datavorous/entrance-exam-dataset huggingface dataset:
Consider the following statements:
P : I have fever
Q : I will not take medicine
R : I will take rest
The statement "If I have fever, then I will take medicine and I will take rest" is equivalent to:

Options:
A ~P∨~Q∧~P∨R
B ~P∨-Q∧~P∨~R
C P∨Q∧~P∨R
D P∨~Q∧P∨~R

The correct answer is:
~P∨~Q∧~P∨R

Solution:
We know the conditional statement:

if A, then B is given by A→B

So the given statements If I have fever, then I will take medicine and I will take rest can be written as,

P→~Q∧R

≡~P∨~Q∧R

≡~P∨~Q∧~P∨R

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2023 (30 Jan Shift 2)
Consider the following statements:
P: Suman is brilliant
Q: Suman is rich
R: Suman is honest
The negation of the statement, "Suman is brilliant and dishonest if and only if Suman is rich" can be equivalently expressed as

Options:
A ~Q↔~P∨R
B ~Q↔P∨~R
C ~Q↔P∧~R
D ~Q↔~P∧R

The correct answer is:
~Q↔P∧~R

Solution:
P : Suman is brilliant
~ P : Suman is not brilliant
Q = Suman is rich
~ Q : Suman is not rich
R : Suman is honest

~ R : Suman is not honest

Also, ~p⇔q≡~p⇔q≡p⇔~q

Hence ~(P∧∼R)⟷Q is ∼Q⟷P∧∼R

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2015 (11 Apr Online)
Consider the following statements
$\mathrm{P}$ : Suman is brilliant
Q : Suman is rich
$\mathrm{R}$ : Suman is honest
The negation of the statement "Suman is brilliant and dishonest if and only if Suman is rich" can be expressed as

Options:
A
$\sim(\mathrm{Q} \leftrightarrow(\mathrm{P} \wedge \sim \mathrm{R}))$

B
$\sim Q \leftrightarrow \sim P \wedge R$

C
$\sim(P \wedge \sim R) \leftrightarrow Q$

D
$\sim P \wedge(Q \leftrightarrow \sim R)$

The correct answer is:
$\sim(\mathrm{Q} \leftrightarrow(\mathrm{P} \wedge \sim \mathrm{R}))$

Solution:
$\sim\left\{\left(P_{\wedge} \sim R\right) \leftrightarrow Q\right\}=\sim\left\{Q \leftrightarrow\left(P_{\wedge} \sim R\right)\right\}$

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2011
Consider the following statements regarding compounds which cause global warming X is a hydrocarbon, A and B are neutral oxides of nitrogen, C is a blue coloured gas and D is released when H2S is reacted with oxygen. Identify the correct statements about X, A, B, C and D.

(I) X is mainly present in natural gas

(II) In A and B one is diamagnetic and another one is paramagnetic

(III) C can be identified by using liquid element in d-block

(IV) D causes acid rain

Options:
A All are correct
B I, II, III
C II, III, IV
D I, III, IV

The correct answer is:
All are correct

Solution:
X : CH 4 ,A and B :NO, N 2 O, C : O 3 , D : SO 2

Topic: Chemistry, Environmental Chemistry, NEET
Consider the following statements regarding rectifier circuits
1. Ripple factor for single-phase half-wave rectifier is 1.414
2. Ripple factor single phase full wave rectifier is 0.482
3. Ripple factor for the three-phase half-wave rectifier is 0.17
Which of the above statement(s) correct?

Options:
A 1,2 and 3
B 2,3 only
C 1,2 only
D 1 only

The correct answer is:
2,3 only

Solution:
Explanation:
The ripple factor indicates the number of ripples present in the DC output.
The output of the power supply is given by:
R. F. $=\frac{I_{r m s} \text { of } A C \text { component }}{I_{D C} \text { component }}$
Thus, if the ripple factor is less, the power supply has fewer AC components and power supply output is purer (i.e more DC without much fluctuations)
Thus ripple factor is an indication of the purity of output of the power supply.
Notes:
- Ripple factor for single-phase half-wave rectifier $=1.21$
- Ripple factor single phase full wave rectifier $=0.482$
- Ripple factor for three-phase half-wave rectifier $=0.17$
We can, therefore, conclude that the value of the ripple factor in a three-phase rectifier is very less in comparison to a single-phase rectifier.

Topic: Physics, Semiconductors, JEE Main
Consider the following statements related to a variable $X$ having a binomial distribution $b_{x}(n, p)$
$1.$ If $p=\frac{1}{2}$, then the distribution is symmetrical.
$2.$ $\quad p$ remaining constant, $P(X=r)$ increases as $n$ increases. Which of the statements given above is/are correct?

Options:
A 1
B 2
C Both 1 and 2 .
D Neither 1 nor 2

The correct answer is:
Both 1 and 2 .

Solution:
Both (1) and (2) statements are correct.

Topic: Mathematics, Probability, NDA, NDA 2010 (Phase 1)
Consider the following statements related to the nature of Bayes' theorem
$1.$ Bayes' theorem is a formula for computation of a conditional probability.
$2.$ Bayes' theorem modifies an assumed probability of an event in the light of a related event which is observed. Which of the statements given above is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
Both 1 and 2

Solution:
Baye's theorem says If $A_{1}, A_{2}, \ldots, A_{n}$ are $(n)$ mutuallyexclusive and exhaustive events in sample space $S$ and $E$ is any event is $S$ intersecting events $A_{i}($ iz $1,2, \ldots, n)$ such that $\mathrm{P}(\mathrm{E}) \neq 0$ then $P\left(A_{i} / E\right)=\frac{P\left(E / A_{i}\right)}{P(E)}$
$=\frac{P\left(A_{i}\right) P\left(E / A_{i}\right)}{\sum_{|2|}^{n} P\left(A_{i}\right) P\left(E / A_{i}\right)}$
Thus, both statement 1 and 2 are correct.

Topic: Mathematics, Probability, JEE Main
Consider the following structures (I), (II), (III) and (IV).
(I)
(II)

(III)

(IV)

Which of the following statements is not correct?

Options:
A II and III are identical
B I and II are enantiomers
C I and III are enantiomers
D II and IV are identical..

The correct answer is:
II and IV are identical..

Solution:
(I) Configuration: S
(II) Configuration: R
(III) Configuration: R
(IV) Configuration: R
II and IV are identical.

Topic: Chemistry, General Organic Chemistry, JEE Main
Consider the following substances
1. CH3C≡CCH3
2. CH3CH2C≡CH
3. CH3CH=CHCH3
4. CH3CH2CH=CH2
Which of the following reagent can be used to distinguish the compound (2) from the rest of the compounds?

Options:
A Bromine/ CCl4
B Bromine/ CH3COOH
C Alk. KMnO4
D Ammonical AgNO3 or ammoniacal cuprous chloride

The correct answer is:
Ammonical AgNO3 or ammoniacal cuprous chloride

Solution:
Compound (2) has acidic hydrogen (acetylinic hydrogen), which can be replaced.
CH3-CH2-C≡CH+AgNH32+OH-→CH3-CH2C≡CAgWhite ppt+H2O+2NH3
CH3CH2C≡CH+CuNH32+OH-→CH3CH2C≡CCuRed ppt.+H2O+2NH3
Compounds 1, 3 and 4 have no acidic hydrogen. Therefore, they do not react with reagent option D.

Topic: Chemistry, Hydrocarbons, JEE Main
Consider the following substances

The decreasing order of basicity is

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

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

Solution:
The decreasing order of basicity is 3 > 1 > 2 > 4.
The β -carbon in uses sp3 -hybrid orbitals; while the β -carbon in benzylamine uses sp2 -hybrid orbitals. The electron-withdrawing p-CN makes the phenyl ring even more electron-withdrawing and base-weakening.

Topic: Chemistry, General Organic Chemistry, JEE Main
Consider the following synthesis. What is wrong with this transformation?

Options:
A \( \mathrm{BH}_{3} \) is actually unreactive towards alkyne.
B \( \mathrm{BH}_{3} \) should be added to the adjacent carbon rather than the same carbon.
C Borane can be used as an electrophilic only with non-terminal alkyne.
D The borane addition follows regiochemistry opposite to that shown.

The correct answer is:
The borane addition follows regiochemistry opposite to that shown.

Solution:
The anti-Markovnikov's addition of H and BH2 takes place.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Consider the following system of equations in matrix form 12λ12λxyz=0. Then which one of the following statements is true?

Options:
A ∀λ∈-∞, ∞, the given system has non-trivial solution
B ∀λ∈-∞, ∞, the given system has only trivial solution
C For λ≠0, the given system does not have any solution
D For λ=0, the given system is inconsistent

The correct answer is:
∀λ∈-∞, ∞, the given system has non-trivial solution

Solution:
Given system of equations 12λ12λxyz=0

⇒1×11×21×λ2×12×22×λλ×1λ×2λ×λxyz=0

⇒12λ242λλ2λλ2xyz=0

Now, 12λ242λλ2λλ2=14λ2-2λ2λ-22λ2-λ2λ+λ22λ-λ4=0

Here, the determinant is zero, so the given system of equations has infinitely many solutions.

We know that for a system of equations of form AX=0, it has non-trivial solution if A=0.

So, the given system of equations 12λ12λxyz=0 has non-trivial solution for any real value of λ.

i.e. ∀λ∈-∞, ∞, the given system has non-trivial solution.

Topic: Mathematics, Determinants, JEE Main
Consider the following system of questions

αx+2y+z=1

2αx+3y+z=1

3x+αy+2z=β

For some α,β∈ℝ. Then which of the following is NOT correct.

Options:
A It has no solution if α=-1 and β≠2
B It has no solution for α=-1 and for all β∈ℝ
C It has no solution for α=3 and for all β≠2
D It has a solution for all α≠-1 and β=2

The correct answer is:
It has no solution for α=-1 and for all β∈ℝ

Solution:
Given,

System of questions

αx+2y+z=1

2αx+3y+z=1

3x+αy+2z=β

Now finding, D=α212α313α2=0

⇒α(6-α)-2(4α-3)+1(2α2-9)=0

⇒α-3(α+1)=0

⇒α=-1,3

Dx=211311α2β=0⇒β=2

⇒2β-2-1(3β-α)+(6-α)=0

⇒β=2

Dy=α112α1132β=0

⇒α(β-2)-1(2αβ-3)+1(4α-3)=0

⇒α(2-β)=0

⇒α=0,β=2

Dz=α212α313αβ=0

⇒α2-αβ-3=0

Now when β=2,α=-1 the system of equations will have infinite solutions,

When α=3, β=2 the system has infinite solution,

And when α=3, β≠2 the system has no solution as Dx & Dy≠0 for β≠2,

Hence, only option B is not correct

Topic: Mathematics, Determinants, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Consider the following table:

Gas	a/k Pa dm6mol-1	b/dm3mol-1

A	642.32	0.05196

B	155.21	0.04136

C	431.91	0.05196

D	155.21	0.4382

a and b are vander Waals constants. The correct statement about the gases is:

Options:
A Gas C will occupy more volume than gas A ; gas B will be lesser compressible than gas D
B Gas C will occupy lesser volume than gas A ; gas B will be lesser compressible than gas D
C Gas C will occupy more volume than gas A ; gas B will be more compressible than gas D
D Gas C will occupy lesser volume than gas A ; gas B will be more compressible than gas D

The correct answer is:
Gas C will occupy more volume than gas A ; gas B will be more compressible than gas D

Solution:
* Gas A and C have same value of ‘b’ but different value of a so gas having higher value of a have more force of attraction so molecules will be more closer hence occupy less volume.
* Gas B and D have same value of ‘a’ but different value of b so gas having lesser value of b will be more compressible.

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2019 (10 Apr Shift 1)
Consider the following test for a group-IV cation.
\(\begin{aligned}
& \mathrm{M}^{2+}+\mathrm{H}_2 \mathrm{~S} \rightarrow \mathrm{A} \text { (Black precipitate) }+ \text { by product } \\
& \mathrm{A}+\text { aqua regia } \rightarrow \mathrm{B}+\mathrm{NOCl}+\mathrm{S}+\mathrm{H}_2 \mathrm{O} \\
& \mathrm{B}+\mathrm{KNO}_2+\mathrm{CH}_3 \mathrm{COOH} \rightarrow \mathrm{C}+\text { by product }
\end{aligned}\)
The spin-only magnetic moment value of the metal complex \(\mathrm{C}\) is ______ BM (Nearest integer)

Options:

The correct answer is:
0

Solution:
If \(M^{+2}=C o^{+2}\)
then
\(\begin{aligned}
& \mathrm{Co}^{+2}+\mathrm{H}_2 \mathrm{~S} \rightarrow \underset{\substack{\text { Black } \\
\text { [A] }}}{\mathrm{CoS}}+\text { byproduct } \\
& \mathrm{CoS}+\text { aqua regia } \rightarrow \underset{\text{[B]}}{\mathrm{CoCl}_2}+\mathrm{NOCl}+\mathrm{S}+\mathrm{H}_2 \mathrm{O}
\end{aligned}\)
\(\mathrm{CoCl}_2+\mathrm{KNO}_2 \xrightarrow{\mathrm{CH}_3 \mathrm{COOH}} k_3\left[\mathrm{CO}\left(\mathrm{NO}_2\right)_6\right]+\text { byproduct }\)
\([C]=C o^{+3} \rightarrow 3 d^6 \Rightarrow N O_2\) is SFL so Co +3 has zero unpaired \(e^{-}\)
\(\Rightarrow n=0 \Rightarrow u=\sqrt{n(n+2)}=0 B . M\)

Topic: Chemistry, Practical Chemistry, JEE Main, JEE Main 2024 (09 Apr Shift 2)
Consider the following

The above process in which α and β form remain in equilibrium with acyclic form and a change in optical rotation is observed, is called as-

Options:
A Mutarotation
B Epimerisation
C Condensation
D Inversion

The correct answer is:
Mutarotation

Solution:
The above phenomenon is called as mutarotation.
Mutarotation is the change in the optical rotation because of the change in the equilibrium between two anomers, when the corresponding stereocenters interconvert. Cyclic sugars show mutarotation as α and β anomeric forms interconvert.

Topic: Chemistry, Biomolecules, JEE Main
Consider the following three halides
(1) CH3–CH2–Cl (2) CH2=CH–Cl (3) C6H5–Cl
Arrange C–Cl bond length of these compounds in decreasing order

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

The correct answer is:
1 > 2 > 3

Solution:
C-Cl bond length of these compounds in decreasing order is as follows
CH3-CH2-Cl>CH2=CH-Cl>C6H5-Cl
C−Cl bond length =  1 no. of resonating structures
CH3-CH2-Cl no conjugation
CH2=CH-Cl Two resonating structures
C6H5-Cl Four resonating structures

Topic: Chemistry, General Organic Chemistry, JEE Main
Consider the following three halides :

(1) CH3-CH2-Cl

(2) CH2=CH-Cl

(3) C6H5-Cl

Arrange C-Cl bond length of these compounds in decreasing order -

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

The correct answer is:
1 > 2 > 3

Solution:
(1) CH3-CH2-Cl
R.S.=0
(2) CH2=CH-Cl
R.S.=2
(3) C6H5-Cl
R.S.=4
R.S. ∝1Bond length of single bond

Topic: Chemistry, General Organic Chemistry, JEE Main
Consider the following three statements:

A If 3+3=7 then 4+3=8

B If 5+3=8 then earth is flat.

C If both A and B are true then 5+6=17.

Then, which of the following statements is correct?

Options:
A A is false, but B and C are true
B A and C are true while B is false
C A is true while B and C are false
D A and B are false while C is true

The correct answer is:
A and C are true while B is false

Solution:
Truth table for if-then statement is

p	q	p→q

T	T	T

T	F	F

F	T	T

F	F	T

A 3+3=7 (false), and 4+3=8 (false), therefore A is true.

B 5+3=8 (true) and earth is flat (false), hence B is false.

C Since, B is false and 5+6=17 (false), hence C is true.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2021 (20 Jul Shift 2)
Consider the following three statements:

P:5 is a prime number

Q:7 is a factor of 192

R: LCM of 5 and 7 is 35

Then the truth value of which one of the following statements is true?

Options:
A P∨~Q∧R
B P∧Q∨~R
C ~P∨Q∧R
D ~P∧~Q∧R

The correct answer is:
P∨~Q∧R

Solution:
P is true

Q is False

R is True

1 P∨~Q∧R=T∨T∧T=T

2 T∧F∨F =F∨F=F

3 F∨F∧T=F∨F=F

4 F∧T∧T=F∧T=F

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2019 (10 Jan Shift 2)
Consider the following transformation involving first order elementary reaction in each step at constant temperature as shown below.
\(A+B \underset{\text { Step 3 }}{\text { Step } 1} C \xrightarrow{\text { Step 2 }} P\)
Some details of the above reactions are listed below.
\(\begin{array}{ccc}
\text { Step } & \text { Rate constant }\left(\mathbf{s e c}^{-\mathbf{1}}\right) & \text { Activation energy }\left(\mathbf{k J ~ m o l}^{-\mathbf{1}}\right) \\
1 & \mathrm{k}_1 & 300 \\
2 & \mathrm{k}_2 & 200 \\
3 & \mathrm{k}_3 & \mathrm{Ea}_3
\end{array}\)
If the overall rate constant of the above transformation \((k)\) is given as \(k=\frac{k_1 k_2}{k_3}\) and the overall activation energy \(\left(\mathrm{E}_{\mathrm{a}}\right)\) is \(400 \mathrm{~kJ} \mathrm{~mol}^{-1}\), then the value of \(\mathrm{Ea}_3\) is \(\qquad\) \(\mathrm{kJ} \mathrm{mol}^{-1}\) (nearest integer)

Options:

The correct answer is:
100

Solution:
$\begin{aligned} & \mathrm{K}=\frac{\mathrm{K}_1 \mathrm{~K}_2}{\mathrm{~K}_3} \\ & \mathrm{Ae}^{\frac{-\mathrm{E}_3}{\mathrm{RT}}}=\frac{\mathrm{A}_1 \mathrm{e}^{\frac{-\mathrm{E}_{\mathrm{a}_1}}{\mathrm{RT}}} \mathrm{A}_2 \mathrm{e}^{\frac{-\mathrm{E}_{3_2}}{\mathrm{RT}}}}{\mathrm{A}_3 \mathrm{e}^{\frac{-\mathrm{E}_{\mathrm{k}_3}}{\mathrm{RT}}}} \\ & \mathrm{Ae}^{\frac{-\mathrm{E}_3}{\mathrm{RT}}}=\frac{\mathrm{A}_1 \mathrm{~A}_2}{\mathrm{~A}_3} \mathrm{e}^{\frac{-\left(\mathrm{E}_{\mathrm{a}_1}-\mathrm{E}_{\mathrm{a}_2}-\mathrm{E}_{3_3}\right)}{\mathrm{RT}}} \\ & \mathrm{E}_{\mathrm{a}}=\mathrm{E}_{\mathrm{a}_1}+\mathrm{E}_{\mathrm{a}_2}-\mathrm{E}_{\mathrm{a}_3} \\ & 400=300+200-\mathrm{E}_{\mathrm{a}_3} \\ & \mathrm{E}_{\mathrm{a}_3}=100 \mathrm{~kJ} / \mathrm{mole}\end{aligned}$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2024 (04 Apr Shift 1)
Consider the following two binary relations on the set A=a, b, c : R1=c, a, b, b, a, c, c, c, b, c, a, a and R2=a, b, b, a, c, c, c, a, a, a, b, b, a, c, then :

Options:
A R2 is symmetric but it is not transitive
B both R1 and R2 are not symmetric
C both R1 and R2 are transitive
D R1 is not symmetric but it is transitive

The correct answer is:
R2 is symmetric but it is not transitive

Solution:
Since, (b, c)∈R1 but (c, b)∉R1

Hence, R1 is not symmetric.

In R1 : b, c∈R1 and c, a∈R1 but b, a∉R1

So, R1 is not transitive.

R2 is symmetric.

In R2 : b, a∈R2 and a, c∈R1 but b, c∉R2

So, R2 is not transitive.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2018 (15 Apr)
Consider the following two binary relations on the set $A=\{a, b, c\}: R_1=\{(\mathrm{c}, a)(b, b),(\mathrm{a}, c),(c$, $c),(b, c),(a, a)\}$ and $\mathrm{R}_2=\{(\mathrm{a}, \mathrm{b}),(\mathrm{b}, \mathrm{a}),(\mathrm{c}, \mathrm{c})$, (c, a), (a, a), (b, b), (a, c). Then

Options:
A
$R_2$ is symmetric but it is not transitive

B
Both $R_1$ and $R_2$ are transitive

C
Both $R_1$ and $R_2$ are not symmetric

D
$R_1$ is not symmetric but it is transitive

The correct answer is:
$R_2$ is symmetric but it is not transitive

Solution:
Both $R_1$ and $R_2$ are symmetric as For any $(x, y) \in R_1$, we have $(y, x) \in R_1$ and similarly for $R_2$ Now, for $R_2,(b, a) \in R_2,(a, c) \in R_2$ but $(b, c) \notin R_2$. Similarly, for $R_1,(b, c) \in R_1,(c, a) \in R_1$ but $(b, a) \notin R_1$.
Therefore, neither $R_1$ nor $R_2$ is transitive.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
Consider the following two methods:
Methodâ¡ 1:(CH3â¡)3CBr+NaOMeâ
Methodâ2: CH 3 Br+t-BuONaâ
which of the following is correct for the formation of ether

Options:
A method 1 should be used
B method 2 should be used
C any of the two methods can be used
D none of the method can be used

The correct answer is:
method 2 should be used

Solution:
In method 1, E2 reaction will takes place and alkene will be formed. While in method 2, substitution (S N 1) will takes place and ether will be formed.
CH 3 Br + t-BuONa → S N 1 t-BuOCH 3

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Consider the following two statements $A$ and $B$ and identify the correct choice in the given answers
A. Paramagnetism is explained by domain theory.
B. Susceptibility of a diamagnetic substance is independent of temperature.

Options:
A Both $A$ and $B$ are correct
B Both A and B are wrong
C $\mathrm{A}$ is correct and $\mathrm{B}$ is wrong
D A is wrong and B is correct

The correct answer is:
A is wrong and B is correct

Solution:
Only ferromagnetism is explained by domain theory, so statement $(A)$ is wrong susceptibility of diamagnetic substance does not depend on temperature, so statement $(B)$ is correct.

Topic: Physics, Magnetic Properties of Matter, AP EAMCET, AP EAMCET 2002
Consider the following two statements $A$ and $B$ and identify the correct choice in the given answers
A. Paramagnetism is explained by domain theory.
B. Susceptibility of a diamagnetic substance is independent of temperature.

Options:
A Both $A$ and $B$ are correct
B Both A and B are wrong
C $\mathrm{A}$ is correct and $\mathrm{B}$ is wrong
D A is wrong and B is correct

The correct answer is:
A is wrong and B is correct

Solution:
Only ferromagnetism is explained by domain theory, so statement $(A)$ is wrong susceptibility of diamagnetic substance does not depend on temperature, so statement $(B)$ is correct.

Topic: Physics, Magnetic Properties of Matter, TS EAMCET, TS EAMCET 2002
Consider the following two statements:
(A) Linear momentum of a system of particles is zero
(B) Kinetic energy of a system of particles is zero
Then

Options:
A
$A$ does not imply $B$ and $B$ does not imply $A$

B
$A$ implies $B$ but $B$ does not imply $A$

C
$A$ does not imply $B$ but $B$ implies $A$

D
$A$ implies $B$ and $B$ implies $A$

The correct answer is:
$A$ does not imply $B$ but $B$ implies $A$

Solution:
No Solution Available

Topic: Physics, Work Power Energy, JEE Main
Consider the following two statements:
(A) Linear momentum of a system of particles is zero
(B) Kinetic energy of a system of particles is zero
Then

Options:
A
$A$ does not imply $B$ and $B$ does not imply $A$

B
$A$ implies $B$ but $B$ does not imply $A$

C
$A$ does not imply $B$ but $B$ implies $A$

D
$A$ implies $B$ and $B$ implies $A$

The correct answer is:
$A$ does not imply $B$ but $B$ implies $A$

Solution:
No Solution Available

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2003
Consider the following two statements P and Q.
P: cos-1⁡cos⁡4π3=4π3
Q: sec2⁡cot-1⁡12+cosec2tan-1⁡13=15
Then, which of the following is true?

Options:
A Both P and Q are true
B P is true, but Q is false
C P is false, but Q is true
D Both P and Q are false

The correct answer is:
P is false, but Q is true

Solution:
We have, cos⁡4π3=-cos⁡π3=-12

Hence, cos-1⁡-12=2π3
Therefore, P is false.

Now, sec2⁡cot-1⁡12+cosec2tan-1⁡13

=1+tan2⁡tan-1⁡2+1+cot2⁡cot-1⁡3
=1+4+1+9=15
Therefore, Q is true.

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
Consider the following two statements:
P : If 7 is an odd number, then 7 is divisible by 2.
Q : If 7 is a prime number, then 7 is an odd number.
If V1 is the truth value of the contrapositive of P and V2 is the truth value of contrapositive of Q, then the ordered pair V1,V2 equals

Options:
A F,T
B F,F
C T,F
D T,T

The correct answer is:
F,T

Solution:
Contrapositive of P:
7 is not divisible by 2⇒7 is not odd number
T ⇒F:FV1
Contrapositive of Q:
7 is not odd number ⇒7 is not a prime number
F⇒ F:T V2

Thus, V1,V2≡F,T

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2016 (09 Apr Online)
Consider the following two statements.
Statement $\boldsymbol{p}$ :
The value of $\sin 120^{\circ}$ can be divided by taking $\theta=240^{\circ}$ in the equation
$$
2 \sin \frac{\theta}{2}=\sqrt{1+\sin \theta}-\sqrt{1-\sin \theta} .
$$
Statement $\boldsymbol{q}$ :
The angles $A, B, C$ and $D$ of any quadrilateral $A B C D$ satisfy the equation
$$
\cos \left(\frac{1}{2}(A+C)\right)+\cos \left(\frac{1}{2}(B+D)\right)=0
$$
Then the truth values of $p$ and $q$ are respectively.

Options:
A
$\mathrm{F}, \mathrm{T}$

B
$\mathrm{T}, \mathrm{T}$

C
F, F

D
T, F

The correct answer is:
$\mathrm{F}, \mathrm{T}$

Solution:
Statement $p$ :
$$
\sin 120^{\circ}=\cos 30^{\circ}=\frac{\sqrt{3}}{2} \Rightarrow 2 \sin 120^{\circ}=\sqrt{3}
$$
So, $\sqrt{1+\sin 240^{\circ}}-\sqrt{1-\sin 240^{\circ}}=$
$$
\sqrt{\frac{1-\sqrt{3}}{2}}-\sqrt{\frac{1+\sqrt{3}}{2}} \neq \sqrt{3}
$$
Statement $\mathbf{q}$ :
So, $A+B+C+D=2 \pi$
$$
\begin{aligned}
&\Rightarrow \frac{A+C}{2}+\frac{B+D}{2}=\pi \\
&\Rightarrow \cos \left(\frac{A+C}{2}\right)+\cos \left(\frac{B+D}{2}\right) \\
&=\cos \left(\frac{A+C}{2}\right)-\cos \left(\frac{A+C}{2}\right)=0
\end{aligned}
$$
Therefore, statement $p$ is false and statement $q$ is true.

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
Consider the fourteen lines in the plane given by $y=x+r, \quad y=-x+r, \quad$ where $r \in\{0,1,2,3,4,5,6\}$. The number of squares formed by these lines, whose sides are of length $\sqrt{2}$, is :

Options:
A 9
B 16
C 25
D 36

The correct answer is:
25

Solution:
We have,
$y=x+r$ $\ldots$ (i)
and $y=-x+r$ $\ldots$ (ii)
The given lines are perpendicular to each other.
perpendicular distance
$=\frac{\left|r_1-r_2\right|}{\sqrt{2}}=\sqrt{2}$
$\Rightarrow \quad r_1-r_2=2$
$\Rightarrow$ The difference between the $y$-intercepts $=2$
This can happen for five combinations $\{(0,2),(1,3),(2,4),(3,5),(4,6)\}$
The difference between the $x$-intercepts $=2$
This can happen for five combination the total number of squares $=5 \times 5=25$

Topic: Mathematics, Permutation Combination, JEE Main