datavorous/entrance-exam-dataset huggingface dataset:
The entropy of any system is given by,
S=α2βlnμkRJβ2+3
where α and β are the constants. μ,J,k and R are number of moles, mechanical equivalent of heat, Boltzmann's constant and gas constant, respectively.
Take S=dQT
Choose the incorrect option from the following:

Options:
A α and J have the same dimensions.
B S,β,k and μR have the same dimensions.
C S and α have different dimensions.
D α and k have the same dimensions.

The correct answer is:
α and k have the same dimensions.

Solution:
S=α2βℓnμkRJβ2+3
S=QT=Joule k-1
α2β=Joule k-1
PV=nRT μkRJβ2=1

R= Joule k
⇒R= Joule k,k= Joule R
⇒β= Joule k
α2β= Joule k
⇒α= dimensionless

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2021 (20 Jul Shift 1)
The entropy values (in $J K^{-1} \mathrm{~mol}^{-1}$ ) of $H_{2(g)}=130.6, C l_{2(g)}=223.0$ and $H C l_{[g)}=186.7$ at $298 \mathrm{~K}$ and 1 gtm pressure. Then entropy change for the reaction
$\mathrm{H}_{2(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{HCl}_{(\mathrm{g})}$ is

Options:
A +540.3
B +727.3
C -166.9
D +19.8

The correct answer is:
+19.8

Solution:
$\Delta S^o=2 S_{H Cl}^o-\left(S_{H_2}^o+S_{\mathrm{Cl}_2}^o\right)$
$=2 \times 186.7-(130.6+223.0)=19.8 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$

Topic: Chemistry, Thermodynamics (C), JEE Main
The enzyme produced by Streptococcus is used to

Options:
A Maintain blood cholesterol level
B Strengthen tissues
C Acts as immunosuppressive agent
D Dissolve blood clots

The correct answer is:
Dissolve blood clots

Solution:
Streptokinase produced by the bacterium Streptococcus and modified by genetic engineering is used as a ‘clot buster’ for removing clots from the blood vessels of patients who have undergone myocardial infarction leading to the heart attack.

Topic: Biology, Microbes in Human Welfare, NEET
The enzyme responsible for atmospheric nitrogen fixation is :

Options:
A nitrogenase
B hydrogenase
C oxygenase
D carboxylase

The correct answer is:
nitrogenase

Solution:
Enzyme nitrogenase plays a key role in $\mathrm{N}_2$ fixations and remains active only during anaerobic condition. It is made up of two protein subunits :
(i) non-heme iron protein or Fe-protein or dinitrogen reductase.
(ii) Iron molybdenum protein or Mo-Fe protein or dinitrogenase.

Topic: Biology, Mineral Nutrition, JIPMER, JIPMER 2006
The epithelial lining in the alveoli of the frogs facing lung cavity is

Options:
A squamous ciliated
B columnar ciliated
C columnar nonciliated
D squamous nonciliated

The correct answer is:
columnar ciliated

Solution:
Alveoli is highly vascularised with an internal lining of very thin squamous epithelium, which always remain moist by mucus. The epithelial cells at the linar most margins of septa facing lung cavity are columnar and ciliated.

Topic: Biology, Structural Organisation in Animals, JIPMER, JIPMER 2004
The epithelial tissue present on the inner surface of bronchioles and fallopian tubes is :

Options:
A Squamous
B Cuboidal
C Glandular
D Ciliated

The correct answer is:
Ciliated

Solution:
No Solution Available

Topic: Biology, Structural Organisation in Animals, NEET, NEET 2009 (Mains)
The epithelium found in hollow organs (like bronchioles) are originally ______________ epithelium bearing fine brush-like structures on its free surface.

Options:
A columnar
B glandular
C squamous
D compound

The correct answer is:
columnar

Solution:
If the columnar or cuboidal cells bear cilia on their free surface they are called ciliated epithelium. Their function is to move particles or mucus in a specific direction over the epithelium. They are mainly present in the inner surface of hollow organs like bronchioles and fallopian tubes.

Topic: Biology, Structural Organisation in Animals, NEET
The equation λ=1.227x nm can be used to find the de-Broglie wavelength of an electron. In this equation x stands for :
Where,
m= mass of electron
P= momentum of electron
K= Kinetic energy of electron
V= Accelerating potential in volts for electron

Options:
A mK
B P
C K
D V

The correct answer is:
V

Solution:
The de-Broglie's wavelength is given by λ=hmv, where, h is Planck's constant and v is velocity.

Using relation between momentum and kinetic energy, we have λ=h2 mKE

Here, kinetic energy of electron is stated as KE=qV.

Then, we have λ=h2mqV

Putting the values of m=9.1×10-31 kg and q=1.62×10-19 C of electron

We get, λ=1.22V nm.

So, x stand for V.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2022 (28 Jul Shift 1)
The equation $\frac{x^{2}}{2-\lambda}-\frac{y^{2}}{\lambda-5}-1=0$ represents an ellipse if

Options:
A $\lambda>5$
B $\lambda < 2$
C $2 < \lambda < 5$
D $2>\lambda>5$

The correct answer is:
$\lambda < 2$

Solution:
We have,
$\begin{aligned} \frac{x^{2}}{2-\lambda}-\frac{y^{2}}{\lambda-5}-1 &=0 \\ \Rightarrow \quad \frac{x^{2}}{2-\lambda}+\frac{y^{2}}{5-\lambda} &=1 \end{aligned}$
It will represent an ellipse, when
$$
\begin{array}{ll}
& 2-\lambda>0 \text { and } 5-\lambda>0 \\
\Rightarrow \quad & \lambda < 2 \text { and } \lambda < 5 \\
\Rightarrow \quad & \lambda < 2
\end{array}
$$

Topic: Mathematics, Ellipse, JEE Main
The equation $\sqrt{3} \sin x+\cos x=4$, has

Options:
A only one solution
B two solution
C infinitely many solution
D no solution

The correct answer is:
no solution

Solution:
We have,
$\begin{array}{cc} & \sqrt{3} \sin x+\cos x=4 \\ \Rightarrow & \frac{\sqrt{3}}{2} \sin x+\frac{1}{2} \cos x=2 \\ \Rightarrow & \sin x \cos \frac{\pi}{6}+\cos x \sin \frac{\pi}{6}=2 \\ \Rightarrow & \sin \left(x+\frac{\pi}{6}\right)=2, \text { which is not possible. }\end{array}$
Therefore, this equation has no solution.

Topic: Mathematics, Trigonometric Equations, JEE Main
The equation 3x2+x+5=x-3, where x⁡ is real, has

Options:
A no solution
B exactly four solutions
C exactly one solution
D exactly two solutions

The correct answer is:
no solution

Solution:
3x2+x+5=x-3 ...1

We know that for a quadratic expression ax2+bx+c, a>0, if discriminant i.e. D=b2-4ac<0, then the quadratic is always positive for all real values of the variable.

Now, consider 3x2+x+5

We have, discriminant =b2-4ac=1-4×3×5<0

∴ 3x2+x+5>0, ∀ x∈R

Squaring 1, we get,

3x2+x+5=x-32

⇒3x2+x+5=x2-6x+9

⇒2x2+7x-4=0

⇒x+42x-1=0

⇒x=-4, x=12

But x-3<0 for x=-4, 12

Hence, the equation has no solution.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2014 (19 Apr Online)
The equation \( 4{ }_{1}^{1} \mathrm{H}^{+} \rightarrow{ }_{2}^{4} \mathrm{He}^{2+}+2 \mathrm{e}^{-1}+26 \mathrm{MeV} \) represents.

Options:
A \( \beta \)-decay
B \( \gamma \)-decay
C fusion
D fission

The correct answer is:
fusion

Solution:
the equation

411H+→He2+24+2e-1+26 MeV

Represents the fusion of hydrogen to helium.

Topic: Physics, Nuclear Physics, JEE Main
The equation $13\left[(x-1)^2+(y-2)^2\right]=3(2 x+3 y-2)^2$ represents

Options:
A Parabola
B Ellipse
C Hyperbola
D None of these

The correct answer is:
Hyperbola

Solution:
Here coefficient of $x^2$ is $+v e$ and that of $y^2$ is -ve i.e., a hyperbola.

Topic: Mathematics, Hyperbola, JEE Main
The equation argz-1z+1=π4 represents a circle with:

Options:
A centre at (0,0) and radius 2
B centre at (0,1) and radius 2
C centre at (0,-1) and radius 2
D centre at (0,1) and radius 2

The correct answer is:
centre at (0,1) and radius 2

Solution:
We have,

argz-1z+1=π4

argz0-1z0+1=π2

⇒z0-1z0+1=eiπ2

⇒z0-1z0+1=i

⇒z0=1+i1-i

⇒z0=(1+i)22

⇒z0=i

⇒ Centre 0,1

Radius is the distance between z0 and -1,0

=-1-02+0-12

=2 units

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2021 (26 Aug Shift 1)
The equation $a x^2+b y^2+2 h x y+2 g x+2 f y+c=0$ will represent a circle, if

Options:
A $a=b=0$ and $c=0$
B $f=g$ and $h=0$
C $a=b \neq 0$ and $h=0$
D $f=g$ and $c=0$

The correct answer is:
$a=b \neq 0$ and $h=0$

Solution:
The given equation is $a x^2+b y^2+2 h x y+2 g x+2 f y+c=0$.
For this equation to represent a circle, the following conditions must be satisfied:
1. The coefficients of $x^2$ and $y^2$ must be equal (i.e., $a=b$ ).
2. The coefficient of the $x y$ term must be zero (i.e., $h=0$ ).
These conditions ensure that the conic section represented by the equation is a circle rather than an ellipse or any other conic section.
Let's check the options against these conditions:
- (a) $a=b=0$ and $c=0$ - This would not define a circle; it actually represents the degenerate case of a point at the origin if $a$ and $b$ are zero.
- (b) $f=g$ and $h=0$ - This is not correct because it addresses the coefficients of linear terms $x$ and $y$ and does not address the quadratic terms which are crucial for identifying the circle.
- (c) $a=b \neq 0$ and $h=0$ - This matches our required conditions for a circle.
- (d) $f=g$ and $c=0$-Again, this is incorrect as it deals with the constant term and the linear terms but ignores the quadratic terms.
Therefore, the correct answer is (c) $a=b \neq 0$ and $h=0$.

Topic: Mathematics, Circle, JEE Main
The equation \( (a+2) x^{2}+(a-3) x=2 a-1, a \neq-2 \) has roots rational for

Options:
A all rational values of \( a \) except \( a=-2 \)
B all real values of \( a \) except \( a=-2 \)
C rational values of \( a>\frac{1}{2} \)
D irrational values

The correct answer is:
all rational values of \( a \) except \( a=-2 \)

Solution:
Given:

a+2x2+a-3x=2a-1,a≠-2

The sum of the coefficients

=a+2+a-3-2a-1=0

So, x=1 is a root.

Product of roots =-2a-1a+2

Therefore, other root =-2a-1a+2 Rational, ∀a,a≠-2

Also, discriminant of the given equation is

D=a-32-4a+21-2a

⇒D=a2+9-6a-4-3a-2a2+2

⇒D=9a2+6a+1

⇒D=3a+12

Clearly, discriminant is a perfect square. Hence, given equation has rational roots for all values of a except a=-2

Topic: Mathematics, Quadratic Equation, JEE Main
The equation cos4⁡x-sin4⁡x+cos⁡2x+α2+α=0 will have at least one solution, if

Options:
A -2≤α≤2
B -3≤α≤1
C -2≤α≤1
D -1≤α≤2

The correct answer is:
-2≤α≤1

Solution:
cos2⁡x-sin2⁡xcos2⁡x+sin2⁡x+cos⁡2x+α2+α=0
⇒cos⁡2x×1+cos⁡2x+α2+α=0
⇒α2+α=-2cos⁡2x
RHS ∈-2,2
for a solution, -2≤α2+α≤2
α2+α+2≥0 and α2+α-2≤0
Now, α2+α+2≥0 is always true

and α+2α-1≤0
⇒α∈-2,1

Topic: Mathematics, Trigonometric Equations, JEE Main
The equation $e^{\sin x}-e^{-\sin x}-4=0$ has

Options:
A
infinite number of real roots

B
no real roots

C
exactly one real root

D
exactly four real roots

The correct answer is:
no real roots

Solution:
$e^{\sin x}-e^{-\sin x}=4 \quad \Rightarrow e^{\sin x}=t$
$t-\frac{1}{t}=4$
$\begin{array}{ll}t^2-4 t-1=0 & \Rightarrow t=\frac{4 \pm \sqrt{16+4}}{2} \\ \Rightarrow t=\frac{4 \pm 2 \sqrt{5}}{2} & \Rightarrow t=2 \pm \sqrt{5}\end{array}$
$e^{\sin x}=2 \pm \sqrt{5} \quad \quad-1 \leq \sin x \leq 1 \quad \quad \frac{1}{e} \leq e^{\sin x} \leq e$
$e^{\sin x}=2+\sqrt{5}$ not possible
$e^{\sin x}=2-\sqrt{5}$ not possible
$\therefore$ hence no solution

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2012 (Offline)
The equation for Freundlich adsorption isotherm is

Options:
A $\frac{x}{m}=k p^{1 / n}$
B $x=m k p^{1 / n}$
C $x / m=k p^{-n}$
D All of these

The correct answer is:
All of these

Solution:
$\frac{x}{m}=k p^{1 / n} \text { or } x=m \cdot k p^{1 / n} \text { or } x / m=k p^{-n}$
All of these equation represent freundlich adsorption isotherm.

Topic: Chemistry, Surface Chemistry, JEE Main
The equation formed by decreasing each root of $a x^2+b x+c=0$ by 1 is $2 x^2+8 x+2=0$, then

Options:
A $a=-b$
B $b=-c$
C $c=-a$
D $b=a+c$

The correct answer is:
$b=-c$

Solution:
$\alpha, \beta$ be the roots of $a x^2+b x+c=0$
$\therefore \alpha+\beta=-b / a, \quad \alpha \beta=c / a$
Now roots are $\alpha-1, \beta-1$
Their sum, $\alpha+\beta-2=(-b / a)-2=-8 / 2=-4$
Their product, $(\alpha-1)(\beta-1)=\alpha \beta-(\alpha+\beta)+1$ $=c / a+b / a+1=1$
New equation is $2 x^2+8 x+2=0$
$\therefore b / a=2$ i.e. $b=2 a$, also $c+b=0 \Rightarrow b=-c$

Topic: Mathematics, Quadratic Equation, JEE Main
The equation Imiz-2z-i+1=0, z∈C, z≠i represents a part of a circle having radius equal to :

Options:
A 1
B 2
C 34
D 12

The correct answer is:
34

Solution:
Let z=x+iy, then
Im ix-y-2x+y-1i x-y-1ix-y-1i+1=y-1 y+2+x2x2+y-12+1=0
⇒2x2+2y2-y-1=0
⇒x2+y2-y2-12=0
Center (0, 14)

Radius =g2+f2-c
Radius= 116+12= 916=34

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2017 (09 Apr Online)
The equation k=6.5×1012 s-1e-26000 K/T is followed for the decomposition of compound A. The activation energy for the reaction is ____ KJ mol-1. [nearest integer]

(Given: R=8.314 J K-1 mol-1)

Options:

The correct answer is:
216

Solution:
K=Ae-Ea/RT=6.5×1012 s-1e-26000 K/T

Ea8.314=26000

Ea=216.164 kJ/mol.

Hence,the activation energy of the given reaction is 216KJ mol-1

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2022 (29 Jun Shift 2)
The equation kcos⁡x-3sin⁡x=k+1 is solvable only if

Options:
A k∈-∞,4
B k∈-∞,4
C k∈4,∞
D k∈4,∞

The correct answer is:
k∈-∞,4

Solution:
kcos⁡x-3sin⁡x=k+1
For a solution k+1≤k2+-32
⇒k+1≤k2+9
On squaring both sides, we get,
k2+1+2k≤k2+9
⇒2k≤8⇒k≤4
⇒k∈-∞,4

Topic: Mathematics, Trigonometric Equations, JEE Main
The equation of a circle is given by x2+y2=a2, where a is the radius. If the equation is modified to change the origin other than (0, 0), then find out the correct dimensions of A and B in a new equation :

x-At2+y-tB2=a2

The dimensions of t is given as T-1

Options:
A A = [L–1T], B = [LT–1]
B A=LT, B=[L–1T –1]
C A=[L–1T –1 ], B = [LT–1]
D A=[L–1T –1], B=[LT]

The correct answer is:
A=LT, B=[L–1T –1]

Solution:
Principle of Homogeneity states that dimensions of each of the terms of a dimensional equation on both sides should be the same. Then,

x-At2+y-tB2=a2

For first term,

At=x⇒A×1T=L

∴A=T1L1

For the second term,

tB=y

⇒1TB=L

⇒B=T-1L-1

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2023 (29 Jan Shift 2)
The equation of a circle passing through origin and co-axial to circles $x^2+y^2=a^2$ and $x^2+y^2+2 a x=2 a^2$, is

Options:
A $x^2+y^2=1$
B $x^2+y^2+2 a x=0$
C $x^2+y^2-2 a x=0$
D $x^2+y^2=2 a^2$

The correct answer is:
$x^2+y^2-2 a x=0$

Solution:
Equation of the circle which passes through origin is $x^2+y^2+2 g x+2 f y=0$.

Topic: Mathematics, Circle, JEE Main
The equation of a circle passing through the point $(4,5)$ and having the centre at $(2,2)$ is

Options:
A $x^2+y^2+4 x+4 y-5=0$
B $x^2+y^2-4 x-4 y-5=0$
C $x^2+y^2-4 x=13$
D $x^2+y^2-4 x-4 y+5=0$

The correct answer is:
$x^2+y^2-4 x-4 y-5=0$

Solution:
Given that centre of circle is $(2,2)$
Since circle is passing from point $(4,5)$
So, radius of circle $=\sqrt{(4-2)^2+(5-2)^2}=\sqrt{(4+9)}=\sqrt{13}$
Therefore, equation of circle is
$(x-2)^2+(y-2)^2=(\sqrt{13})^2$
or $x^2-4 x+4+y^2-4 y+4=13$
or $x^2+y^2-4 x-4 y-5=0$.

Topic: Mathematics, Circle, JEE Main
The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length 3a is

Options:
A
$x^2+y^2=9 a^2$

B
$x^2+y^2=16 a^2$

C
$x^2+y^2=4 a^2$

D
$x^2+y^2=a^2$

The correct answer is:
$x^2+y^2=4 a^2$

Solution:
Let $A B C$ be an equilateral triangle, whose median is $A D$.
Given $A D=3 a$
In $\triangle \mathrm{ABD}, \mathrm{AB}^2=\mathrm{AD}^2+\mathrm{BD}^2$
$\Rightarrow x^2=9 a^2+\left(x^2 / 4\right)$ where $A B=B C=A C=x . \frac{3}{4} x^2=9 a^2 \Rightarrow x^2=12 a^2$
In $\triangle \mathrm{OBD}, \mathrm{OB}^2=\mathrm{OD}^2+\mathrm{BD}^2$
$\Rightarrow r^2=(3 a-r)^2+\frac{x^2}{4} \Rightarrow r^2=9 a^2-6 a r+r^2+3 a^2 ; \Rightarrow 6 a r=12 a^2 \Rightarrow r=2 a$
So equation of circle is $x^2+y^2=4 a^2$

Topic: Mathematics, Properties of Triangles, JEE Main, JEE Main 2002
The equation of a circle with the origin as the centre and passing through the vertices of an equilateral triangle whose altitude is of length 3 units is

Options:
A x2+y2=9
B x2+y2=16
C x2+y2=4
D x2+y2=1

The correct answer is:
x2+y2=4

Solution:
Circle circumscribes the triangle ABC and has center at 0,0.
AD=3
∴OA=23×3=2
∴ radius of circle, OA=2
Thus, equation of circle is x2+y2=4

Topic: Mathematics, Circle, JEE Main
The equation of a curve passing through $\left(2, \frac{7}{2}\right)$ and having gradient $1-\frac{1}{x^2}$ at $(x, y)$ is

Options:
A $y=x^2+x+1$
B $x y=x^2+x+1$
C $x y=x+1$
D None of these

The correct answer is:
$x y=x^2+x+1$

Solution:
We have $\frac{d y}{d x}=1-\frac{1}{x^2} \Rightarrow y=x+\frac{1}{x}+c$
This passes through $\left(2 \frac{7}{2}\right), \therefore \frac{7}{2}=2+\frac{1}{2}+c \Rightarrow c=1$
Thus the equation of the curve is $y=x+\frac{1}{x}+1$ or $x y=x^2+x+1$

Topic: Mathematics, Differential Equations, JEE Main
The equation of a given straight line is $\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}=\gamma$. If the equation of the line perpendicular to the given line and passing through $(\alpha, \beta)$ is $\frac{x}{a}+\frac{y}{b}=1$, then $\frac{b}{a}$ is equal to

Options:
A $\tan \theta$
B $-\tan \theta$
C $\cot \theta$
D $-\cot \theta$

The correct answer is:
$\cot \theta$

Solution:
Given equation of line is
$\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}=\gamma$
Slope of Eq. (i) is
$m=\tan \theta$
So, slope of a line perpendicular to Eq. (i) is $(-\cot \theta)$. According to the question, equation of required line is $\frac{x}{a}+\frac{y}{b}=1$
Slope of line (ii) is $\left(\frac{-b}{a}\right)$
So, $\frac{-b}{a}=-\cot \theta \Rightarrow \frac{b}{a}=\cot \theta$

Topic: Mathematics, Straight Lines, JEE Main
The equation of a hyperbola is x2a2-y2b2=1. If P2,5 is a point from which perpendicular tangents can be drawn to the hyperbola and distance between both the foci of the hyperbola is 10, then its eccentricity is

Options:
A 32
B 54
C 169
D 43

The correct answer is:
54

Solution:
âµP lies on director circle x2+y2=a2-b2
â´ a2-b2=7 â¦(i)
Also, 2ae=10âae=5 â¦(ii)
Now, e2=a2+b2a2âa2+b2=25 â¦(iii)
From (i) & (iii), we get,
a2=16,b2=9
e=2516=54

Topic: Mathematics, Hyperbola, JEE Main
The equation of a line passing through the point $(-3,2,-4)$ and equally inclined to the axes, are

Options:
A $x-3=y+2=z-4$
B $x+3=y-2=z+4$
C $\frac{x+3}{1}=\frac{y-2}{2}=\frac{z+4}{3}$
D None of these

The correct answer is:
$x+3=y-2=z+4$

Solution:
Since, line equally inclined to the axes.
\(\therefore l=m=n\) ...(i)
The required equation of line is
\(\begin{aligned}
& \frac{x+3}{l}=\frac{y-2}{l}=\frac{z+4}{l} \quad \text { [using (1)] } \\
& \Rightarrow \frac{x+3}{1}=\frac{y-2}{1}=\frac{z+4}{1} \\
& \Rightarrow x+3=y-2=z+4
\end{aligned}\)

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of a line passing through the point of intersection of the lines $x+5 y+7=0,3 x+2 y-5=0$, and perpendicular to the line $7 x+2 y-5=0$, is given by

Options:
A $2 x-7 y-20=0$
B $2 x+7 y-20=0$
C $-2 x+7 y-20=0$
D $2 x+7 y+20=0$

The correct answer is:
$2 x-7 y-20=0$

Solution:
Point of intersection of the lines is $(3,-2)$. Hence the equation is $2 x-7 y=2(3)-7(-2)=20$

Topic: Mathematics, Straight Lines, JEE Main
The equation of a line through $(3,-4)$ and perpendicular to the line $3 x+4 y=5$ is

Options:
A $4 x+3 y=24$
B $y-4=(x+3)$
C $3 y-4 x=24$
D $y+4=\frac{4}{3}(x-3)$

The correct answer is:
$y+4=\frac{4}{3}(x-3)$

Solution:
The required equation, which passes through $(3,-4)$ and its gradient is $\frac{4}{3}$, is $(y+4)=\frac{4}{3}(x-3)$

Topic: Mathematics, Straight Lines, JEE Main
The equation of a line through the intersection of lines $x=0$ and $y=0$ and through the point $(2,2)$, is

Options:
A $y=x-1$
B $y=-x$
C $y=x$
D $y=-x+2$

The correct answer is:
$y=x$

Solution:
Obviously the point of intersection is $(0,0)$. Thus the equation of line passing through the points $(0,0)$ and $(2,2)$ is $y=x$

Topic: Mathematics, Straight Lines, JEE Main
The equation of a normal to the curve, sin⁡y=xsin⁡π3+y at x=0, is:

Options:
A 2x-3 y=0
B 2y-3Â x=0
C 2y+3Â x=0
D 2x+3 y=0

The correct answer is:
2x+3 y=0

Solution:
sin⁡y=xsin⁡π3+y
If x=0, then y=0
So point is 0, 0
Slope of tangent =dydx=+sin⁡π3+ycos⁡y-xcos⁡π3+y
at 0, 0 dydx=sin⁡π3=32
Slope of tangent =32
Slope of Normal =-23
∴ equation of normal 2x+3y=0

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2015 (11 Apr Online)
The equation of a normal to the parabola y=x2-6x+6 which is perpendicular to the line joining the origin to the vertex of the parabola is

Options:
A 4x-4y-11=0
B 4x-4y+1=0
C 4x-4y-21=0
D 4x-4y+21=0

The correct answer is:
4x-4y-21=0

Solution:
4x-4y-21=0
Here y=x-32-3
⇒y+3=x-32
Hence, the vertex is 3,-3
Slope of the line joining origin 0,0 and vertex 3,-3 is –1
Hence, the slope of the normal is 1
Equation of the normal is x-3=1y+3-241-1413
⇒4x-4y-21=0

Topic: Mathematics, Parabola, JEE Main
The equation of a parabola is \( y^{2}=4 x . \) If \( P(1,3) \) and \( Q(1,1) \) are two points in the \( X Y \)-plane. Then, for the parabola

Options:
A \( P \) and \( Q \) are exterior points
B \( P \) is an interior point while \( Q \) is an exterior point
C \( P \) and \( Q \) are interior points
D \( P \) is an exterior point while \( Q \) is an interior point

The correct answer is:
\( P \) is an exterior point while \( Q \) is an interior point

Solution:
Here the given parabola is

S≡y2-4x=0.

To check the position of a point with respect to a parabola, we put the point in the equation of the parabola and check for the sign of the expression.

If, the sign of the expression is positive, then the point is outside the parabola.

If, the sign of the expression is zero, then the point is on the parabola.

If, the sign of the expression is negative, then the point is inside the parabola.

S1,3=32-4·1>0,

So, P1,3 is an exterior point.

S1,1=12-4·1<0,

Thus, Q1,1 is an interior point.

Topic: Mathematics, Parabola, JEE Main
The equation of a particle executing simple harmonic motion is given by x=sinπt+13 m. At t=1 s, the speed of particle will be (Given: π=3.14)

Options:
A 157 cm s-1
B 0 cm s-1
C 272 cm s-1
D 314 cm s-1

The correct answer is:
157 cm s-1

Solution:
Given, x=sinπt+π3

Velocity v=dxdt=πcosπt+π3

Now, at t=1 s, v=πcosπ+π3

v=-π2=-3.142=-1.57 m s-1⇒v=157 cm s-1

Topic: Physics, Oscillations, JEE Main, JEE Main 2022 (27 Jun Shift 2)
The equation of a plane containing the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and the point $(0,7,-7)$ is

Options:
A
$x+y+z=0$

B
$x+2 y+z=21$

C
$3 x-2 y+5 z+35=0$

D
$3 x+2 y+5 z+21=0$

The correct answer is:
$x+y+z=0$

Solution:
The equation of the plane containing the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ is a $(x+1)+b$ $(y-3)+c(z+2)=0$ where $-3 a+2 b+c=0$
This passes through $(0,7,-7)$
$$
\begin{aligned}
& \therefore a(0+1)+b(7-3)+c(-7+2)=0 \\
& \Rightarrow a+4 b-5 c=0 \quad \text {...(B) }
\end{aligned}
$$
On solving equation (A) and (B) we get $a=1, b=1, c=1$
$\therefore$ Required plane is
$$
\begin{aligned}
& x+1+y-3+z+2=0 \\
& \Rightarrow x+y+z=0
\end{aligned}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2012 (26 May Online)
The equation of a plane containing the line formed by 2x-y+z-3=0, 3x+y+z=5 and at a distance of 16 units from the point 2,1,-1 is

Options:
A x+y+z-3=0
B 2x-y-z-3=0
C 2x-y+z+3=0
D 62x+29y+19z-105=0

The correct answer is:
62x+29y+19z-105=0

Solution:
Equation of the required plane is
2x-y+z-3+λ3x+y+z-5=0
⇒2+3λx+λ-1y+λ+1z-3-5λ=0
Distance of the plane from 2,1,-1=16
4+6λ+λ-1-λ-1-3-5λ2+3λ2+λ-12+λ+12=16
λ-111λ2+12λ+6=16
6λ-12=11λ2+12λ+6
5λ2+24λ=0⇒λ=0,-245
Hence, the equation of the plane is
62x+29y+19z-105=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of a plane containing the line of intersection of the planes 2x-y-4=0 and y+2z-4=0 and passing through the point 1,1,0 is

Options:
A x-3y-2z=-2
B x+3y+z=4
C x-y-z=0
D 2x-z=2

The correct answer is:
x-y-z=0

Solution:
Equation of plane passing through the intersection of two given Plane 2x-y-4=0 and y+2z-4=0 is,

2x-y-4+λ y+2z-4=0

⇒2x+y λ-1+2λz-4-4λ=0

∵ It passes through (1, 1, 0)

∴ 2×1+λ-1+0-4-4λ=0

⇒-3λ-3=0⇒λ=-1

∴ Equation of plane 2x-2y-2z=0

⇒x-y-z=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (08 Apr Shift 1)