is:
SnO2, PbO2

Solution:
All members of group 14 when heated in oxygen form oxides. There are mainly two types of oxides, i.e., monoxide and dioxide of formula MO and MO2 respectively. SiO only exists at high temperature. Oxides in higher oxidation states of elements are generally more acidic than those in lower oxidation states. The dioxides CO2, SiO2 and GeO2 are acidic, whereas SnO2 and PbO2 are amphoteric in nature. Amongst monoxides, CO is neutral.

Hence option C is the answer.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2024 (31 Jan Shift 1)
Consider the oxy acids HClOn series, here value of n is 1 to 4. Then incorrect statement regarding these oxy acids is

Options:
A Acidic character of oxy acids increases with increasing value of n
B Oxidizing power of oxy acids increases with decreasing value of n
C Thermal stability of oxy acids decreases with increasing value of n
D Cl-O bond order decreases with decreasing value of n

The correct answer is:
Thermal stability of oxy acids decreases with increasing value of n

Solution:
Here statement given in option C is incorrect while rest other are correct as follows
Acidic character: HOCl<HClO2<HClO3<HClO4
Oxidizing power: HOCl>HClO2>HClO3>HClO4
Thermal stability: HOCl<HClO2<HClO3<HClO4
Cl-O bond order: HOCl<HClO2<HClO3<HClO4 .

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Consider the parabola with vertex 12, 34 and the directrix y=12. Let P be the point where the parabola meets the line x=-12. If the normal to the parabola at P intersects the parabola again at the point Q. then (PQ)2 is equal to :

Options:
A 252
B 758
C 12516
D 152

The correct answer is:
12516

Solution:
Equation of the parabola is x-122=y-34 ...1

Put x=-12 in equation 1, we get y=74

Hence, point P is P-12,74

Now, differentiate equation 1 on both sides, we get y'=2x-12

At point P-12,74, the slope of tangent is -2.

Therefore, the slope of normal is 12

Equation of normal at P-12, 74 is x=2 y-4 ...2

On solving equations 1 & 2, we get the intersection point Q2, 3

∴(PQ)2=2+122+3-742

=254+2516=12516

Topic: Mathematics, Parabola, JEE Main, JEE Main 2021 (01 Sep Shift 2)
Consider the planes P1: 2x-y+z=6 and P2: x+2y-z=4 having normals Nâ1 and Nâ2 respectively. The distance of the origin from the plane passing through the point 1,1,1 and whose normal is perpendicular to N1 and N2 is

Options:
A 75 units
B 75 units
C 35 units
D 1435 units

The correct answer is:
75 units

Solution:
Nâ1=2i^-j^+k^ (given)
Nâ2=i^+2j^-k^ (given)
âµ normal of required plane is Nâ1ÃNâ2
=i^j^k^2-1112-1
=i^-1-j^-3+k^5
=-i^+3j^+5k^
So, equation of the required plane is
-1x-1+3y-1+5z-1=0
-x+3y+5z-7=0
x-3y-5z+7=0
So, distance from 0,0,0 is =0-0+0+735
=75

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Consider the quadratic equation $2 x^2-5 x+2=0$. What can you conclude about the graph of this equation?

Options:
A The graph opens upwards.
B The graph opens downwards.
C The graph intersects the $x$-axis at two distinct points.
D The graph intersects the $x$-axis at a single point.

The correct answer is:
The graph intersects the $x$-axis at two distinct points.

Solution:
The given quadratic equation is $2 x^2-5 x+2=0$. To determine the graph's behavior and the number of its intersections with the $\mathrm{x}$-axis, we can look at the discriminant, $b^2-4 a c$, where $a=2$, $b=-5$, and $c=2$.
Discriminant $=(-5)^2-4(2)(2)=25-16=9$
Since the discriminant is positive $(\gt0)$, the graph intersects the $x$-axis at two distinct points.
Therefore, the correct answer is option 3) The graph intersects the $x$-axis at two distinct points.

Topic: Mathematics, Quadratic Equation, JEE Main
Consider the quadratic equation c-5x2-2cx+c-4=0, c≠5. Let S be the set of all integral values of c for which one root of the equation lies in the interval 0, 2 and its other root lies in the interval 2, 3. Then the number of elements in S is

Options:
A 11
B 12
C 18
D 10

The correct answer is:
11

Solution:
Case -I

c-5>0

⇒c>5 ...i

Then, the graph of the quadratic is shown below

Hence, we have f0>0
⇒c-4>0

⇒c>4 ...ii

And f2<0

⇒4c-5-4c+c-4<0

⇒4c-20-4c+c-4<0
⇒c<24 ...iii

And f3>0

⇒9c-5-6c+c-4>0

⇒9c-45-6c+c-4>0

⇒4c-49>0

⇒c>494 ...iv

Hence, on taking the intersection of all the above inequalities, we get c∈494, 24.
Case -II

c-5<0

⇒c<5 ...v

Then, the graph of the quadratic is shown below

Hence, we have f0<0

⇒c<4 ...vi

And, f2>0

⇒c>24 ...vii

Also, f3<0

⇒c<494 ...viii

Hence, on taking the intersection of all the above inequalities, we get ⇒c∈ϕ

Hence, c∈494, 24

Hence, the total number of element in S are 11.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2019 (10 Jan Shift 1)
Consider the quadratic function $f(x)=a x^2+b x+c$, where $a, b$, and $c$ are real numbers with $a \neq 0$. Which of the following statements correctly describes the range of $f(x)$ ?

Options:
A The range of $f(x)$ is all real numbers.
B The range of $f(x)$ is all real numbers except for a single excluded value.
C The range of $f(x)$ is bounded above or below, but not both.
D The range of $f(x)$ is bounded above and below.

The correct answer is:
The range of $f(x)$ is bounded above or below, but not both.

Solution:
Case 1: $a\gt0$
When $a\gt0$, the parabola opens upwards. In this case, the range of the function is bounded below by the $y$-coordinate of the vertex and extends upward indefinitely. The minimum value of the quadratic function occurs at the vertex. Hence, the range of $f(x)$ is bounded below, but not above.
Case 2: $a \lt 0$
When $a \lt 0$, the parabola opens downwards. In this case, the range of the function is bounded above by the $y$-coordinate of the vertex and extends downward indefinitely. The maximum value of the quadratic function occurs at the vertex. Hence, the range of $f(x)$ is bounded above, but not below.
Given these cases, the correct option is:
3) The range of $f(x)$ is bounded above or below, but not both.

Topic: Mathematics, Quadratic Equation, JEE Main
Consider, the radioactive disintegration ${ }_{82} A^{210} \rightarrow B \rightarrow C \rightarrow{ }_{82} D^{206}$, The sequence of emission can be

Options:
A $\beta, \beta, \beta$
B $\alpha, \alpha, \beta$
C $\beta, \beta, \gamma$
D $\beta, \beta, \alpha$

The correct answer is:
$\beta, \beta, \alpha$

Solution:
Given, $82^{A^{210}} \longrightarrow B \longrightarrow C \longrightarrow{ }_{82} D^{206}$
i.e. difference in mass no. between $A$ and $D$ is of 4 units i.e., mass no. is decreased by 4 units while atomic no. remains the same, i.e. 82 for $A$ and $D$.
Also,
(i) On emission of each $\beta$ -particle, mass no. remains the same but atomic no. is increased by one unit.
(ii) On emission of each $\alpha$ -particle, mass no. is decreased by 4 units, while atomic no. is decreased by 2 units.
Thus, on emission of $2-\beta$ and $1-\alpha$ particle, we get $_{82} D^{206}$
i.e. ${ }_{82} A^{210} \stackrel{-\beta}{\longrightarrow}{ }_{83} A^{210} \stackrel{-\beta}{\longrightarrow}{ }_{84} A^{210} \stackrel{-a}{\longrightarrow}{ }_{82} A^{206}$
Thus, option (d) is the correct answer.

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2019
Consider the reaction,
$2 \mathrm{A}+\mathrm{B} \rightarrow$ Products
When concentration of $\mathrm{B}$ alone was doubled, the half-life did not change. When the concentration of $\mathrm{A}$ alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is

Options:
A
$\mathrm{L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}$

B
no unit

C
$\mathrm{mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$

D
$\mathrm{s}^{-1}$

The correct answer is:
$\mathrm{L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}$

Solution:
$2 \mathrm{A}+\mathrm{B} \rightarrow$ Product
When conc. of $\mathrm{B}$ is doubled, the half life did not change, hence reaction is of first order w.r.t. $\mathrm{B}$.
When concentration of $\mathrm{A}$ is doubled, reaction rate is doubled, hence reaction is of first order w.r.t. $\mathrm{A}$.
Hence over all order of reaction is $1+1=2$
So, unit of rate constant $\mathrm{mol}^{-1} \mathrm{~lit~s}^{-1}$.
Hence, (A) is correct.

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2007
Consider the reaction 2 NO(g) +  O 2 (g)→2  NO 2 (g) ; Predict whether the reaction is spontaneous at 298 K. (Given:   Δ f G NO  = 86.69 kJ/mol,   Δ f G  NO 2  = 51.84 kJ/mol )

Options:
A Yes, spontaneous
B No, the reaction is Non-spontaneous
C Equilibrium
D Cannot predict

The correct answer is:
Yes, spontaneous

Solution:
For the reaction :
2âNO(g)â+â O 2 (g)âââ2â NO 2 (g)
ΔG =  2Δ f G NO 2 −  2Δ f G NO  +  Δ f G O 2
= 2 Ã 5 1 . 8 4 - 2 Ã 8 6 . 6 9 + 0
= 1 0 3 . 6 8 - 1 7 3 . 3 8
= - 6 9 . 7 kJ/mol
âµÎG is negative
so, the reaction is spontaneous

Topic: Chemistry, Thermodynamics (C), JEE Main
Consider the reaction A â 2 B + C, Î H = - 1 5 kcal . The energy of activation of backward reaction is 20 kcal mol-1. In presence of catalyst, the energy of activation of forward reaction is 3 kcal mol-1. At 400 K the catalyst causes the rate of the forward reaction to increase by the number of times equal to-

Options:
A e3.5
B e2.5
C e-2.5
D e2.303

The correct answer is:
e2.5

Solution:
E a f - E a b = Î H = - 1 5 kcal
â E a f = - 1 5 + 2 0 = 5 kcal
k catalyst k a = e E a - E a catalyst / RT
= e 5 - 3 Ã 1 0 3 / 2 Ã 4 0 0
= e 2 . 5

Topic: Chemistry, Chemical Kinetics, JEE Main
Consider the reaction, CH33CâCH|CH3âOH+Conc. HClâX. The product (X) is

Options:
A CH33CCHCH3Cl
B CH32CClCHCH32
C Mixture of both A and B
D CH32CHClClCH32

The correct answer is:
CH32CClCHCH32

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Consider the reaction :
$$
\mathrm{Cl}_2(\mathrm{aq})+\mathrm{H}_2 \mathrm{~S}(\mathrm{aq}) \rightarrow \mathrm{S}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq})
$$
The rate equation for this reaction is rate $=\mathrm{k}\left[\mathrm{Cl}_2\right]\left[\mathrm{H}_2 \mathrm{~S}\right]$ Which of these mechanisms is/are consistent with this rate equation?
(A) $\mathrm{Cl}_2+\mathrm{H}_2 \rightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-}+\mathrm{Cl}^{+}+\mathrm{HS}^{-} \quad$ (slow)
$$
\mathrm{Cl}^{+}+\mathrm{HS}^{-} \rightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-}+\mathrm{S} \text { (fast) }
$$
(B) $\mathrm{H}_2 \mathrm{~S} \Leftrightarrow \mathrm{H}^{+}+\mathrm{HS}^{-} \quad$ (fast equilibrium) $\mathrm{Cl}_2+\mathrm{HS}^{-} \rightarrow 2 \mathrm{Cl}^{-}+\mathrm{H}^{+}+\mathrm{S} \quad$ (slow)

Options:
A
B only

B
Both A and B

C
Neither A nor B

D
A only

The correct answer is:
A only

Solution:
Rate equation is to be derived wrt slow
Step $\therefore$ from mechanism (A)
$$
\text { Rate }=\mathrm{k}\left[\mathrm{Cl}_2\right]\left[\mathrm{H}_2 \mathrm{~S}\right]
$$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2010