datavorous/entrance-exam-dataset huggingface dataset:
If a diamagnetic solution is poured into a U-tube and one arm of this U-tube is placed between the poles of a strong magnet, with the meniscus in line with the field, then the level of the solution in this arm will

Options:
A rise
B fall
C oscillate slowly
D remain as such

The correct answer is:
fall

Solution:
A diamagnetic liquid moves from the stronger parts of a magnetic field towards the weaker parts.

Topic: Physics, Magnetic Properties of Matter, NEET
If a diamagnetic substance is brought near north or south pole of a bar magnet, it is

Options:
A attracted by poles
B repelled by poles
C repelled by the north pole and attracted by the south pole
D attracted by the north pole and repelled by the south pole

The correct answer is:
repelled by poles

Solution:
Diagmagnetic substances are repelled by the poles of a bar magnet.

Topic: Physics, Magnetic Properties of Matter, JEE Main
If a diamagnetic substance is brought near the north or the south pole of a bar magnet, it is :

Options:
A Attracted by both the poles
B Repelled by both the poles
C Repelled by the north pole and attracted by the south pole
D Attracted by the north pole and repelled by the south pole

The correct answer is:
Repelled by both the poles

Solution:
As diamagnetic substances have negative intensity of magnetisation, they are weekly repelled by the external field.

Topic: Physics, Magnetic Properties of Matter, NEET, NEET 2009 (Mains)
If a diamagnetic substance is brought near the north or the south pole of a bar magnet, it is :

Options:
A Attracted by both the poles
B Repelled by both the poles
C Repelled by the north pole and attracted by the south pole
D Attracted by the north pole and repelled by the south pole

The correct answer is:
Repelled by both the poles

Solution:
As diamagnetic substances have negative intensity of magnetisation, they are weekly repelled by the external field.

Topic: Physics, Magnetic Properties of Matter, JEE Main
If a diameter of a hyperbola meets the hyperbola in real points then

Options:
A it meets the conjugate hyperbola in imaginary points
B the conjugate diameter meets the given hyperbola in real point
C the conjugate diameter meets the conjugate hyperbola in imaginary points.
D none of these

The correct answer is:
it meets the conjugate hyperbola in imaginary points

Solution:
Equation of the hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ so equation of the conjugate hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$
Equation of a diameter of the hyperbola is $y=\left(b^2 / a^2 m\right) x$ which meets the hyperbola at points given by
$\frac{x^2}{a^2}-\frac{b^4 x^2}{b^2 a^4 m^2}=1 \Rightarrow x^2=\frac{a^2 m^2}{m^2-b^2}$
These points are real if $m^2 \gt b^2$ and for this value of $\mathrm{m}$, the diameter will meet the conjugate hyperbola at point for which $x^2=\frac{-a^2 m^2}{m^2-b^2}$ and these points are imaginary.
Note: Conjugate diameter will meet the conjugate hyperbola in real points and the hyperbola in imaginary points

Topic: Mathematics, Hyperbola, JEE Main
If a dice is thrown 7 times, then the probability of obtaining 5 , exactly 4 times is

Options:
A ${ }^7 C_4\left(\frac{1}{6}\right)^4\left(\frac{5}{6}\right)^3$
B ${ }^7 C_4\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^4$
C $\left(\frac{1}{6}\right)^4\left(\frac{5}{6}\right)^3$
D $\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^4$

The correct answer is:
${ }^7 C_4\left(\frac{1}{6}\right)^4\left(\frac{5}{6}\right)^3$

Solution:
Required probability
$={ }^7 C_4\left(\frac{1}{6}\right)^4\left(\frac{5}{6}\right)^3$.

Topic: Mathematics, Probability, JEE Main
If a double stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA.

Options:

None

Solution:
In a DNA molecule, the number of cytosine molecule is equal to guanine molecules \& the number of adenine molecules are equal to thymine molecules. As a result, if a double stranded DNA has $20 \%$ of cytosine, it has $20 \%$ of guanine. The remaining $60 \%$ includes both adenine \& thymine which are in equal amounts. So, the percentage of adenine is $30 \%$.

Topic: Biology, Molecular Basis of Inheritance
If a double-stranded DNA has 30% of guanine, what will be the percentage of adenine and thymine base pairs?

Options:
A 20%
B 30%
C 40%
D 60%

The correct answer is:
40%

Solution:
If a double stranded DNA consists of 30% guanine, then according to Chargaff's rule, cytosine is also 30%. Remaining 40% includes adenine and thymine base pairs.

Topic: Biology, Molecular Basis of Inheritance, NEET
If a drop of liquid breaks into smaller droplets, it results in lowering of temperature of the droplets. Let a drop of radius $R$, break into $N$ small droplets each of radius $r$. Estimate the drop in temperature.

Options:

None

Solution:
When a big drop of radius $R$, breaks into $N$ droplets each of radius $r$, the volume remains constant. So by the law of conservation of mass, final volume $=$ initial volume. Volume of big drop $=N \times$ Volume of each small drop $\frac{4}{3} \pi R^3=N \times \frac{4}{3} \pi r^3$ or $R^3=N r^3$
or $N=\frac{R^3}{r^3}$,
$\Delta E=\mathrm{T}$ [Initial area of big drop-final area of $N$ drops of radius $(R)$ ]
Now, $\triangle A$ (change in surface area)
$$
=4 \pi R^2-N 4 \pi r^2
$$
$\Delta E($ Energy released $)=\mathrm{T} \times \Delta A$
$\Delta A=4 \pi\left(R^2-N r^2\right)$
$=\mathrm{T} \times 4 \pi\left(R^2-N r^2\right)$
Due to releasing of this energy, the temperature is lowered. If $\rho$ is the density and $s$ is specific heat of liquid and its temperature is lowered by $\Delta t$, then energy released $=m s \Delta t$ [ $s=$ specific heat $\Delta t=$ change in temperature]
So, $\Delta E=m s \Delta t$
Equating both equation,
$$
\begin{aligned}
&\mathrm{T} \times 4 \pi\left(R^2-N r^2\right)=\left(\frac{4}{3} \pi \times R^3 \times \rho\right) s \Delta t \\
&{\left[\because m=V \rho=\frac{4}{3} \pi R^3 \rho\right]} \\
&\Rightarrow \Delta t=\frac{T \times 4 \pi\left(R^2-N r^2\right)}{\frac{4}{3} \pi R^3 \rho \times s}=\frac{3 T}{\rho s}\left[\frac{R^2}{R^3}-\frac{N r^2}{R^3}\right] \\
&=\frac{3 T}{\rho s}\left[\frac{1}{R}-\frac{\left(R^3 / r^3\right) \times r^2}{R^3}\right] \quad=\frac{3 T}{\rho s}\left[\frac{1}{R}-\frac{1}{r}\right] \\
&
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids
If A=ete-tcos te-tsin⁡ tet-e-tcos⁡t-e-tsin⁡t-e-tsin⁡t+e-tcos⁡tet2e-tsin⁡t-2e-tcos⁡t, then A is:

Options:
A Invertible only if t=π
B Not invertible for any t∈R
C Invertible only if t=π2
D Invertible for all t∈R

The correct answer is:
Invertible for all t∈R

Solution:
Since given matrix A is invertible ⇒A≠0

A=e-t1cos tsin t1-cos t-sin t-sin t+cos t12 sin t-2 cos t

R2→R2-R1

R3→R3-R1

=e-t1cos tsin t0-2 cos t-sin t-2 sin t+cos t02 sin t-cos t-2 cos t-sin t

=e-t2cost+sint2+2sint-cost2

=5e-t

A=5 e-t≠0 ∀ t∈R
Hence, given matrix is always invertible.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2019 (09 Jan Shift 2)
If a ferromagnetic material is heated upto Curie temperature, it becomes

Options:
A Permanent magnet
B Anti-ferromagnetic material
C Diamagnetic material
D Paramagnetic material

The correct answer is:
Paramagnetic material

Solution:
When ferromagnetic material heated beyond curie temperature, turns into paramagnetic material, as the ferromagnetic domains become random.

Topic: Physics, Magnetic Properties of Matter, JEE Main
If a flagstaff subtends equal angles at the points \( A, B, C \) and \( D \) on the horizontal ground through the foot of flagstaff, then the points \( A, B, C \) and \( D \) necessarily form \( a \)

Options:
A Rectangle
B Parallelogram
C Square
D None of these

The correct answer is:
None of these

Solution:
Let PQ be the vertical tower, having Q as its foot. We have, AQ=BQ=CQ=DQ, which implies that the points A,B,C and D are concylic having Q as its centre.

Topic: Mathematics, Heights and Distances, JEE Main
If a force $F$ is applied on a body and it moves with a velocity $v$, the power will be

Options:
A $F \times v$
B $F / v$
C $F / v^2$
D $F \times v^2$

The correct answer is:
$F \times v$

Solution:
Given, Velocity $=\mathrm{v}$ force $=\mathrm{F}$
We know that work done is equal to the product of force and displacement
$W=f \times x$
Divided by $t$ in both sides then we get,
$\frac{w}{t}=\frac{f x}{t} \Rightarrow \frac{x}{t}=v ~\&~ \frac{w}{t}=p$
Therefore $\mathrm{p}=\mathrm{fv}$

Topic: Physics, Work Power Energy, JEE Main
If a full wave rectifier circuit is operating from $50 \mathrm{~Hz}$ mains, the fundamental frequency in the ripple will be:

Options:
A $25 \mathrm{~Hz}$
B $50 \mathrm{~Hz}$
C $70.7 \mathrm{~Hz}$
D $100 \mathrm{~Hz}$

The correct answer is:
$100 \mathrm{~Hz}$

Solution:
In full wave rectifier the fundamental frequency in ripple is twice of input frequency.

Topic: Physics, Semiconductors, NEET, NEET 2003
If a full wave rectifier circuit is operating from $50 \mathrm{~Hz}$ mains, the fundamental frequency in the ripple will be:

Options:
A $25 \mathrm{~Hz}$
B $50 \mathrm{~Hz}$
C $70.7 \mathrm{~Hz}$
D $100 \mathrm{~Hz}$

The correct answer is:
$100 \mathrm{~Hz}$

Solution:
In full wave rectifier the fundamental frequency in ripple is twice of input frequency.

Topic: Physics, Semiconductors, JEE Main
If a full wave rectifier circuit is operating from $50 \mathrm{~Hz}$ mains, the fundamental frequency in the ripple will be

Options:
A $70.7 \mathrm{~Hz}$
B $100 \mathrm{~Hz}$
C $25 \mathrm{~Hz}$
D $59 \mathrm{~Hz}$

The correct answer is:
$100 \mathrm{~Hz}$

Solution:
For full wave rectifier, ripple frequency
$=2 \times$ input frequency
$$
\begin{array}{l}
=2 \times 50 \\
=100 \mathrm{~Hz}
\end{array}
$$
Note A full wave rectifier consists of two junction diodes, so, its efficiency is twice that of half wave rectifier.

Topic: Physics, Magnetic Effects of Current, MHT CET, MHT CET 2007
If a full-wave rectifier circuit is operating from 50 Hz mains, the fundamental frequency in the ripple will be

Options:
A 70.7 Hz
B 100 Hz
C 25 Hz
D 59 Hz

The correct answer is:
100 Hz

Solution:
For full-wave rectifier, ripple frequency
= 2× input frequency
= 2×50=100 Hz

Topic: Physics, Semiconductors, NEET
If a fully charged capacitor C with initial charge q0 is connected to a coil of self inductance L at t=0 . The time at which the energy is stored equally between the electric field and magnetic field is

Options:
A πLC
B π4LC
C π2LC
D π6LC

The correct answer is:
π4LC

Solution:
In LC oscillation, energy is transferred C to L or L to C , maximum energy in L is12LImax2
Maximum energy in C isqmax22C
Energy will be equal when,
12LI2=1212LImax2⇒I=Imax2
I=Imaxsin⁡ωt=12Imax
ωt=π4⇒2πT t=π4⇒t=T8
t=182πLC=π4LC
⇒ t=π4LC

Topic: Physics, Electrostatics, JEE Main
If a function $f:(-1,1) \rightarrow B(\subseteq R)$ is defined as $f(x)=x+x^2+x^3+\ldots \infty$, then in order to have the inverse function of $f, B$ is equal to

Options:
A $\left(-\infty, \frac{1}{2}\right)$
B $\left(\frac{-1}{2}, \infty\right)$
C ( -1,1 )
D R

The correct answer is:
$\left(\frac{-1}{2}, \infty\right)$

Solution:
Given, $f(x)=x+x^2+x^3+\ldots \infty$
Let $y=f(x)$
[sum of infinite GP series]
$\Rightarrow y=\frac{x}{1-x}=\frac{1}{1-x}-1$
Given, Domain of $f=(-1,1)$
$\begin{array}{ll}
\text { i.e. } & -1 < x < 1 \Rightarrow-1 < -x < 1 \\
\Rightarrow & 0 < 1-x < 2 \\
& \quad \frac{1}{2} < \frac{1}{1-x} < \infty \\
& \frac{1}{2}-1 < \frac{1}{1-x}-1 < \infty-1 \\
\Rightarrow & -\frac{1}{2} < y < \infty
\end{array}$
So, range $=y \in\left(\frac{-1}{2}, \infty\right)=B$ (according to question $)$.
Hence, $\rho([A: B])=2$
for this $-c+2 a-b=0$
$\Rightarrow \quad a=\frac{b+c}{2}$

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2021 (06 Aug Shift 2)
If a function $\mathrm{f}: \mathbf{R}-\{l\} \mathbf{R}-\{m\}$ defined by $\mathrm{f}(\mathrm{x})=\frac{x+3}{x-2}$ is a bijection, then $3 l-2 m=$

Options:
A 10
B 12
C 8
D 14

The correct answer is:
8

Solution:
$f(x)=\frac{x+3}{x-2}$
$\because \quad f(x)$ is not defined for $x=2$
i.e. domain of $f(x)$ is $\mathrm{R}-\{2\}$
$\therefore l=2$
Now, $y=\frac{x+3}{x-2}$
$\begin{aligned} & x y-2 y=x+3 \\ & x(y-1)=2 y+3 \\ & x=\frac{2 y+3}{y-1} \\ & \because y \text { can take any value except } \\ & y=1 \\ & \therefore \text { co-domain }=\mathrm{R}-\{1\}\end{aligned}$
$m=1$
$3 l+2 m=3(2)+2(1)=8$

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2022 (05 Jul Shift 2)
If a function f: RâR is defined as fx=â«x8+4x4-2x2+2dx and f0=1, then which of the following is correct?

Options:
A fx is an even function
B fx is an onto function
C fx is an odd function
D fx is many one function

The correct answer is:
fx is an onto function

Solution:
fx=∫x8+4+4x4-4x4x4-2x2+2dx
=∫x4+22-2x22x4-2x2+2dx
=∫x4+2x2+2x4-2x2+2x4-2x2+2dx
fx=x55+2x33+2x+C
f0=0+0+0+C=1⇒C=1
fx=x55+2x33+2x+1
Range of fx is R, so f(x) is an onto function
f-x≠fx, so not even
f-x≠-fx, so not odd
f'x>0∀x∈R, so f(x) is one-one

Topic: Mathematics, Indefinite Integration, JEE Main
If a function $f: R \rightarrow R$ is defined by $f(x)=\frac{4 x}{5}+3$, then $f^{-1}(x)=$

Options:
A $\frac{5(x+3)}{4}$
B $\frac{5(x-3)}{4}$
C $\frac{4(x+3)}{5}$
D $\frac{4(x-3)}{5}$

The correct answer is:
$\frac{5(x-3)}{4}$

Solution:
(A)
Let $f(x)=\frac{4 x}{5}+3=y$
$\therefore 4 x=5 y-15 \Rightarrow x=\frac{5 y-15}{4}$
$\therefore f^{-1}(y)=\frac{5 y-15}{4} \Rightarrow f^{-1}(x)=\frac{5 x-15}{4}=\frac{5(x-3)}{4}$

Topic: Mathematics, Functions, JEE Main
If a function \( f: R \rightarrow R \) is defined as \( f(x)=\int \frac{x^{8}+4}{x^{4}-2 x^{2}+2} d x \) and \( f(0)=1 \), then which of the following is correct?

Options:
A \( f(x) \) is an even function
B \( f(x) \) is an onto function
C \( f(x) \) is an odd function
D \( f(x) \) is many one function

The correct answer is:
\( f(x) \) is an onto function

Solution:
We have, fx=∫x8+4x4-2x2+2dx

Now, fx=∫x8+4+4x4-4x4x4-2x2+2dx
=∫x4+22-2x22x4-2x2+2dx
=∫x4+2x2+2x4-2x2+2x4-2x2+2dx
Therefore, fx=x55+2x33+2x+C
⇒f0=0+0+0+C=1

⇒C=1
⇒fx=x55+2x33+2x+1
Range of fx is R,

So, f(x) is an onto function
f-x≠fx,

So, fx is not even
f-x≠-fx,

So, fx is not odd
f'x>0∀x∈R,

So, f(x) is one-one

Topic: Mathematics, Indefinite Integration, JEE Main
If a function fx defined by fx=aex+be−x,−1≤x<1cx2,1≤x≤3ax2+2cx,3<x≤4be continuous for some a,b,c∈R and f'0+f'2=e, then the value of a is

Options:
A 1e2-3e+13
B ee2-3e-13
C ee2+3e+13
D ee2-3e+13

The correct answer is:
ee2-3e+13

Solution:
Continuous at x=1,3

f1-=f1 ⇒ae+be-1=c ...(1)

f3=f3+ ⇒9c=9a+6c ⇒c=3a ...2

From 1 and 2

b=ae(3-e) ...3

f'x=aex-be−x,−1<x<12cx,1<x<32ax+2c,3<x<4

f'0=a-b, f'2=4c

Given f'(0)+f'(2)=e

a-b+4c=e ...4

By using eq. 1, 2, 3 & 4

a=ee2-3e+13

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2020 (02 Sep Shift 1)
If a gas expands at constant temperature indicates that

Options:
A kinetic energy of molecules decreases
B pressure of the gas increases
C kinetic energy of molecules remains as same
D number of the molecules of gas increase

The correct answer is:
kinetic energy of molecules remains as same

Solution:
$\begin{gathered}\because \quad \mathrm{KE}=\frac{3}{2} R T \\ \mathrm{KE} \propto T\end{gathered}$
Here, the temperature of gas remains constant hence kinetic energy of molecules remains the same.

Topic: Chemistry, Thermodynamics (C), JIPMER, JIPMER 2014
If a gas expands at constant temperature, it indicates that

Options:
A kinetic energy of molecules decreases
B pressure of the gas increases
C kinetic energy of molecules remains the same
D number of the molecules of gas increases

The correct answer is:
kinetic energy of molecules remains the same

Solution:
Key Idea :
$\begin{gathered}
K E=\frac{3}{2} R T \\
\mathrm{KE} \propto \mathrm{T}
\end{gathered}$
Here, the temperature of gas remains constant, hence kinetic energy of molecules remains the same.

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2008 (Screening)
If a gas expands at constant temperature, it indicates that

Options:
A Number of the molecules of gas increases
B Kinetic energy of molecules decreases
C Pressure of the gas increases
D Kinetic energy of molecules remains the same

The correct answer is:
Kinetic energy of molecules remains the same

Solution:
Kinetic energy of gaseous molecules depends on temperature only.

Topic: Chemistry, States of Matter, NEET, NEET 2008 (Mains)
If a gas has \( n \) degrees of freedom, ratio of specific heats of gas is

Options:
A \( \frac{1+n}{2} \)
B \( 1+\frac{1}{n} \)
C \( -1+\frac{n}{2} \)
D \( 1+\frac{2}{n} \)

The correct answer is:
\( 1+\frac{2}{n} \)

Solution:
The given degrees freedom for a gas= n

The total energy of the system, U=n2RT

The molar specific heat at constant volume for given gas, CV=n2R

The molar specific heat at constant pressure for given gas, CP=CV+R

CP=n2R+R

CP=n2+1R

The ratio of specific heats of gas, CPCV=n2+1Rn2R

CPCV=1+2n

Topic: Physics, Kinetic Theory of Gases, JEE Main
If a gas has ' \(n\) ' degrees of freedom, the ratio of the specific heats \(\gamma\) of the gas is

Options:
A \(1+\frac{2}{n}\)
B \(1+\frac{\mathrm{n}}{2}\)
C \(1+\frac{1}{n}\)
D \(\frac{1+\mathrm{n}}{2}\)

The correct answer is:
\(1+\frac{\mathrm{n}}{2}\)

Solution:
$\gamma=\frac{C_P}{C_v}=\frac{\frac{f}{2} R+R}{\frac{f}{2} R}=1+\frac{2}{f}=1+\frac{2}{n}$

Topic: Physics, Kinetic Theory of Gases, JEE Main
If a gas is compressed isothermally then the r.m.s. velocity of the molecules

Options:
A decreases.
B increases.
C remains the same.
D first decreases and then increases.

The correct answer is:
remains the same.

Solution:
The root-mean-square (rms) velocity of gas molecules is,
$$
\mathrm{v}_{\mathrm{r} . \mathrm{m} . \mathrm{s}}=\sqrt{\frac{3 \mathrm{kT}}{\mathrm{m}}}
$$
$\Rightarrow \mathrm{v}_{\mathrm{rms}} \propto \sqrt{\mathrm{T}}$ for a given gas.
As the temperature is constant in isothermal process, the r.m.s velocity remains the same.

Topic: Physics, Kinetic Theory of Gases, MHT CET, MHT CET 2023 (11 May Shift 1)
If a gas is compressed isothermally then the r.m.s. velocity of the molecules

Options:
A remains the same.
B increases.
C decreases.
D first decreases and then increases.

The correct answer is:
remains the same.

Solution:
No Solution

Topic: Physics, Kinetic Theory of Gases, MHT CET, MHT CET 2020 (15 Oct Shift 2)
If a green plant is kept in dark with proper ventilation, can this plant carry out photosynthesis? Can anything be given as supplement to maintain its growth or survival?

Options:

None

Solution:
The plant in given conditions cannot carry out photosynthesis. Light is necessary for any green plant to make its own food. The plant should be watered properly for its survival.

Topic: Biology, Photosynthesis in Higher Plants
If a growing plant is decapitated

Options:
A axillary buds are inactivated
B axillary buds are activated
C leaves become yellow and have a
D growth stops

The correct answer is:
axillary buds are activated

Solution:
The hormones which are produced in the plant are called phytohormone. Auxin is a phytohormone. An apical bud is present at the top of the plant and an axillary bud or lateral bud is present on the lateral side. The phytohormone which is synthesized in the apical bud is auxin that helps in maintaining the apical dominance in which the growth of the lateral bud stop but if the apical bud is decapitated, means removed then the concentration of the auxin gets low and this causes the activation of the axillary bud which helps in increasing the plant width.

Topic: Biology, Plant - Growth and Development, NEET
If a hyperbola has length of its conjugate axis equal to 5 and the distance between its foci is 13 , then the eccentricity of the hyperbola is:

Options:
A $\frac{13}{12}$
B 2
C $\frac{13}{6}$
D $\frac{13}{8}$

The correct answer is:
$\frac{13}{12}$

Solution:
$\therefore$ Conjugate axis $=5$

$\therefore 2 b=5$

Distance betwecn foci $=13$

$2 a e=13$

Then, $b^{2}=a^{2}\left(e^{2}-1\right)$

$\Rightarrow a^{2}=36$

$\therefore a=6$

$a e=\frac{13}{2} \Rightarrow e=\frac{13}{12}$

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2019 (11 Jan Shift 2)
If a hyperbola passes through the point \(\mathrm{P}(\sqrt{2}, \sqrt{3})\) and foci at \(( \pm 2,0)\) then the tangent to this hyperbola at \(\mathrm{P}\) is

Options:
A \(\mathrm{y}=\mathrm{x} \sqrt{6}-\sqrt{3}\)
B \(\mathrm{y}=\mathrm{x} \sqrt{3}-\sqrt{6}\)
C \(\mathrm{y}=\mathrm{x} \sqrt{6}+\sqrt{3}\)
D \(y=x \sqrt{3}+\sqrt{6}\)

The correct answer is:
\(\mathrm{y}=\mathrm{x} \sqrt{6}-\sqrt{3}\)

Solution:
Hint: \(a e=2, a^2+b^2=4\)
\(\Rightarrow \frac{2}{a^2}-\frac{3}{b^2}=1 \Rightarrow b^2=3 ; a^2=1\)
\(\frac{x^2}{1}-\frac{y^2}{3}=1\). Tangent at \(P(\sqrt{2}, \sqrt{3})\) is
\(\sqrt{6 x}-y=\sqrt{3}\)

Topic: Mathematics, Hyperbola, JEE Main
If a hyperbola passes through the point P10, 16, and it has vertices at ±6, 0, then the equation of the normal to it at P, is.

Options:
A 3x+4y=94
B 2x+5y=100
C x+2y=42
D x+3y=58

The correct answer is:
2x+5y=100

Solution:
Let the equation of hyperbola is x2a2-y2b2=1.

As, vertex is at ±6,0, and it passes through 10, 16,
So, the equation of hyperbola is x236-y2144=1.
Equation of normal to the hyperbola at x1, y1 is a2xx1+b2yy1=a2+b2

Thus, on putting values, equation of normal is 2x+5y=100.

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2020 (08 Jan Shift 2)
If a hyperbola passes through the point \( P(10,16) \), and it has vertices at \( (\pm 6,0) \), then the equation of the normal to it at \( P \), is.

Options:
A \( 3 x+4 y=94 \)
B \( 2 x+5 y=100 \)
C \( x+2 y=42 \)
D \( x+3 y=58 \)

The correct answer is:
\( 2 x+5 y=100 \)

Solution:
Let the equation of hyperbola is x2a2-y2b2=1.

As, vertex is at ±6,0, and it passes through 10, 16,
So, the equation of hyperbola is x236-y2144=1.
Equation of normal to the hyperbola at x1, y1 is a2xx1+b2yy1=a2+b2

Thus, on putting values, equation of normal is 2x+5y=100.

Topic: Mathematics, Hyperbola, JEE Main
If $A=\left[\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right]$, $B=\left[\begin{array}{ll}0 & i \\ i & 0\end{array}\right]$
where $i=\sqrt{-1}$, then the correct relation is

Options:
A $A+B=0$
B $A^2=B^2$
C $A-B=0$
D $A^2+B^2=0$

The correct answer is:
$A^2=B^2$

Solution:
Relation $A^2=B^2$ is true because $A^2=\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]$ and $B^2=\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]$ have same matrices.

Topic: Mathematics, Matrices, JEE Main
If $\mathbf{a}=\mathbf{i}+2 \mathbf{j}-2 \mathbf{k}, \mathbf{b}=2 \mathbf{i}-\mathbf{j}+\mathbf{k}$ and $\mathbf{c}=\mathbf{i}+3 \mathbf{j}-\mathbf{k}$, then $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})$ is equal to

Options:
A $20 \mathbf{i}-3 \mathbf{j}+7 \mathbf{k}$
B $20 \mathbf{i}-3 \mathbf{j}-7 \mathbf{k}$
C $20 \mathbf{i}+3 \mathbf{j}-7 \mathbf{k}$
D None of these

The correct answer is:
$20 \mathbf{i}-3 \mathbf{j}+7 \mathbf{k}$

Solution:
$\begin{aligned} \mathbf{a} \times(\mathbf{b} \times \mathbf{c})=\mathbf{a} \times\left|\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ 1 & 3 & -1\end{array}\right| & =\mathbf{a} \times(-2 \mathbf{i}+3 \mathbf{j}+7 \mathbf{k}) \\ & =\left|\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -2 \\ -2 & 3 & 7\end{array}\right|=20 \mathbf{i}-3 \mathbf{j}+7 \mathbf{k} .\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main
If $\mathbf{a}=\mathbf{i}+2 \mathbf{j}-3 \mathbf{k}$ and $\mathbf{b}=3 \mathbf{i}-\mathbf{j}+2 \mathbf{k}$, then the angle between the vectors $\mathbf{a}+\mathbf{b}$ and $\mathbf{a}-\mathbf{b}$ is

Options:
A $30^{\circ}$
B $60^{\circ}$
C $90^{\circ}$
D $0^{\circ}$

The correct answer is:
$90^{\circ}$

Solution:
We have $\mathbf{a + b = 4 i + j - \mathbf { k }}$ and $\mathbf{a - b}=-2 \mathbf{i}+3 \mathbf{j}-5 \mathbf{k}$.
Clearly $(\mathbf{a}+\mathbf{b}) .(\mathbf{a}-\mathbf{b})=0 . \quad$ Hence $(\mathbf{a}+\mathbf{b}) \perp(\mathbf{a}-\mathbf{b})$.

Topic: Mathematics, Vector Algebra, JEE Main