datavorous/entrance-exam-dataset huggingface dataset:
Induction of flowering by low temperature treatment is

Options:
A vernalization
B cryobiology
C photoperiodism
D prunning

The correct answer is:
vernalization

Solution:
Vernalization is chilling or cold treatment of the young plants or seeds to induce flowering. It is a process of shortening of the juvenile or vegetative phase and fastening the flowering by a process of cold treatment. It was first reported by a Russion worker, Lysenko (1928) while working on cold requiring biennial plants. Common examples of plants requiring vernalization are winter rye, winter wheat, winter barley, pea, Chrysanthemum etc.Cryobiology is the study of life at low temperature, usually at liquid nitrogen (ie $-196^{\circ} \mathrm{C}$ ) and preservation of organisms in liquid by chemical methods.Photoperiodism is the effect of photoperiods or daily duration of light hours (and dark periods) on the growth and development mainly flowering in plants. It was first studied by 'Garner and Allard' (1920) on 'Maryland Mammoth' variety of tobacco.

Topic: Biology, Plant - Growth and Development, JIPMER, JIPMER 2009
Indusium is found in

Options:
A algae
B ferns
C moss
D Cycas

The correct answer is:
ferns

Solution:
Indusium is a membranous structure which covers a group of sporangia below it. It is differentiated, into a kidney-shaped dorsal plate called hood and a short stalk through which it remains attached with the undersurface of the pinnule at a swollen base called placenta. The sporangia attached with a placenta form group called sorus (pl. sori). Different types of sorus (simple, gradate, mixed) are seen to be present in a box like structure called sporocarp formed on the vegetative leaf, $i e$, frond of ferns.
Such spore producing structure, sorus, covered by a membranous structure is not found in any group of plants (Algae, Bryophyta, Gymnosperms) except, ferns of Pteridophyta.

Topic: Biology, Plant Kingdom, JIPMER, JIPMER 2009
Inert gases have positive electron gain enthalpy. Its correct order is

Options:
A Xe<Kr<Ne<He
B He<Ne<Kr<Xe
C He<Xe<Kr<Ne
D He<Kr<Xe<Ne

The correct answer is:
He<Xe<Kr<Ne

Solution:
All the given elements belong to the group - 18. All noble gases have general electronic configuration ns2np6 except helium which has 1s2. Since noble gases have stable electronic configuration, they have no tendency to accept the electron and therefore, have large positive values of electron gain enthalpy.

Values of electron gain enthalpy of all the elements are given in the below table

Element	ΔegH[KJ/mol]

He	+48

Ne	+116

Kr	+96

Xe	+77

So, the order is He < Xe < Kr < Ne.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2023 (25 Jan Shift 1)
Infinite number of bodies, each of mass $2 \mathrm{~kg}$ are situated on $x$-axis at distance $1 \mathrm{~m}, 2 \mathrm{~m}, 4 \mathrm{~m}, 8 \mathrm{~m}$, respectively from the origin. The resulting gravitational potential due to this system at the origin will be

Options:
A $-G$
B $-\frac{8}{3} G$
C $-\frac{4}{3} G$
D $-4 G$

The correct answer is:
$-4 G$

Solution:
The resulting gravitational potential,
$\begin{array}{ll}
& V=-2 G\left[\frac{1}{1}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots\right] \\
\Rightarrow & V=-2 G\left[1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3} \ldots\right] \\
\Rightarrow & V=-2 G\left(1+\frac{1}{2}\right)^{-1} \\
\Rightarrow & V=-\frac{2 G}{\left(1-\frac{1}{2}\right)}=-\frac{2 G}{\left(\frac{1}{2}\right)}=-4 G
\end{array}$

Topic: Physics, Gravitation, JEE Main
Infinite number of masses, each 1 kg, are placed along the x-axis at x=Â±1m, Â±2m, Â±4m, Â±8m, Â±16m ..... The magnitude of the resultant gravitational potential in terms of gravitational constant G at the origin (x=0) is

Options:
A G
B 3G
C 2G
D 8G

The correct answer is:
2G

Solution:
Gravitational potential at a distance r from a mass = V=GMr
Net Gravitational potential = âGMri
V=GM1r1+1r2+1r3+â¦
=GÃ111+12+14+18+116+â¦.
â´sum of GP=a1-r
V=2G

Topic: Physics, Gravitation, JEE Main
Infinite springs with force constants k,2k,4k and 8kâ¦.. respectively are connected in series. The effective force constant of the spring will be

Options:
A 2k
B k
C k/2
D k/4

The correct answer is:
k/2

Solution:
Effective force constant is equal to the reciprocal of the sum of individual force constant, hence
1ke=1k1+1k2+1k3+…
Given, k1=k, k2=2k, k3=3k,…..
∴ 1ke=1k+12k+14k+18k+…
The given series a geometric progression series, hence sum is
S∞=a1-r
where a is first term of series and r the common difference.
⇒ 1ke=1k×11-12=2k⇒ke=k2

Topic: Physics, Oscillations, JEE Main
Infra red radiation is detected by

Options:
A
spectrometer

B
pyrometer

C
nanometer

D
photometer

The correct answer is:
pyrometer

Solution:
No Solution Available

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2002
Inhibin is composed of

Options:
A glycoprotein
B lipoprotein
C steroid
D amino acid derivative

The correct answer is:
glycoprotein

Solution:
Inhibins are dimeric glycoproteins composed of an $\alpha$-subunit and either a $\beta_{\mathrm{A}}$ (inhibin A) or a $\beta_{\mathrm{B}}$ subunit (inhibin B). They are mainly produced in gonads and provide negative regulation of FSH secretion.

Topic: Biology, Chemical Coordination and Integration, JIPMER, JIPMER 2018
Initial angular velocity of a circular disc of mass $M$ is $\omega_1$. Then two small spheres of mass $m$ are attached gently to diametrically opposite points on the edge of the disc. What is the final angular velocity of the disc?

Options:
A
$\left(\frac{M+m}{M}\right) \omega_1$

B
$\left(\frac{\mathrm{M}+\mathrm{m}}{\mathrm{m}}\right) \omega_1$

C
$\left(\frac{\mathrm{M}}{\mathrm{M}+4 \mathrm{~m}}\right) \omega_1$

D
$\left(\frac{M}{M+2 m}\right) \omega_1$

The correct answer is:
$\left(\frac{\mathrm{M}}{\mathrm{M}+4 \mathrm{~m}}\right) \omega_1$

Solution:
Angular momentum = conserved
$\frac{1}{2} \mathrm{MR}^2 \omega_1=2 m R^2 \omega+\frac{1}{2} \mathrm{MR}^2 \omega \Rightarrow \omega=\frac{\mathrm{M} \omega_1}{\mathrm{M}+4 \mathrm{~m}}$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2002
Initially 3 moles of 'A' was taken in a 1 L container. The approx. moles of A left in the container when the following equilibrium established is x×10-6 .The value of ' x ' is.
3A⇌B;KC=8×1015

Options:

The correct answer is:
5

Solution:
Now,
1-x(3x⁡)3=8×1015
3x=5×10-6

Topic: Chemistry, Chemical Equilibrium, JEE Main
Initially, 0.8 mol of PCl5 and 0.2 mol of PCl3 are mixed in a one litre vessel. At equilibrium, 0.4 mol of PCl3 is present. The value of Kc for the reaction, PCl5g⇌PCl3g+Cl2g, would be

Options:
A 0.13 mol L-1
B 0.66Â molÂ L-1
C 0.013Â molÂ L-1
D 0.05Â molÂ L-1

The correct answer is:
0.13 mol L-1

Solution:
So x=0.2
Kc=PCl3Cl2PCl5=0.4Ã0.20.6
=0.13 mol L-1

Topic: Chemistry, Chemical Equilibrium, JEE Main
Initially, a beaker has 100 g of water at temperature 90 Â°C. Later another 600 g of water at temperature 20 Â°C was poured into the beaker. The temperature, T of the water (in Â°C) after mixing is

Options:

The correct answer is:
30

Solution:
Given,
Mass of water at 90o C=100 g
=100×10-3 kg
Mass of water at 20 °C=600 g
=600×10-3 kg
From calorimetry
100×10−3Sw(90 °C−T °C)=600×10−3Sw(T °C−20 °C)
90−T=6T−120
T=2107=30 °C

Topic: Physics, Thermodynamics, JEE Main
Initially, a block of mass m is at rest on a frictionless floor and the spring is in a relaxed condition. One end of the spring is rigidly attached to the block and the other end is fixed to a wall. A constant force is applied on the block as shown in the figure. The maximum velocity of the block is

Options:
A FmK
B F2mK
C F2mK
D 2FmK

The correct answer is:
FmK

Solution:
The block attains maximum speed when it passes through its equilibrium position.

F=Kx

x=FK

From work-energy theorem

-12 K KF2+FFK=12mvmax2

F22K=12mvmax2

vmax=FmK

Topic: Physics, Work Power Energy, JEE Main
Initially a gas of diatomic molecules is contained in a cylinder of volume V1 at a pressure P1 and temperature 250 K. Assuming that 25% of the molecules get dissociated causing a change in number of moles. The pressure of the resulting gas at temperature 2000 K, when contained in a volume 2V1 is given by P2. The ratio P2/P1 is -

Options:

The correct answer is:
5

Solution:
P1=P1
v1=v1
T1=250
n1=n0
n2=0.75n0+2×0.25n0=1.25no
v2=2V1
T2=2000
P1V1=n1RT1
P2V2=n2RT2
P1V1=n0R×250
P22V1=.25n0R(2000)
equation (1) /(2)
P12P2=2501.25×2000
P1P2=15
P2P1=5

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2020 (06 Sep Shift 1)
Initially, both the blocks are at rest on horizontal surface as shown in the figure. Find the minimum value of force F in N so that sliding starts between the blocks g=10 m s-2

Options:

The correct answer is:
80

Solution:
F-10μg-20μg=10a .......(i)
μ10g=10a ..................(ii)

From (i) and (ii)

F=40μg
F=80 N

Topic: Physics, Laws of Motion, JEE Main
Initially charge on \( C_{1} \) is \( q_{0} \) and all other capacitors are uncharged. If switch is closed at \( t=0 \), then the charge on \( C_{4} \) after along time of closing the switch is,

Options:
A \( q_{0} \)
B \( \frac{q 0}{2} \)
C \( \frac{q 0}{3} \)
D \( \frac{q 0}{4} \)

The correct answer is:
\( \frac{q 0}{3} \)

Solution:
In parallel, total charge distribution is the direct ratio of capacitance.

Hence, q4=q03.

Topic: Physics, Capacitance, JEE Main
Initially $n$ identical capacitors are joined in parallel and are charged to potential $V$. Now they are separated and joined in series. Then

Options:
A Potential difference becomes $n V$ and energy remains the same.
B potential difference is $n V$ and energy increases $n$ times
C potential difference total energy of combination remains same
D potential difference remains the same and the energy increases $n$ times.

The correct answer is:
Potential difference becomes $n V$ and energy remains the same.

Solution:
Charge will not flow and potential difference across each capacitor will be $V$. Hence, total potential difference across the series combination will be $n V$.
Total energy will remain same as charge is not flowing.

Topic: Physics, Capacitance, MHT CET, MHT CET 2022 (11 Aug Shift 1)
Initially, the air in an open vessel is maintained at 60 Â°C. Later on, the vessel is heated to a temperature T, so that one-fourth mass of air escapes from the vessel. If the volume of the vessel remains constant during heating, the value of T is

Options:
A 80 °C
B 444 °C
C 333 °C
D 171 Â°C

The correct answer is:
171 Â°C

Solution:
Since it is an open vessel of constant volume, the pressure and volume of the gas remain constant but the number of moles and the temperature changes.

PV=n1RT1=n2RT2

âT2T1=n1n2=134

âT2=43T1=43Ã60+273

âT2=444 Kâ¡171 Â°C

Topic: Physics, Thermodynamics, JEE Main
Initially, the root-mean-square (RMS) velocity of N2 molecules at certain temperature is u. If this temperature is doubled and all the nitrogen molecules dissociate into nitrogen atoms, then the new RMS velocity will be:

Options:
A 2 u
B 14 u
C 4 u
D u/2

The correct answer is:
2 u

Solution:
u=3RT28
u′=3R×2T14
uu′=14
uu′=12
u′=2u

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2016 (10 Apr Online)
Injury to vagus nerve in humans is not likely to affect:

Options:
A tongue movements
B gastrointestinal movements
C pancreatic secretion
D cardiac movements.

The correct answer is:
tongue movements

Solution:
Vagus nerve, cranial nerve $\mathrm{X}$ is the longest cranial nerve which runs from the brain through the face and thorax to the abdomen. It is a mixed nerve that contains parasympathetic fibres.
Related Theory
The Hypoglossal Nerve (Cranial Nerve XII) is mainly an efferent nerve for the tongue musculature. There are two cranial nerves that innervate the tongue and are used for taste: the facial nerve (cranial nerve VII) and the glossopharyngeal nerve (cranial nerve $I X$ ).
Caution
Mnemonics - SAME : Sensory nerve Afferent nerves; Motor nerves Efferent nerves

Topic: Biology, Neural Control and Coordination, NEET, NEET 2004
Inner layer of male gametophyte in Angiosperm is mainly composed of

Options:
A Polymer of amino acids
B Polymer of nucleotide
C Polymer of Lipid
D Polymer of carbohydrate

The correct answer is:
Polymer of carbohydrate

Solution:
The intine, or inner layer, consists primarily of cellulose and pectin. Cellulose and pectin both are carbohydrates in nature. Pollen tube forms by extension of intine through germ pore.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Inside a bar magnet, the magnetic field lines

Options:
A are not present
B are parallel to the cross sectional area
the magnemt
C are in the direction from \( \mathrm{N} \) pole to \( \mathrm{S} \) pole
D are in the direction from \( \mathrm{S} \) pole to \( \mathrm{N} \) pole

The correct answer is:
are in the direction from \( \mathrm{S} \) pole to \( \mathrm{N} \) pole

Solution:
The magnetic lines form a closed loop outside the magnet, these lines are from N-pole to S-pole and inside the magnet, they reverse i.e. form S-pole to N-pole. Magnetic field lines are a visual tool used to represent magnetic fields. They describe the direction of the magnetic force on a north monopole at any given position. Because monopoles are not found to exist in nature, we also discuss alternate means to describe the field lines.

Topic: Physics, Magnetic Properties of Matter, JEE Main
Inside a bar magnet, the magnetic lines of force

Options:
A are from S-pole to $\mathrm{N}$ -pole of the magnet.
B do not exist
C depend upon area of cross-section of bar magnet.
D are from $\mathrm{N}$ -pole to $\mathrm{S}$ -pole of the magnet.

The correct answer is:
are from S-pole to $\mathrm{N}$ -pole of the magnet.

Solution:
Inside a bar magnet, the magnetic field lines(A) are not present(B) are parallel to the cross-sectional area of the magnet(C) are in the direction from $\mathrm{N}$-pole to S-pole(D) are in the direction from S-pole to $N$-pole

Topic: Physics, Magnetic Properties of Matter, JEE Main
Inside a cyclotron, a charged particle is subjected to both an electric field and a magnetic field. It gains kinetic energy due to the

Options:
A electric field
B magnetic field
C Both (a) and (b)
D None of these

The correct answer is:
electric field

Solution:
Inside a cyclotron, charged particle gains kinetic energy only due to electric field, because in magnetic field, its kinetic energy remains unchanged.

Topic: Physics, Magnetic Effects of Current, COMEDK, COMEDK 2015
Inside a horizontally moving box, an experimenter finds that when an object is placed on a smooth horizontal table and is released, it moves with an acceleration of 10 m s-2. In this box, if 1 kg body is suspended with a light string, the tension in the string in equilibrium position (with respect to experimenter) will be (Take g=10 m s-2)

Options:
A 10 N

B 1 0 2 N
C 20 N
D Zero

The correct answer is:
1 0 2 N

Solution:
Acceleration of box = 10 m s-2
Inside the box forces acting on bob are shown in the figure.

T = mg 2 + ma 2 = 1 0 2 N

Topic: Physics, Laws of Motion, JEE Main
Inside a uniform spherical shell :

(a) The gravitational field is zero.

(b) The gravitational potential is zero.

(c) The gravitational field is the same everywhere.

(d) The gravitation potential is the same everywhere.

(e) All the above.

Choose the most appropriate answer from the options given below:

Options:
A (a), (c) and (d) only
B (a), (b) and (c) only
C (b), (c) and (d) only
D (e) only

The correct answer is:
(a), (c) and (d) only

Solution:
By using graphical representation of gravitational potential due to a uniform spherical shell and radius of the shell, we can see inside the shell the gravitational potential is not zero, and the gravitational potential is same everywhere inside the shell.

By using graphical representation of gravitational field due to a uniform spherical shell and radius of the shell, we can see the gravitational field inside the shell is zero.

Topic: Physics, Gravitation, JEE Main, JEE Main 2021 (26 Aug Shift 1)
Inspiration occurs when

Options:
A the volume of the lung increases while the pressure in the lung decreases
B the volume of the lung decreases while the pressure in the lung increases
C the volume of the lung increases while the pressure in the lung increases
D the volume of the lung decreases while the pressure in the lung decreases

The correct answer is:
the volume of the lung increases while the pressure in the lung decreases

Solution:
Breathing constitutes inspiration and expiration processes. This depends on Boyle's law. It states that pressure is inversely proportional to volume when the temperature remains constant (P α 1/V). Pressure here is the partial pressure as it involves the pressure of one gas in the atmosphere. During inspiration, oxygen from the atmosphere enters the lungs thus volume of the lung cavity should be more which means pressure should be less. External intercostal muscles and diaphragm causes the volume to increase and pressure to decrease and thus inspiration takes place.

Topic: Biology, Breathing and Exchange of Gases, NEET
Insulin and glucagon, which are responsible to maintain the blood glucose level, fall under

Options:
A Antibodies
B Harmones
C Enzymes
D Transport agents

The correct answer is:
Harmones

Solution:
Insulin and glucagon, which are responsible to maintain the blood glucose level, fall under harmones. The insulin and glucagon hormone signal your liver and muscle cells to change the stored glycogen back into the glucose.

Topic: Chemistry, Biomolecules, TS EAMCET, TS EAMCET 2020 (09 Sep Shift 1)
Insulin cannot be orally administered to a diabetic patient because

Options:
A It is bitter in taste
B It is a glycoprotein
C It is proteinaceous
D It is very expensive to get oral form of insulin

The correct answer is:
It is proteinaceous

Solution:
Insulin is a hormone secreted by beta cells of Islets of Langerhans of Pancreas. Its main function is to transport glucose from blood to the cells. In some individuals the gland is not able to secrete sufficient amounts of Insulin as a result of which the cellular uptake of glucose is affected. This is known as Insulin dependant Diabetes mellitus. In such individuals artificial Insulin is given so that the cells do not starve of glucose. This cannot be given orally as Insulin is proteinaceous in nature. Being a protein it is easily digestible in stomach due to enzyme Pepsin and in small intestine due to Trypsin and Dipeptidases. Thus insulin breaks before entering the bloodstream and this makes it ineffective.

Topic: Biology, Chemical Coordination and Integration, NEET
Insulin is

Options:
A An amino acid
B Protein
C A carbohydrate
D A lipid

The correct answer is:
Protein

Solution:
Insulin is a protein consists of 51 amino acids in two chains $\alpha$ and $\beta$. $\alpha-21$ amino acids, $\beta-30$ amino acids. It is secreted by pancreas for controlling the sugar level in blood.

Topic: Chemistry, Biomolecules, JEE Main
Insulin production and its action in human body are responsible for the level of diabetes. This compound belongs to which of the following categories?

Options:
A
A co- enzyme

B
An antibiotic

C
An enzyme

D
A hormone

The correct answer is:
A hormone

Solution:
Insulin

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2004
\( \int e^{x \sin x}\left(x^{2} \cos x+x \sin x+1\right) d x \), equals (where \( C \) is constant of integration)

Options:
A \( x e^{x^{2} \sin x}+C \)
B \( x^{2} e^{x \sin x}+C \)
C \( x e^{x \sin x}+C \)
D \( 2 x^{2} e^{x \sin x}+C \)

The correct answer is:
\( x e^{x \sin x}+C \)

Solution:
Let I=∫exsinxx2cosx+xsinx+1dx

⇒I=∫x.xcosx+sinxexsinx+1.exsinxdx
Let exsinx=t⇒exsinxsinx+xcosxdx=dt

Using integration by parts,

I=∫xdtdx+tdx=x t+C=x exsinx+C, where C is the constant of integration.

Topic: Mathematics, Indefinite Integration, JEE Main
\( \int e^{-x} \tan ^{-1}\left(e^{x}\right) d x=f(x)-\frac{1}{2} \log \left(1+e^{2 x}\right)+c \Rightarrow f(x)= \)

Options:
A \( e^{x}-e^{-x} \tan ^{-1}\left(e^{x}\right) \)
B \( x^{2}+e^{-x} \tan ^{-1}\left(e^{x}\right) \)
C \( -e^{x} \tan ^{-1}\left(e^{x}\right) \)
D \( x-e^{-x} \tan ^{-1}\left(e^{x}\right) \)

The correct answer is:
\( x-e^{-x} \tan ^{-1}\left(e^{x}\right) \)

Solution:
Put ex=t, exdx=dt, dx=1t dt
=∫tan-1⁡tt2dt =tan-1⁡t-1t-∫-1t11+t2dt
=∫tan-1⁡tt2+12 ∫2tt21+t2dt t2=z, 2tdt=dz
=-tan-1⁡tt+12 ∫z+1-zz1+zdz =-tan-1⁡tt+12∫1z+1dz
=-tan-1⁡tt+12logt21+t2+c =-tan-1⁡tt+12log⁡e2x1+e2x+c
=-e-xtan-1⁡ex-12log⁡1+e2x+12 2x+c =x-e-xtan-1⁡ex-12log⁡1+e2x+c

Topic: Mathematics, Indefinite Integration, JEE Main
\( \int e^{x}\left[\frac{2+\sin 2 x}{1+\cos 2 x}\right] d x= \)

Options:
A \( e^{x} \tan x+C \)
B \( e^{x}+\tan x+C \)
C \( 2 e^{x} \tan x+C \)
D \( e^{x} \tan 2 x+C \)

The correct answer is:
\( e^{x} \tan x+C \)

Solution:
Let, I=∫ex2+sin⁡2x1+cos⁡2x dx

We know that sin2x=2sinxcosx & cos2x=2cos2x-1

Using the above formulas we can write

I=∫ex2+2sinxcosx2cos2xdx

I=∫ex21+sinxcosx2cos2xdx

I=∫ex1cos2x+sinxcosxcos2xdx
I= ∫ e x ( tanx+ sec 2 x )dx

We know that ∫exfx+f'xdx=ex·f+C

Using the above formula we can write

I= e x tanx+C

Topic: Mathematics, Indefinite Integration, JEE Main
\( \int \frac{\cos x-\sin x}{7-9 \sin 2 x} d x= \)

Options:
A \( \frac{1}{24} \log \left|\frac{4+3(\sin x+\cos x)}{4-3(\sin x+\cos x)}\right|+c \)
B \( \frac{1}{24} \log \left|\frac{4-3(\sin x+\cos x)}{4+3(\sin x+\cos x)}\right|+c \)
C \( \frac{1}{24} \log \left|\frac{4-(\sin x-\cos x)}{4+(\sin x-\cos x)}\right|+c \)
D \( \frac{1}{24} \log \left|\frac{4+(\sin x-\cos x)}{4-(\sin x-\cos x)}\right|+c \)

The correct answer is:
\( \frac{1}{24} \log \left|\frac{4+3(\sin x+\cos x)}{4-3(\sin x+\cos x)}\right|+c \)

Solution:
I=∫cosx-sinx7-9sin2xdx
⇒ I=∫cosx-sinx7-91+sin2x-1dx

⇒ I=∫cosx-sinx7-9sin2x+cos2x+2sinxcosx-1

⇒ I=∫cosx-sinx7-9[sinx+cosx2-1]dx
Let sinx+cosx=t⇒cosx-sinxdx=dt

⇒ I=∫dt7-9t2-1
⇒ I=∫dt42-3t2
⇒ I=12·4·13log4+3t4-3t+c
⇒ I=124log4+3sinx+cosx4-3sinx+cosx+c

Topic: Mathematics, Indefinite Integration, JEE Main
\( \int \sqrt{\frac{1-x}{1+x}} \mathrm{dx}= \)

Options:
A \( \sin ^{-1} \mathrm{x}-\frac{1}{2} \sqrt{1-\mathrm{x}^{2}}+\mathrm{c} \)
B \( \sin ^{-1} \mathrm{x}+\frac{1}{2} \sqrt{1-\mathrm{x}^{2}}+\mathrm{c} \)
C \( \sin ^{-1} x-\sqrt{1-x^{2}}+c \)
D \( \sin ^{-1} \mathrm{x}+\sqrt{1-\mathrm{x}^{2}}+\mathrm{c} \)

The correct answer is:
\( \sin ^{-1} \mathrm{x}+\sqrt{1-\mathrm{x}^{2}}+\mathrm{c} \)

Solution:
∫1-x1+xdx

Multiply and divide by 1-x
=∫1-x1-x2 dx=∫dx1-x2-∫x1-x2dx
=sin-1⁡x+12∫-2x1-x2dx
=sin-1⁡x+1-x2+c

Topic: Mathematics, Indefinite Integration, JEE Main
\( \int_{0}^{2 n \pi}\left\{|\sin x|-\left|\frac{1}{2} \sin x\right|\right\} d x \) equals:

Options:
A \( n \)
B \( 2 n \)
C \( -2 n \)
D None of these

The correct answer is:
\( 2 n \)

Solution:
Let I=∫02nπsin x-12sin xdx

=12∫02nπsin xdx

=12n×∫02πsinxdx

Now, I1=∫02πsin xdx

I1=∫0πsin x dx-∫π2πsin x dx

=-cos x0π+cos xπ2π=--1-1+1+1

=2+2=4

∴ I=124n=2n

Topic: Mathematics, Definite Integration, JEE Main
\( \int_{0}^{4}\{\sin \{x\}\} d x=N \), where \( \{x\} \) represents fractional part of \( x . \) Then \( \frac{N}{4} \) is

Options:
A \( -\cos 1 \)
B \( 1-\cos 1 \)
C \( 1-\sin 1 \)
D None of these

The correct answer is:
\( 1-\cos 1 \)

Solution:
Let I=∫04{sinx}dx

Using ∫0nTfxdx=n∫0Tfxdx, n∈Z if fx is a periodic function with period T.

We know that x has period equals to 1.

⇒I=4∫01sinxdx

Also, for x∈0, 1, sinx=sinx
⇒I=4∫01sinxdx

⇒I=-4cosx01

⇒I=41-cos1.

Topic: Mathematics, Definite Integration, JEE Main
Integrating Factor (I.F.) of the defferential equation
\(\frac{d y}{d x}-\frac{3 x^2 y}{1+x^3}=\frac{\sin ^2(x)}{1+x} \text { is }\)

Options:
A \(\mathrm{e}^{1+\mathrm{x}^3}\)
B \(\log \left(1+x^3\right)\)
C \(1+x^3\)
D \(\frac{1}{1+x^3}\)

The correct answer is:
\(\frac{1}{1+x^3}\)

Solution:
\(\begin{aligned}
& \text {Hints: If } e^{\int p d x}=e^{-\int \frac{3 x^2 d x}{1+x^3}}=e^{-\log \left(1+x^3\right)}=e^{\log \left(1+x^3\right)^{-1}} \\
& =\left(1+x^3\right)^{-1}=\frac{1}{1+x^3}
\end{aligned}\)

Topic: Mathematics, Differential Equations, JEE Main
Integrating factor of differential equation $\cos x \frac{d y}{d x}+y \sin x=1$ is

Options:
A $\cos x$
B $\tan x$
C $\sec x$
D $\sin x$

The correct answer is:
$\sec x$

Solution:
$\frac{d y}{d x}+y \tan x=\sec x$
I.F. $=e^{\int \tan x d x}=e^{\log \sec x}=\sec x$

Topic: Mathematics, Differential Equations, JEE Main