datavorous/entrance-exam-dataset huggingface dataset:
Lack of relaxation between successive stimuli in sustained muscle contraction is known as:

Options:
A Spasm
B Fatigue
C Tetanus
D Tonus

The correct answer is:
Tetanus

Solution:
Tonus→Low-level activity of muscles at rest to maintain posture
Spasm→Sudden involuntary muscle contraction
Fatigue→Decline in muscle activity
Tetanus→Sustained muscle contraction in response to successive stimuli. Tetanus is also known as lockjaw disease and is caused by the bacterium Clostridium tetani.

Topic: Biology, Human Health and Diseases, NEET, NEET 2016 (Phase 1)
Lanthanoid contraction is caused due to

Options:
A
the appreciable shielding on outer electrons by 4f electrons from the nuclear charge

B
the appreciable shielding on outer electrons by 5d electrons from the nuclear charge

C
the same effective nuclear charge from Ce to Lu

D
the imperfect shielding on outer electrons by 4f electrons from the nuclear charge

The correct answer is:
the imperfect shielding on outer electrons by 4f electrons from the nuclear charge

Solution:
No Solution Available

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2006
Laplace's correction in the expression for the velocity of sound given by Newton's is needed because

Options:
A sound waves are longitudinal
B sound waves propagate adiabatically
C sound waves propagate isothermally.
D sound waves are of long wavelength.

The correct answer is:
sound waves propagate adiabatically

Solution:
Laplace correction in the expression for the velocity of sound given by Newton's is needed because sound waves propagate adiabatically.

After correction, the velocity of sound formula,

v=γPρWhere, γ= Adiabatic indexP=Pressureρ=Density

Topic: Physics, Waves and Sound, JEE Main
Larger number of oxidation states are exhibited by the actinoids than those by the lanthanoids, the main reason being

Options:
A
4 f orbitals more diffused than the $5 f$ orbitals

B
lesser energy difference between $5 f$ and $6 d$ than between $4 f$ and $5 d$ orbitals

C
more energy difference between $5 f$ and $6 d$ than between $4 f$ and $5 d$ orbitals

D
more reactive nature of the actinoids than the lanthanoids

The correct answer is:
lesser energy difference between $5 f$ and $6 d$ than between $4 f$ and $5 d$ orbitals

Solution:
Being lesser energy difference between $5 f$ and $6 d$ than $4 f$ and $5 d$ orbitals.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2008
Lassaigne's test is used for detection of:

Options:
A Nitrogen and Sulphur only
B Nitrogen, Sulphur and Phosphorous Only
C Phosphorous and halogens only
D Nitrogen, Sulphur, and halogens

The correct answer is:
Nitrogen, Sulphur, and halogens

Solution:
Lassaigne's test helps to detect Nitrogen, Sulphur, and Halogens in organic compounds. Here, sodium metal Na is heated in the presence of the organic compound inside a fusion tube. The primary objective of this test is that all the available elements are converted by Sodium metal Na into the ionic state. Lassaigne test reaction for sodium is given below:

N+Na+C→NaCN

S+2Na→Na2S

X+Na→NaX

In the third experiment, X can be anything such as Cl, Br or I.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (01 Feb Shift 2)
Last element of group-IV is found to be

Options:
A Strong metallic
B Weak metallic
C Strong non-metallic
D Weak non-metallic

The correct answer is:
Weak metallic

Solution:
Correct option is (2). Weak metallic
Down the group metallic character increases. Group IV shows variation from non-metallic character to weak metallic character as we move down the group.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Lateral branches originate from the basal & underground portion of the main stem, grow horizontally beneath the soil & then come out obliquely upward giving rise to leafy shoots. This condition does not belong to which plant?

Options:
A Banana
B Pineapple
C Chrysanthemum
D Oxalis

The correct answer is:
Oxalis

Solution:
In banana, Chrysanthemum and pineapple, lateral branches originate from the basal & underground portion of the main stem, grow horizontally beneath the soil & then come out obliquely upward giving rise to leafy shoots.

Topic: Biology, Morphology of Flowering Plants, NEET
Lattice energy of an ionic compound depends upon

Options:
A Charge on the ion only
B Size of the ion only
C Packing of ions only
D Charge on the ion and size of the ion

The correct answer is:
Charge on the ion and size of the ion

Solution:
The value of lattice energy depends on the charges present on the two ions and distance between them. It shell be high if charges are high and ionic radii are small.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Lattice energy of an ionic compound depends upon :

Options:
A Charge on the ion only
B Size of the ion only
C Packing of ions only
D Charge and size of the ion

The correct answer is:
Charge and size of the ion

Solution:
Lattice Energyâq1q2rc+ra
Where q represents charge of ion and r represent radius of ion.
Lattice energy does not dependent packing of ions.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Lattice energy of an ionic compounds depends upon

Options:
A
Charge on the ion only

B
Size of the ion only

C
Packing of ions only

D
Charge on the ion and size of the ion

The correct answer is:
Charge on the ion and size of the ion

Solution:
No Solution Available

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2005
Lattice enthalpy and enthalpy of solution of NaCl are 788kJmol-1and 4kJmol -1, respectively. The hydration enthalpy of NaCl is:

Options:
A -780kJmol-1

B 780kJmol-1

C -784kJmol-1

D 784kJmol-1

The correct answer is:
-784kJmol-1

Solution:
Enthalpy of solution= Lattice enthalpy + enthalpy of hydration

4=788+ ∆Hhydration⇒∆Hhydration=-784 KJ/mol

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2020 (05 Sep Shift 2)
Lead storage battery contains 38% by weight solution of H2SO4. The van't Hoff factor is 2.67 at this concentration. The temperature in Kelvin at which the solution in the battery will freeze is _____ (Nearest integer).

Given Kf=1.8 K kg mol-1

Options:

The correct answer is:
243

Solution:
The depression in freezing point of the solution can be written as follows,

ΔTf=i·Kf·m

molality (m) =w/w%Molar mass of solute×1000mass of solvent in grams

⇒ΔTf=2.67×1.8×3898×100062

⇒ΔTf=30.05

∴ Freezing point of solution at which battery freeze=243 K

Topic: Chemistry, Solutions, JEE Main, JEE Main 2023 (30 Jan Shift 2)
Lead sulphate is soluble

Options:
A In conc. nitric acid
B In conc. hydrochloric acid
C In a solution of ammonium acetate
D In water

The correct answer is:
In a solution of ammonium acetate

Solution:
Lead sulphate is solube in a solution of ammonium acetate.
because the reaction,
$\mathrm{Pb}\left(\mathrm{SO}_4\right)_2 \xrightarrow[\text { acetate }]{\text { Ammonium }}$ lead acetate + Ammonium sulphate

Topic: Chemistry, Practical Chemistry, JEE Main
Lead sulphate is soluble

Options:
A In conc. nitric acid
B In conc. hydrochloric acid
C In a solution of ammonium acetate
D In water

The correct answer is:
In conc. nitric acid

Solution:
Lead sulphate is soluble In conc. nitric acid.

Topic: Chemistry, Practical Chemistry, JEE Main
Least count of a vernier caliper is $\frac{1}{20 \mathrm{~N}} \mathrm{~cm}$. The value of one division on the main scale is $1 \mathrm{~mm}$. Then the number of divisions of main scale that coincide with $\mathrm{N}$ divisions of vernier scale is :

Options:
A $(2 \mathrm{~N}-1)$
B $\left(\frac{2 \mathrm{~N}-1}{2 \mathrm{~N}}\right)$
C $\left(\frac{2 \mathrm{~N}-1}{2}\right)$
D $\left(\frac{2 \mathrm{~N}-1}{20 \mathrm{~N}}\right)$

The correct answer is:
$\left(\frac{2 \mathrm{~N}-1}{2}\right)$

Solution:
Least count of vernier calipers $=\frac{1}{20 \mathrm{~N}} \mathrm{~cm}$
$\because$ Least count $=1 \mathrm{MSD}-1 \mathrm{VSD}$
let $\mathrm{x}$ no. of divisions of main scale coincides with $\mathrm{N}$ division of vernier scale, then
$\begin{aligned} & 1 \mathrm{VSD}=\frac{\mathrm{x} \times 1 \mathrm{~mm}}{\mathrm{~N}} \\ & \therefore \frac{1}{20 \mathrm{~N}} \mathrm{~cm}=1 \mathrm{~mm}-\frac{\mathrm{x} \times 1 \mathrm{~mm}}{\mathrm{~N}} \\ & \frac{1}{2 \mathrm{~N}} \mathrm{~mm}=1 \mathrm{~mm}-\frac{\mathrm{x}}{\mathrm{N}} \mathrm{mm} \\ & \mathrm{x}=\left(1-\frac{1}{2 \mathrm{~N}}\right) \mathrm{N} \\ & \mathrm{x}=\frac{2 \mathrm{~N}-1}{2}\end{aligned}$

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2024 (08 Apr Shift 2)
Least stable peroxide among the following:

Options:
A MgO2
B CaO2
C SrO2
D BaO2

The correct answer is:
MgO2

Solution:
Stability of peroxides of group-2 elements increases down the group

Topic: Chemistry, s Block Elements, JEE Main
Le-chatelier principle is applicable

Options:
A Both for physical and chemical equilibrium
B Only for chemical equilibrium
C Only for physical equilibrium
D Neither for (b) nor for $s$

The correct answer is:
Both for physical and chemical equilibrium

Solution:
Both for physical and chemical equilibrium

Topic: Chemistry, Chemical Equilibrium, JEE Main
lectronic configuration of \(\mathrm{Cu}(\mathrm{II})\) is \(3 \mathrm{~d}^9\) whereas that of \(\mathrm{Cu}(\mathrm{I})\) is \(3 \mathrm{~d}^{10}\). Which of the following is correct?

Options:
A Stability of \(\mathrm{Cu}(\mathrm{I})\) and \(\mathrm{Cu}(\mathrm{II})\) depends on nature of copper salts
B \(\mathrm{Cu}(\mathrm{II})\) is more stable
C \(\mathrm{Cu}(\mathrm{I})\) and \(\mathrm{Cu}(\mathrm{II})\) are equally stable
D \(\mathrm{Cu}\) (II) is less stable

The correct answer is:
\(\mathrm{Cu}(\mathrm{II})\) is more stable

Solution:
$\mathrm{Cu}(\mathrm{II})$ is more stable than $\mathrm{Cu}(\mathrm{I})$ because hydration energy of $\mathrm{Cu}^{+2}$ ion compensate $\mathrm{IE}_2$ of $\mathrm{Cu}$.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (09 Apr Shift 1)
LED is constructed from Ga-As-P semiconducting material. The energy gap of this LED is 1.9 eV. Calculate the wavelength of light emitted and its colour. h=6.63×10-34 J-s and c=3×108 m s-1

Options:
A 654 nm and red colour
B 654 nm and orange colour
C 1046 nm and red colour
D 1046 nm and blue colour

The correct answer is:
654 nm and red colour

Solution:
λ=hcE=6.6×10-34×3×1081.9×1.6×10-19=6.54×10-7

=654 nm

Red color

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (26 Feb Shift 1)
Leguminous plants are able to fix atmospheric nitrogen through the process of symbiotic nitrogen fixation. Which one of the following statement is not correct during the process of nitrogen fixation?

Options:
A Leghaemoglobin scavenges oxygen and is pinkish in colour
B Nodules act as a site for nitrogen fixation
C The enzyme nitrogenase catalyses the conversion of atmospheric $\mathrm{N}_2$ to $\mathrm{NH}_3$
D Nitrogenase is insensitive to oxygen

The correct answer is:
Nitrogenase is insensitive to oxygen

Solution:
Root nodules of leguminous plants harbour symbiotic bacteria, Rhizobium, which fixes atmospheric nitrogen and provides the plant with most or all of its nitrogen requirements. In return, they have access to a rich supply of carbohydrate. The fixation of atmospheric nitrogen $\left(\mathrm{N}_2\right)$ to ammonia $\left(\mathrm{NH}_3\right)$ occurs within the nodules, using the enzyme nitrogenase but this enzyme is very sensitive to oxygen and gets inhibited in oxygen rich environment. This problem is solved by leghaemoglobin, a pinkish-red, oxygen-binding, iron containing protein present in root nodules. It functions as oxygen buffer, scavanges oxygen and provides the enzyme nitrogenase, a low oxygen environment in which nitrogen fixation can occur.

Topic: Biology, Mineral Nutrition, NEET, NEET 2010 (Mains)
Length cannot be measured by

Options:
A Fermi
B Debye
C Micron
D Light year

The correct answer is:
Debye

Solution:
The length can be measured in terms of Fermi, Micron and Light year. whereas a unit used to express electric dipole moments of molecules is debye.
So, option 2 is correct.

Topic: Physics, Units and Dimensions, JEE Main
Length of a string tied to two rigid supports is $40 \mathrm{~cm}$. Maximum length (wave length in $\mathrm{cm}$ ) of a stationary wave produced on it is

Options:
A
20

B
80

C
40

D
120

The correct answer is:
80

Solution:
$\lambda_{\max } / 2=40 \Rightarrow \lambda_{\max }=80$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2002
Length of the horizontal arm of a uniform cross-section U-tube is l=5 cm and both ends of the vertical arms are open to the surrounding pressure of 7700 N m-2. A liquid of density Ï=103 kg m-3 is poured into the tube such that the liquid just fills the horizontal part of the tube. Now, one of the open ends is sealed and the tube is then rotated about a vertical axis passing through the other vertical arm with angular velocity Ï due to which the liquid rises up to half the length of the vertical arm. If the length of each vertical arm is a=6 cm. What is the value of Ï (in rad s-1)?

Options:

The correct answer is:
100

Solution:
Let cross-section area of the tube =A
In the right limb for compressed air

x=3 cm
p1V1=p2V2
p0AÃ6=p2AÃ3
p2=2p0........(1)
â
Force at the corner of the right limb due to liquid above,
F1=2p0+xÏgA
Mass of the liquid in the horizontal arm, m=Ï(l-x)A
It is rotated about the left limb, then centripetal force,
mÏ2r=Ïl-xAÏ2Ãl+x2=ÏAÏ22Ãl2-x2

ÏAÏ2l2-x22=2p0+xÏgA-p0A

Ï=2p0+xÏgÏl2-x2=100 rad s-1

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Length, width and mass of a rectangular plate are \( \ell, \mathrm{b} \) and \( \mathrm{m} \) respectively. The radius of
gyration about the axis passing through centre and perpendicular to the plane is :

Options:
A \( \sqrt{\frac{\ell^{2}+\mathrm{b}^{2}}{2}} \)
B \( \sqrt{\frac{\ell^{2}+\mathrm{b}^{2}}{8}} \)
C \( \sqrt{\frac{\ell^{2}+\mathrm{b}^{2}}{12}} \)
D \( \sqrt{\frac{\ell^{2}+b^{3}}{12}} \)

The correct answer is:
\( \sqrt{\frac{\ell^{2}+\mathrm{b}^{2}}{12}} \)

Solution:
Moment of inertia of a rectangular plate about an axis passing through its centre and perpendicular to the plane is given by

I=ml2+b212 ....1

As radius of gyration k is given by using the expression

I=mk2

So, k=Im

⇒k=l2+b212.

Topic: Physics, Rotational Motion, JEE Main
Lenticels are involved in

Options:
A transpiration
B gaseous exchange
C food transport
D photosynthesis

The correct answer is:
gaseous exchange

Solution:
Plants have stomata and lenticles involved in gaseous exchange. In stems the living cells are organised in thin layers inside and beneath the bark. They have openings called lenticles.
Transpiration is the evaporative loss of water by plants. It occurs mainly through stomata in leaves Food transport accurs by the vascular tissue phloem from a source to a sink. Photosynthesis is a physico-chemical process by which they use light energy to derive re synthesis of organic compounds.

Topic: Biology, Transport in Plants, NEET, NEET 2013 (All India)
Lenz's law gives

Options:
A The magnitude of the induced e.m.f.
B The direction of the induced current
C Both the magnitude and direction of the induced current
D The magnitude of the induced current

The correct answer is:
The direction of the induced current

Solution:
Option (2): the direction of the induced current
1. Lenz's law is known as a law that basically states that the current induced in the circuit because of a change in the magnetic field is directed to oppose the change in the flux, which exerts a mechanical force that opposes the motion.
2. Lenz's law can be given by an equation : $\varepsilon=-\frac{\partial \varnothing_B}{\partial t}$
3.The given equation basically indicates that the induced electromotive force $\varepsilon$ and the rate of magnetic flux have opposite signs.
4. Which refers to the direction of the back EMF of an induced field that opposes the changing current.
5. Thus Lenz's law generally helps us to give the direction of the induced current.

Topic: Physics, Electromagnetic Induction, JEE Main
Lenz's law is a consequence of law of conservation of

Options:
A energy only
B charge only
C momentum only
D energy and momentum

The correct answer is:
energy only

Solution:
Lenz's law is based upon the law of conservation of energy.

Topic: Physics, Electromagnetic Induction, JEE Main
Lenz's law is consequence of the law of conservation of

Options:
A Charge
B Momentum
C Mass
D Energy

The correct answer is:
Energy

Solution:
The energy of the field increases with the magnitude of the field. Lenz's law infers that there is an opposite field created due to increase or decrease of magnetic flux around a conductor so as to hold the law of conservation of energy.

Topic: Physics, Electromagnetic Induction, JEE Main
Lenz's law is expressed by the following formula (here $e=$ induced e.m.f., $\varphi=$ magnetic flux in one turn and $N=$ number of turns)

Options:
A $e=-\phi \frac{d N}{d t}$
B ${ }^{e=-N \frac{d \phi}{d t}}$
C ${ }^{e=-\frac{d}{d t}\left(\frac{\phi}{N}\right)}$
D $e=N \frac{d \phi}{d t}$

The correct answer is:
${ }^{e=-N \frac{d \phi}{d t}}$

Solution:
Correct option is (2) $e=-N \frac{d \phi}{d t}$
An induced electromotive force (emf) always gives rise to a current whose magnetic field opposes the change in original magnetic flux.
$e=-N \frac{d \phi}{d t}$

Topic: Physics, Electromagnetic Induction, JEE Main
Less polluted water sample is

Options:
A \( 2 \) milliequivalents of \( K_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) (in \( 50 \% \mathrm{H}_{2} \mathrm{SO}_{4} \) ) consumed for \( 100 \mathrm{~kg} \) water sample on oxidizing the pollutan
B \( 2 \) milliequivalents of oxygen consumed for oxidizing the pollutant in \( 1000 \) litre of water sample
C \( 0.2 \mathrm{~g} \) of oxygen consumed for oxidizing the pollutant million parts of water
D \( 49 \times 10^{-3} \mathrm{~g} \) of \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) units \( 50 \% \mathrm{H}_{2} \mathrm{SO}_{4} \) is required to oxidize the pollutant in \( 10^{6} \mathrm{~mL} \) of water sample

The correct answer is:
\( 2 \) milliequivalents of oxygen consumed for oxidizing the pollutant in \( 1000 \) litre of water sample

Solution:
2 milliequivalents of oxygen consumed for oxidizing the pollutant in 1000 litre of water sample

Topic: Chemistry, Environmental Chemistry, JEE Main
Let 0<Î¸1<Î¸2<Î¸3<â¦ denotes the positive solutions of the equation 3+3cosâ¡Î¸=2sin2â¡Î¸. If Î¸3+Î¸7=aÏ, where a is an integer, then the value of a is equal to

Options:
A 6
B 7
C 8
D 4

The correct answer is:
6

Solution:
3+3cosâ¡Î¸=2sin2â¡Î¸=2-2cos2â¡Î¸.
â2cos2â¡Î¸+3cosâ¡Î¸+1=0
âcosâ¡Î¸=-1,-12
âÎ¸=2n+1Ï or 2nÏÂ±2Ï3
âÎ¸=2Ï3,Ï,4Ï3,8Ï3,3Ï,10Ï3,14Ï3,5Ï
i.e. Î¸3+Î¸7=4Ï3+14Ï3=18Ï3=6Ï
âa=6

Topic: Mathematics, Trigonometric Equations, JEE Main
Let α∈0,π2 and fx=x2+x+tan2⁡αx2+x,x>0. If the least value of fx is 23, then α is equal to

Options:
A π3
B π8
C π6
D π4

The correct answer is:
π3

Solution:
By AM-GM inequality
x2+x+tan2αx2+x≥2⋅x2+x⋅tan2αx2+x=2tanα
Since the least value of fx is 2tan⁡α=23 (given)
Hence, α=π3

Topic: Mathematics, Sequences and Series, JEE Main
Let α∈(0,π2) , be constant.If the integral ∫tanx+tanαtanx-tanαdx=Axcos2α+Bxsin2α+C, where C is a constant of integration, then the functions A(x) and B(x) are respectively

Options:
A x-α and logesinx-α
B x+Î± and logecosx-Î±Â
C x+Î± and logesinx+Î±
D x-Î± and logecosx-Î±

The correct answer is:
x-α and logesinx-α

Solution:
I=∫tanx+tanαtanx-tanαdx=∫sinxcosx+sinαcosαsinxcosx-sinαcosαdx
=∫sinxcosα+cosxsinαsinxcosα-cosxsinαdx=∫sinx+αsinx-αdx
=∫sinx-α+2αsin(x-α)dx=∫sinx-αcos2α+cosx-αsin2αsin(x-α)dx
=∫cos2α+sin2αcotx-αdx
=xcos2α+sin2αlogesinx-α+C
=(x-α)cos2α+sin2αlogesinx-α+C
Hence, Ax=x-α and Bx=logesinx-α

Topic: Mathematics, Indefinite Integration, JEE Main
Let α∈(0,π2) , be constant.If the integral ∫tanx+tanαtanx-tanαdx=Axcos2α+Bxsin2α+C, where C is a constant of integration, then the functions A(x) and B(x) are respectively

Options:
A x-α and logesinx-α
B x+Î± and logecosx-Î±Â
C x+Î± and logesinx+Î±
D x-Î± and logecosx-Î±

The correct answer is:
x-α and logesinx-α

Solution:
I=∫tanx+tanαtanx-tanαdx=∫sinxcosx+sinαcosαsinxcosx-sinαcosαdx
=∫sinxcosα+cosxsinαsinxcosα-cosxsinαdx=∫sinx+αsinx-αdx
=∫sinx-α+2αsin(x-α)dx=∫sinx-αcos2α+cosx-αsin2αsin(x-α)dx
=∫cos2α+sin2αcotx-αdx
=xcos2α+sin2αlogesinx-α+C
=(x-α)cos2α+sin2αlogesinx-α+C
Hence, Ax=x-α and Bx=logesinx-α

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2019 (12 Apr Shift 2)
Let 0<θ<π2. If the eccentricity of the hyperbola x2cos2⁡θ-y2sin2⁡θ=1 is greater than 2, then the length of its latus rectum lies in the interval:

Options:
A 3, ∞
B 1, 32
C 2, 3
D 32, 2

The correct answer is:
3, ∞

Solution:
For given hyperbola the eccentricity is given as
e2=1+sin2θcos2θ=1+tan2θ=sec2θ
⇒e=secθ
∴ Length of latus rectum l=2sin2θcosθ=2tan2θsecθ
⇒l=2(e2-1)e=2e-1e
On differentiating w.r.t. e, we get

dlde=21+1e2>0
∴ l is an increasing function
∴lmin=22-12=3
∴ Range of latus rectum is 3,∞.

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2019 (09 Jan Shift 1)
Let $\theta \in\left(0, \frac{\pi}{4}\right)$ and $t_1=(\tan \theta)^{\tan \theta}, t_2=(\tan \theta)^{\cot \theta}$ and $t_4=(\cot \theta)^{\tan \theta}$, then

Options:
A
$t_1>t_2>t_3>t_4$

B
$t_4>t_3>t_1>t_2$

C
$t_3>t_1>t_2>t_4$

D
$t_2>t_3>t_1>t_4$

The correct answer is:
$t_4>t_3>t_1>t_2$

Solution:
As when $\theta \in\left(0, \frac{\pi}{4}\right)$
$\tan \theta < \cot \theta$
Since, $\tan \theta < 1$ and $\cot \theta>1$
$\therefore(\tan \theta)^{\cot \theta} < 1$ and $(\cot \theta)^{\tan \theta}>1$
$\therefore t_4>t_1$ which only holds in option (b).

Topic: Mathematics, Trigonometric Equations, JEE Main
Let Î=â«024Ïsinâ¡xdx, then the value of 2I is equal to (where, . denotes the fractional part function)

Options:
A 10π
B 24π
C 12π
D 4π

The correct answer is:
24π

Solution:
Applying a+b-x property,
Î=â«024Ïsin24Ï-xdx
=â«024Ï-sinâ¡xdx
Adding the two integrals, we get,
2Î=â«024Ïsinâ¡x+-sinâ¡xdx
Now, Î±+-Î±=0;Î±âÎ1;Î±âÎ
As, sinâ¡x takes integral values only at discrete points,
Hence,
2Î=â«1024Ïdx=x024Ï=24Ï-0
=24Ï

Topic: Mathematics, Definite Integration, JEE Main
Let $\alpha \beta \neq 0$ and $A=\left[\begin{array}{rrr}\beta & \alpha & 3 \\ \alpha & \alpha & \beta \\ -\beta & \alpha & 2 \alpha\end{array}\right]$. If $B=\left[\begin{array}{rrr}3 \alpha & -9 & 3 \alpha \\ -\alpha & 7 & -2 \alpha \\ -2 \alpha & 5 & -2 \beta\end{array}\right]$ is the matrix of cofactors of the elements of $A$, then $\operatorname{det}(A B)$ is equal to :

Options:
A 64
B 216
C 343
D 125

The correct answer is:
216

Solution:
Equating co-factor fo $\mathrm{A}_{21}$
$\begin{aligned}
& \left(2 \alpha^2-3 \alpha\right)=\alpha \\
& \alpha=0,2 \text { (accept) }
\end{aligned}$
Now, $2 \alpha^2-\alpha \beta=3 \alpha$
$\begin{aligned}
& \alpha=2 \quad \beta=1 \\
& |\mathrm{AB}|=|\mathrm{A} \operatorname{cof}(\mathrm{A})|=|\mathrm{A}|^3 \\
& \mathrm{~A}=\left|\begin{array}{lll}
1 & 2 & 3 \\
2 & 2 & 1 \\
-1 & 2 & 4
\end{array}\right|=6-2(9)+3(6)=6
\end{aligned}$

Topic: Mathematics, Matrices, JEE Main, JEE Main 2024 (05 Apr Shift 2)
Let λ≠0 be in R. If α and β are the roots of the equation x2-x+2λ=0, and α and γ are the roots of the equation 3x2-10x+27λ=0, then βγλ is equal to ________.

Options:

The correct answer is:
18

Solution:
α+β=1 ...i

αβ=2λ ...ii
α+γ=103 ...iii
αγ=9λ ...iv
from iv & ii
γβ=92 ...v

from iii-i

γ-β=103-1=73 ...vi

from v and vi

β=23,γ=3
from i α=1-β=1-23=13
from ii λ=αβ2=13×232=19
So βγλ=23×319=18

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2021 (26 Aug Shift 2)
Let λ≠0 be in R. If α and β are the roots of the equation, x2-x+2λ=0 and α and γ are the roots of the equation, 3x2-10x+27λ=0, then βγλ is equal to:

Options:
A 27
B 18
C 9
D 36

The correct answer is:
18

Solution:
Satisfying α we have,

3α2-10α+27λ=0 …i and
3α2-3α+6λ=0 …ii
Subtract

-7α+21λ=0
3λ=α
by ii, 9λ2-3λ+2λ=0
⇒λ=0,19 but λ≠0
∴ Given equation are x2-x+29=0 and 3x2-10x+3=0
∴α=13,β=23,α=13,γ=3
∴βγλ=23·319=18

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2020 (04 Sep Shift 2)