datavorous/entrance-exam-dataset huggingface dataset:
If $\tan x+\tan \left(x+\frac{\pi}{3}\right)+\tan \left(x+\frac{2 \pi}{3}\right)=3$ then $\tan 3 x$ is equal to

Options:
A $3$
B $2$
C $1$
D $0$

The correct answer is:
$1$

Solution:
$\begin{aligned} & \text { LHS }=\tan x+\tan \left(x+\frac{\pi}{3}\right)+\tan \left(x+\frac{2 \pi}{3}\right) \\ & =\tan x+\frac{\tan x+\sqrt{3}}{1-\sqrt{3} \tan x}+\frac{\tan x-\sqrt{3}}{1+\sqrt{3} \tan x} \\ & =\tan x+\frac{\left[\begin{array}{c}(\tan x+\sqrt{3})(1+\sqrt{3} \tan x) \\ +(\tan x-\sqrt{3})(1-\sqrt{3} \tan x)\end{array}\right]}{1-3 \tan ^2 x} \\ & =\frac{\left[\begin{array}{c}\tan x\left(1-3 \tan ^2 x\right)+\tan x+\sqrt{3} \\ +\sqrt{3} \tan ^2 x+3 \tan x+\tan x-\sqrt{3} \\ -\sqrt{3} \tan ^2 x+3 \tan x\end{array}\right]}{1-3 \tan ^2 x} \\ & \Rightarrow \frac{\tan x\left(1-3 \tan ^2 x\right)+8 \tan x}{1-3 \tan ^2 x}=3 \quad \text { (given) } \\ & \Rightarrow \tan x\left(1-3 \tan ^2 x\right)+8 \tan x \\ & =3\left(1-3 \tan ^2 x\right) \\ & \end{aligned}$
$\begin{aligned} & \Rightarrow \quad \tan x\left(9-3 \tan ^2 x\right)=3\left(1-3 \tan ^2 x\right) \\ & \Rightarrow \quad \tan x\left(3-\tan ^2 x\right)=1-3 \tan ^2 x \\ & \Rightarrow \quad \frac{3 \tan x-\tan ^3 x}{1-3 \tan ^2 x}=1 \\ & \Rightarrow \quad \tan 3 x=1\end{aligned}$

Topic: Mathematics, Trigonometric Equations, JEE Main
If $\tan (x+y)+\tan (x-y)=1$, then $\frac{d y}{d x}=$

Options:
A $\frac{\sec ^2(x+y)+\sec ^2(x-y)}{\sec ^2(x+y)-\sec ^2(x-y)}$
B $\frac{\sec ^2(x+y)+\sec ^2(x-y)}{\sec ^2(x-y)-\sec ^2(x+y)}$
C $\frac{\sec ^2(x+y)-\sec ^2(x-y)}{\sec ^2(x+y)+\sec ^2(x-y)}$
D None of these

The correct answer is:
$\frac{\sec ^2(x+y)+\sec ^2(x-y)}{\sec ^2(x-y)-\sec ^2(x+y)}$

Solution:
$\tan (x+y)+\tan (x-y)=1$
Differentiating w.r.t. $x$ of $y$, we get
$\begin{aligned} & \Rightarrow \sec ^2(x+y)\left(1+\frac{d y}{d x}\right)+\sec ^2(x-y)\left(1-\frac{d y}{d x}\right)=0 \\ & \Rightarrow \frac{d y}{d x}=\frac{\sec ^2(x+y)+\sec ^2(x-y)}{\sec ^2(x-y)-\sec ^2(x+y)}\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
If tangents are drawn from any point on the circle $x^2+y^2=25$ to the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$, then the angle between the tangents is

Options:
A $\frac{2 \pi}{3}$
B $\frac{pi}{4}$
C $\frac{pi}{3}$
D $\frac{pi}{2}$

The correct answer is:
$\frac{pi}{2}$

Solution:
Given circle is a director circle to the given ellipse.
$\therefore$ Angle between tangelt is $\theta=\frac{\pi}{2}$

Topic: Mathematics, Ellipse, JEE Main
If tangents are drawn to the ellipse $x^{2}+2 y^{2}=2$ at all

points on the ellipse other than its four vertices then the mid points of the tangents intercepted between the coordinate axes lie on the curve :

Options:
A $\frac{1}{4 x^{2}}+\frac{1}{2 y^{2}}=1$
B $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$
C $\frac{1}{2 x^{2}}+\frac{1}{4 y^{2}}=1$
D $\frac{x^{2}}{2}+\frac{y^{2}}{4}=1$

The correct answer is:
$\frac{1}{2 x^{2}}+\frac{1}{4 y^{2}}=1$

Solution:
Given the equation of ellipse,

$\frac{x^{2}}{(\sqrt{2})^{2}}+y^{2}=1$

$\frac{\sqrt{2} \cos \theta x}{2}+y \sin \theta=1$

$P\left(\frac{\sqrt{2}}{\cos \theta}, 0\right)$ and $Q\left(0, \frac{1}{\sin \theta}\right)$

Let mid point be $(h, k)$

$\Rightarrow h=\frac{1}{\sqrt{2} \cos \theta}, k=\frac{1}{2 \sin \theta}$

As $\cos ^{2} \theta+\sin ^{2} \theta=1$

$\therefore \quad \frac{1}{2 h^{2}}+\frac{1}{4 k^{2}}=1$

Locus is $\frac{1}{2 x^{2}}+\frac{1}{4 y^{2}}=1$

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2019 (11 Jan Shift 1)
If temperature scale is changed from °C to °F , the numerical value of specific heat will

Options:
A increase
B decrease
C remains unchanged
D none of these

The correct answer is:
decrease

Solution:
ÎC5=ÎF9
â´ ÎF=1.8ÎC
â´ Temperature change 1Â°C is equal to 1.8Â°F .
As unit of specific heat is J/kg/ oC , therefore If temperature scale is changed from Â°C to Â°F , the numerical value of specific heat will decrease.

Topic: Physics, Thermal Properties of Matter, NEET
If the $10^{\text {th }}$ term of a geometric progression is 9 and $4^{\text {th }}$ term is 4 , then its $7^{\text {th }}$ term is

Options:
A $6$
B $36$
C $\frac{4}{9}$
D $\frac{9}{4}$

The correct answer is:
$6$

Solution:
Accordingly, $a r^9=9$ and $a r^3=4$
$\Rightarrow r^3=\frac{3}{2}$ and $a=\frac{8}{3}$
$\therefore 7^{\text {th }}$ term i.e. $a r^6=\frac{8}{3}\left(\frac{3}{2}\right)^2=6$
Trick : $7^{\text {th }}$ term is equidistant from $10^{\text {th }}$ and $4^{\text {th }}$ so it will be $\sqrt{9 \times 4}=6$.

Topic: Mathematics, Sequences and Series, JEE Main
If the 2nd, 5th and 9th terms of a non-constant arithmetic progression are in geometric progression, then the common ratio of this geometric progression is

Options:
A 1
B 74
C 85
D 43

The correct answer is:
43

Solution:
Let the A.P. be a, a+d, a+2d, .....
Given, a+d.a+8d=a+4d2
a2+9ad+8d2=a2+8ad+16d2
8d2-ad=0
d8d-a=0
∵ d ≠0
d=a8
So, 2nd term, 5th term, 9th term
would be a+a8, a+a2, a+a
9a8, 3a2, 2a
Common Ratio =2nd term1st term=3a×82×9a =43

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2016 (03 Apr)
If the 2nd, 5th and 9th terms of a non-constant arithmetic progression are in geometric progression, then the common ratio of this geometric progression is

Options:
A 1
B 74
C 85
D 43

The correct answer is:
43

Solution:
Let the A.P. be a, a+d, a+2d, .....
Given, a+d.a+8d=a+4d2
a2+9ad+8d2=a2+8ad+16d2
8d2-ad=0
d8d-a=0
∵ d ≠0
d=a8
So, 2nd term, 5th term, 9th term
would be a+a8, a+a2, a+a
9a8, 3a2, 2a
Common Ratio =2nd term1st term=3a×82×9a =43

Topic: Mathematics, Sequences and Series, JEE Main
If the 4th term in the expansion of x1log10â¡x+1+x1126 is 200, then the value of x is (where, x>1)

Options:

The correct answer is:
10

Solution:
T4=C36x1log10x+13x1123

⇒200=20x1log10x+132x14
⇒10=x321log10x+1+14
⇒1t=321t+1+14 ….. [Take t=log10x]
⇒t2+3t-4=0
⇒t-1t+4=0
⇒t=1 or t=-4
⇒log10x=1 or log10x=-4
⇒x=10 … [as x>1]

Topic: Mathematics, Binomial Theorem, JEE Main
If the $4^{\text {th }}, 7^{\text {th }}$ and $10^{\text {th }}$ terms of a G.P. be ${ }^{a, b, c}$ respectively, then the relation between $a, b, c$ is

Options:
A $b=\frac{a+c}{2}$
B $a^2=b c$
C $b^2=a c$
D $c^2=a b$

The correct answer is:
$b^2=a c$

Solution:
Let first term of G.P. $=A$ and common ratio $=r$
We know that $n^{\text {th }}$ term of G.P. $=A r^{n-1}$
Now $t_4=a=A r^3, t_7=b=A r^6$ and $t_{10}=c=A r^9$
Relation $b^2=a c$ is true because $b^2=\left(A r^6\right)^2=A^2 r^{12}$ $a c=\left(A r^3\right)\left(A r^9\right)=A^2 r^{12}$
Aliter : As we know, if $p, q, r$ in A.P., then $p^{\text {th }}, q^{\text {th }}, r^{\text {th }}$ terms of a G.P. are always in G.P., therefore, $a, b, c$ will be in G.P. i.e. $b^2=a c$.

Topic: Mathematics, Sequences and Series, JEE Main
If the $5^{\text {th }}$ term of a G.P. is $\frac{1}{3}$ and $9^{\text {th }}$ term is $\frac{16}{243}$, then the $4^{\text {th }}$ term will be

Options:
A $\frac{3}{4}$
B $\frac{1}{2}$
C $\frac{1}{3}$
D $\frac{2}{5}$

The correct answer is:
$\frac{1}{2}$

Solution:
$T_5=a r^4=\frac{1}{3} \ldots(i)$
and $T_9=a r^8=\frac{16}{243} \ldots(ii)$
Solving (i) and (ii), we get$r=\frac{2}{3}$ and $a=\frac{27}{16}$
Now $4^{\text {th }}$ term $=a r^3=\frac{3^3}{2^4} \cdot \frac{2^3}{3^3}=\frac{1}{2}$

Topic: Mathematics, Sequences and Series, JEE Main
If the 6th term in the expansion of 1x83+x2log10x8 is 5600, then the value of x is

Options:

The correct answer is:
10

Solution:
T6= 8C51x838-5x2log10â¡x5
=561x8x2log10â¡x5
=56x2log10â¡x5
T6=5600
â 56x2log10â¡x5=5600
âx2log10â¡x5=100
âx2log10â¡x5=102
âx=10

Topic: Mathematics, Binomial Theorem, JEE Main
If the 7th term in the binomial expansion of $\left(\frac{3}{\sqrt[3]{84}}+\sqrt{3} \ln x\right)^9, x>0$, is equal to 729 , then $x$ can be:

Options:
A
$e^2$

B
$e$

C
$\frac{e}{2}$

D
$2 e$

The correct answer is:
$e$

Solution:
Let $r+1=7 \Rightarrow r=6$
Given expansion is
$$
\left(\frac{3}{\sqrt[3]{84}}+\sqrt{3} \ln x\right)^9, x>0
$$
We have
$$
\mathrm{T}_{r+1}={ }^n \mathrm{C}_r(x)^{n-r} a^r \text { for }(x+a)^n \text {. }
$$
$\therefore$ According to the question
$$
\begin{aligned}
& 729={ }^9 \mathrm{C}_6\left(\frac{3}{\sqrt[3]{84}}\right)^3 \cdot(\sqrt{3} \ln x)^6 \\
& \Rightarrow 3^6=84 \times \frac{3^3}{84} \times 3^3 \times(6 \ln x) \\
& \Rightarrow(\ln x)^6=1 \Rightarrow(\ln x)^6=(\ln e)^6 \\
& \Rightarrow x=\mathrm{e}
\end{aligned}
$$

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2013 (22 Apr Online)
If the $9^{\text {th }}$ term of an A.P. is 35 and $19^{\text {th }}$ is 75 , then its $20^{\text {th }}$ terms will be

Options:
A $78$
B $79$
C $80$
D $81$

The correct answer is:
$79$

Solution:
$T_9=a+8 d=35$ and $T_{19}=a+18 d=75$
Solving the equations, we get $d=4$ and $a=3$
Hence $20^{\text {th }}$ term of A.P. $a+19 d=3+19 \times 4=79$.

Topic: Mathematics, Sequences and Series, JEE Main
If the 10th, term of an A.P. is 120, and its 20th, term is 110, then the sum of its first 200, terms is.

Options:
A 50
B 5014
C 100
D 10012

The correct answer is:
10012

Solution:
Let, a & d is the first term and common difference of arithmetic progression respectively, then

T10=120=a+9d ...i
T20=110=a+19d ...ii

After solving i and ii, we get
a=1200, d=1200

Now

S200=20022200+199200=2012=10012.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2020 (08 Jan Shift 2)
If the 1011th term from the end in the binomial expansion of 4x5-52x2022 is 1024 times 1011th term from the beginning, then 32x is equal to

Options:
A 15

B 10
C 12
D 8

The correct answer is:
10

Solution:
Given,

The 1011th term from the end in the binomial expansion of 4x5-52x2022 is 1024 times 1011th term from the beginning,

So, according to question we have, T1013T1011=1024
⇒C101220224x51010·-52x1012C101020224x51012-52x1010=1024

⇒-52x24x52=1024

⇒5426x4=210

⇒5x4=216

⇒x=516

⇒32x=10

Note this question was bonus in Jee Main 2023 April session as answer key was not matching, so we have modified the question.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (11 Apr Shift 2)
If the absorbate is held on a surface by weak Vander Waal's forces, the absorption process is called

Options:
A Physical adsorption
B Chemical adsorption
C Heat of adsorption
D Enthalpy of adsorption

The correct answer is:
Physical adsorption

Solution:
The adsorption in which molecules of adsorbate are held on the surface of adsorbent by Van der Waals forces or weak physical forces is called as physical adsorption or Van der Waals adsorption or physisorption.

Topic: Chemistry, Surface Chemistry, JEE Main
If the acceleration due to gravity experienced by a point mass at a height h above the surface of earth is same as that of the acceleration due to gravity at a depth αhh≪Re ) from the earth surface. The value of α will be _____ .

(use Re=6400 km)

Options:

The correct answer is:
2

Solution:
Acceleration due to gravity at a height h above the earth's surface is given by g1-2hRe

And acceleration due to gravity at depth d below the earth surface is given by g1-dRe.

As gabove=gbelow

So, g1-2hRe=g1-dRe

⇒2hRe=dRe

⇒αh=d

Thus, the value of α=2.

Topic: Physics, Gravitation, JEE Main, JEE Main 2022 (29 Jul Shift 1)
If the activation energy of a reaction is 80.9 kJ mol-1, the fraction of molecules at 700 K, having enough energy to react to form products is e-x. The value of x is (Rounded off to the nearest integer) [Use R=8.31 J K-1 mol-1]

Options:

The correct answer is:
14

Solution:
Fraction of molecules to have enough energy to react =e-Ea/RT

So, x=EaRT

=80.9×1038.31×700

=13.9≈14

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2021 (26 Feb Shift 2)
If the acute angle formed between y-axis and the tangent drawn to the curve y=x2+4x-17 at the point P52,-34 is θ, then the value of cot⁡θ is equal to

Options:

The correct answer is:
9

Solution:
y ′ =2x+4
At P,dydx=5+4=9=tanâ¡Î±
As Î¸=Ï2-Î±
âcotâ¡Î¸=tanâ¡Î±=9

Topic: Mathematics, Application of Derivatives, JEE Main
If the algebraic sum of deviations of $20$ observations from $30$ is $20$ , then the mean of observations is

Options:
A $30$
B $30.1$
C $29$
D $31$

The correct answer is:
$31$

Solution:
$\sum_{i=1}^{20}\left(x_i-30\right)=20 \Rightarrow \sum_{i=1}^{20} x_i-20 \times 30=20$
$\Rightarrow \sum_{i=1}^{20} x_i=620$.
Mean $=\frac{\sum_{i=1}^{20} x_i}{20}=\frac{620}{20}=31$.

Topic: Mathematics, Statistics, JEE Main
If the altitude of a triangle are in arithmetic progression, then the sides of the triangles are in

Options:
A $\mathrm{AP}$
B $\mathrm{HP}$
C $\mathrm{GP}$
D $\mathrm{AGP}$

The correct answer is:
$\mathrm{HP}$

Solution:
In $\triangle A B C$
$$
\Delta=\frac{1}{2} a p_1=\frac{1}{2} b p_2=\frac{1}{2} c p_3
$$
where $a, b, c$ length of the sides of the triangle and $p_1, p_2, p_3$ are respectively altitudes
Give that $p_1, p_2, p_3$ are in AP
$\Rightarrow \quad \frac{2 \Delta}{a}, \frac{2 \Delta}{b}, \frac{2 \Delta}{c}$ are in $\mathrm{AP}$
$\Rightarrow \quad \frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $\mathrm{AP}$
$\Rightarrow \quad a, b, c$ are in HP

Topic: Mathematics, Sequences and Series, JEE Main
If the A.M. between $p^{\text {th }}$ and $q^{\text {th }}$ terms of an A.P. is equal to the A.M. between $r^{\text {th }}$ and $s^{\text {th }}$ terms of the same A.P., then $p+q$ is equal to

Options:
A
$r+s-1$

B
$r+s-2$

C
$r+s+1$

D
$r+s$

The correct answer is:
$r+s$

Solution:
Given: $\frac{a_p+a_q}{2}=\frac{a_r+a_s}{2}$
$$
\begin{aligned}
& \Rightarrow a+(p-1) d+a+(q-1) d \\
& \quad=a+(r-1) d+a+(s-1) d \\
& \Rightarrow 2 a+(p+q) d-2 d=2 a+(r+s) d-2 d \\
& \Rightarrow(p+q) d=(r+s) d \Rightarrow p+q=r+s .
\end{aligned}
$$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2012 (26 May Online)
If the A.M. is twice the G.M. of the numbers $a$ and $b$, then $a: b$ will be

Options:
A $\frac{2-\sqrt{3}}{2+\sqrt{3}}$
B $\frac{2+\sqrt{3}}{2-\sqrt{3}}$
C $\frac{\sqrt{3}-2}{\sqrt{3}+2}$
D $\frac{\sqrt{3}+2}{\sqrt{3}-2}$

The correct answer is:
$\frac{2+\sqrt{3}}{2-\sqrt{3}}$

Solution:
Given A.M. $=2$ (G.M.) or $\frac{1}{2}(a+b)=2 \sqrt{a b}$
$\frac{a+b}{2 \sqrt{a b}}=\frac{2}{1}$ $\Rightarrow \frac{a+b+2 \sqrt{a b}}{a+b-2 \sqrt{a b}}$ $=\frac{2+1}{2-1}=\frac{3}{1}$
$\Rightarrow \frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2}=\frac{3}{1} \Rightarrow$ $\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{3}}{1}$
$\Rightarrow \frac{a}{b}=\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)^2$
$\Rightarrow \quad \frac{a}{b}=\frac{2+\sqrt{3}}{2-\sqrt{3}}$ $a: b=(2+\sqrt{3}):(2-\sqrt{3})$

Topic: Mathematics, Sequences and Series, JEE Main
If the amplitude of sound is doubled and the frequency is reduced to one fourth, the intensity of sound at the same point will be

Options:
A increasing by a factor of 2
B dncreasing by a factor of 2
C decreasing by a factor of 4
D remain unchanged

The correct answer is:
decreasing by a factor of 4

Solution:
Intensity is proportional to (amplitude)2 and also intensity is proportional to (frequency)2. Therefore, intensity becomes 2242=14th

Topic: Physics, Waves and Sound, NEET
If the angle $\theta$ between the line $\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$ and the plane $2 x-y+\sqrt{\lambda} z+4=$ 0 is such that $\sin \theta=\frac{1}{3}$ the value of $\lambda$ is

Options:
A
$\frac{5}{3}$

B
$\frac{-3}{5}$

C
$\frac{3}{4}$

D
$\frac{-4}{3}$

The correct answer is:
$\frac{5}{3}$

Solution:
Angle between line and normal to plane is
$\cos \left(\frac{\pi}{2}-\theta\right)=\frac{2-2+2 \sqrt{\lambda}}{3 \times \sqrt{5+\lambda}}$ where $\theta$ is angle between line \& plane
$$
\Rightarrow \sin \theta=\frac{2 \sqrt{\lambda}}{3 \sqrt{5+\lambda}}=\frac{1}{3}
$$
$\Rightarrow \lambda=\frac{5}{3}$.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2005
If the angle between the line 2x⁡+1=y⁡=z⁡+4 and the plane 2x⁡-y⁡+λz⁡+4=0 is π 6 , then the value of λ is

Options:
A 4 5 7
B 1 3 5 1 1
C 1 3 5 7
D 4 5 1 1

The correct answer is:
4 5 7

Solution:
Given line is L:x+11=y2=z+42 and plane is P: 2x-y+λz+4=0.

The angle between a line x-x1l=y-y1m=z-z1n and a plane ax+by+cz+d=0 is

sinθ=al+bm+cna2+b2+c2l2+m2+n2

sinπ6=2×1+-1×2+2λ12+22+2222+12+λ

⇒12=2-2+2λ1+4+44+1+λ

⇒12=2λ35+λ

⇒4λ=35+λ

Squaring both sides, we get

⇒16λ=95+λ

⇒16λ=45+9λ

⇒7λ=45

⇒λ=457.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2014 (19 Apr Online)
If the angle between the line $x=\frac{y-1}{2}=\frac{z-3}{\lambda}$ and the plane $x+2 y+3 z=4$ is $\cos ^{-1}\left(\sqrt{\frac{5}{14}}\right)$, then $\lambda$ equals

Options:
A
$\frac{3}{2}$

B
$\frac{2}{5}$

C
$\frac{5}{3}$

D
$\frac{2}{3}$

The correct answer is:
$\frac{2}{3}$

Solution:
$\cos \theta=\sqrt{\frac{5}{14}}$
$\sin \theta=\frac{3}{\sqrt{14}}$
$\sin \theta=\frac{1+4+3 \lambda}{\sqrt{1+4+\lambda^2} \sqrt{1+4+9}}$
$\frac{3}{\sqrt{14}}=\frac{5+3 \lambda}{\sqrt{5+\lambda^2} \sqrt{14}} \Rightarrow \lambda=\frac{2}{3}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2011
If the angle between the lines x2=y2=z1 and 5-x-2=7y-14P=z-34 is cos-123, then P is equal to

Options:
A 27
B 72
C -47
D -74

The correct answer is:
72

Solution:
Given lines are x2=y2=z1 and 5-x-2=7y-14p=z-34

⇒x-52=y-2P7=z-34

We know that the angle between the lines x-x1a1=y-y1b1=z-z1c1 and x-x2a2=y-y2b2=z-z2c2 is given by θ=cos-1a1a2+b1b2+c1c2a12+b12+c12×a22+b22+c22

Given, the angle between both lines is cos-123

⇒cos-123=cos-12×2+2×P7+1×422+22+12×22+P72+42

⇒23=4+2P7+43×4+P249+16

⇒23=56+2P3×P2+980

⇒P2+980=P+28

⇒P2+980=P+282

⇒P2+980=P2+56P+784

⇒56P=196

⇒P=72.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2018 (16 Apr Online)
If the angle between the plane x-3y+2z=1 and the line x-12=y-1-1=z-1-3 is θ, then sec⁡2θ is equal to

Options:
A 10711
B 9897
C 1009
D 8779

The correct answer is:
9897

Solution:
A normal vector to the plane n→=<1,-3,2>
A vector parallel to the line b→=<2,-1,-3>
sinθ= b → . n → b → ⋅ n → = 2+3−6 14 14 = −1 14
sec⁡2θ=1cos⁡2θ=11-2sin2θ=11-2196=196194=9897

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the angle between the planes r â .( m i ^ â j ^ +2 k ^ )+3=0ââand r â .( 2 i ^ â m j ^ â i ^ )â5=0 is Ï3 then m=

Options:
A 2
B ± 3
C 3
D -2

The correct answer is:
3

Solution:
Angle between given planes is
Cos Î¸=n-1 â n-2n-1 n-2 â12=2m+m-2m2 + 5 m2 + 5
12=3m-2m2+5âm2+5=Â± 6m-4
â m2+5=6m-4,m2+5=-6m+4
m2-6m+9=0,m2+6m+1=0
m-32=0
m=3

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the angle made by the tangent at the point x0,y0 on the curve x=12t+sintcost,
y=121+sint2,0<t<π2, with the positive x-axis is π3, then y0 is equal to

Options:
A 63+22
B 37+43
C 27
D 48

The correct answer is:
27

Solution:
Given,

Curve x=12t+sintcost

y=121+sint2

Now dydx=dydtdxdt

tanπ3=ddt121+sint2ddt12t+sintcost

3=12×21+sintcost121+cos2t

3=21+sint·cost2cos2t

3=1+sintcost=cost2+sint22cos2t2-sin2t2

3=tant2+π4

π3=t2+π4 ⇒t2=π12 ⇒ t=π6

Now putting t=π6 in y=121+sint2

as x0,y0 lie on curve

So y0=121+sinπ62

=12322=27

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2022 (25 Jun Shift 2)
If the angle of elevation of a cloud from a point P which is 25m above a lake be 30o and the angle of depression of reflection of the could in the lake from P be 60o, then the height of the cloud (in meters) from the surface of the lake is :

Options:
A 50
B 60
C 45
D 42

The correct answer is:
50

Solution:
tan⁡30o=x-25d

and tan⁡60o=x+25d

⇒tan⁡30otan⁡60o=x-25x+25⇒13=x-25x+25

⇒x+25=3x-75

⇒2x=100⇒x=50m

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2019 (12 Jan Shift 2)
If the angle of intersection at a point where the two circles with radii 5 cm and 12 cm intersect is 90°, then the length (in cm) of their common chord is:

Options:
A 12013
B 6013
C 135
D 132

The correct answer is:
12013

Solution:
∵C1P=5
C2P=12 & ∠C1PC2=90o
∴C1C2=13
∴ ΔCC2P
tan⁡θ=125⇒sin⁡θ=1213
C1C2 biscets the PQ at R & C1C2⊥PQ
in ΔPC1R, sin θ=PR5
⇒1213=PR5⇒PR=6013
& PQ=2PR=12013

Topic: Mathematics, Circle, JEE Main, JEE Main 2019 (12 Apr Shift 1)
If the angle of triangle $A, B$ and $C$ are in $A P$ and $b: a=\sqrt{3}: 1$, then what is the value of $A$ ?

Options:
A $30^{\circ}$
B $45^{\circ}$
C $60^{\circ}$
D $90^{\circ}$

The correct answer is:
$30^{\circ}$

Solution:
Here, $A, B$ and $C$ are in $A P$
So, $2 B=A+C$
And sum of angles of triangle, $A+B+C=180^{\circ}$
$\Rightarrow 2 B+B=180$
$\begin{aligned} & \Rightarrow 3 B=180 \\ & \Rightarrow B=60^{\circ}\end{aligned}$
Now, by sine rule,
$\frac{a}{\sin A}=\frac{b}{\sin B}$
$\Rightarrow \frac{b}{\mathbf{a}}=\frac{\sin B}{\sin A}$
$\Rightarrow \frac{\sqrt{3}}{1}=\frac{\sin 60^{\circ}}{\sin A}$
$\Rightarrow \sin A=(\sqrt{3} / 2) \times(1 / \sqrt{ } 3)$
$\Rightarrow \sin A=1 / 2$
$\Rightarrow A=30^{\circ}$
Hence, option (1) is correct.

Topic: Mathematics, Properties of Triangles, JEE Main
If the angles between the vectors aâ and bâ,bâ and câ,câ and aâ be Ï4,Ï3,Ï3 respectively, then the angle which aâ makes with the plane containing bâ and câ is

Options:
A sin-1⁡23
B sin-1⁡23
C sin-1⁡14
D sin-1⁡23

The correct answer is:
sin-1⁡23

Solution:
A normal vector perpendicular to the plane is b→×c→
sin⁡θ=a→⋅b→×c→a→b→×c→=a→b→c→a→b→×c→…i
b→×c→=b→c→sin⁡π3=32b→c→…ii
a→b→c→2=a→⋅a→a→⋅b→a→⋅c→b→⋅a→b→⋅b→b→⋅c→c→⋅a→c→⋅b→c→⋅c→
=a→2a→b→2a→c→2a→b→2b→2b→c→2a→c→2b→c→2c→2
=a234b→2c→2-a→b→2a→b→c→212-14+a→c→2a→b→2c→122-12
=122a→2b→2c→2
⇒a→b→c→=122a→b→c→…iii
From i,ii and iii
sin⁡θ=12232=23⇒θ=sin-1⁡23

Topic: Mathematics, Vector Algebra, JEE Main
If the angles of elevation of the top of a tower from three collinear points A, B and C on a line leading to the foot of the tower are 30°, 45° and 60° respectively, then the ratio AB:BC, is

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

The correct answer is:
3 : 1

Solution:
Let, the height of the tower=h

Let, AB=x, BC=y, CD=z

In ∆ADE, tan⁡30o= hx+y+z
⇒x+y+z=3h ...i
In ∆BDE, tan45°=hy+z

⇒y+z=h ...ii

In ∆CDE, tan60°=hz
⇒z=h3 ...iii
From ii & iii,
y=h1- 13 ...iv
From i & ii,
x=3-1h ...v

∴ ABBC=xy= 3-1hh3-13= 31

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2015 (04 Apr)
If the angles of elevation of the top of a tower from three collinear points A, B and C on a line leading to the foot of the tower are 30Â°, 45Â° and 60Â°respectively, then the ratio AB : BC, is

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

The correct answer is:
3 : 1

Solution:
tan⁡30o= hx+y+z, tan⁡450=hy+z, tan⁡600=hz
⇒ x+y+z= 3 h ... (i)
y+z=h ... (ii)
z=h3 ... (iii)
from (ii) & (iii)
y=h1- 13 ... (iv)
from (i) & (ii)
x=3-1h ... (v)
xy= 3-1hh3-13
= 31

Topic: Mathematics, Heights and Distances, JEE Main
If the angular momentum of a particle of mass $m$ rotating along a circular path of radius $r$ with uniform speed is $L$, the centripetal force acting on the particle is

Options:
A $\frac{L^2}{m r^2}$
B $\frac{L^2}{m r}$
C $\frac{L}{m r}$
D $\frac{L^2 m}{r}$

The correct answer is:
$\frac{L^2}{m r^2}$

Solution:
From the relation, angular momentum, $L=m v r$
$\mid \left.v=\frac{L}{m r} \right\rvert\,$ Centripetal force acting on the particle $F=\frac{m v^2}{r}=\frac{m\left(\frac{L}{m r}\right)^2}{r}=\frac{L^2}{m r^3}$

Topic: Physics, Rotational Motion, JEE Main
If the angular momentum of a planet of mass m, moving around the Sun in a circular orbit is L, about the center of the Sun, its areal velocity is:

Options:
A Lm
B 4Lm
C L2m
D 2Lm

The correct answer is:
L2m

Solution:
dA→=12 r→×v→dt
dA→dt= areal velocity =12 r→×v→
=12r→×mv→m=12m L
dAdt=L2m

Topic: Physics, Gravitation, JEE Main, JEE Main 2019 (09 Jan Shift 1)
If the angular velocity of earth's spin is increased such that the bodies at the equator start floating, the duration of the day would be approximately :
(Take : g=10 ms-2, the radius of earth, R=6400×103 m, Take π=3.14)

Options:
A 60 minutes
B does not change
C 1200 minutes
D 84 minutes

The correct answer is:
84 minutes

Solution:
For objects to float,

mg=mω2R

ω= angular velocity of the earth.

R= Radius of earth

ω=gR ...1

Duration of day =T

T=2πω ...2

⇒T=2πRg

=2π6400×10310

⇒T60=83.775 minutes

≃84 minutes

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2021 (18 Mar Shift 2)