datavorous/entrance-exam-dataset huggingface dataset:
The shortest wavelength of H atom in the Lyman series is λ1. The longest wavelength in the Balmer series of He+ is :

Options:
A 36λ15
B 5λ19
C 9λ15
D 27λ15

The correct answer is:
9λ15

Solution:
For hydrogen atom :

For Lyman series n1=1 & n2=∞

1λH=Rr11-1∞ So, λ=1RH

For He+ion

Balmer series n1=2 & n2=3

1λHe+=RH×Z214-19

1λHe+=RH×4×536

1λHe+=59RH=591λ

λHe+=95λ

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2020 (04 Sep Shift 2)
The shortest wavelength of hydrogen atom in Lyman series is λ. The longest wavelength in Balmer series of He+ is

Options:
A 59λ
B 9λ5
C 36λ5
D 5λ9

The correct answer is:
9λ5

Solution:
For Hydrogen atom (H) (Z = 1) : 1λH=RH×12112-1∞2 ...(1)

For He+ atom (Z = 2) : 1λHe+=RH×22×122-132 = RH 1 - 49 = RH 59 ..(2)

From (1) & (2)

λHe+λH=95

λHe+=λ×95

λHe+=9λ5.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2023 (29 Jan Shift 1)
The shortest wavelength of Lyman series of hydrogen atom is equal to the shortest wavelength of Balmer series of an hydrogen like atom of atomic number Z. The value of Z is equal to

Options:
A 2
B 3
C 4
D 6

The correct answer is:
2

Solution:
13.6Z2122-1∞=13.6112-1∞2
⇒Z=2

Topic: Physics, Atomic Physics, JEE Main
The shortest wavelength of the Brackett series of a hydrogen like atom (Atomic number =Z ) is the same as the shortest wavelength of the Balmer series of hydrogen atom. The value of Z is

Options:
A 2
B 3
C 4
D 6

The correct answer is:
2

Solution:
Shortest wavelength corresponds to maximum energy therefore transitions are from infinity to n=4 and n=2 respectively.
Z213.616=13.64
Z2=4 or Z=2

Topic: Physics, Atomic Physics, JEE Main
The shortest wavelength of the Brackett series of a hydrogen like atom of atomic number Z is same as the shortest wavelength of the Balmer series of hydrogen atom, then the value of Z is

Options:

The correct answer is:
2

Solution:
Shortest wavelength of Brackett corresponds to n = 4 and n = ∞ and shortest wavelength of Balmer series corresponds to n = 2 and n = ∞
∴Z213.616=13.616⇒Z=2

Topic: Physics, Atomic Physics, JEE Main
The shortest wavelength of the spectral lines in the Lyman series of hydrogen spectrum is $915 Ã…$. The longest wavelength of spectral lines in the Balmer series will be _____ $Ã…$.

Options:

The correct answer is:
6588

Solution:
Shortest, $\frac{\mathrm{hc}}{\lambda}=-13.6\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)$
$\lambda \downarrow \mathrm{E} \uparrow ; \frac{\mathrm{hc}}{\lambda_0}=-13.6(1)$
Balmer Series :
$\begin{aligned}
& \frac{\mathrm{hc}}{\lambda_1}=-13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\
& \frac{\mathrm{hc}}{\lambda_1}=-13.6\left(\frac{1}{4}-\frac{1}{9}\right)
\end{aligned}$
$\begin{aligned} & \frac{\mathrm{hc}}{\lambda_1}=-13.6 \times\left(\frac{5}{36}\right) \\ & \Rightarrow \frac{-13.6 \lambda_0}{\lambda_1}=-13.6 \times \frac{5}{36} \\ & \lambda_1=\frac{\lambda_0 \times 36}{5}=\frac{915 \times 36}{5}=6588\end{aligned}$

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2024 (05 Apr Shift 2)
The shortest wavelengths of Paschen, Balmer and Lyman series are in the ratio

Options:
A $9: 1: 4$
B $1: 4: 9$
C $9: 4: 1$
D $1: 9: 4$

The correct answer is:
$9: 4: 1$

Solution:
Wavelength $\lambda$ of spectral lines in $\mathrm{H}$-atom is given as
$$
\frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \quad \text{...(i)}
$$
For shortest wavelength of Paschen series,
$$
n_{1}=3 \text { and } n_{2}=\infty
$$
$\therefore$ From Eq. (i), we get
$$
\begin{aligned}
& \frac{l}{\lambda_{P S}}=R\left(\frac{1}{3^{2}}-\frac{1}{\infty}\right)=\frac{R}{9} \\
\Rightarrow \quad & \lambda_{P S}=\frac{9}{R} \quad \text{...(ii)}
\end{aligned}
$$
For shortest wavelength in Balmer series,
$$
n_{1}=2, n_{2}=\infty
$$
$\therefore$ From Eq. (i), we get
$\frac{1}{\lambda_{B S}}=R\left(\frac{1}{2^{2}}-\frac{1}{\infty}\right)=\frac{R}{4}$
$\Rightarrow \quad \lambda_{B S}=\frac{4}{R} \quad \text{...(iii)}$
For shortest wavelength of Lyman series,
$$
n_{1}=1 \text { and } n_{2}=\infty
$$
$\therefore$ From Eq. (i), we get
$$
\begin{aligned}
\frac{1}{\lambda_{L S}} &=R\left(\frac{1}{1^{2}}-\frac{1}{\infty}\right)=R \\
\Rightarrow \quad \lambda_{L S} &=\frac{1}{R}
\end{aligned}
$$
Hence, from Eq. (i), (ii) and (iii), we have
$$
\begin{aligned}
\lambda_{P S}: \lambda_{B S}: \lambda_{L S}=& \frac{9}{R}: \frac{4}{R}: \frac{1}{R} \\
&=9: 4: 1
\end{aligned}
$$

Topic: Physics, Atomic Physics, JEE Main
The _______ shows adaptive radiation resulting from competition for food in a population.

Options:
A Sickle cell anaemia
B Peppered moth of England
C DDT resistant mosquitoes
D Darwin finches

The correct answer is:
Darwin finches

Solution:
Darwin finches although nearly identical in all other ways to mainland finches, had different beaks. Their beaks had adapted to the type of food they ate in order to fill different niches on the Galapagos Islands. Their isolation on the islands over long periods of time made them undergo speciation.

Topic: Biology, Evolution, NEET
The SI unit of a physical quantity is Pascal‐sec. The dimensional formula of this quantity will be

Options:
A ML2T-1
B M-1L3T0
C ML-1T-1
D ML-1T-2

The correct answer is:
ML-1T-1

Solution:
Unit of Pascal is N m-2.

So, dimension of Pascal‐sec is

N m-2s=MLT-2L2×T=ML-1T-1

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2022 (27 Jun Shift 2)
The side of a copper cube is $1 \mathrm{~m}$ at $0^{\circ} \mathrm{C}$. What will be the change in its volume, when it is heated to $100^{\circ} \mathrm{C}$ ? $\left[\alpha_{\text {copper }}=18 \times 10^{-6} /{ }^{\circ} \mathrm{C}\right]$

Options:
A $45 \times 10^{-4} \mathrm{~m}^3$
B $54 \times 10^{-4} \mathrm{~m}^3$
C $34 \times 10^{-4} \mathrm{~m}^3$
D $64 \times 10^{-4} \mathrm{~m}^3$

The correct answer is:
$54 \times 10^{-4} \mathrm{~m}^3$

Solution:
In volumetric expansion for a cube, Change in volume $\Delta \mathrm{V}=\mathrm{V} 3 \alpha \Delta \mathrm{T}$
$\begin{aligned}
& \Delta V=3 \times 18 \times 10^{-6} \times 100 \\
& \Delta V=54 \times 10^{-4} \mathrm{~m}^3
\end{aligned}$

Topic: Physics, Thermal Properties of Matter, JEE Main
The sides $a, b, c$ of a triangle are in AP. If $\cos \alpha=\frac{a}{b+c}, \cos \beta=\frac{b}{c+a}, \cos \gamma=\frac{c}{a+b}$ then $\tan ^{2} \frac{\alpha}{2}+\tan ^{2} \frac{\gamma}{2}=$

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

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

Solution:
$\frac{\cos \alpha}{1}=\frac{a}{b+c} \Rightarrow \frac{1}{\cos \alpha}=\frac{b+c}{a}$
By componendo and dividendo,we have
$\frac{1-\cos \alpha}{1+\cos \alpha}=\frac{b+c-a}{b+c+a}$
$\Rightarrow \quad \tan ^{2} \frac{\alpha}{2}=\frac{b+c-a}{3 b}\left\{\begin{array}{l}\because a, b, c \text { are in A.P. } \\ \Rightarrow 2 b=a+c\end{array}\right\}$
$\Rightarrow \quad \frac{1-\cos \gamma}{1+\cos \gamma}=\frac{a+b-c}{a+b-c}$
$\Rightarrow \quad \tan ^{2} \frac{\gamma}{2}=\frac{a+b-c}{3 b}$
$\therefore \tan ^{2} \frac{\alpha}{2}+\tan ^{2} \frac{\gamma}{2}=\frac{2 b}{3 b}=\frac{2}{3}$

Topic: Mathematics, Trigonometric Equations, JEE Main
The sides of a rectangle are 7.01ââm and 1.2Ã101 m. Taking the significant figures into account, the area of the rectangle is

Options:
A 84.1Â m2
B 84.00Â m2
C 84.12Â m2
D 84 m2

The correct answer is:
84 m2

Solution:
Given,
Side of rectangle are 7.01ââm and 12ââm .
Area of rectangle=LengthÃBreadth
=7.01Ã12=84.12 m2
Rounded of two significant figures.
Area of rectangle =84 m2

Topic: Physics, Units and Dimensions, JEE Main
The sides of a rhombus $A B C D$ are parallel to the lines, $x-y+2=0$ and $7 x-y+3=0$. If the diagonals of the rhombus intersect at $P(1,2)$ and the vertex $A$ (different from the origin) is on the $y$ axis, then the ordinate of $A$ is

Options:
A
2

B
$\frac{7}{4}$

C
$\frac{7}{2}$

D
$\frac{5}{2}$

The correct answer is:
$\frac{5}{2}$

Solution:
Let the coordinate $A$ be $(0, c)$
Equations of the given lines are
$x-y+2=0$ and
$7 x-y+3=0$
We know that the diagonals of the rhombus will be parallel to the angle bisectors of the two given lines; $y=x+2$ and $y=7 x+3$
$\therefore$ equation of angle bisectors is given as:
$$
\begin{aligned}
&\frac{x-y+2}{\sqrt{2}}=\pm \frac{7 x-y+3}{5 \sqrt{2}} \\
&5 x-5 y+10=\pm(7 x-y+3)
\end{aligned}
$$
$\therefore$ Parallel equations of the diagonals are $2 x$ $+4 y-7=0$ and $12 x-6 y+13=0$
$\therefore$ slopes of diagonals are $\frac{-1}{2}$ and 2 .
Now, slope of the diagonal from $A(0, \mathrm{c})$ and passing through $P(1,2)$ is $(2-c)$
$$
\therefore 2-c=2 \Rightarrow c=0 \text { (not possible) }
$$
$$
\therefore 2-c=\frac{-1}{2} \Rightarrow c=\frac{5}{2}
$$
$$
\therefore \text { ordinate of A is } \frac{5}{2} \text {. }
$$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
The sides of a right-angled triangle are in arithmetic progression. If the triangle has area \( 24 \) square units, then what is the length of its smallest side?

Options:

The correct answer is:
6

Solution:
Let three sides of triangle be: a-d, a, a+d.

Where, d>0, a>0.

⇒ length of smallest side =a-d units

Now, ( a+d ) 2 = a 2 + ( a−d ) 2

⇒ aa-4d=0

∴ a=4d .....(i)

(As a = 0 is rejected)

Also, 12a.a-d=24

⇒ aa-d=48 .....(ii)

∴ From (i) and (ii), we get a=8, d=2

Hence, length of smallest side is

a-d=8-2=6units

Topic: Mathematics, Sequences and Series, JEE Main
The sides of a triangle are 3x + 4y, 4x+37 and 5x + 57 where x, y > 0 then the triangle is

Options:
A
right angled

B
obtuse angled

C
equilateral

D
none of these

The correct answer is:
right angled

Solution:
No Solution Available

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2002
The sides of a triangle are $\sin \alpha, \cos \alpha$ and $\sqrt{1+\sin \alpha \cos \alpha}$ for some $0 < \alpha < \frac{\pi}{2}$. Then the greatest angle of the triangle is

Options:
A
$60^{\circ}$

B
$90^{\circ}$

C
$120^{\circ}$

D
$150^{\circ}$

The correct answer is:
$120^{\circ}$

Solution:
Greatest side is $\sqrt{1+\sin \alpha \cos \alpha}$, by applying cos rule we get greatest angle $=120^{\circ}$.

Topic: Mathematics, Properties of Triangles, JEE Main, JEE Main 2004
The sides of an equilateral triangle, a square and a regular hexagon circumscribed about a circle are in

Options:
A $A.P.$
B $G.P.$
C $H.P.$
D $None$

The correct answer is:
$G.P.$

Solution:
Let $r$ be the radius of the circle. We know that $r=\frac{1}{2}$ a cot $\left(\frac{\pi}{n}\right)$ for a regular polygon of $n$ sides of length a circumscribed about this circle.
Therefore, $a=2 r \tan \left(\frac{\pi}{n}\right)$. Now if $a_1, a_2$ and $a_3$ are the sides of the equilateral triangle, square and regular hexagon respectively, then
$a_1=2 r \tan \frac{\pi}{3}=2 \sqrt{3} r, a_2=2 r \tan \frac{\pi}{4}=2 r$ and $a_3=2 r \tan \frac{\pi}{6}=\frac{2}{\sqrt{3}} r$
we see that $a_1 a_3=(2 r)^2=a_2{ }^2$, from which it follows that $a_1, a_2$ and $a_3$ are in G.P.

Topic: Mathematics, Properties of Triangles, JEE Main
The sides of the rectangle of greatest area that can be inscribed in the ellipse $x^2+4 y^2=64$ are :

Options:
A $(6 \sqrt{2}, 4 \sqrt{2})$
B $(8 \sqrt{2}, 4 \sqrt{2})$
C $(8 \sqrt{2}, 8 \sqrt{2})$
D $(16 \sqrt{2}, 4 \sqrt{2})$

The correct answer is:
$(8 \sqrt{2}, 4 \sqrt{2})$

Solution:
No Solution Available

Topic: Mathematics, Ellipse, JEE Main
The silicate anion in the mineral wollastonite is a chain of three SiO44- tetrahedra, that share corners with adjacent tetrahedra. The charge of the silicate anion is

Options:
A -4
B -8
C -6
D -2

The correct answer is:
-6

Solution:
The silicate anion in the mineral wollastonite is Si3O96- . Hence, charge on it is -6.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
The simple pendulum A of mass mA and length l is suspended from the trolley B of mass mB. If the system is released from rest at Î¸=0, determine the velocity vB of the trolley. Friction is negligible.

Options:
A v B = m A m B 2 g ℓ 1 + m A / m B
B v B = m A m B 4 g ℓ 1 + m A / m B
C v B = m A m B 2 g ℓ 1 - m A / m B
D v B = m A m B 4 g ℓ 1 - m A / m B

The correct answer is:
v B = m A m B 2 g ℓ 1 + m A / m B

Solution:
Linear momentum will be conserved along the horizontal axis.
P→trolly+P→ball=0⇒P→trolly=-P→ball⇒Ptrolly=Pball=P...(1)
so linear momentum of trolly at any instant will be equal to linear momentum of the ball but in opposite direction.
From energy conservation Loss of PE of ball = gain of KE (trolly + ball)
mAgℓ=P22mB+P22mA....(2)
from (1) & (2) P 2 = 2m A g ℓ × m B m A m A + m B ⇒ P = 2m A 2 g ℓ × m B m A + m B
P =mA 2m B g ℓ m A + m B ⇒ m A v A = m A 2m B g ℓ m A + m B ⇒ v A = 2m B g ℓ m A + m B
&mBvB=mA2gℓ1+mA/mB⇒vB=mAmB2gℓ1+mA/mB

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
The simplest formula of a compound containing $50 \%$ of element $X$ (atomic mass 10 ) and $50 \%$ of element $Y$ (atomic mass 20 ) is

Options:
A $X Y$
B $X_2 Y$
C $X Y_3$
D $X_2 Y_3$

The correct answer is:
$X_2 Y$

Solution:
\(\begin{array}{lllll}
\text {Element } & \%(\mathrm{a}) & \text {At.wt.() }& \mathrm{a} / \mathrm{b} & \text {Ratio } \\
X & 50 & 10 & 5 & 2 \\
Y & 50 & 20 & 2.5 & 1
\end{array}\)
Simplest formula \(=X_2 Y\)

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The simultaneous solubility of AgCN Ksp=2.5×10-6 and AgCl Ksp=1.6×10-10 in 1 M NH3 (aq) are respectively
Given Kf[Ag(NH3)2]+=107

Options:
A 1.58×10-3,1.26×10-5
B 0.04,6.25×10-8
C 5.58×10-8,0.037
D 0.037,5.78×10-8

The correct answer is:
0.037,5.78×10-8

Solution:
AgCl(s)+2NH30-2x(aq)⇌Ag(NH3)2+x+Cl-(aq)x
K1=1.6×10-10×107=1.6×10-3
K1=[Ag(NH3)2+[Cl-][NH3]2
[Cl-][CN-]=1.6×10-32.5×10-9=6.4×105
K1=x2(1-2x)2=1.6×10-3
x(1-2x)=0.04
x=0.037
[CN-]=0.0376.4×105=5.78×10-8

Topic: Chemistry, Ionic Equilibrium, JEE Main
The single largest industrial application of dihydrogen is:

Options:
A Manufacture of metal hydrides
B Rocket fuel in space research
C In the synthesis of ammonia
D In the synthesis of nitric acid

The correct answer is:
In the synthesis of ammonia

Solution:
Informative, according to ncert uses of di hydrogen.

In fact NH3 largest production in used to manufacture nitrogenous fertilisers.

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2021 (20 Jul Shift 2)
The site of action of insulin is

Options:
A mitochondria
B nucleus
C plasma membrane
D DNA

The correct answer is:
plasma membrane

Solution:
Insulin is an hormone, secreted by pancreas. It is transported to different body parts by the blood stream. It maintains glucose level in blood and glucose metabolism. Thus, its site of action is plasma membrane.

Topic: Chemistry, Surface Chemistry, AP EAMCET, AP EAMCET 2012
The site of action of insulin is

Options:
A mitochondria
B nucleus
C plasma membrane
D DNA

The correct answer is:
plasma membrane

Solution:
Insulin is an hormone, secreted by pancreas. It is transported to different body parts by the blood stream. It maintains glucose level in blood and glucose metabolism. Thus, its site of action is plasma membrane.

Topic: Chemistry, Surface Chemistry, TS EAMCET, TS EAMCET 2012
The site of production of $A D A$ in the body is

Options:
A
Bone marrow

B
Iymphocytes

C
blood plasma

D
Monocytes

The correct answer is:
Iymphocytes

Solution:
ADA gene is responsible for producing the enzyme adenosine deaminase, that is primarily involved in the development and maintenance of immune system.
It is produced in all cells, but the highest level of adenosine deaminase occurs in the cells of immune system called lymphocytes. These develop in lymphoid tissues. ADA converts deoxyadenosine (toxic to lymphocytes) to deoxyinosine (non-toxic form).

Topic: Biology, Biotechnology and Its Applications
The sixth H.M. between 3 and $\frac{6}{13}$ is

Options:
A $\frac{63}{120}$
B $\frac{63}{12}$
C $\frac{126}{105}$
D $\frac{120}{63}$

The correct answer is:
$\frac{63}{120}$

Solution:
$x_n=\frac{(n+1) a b}{n a+b}$
Sixth H.M.
$x_6=\frac{7.3 .6 / 13}{\left(6.3+\frac{6}{13}\right)}$ $=\frac{126}{240}=\frac{63}{120}$

Topic: Mathematics, Sequences and Series, JEE Main
The size of a raw mango shrinks to a much smaller size when kept in a concentrated salt solution. Which one of the following process can explain this?

Options:
A Osmosis
B Dialysis
C Diffusion
D Reverse osmosis

The correct answer is:
Osmosis

Solution:
When mango kept in concentrate salt solution then solvent (water) flow from mango to concentrate solution that's why mango shrinks this is called. "Osmosis"

Topic: Chemistry, Solutions, JEE Main, JEE Main 2020 (02 Sep Shift 2)
The size of colloidal particles may range from:

Options:
A \( 1 \) to \( 1000 \mathrm{~nm} \)
B \( 10 \) to \( 100 \mathrm{pm} \)
C \( 1 \) to \( 100 \mu \mathrm{m} \)
D \( 1 \) to \( 10 \mathrm{~mm} \)

The correct answer is:
\( 1 \) to \( 1000 \mathrm{~nm} \)

Solution:
A colloid is a heterogeneous system in which one substance is dispersed (dispersed phase) as very fine particles in another substance called dispersion medium. The colloid and solution must differ from one another by particle size.

In a colloid, the dispersed phase may consist of particles of a single macromolecule or an aggregate of many atoms, ions or molecules. Colloidal particles are larger than simple molecules but small enough to remain suspended. They have a range of diameters between 1 to 1000 nm(10-9 to 10-6m).

Topic: Chemistry, Surface Chemistry, JEE Main
The size of the iso-electronic species Cl- , Ar and Ca2+ is affected by:

Options:
A nuclear charge
B azimuthal quantum number of valence shell
C electron –electron interaction in the outer orbitals
D Principal quantum number of valence shell

The correct answer is:
nuclear charge

Solution:
For isoelectronic species the size is compared by nuclear charge.
Size of isoelectronic species is compared on the basis of their nuclear charge.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2019 (08 Apr Shift 1)
The slag obtained during the extraction of copper from copper pyrites is composed mainly of

Options:
A Cu2S
B FeSiO3
C CuSiO3
D SiO2

The correct answer is:
FeSiO3

Solution:
The reactions are
2Cu2S+3O2→Cu2O+2SO2
Cu2O+FeS→FeO+Cu2S
2FeS+3O2→2FeO+2SO2
FeO+SiO2→FeSiO3
(Impurity) (Flux) (Iron silicate slag)

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
The slope at any point of a curve \( y=f(x) \) is given by \( \frac{d y}{d x}=3 x^{2} \) and it passes through \( (-1,1) \) The equation of the curve is

Options:
A \( y=x^{3}+2 \)
B \( y=-x^{3}-2 \)
C \( y=3 x^{3}+4 \)
D \( y=-x^{3}+2 \)

The correct answer is:
\( y=x^{3}+2 \)

Solution:
We have,

dydx=3x2

⇒dy=3x2dx

Integrating both sides, we have

∫dy=3∫x2dx

⇒y=3x33+c

⇒y=x3+c

It is passing through -1,1. Therefore,

⇒1=-13+c⇒c=2

Hence, the required curve is

y=x3+2

Topic: Mathematics, Differential Equations, JEE Main
The slope of a curve at any point is the reciprocal of twice the ordinate at the point and it passes though the point $(4,3)$. The equation of the curve is

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

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

Solution:
Accordingly $\frac{d y}{d x}=\frac{1}{2 y} \Rightarrow 2 y d y=d x$
Integrating, we get $y^2=x+c$
This passes through $(4,3), \therefore 9=4+c \Rightarrow c=5$
Hence the equation of the curve is $y^2=x+5$

Topic: Mathematics, Differential Equations, JEE Main
The slope of normal at any point P of a curve (lying in the first quadrant) is reciprocal of twice the product of the abscissa and the ordinate of point P. Then, the equation of the curve is (where, c is an arbitrary constant)

Options:
A y2=x+c
B y=c e â x 2
C y=ce-x
D y2=lnâ¡x+c

The correct answer is:
y=c e â x 2

Solution:
Slope of normal at P=-1dydxP
â12xy=-1dydx
or dydx=-2xy
ââ«dyy=â«-2xdx
âlnâ¡y=-x2+k
ây= e â x 2 +k =c e â x 2

Topic: Mathematics, Differential Equations, JEE Main
The slope of tangent at any point x, y on a curve y=yx is x2+y22xy, x>0. If y2=0, then a value of y8 is

Options:
A -42
B 23
C -23
D 43

The correct answer is:
43

Solution:
Given:

dydx=x2+y22xy

Put y=vx⇒dydx=v+xdvdx

v+xdvdx=1+v22v

⇒xdvdx=1-v22v

⇒∫2vv2-1dv=-∫dxx

⇒logev2-1=logeCx

⇒y2-x2x2=Cx

⇒y2-x2=Cx

Put x=2 and y=0 we get,

0-22=2C⇒C=-2

⇒y2=x2-2x

⇒y8=82-16

⇒y8=48=43

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2023 (10 Apr Shift 1)
The slope of the line touching both the parabolas y2=4x and x2=-32y is

Options:
A 1 8
B 2 3
C 1 2
D 3 2

The correct answer is:
1 2

Solution:
y2=41x
x2=4-8y
The equation of the tangent to y2=4x is y= mx+am
The equation of the tangent to x2=4by is y= mx-bm2
Since, they have a common tangent,
am=-bm2
∴ 1m=8m2
∴ m3=18
∴ m=12
∴ The slope of the common tangent is 1 2 .

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2014 (06 Apr)
The slope of the normal at the point $\left(a t^2, 2 a t\right)$ of the parabola $y^2=4 a x$, is

Options:
A $\frac{1}{t}$
B $t$
C $-t$
D $-\frac{1}{t}$

The correct answer is:
$-t$

Solution:
Normal is $y-2 a t_1=\frac{-2 a t}{2 a}\left(x-a t^2\right)$
Therefore, slope $=-t$.

Topic: Mathematics, Parabola, JEE Main
The slope of the tangent at any arbitrary point of a curve is twice the product of the abscissa and square of the ordinate of the point. Then, the equation of the curve is (where c is an arbitrary constant)

Options:
A x2y+y+c=0
B x2y+cy+1=0
C xy+y+c=0
D xy2+cy+y=0

The correct answer is:
x2y+cy+1=0

Solution:
Given, dydx=2y2x
⇒∫dyy2=∫2x dx
or -1y=x2+c
or x2y+cy+1=0

Topic: Mathematics, Differential Equations, JEE Main
The slope of the tangent (other than the x -axis) drawn from the origin to the curve y=x-16 is

Options:
A 6554
B -6555
C 6555
D -6655

The correct answer is:
-6655

Solution:
Slope of OA=α-16-0α-0=dydx
⇒α-16α=6α-15
⇒α-1=0 or α-1=6α
α=1 or α=-15
⇒A-15,6656
∴ Slope of OA=6656-0-15-0
=-6655

Topic: Mathematics, Application of Derivatives, JEE Main
The slope of the tangent to a curve C: y=yx at any point [x,y) on it is 2e2x-6e-x+92+9e-2x. If C passes through the points 0,12+π22 and α,12e2α then eα is equal to

Options:
A 3+23-2
B 323+23-2
C 122+12-1
D 2+12-1

The correct answer is:
323+23-2

Solution:
Given,

dydx=2e2x-6e-x+92+9e-2x

On rearranging we get,

dydx=e2x-6ex2e2x+9

Integrating both side we get,

y=e2x2-2tan-12ex3+c

If the curve passes through the point 0,12+π22

Then c=2π4+tan-123

So, curve will be y=e2x2-2tan-12ex3-π4-tan-123

Again curve passes through the point

α,12e2α

Putting the value in curve equation we get,

e2α2=e2α2-2tan-12eα3-π4-tan-123

⇒tan-12eα3=π4+tan-123

⇒2eα3=tanπ4+tan-123

⇒2eα3=tanπ4+tantan-1231-tanπ4×tantan-123

⇒2eα3=1+231-23

⇒eα=323+23-2

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2022 (25 Jul Shift 1)
The slope of the tangent to the curve $x=3 t^2+1, y=t^3-1$ at $x=1$ is

Options:
A $0$
B $\frac{1}{2}$
C $\infty$
D $-2$

The correct answer is:
$0$

Solution:
$\begin{aligned}& x=3 t^2+1, y=t^3-1 \\
& \frac{d x}{d t}=6 t, \quad \frac{d y}{d t}=3 t^2 \\
& \frac{d y}{d x}=\left(\frac{\frac{d y}{d t}}{\frac{d x}{dt}}\right) \left(\frac{3 t^2}{d t}\right)=\frac{t}{2 t}\end{aligned}$
For $x=1,3 t^2+1=1 \Rightarrow t=0 \Rightarrow$ Slope $=\frac{0}{2}=0$

Topic: Mathematics, Application of Derivatives, JEE Main
The slope of the tangent to the curve \( x=t^{2}+3 t-8, y=2 t^{2}-2 t-5 \) at the point \( (2,-1) \) is

Options:
A \( \frac{22}{7} \)
B \( \frac{6}{7} \)
C \( \frac{7}{6} \)
D \( -\frac{6}{7} \)

The correct answer is:
\( \frac{6}{7} \)

Solution:
Given that, $x=t^{2}+3 t-8 \rightarrow(1)$
$y=2 t^{2}-2 t-5 \rightarrow(2)$
At point $(2,-1)$, Eq. (2) becomes
$-1=2 t^{2}-2 t-5$
$\Rightarrow 2 t^{2}-2 t-4=0$
$\Rightarrow t^{2}-t-2=0$
$\Rightarrow(t-2)(t+1)=0$
$\Rightarrow t=2,-1 \rightarrow(3)$
Similarly, at point $(2,-1)$ Eq. (1) becomes
$2=t^{2}+3 t-8$
$\Rightarrow t^{2}+3 t-10=0$
$\Rightarrow(t+5)(t-2)=0$
$\Rightarrow t=2,-5 \rightarrow(4)$
From Eqs. (3) and (4), we have common value of $t=2$
Now, $\frac{d y}{d t}=4 t-2$ and $\frac{d x}{d t}=2 t+3$
So, slope of the tangent to the curve is
$\frac{d y}{d x}=\frac{4 t-2}{2 t+3}$
$\Rightarrow\left(\frac{d y}{d x}\right)_{t=2}=\frac{4(2)-2}{(2)+3}=\frac{6}{7}$

Topic: Mathematics, Differentiation, JEE Main