datavorous/entrance-exam-dataset huggingface dataset:
By the application of Simpson's one-third rule for numerical integration, with two subintervals, the value of $\int_{0}^{1} \frac{d x}{1+x}$ is

Options:
A $\frac{17}{36}$
B $\frac{17}{25}$
C $\frac{25}{36}$
D $\frac{17}{24}$

The correct answer is:
$\frac{25}{36}$

Solution:
Since, the given integration is divided into two subintervals.
ie,
$$
h=\frac{1-0}{2}=\frac{1}{2}
$$
$$
\therefore \quad \int_{0}^{1} \frac{1}{1+x} d x=\frac{h}{3}\left[\left(y_{0}+y_{2}\right)+4\left(y_{1}\right)\right]
$$
At
$$
\begin{aligned}
x=0, y &=1 \\
x &=\frac{1}{2}, y_{1}=\frac{2}{3}
\end{aligned}
$$
and
$$
x=1, y_{2}=\frac{1}{2}
$$
$$
\begin{aligned}
\therefore \int_{0}^{1} \frac{1}{1+x} d x &=\frac{1}{2 \cdot 3}\left[\left(1+\frac{1}{2}\right)+4\left(\frac{2}{3}\right)\right] \\
&=\frac{1}{6}\left[\frac{3}{2}+\frac{8}{3}\right]=\frac{25}{36}
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Main
By using the non-zero digits, the number of 5 digit numbers that can be formed so that each number has largest digit in its middle place and the digits in the number are different is

Options:
A $\sum_{r=4}^9{ }^r P_4$
B $\sum_{r=4}^8{ }^r P_4-\sum_{r=4}^8{ }^r P_3$
C $\sum_{r=4}^8{ }^r P_3$
D $\sum_{r=4}^8{ }_r^r P_4$

The correct answer is:
$\sum_{r=4}^8{ }_r^r P_4$

Solution:
Since, the largest digit is in the middle, the middle digit is greater than or equal to 5 , the number of numbers with 5 in the middle $={ }^4 P_4$ ( $\because$ the four places are to be filled by $1,2,3,4$ ) Similarly, the number of numbers with 6 in the middle $={ }^5 P_4$
The number of numbers with 7 in the middle $={ }^6 P_4$
The number of numbers with 8 in the middle $={ }^7 P_4$
The number of numbers with 9 in the middle $={ }^8 P_4$
$\therefore$ Total number of such numbers
$$
={ }^4 P_4+{ }^5 P_4+{ }^6 P_4+{ }^7 P_4+{ }^8 P_4=\sum_{r=4}^8{ }^r P_4
$$

Topic: Mathematics, Permutation Combination, JEE Main
By what factor the r.m.s. velocity will change, if the temperature is raised from $327^{\circ} \mathrm{C}$

Options:
A $\sqrt{2}$ times
B $2$
C $2 \sqrt{2}$ times
D $1$

The correct answer is:
$\sqrt{2}$ times

Solution:
$v_{\text {rms }} \propto \sqrt{T} \Rightarrow \frac{v_1}{v_2}=\sqrt{\frac{T_1}{T_2}}=\sqrt{\frac{(273+27)}{(273+327)}}=\sqrt{\frac{300}{600}}=\frac{1}{\sqrt{2}}$
$\Rightarrow v_2=\sqrt{2} v_1$

Topic: Physics, Kinetic Theory of Gases, JEE Main
By what percentage will the illumination of the lamp decrease if the current drops by 20%?

Options:
A 46%
B 26%
C 36%
D 56%

The correct answer is:
36%

Solution:
Power through the resistor is given by, P=I2R.

Now, Pint=I2intR

and Pfinal=0.8Iint2R

Therefore, % change in power =

Pfinal−PintPint×100=0.64−1×100=−36%

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (31 Jan Shift 2)
$C_0-C_1+C_2-C_3+\ldots . .+(-1)^n C_n$ is equal to

Options:
A $2^n$
B $2^n-1$
C $0$
D $2^{n-1}$

The correct answer is:
$0$

Solution:
We know that
$(1+x)^n={ }^n C_0+{ }^n C_1 x+{ }^n C_2 x^2+\ldots .+{ }^n C_n x^n$
Putting $x=-1$, we get
$(1-1)^n={ }^n C_0-{ }^n C_1+{ }^n C_2-\ldots . .(-1)^n{ }^n C_n$
Therefore $C_0-C_1+C_2-C_3+\ldots .(-1)^n C_n=0$

Topic: Mathematics, Binomial Theorem, JEE Main
$C_0 C_r+C_1 C_{r+1}+C_2 C_{r+2}+\ldots .+C_{n-r} C_n=$

Options:
A $\frac{(2 n)!}{(n-r)!(n+r)!}$
B $\frac{n!}{(-r)!(n+r)!}$
C $\frac{n!}{(n-r)!}$
D None of these

The correct answer is:
$\frac{(2 n)!}{(n-r)!(n+r)!}$

Solution:
$\begin{aligned} & (1+x)^n=C_0+C_1 x+C_2 x^2+\ldots+C_r x^r+\ldots ..(i)\\ & \left(1+\frac{1}{x}\right)^n=C_0+C_1 \frac{1}{x}+C_2 \frac{1}{x^2}+\ldots .+C_r \frac{1}{x^r}+\ldots(ii)\end{aligned}$
Multiplying both sides and equating coefficient of $x^r$ in $\frac{1}{x^n}(1+x)^{2 n}$ or the coefficient of $x^{n+r}$ in $(1+x)^{2 n}$ we get the value of required expression
$={ }^{2 n} C_{n+r}=\frac{(2 n)!}{(n-r)!(n+r)!}$
Trick : Solving conversely.
Put $n=1$ and $r=0$ in first term, (given condition)
(i) $\left.{ }^1 C_0{ }^1 C_0+{ }^1 C_1{ }^1 C_1=1+1=2 \quad, (r \leq n\right)$
Put $n=2, r=1$, then
(ii) ${ }^2 C_0{ }^2 C_1+{ }^2 C_1{ }^2 C_2=2+2=4$
Now check the options
(a) (i) Put $n=1, r=0$, we get $\frac{2!}{(1)!(1)!}=2$
(ii) Put $n=2, r=1$, we get $\frac{4!}{(1)!(3)!}=4$
Note : Students should remember this question as an identity.

Topic: Mathematics, Binomial Theorem, JEE Main
C1 and C2 are two hollow concentric cubes enclosing charges 2Q and 3Q respectively as shown in figure. The ratio of electric flux passing through C1 and C2 is:

Options:
A 2 : 5
B 5 : 2
C 2 : 3
D 3 : 2

The correct answer is:
2 : 5

Solution:
Gauss's law states that the net flux of an electric field in a closed surface is directly proportional to the enclosed electric charge. Therefore,

ϕsmaller cube =2Qε0

Since the smaller cube is inside the bigger cube, so total charge contained by the bigger cube will be 2Q+3Q=5Q.

ϕbigger cube =5Qε0

Therefore, required ratio will be,

ϕsmaller cube ϕbigger cube =25.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2024 (01 Feb Shift 2)
C2 -epimer of D-Glucose is

Options:
A D-glucose
B D-allose
C D-altrose
D D-mannose

The correct answer is:
D-mannose

Solution:
When only second carbon have opposite stereo-chemistry and remaining compound is exactly same then two compound is known as C2 -epimer.

Topic: Chemistry, Biomolecules, NEET
$\mathrm{C}_2 \mathrm{H}_2 \xrightarrow[\mathrm{H}_2 \mathrm{SO}_4] {\mathrm{H}_2 \mathrm{O}} A \xrightarrow{[0]} B$, B is

Options:
A An acid
B An aldehyde
C A ketone
D Ethanol

The correct answer is:
An acid

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
C 2 H 2 → H 2 SO 4 1%  Hg 2 (OH) 2 A → [O] B, , B is

Options:
A An acid
B An aldehyde
C A ketone
D Ethanol

The correct answer is:
An acid

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{Br} \stackrel{\mathrm{AgCN}}{\longrightarrow} \mathrm{X} \underset{\mathrm{Zn}-\mathrm{Rg} / \mathrm{HCl}}{\stackrel{\text { Reduction }}{\longrightarrow}} \mathrm{Y}$, Here $\mathrm{Y}$ is

Options:
A
Ethyl methyl amine

B
n-propylamine

C
Isopropylamine

D
Ethylamine

The correct answer is:
Ethyl methyl amine

Solution:

Topic: Chemistry, Amines, JEE Main
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{Br} \stackrel{\mathrm{AgCN}}{\longrightarrow} \mathrm{X} \underset{\mathrm{Zn}-\mathrm{Rg} / \mathrm{HCl}}{\stackrel{\text { Reduction }}{\longrightarrow}} \mathrm{Y}$, Here $\mathrm{Y}$ is

Options:
A
Ethyl methyl amine

B
n-propylamine

C
Isopropylamine

D
Ethylamine

The correct answer is:
Ethyl methyl amine

Solution:

Topic: Chemistry, Amines, JEE Main, JEE Main 2012 (07 May Online)
$\mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})+\mathrm{nO}_2 \rightarrow \mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})$
In this equation, the ratio of the coefficients of $\mathrm{CO}_2$ and $\mathrm{H}_2 \mathrm{O}$ is

Options:
A $1: 1$
B $2: 3$
C $3: 2$
D $1: 3$

The correct answer is:
$2: 3$

Solution:
The balanced equation is \(2 \mathrm{C}_2 \mathrm{H}_6+7 \mathrm{O}_2 \rightarrow 4 \mathrm{CO}_2+6 \mathrm{H}_2 \mathrm{O}\).
Ratio of the coefficients of \(\mathrm{CO}_2\) and \(\mathrm{H}_2 \mathrm{O}\) is \(4: 6\) or \(2: 3\).

Topic: Chemistry, Redox Reactions, JEE Main
C5H10→H2/NiX→Cl2/hv3 monochloro structural isomers for compound C5H10 how many pairs of geometrical isomers are possible?

Options:
A 1
B 2
C 3
D None

The correct answer is:
1

Solution:
C5H10→H2/NiX→Cl2/hv3 monochloro structural isomers f(X) must have 3-type of H, so it is n-pentane
Possible structure of C5H10 is
CH3–CH2–CH2–CH=CH2 no G.I
CH3–CH=CH–CH2–CH3 Geometrical Isomer
Only one pair of G.I is possible (cis/trans)

Topic: Chemistry, General Organic Chemistry, JEE Main
C5H10O is a carbonyl compound. The number of structural isomers possible for this molecular formula are

Options:
A 5
B 8
C 6
D 7

The correct answer is:
7

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{Cl}+\mathrm{KCN}(\mathrm{aq}.) \rightarrow X+Y$ Compounds X and Y are

Options:
A $\mathrm{C}_6 \mathrm{H}_6+\mathrm{KCl}$
B $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CN}+\mathrm{KCl}$
C $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_3+\mathrm{KCl}$
D None of these

The correct answer is:
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CN}+\mathrm{KCl}$

Solution:
$\mathrm{C}_6 \mathrm{H}_5-\mathrm{CH}_2-\mathrm{Cl}+\mathrm{KCN}(\mathrm{aq}) \rightarrow \mathrm{C}_6 \mathrm{H}_5-\mathrm{CH}_2-\mathrm{C} \equiv \mathrm{N}+\mathrm{KCl}$

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
$\mathrm{C}_6 \mathrm{H}_5-\mathrm{CH}=\mathrm{CHCHO} \xrightarrow{x} \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}=\mathrm{CHCH}_2 \mathrm{OH}$. In the above sequence $X$ can be

Options:
A $\mathrm{H}_2 / \mathrm{Ni}$
B $\mathrm{NaBH}_4$
C $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 / \mathrm{H}^{+}$
D Both (a) and (b)

The correct answer is:
$\mathrm{NaBH}_4$

Solution:
$\mathrm{NaBH}_4$ and $\mathrm{LiAlH}_4$ attacks only carbonyl group and reduce it into alcohol group. They do not attack on double bond.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{Cl}$ prepared by aniline with

Options:
A $\mathrm{HCl}$
B $\mathrm{Cu}_2 \mathrm{Cl}_2$
C $\mathrm{Cl}_2$ in presence of anhydrous $\mathrm{AlCl}_3$
D $\mathrm{HNO}_2$ and then heated with $\mathrm{Cu}_2 \mathrm{Cl}_2$

The correct answer is:
$\mathrm{HNO}_2$ and then heated with $\mathrm{Cu}_2 \mathrm{Cl}_2$

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
$\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_3 \mathrm{Cl} \xrightarrow[\mathrm{AlCl}_3]{\text { anhydrous }} \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_3+\mathrm{HCl}$ is an example of

Options:
A Friedel-Craft's reaction
B Kolbe's synthesis
C Wurtz reaction
D Grignard reaction

The correct answer is:
Friedel-Craft's reaction

Solution:
it is a Friedel-craft’s reaction.

Topic: Chemistry, Hydrocarbons, JEE Main
$\mathrm{C}_6 \mathrm{H}_6+\mathrm{Cl}_2 \xrightarrow{\text { UV Light }}$ Product. In above reaction product is

Options:
A $\mathrm{CCl}_3 \mathrm{CHO}$
B $\mathrm{C}_6 \mathrm{H}_6 \mathrm{Cl}_6$
C $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{Cl}_6$
D $\mathrm{C}_6 \mathrm{H}_9 \mathrm{Cl}_2$

The correct answer is:
$\mathrm{C}_6 \mathrm{H}_6 \mathrm{Cl}_6$

Solution:
$\underset{\text { Benzene }}{\mathrm{C}_6 \mathrm{H}_6}+3 \mathrm{Cl}_2 \xrightarrow{\text { u.V light }} \underset{\text { BHC }}{\mathrm{C}_6 \mathrm{H}_6 \mathrm{Cl}_6}$

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
C6H6 freezes at 5.5°C. The temperature at which a solution of 10 g of C4H10 in 200 g of C6H6 freeze is _______C∘. (nearest integer value), (The molal freezing point depression constant of C6H6 is 5.12°C/m.)

Options:

The correct answer is:
1

Solution:
Pure Solvent: C6H6ℓ

Given: Tf°=5.5°C

Kf=5.12°C/m

:Solute is non dissociative

∵ ΔTf=kf×m

⇒Tf0-Tf'=5.12×10582001000kgmol

⇒5.5-Tf'=5.12×5×1058

⇒Tf'=1.086°C≃1C∘

Topic: Chemistry, Solutions, JEE Main, JEE Main 2021 (24 Feb Shift 2)
$\mathrm{C}_6 \mathrm{H}_6 \xrightarrow[\mathrm{H}_2 \mathrm{SO}_4]{\mathrm{HNO}_3} X \xrightarrow[\mathrm{FeCl}_3]{\mathrm{Cl}} Y$
. In the above sequence $Y$ is

Options:
A 1-nitrochloro benzene
B 3-nitrochlorobenzene
C 4-nitrochlorobenzene
D 1,2-nitrochlorobenzene

The correct answer is:
3-nitrochlorobenzene

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
$\mathrm{C}_6 \mathrm{H}_6(\mathrm{liq})+\frac{15}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 6 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{liq})$
Benzene burns in oxygen according to the above equation. What is the volume of oxygen (at STP) needed for complete combustion of 39 gram of liquid benzene?

Options:
A $11.2$ litre
B $22.4$ litre
C 84 litre
D 168 litre

The correct answer is:
84 litre

Solution:
$\mathrm{n}_{\mathrm{C}_6 \mathrm{H}_6}=\frac{39}{78}=\frac{1}{2}$ mole
From Balanced equation,
$\begin{aligned}
&\mathrm{n}_{\mathrm{C}_6 \mathrm{H}_{\mathrm{G}}}=\frac{2}{15} \mathrm{n}_{\mathrm{O}_2} \\
&\frac{1}{2}=\frac{2}{15} \times \frac{\mathrm{xL}}{22.4 \mathrm{~L}} \\
&\therefore \mathrm{x}=84 \mathrm{~L}
\end{aligned}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
C6H12 (A) has chirality but on hydrogenation, A is converted into C6H14 (B) in which chirality disappears. Hence, A is:

Options:
A 3-Methyl-1-pentene
B 2-Methyl-2-pentene
C 2, 3-dimethyl-2-buene
D 3, 3-dimethyl-1-butene

The correct answer is:
3-Methyl-1-pentene

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
C6H12O6⟶ZymaseA→∆NaOIB+CHI3

The number of carbon atoms present in the product B is

Options:

The correct answer is:
1

Solution:
no. of carbon atoms present in B is 1

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2022 (29 Jun Shift 1)
C7H5O22⟶hvX→2C˙6H5+2CO2

Consider the above reaction and identify the intermediated 'X'

Options:
A
B
C
D

The correct answer is:

Solution:
In presence of sunlight homolytic fission of peroxide linkage will take place.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2022 (26 Jun Shift 1)
C 12 H 16 A â i) O 3 ii) Zn, H 2 O 2 2H 3 Câ C || O â CH 3 â+â2HOOCâ C || O âCOOH
What is degree of unsaturation of compound A?

Options:

The correct answer is:
5

Solution:
Here degree of unsaturation is 5 (4 double bond and 1 cyclic ring)

Topic: Chemistry, Aldehydes and Ketones, JEE Main
C14 has half-life 5700 year. At the end of 11400 years, the actual amount left is

Options:
A 0.5 times of original amount
B 0.25 times of original amount
C 0.125 times of original amount
D 0.0625 times of original amount

The correct answer is:
0.25 times of original amount

Solution:
N = N 0 Ã 1 2 1 1 4 0 0 / 5 7 0 0
= N0122=0.25 N0

Topic: Physics, Nuclear Physics, JEE Main
C60, an allotrope of carbon contains

Options:
A 12 hexagons and 20 pentagons.
B 16 hexagons and 16 pentagons.
C 18 hexagons and 14 pentagons.
D 20 hexagons and 12 pentagons.

The correct answer is:
20 hexagons and 12 pentagons.

Solution:
Fullerene is an Isotope of carbon which consist 20 hexagonal and 12 pentagonal rings, it has sp2 hybridization and forms a cage-like structure.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2019 (09 Apr Shift 1)
' $c-c^{\prime}$ bond length in benzene lies between single and double bond. The reason is

Options:
A Resonance
B Isomerism
C Metamerism
D Inductive effect

The correct answer is:
Resonance

Solution:
Due to delocalisation of $\pi$ electrons benzene has resonance.

Topic: Chemistry, General Organic Chemistry, JEE Main
CH32C=CHCOCH3 can be oxidized to CH32C=CHCOOH by

Options:
A chromic acid
B NaOI, followed by acidification
C Cu at 573 K
D KMnO4+H2SO4

The correct answer is:
NaOI, followed by acidification

Solution:
CH32C=CHCOCH3 is a methyl ketone and therefore, it can be easily oxidized to CH32CCHCOOH by haloform or iodoform reaction (NaOI).
(CH3)2C=CHCOCH3→NaOI(CH3)2C=CHCOCI3→NaOH
CHI3+(CH3)2C=CHCOONa→(CH3)2C=CHCOOH
Chromic acid and KMnO4 , being strong oxidizing agents, will also attack the multiple bond and cause the cleavage giving the mixture of carboxylic acids.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
CH33 SiCl is used during polymerisation of organosilicons because

Options:
A the chain length of organosilicon polymers can be controlled by adding CH33SiCl
B CH33SiCl improves the quality and yield of the polymer
C CH33SiCl does not block the end terminal of silicone polymer
D CH33SiCl acts as a catalyst during polymerisation

The correct answer is:
the chain length of organosilicon polymers can be controlled by adding CH33SiCl

Solution:
CH33SiCl is used during polymerisation of organosilicons because the chain length of organosilicon polymers can be controlled by adding CH33SiCl .

Topic: Chemistry, Coordination Compounds, JEE Main
CH3-CH2-N=O and CH3-CH=N-OH are:

Options:
A Functional group isomer
B Tautomer
C Position isomer
D Not an isomer

The correct answer is:
Tautomer

Solution:
Tautomers are constitutional isomers of organic compounds that readily interconvert. These two show tautomerism

Topic: Chemistry, General Organic Chemistry, JEE Main
C is with the product

Options:
A $\mathrm{CO}_{2}$
B
C Both (a) and (b)
D None of the above

The correct answer is:
$\mathrm{CO}_{2}$

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
CO2 gas is liquefied at 27oC . Gas follows the following graph during liquefaction for 1 mole of CO2 .

Then which statement is (are) correct in gaseous phase will
(i) The maximum density of gas is 0.1 gm/ml
(ii) The density of liquid CO2 is 1 g/ml at 60 atm.
(iii) At point C 50% of CO2 is liquefied.
(iv) The compressibility factor of gas at 27oC is always less than 1.
Which of the above is/are correct

Options:
A only i
B i and ii
C i,ii and iii
D all

The correct answer is:
i,ii and iii

Solution:
A Maximum density of CO2 in gaseous phase will be at point 'A'
Now: d=PMRT=60Ã44RÃ300=107 gm/lt=0.107 gm/cc
Or d=mass of gasVolume of gas=44 gm440 cc=0.1
B Density of liquid CO2 at 60 atm (at point B)
At point B there will be only liquid CO2 . So we can find out the density of density of liquid CO2 only at point B.
d= M V = 44 0.044 =1 gm/cc
C at point C the volume will be 0.44+0.044 2 ≈0.22 L
And pressure is same so obviously approx. half of CO2 is liquified
D Z can be greater than 1 at 27Î¿C. For example at point 'A' Z is 1.07, So wrong.

Topic: Chemistry, States of Matter, JEE Main
C || O − NH 2 →   P 2 O 5   A → (i)CH 3 MgBr ii H 3 O +   B → i I 2 +Ca (OH) 2 ii H 2 O C; Product is

Options:
A C||O-CH3
B C||O-OH
C C||O
D C||O-CH2CH3

The correct answer is:
C||O-OH

Solution:
C||O-NH2âP2O5 Câ¡N âiCH3MgBriiH3O+ C||O-CH3 → i I 2 +Ca OH 2 ii H 2 O C||O-OH

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
$\int\left[\frac{(1+\log x)}{\cos ^{2}(x \log x)}\right] d x=$

Options:
A $\sin (x \log x)+c$
B $\sin ^{2}(x \log x)+c$
C $\log (x \log x)+c$
D $\tan (x \log x)+c$

The correct answer is:
$\tan (x \log x)+c$

Solution:
$I=\int \frac{(1+\log x)}{\cos ^{2}(x \log x)} d x$
Put $x \log x=t \Rightarrow\left[x \cdot \frac{1}{x}+\log x(1)\right] d x=d t \Rightarrow(1-\log x) d x=d t$
$1=\int \frac{1}{\cos ^{2} t} d t=\int \sec ^{2} t d t=\tan t+c$
$=\tan (x \log x)-c$

Topic: Mathematics, Indefinite Integration, JEE Main
$\int \frac{\sec x d x}{\sqrt{\cos 2 x}}=$

Options:
A $\sin ^{-1}(\tan x)$
B $\tan x$
C $\cos ^{-1}(\tan x)$
D $\frac{\sin x}{\sqrt{\cos x}}$

The correct answer is:
$\sin ^{-1}(\tan x)$

Solution:
$\begin{aligned}
& \int \frac{\sec x d x}{\sqrt{\cos 2 x}}=\int \frac{\sec x}{\sqrt{\cos ^2 x-\sin ^2 x}} d x \\
& \left.=\int \frac{\sec ^2 x d x}{\sqrt{1-\tan ^2 x}} \text { \{Multiplying } N^{\prime} r \text { and } D^{\prime} r \text { by } \sec x\right\}
\end{aligned}$
Now putting $\tan x=t \Rightarrow \sec ^2 x d x=d t$, we get the integral $=\sin ^{-1} t=\sin ^{-1}(\tan x)$.
Trick : Since $\frac{d}{d x}\left\{\sin ^{-1}(\tan x)\right\}=\frac{\sec ^2 x}{\sqrt{1-\tan ^2 x}}$
$=\frac{\sec ^2 x \cdot \cos x}{\sqrt{\cos ^2 x-\sin ^2 x}}=\frac{\sec x}{\sqrt{\cos 2 x}} .$

Topic: Mathematics, Indefinite Integration, JEE Main
$\frac{\sin 3 \theta+\sin 5 \theta+\sin 7 \theta+\sin 9 \theta}{\cos 3 \theta+\cos 5 \theta+\cos 7 \theta+\cos 9 \theta}=$

Options:
A $\tan 30$
B $\cot 3 \theta$
C $\tan 60$
D $\cot 6 \theta$

The correct answer is:
$\tan 60$

Solution:
$\begin{aligned}
& \frac{\sin 3 \theta+\sin 5 \theta+\sin 7 \theta+\sin 9 \theta}{\cos 3 \theta+\cos 5 \theta+\cos 7 \theta+\cos 9 \theta} \\
& =\frac{(\sin 3 \theta+\sin 9 \theta)+(\sin 5 \theta+\sin 7 \theta)}{(\cos 3 \theta+\cos 9 \theta)+(\cos 5 \theta+\cos 7 \theta} \\
& =\frac{2 \sin 6 \theta \cos 3 \theta+2 \sin 6 \theta \cos \theta}{2 \cos 6 \theta \cos 3 \theta+2 \cos 6 \theta \cos \theta} \\
& =\frac{2 \sin 6 \theta(\cos 3 \theta+\cos \theta)}{2 \cos 6 \theta(\cos 3 \theta+\cos \theta)}=\tan 6 \theta \\
&
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
$\int \frac{\cos 2 x \cdot \sin 4 x}{\cos ^4 x\left(1+\cos ^2 2 x\right)} d x=$

Options:
A $\log \left[\frac{1+\cos ^2 2 x}{1+\cos 2 x}\right]-\tan ^2 x+c$
B $\log \left(\frac{1+\cos ^2 2 x}{1+\cos 2 x}\right)+\tan ^2 x+c$
C $\log \left(\frac{1+\cos 2 x}{1+\cos ^2 2 x}\right)+\sec ^2 x+c$
D $\log \left(\frac{(1+\cos 2 x)^2}{1+\cos ^2 2 x}\right)+\sec ^2 x+c$

The correct answer is:
$\log \left(\frac{(1+\cos 2 x)^2}{1+\cos ^2 2 x}\right)+\sec ^2 x+c$

Solution:
$$
\begin{aligned}
& \text { Let } I=\int \frac{\cos 2 x \sin 4 x}{\cos ^4 x\left(1+\cos ^2 2 x\right)} d x \\
& =4 \int \frac{\cos 2 x(2 \sin 2 x \cos 2 x)}{(1+\cos 2 x)^2\left(1+\cos ^2 2 x\right)} d x
\end{aligned}
$$
Put $\cos 2 x=t \Rightarrow-2 \sin 2 x d x=d t$
$$
\begin{aligned}
\therefore I & =-4 \int \frac{t^2}{(1+t)^2\left(1+t^2\right)} d t \\
& =-4 \int\left(\frac{1}{2(1+t)^2}-\frac{1}{2(1+t)}+\frac{t}{2\left(1+t^2\right)}\right) d t \\
& =-2\left[\frac{-1}{1+t}-\log (1+t)+\frac{1}{2} \operatorname{Iog}\left(1+t^2\right)\right]+c \\
& =\frac{2}{1+\cos 2 x}+2 \log (1+\cos 2 x) \\
& =\frac{2}{2 \cos ^2 x}+\log \frac{\left(1+\cos ^2 x\right)^2}{\left(1+\cos ^2 2 x\right)}+c \\
& =\log \left(\frac{\left(1+\cos ^2 x\right)^2}{\left(1+\cos ^2 2 x\right)}\right)+\sec ^2 x+c
\end{aligned}
$$

Topic: Mathematics, Indefinite Integration, JEE Main
cosec2Î¸cos2Î¸-3cosâ¡Î¸+2â¥1, if Î¸ belongs to

Options:
A 0,Ï3
B π2,π
C Ï3,Ï2
D 0,Ï4

The correct answer is:
π3,π2

Solution:
cos2⁡θ-3cos⁡θ+2≥1cosec2θ=sin2⁡θ
cos2⁡θ-3cos⁡θ+2≥1-cos2⁡θ
2cos2⁡θ-3cos⁡θ+1≥0
2cos⁡θ-1cos⁡θ-1≥0
⇒cos⁡θ≤12 or cos⁡θ≥1
So, θ∈π3,π2 of the given intervals.

Topic: Mathematics, Trigonometric Equations, JEE Main
$\int \frac{1}{\cos x(1+\cos x)} d x=$

Options:
A $\log (\sec x+\tan x)+2 \tan \frac{x}{2}+c$
B $\log (\sec x+\tan x)-2 \tan \frac{x}{2}+c$
C $\log (\sec x+\tan x)+\tan \frac{x}{2}+c$
D $\log (\sec x+\tan x)-\tan \frac{x}{2}+c$

The correct answer is:
$\log (\sec x+\tan x)-\tan \frac{x}{2}+c$

Solution:
$\begin{aligned} \int \frac{1}{\cos x(1+\cos x)} d x & =\int \frac{d x}{\cos x}-\int \frac{d x}{1+\cos x} \\ & =\int \sec x d x-\frac{1}{2} \int \sec ^2 \frac{x}{2} d x \\ & =\log (\sec x+\tan x)-\tan \frac{x}{2}+c .\end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main
$\int \frac{\cos x-1}{\cos x+1} d x=$

Options:
A $2 \tan \frac{x}{2}-x+c$
B $\frac{1}{2} \tan \frac{x}{2}-x+c$
C $x-\frac{1}{2} \tan \frac{x}{2}+c$
D $x-2 \tan \frac{x}{2}+c$

The correct answer is:
$x-2 \tan \frac{x}{2}+c$

Solution:
$\begin{aligned} & \int \frac{\cos x-1}{\cos x+1} d x=-\int \tan ^2 \frac{x}{2} d x \\ & =-\int\left(\sec ^2 \frac{x}{2}-1\right) d x=\int\left(1-\sec ^2 \frac{x}{2}\right) d x=x-2 \tan \frac{x}{2}+c\end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main
$\int \frac{d x}{\cos x+\sqrt{3} \sin x}$ equals

Options:
A
$\frac{1}{2} \log \tan \left(\frac{x}{2}+\frac{\pi}{12}\right)+c$

B
$\frac{1}{2} \log \tan \left(\frac{x}{2}-\frac{\pi}{12}\right)+c$

C
$\log \tan \left(\frac{\mathrm{x}}{2}+\frac{\pi}{12}\right)+\mathrm{c}$

D
$\log \tan \left(\frac{\mathrm{x}}{2}-\frac{\pi}{12}\right)+\mathrm{c}$

The correct answer is:
$\frac{1}{2} \log \tan \left(\frac{x}{2}+\frac{\pi}{12}\right)+c$

Solution:
$\int \frac{d x}{\cos x+\sqrt{3} \sin x}$
$=\frac{1}{2} \int \sec \left(x-\frac{\pi}{3}\right) d x$
$=\frac{1}{2} \log \tan \left(\frac{x}{2}-\frac{\pi}{6}+\frac{\pi}{4}\right)+c$
$=\frac{1}{2} \log \tan \left(\frac{x}{2}+\frac{\pi}{12}\right)+c$.

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2007
$\int \frac{d x}{\cos x \sqrt{\cos 2 x}}=$

Options:
A $\sin ^{-1}(\tan x)+c$
B $\frac{1}{2} \log \left|\tan \left(\frac{\pi}{4}+\mathrm{x}\right)\right|+\mathrm{c}$
C $2 \log \left|\frac{1+\tan x}{1-\tan x}\right|+c$
D $\frac{1}{2} \log \left|\frac{1-\tan x}{1+\tan x}\right|+c$

The correct answer is:
$\sin ^{-1}(\tan x)+c$

Solution:
$$
\begin{aligned}
& \text { Let } I=\int \frac{d x}{\cos x \sqrt{\cos 2 x}} \\
& =\int \frac{d x}{\cos x \cdot \cos x \sqrt{\frac{\cos ^2 x-\sin ^2 x}{\cos ^2 x}}}=\int \frac{d x}{\cos ^2 x \sqrt{1-\tan ^2 x}} \\
& I=\int \frac{\sec ^2 x}{\sqrt{1-\tan ^2 x}} d x
\end{aligned}
$$
Put $\tan x=t \Rightarrow \sec ^2 d x=d t$
$$
\therefore \mathrm{I}=\int \frac{\mathrm{dt}}{\sqrt{1-\mathrm{t}^2}}=\sin ^{-1}(\mathrm{t})+\mathrm{c}=\sin ^{-1}(\tan \mathrm{x})+\mathrm{c}
$$

Topic: Mathematics, Indefinite Integration, JEE Main