datavorous/entrance-exam-dataset huggingface dataset:
The life cycle of some plants do not involve a free-living sporophyte and the dominant, photosynthetic phase is the free-living gametophyte. This type of life cycle is called

Options:
A the Haplo-diplontic life cycle and it is seen in kelp.
B the Haplontic life cycle which is seen in Volvox and some species of Chlamydomonas.
C the Diplontic life cycle which is seen in seed-bearing plants.
D the Haplo-diplontic life cycle which is seen in bryophytes and pteridophytes.

The correct answer is:
the Haplontic life cycle which is seen in Volvox and some species of Chlamydomonas.

Solution:
In plants mainly three types of life cycle are seen (a) Haplontic (b) Diplontic (c) Haplo-diplontic. Haplontic life cycle is seen in most algae (like Volvox and some species of Chlamydomonas) and fungi. In which dominant phase is haploid gametophyte which is free-living and photosynthetic. Its alternate phase is sporophyte which is short living and is dependent on gametophyte. In Kelp and seed bearing plants diplontic life cycle takes place. In bryophytes and pteridophytes haplo-diplontic life cycle takes place.

Topic: Biology, Plant Kingdom, NEET
The light bulbs A & B in the following circuits are identical. When the switch is closed

Options:
A Intensity of bulb A increase
B Intensity of bulb A decrease
C Intensity of bulb B increase
D Nothing changes

The correct answer is:
Nothing changes

Solution:
After closing of switch voltage across A & B
remain same
∴ Nothing change

Topic: Physics, Current Electricity, NEET
The light of two different frequencies whose photons have energies 3.8eV and 1.4eV respectively, illuminate a metallic surface whose work function is 0.6eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be :

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

The correct answer is:
2:1

Solution:
Expression for maximum kinetic energy can be written as, KEmax=E-ϕ

KEmax, 1=E1-ϕ=3.8-0.6=3.2 eV, KEmax, 2=E2-ϕ=1.4-0.6=0.8 eV

KEmax, 1KEmax, 2=3.20.8=4⇒4=12 mv1212 mv22

⇒v12v22=4 ⇒v1v2=2:1

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2022 (24 Jun Shift 2)
The light ray is incidence at angle of $60^{\circ}$ on a prism of angle $45^{\circ}$. When the light ray falls on the other surface at $90^{\circ}$, the refractive index of the material of prism $\mu$ and the angle of deviation $\delta$ are given by

Options:
A $\mu=\sqrt{2}, \delta=30^{\circ}$
B $\mu=1.5, \delta=15^{\circ}$
C $\mu=\frac{\sqrt{3}}{2}, \delta=30^{\circ}$
D $\mu=\sqrt{\frac{3}{2}}, \delta=15^{\circ}$

The correct answer is:
$\mu=\sqrt{\frac{3}{2}}, \delta=15^{\circ}$

Solution:
From figure it is clear that $\angle e=\angle r_2=0$
From $A=r_1+r_2$
$\Rightarrow r_1=A=45^{\circ}$
$\therefore \mu=\frac{\sin i}{\sin r_1}=\frac{\sin 60}{\sin 45}=\sqrt{\frac{3}{2}}$
Also from $i+e=A+\delta \Rightarrow 60+0=45+\delta \Rightarrow \delta=15^{\circ}$

Topic: Physics, Ray Optics, JEE Main
The light rays from an object have been reflected towards an observer from a standard flat mirror, the image observed by the observer are :-

A. Real

B. Erect

C. Smaller in size than object

D. Laterally inverted

Choose the most appropriate answer from the options given below:

Options:
A B and D only
B B and C only
C A and D only
D A, C and D only

The correct answer is:
B and D only

Solution:
Image formed by a plane mirror is always virtual, laterally inverted, erect image and of the same size as the object.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (25 Jan Shift 2)
The light waves from two coherent sources have same intensity I1=I2=I0. In interference pattern the intensity of light at minima is zero. What will be the intensity of light at maxima?

Options:
A I0
B 2I0
C 4I0
D 5I0

The correct answer is:
4I0

Solution:
Iminimum =0

I1-I22=0

⇒ I1=I2

Imax=I1+I22=4I0

Topic: Physics, Wave Optics, JEE Main, JEE Main 2021 (27 Aug Shift 2)
The $\lim _{x \rightarrow \infty}\left(\frac{3 x-1}{3 x+1}\right)^{4 x}$ equals

Options:
A 1
B 0
C $\mathrm{e}^{-8 / 3}$
D $\mathrm{e}^{-4 / 9}$

The correct answer is:
$\mathrm{e}^{-8 / 3}$

Solution:
$\lim _{x \rightarrow \infty}\left(\frac{3 x-1}{3 x+1}\right)^{4 x}=1^{\infty}=e^{L}=e^{-\frac{8}{3}}$
$L=\lim _{x \rightarrow \infty} 4 x\left(\frac{3 x-1}{3 x+1}-1\right)=\lim _{x \rightarrow \infty} 4 x\left(\frac{-2}{3 x+1}\right)=\frac{-8}{3}$

Topic: Mathematics, Limits, JEE Main
The limit L=limn→∞⁡∑r=1nnn2+r2 satisfies

Options:
A πL>6
B πL<2
C πL>3
D πL<1

The correct answer is:
πL>3

Solution:
Given limit is L= lim n→∞ ∑ r=1 n 1 1+ r n 2 ⋅ 1 n
=â«01dx1+x2=tan-1â¡x01
=Ï4

Topic: Mathematics, Definite Integration, JEE Main
The limit L=limn→∞⁡∑r=4n-4nn2+r2 satisfies the relation

Options:
A eL>e
B eL>3
C etan⁡L<2e
D πL<1

The correct answer is:
etan⁡L<2e

Solution:
Given limits is L=limn→∞⁡∑r=4n-411+rn2⋅1n
=∫01dx1+x2=tan-1⁡x01=π4

Hence, etanπ4=e<2e

Topic: Mathematics, Definite Integration, JEE Main
The limit $\lim _{x \rightarrow 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1}$

Options:
A exists and is equal to $\sqrt{2}$
B exists and is equal to $-\sqrt{2}$
C does not exist
D exists and is equal to $\left(\frac{1}{2}\right)$

The correct answer is:
does not exist

Solution:
$$
\text { Let } \begin{array}{rlr}
L & =\lim _{x \rightarrow 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1} & \\
& =\lim _{x \rightarrow 1} \frac{\sqrt{2 \sin ^2(x-1)}}{x-1} \quad\left[\because 1-\cos 2 \theta=2 \sin ^2 \theta\right] \\
& =\sqrt{2} \lim _{x \rightarrow 1} \frac{|\sin (x-1)|}{x-1} & {\left[\frac{0}{0}\right. \text { form ] }} \\
& =\sqrt{2} \lim _{(x-1) \rightarrow 0} \frac{|\sin (x-1)|}{x-1} & \\
& =\sqrt{2} \lim _{z \rightarrow 0} \frac{|\sin z|}{z} & \text { [Let } x-1=z]
\end{array}
$$
[Let $x-1=z$ ]
$$
\begin{aligned}
& \text { NHL }=\sqrt{2} \lim _{z \rightarrow 0^{+}} \frac{\sin z}{z}=\sqrt{2} \\
& \text { LHL }=\sqrt{2} \lim _{z \rightarrow 0^{-}}-\frac{\sin z}{z}=-\sqrt{2}
\end{aligned}
$$
$\because$ RHL $\neq$ LH
$\therefore$ Limit does not exist.

Topic: Mathematics, Limits, JEE Main
The limit of resolution of a 100 cm telescope for Î»=5000 â« is equal to

Options:
A 61×10-8 rad
B 6.1×10-8 rad
C 12.2×10-8 rad
D 122×10-8 rad

The correct answer is:
61×10-8 rad

Solution:
dθ=1.22λd=1.22×5×10-71=61×10-8 rad

Topic: Physics, Wave Optics, JEE Main
The limit of $\left\{\frac{1}{x} \sqrt{1+x}-\sqrt{1+\frac{1}{x^{2}}}\right\}$ as $x \rightarrow 0$

Options:
A does not exist
B is equal to $\frac{1}{2}$
C is equal to 0
D is equal to 1

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

Solution:
$\lim _{x \rightarrow 0}\left\{\frac{\sqrt{1+x}}{x}-\sqrt{1+\frac{1}{x^{2}}}\right\}$
$=\lim _{x \rightarrow 0}\left\{\frac{\sqrt{1+x}-\sqrt{1+x^{2}}}{x}\right\}\left(\frac{0}{0}\right.$ form $)$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{2 \sqrt{1+x}}-\frac{x}{\sqrt{1+x^{2}}}}{1}$ (Use L-Hospital Rule)
$=\frac{1}{2 \sqrt{1+0}}-0=\frac{1}{2}$

Topic: Mathematics, Limits, JEE Main
The limit of $x \sin \left(e^{\frac{1}{x}}\right)$ as $x \rightarrow 0$

Options:
A is equal to 0
B is equal to 1
C is equal to $\frac{e}{2}$
D does not exist

The correct answer is:
is equal to 0

Solution:
$\lim _{x \rightarrow 0} x \sin e^{(1 / x)}$
$\begin{aligned} L H L &=f(0-0)=\lim _{h \rightarrow 0}(-h) \sin e^{(-1 / h)} \\ &=-0 \times \sin \left(e^{-\infty}\right)=-0 \times \sin (0)=0 \end{aligned}$
$=0 \times(\mathrm{a}$ finite number between $-1 \text{ to}+1)$
$$
=0 \quad(\because-1 \leq \sin x \leq 1)
$$
$\because \mathrm{LHL}=\mathrm{RHL}$
$\therefore \lim _{x \rightarrow 0} x \sin \left(e^{1 / x}\right)$ exist and equal to 0

Topic: Mathematics, Limits, JEE Main
The limiting line in Balmer series will have a frequency of (Rydberg constant, $R_{\infty}=3.29 \times 10^{15}$ cycles $/ \mathrm{s}$ )

Options:
A
$8.22 \times 10^{14} \mathrm{~s}^{-1}$

B
$3.29 \times 10^{15} \mathrm{~s}^{-1}$

C
$3.65 \times 10^{14} \mathrm{~s}^{-1}$

D
$5.26 \times 10^{13} \mathrm{~s}^{-1}$

The correct answer is:
$8.22 \times 10^{14} \mathrm{~s}^{-1}$

Solution:
$\bar{v}=\frac{1}{\lambda}=R_H Z\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
In Balmer series $n_1=2 \& n_2=3,4,5 \ldots$... Last line of the spectrum is called series limit.
Limiting line is the line of shortest wavelength and high energy when $n_2=\infty$
$$
\begin{aligned}
\therefore \quad \bar{v}= & \frac{1}{\lambda}=\frac{R_H}{n_1^2}=\frac{3.29 \times 10^{15}}{2^2}=\frac{3.29 \times 10^{15}}{4} \\
& =8.22 \times 10^{14} \mathrm{~s}^{-1}
\end{aligned}
$$

Topic: Chemistry, Structure of Atom, JEE Main
The limiting line in Balmer series will have a frequency of (Rydberg constant, $R_{\infty}=3.29 \times 10^{15}$ cycles $/ \mathrm{s}$ )

Options:
A
$8.22 \times 10^{14} \mathrm{~s}^{-1}$

B
$3.29 \times 10^{15} \mathrm{~s}^{-1}$

C
$3.65 \times 10^{14} \mathrm{~s}^{-1}$

D
$5.26 \times 10^{13} \mathrm{~s}^{-1}$

The correct answer is:
$8.22 \times 10^{14} \mathrm{~s}^{-1}$

Solution:
$\bar{v}=\frac{1}{\lambda}=R_H Z\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
In Balmer series $n_1=2 \& n_2=3,4,5 \ldots$... Last line of the spectrum is called series limit.
Limiting line is the line of shortest wavelength and high energy when $n_2=\infty$
$$
\begin{aligned}
\therefore \quad \bar{v}= & \frac{1}{\lambda}=\frac{R_H}{n_1^2}=\frac{3.29 \times 10^{15}}{2^2}=\frac{3.29 \times 10^{15}}{4} \\
& =8.22 \times 10^{14} \mathrm{~s}^{-1}
\end{aligned}
$$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2012 (07 May Online)
The limiting molar conductivities $\wedge^0$ for $\mathrm{NaCl}, \mathrm{KBr}$ and $\mathrm{KCl}$ are 126,152 and $150 \mathrm{Sm}^2 \mathrm{~mol}^{-1}$ respectively. The $\wedge^0$ for $N a B r$ is

Options:
A $278 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
B $176 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
C $128 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
D $302 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

The correct answer is:
$128 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

Solution:
$\left(152 \mathrm{scm}^2\right) \wedge_{K B r}^0=\wedge_{X^{+}}^0+\wedge_{\zeta \zeta^{-}}^0$ ...(i)
$\left(126 \mathrm{scm}^2\right) \wedge_{\mathrm{NGCl}}^0=\wedge_{\mathrm{Nb}^{+}}^0+\wedge_{\mathrm{cr}}^0$ ...(ii)
$\left(150 \mathrm{scm}^2\right) \wedge_{\mathrm{KCl}}^0=\wedge_{\mathrm{K}^{+}}^0+\wedge_{\mathrm{Cl}}^0$ ...(iii)
By equation (i)+(ii)-(iii)
$=126+152-150=128 \mathrm{Scm}^2 \mathrm{~mol}^{-1}$
\(\begin{array}{|l|l|l|l|l|l|} \hline \text { Electrolyte: } & \mathrm{KCl} & \mathrm{KNO}_3 & \mathrm{HCl} & \mathrm{NaOAc} & \mathrm{NaCl} \\ \hline \Lambda^*\left(\mathrm{Scm}^2 \mathrm{~mol}^{-1}\right): & 149.9 & 145.0 & 426.2 & 91.0 & 126.5 \\ \hline \end{array}\)

Topic: Chemistry, Electrochemistry, JEE Main
The limiting molar conductivities $\Lambda^{\circ}$ for $\mathrm{NaCl}, \mathrm{KBr}$ and $\mathrm{KCl}$ are 126,152 and $150 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ respectively. The $\Lambda^{\circ}$ for $\mathrm{NaBr}$ is

Options:
A
$128 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

B
$302 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

C
$278 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

D
$176 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

The correct answer is:
$128 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

Solution:
$\Lambda_{\mathrm{NaCl}}^{\circ}=\lambda_{\mathrm{Na}}^{\circ}+\lambda_{\mathrm{Cl}}^{\circ}=126 \ldots(1)$
$\Lambda_{\mathrm{KBr}}^{\circ}=\lambda_{\mathrm{K}^{+}}^{\circ}+\lambda_{\mathrm{Br}^{-}}^{\circ}=152$
$\Lambda_{\mathrm{KCl}}^{\circ}=\lambda_{\mathrm{K}^{+}}^{\circ}+\lambda_{\mathrm{Cl}^{-}}^{\circ}=150$
$\Lambda_{\mathrm{NaBr}}^{\circ}=\lambda_{\mathrm{Na}}^{\circ}+\lambda_{\mathrm{Br}^{-}}^{\circ}$
$\Lambda_{\mathrm{NaBr}}^{\circ}=126+152-150=128$

Topic: Chemistry, Electrochemistry, JEE Main
The limiting molar conductivities $\Lambda^{\circ}$ for $\mathrm{NaCl}, \mathrm{KBr}$ and $\mathrm{KCl}$ are 126,152 and $150 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ respectively. The $\Lambda^{\circ}$ for $\mathrm{NaBr}$ is

Options:
A
$128 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

B
$302 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

C
$278 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

D
$176 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

The correct answer is:
$128 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

Solution:
$\Lambda_{\mathrm{NaCl}}^{\circ}=\lambda_{\mathrm{Na}}^{\circ}+\lambda_{\mathrm{Cl}}^{\circ}=126 \ldots(1)$
$\Lambda_{\mathrm{KBr}}^{\circ}=\lambda_{\mathrm{K}^{+}}^{\circ}+\lambda_{\mathrm{Br}^{-}}^{\circ}=152$
$\Lambda_{\mathrm{KCl}}^{\circ}=\lambda_{\mathrm{K}^{+}}^{\circ}+\lambda_{\mathrm{Cl}^{-}}^{\circ}=150$
$\Lambda_{\mathrm{NaBr}}^{\circ}=\lambda_{\mathrm{Na}}^{\circ}+\lambda_{\mathrm{Br}^{-}}^{\circ}$
$\Lambda_{\mathrm{NaBr}}^{\circ}=126+152-150=128$

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2004
The limiting molar conductivities of the given electrolytes at $298 \mathrm{~K}$ follow the order
$$
\begin{array}{l}
{\left[\lambda^{0}\left(\mathrm{~K}^{+}\right)=73.5, \lambda^{0}\left(\mathrm{Cl}^{-}\right)=76.3\right.} \\
\left.\lambda^{0}\left(\mathrm{Ca}^{2+}\right)=119.0, \lambda^{0}\left(\mathrm{SO}_{4}^{2-}\right)=160.0 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\right]
\end{array}
$$

Options:
A $\mathrm{KCl} < \mathrm{CaCl}_{2} < \mathrm{K}_{2} \mathrm{SO}_{4}$
B $\mathrm{KCl} < \mathrm{K}_{2} \mathrm{SO}_{4} < \mathrm{CaCl}_{2}$
C $\mathrm{K}_{2} \mathrm{SO}_{4} < \mathrm{CaCl}_{2} < \mathrm{KCl}$
D $\mathrm{CaCl}_{2} < \mathrm{K}_{2} \mathrm{SO}_{4} < \mathrm{KCl}$

The correct answer is:
$\mathrm{KCl} < \mathrm{CaCl}_{2} < \mathrm{K}_{2} \mathrm{SO}_{4}$

Solution:
$\lambda^{\circ}{ }_{\mathrm{KCl}}=73.5+76.3=149.8$
$\lambda_{\mathrm{CaCl}_{2}}^{\circ}=119+2 \times 76.3=271.6$
$\lambda^{\circ}{ }_{\mathrm{K}_{2} \mathrm{SO}_{4}}=2 \times 73.5+160=307$

Topic: Chemistry, Electrochemistry, JEE Main
The limiting points of the co-axial system containing the two circles
$\begin{aligned} & x^2+y^2+2 x-2 y+2=0 \text { and } \\ & 25\left(x^2+y^2\right)-10 x-80 y+65=0 \text { are }\end{aligned}$

Options:
A $(1,-1),(-3,-40)$
B $(1,-1),\left(-\frac{1}{5}, \frac{8}{5}\right)$
C $(-1,1),\left(\frac{1}{5}, \frac{8}{5}\right)$
D $\left(-\frac{1}{5},-\frac{8}{5}\right)$

The correct answer is:
$(-1,1),\left(\frac{1}{5}, \frac{8}{5}\right)$

Solution:
Equation of circles are
$x^2+y^2+2 x-2 y+2=0$
and $x^2+y^2-\frac{2}{5} x-\frac{16}{5} y+\frac{13}{5}=0$
$\begin{aligned} & \Rightarrow \quad\left(x^2+y^2+2 x-2 y+2\right) \\ & +\lambda\left[x^2+y^2-\frac{2}{5} x-\frac{16}{5} y+\frac{13}{5}\right]=0 \\ & \Rightarrow \quad(1+\lambda) x^2+(1+\lambda) y^2+2\left(1-\frac{\lambda}{5}\right) x \\ & \quad-2\left(1+\frac{8 \lambda}{5}\right) y+\left(2+\frac{13 \lambda}{5}\right)=0 \\ & \Rightarrow \quad\left(1-\frac{\lambda}{5}\right) \\ & \quad x^2+y^2+2 \frac{1+\lambda}{1+\lambda} x \\ & \quad-\frac{2\left(1+\frac{8 \lambda}{5}\right)}{1+\lambda} y+\frac{2+\frac{13 \lambda}{5}}{1+\lambda}=0\end{aligned}$
At
$g=0 \Rightarrow \frac{1-\frac{\lambda}{5}}{1+\lambda}=0 \Rightarrow \lambda=5$
Then $\quad x^2+y^3-3 y+\frac{5}{2}=0$
$\therefore$ Centre is $\left(0, \frac{3}{2}\right)$.
The limiting points are $(-1,1),\left(\frac{1}{5}, \frac{8}{5}\right)$.

Topic: Mathematics, Circle, JEE Main
The line $1 x+m y+n=0$ is normal to the circle $x^2+y^2+2 g x+2 f y+c=0$, if

Options:
A $l g+m f-n=0$
B $l g+m f+n=0$
C $l g=m f-n=0$
D $l g-m f+n=0$

The correct answer is:
$l g+m f-n=0$

Solution:
Since normal always passes through centre of circle, therefore ${(-g,-f)}$ must lie on $l x+m y+n=0$. Hence $\lg +m f-n=0$.

Topic: Mathematics, Circle, JEE Main
The line $1 x+m y+n=0$ will touch the parabola $y^2=4 a x$, if

Options:
A $m n=a l^2$
B $l m=a n^2$
C $l n=a m^2$
D $m n=a l$

The correct answer is:
$m n=a l$

Solution:
$y=-\frac{l}{m} x-\frac{n}{m}$
Condition for above line to be tangent to $y^2=4 a x$ is $-\frac{n}{m}=\frac{a m}{-l}$ or $n l=a m^2$.

Topic: Mathematics, Parabola, JEE Main
The line 2x+3y=12 meets the coordinates axes at A and B respectively. The line through 5,5 perpendicular to AB meets the coordinate axes and the line AB at C, D and E respectively. If O is the origin, then the area (in sq. units) of the figure OCEB is equal to

Options:
A 133
B 233
C 11
D 7

The correct answer is:
233

Solution:
The equation of a line passing through P5,5 and perpendicular to 2x+3y=12 is
y-5=32x-5
or, 3x-2y-5=0
The coordinates of the point of intersection of 2x+3y-12=0 and 3x-2y-5=0 are E3,2.
The line 3x-2y-5=0 meets x-axis at C53,0.
Now,
Area of figure OCEB
=Area of ΔAOB-Area of ΔACE
=12OA×OB-12AC×EL
=12×6×4-12×133×2
=233 sq. units

Topic: Mathematics, Straight Lines, JEE Main
The line 2x-y+1=0 is a tangent to the circle at the point 2, 5 and the centre of the circle lies on x-2y=4. Then, the radius of the circle is:

Options:
A 35
B 53
C 54
D 45

The correct answer is:
35

Solution:
Given

Slope of OA=5-h-422-h=m1Slope of the given tangent equation=m2=2

⇒5-(h-4)22-h2=-1

⇒h=8

Center=8, 2

Radius=(8-2)2+(2-5)2=35

Topic: Mathematics, Circle, JEE Main, JEE Main 2021 (17 Mar Shift 1)
The line 2x-y+1=0 is tangent to the circle at the point 2,5 and the centre of the circle lies on x-2y=4. The square of the radius of the circle is

Options:
A 45
B 75
C 20
D 50

The correct answer is:
45

Solution:
2x-y+1=0 is tangent to the circle
Slope of line OA=-12
Equation of OA, y-5=-12x-2
2y-10=-x+2
x+2y=12
∴ intersection with x-2y=4 will give coordinates of centre

On solving, we get, the centre of the circle is (8,2)
The radius OA=8-22+2-52=36+9
=45=35 units

Hence, OA2=45

Topic: Mathematics, Circle, JEE Main
The line 2x-y+1=0 touches a circle at the point 2,5 and the centre of the circle lies on x-2y=4. The diameter (in units) of the circle is

Options:
A 6 5
B 103
C 45
D 102

The correct answer is:
6 5

Solution:
tan⁡θ=2-121+212=34
r82+42=tan⁡θ=34=r45
⇒r=35⇒ diameter =65 units

Topic: Mathematics, Circle, JEE Main
The line $2 x+y=1$ is a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1(a>b)$. If this line passes through the point of intersection of a directrix and the positive $X$-axis, then the eccentricity of that hyperbola is

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

The correct answer is:
2

Solution:
The line $2 x+y-1=0$ is a tangent to the hyperbola
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and also passes through the point of intersection of directrix and the positive $x$-axis
Equation of direction $x=\frac{a}{e}$
$\therefore$ Point $\left(\frac{a}{e}, 0\right)$ passes through $2 x+y=1$
$\Rightarrow \quad 2\left(\frac{a}{e}\right)+0=1$
$\Rightarrow \quad 2 a=e$
$\Rightarrow \quad 2 a^2=a e$
We know, $(a e)^2=a^2+b^2$
$\therefore \quad 4 a^4=a^2+b^2$
Equation of tangent of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
is $y=m x \pm \sqrt{a^2 m^2-b^2}$
$\therefore \quad a^2(2)^2-b^2=1$
$$
\Rightarrow \quad 4 a^2-b^2=1
$$
From Eqs. (i) and (ii), we get
$$
\begin{array}{ll}
& 4 a^4=a^2+4 a^2-1 \\
\Rightarrow & 4 a^4-5 a^2+1=0 \\
& \left(4 a^2-1\right)\left(a^2-1\right)=0 \\
\Rightarrow & a^2=1, \frac{1}{4} \Rightarrow a=1, \frac{1}{2} \\
\therefore & e=2 a \\
& e=2(1)=2
\end{array}
$$

Topic: Mathematics, Hyperbola, JEE Main
The line 2x+y=3 cuts the ellipse 4x2+y2=5 at P and Q . If θ is the acute angle between normal at P and tangent at Q, then θ is equal to

Options:
A tan-1⁡45
B sin-1⁡334
C cos-1⁡334
D cot-1⁡34

The correct answer is:
cos-1⁡334

Solution:
Solving y=3-2x and 4x2+y2=5 , we get the intersection point P12,2 and Q1,1
The equation of tangent at these points are 4x12+2y=5 and 4x1+y1=5
⇒ Slope of tangents at P and Q are -1 and -4 respectively
⇒ Slope of normal at P is 1
⇒tan⁡θ=1--41+1-4=53⇒cos⁡θ=334

Topic: Mathematics, Ellipse, JEE Main
The line 2x+y=3 cuts the ellipse 4x2+y2=5 at points P and Q. If Î¸ is the acute angle between the normals at P and Q, then Î¸ is equal to

Options:
A tan-1⁡53
B sin-1⁡334
C cos-1⁡334
D cot-1⁡34

The correct answer is:
sin-1⁡334

Solution:
Solving y=3-2x and 4x2+y2=5, we get the intersection points P12,2 and Q1,1
The equations of the tangents at these points are 4x12+2y=5 and 4x1+y1=5
â The slopes of the tangents at P & Q are -1 & -4 respectively
â The slopes of the normals at P & Q are 1 & 14 respectively
âtanâ¡Î¸=1-141+114=35âÎ¸=tan-1â¡35
âcosâ¡Î¸=534, sinâ¡Î¸=334âÎ¸=sin-1â¡334

Topic: Mathematics, Ellipse, JEE Main
The line 2x+y=3 interesects the ellipse 4x2+y2=5 at two points. The point of intersection of the tangents to the ellipse at these points is

Options:
A 56,56
B 53,56
C 56,53
D 53,53

The correct answer is:
56,53

Solution:
Let the required point be h,k.

Now from this point, the equation of chord of contact to the ellipse is T=0

⇒4hx+ky=5 which is same as 2x+y=3
⇒4h2=k1=53⇒h=56, k=53
⇒ Point of intersection of the tangents is 56,53

Topic: Mathematics, Ellipse, JEE Main
The line $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{0}$ is parallel to

Options:
A $x y$-plane
B $y z$-plane
C $z x$-plane
D None of these

The correct answer is:
$x y$-plane

Solution:
The given line is perpendicular to $z$-axis. Hence parallel to $x y$-plane.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The line 3x+2y=24 meets the y-axis at A & the x-axis at B. The perpendicular bisector of AB meets the line through (0, -1) parallel to x-axis at C. Then the area of the triangle ABC is

Options:

The correct answer is:
91

Solution:
Equation of CD ⇒
2 x - 3 y = λ .....(1)
it passes through (4, 6) then λ = - 1 0
2 x - 3 y + 1 0 = 0 ....(2)
coordinates of C - 1 3 2 , - 1
Now, the area of the triangle,
Δ = 1 2 0 1 2 1 8 0 1 - 1 3 2 - 1 1
= 1 2 0 - 1 2 8 + 1 3 2 + 1 - 8
= 1 2 - 6 2 9 - 8 = 9 1

Topic: Mathematics, Straight Lines, JEE Main
The line $3 x-4 y=5$ is a tangent to the hyperbola $x^2-4 y^2=5$. The point of contact is

Options:
A $(3,1)$
B $(2,1 / 4)$
C $(1,3)$
D None of these

The correct answer is:
$(3,1)$

Solution:
Suppose point of contact be $(h, k)$, then tangent is $h x-4 k y-5=0 \equiv 3 x-4 y-5=0$ or $h=3, k=1$
Hence the point of contact is $(3,1)$

Topic: Mathematics, Hyperbola, JEE Main
The line 3x-4y+7=0 is rotated through an angle Ï4 in the clockwise direction about the point -1, 1. The equation of the line in its new position is

Options:
A 7y+x-6=0
B 7y-x-6=0
C 7y+x+6=0
D 7y-x+6=0

The correct answer is:
7y+x-6=0

Solution:
As (-1, 1) is a point on 3x – 4y + 7 = 0, the rotation is possible.
Slope of the given line =34
Slope of the line in its new position =34-11+34=-17
Therefore, the required equation is y-1=-17 x+1 or 7y+x-6=0

Topic: Mathematics, Straight Lines, JEE Main
The line 12x cosθ+5y sinθ=60 is tangent to which of the following curves ?

Options:
A x2+y2=30
B 144x2+25y2=3600
C x2+y2=169
D 25x2+12y2=3600

The correct answer is:
144x2+25y2=3600

Solution:
Given that 12x cosθ+5y sinθ=60

Divide the above equation with 60 on both sides

x cosθ5+y sinθ12=1…i

Let ellipse x2a2+y2b2=1

⇒ Equation of tangent at point (acosθ,bsinθ)
xa cosθa2+yb sinθb2=1

xcosθa+ysinθb=1…ii

Using equation i and ii⇒a=5
b=12
Equation of ellipse is x225+y2144=1

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2021 (31 Aug Shift 1)
The line $21 x+5 y=116$ is a tangent to the hyperbola $7 x^{2}-5 y^{2}=232$. Its point of contact is

Options:
A $(-6,-2)$
B $(6,2)$
C $(6,-2)$
D $(-6,2)$

The correct answer is:
$(6,-2)$

Solution:
The line $21 x+5 y=116$ is a tangent to $7 x^{2}-5 y^{2}=232$. If $P\left(x_{1}, y_{1}\right)$ is the point of contact, then $S_{1}=0$.
$\begin{aligned}
&\Rightarrow 7 x_{1} x-5 y_{1} y=232 \text { It is same as } 21 x+5 y=116 \\
&\therefore \quad \frac{7 x_{1}}{21}=-\frac{5 y_{1}}{5}=\frac{232}{116}=2 \\
&\therefore \quad x_{1}=6, y_{1}=-2 . \text { So, } P \text { is }(6,-2) .
\end{aligned}$

Topic: Mathematics, Hyperbola, JEE Main