datavorous/entrance-exam-dataset huggingface dataset:
If the differential equation given by
\[
\frac{d^{2} y}{d t^{2}}+2 k \frac{d y}{d t}+\omega^{2} y=F_{0} \sin p t
\]
Describes the oscillatory motion of body in a dissipative medium under the influence of a periodic force, then the state of
maximum amplitude of the oscillation is a phenomena of

Options:
A Free vibration
B Damped vibration
C Forced vibration
D Resonance

The correct answer is:
Resonance

Solution:
Here, the state of maximum amplitude of the oscillation is a measure of resonance.

Topic: Physics, Electromagnetic Induction, JEE Main
If the differential equation representing the family of all circles touching x-axis at the origin is x2-y2dydx=gxy, then gx equals

Options:
A 12x2
B 2x
C 12x
D 2x2

The correct answer is:
2x

Solution:
Equation of circle touching x-axis in origin is x-02+y-k2=k2
⇒x2+y2-2ky=0 ...1, where k is the radius of the circle.

Differentiating the above equation w.r.t. x, we get,

2x+2ydydx-2kdydx=0

⇒x+y-kdydx=0

From equation 1,

⇒x+y-x2+y22ydydx=0

⇒x2-y2dydx=2xy=gxy

⇒gx=2x

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2014 (09 Apr Online)
If the digits are not allowed to repeat in any number formed by using the digits 0, 2, 4, 6, 8, then the number of all numbers greater than 10,000 is equal to __________.

Options:

The correct answer is:
96

Solution:
We have to form a number greater than 10,000 by using the digits 0, 2, 4, 6, 8, hence, the first place of the number can't be zero.

Thus, for the first place we have 4 choices i.e. 2, 4, 6, 8 and the remaining places can be filled with any of the digits, thus we have respectively 4, 3, 2, 1 choices.

Hence, using the fundamental principle of multiplication, the total numbers which are greater than 10,000 are

=4×4×3×2=96.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2021 (22 Jul Shift 1)
If the dimensions of a physical quantity are given by $\left[\mathrm{M}^{\mathrm{a}} \mathrm{L}^{\mathrm{b}} \mathrm{T}^{\mathrm{c}}\right]$, then the physical quantity will be :

Options:
A Force if $a=0, b=-1, c=-2$
B Pressure if $\mathrm{a}=1, \mathrm{~b}=-1, \mathrm{c}=-2$
C Velocity if $a=1, b=0, c=-1$
D Acceleration if $a=1, b=1, c=-2$

The correct answer is:
Pressure if $\mathrm{a}=1, \mathrm{~b}=-1, \mathrm{c}=-2$

Solution:
$$
[\mathrm{P}]=\frac{\mathrm{F}}{\mathrm{A}}=\left[\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2}\right]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]
$$

Topic: Physics, Units and Dimensions, JEE Main
If the dimensions of a physical quantity are given by $\left[\mathrm{M}^{\mathrm{a}} \mathrm{L}^{\mathrm{b}} \mathrm{T}^{\mathrm{c}}\right]$, then the physical quantity will be :

Options:
A Force if $a=0, b=-1, c=-2$
B Pressure if $\mathrm{a}=1, \mathrm{~b}=-1, \mathrm{c}=-2$
C Velocity if $a=1, b=0, c=-1$
D Acceleration if $a=1, b=1, c=-2$

The correct answer is:
Pressure if $\mathrm{a}=1, \mathrm{~b}=-1, \mathrm{c}=-2$

Solution:
$$
[\mathrm{P}]=\frac{\mathrm{F}}{\mathrm{A}}=\left[\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2}\right]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]
$$

Topic: Physics, Units and Dimensions, NEET, NEET 2009 (Mains)
If the dipole moment of AB molecule is given by 1.2 D and A-B the bond length is 1 â« then % ionic character of the bond is
[Given: 1debye = 10 â18 âesu.âcm ]

Options:
A 75
B 50
C 60
D 25

The correct answer is:
25

Solution:
Î¼Cal=4.8Ã10-10e.s.uÃ10-8 cm
=4.8Ã10-18esu cm
=4.8 D
Hence % ionic character=â Î¼ experimental Î¼ calculated Ã100
= 1.2 4.8 Ã100=25%

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
If the dipole moment of Toluene and Nitro-benzene are 0.43 D and 3.93 D respectively, then what is the expected dipole moment of p-Nitrotoluene?

Options:
A 3.50 D
B 2.18 D
C 4.36 D
D 5.30 D

The correct answer is:
4.36 D

Solution:
Methyl group has +I effect and -NO2 group has -I effect. Therefore, in p-nitro toluene the dipole moments of âCH3 and -NO2 groups act in the same direction. So, the resultant dipole moment is additive.
i.e., 3.93 + 0.43 = 4.36 debye

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
If the direction ratios of a line are 1+Î», 2-Î», 4 and if it makes an angle of 60o with the y-axis, then the sum of the values of Î» is

Options:

The correct answer is:
7

Solution:
The direction ratios of the line are <1+Î»,2-Î»,4>
direction ratios of the y-axis are <0,1,0>
So, cosâ¡60o=2-Î»1+Î»2+2-Î»2+16=12
â4-2Î»=2Î»2-2Î»+21
â4Î»2+16-16Î»=2Î»2-2Î»+21
â2Î»2-14Î»-5=0
Hence, the sum of values of Î» =7

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the dispersed phase is a liquid and the dispersion medium is a solid, the colloid is known as

Options:
A A sol
B An emulsion
C A gel
D A foam

The correct answer is:
A gel

Solution:
$\underset{\substack{\text { (Dispersed } \\ \text { phase) }}}{\text { Liquid }}+\underset{\substack{\text { (Dispersion } \\ \text { medium }}}{\text { Solid }}=\underset{\text { (Colloid) }}{\mathrm{Gel}}$ (e.g. Butter)

Topic: Chemistry, Surface Chemistry, JEE Main
If the dispersed phase is a liquid and the dispersion medium is a solid, the colloid is known as

Options:
A A sol
B An emulsion
C A gel
D A foam

The correct answer is:
A gel

Solution:
$\underset{\substack{\text { (Dispersed } \\ \text { phase) }}}{\text { Liquid }}+\underset{\substack{\text { (Dispersion } \\ \text { medium) }}}{\text { Solid }}=\underset{\text { (Colloid) }}{\mathrm{Gel}}$ (e.g. Butter)

Topic: Chemistry, Surface Chemistry, JEE Main
If the displacement of a particle executing SHM is given by $y=0.30 \sin (220 t+0.64)$ in metre, then the frequency and maximum velocity of the particle is

Options:
A $35 \mathrm{~Hz}, 66 \mathrm{~m} / \mathrm{s}$
B $45 \mathrm{~Hz}, 66 \mathrm{~m} / \mathrm{s}$
C $58 \mathrm{~Hz}, 113 \mathrm{~m} / \mathrm{s}$
D $35 \mathrm{~Hz}, 132 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$35 \mathrm{~Hz}, 66 \mathrm{~m} / \mathrm{s}$

Solution:
$\begin{aligned} & n=\frac{\omega}{2 \pi}=\frac{220}{2 \pi}=35 \mathrm{~Hz} \\ & v_{\text {max }}=\omega a=220 \times 0.30 \mathrm{~m} / \mathrm{s}=66 \mathrm{~m} / \mathrm{s}\end{aligned}$

Topic: Physics, Oscillations, JEE Main
If the displacement (x) and velocity (v) of a particle executing SHM are related through the expression 3v2=30-x2. If the time period of the particle is T=Ïn, then what is the value of n?

Options:

The correct answer is:
12

Solution:
Given equation,

3v2=30-x2
Differentiating both sides, we get
6vdvdx=-2x

⇒vdvdx=-13x

⇒ax=-13 …∵a=dvdt=dvdxdxdt=vdvdx

T=2π acceleration displacement =2π-13=2π3=π12

n=12

Topic: Physics, Oscillations, JEE Main
If the displacement (x) and velocity (v) of a particle executing SHM are related through the expression 3v2=30-x2. If the time period of the particle is T=Ïn, then what is the value of n?

Options:

The correct answer is:
12.00

Solution:
Given equation,

3v2=30-x2
Differentiating both sides, we get
6vdvdx=-2x

⇒vdvdx=-13x

⇒ax=-13 …∵a=dvdt=dvdxdxdt=vdvdx

T=2π acceleration displacement =2π-13=2π3=π12

n=12

Topic: Physics, Oscillations, JEE Main
If the dissociation energy of the molecule is \( D=\left[U_{(x=\infty)}-U_{\text {at }}\right. \) equilibrium \( ] . \) The potential energy function for the force between two atoms in a diatomic molecule is approximately given by \( U(x)=\frac{a}{x^{12}}-\frac{b}{x^{6}} \) where a and b are constants and x is the distance between the atoms.Find the dissociation energy.

Options:
A \(\frac{b^{2}}{2 a}\)
B \(\frac{b^{2}}{12 a}\)
C \(\frac{b^{2}}{4 a}\)
D \(\frac{b^{2}}{6 a}\)

The correct answer is:
\(\frac{b^{2}}{4 a}\)

Solution:
The potential energy function for the force between two atoms in a diatomic molecule is approximately given by

U=ax12-bx6
If Force is denoted by F, then

F=-dUdx=-ddxax12-bx6=--12ax13+6bx7=12ax13-6bx7

At equilibrium, F=0.
∴ 12ax13-6bx7=0 or x6=2ab

By using the above equation we can writeUat equilibrium =a2ab2-b2ab
=ab24a2-b22a=b24a-b22a=-b24a

Since, U(x=∞)=0

Therefore, th dissociation energy is given by
D=[U(x=∞)-Uat equilibrium
=0--b24a=b24a

Topic: Physics, Work Power Energy, JEE Main
If the distance between object and its two times magnified virtual image produced by a curved mirror is 15 cm, the focal length of the mirror must be :

Options:
A 15 cm
B -12 cm
C -10 cm
D 103 cm

The correct answer is:
-10 cm

Solution:
Let object be at a distance x cm from the pole. Then, distance of image from the pole will be =15-x.

Given:

Magnification, m=2=-vu

⇒2=-(15-x)-x

⇒2x=15-x

⇒3x=15⇒x=5 cm

Now,

Distance of the image from the pole =15-x=15-5=10 cm

Therefore, v=10 cm

From mirror formula:

1f=1v+1u

=110+1(-5)=1-210=-110

⇒f=-10 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2024 (29 Jan Shift 2)
If the distance between parallel plates of a capacitor is halved and dielectric constant is doubled then the capacitance

Options:
A Decreases two times
B Increases two times
C Increases four times
D Remain the same

The correct answer is:
Increases four times

Solution:
We know,
\(\mathrm{C}=\frac{\mathrm{kl}_{\mathrm{O}} \mathrm{A}}{\mathrm{d}}\)
Now distance is halved \(\mathrm{k}\) id doubled
\(\begin{aligned}
& \mathrm{C}_1=\frac{(2 \mathrm{k}) \mathrm{l}_0 \mathrm{~A}}{(\mathrm{~d} / 2)} \\
& \mathrm{C}_1=4 \mathrm{C}
\end{aligned}\)

Topic: Physics, Capacitance, JEE Main
If the distance between planes, $4 x-2 y-4 z+1=$ 0 and $4 x-2 y-4 z+d=0$ is 7 , then $d$ is:

Options:
A
41 or $-42$

B
42 or $-43$

C
$-41$ or 43

D
$-42$ or 44

The correct answer is:
$-41$ or 43

Solution:
Given planes are
$$
4 x-2 y-4 z+1=0
$$
and $4 x-2 y-4 z+d=0$
They are parallel.
Distance between them is $\pm 7$
$=\frac{d-1}{\sqrt{16+4+16}}$
$$
\begin{aligned}
&\Rightarrow \frac{d-1}{6}=\pm 7 \\
&\Rightarrow d=42+1 \\
&\text { or }-42+1 \\
&\text { i.e. } d=-41 \text { or } 43 .
\end{aligned}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2014 (12 Apr Online)
If the distance between the earth and the sun were half of the present distance, then the duration of a year would have been nearly

Options:
A 45 days
B 100 days
C 182 days
D 129 days

The correct answer is:
129 days

Solution:
According to the Kepler's law
T 1 T 2 2 = r 1 r 2 3
â´ T 2 T 1 = r 2 r 1 3 / 2
â´ T 2 = T 1 r 2 r 1 3 / 2 = 3 6 5 1 2 3 / 2 = 129 days

Topic: Physics, Gravitation, JEE Main
If the distance between the foci and the distance between the directrices of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ are in the ratio $3: 2$, then $a: b$ is

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

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

Solution:
Equation of hyperbola is
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
Distance between foci $=2 a e$ and distance between directrices $=\frac{2 a}{e}$ According to question, we have
$\frac{2 a e}{2 a / e}=\frac{3}{2}$
$\begin{array}{ll}\Rightarrow & e^{2}=\frac{3}{2} \\ \because & b^{2}=a^{2}\left(e^{2}-1\right) \\ \Rightarrow & \frac{b^{2}}{a^{2}}=\frac{3}{2}-1=\frac{1}{2} \\ \Rightarrow & \frac{b}{a}=\frac{1}{\sqrt{2}} \\ \Rightarrow & a: b=\sqrt{2}: 1\end{array}$

Topic: Mathematics, Hyperbola, JEE Main
If the distance between the foci of an ellipse is 6 and the distance between its directrix is 12, then the length of its latus rectum is

Options:
A 3
B 32
C 32
D 23

The correct answer is:
32

Solution:
Given, 2ae=6 and 2ae=12, directrices are x=±ae

So, ae=6
Now, using the formula for eccentricity of an ellipse

⇒b2=a2-a2e2=18-9=9

⇒L.R.=2b2a=2×932=32

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2020 (07 Jan Shift 1)
If the distance between the foci of an ellipse is equal to the length of the latusrectum, then its eccentricity is

Options:
A $\frac{1}{4}(\sqrt{5}-1)$
B $\frac{1}{2}(\sqrt{5}+1)$
C $\frac{1}{2}(\sqrt{5}-1)$
D $\frac{1}{4}(\sqrt{5}+1)$

The correct answer is:
$\frac{1}{2}(\sqrt{5}-1)$

Solution:
Given,
In ellipse the distance between the foci
= Length of the latusrectum
$\begin{array}{ll}\Rightarrow \quad & \text { 2ae }=\frac{2 b^{2}}{a} \\ \Rightarrow \quad & \quad a^{2} e=b^{2} \Rightarrow e=\frac{b^{2}}{a^{2}}\end{array}$
$\because \quad e^{2}=1-\frac{b^{2}}{a^{2}}$
$\Rightarrow \quad e^{2}=1-e \quad$ [from Eq.
$\Rightarrow e^{2}+e-1=0$
$\Rightarrow \quad e=\frac{-1 \pm \sqrt{1+4}}{2}$
(by quadratic formula) $\Rightarrow \quad e=\frac{-1 \pm \sqrt{5}}{2}$
$\Rightarrow$
$$
e=\frac{\sqrt{5}-1}{2} \quad(\because 1>e>0)
$$

Topic: Mathematics, Ellipse, JEE Main
If the distance between the foci of an ellipse is half the length of its latus rectum, then the eccentricity of the ellipse is:

Options:
A 12
B 2-1
C 2-12
D 22-12

The correct answer is:
2-1

Solution:
2ae=12×2b2a
2e=b2a2
We know that
e=1-b2a2
e=1-2e
e2=1-2e
e2+2e-1=0
-2±4+42=-2±222
-1+2 & -1-2
∴ e=2-1

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2015 (11 Apr Online)
If the distance between the plane, 23x-10y-2z+48=0 and the plane containing the lines x+12=y-34=z+13 and x+32=y+26=z-1λλ∈R is equal to k633, then k is equal to ____________.

Options:

The correct answer is:
3

Solution:
Lines must be intersecting

⇒2s-1,4s+3,3s-1=2t-3,6t-2,λt+1
2s-1=2t-3,4s+3=6t-2,3s-1=λt+1⇒t=12,s=-12,λ=-7

Distance of plane contains given lines from given plane is same as distance between point -3,-2,1 from given plane.

Required distance equal to

-69+20-2+48529+100+4=3633=k633⇒k=3

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2020 (09 Jan Shift 2)
If the distance between the points $(a \cos \theta, a \sin \theta)$ and $(a \cos \phi, a \sin \phi)$ is $2 a$, then $\theta$ is equal to

Options:
A $2 n \pi \pm \pi+\phi, n \in Z$
B $n \pi+\frac{\pi}{2}+\phi, n \in Z$
C $n \pi-\phi, n \in Z$
D $2 n \pi+\phi, n \in Z$

The correct answer is:
$2 n \pi \pm \pi+\phi, n \in Z$

Solution:
Let the points be $A=(a \cos \theta, a \sin \theta)$ and
$$
\begin{aligned}
& B=(a \cos \phi, a \sin \phi) \\
& \therefore \quad A B=\sqrt{(a \cos \theta-a \cos \phi)^2}+(a \sin \theta-a \sin \phi)^2 \\
& =\sqrt{a^2 \cos ^2 \theta+a^2 \cos ^2 \phi-2 a^2 \cos \theta \cos \phi} \\
& \quad \quad+a^2 \sin ^2 \theta+a^2 \sin ^2 \phi-2 a^2 \sin \theta \sin \phi
\end{aligned}
$$
$$
\begin{aligned}
& =\sqrt{2 a^2-2 a^2(\cos \theta \cos \phi+\sin \theta \sin \phi)} \\
& =\sqrt{2} a(\sqrt{1-\cos (\theta-\phi)} \\
& \Rightarrow \quad 2 a=\sqrt{2} a \sqrt{2} \sin \left(\frac{\theta-\phi}{2}\right) \\
& \Rightarrow \quad \sin \left(\frac{\theta-\phi}{2}\right)=1 \Rightarrow \frac{\theta-\phi}{2}=n \pi \pm \frac{\pi}{2} \\
& \Rightarrow \quad \sin \left(\frac{\theta-\phi}{2}\right)=1 \Rightarrow \frac{\theta-\phi}{2}=n \pi \pm \frac{\pi}{2} \\
& \Rightarrow \quad \theta-\phi=2 n \pi \pm \pi \\
& \Rightarrow \quad \theta=2 n \pi \pm \pi+\phi \\
&
\end{aligned}
$$
where, $n \in Z$

Topic: Mathematics, Straight Lines, JEE Main
If the distance between two current carrying parallel wire is made $\left(\frac{1}{3}\right)^{r d}$ then the magnitude of the force between them (the same current is passing through both wires)

Options:
A is tripled.
B becomes 6 times.
C becomes q times.
D remains same.

The correct answer is:
is tripled.

Solution:
Force between wires is inversely proportional to the separation between them,
$F \propto \frac{1}{d}$
Therefore, if the separation between wires becomes $\left(\frac{1}{3}\right)^{r d}$, the magnitude of force is tripled.

Topic: Physics, Magnetic Effects of Current, MHT CET, MHT CET 2022 (10 Aug Shift 1)
If the distance between two masses is doubled, the gravitational attraction between them

Options:
A Is doubled
B Becomes four times
C Is reduced to half
D Is reduced to a quarter

The correct answer is:
Is reduced to a quarter

Solution:
$F \propto \frac{1}{r^2}$. If $r$ becomes double then $F$ reduces to $\frac{F}{4}$

Topic: Physics, Gravitation, JEE Main
If the distance of point P3,2,6 from the line x-12=y-23=z-34 measured parallel to the plane 3x-5y+2z=5 is k, then the value of k2 is equal to

Options:

The correct answer is:
4573

Solution:
Coordinates of the point Q are 2Î»+1,3Î»+2,4Î»+3
PQâ is perpendicular to the normal vector of the plane,
PQâ=2Î»-2i^+3Î»j^+4Î»-3k^
Normal vector =3i^-5j^+2k^
Hence, 32Î»-2-53Î»+24Î»-3=0
â6Î»-6-15Î»+8Î»-6=0â-Î»-12=0
âÎ»=-12
So, the coordinates of the point Q are -23,-34,-45
Hence, PQ=262+362+512=4573

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the distance of the earth from Sun is 1.5×106 km, then the distance of an imaginary planet from Sun, if its period of revolution is 2.83 years is:

Options:
A 6 × 107 km
B 6 × 106 km
C 3 × 106 km
D 3 × 107 km

The correct answer is:
3 × 106 km

Solution:
From Kepler's law of periods, we know that

T2∝R3⇒T1T22=R1R23

⇒12.832=1.5×106R23

⇒R2=2.832×1.5×106313

=813×1.5×106=3×106 km

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (24 Jan Shift 2)
If the distance of the point 1, -2, 3 from the plane x+2y-3z+10=0 measured parallel to the line, x-13=2-ym=z+31 is 72, then the value of m is equal to _______.

Options:

The correct answer is:
2

Solution:
DC of line ≡3m2+10,-mm2+10,1m2+10

Q≡1+3rm2+10,-2+-mrm2+10,3+rm2+10

Q lies on x+2y-3z+10=0

1+3rm2+10-4-2mrm2+10-9-3rm2+10+10=0

⇒rm2+103-2 m-3=2

⇒rm2+10-2 m=2

r2m2=m2+10

72 m2=m2+10⇒52 m2=10⇒m2=4

m=2

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (16 Mar Shift 2)
If the distribution of molecular speeds of a gas is as per the figure shown below, then the ratio of the most probable, the
average, and the root mean square speeds, respectively, is

Options:
A \( 1: 1: 1 \)
B \( 1: 1: 1.224 \)
C \( 1: 1.128: 1.224 \)
D \( 1: 1.128: 1 \)

The correct answer is:
\( 1: 1: 1.224 \)

Solution:
Due to symmetrical graph given, the average value of velocities will be at the middle, symmetrical velocity
which is also most probable velocity.

While calculating rms speed, we have to take average of square of speeds and its root. In square of
speeds, higher speeds will give more contribution so the value will be more than the average value.

Topic: Physics, Kinetic Theory of Gases, JEE Main
If the DNA percentage of thymine is $20 \%$, then what is the percentage of guanine?

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

The correct answer is:
$30 \%$

Solution:
According to Chargaff's Rule, the DNA molecule should have an equal ratio of pyrimidine (cytosine and thymine) and purine (adenine and guanine). It means that the number of adenine molecules is equal to thymine molecules and the number of guanine molecules is equal to cytosine molecules.
A complement with $\mathrm{T}$, and $\mathrm{G}$ complement with $\mathrm{C}$.
Thus, $\mathrm{T}=\mathrm{A}=20 \%$.
And, $\mathrm{A}+\mathrm{T}=40 \%$
Since, $\mathrm{G}=\mathrm{C}$
and, remaining of $\mathrm{G}+\mathrm{C}=60 \%$
so, $\mathrm{G}=\mathrm{C}=30 \%$

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2002
If the domain of function $f(x)=x^2-6 x+7$ is $(-\infty, \infty)$, then the range of function is

Options:
A $(-\infty, \infty)$
B $[-2, \infty)$
C $(-2,3)$
D $(-\infty,-2)$

The correct answer is:
$[-2, \infty)$

Solution:
$x^2-6 x+7=(x-3)^2-2$
Obviously, minimum value is -2 and maximum $\infty$. Hence range of function is $[-2, \infty]$.

Topic: Mathematics, Functions, JEE Main
If the domain of \( f(x) \) is \( [1,3] \), then the domain of \( f\left(\log _{2}\left(x^{2}+3 x-2\right)\right) \) is

Options:
A \( [-5,-4] \cup[1,2] \)
B \( [-13,-2] \cup\left[\frac{3}{5}, 5\right] \)
C \( [4,1] \cup[2,7] \)
D \( [-3,2] \)

The correct answer is:
\( [-5,-4] \cup[1,2] \)

Solution:
Domain of fx is 1,3

1≤log2x2+3x-2≤3

⇒2≤x2+3x-2≤8

It is the same as solving two inequalities.

Case 1: x2+3x-2≥2

or, x2+3x-4≥0

or, x+4x-1≥0

∴ x∈(-∞,-4]∪[1,∞)

Case 2: x2+3x-2≤8

or, x2+3x-10≤0

or, x+5x-2≤0

∴ x∈-5,2

Considering both Case 1 and Case 2 we have, x∈-5,-4∪1,2

Topic: Mathematics, Functions, JEE Main
If the domain of the function fx=cos-12-|x|4+loge(3-x)-1 is [-α,β)-{γ}, then α+β+γ is equal to :

Options:
A 12
B 9
C 11
D 8

The correct answer is:
11

Solution:
Given:

y=cos-12-x4+log3-x-1

⇒-1≤2-x4≤1

⇒-4≤2-x≤4

⇒-6≤-x≤2

⇒-2≤x≤6

⇒x≤6

⇒-6≤x≤6 ...i

Also, 3-x>0 and 3-x≠1

⇒x≠2 ...ii and x<3 ...iii

Using i, ii and iii we get,

⇒x∈[-6,3)-2

So, on comparing with x∈[-α,β)-γ we get,

⇒α+β+γ=6+3+2=11

Topic: Mathematics, Functions, JEE Main, JEE Main 2024 (30 Jan Shift 1)
If the domain of the function fx=cos-1x2-x+1sin-12x-12 is the interval (α, β], then α+β is equal to:

Options:
A 32
B 2
C 12
D 1

The correct answer is:
32

Solution:
Given function is fx=cos-1x2-x+1sin-12x-12

For, sin-1x & cos1x to be defined -1≤x≤1 and for fx to be defined fx≥0 and the denominator of any expression can never be zero.

Thus, to define the domain of the given function, we have

-1≤x2-x+1≤1, x2-x+1≥0 and sin-12x-12>0

Taking the common values and using the range of sin-1x i.e. for positive values 0<sin-1x≤π2, we get

0≤x2-x+1≤1 and 0<sin-12x-12≤π2

⇒ x2-x≤0 and 0<2x-12≤1

⇒ xx-1≤0 and 0<2x-1≤2

⇒ x∈[0, 1] and 1<2x≤3

⇒ x∈0, 1 and 12<x≤32

Now, taking intersection, we get x∈12, 1

Given, domain of the function is α, β=12, 1

⇒ α=12, β=1

⇒ α+β=32.

Topic: Mathematics, Functions, JEE Main, JEE Main 2021 (22 Jul Shift 1)
If the domain of the function fx=loge2x+34x2+x-3+cos-12x-1x+2 is (α,β], then the value of 5β-4α is equal to

Options:
A 10
B 12
C 11
D 9

The correct answer is:
12

Solution:
Given,

fx=loge2x+34x2+x-3+cos-12x-1x+2

Now, using the definition of log function and cos-1 function we get,

2x+34x2+x-3>0 and -1≤2x-1x+2≤1

⇒2x+3(4x-3)(x+1)>0, 3x+1x+2≥0 & x-3x+2≤0

Now, solving 2x+3(4x-3)(x+1)>0 we get,

⇒x∈-32,-1∪34,∞ ....i

Similarly for 3x+1x+2≥0 we get,

x∈(-∞,-2)∪-13,∞ ...ii

And for x-3x+2≤0 we get,

x∈(-2,3] ... iii

Now, taking intersection of i∩ii∩iii we get,

(34,3]

So, on comparing with (α,β] we get, α=34 & β=3

Hence, 5β-4α=15-3=12

Topic: Mathematics, Functions, JEE Main, JEE Main 2024 (30 Jan Shift 2)
If the domain of the function fx=loge4x2+11x+6+sin-14x+3+cos-110x+63 is (α, β], then 36α+β is equal to

Options:
A 54
B 72
C 63
D 45

The correct answer is:
45

Solution:
Given that fx=log4x2+11x+6+sin-14x+3+cos-110x+63

We know that the domain of Ax+Bx+Cx is DA∩DB∩DC.

Let us find the domain of log4x2+11x+6.

We know that domain of logfx is fx>0.

⇒4x2+11x+6>0

⇒4x2+8x+3x+6>0

⇒4x+3x+2>0

⇒x∈-∞,-2∪-34,∞

Domain of log4x2+11x+9 is x∈-∞,-2∪-34,∞.

Let us find the domain of sin-14x+3

⇒-1≤4x+3≤1

⇒-1≤x≤-12

⇒x∈-1,-12

Let us find the domain of cos-110x+63.

⇒-1≤10x+63≤1

⇒-3≤10x+6≤3

⇒-910≤x≤-310.

Now the common domain is -34≤x≤-12.

⇒x∈-34,-12

But given that the domain of fx is x∈α,β.

⇒α=-34, β=-12.

⇒36α+β=36-34+-12

⇒36α+β=36×-54

⇒36α+β=45

Therefore, the required answer is 45.

Topic: Mathematics, Functions, JEE Main, JEE Main 2023 (15 Apr Shift 1)
If the domain of the function fx=sec-12x5x+3 is [α,β)∪(γ,δ], then 3α+10β+γ+21δ is equal to __________

Options:

The correct answer is:
24

Solution:
We know that,

Domain of sec-1x is (-∞,-1]∪[1,∞).

Now, fx=sec-12x5x+3

So,

2x5x+3≤-1 or 2x5x+3≥1

⇒2x5x+3+1≤0 or 2x5x+3-1≥0

⇒7x+35x+3≤0 or -3x-35x+3≥0

⇒7x+35x+3≤0 or x+15x+3≤0

⇒x∈[-1,-35)∪(-35,-37]

So,

α=-1, β=-35, γ=-35, δ=-37

Now,

3α+10β+γ+21δ=-3-6-6-9=24

Hence this is the required answer.

Topic: Mathematics, Functions, JEE Main, JEE Main 2023 (10 Apr Shift 2)
If the domain of the function $f(x)=\sin ^{-1}\left(\frac{x-1}{2 x+3}\right)$ is $\mathbf{R}-(\alpha, \beta)$, then $12 \alpha \beta$ is equal to :

Options:
A 32
B 40
C 24
D 36

The correct answer is:
32

Solution:
Domain of $f(x)=\sin ^{-1}\left(\frac{x-1}{2 x+3}\right)$ is
$\begin{aligned}
& 2 x+3 \neq 0 \& x \neq \frac{-3}{2} \text { and }\left|\frac{(x-1)}{2 x+3}\right| \leq 1 \\
& |x-1| \leq|2 x+3|
\end{aligned}$
$\begin{aligned} & \text { For }|2 x+3| \geq|x-1| \\ & x \in(-\infty,-4] \cup\left(-\frac{2}{3}, \infty\right) \\ & \alpha=-4 \& \beta=-\frac{2}{3}: 12 \alpha \beta=32\end{aligned}$

Topic: Mathematics, Functions, JEE Main, JEE Main 2024 (09 Apr Shift 1)
If the domain of the function fx=x1+x2, where x is greatest integer ≤x, is [2,6), then its range is

Options:
A 526,25-929,27109,1889,953
B 526,25
C 537,25-929,27109,1889,953
D 537,25

The correct answer is:
537,25

Solution:
Given:

fx=x1+x2

So,

fx=21+x2; x∈[2,3)fx=31+x2; x∈[3,4)fx=41+x2; x∈[4,5)fx=51+x2; x∈[5,6)

Hence,

fx∈537,25

Topic: Mathematics, Functions, JEE Main, JEE Main 2023 (31 Jan Shift 1)