datavorous/entrance-exam-dataset huggingface dataset:
Consider the statements
(I) If $f(x)=\sin \left(\cot ^{-1}\left(\cos \left(\tan ^{-1} x\right)\right)\right)$, then
$f(0)=\frac{1}{2}$
(II) $\sin \left(4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239}\right)=1$
Then the correct option among the following is

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

The correct answer is:
Both I and II are false

Solution:
(I) We have,
$\begin{aligned} f(x) & =\sin \left(\cot ^{-1}\left(\cos \left(\tan ^{-1} x\right)\right)\right) \\ & =\sin \cot ^{-1} \cos \cos ^{-1} \frac{1}{\sqrt{1+x^2}} \\ & =\sin \cot ^{-1} \frac{1}{\sqrt{1+x^2}} \\ & =\sin \sin ^{-1} \sqrt{\frac{1+x^2}{2+x^2}}=\sqrt{\frac{1+x^2}{2+x^2}}\end{aligned}$
$\therefore \quad f(0)=\sqrt{\frac{1+0}{2+0}}=\frac{1}{\sqrt{2}}$
So, Statement $\mathrm{I}$ is false.
(II) We have,
$\begin{aligned} \sin \left(4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239}\right) & =\sin \left(2 \tan ^{-1}\left(\frac{\frac{2}{5}}{1-\frac{1}{25}}\right)-\tan ^{-1} \frac{1}{239}\right) \\ = & \sin \left(2 \tan ^{-1} \frac{5}{12}-\tan ^{-1} \frac{1}{239}\right) \\ = & \sin \left(\tan ^{-1} \frac{\frac{10}{12}}{1-\frac{25}{144}}-\tan ^{-1} \frac{1}{239}\right) \\ = & \sin \left(\tan ^{-1} \frac{120}{119}-\tan ^{-1} \frac{1}{239}\right) \\ = & \sin \left(\tan ^{-1}\left(\frac{119}{1+\frac{120}{119} \times \frac{1}{239}}\right)\right) \\ = & \sin \tan ^{-1}\left[\frac{28680-119}{28441+120}\right]=\sin \tan ^{-1} \frac{28561}{28561} \\ = & \sin \tan ^{-1} 1=\sin \cdot \frac{\pi}{4}=\frac{1}{\sqrt{2}}\end{aligned}$
Statement II is false.

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
Consider the statements S1 and S2 :
S1 : Conductivity always increases with decreases in the concentration of electrolyte.
S2 : Molar conductivity always increases with decreases in the concentration of electrolyte.
The correct option among the following

Options:
A S1 is wrong and S2 is correct
B S1 is correct and S2 is wrong
C Both S1 and S2 are correct
D Both S1 and S2 are wrong

The correct answer is:
S1 is wrong and S2 is correct

Solution:
Conductivity changes with the concentration of electrolyte. The number of ions per unit volume carrying the current decreases on dilution, so conductivity (K) always decreases with decrease in concentration.
Molar. Conductivity increases with dilution.
∧mmolar conductivity=1000×KC

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2019 (10 Apr Shift 1)
Consider the statistics of two sets of observations as follows:

Size	Mean	Variance

Observation I	10	2	2

Observation II	n	3	1

If the variance of the combined set of these two observations is 179, then the value of n is equal to ________.

Options:

The correct answer is:
5

Solution:
σ2=n1σ12+n2σ22n1+n2+n1n2n1+n2x¯1-x¯22

n1=10,n2=n,σ12=2,σ22=1

x¯1=2,x¯2=3,σ2=179

179=10×2+nn+10+10n(n+10)23-22

⇒179=(n+20)(n+10)+10n(n+10)2

⇒17n2+1700+340n=90n+9n2+30n+200

⇒8n2-20n-100=0

2n2-5n-25=0

⇒2n+5n-5=0⇒n=-52↓Rejected,5

Hence n=5

Topic: Mathematics, Statistics, JEE Main, JEE Main 2021 (16 Mar Shift 2)
Consider the straight lines
$$
\begin{aligned}
& L_1: x-y=1 \\
& L_2: x+y=1 \\
& L_3: 2 x+2 y=5 \\
& L_4: 2 x-2 y=7
\end{aligned}
$$
The correct statement is

Options:
A
$L_1\left\|L_4, L_2\right\| L_3, L_1$ intersect $L_4$.

B
$L_1 \perp L_2, L_1 \| L_3, L_1$ intersect $L_2$.

C
$L_1 \perp L_2, L_2 \| L_3, L_1$ intersect $L_4$.

D
$L_1 \perp L_2, L_1 \perp L_3, L_2$ intersect $L_4$.

The correct answer is:
$L_1 \perp L_2, L_1 \perp L_3, L_2$ intersect $L_4$.

Solution:
Consider the lines
$L_1: x-y=1$
$L_2: x+\mathrm{y}=1$
$L_3: 2 x+2 y=5$
$L_4: 2 x-2 y=7$
$L_1 \perp L_2$ is correct statement
$(\because$ Product of their slopes $=-1)$
$L_1 \perp L_3$ is also correct statement
$(\because$ Product of their slopes $=-1)$
Now, $L_2: x+y=1$
$L_4: 2 x-2 y=7$
$\Rightarrow 2 x-2(1-x)=7$
$\Rightarrow 2 x-2+2 x=7$
$\Rightarrow x=\frac{9}{4}$ and $y=\frac{-5}{4}$
Hence, $L_2$ intersects $L_4$.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2012 (26 May Online)
Consider the sulphides HgS, PbS, CuS, Sb2 S3, As2S3 and CdS. Number of these sulphides soluble in 50% HNO3 is ___________ .

Options:

The correct answer is:
4

Solution:
CdS, PbS, As2S3 and CuS, are soluble in 50% HNO3 while HgS and Sb2S3 are insoluble in 50% HNO3 but HgS soluble in aqua regia.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2021 (31 Aug Shift 1)
Consider the system of equations ax+y+bz=0, bx+y+az=0 and ax+by+abz=0 where a,b∈0,1,2,3,4. The number of ordered pairs a,b for which the system has non-trivial solutions is

Options:

The correct answer is:
6

Solution:
For non-trivial solutions,
a1bb1aabab=0
⇒a-ba-b2=0
⇒ Either a=b or a=b2
When a=b, then ordered pairs are 0,0,1,1,2,2,3,3,4,4
When a=b2, then 4,2
Hence, number of ordered pairs are 6

Topic: Mathematics, Determinants, JEE Main
Consider the system of equations
x+2y+3z=6,
4x+5y+6z=λ,
7x+8y+9z=24.
Then, the value of λ for which the system has infinite solutions is

Options:
A 15
B 20
C 16
D 30

The correct answer is:
15

Solution:
Note that D=123456789=0R3→R3-2R2+R1
D1=623λ562489=-330-2λ
D2=1634λ67249=630-2λ
D3=12645λ7824=-330-2λ
Hence, for λ=15 the system has infinite solutions

Topic: Mathematics, Determinants, JEE Main
Consider the system of equations :
$x+a y=0, y+a z=0$ and $z+a x=0$. Then the set of all real values of ' $a$ ' for which the system has a unique solution is:

Options:
A
$\mathrm{R}-\{1\}$

B
$\mathrm{R}-\{-1\}$

C
$\{1,-1\}$

D
$\{1,0,-1\}$

The correct answer is:
$\mathrm{R}-\{-1\}$

Solution:
Given system of equations is homogeneous which is
$$
\begin{aligned}
&x+a y=0 \\
&y+a z=0 \\
&z+a x=0
\end{aligned}
$$
It can be written in matrix form as
$$
\mathrm{A}=\left(\begin{array}{lll}
1 & a & 0 \\
0 & 1 & a \\
a & 0 & 1
\end{array}\right)
$$
Now, $|\mathrm{A}|=\left[1-a\left(-a^2\right)\right]=1+a^3 \neq 0$
So, system has only trivial solution.
Now, $|\mathrm{A}|=0$ only when $a=-1$
So, system of equations has infinitely many solutions which is not possible because it is given that system has a unique solution.
Hence set of all real values of ' $a$ ' is
$$
\mathrm{R}-\{-1\}
$$

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2013 (25 Apr Online)
Consider the system of equations αx+y+z=p, x+αy+z=q and x+y+αz=r, then the sum of all possible distinct value(s) of α for which system does not possess a unique solution is

Options:
A -2
B 1
C -1
D 0

The correct answer is:
-1

Solution:
Here D=α111α111α=α+2α-12
Hence, for α=-2,1 the system will not have a unique solution.

The sum of values of α=-2+1=-1

Topic: Mathematics, Determinants, JEE Main
Consider the system of linear equation x+y+z= 4μ,x+2y+2λz=10μ,x+3y+4λ2z=μ2+15, where λ,μ∈R. Which one of the following statements is NOT correct?

Options:
A The system has unique solution if λ≠12 and μ≠1,15
B The system is inconsistent if λ=12 and μ≠1
C The system has infinite number of solutions if λ=12 and μ=15
D The system is consistent if λ≠12

The correct answer is:
The system is inconsistent if λ=12 and μ≠1

Solution:
Given:

System of linear equations,

x+y+z=4μ,

x+2y+2λz=10μ,

x+3y+4λz2=μ2+15.

⇒Δ=111122λ134λ2

⇒Δ=8λ2-6λ-4λ2-2λ+3-2

⇒Δ=8λ2-6λ-4λ2+2λ+1

⇒Δ=4λ2-4λ+1

⇒Δ=(2λ-1)2

For unique solution, Δ≠0.

⇒2λ-1≠0

⇒λ≠12

Let, Δ=0, λ=12

We know that, for infinite solutions, Δx=Δy=Δz=0

⇒Δx=4μ1110μ21μ2+1531

⇒Δx=4μ2-3-10μ-μ2-15+30μ-2μ2-30

⇒Δx=-4μ-10μ+μ2+15+30μ-2μ2-30

⇒Δx=-μ2+16μ-15

⇒(μ-15)(μ-1)=0

So, for infinite solution λ=12, μ=1 or 15.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2024 (30 Jan Shift 1)
Consider the system of linear equations:
$$
\begin{aligned}
& x_1+2 x_2+x_3=3 \\
& 2 x_1+3 x_2+x_3=3 \\
& 3 x_1+5 x_2+2 x_3=1
\end{aligned}
$$
The system has

Options:
A
exactly 3 solutions

B
a unique solution

C
no solution

D
infinite number of solutions

The correct answer is:
no solution

Solution:
$$
\begin{aligned}
& D=\left|\begin{array}{lll}
1 & 2 & 1 \\
2 & 3 & 1 \\
3 & 5 & 2
\end{array}\right|=0 \\
& D_1=\left|\begin{array}{lll}
3 & 2 & 1 \\
3 & 3 & 1 \\
1 & 5 & 2
\end{array}\right| \neq 0
\end{aligned}
$$
$\Rightarrow$ Given system, does not have any solution.
$\Rightarrow$ No solution.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2010
Consider the system of linear equations

-x+y+2z=0

3x-ay+5z=1

2x-2y-az=7

Let S1 be the set of all a∈R for which the system is inconsistent and S2 be the set of all a∈R for which the system has infinitely many solutions. If nS1 and nS2 denote the number of elements in S1 and S2 respectively, then

Options:
A nS1=2, nS2=0
B nS1=2,nS2=2
C nS1=0, nS2=2
D nS1=1,nS2=0

The correct answer is:
nS1=2, nS2=0

Solution:
For system to be inconsistent D=-1123-a52-2-a=0

⇒a2-7a+12=0⇒a=3, 4

Dx=0121-a57-2-a=15a+31

Dx≠0 for a=3,4

Similarly we can show Dy≠0, Dz≠0 for a=3,4

⇒nS1=2

Now for infinitely many solutions D=0 also Dx=Dy=Dz=0 which is not possible value of any real of a, since D=0 & Dx=0 have different solutions for a

⇒nS2=0

Hence, nS1=2 & nS2=0

Topic: Mathematics, Determinants, JEE Main, JEE Main 2021 (01 Sep Shift 2)
Consider the system of linear equations x+y+z=5,x+2y+λ2z=9 and x+3y+λz=μ, where λ,μ∈R. Then, which of the following statement is NOT correct ?

Options:
A System has infinite number of solution if λ=1 and μ=13
B System is inconsistent if λ=1 and μ≠13
C System has unique solution if λ≠1 and μ≠13
D System is consistent if λ≠1 and μ=13

The correct answer is:
System has unique solution if λ≠1 and μ≠13

Solution:
Given:

x+y+z=5

x+2y+λ2z=9

x+3y+λz=μ

⇒△=11112λ213λ

⇒△=2λ-3λ2-λ+λ2+1

⇒△=-2λ2+λ+1

We know that for infinite solutions, △=0 & △1=△2=△3=0

⇒-2λ2+λ+1=0

⇒2λ2-λ-1=0

⇒λ-12λ+1=0

⇒λ=1, -12

Now, finding △1 for λ=1 we get,

Δ1=511921μ31

⇒Δ1=μ-13

And for λ=-12 we get,

Δ1=5119214μ3-12=14514921μ3-2=7413-μ

So, for λ≠1 & μ≠13 system has no solution,

Hence, option C is the correct answer.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Consider the system shown in figure. The wall is smooth, but the surfaces of blocks A and B in contact are rough. The friction on B due to A in equilibrium is

Options:
A Upward
B Downward
C Zero
D The system cannot remain in equilibrium

The correct answer is:
The system cannot remain in equilibrium

Solution:
Considering A and B as a system. There is no vertical force in upward direction to support their weight. Therefore, the system cannot remain in equilibrium.

Topic: Physics, Laws of Motion, JEE Main
Consider the three planes

P1:3x+15y+21z=9

P2:x-3y-z=5, and

P3:2x+10y+14z=5

Then, which one of the following is true?

Options:
A P2 and P3 are parallel.
B P1,P2 and P3 all are parallel.
C P1 and P2 are parallel.
D P1 and P3 are parallel.

The correct answer is:
P1 and P3 are parallel.

Solution:
P1:x+5y+7z=3

P2:x-3y-z=5

P3:x+5y+7z=52

For P1 & P3, 11=55=77.

So, P1 and P3 are parallel.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (26 Feb Shift 1)
Consider the three waves $z_1, z_2$ and $z_3$ as $z_1=A \sin (k x-\omega t) \quad z_2=A \sin (k x+\omega t)$ and $z_3=A \sin (k y-\omega t)$. Which of the following represents a standing wave

Options:
A $z_1+z_2$
B $z_2+z_3$
C $z_3+z_1$
D $z_1+z_2+z_3$

The correct answer is:
$z_1+z_2$

Solution:
Waves $Z_1=A \sin (k x-\omega t)$ is travelling towards positive $x$-direction. Wave $Z_2=A \sin (k x+\omega t)$, is travelling towards negative $x$-direction. Wave $z_3=A \sin (k y-\omega t)$ is travelling towards positive $y$ direction.
Since waves $Z_1$ and $Z_2$ are travelling along the same line so they will produce stationary wave.

Topic: Physics, Waves and Sound, JEE Main
Consider the two different first order reactions given below
\(\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}\) (Reaction 1\()\)
\(P \rightarrow Q\) (Reaction 2)
The ratio of the half life of Reaction 1 : Reaction 2 is \(5: 2\). If \(t_1\) and \(t_2\) represent the time taken to complete \(2 / 3^{\text {rd }}\) and \(45^{\text {th }}\) of Reaction 1 and Reaction 2, respectively, then the value of the ratio \(t_1: t_2\) is ______ \(\times 10^{-1}\) (nearest integer).
[Given : \(\log _{10}(3)=0.477\) and \(\log _{10}(5)=0.699\)]

Options:

The correct answer is:
17

Solution:
$\begin{aligned} & \frac{\left(\mathrm{t}_{1 / 2}\right)_1}{\left(\mathrm{t}_{1 / 2}\right)_{\mathrm{II}}}=\frac{\mathrm{K}_2}{\mathrm{~K}_1}=\frac{5}{2} \\ & \therefore \mathrm{K}_1 \mathrm{t}_1=\ln \frac{1}{1-\frac{2}{3}}=\ln 3 \\ & \mathrm{~K}_2 \mathrm{t}_2=\ln \frac{1}{1-\frac{4}{5}}=\ell \mathrm{n} 5 \\ & \Rightarrow \frac{\mathrm{K}_1}{\mathrm{~K}_2} \times \frac{\mathrm{t}_1}{\mathrm{t}_2}=\frac{0.477}{0.699} \\ & \Rightarrow \frac{\mathrm{t}_1}{\mathrm{t}_2}=\frac{0.477}{0.699} \times \frac{5}{2}=1.7=17 \times 10^{-1}\end{aligned}$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2024 (06 Apr Shift 2)
Consider the two following statements I and II, and identify the correct choice given in the answers
1. In photovoltaic cells, the photoelectric current produced is not proportional to the intensity of incident light.
2. In gas-filled photoemissive cells, the velocity of photoelectrons depends on the wavelength of the incident radiation.

Options:
A Both 1 and 2 are true
B Both 1 and 2 are false
C 1 is true but 2 is false
D 1 is false but 2 is true

The correct answer is:
1 is false but 2 is true

Solution:
In photovoltaic cells, photoelectric current produced is proportional to the intensity of incident light, so the statement 1 is false.
The velocity of photoelectrons depends upon the maximum kinetic energy of photoelectrons, which depends on the frequency and hence on the wavelength of incident radiation, so the statement 2 is true.

Topic: Physics, Dual Nature of Matter, JEE Main
Consider the two sets:
A={m∈R: both the roots of x2-(m+1)x+m+4=0 are real } and B=[-3,5)
Which of the following is not true?

Options:
A A-B=(-∞,-3)∪(5,∞)
B A∩B={-3}
C B-A=(-3,5)
D A∪B=R

The correct answer is:
A-B=(-∞,-3)∪(5,∞)

Solution:
Since both roots are real m+12-4m+4≥0

⇒m2-2m-15≥0⇒m-5m+3≥0

⇒m∈(-∞,-3]∪[5,∞)

⇒A=(-∞,-3]∪[5,∞)

A-B=-∞,-3∪[5,∞)

A∩B=-3

B-A=-3,5

A∪B=R

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2020 (03 Sep Shift 1)
Consider the two statements :
S1 :(p→q)∨(~q→p) is a tautology.
S2 :(p∧~q)∧(~p∨q) is a fallacy.
Then :

Options:
A only (S1) is true.
B both (S1) and (S2) are false.
C only (S2) is true.
D both (S1) and (S2) are true.

The correct answer is:
both (S1) and (S2) are true.

Solution:
pq~p~qp→q~q→p(p→q)∨(~q→p)p∧~q~p∨q(p∧~q)∧(~p∨q)TTFFTTTFTFTFFTFTTTFFFTTFTTTFTFFFTTTFTFTF

From table
(p→q)∨(~q→p) is tautology and (p∧~q)∧(~p∨q) is fallacy.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2021 (26 Aug Shift 2)
Consider the van der Waal's constants, a and b, for the following gases.

Gas	Ar	Ne	Kr	Xe

a/(atm dm6 mol-2)	1.3	0.2	5.1	4.1

b/(10-2 dm3 mol-1)	3.2	1.7	1.0	5.0

Which gas is expected to have the highest critical temperature?

Options:
A Kr
B Ar
C Xe
D Ne

The correct answer is:
Kr

Solution:
Critical temperature is :-

TC=8a27Rb
for Kr the value of ab is the highest so, Tc for Kr will be highest.
ab for Ar=1.33.2=0.406
ab for Ne=0.21.7=0.118
ab for Kr=5.11=5.1
ab for Xe=4.15=0.82

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2019 (09 Apr Shift 1)
Consider the vectors
$\overrightarrow{\mathrm{A}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}, \overrightarrow{\mathrm{B}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{C}}=\frac{1}{\sqrt{5}}(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})$. What is the value of $\overrightarrow{\mathrm{C}} .(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}) ?$

Options:
A 1
B 0
C $3 \sqrt{2}$
D $18 \sqrt{5}$

The correct answer is:
0

Solution:
Hint:
$\overrightarrow{\mathrm{C}}(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}})=\left|\begin{array}{ccc}1 & 1 & -1 \\ 2 & -1 & 1 \\ 1 & -2 & 2\end{array}\right|=1(-2+2)-1(4-1)-1(-4+1)=-3+3=0$

Topic: Physics, Mathematics in Physics, JEE Main
Consider the vernier calipers as shown, the instrument has no zero error. What is the length of the rod (in mm) shown, if 1 msd = 1 mm? Use 7 msd = 8 vsd.

Options:

The correct answer is:
4.5

Solution:
7 msd = 8 vsd
â 7 vsd = 7 8 msd
â L.C = 1 - 7 8 mm = 1 8 mm
Were M.SR = 4 mm
V.S.R = 4
value value = 4 + 4 1 8 mm
= 4.5 mm

Topic: Physics, Experimental Physics, JEE Main
Consider the vernier callipers shown below. The instrument has no zero error. If 1 m.s.d=1 mm and 7 m.s.d = 8 v.s.d, then what is the length (in mm) of the rod shown in the figure? [Given; the 4th v.s.d coincides with m.s.d]

Options:

The correct answer is:
4.5

Solution:
7 m.s.d=8 v.s.d
⇒1 v.s.d=78 m.s.d
⇒   L.C=1−78 mm=18mm
Were M.S.R=4 mm
V.S.R=4
value=4+418 mm
=4.5  mm

Topic: Physics, Experimental Physics, JEE Main
Consider three lines L1: 5x-y+4=0, L2: 3x-y+5=0 and L3: x+y+8=0. If these lines form a triangle ABC, then the sum of the squares of the tangent of all the interior angles of the triangle is

Options:
A 40164
B 39964
C 11164
D 11364

The correct answer is:
40164

Solution:
Arranging the lines in descending order of slopes, we get,
m1=5, m2=3 and m3=-1
∴tan⁡A=m1-m21+m1m2=21+15=18
tan⁡B=m2-m31+m2m3=3+11-3=-2
tan⁡C=m3-m11+m3m1=-1-51-5=32
Hence, ∑tan2A=164+4+94=1+256+14464=40164

Topic: Mathematics, Straight Lines, JEE Main
Consider three observations a,b and c such that b=a+c. If the standard deviation of a+2,c+2 is d, then which of the following is true?

Options:
A b2=3a2+c2+9d2
B b2=a2+c2+3d2
C b2=3a2+c2+d2
D b2=3a2+c2-9d2

The correct answer is:
b2=3a2+c2-9d2

Solution:
For a,b,c

mean =a+b+c3=x¯

b=a+c

⇒x¯=2b3 ...(1)

S.D.a+2,b+2,c+2=S.D.a,b,c=d

⇒d2=a2+b2+c23-x¯2

⇒d2=a2+b2+c23-4b29

⇒9d2=3a2+b2+c2-4b2

⇒b2=3a2+c2-9d2

Topic: Mathematics, Statistics, JEE Main
Consider three observations a,b and c such that b=a+c. If the standard deviation of a+2,c+2 is d, then which of the following is true?

Options:
A b2=3a2+c2+9d2
B b2=a2+c2+3d2
C b2=3a2+c2+d2
D b2=3a2+c2-9d2

The correct answer is:
b2=3a2+c2-9d2

Solution:
For a,b,c

mean =a+b+c3=x¯

b=a+c

⇒x¯=2b3 ...(1)

S.D.a+2,b+2,c+2=S.D.a,b,c=d

⇒d2=a2+b2+c23-x¯2

⇒d2=a2+b2+c23-4b29

⇒9d2=3a2+b2+c2-4b2

⇒b2=3a2+c2-9d2

Topic: Mathematics, Statistics, JEE Main, JEE Main 2021 (16 Mar Shift 1)
Consider three points $P=(-\sin (\beta-\alpha),-\cos \beta), Q=(\cos (\beta-\alpha), \sin \beta)$ and $R=(\cos (\beta-\alpha+\theta), \sin (\beta-\theta)$, where $0 < \alpha, \beta, \theta < \frac{\pi}{4}$. Then,

Options:
A
$P$ lies on the line segment $R Q$

B
$Q$ lies on the line segment $P R$

C
$R$ lies on the line segment $Q P$

D
$P, Q, R$ are non- collinear

The correct answer is:
$P, Q, R$ are non- collinear

Solution:
For collinear points
$$
\Delta=\left|\begin{array}{ccc}
-\sin (\beta-\alpha) & -\cos \beta & 1 \\
\cos (\beta-\alpha) & \sin \beta & 1 \\
\cos (\beta-\alpha+\theta) & \sin (\beta-\theta) & 1
\end{array}\right|
$$
Clearly, $\Delta \neq 0$ for any value of $\alpha, \beta, \theta$, hence points are non-collinear.

Topic: Mathematics, Straight Lines, JEE Main
Consider three rods of length L1, L2 and L3 respectively joined in series. Each has same cross-sectional area with Young's moduli Y, 2Y and 3Y respectively and thermal coefficients of linear expansion Î±, 2Î± and 3Î± respectively. They are placed between two rigid fixed walls. The temperature of the whole system is increased and it is found that length of the middle rod does not change with temperature rise. Find the value of 9L1L3 .

Options:

The correct answer is:
5

Solution:
Since the middle rod doesn't expand at all, the strain in the rod will be

strain=2Î±âT

stress=2Y2Î±âT

The stress is same in each rod

4YÎ±âT=YL1Î±âT+xL1=3YL33Î±âT+xL3

On solving for x, we get

x=3L1Î±âT=5L3Î±âT3

â9L1L3=5

Topic: Physics, Thermal Properties of Matter, JEE Main
Consider three square matrices A, B and C of order 3 such that AT=A-2B and BT=B-4C, then the incorrect option is

Options:
A A=0
B B=0
C C=0
D B=2C

The correct answer is:
A=0

Solution:
AT=A-2B⇒2B=A-AT

⇒B is a skew symmetric matrix of order 3⇒B=0
BT=B-4C⇒4C=B-BT

⇒C is a skew symmetric matrix of order 3⇒C=0
Taking transpose of first equation
ATT=A-2BT⇒A=AT-2BT
⇒A=A-2B-2B+8C
⇒B=2C

Topic: Mathematics, Determinants, JEE Main
Consider three statements
p: Aman will come today
q: Aditi will be at her home
r: They will go to party.
Then, the statement qâ§â¼râp is logically equivalent to

Options:
A If Aman will not come today then they will go to party.
B If Aman will not come today then they will not go to party.
C Aman will come today or Aditi will not be at her home or they will got to party.
D Aman will come today or Aditi will go to party.

The correct answer is:
Aman will come today or Aditi will not be at her home or they will got to party.

Solution:
q∧∼r⇒p≡∼q∧∼r∨p
≡∼q∨r∨p≡p∨∼q∨r

Topic: Mathematics, Mathematical Reasoning, JEE Main
Consider three statements

p : Person âAâ passed in mathematics exam

q : Person âAâ passed in physics exam

r : Person âAâ passed in chemistry exam

Then, the statement ~~pâqâ~r is equivalent to

Options:
A Person A passed only in mathematics among mathematics, physics and chemistry.
B Person A failed only in physics among mathematics, physics and chemistry.
C Person A passed in all the three subjects mathematics and physics and chemistry.
D Person A passed in chemistry but failed in mathematics and physics.

The correct answer is:
Person A failed only in physics among mathematics, physics and chemistry.

Solution:
~~p⇒q⇒~r≡~p∧~q⇒~r
≡p∧~q∧r

Topic: Mathematics, Mathematical Reasoning, JEE Main
Consider three statements
p: Person âAâ passed in mathematics exam
q: Person âAâ passed in physics exam
r: Person âAâ passed in chemistry exam
Then, the statement ~~pâqâr is equivalent to

Options:
A Person A passed only in mathematics among mathematics & physics & chemistry.
B Person A failed only in mathematics among mathematics & physics & chemistry.
C Person A passed in all the three subjects mathematics & physics & chemistry.
D Person A passed in chemistry but failed in mathematics & physics.

The correct answer is:
Person A passed only in mathematics among mathematics & physics & chemistry.

Solution:
~~pâqârâ¡~pâ§~qâr
â¡pâ§~qâ§~râ¡pâ§~qâ¨râ¡only p

Topic: Mathematics, Mathematical Reasoning, JEE Main
Consider three statements
p: Suyash will come today evening
q: Shalini will be at his home
r: They will go to party.
Then the statement ~q∧~r⇒p is logically equivalent to

Options:
A If Suyash will not come today then Shalini will go to party.
B If Suyash will not come today then Shalini will not go to party.
C Suyash will come today or Shalini will be at his home or they will go to party.
D Suyash will come today or Shalini will go to party.

The correct answer is:
Suyash will come today or Shalini will be at his home or they will go to party.

Solution:
~qâ§~r⇒pâ¡~~qâ§~râ¨p
â¡qâ¨râ¨pâ¡pâ¨qâ¨r

Topic: Mathematics, Mathematical Reasoning, JEE Main
Consider three vectors $\vec{a}, \vec{b}, \vec{c}$. Let $|\vec{a}|=2,|\vec{b}|=3$ and $\vec{a}=\vec{b} \times \vec{c}$. If $\alpha \in\left[0, \frac{\pi}{3}\right]$ is the angle between the vectors $\vec{b}$ and $\vec{c}$, then the minimum value of $27|\vec{c}-\vec{a}|^2$ is equal to:

Options:
A 110
B 124
C 121
D 105

The correct answer is:
124

Solution:
$\begin{aligned} & |\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|=|\overrightarrow{\mathrm{c}}|^2+|\overrightarrow{\mathrm{a}}|^2-2 \overline{\mathrm{a}} \cdot \overline{\mathrm{c}} \\ & =|\overrightarrow{\mathrm{c}}|^2+4-0 \\ & \because \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}} \\ & |\overrightarrow{\mathrm{a}}|=|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}| \\ & 2=3|\overrightarrow{\mathrm{c}}| \sin \alpha \\ & |\overrightarrow{\mathrm{c}}|=\frac{2}{3} \operatorname{cosec} \alpha \quad \alpha \in\left[0, \frac{\pi}{3}\right] \\ & |\overrightarrow{\mathrm{c}}|_{\min }=\frac{2}{3} \times \frac{2}{\sqrt{3}} \quad \operatorname{cosec} \alpha \in\left[\frac{2}{\sqrt{3}}, \infty\right) \\ & \Rightarrow 27|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|_{\min }^2=27\left(\frac{16}{27}+4\right)=124\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2024 (05 Apr Shift 2)