datavorous/entrance-exam-dataset huggingface dataset:
The methods NOT involved in concentration of ore are

(A) Liquation
(B) Leaching
(C) Electrolysis
(D) Hydraulic washing
(E) Froth floatation

Choose the correct answer from the options given below:

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

The correct answer is:
A and C only

Solution:
Electrolysis is not a concentration method. Electrolytic refining is a process of refining a metal (mainly copper) by the process of electrolysis.

Liquation is not a concentration method. It is a refining method. In this method a low melting metal like tin can be made to flow on a sloping surface. In this way it is separated from higher melting impurities.

The other three methods are concentration methods of ores.

Leaching is used when the ore is soluble in a solvent. The powdered ore is dissolved in a chemical, usually a strong solution of NaOH. The chemical solution dissolves the metal in the ore, and it can be extracted and separated from the gangue by extracting the chemical solution.

Froth flotation is a process for selectively separating hydrophobic materials from hydrophilic.

Hydraulic washing is a technique used when the impurities are lighter and the ore particles are heavier. The lighter impurities are removed by washing in current of water. As gold particles are heavier than the impurities like sand, we can use hydraulic washing for the concentration of ores of gold.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2023 (31 Jan Shift 1)
The metre bridge wire AB shown in the adjoining figure is 100 cm long. When AD=30 cm, no deflection occurs in the galvanometer. What is the value of R (in ohm)?

Options:

The correct answer is:
14

Solution:
Let Ï be the resistance per unit length of the metre bridge wire.
From the condition of balance of Wheatstone bridge.
PQ=RS
â 6R=30Ï100-30Ï=3070=37
â R=7Ã63
â R=14 Î©

Topic: Physics, Current Electricity, JEE Main
The mid- point of the domain of the function $\mathrm{f}(\mathrm{x})=\sqrt{4-\sqrt{2 \mathrm{x}+5}}$ for real $\mathrm{x}$ is $-$

Options:
A $1 / 4$
B $3 / 2$
C $2 / 3$
D $-2 / 5$

The correct answer is:
$3 / 2$

Solution:
$\begin{array}{l}
f(x)=\sqrt{4-\sqrt{2 x+5}} \\
4-\sqrt{2 x+5} \geq 0 \quad 2 x+5 \geq 0 \\
\sqrt{2 x+5} \leq 4 \quad x \geq-5 / 2 \\
x \leq \frac{11}{2} \\
x \in\left[-\frac{5}{2}, \frac{11}{2}\right] \\
\text { mid point }=\frac{-5 / 2+11 / 2}{2}=\frac{3}{2}
\end{array}$

Topic: Mathematics, Functions, KVPY, KVPY 2012 (SB/SX)
The mid points of two small magnetic dipoles of length $d$ in end-on positions, are separated by a distance $x,(x \gt \gt d)$. The force between them is proportional to $x^{-n}$ where $n$ is :

Options:
A 1
B 2
C 3
D 4

The correct answer is:
4

Solution:
Magnetic field due to dipole, $B \propto \frac{1}{x^3}$
The interaction energy of two dipole is $\mathrm{U}=-\mathrm{m} \cdot \mathrm{B}$
$\mathrm{U} \propto \mathrm{x}^{-3}$
The force between them, $F=-\frac{\mathrm{dU}}{\mathrm{dx}}, \mathrm{F} \propto \mathrm{x}^{-4}$

Topic: Physics, Magnetic Properties of Matter, JEE Main
The mid points of two small magnetic dipoles of length d in end-on positions, are separated by a distance xx≫d. The magnitude of force between them is proportional to x−n where n is :

Options:
A 3
B 4
C 2
D 1

The correct answer is:
4

Solution:
B = Î¼ o 4 Ï 2 M x 3 for end-on position
U=â M â . B â
= Î¼o 4Ï 2 M 2 x 3
F=â dU dx = â3Î¼o 4Ï 2 M 2 x 4
therefore force between them â x - 4

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2014 (09 Apr Online)
The mid points of two small magnetic dipoles of length d in end-on positions, are separated by a distance xx≫d. The magnitude of force between them is proportional to x−n where n is :

Options:
A 3
B 4
C 2
D 1

The correct answer is:
4

Solution:
B = Î¼ o 4 Ï 2 M x 3 for end-on position
U=â M â . B â
= Î¼o 4Ï 2 M 2 x 3
F=â dU dx = â3Î¼o 4Ï 2 M 2 x 4
therefore force between them â x - 4

Topic: Physics, Magnetic Properties of Matter, JEE Main
The mid points of two small magnetic dipoles of length d in end-on positions, are separated by a distance x x>>d. The magnitude of force between them is proportional to x ân where n is :

Options:
A 3
B 4
C 2
D 1

The correct answer is:
4

Solution:
B = Î¼ o 4 Ï 2 M x 3 for end-on position
U=â M â . B â
= Î¼o 4Ï 2 M 2 x 3
F=â dU dx = â3Î¼o 4Ï 2 M 2 x 4
therefore force between them â x - 4

Topic: Physics, Magnetic Properties of Matter, JEE Main
The middle term in the expansion of $\left(1-\frac{1}{x}\right)^n\left(1-x^n\right)$ in powers of $x$ is

Options:
A
$-{ }^{2 n} \mathrm{C}_{n-1}$

B
$-{ }^{2 n} \mathrm{C}_n$

C
${ }^{2 n} \mathrm{C}_{n-1}$

D
${ }^{2 n} \mathrm{C}_n$

The correct answer is:
${ }^{2 n} \mathrm{C}_n$

Solution:
Given expansion can be re-written as
$$
\left(\frac{x-1}{x}\right)^n \cdot(1-x)^n=(-1)^n x^{-n}(1-x)^{2 n}
$$
Total number of terms will be $2 n+1$ which is odd ( $\because 2 n$ is always even)
$\therefore$ Middle term $=\frac{2 n+1+1}{2}=(n+1)$ th
Now, $T_{r+1}={ }^n C_r(1)^r x^{n-r}$
So, $\frac{{ }^{2 n} C_n \cdot x^{2 n-n}}{x^n \cdot(-1)^n}={ }^{2 n} C_n \cdot(-1)^n$
Middle term is an odd term. So, $n+1$ will be odd.
So, $n$ will be even.
$\therefore$ Required answer is ${ }^{2 n} \mathrm{C}_n$.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2012 (26 May Online)
The midpoint of A0,0 and B1024, 2048 is A1, midpoint of A1 and B is A2, midpoint of A2 and B is A3 and so on. The coordinates of A10 are

Options:
A 1025,Â 2050
B 1022,Â 2044
C 1023, 2046
D None of these

The correct answer is:
1023, 2046

Solution:
A1B=AB2, A2B=AB4
A3B=AB8,........, A10B=AB210=AB1024
âA A10=AB-AB1024=1023AB1024
â A10 divides the line joining A and B internally in the ratio 1023:1
â A10=0Ã1+1024Ã10231023+1,0Ã1+2048Ã10231023+1
=1023, 2046

Topic: Mathematics, Straight Lines, JEE Main
The mid-point of a chord of the ellipse $x^2+4 y^2-2 x+20 y=0 \quad$ is $(2,-4)$. The equation of the chord is

Options:
A $x-6 y=26$
B $x+6 y=26$
C $6 x-y=26$
D $6 x+y=26$

The correct answer is:
$x-6 y=26$

Solution:
Equation of chord at $(2,-4)$ is
$$
\begin{aligned}
& T=S^{\prime} \\
& \Rightarrow \quad 2 x+4 y(-4)-(x+2)+10(y-4) \\
& =(2)^2+4(-4)^2+2(2)+20(-4) \\
& \Rightarrow \quad 2 x-16 y-x-2+10 y-40 \\
& =4+64-4-80 \\
& \Rightarrow \quad x-6 y=42-16 \Rightarrow 26 \\
&
\end{aligned}
$$

Topic: Mathematics, Ellipse, JEE Main
The midpoints of two small magnetic dipoles of length \( d \) in end-on positions, are separated by a distance \( x,(x>>d) \).
The force between them is proportional to \( x^{-n} \) where \( n \) is

Options:
A \( 3 \)
B \( 4 \)
C \( 2 \)
D \( 1 \)

The correct answer is:
\( 4 \)

Solution:
B = μ o 4 π 2 M x 3 , for end-on position.
∴ dF ∝ dx x 4
Therefore, force between then ∝ x - 4

Topic: Physics, Magnetic Effects of Current, JEE Main
The migration of dispersion medium under the influence of an electric potential is called :

Options:
A
Cataphoresis

B
Electroosmosis

C
Electrophoresis

D
Sedimentation

The correct answer is:
Electroosmosis

Solution:
The motion of a liquid through a membrane under the influence of an applied electric field is known as electro-osmosis.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2013 (09 Apr Online)
The minimum acceleration with which a fireman can slide down a rope, if the breaking strength of the rope is 23 of his weight, is

Options:
A 23g
B g
C 13g
D zero

The correct answer is:
13g

Solution:
If the fireman slides down with some acceleration, his apparent weight decreases. For critical condition, rope can bear only 23 of his weight. If a is the minimum acceleration then,
Tension in the rope = Breaking stress =mg-a
mg-a=23mg âa=g-2g3=g3

Topic: Physics, Laws of Motion, JEE Main
The minimum acceleration with which a fireman can slide down a rope of breaking strength two-third of his weight is

Options:
A zero
B $\frac{g}{3}$
C $3 \mathrm{~g}$
D $g$

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

Solution:
If fireman slides down with some acceleration then its apparent weight decreases. For critical condition rope can bear only $\frac{2}{3}$ of the fireman's weight. Let $a$ is the minimum acceleration then,
Tension in the rope $=m(g-a)=$ Breaking strength
$$
\begin{array}{ll}
\Rightarrow & m(g-a)=\frac{2}{3} m g \\
\Rightarrow &=g-\frac{2 g}{3}=\frac{g}{3}
\end{array}
$$

Topic: Physics, Laws of Motion, JEE Main
The minimum amount of O2g consumed per gram of reactant is for the reaction:
(Given atomic mass: Fe=56, O=16, Mg=24, P=31, C=12,H=1 )

Options:
A P4s+5O2g→P4O10s
B 2 Mgs+O2g→2MgOs
C 4Fes+3O2g→2Fe2O3s
D C3H8g+5O2g→3CO2g+4H2O(l)

The correct answer is:
4Fes+3O2g→2Fe2O3s

Solution:
(i) 5 mole O2 required = 1 mole of P4 =124 gm P4
for per gram, O2 required = 5124 mole
(ii) 1 mole O2 required = 2 mole Mg = 2×24 gm
for per gram, O2 required =148 mole
(iii) 3 mole O2 required = 2 mole Fe = 4×56 gm
for per gram Fe, O2 required =3112×2 mole
(iv) 5 mole O2 required - =1 mole C3H8 =44 gm C3H8
per gram C3H8, O2 required =544 mole

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2019 (10 Apr Shift 2)
The minimum and maximum distances of a planet revolving around the Sun are x1 and x2. If the minimum speed of the planet on its trajectory is v0, then its maximum speed will be:

Options:
A v0x12x22
B v0x22x12
C v0x1x2
D v0x2x1

The correct answer is:
v0x2x1

Solution:
Angular momentum conservation equation,

v0x2=v1x1

v1=v0x2x1

Topic: Physics, Gravitation, JEE Main, JEE Main 2021 (25 Jul Shift 1)
The minimum area of a triangle formed by any tangent to the ellipse $\frac{x^2}{16}+\frac{y^2}{81}=1$ and the co-ordinate axes is:

Options:
A
12

B
18

C
26

D
36

The correct answer is:
36

Solution:
Let $(h, k)$ be the point on ellipse through which tangent is passing. Equation of tangent at
$$
(h, k)=\frac{x h}{16}+\frac{y k}{81}=1
$$
at $y=0, x=\frac{16}{h}$
at $x=0, y=\frac{81}{k}$
Area of $\mathrm{AOB}=\frac{1}{2} \times\left(\frac{16}{h}\right) \times\left(\frac{81}{k}\right)=\frac{648}{h k}$
$$
\mathrm{A}^2=\frac{(648)^2}{h^2 k^2}
$$
$(h, k)$ must satisfy equation of ellipse
$$
\begin{aligned}
&\frac{h^2}{16}+\frac{k^2}{81}=1 \\
&h^2=\frac{16}{81}\left(81-k^2\right)
\end{aligned}
$$
Putting value of $h^2$ in equation (1)
$$
\mathrm{A}^2=\frac{81(648)^2}{16 \times k^2\left(81-k^2\right)}=\frac{\alpha}{81 k^2-k^4}
$$
differentiating w.r. to $k$
$$
\begin{aligned}
&2 \mathrm{AA}^{\prime}=\alpha\left(\frac{-1}{81 k^2-k^4}\right)\left(162 k-4 k^3\right) \\
&2 \mathrm{AA}^{\prime}=-2 \mathrm{~A}\left(81 k-4 k^3\right) \\
&\Rightarrow \mathrm{A}^{\prime}=-81 k-4 k^3 \\
&\text { Put } \mathrm{A}^{\prime}=0 \\
&\Rightarrow 162 k-4 k^3=0, k\left(162-4 k^2\right)=0 \\
&\Rightarrow k=0, k=\pm \frac{9}{\sqrt{2}} \\
&\mathrm{~A}^{\prime \prime}=-\left(81-12 k^2\right)
\end{aligned}
$$
For both value of k, A" $=405>0$
Area will be minimum for $k=\pm \frac{9}{\sqrt{2}}$
$$
\begin{aligned}
&h^2=\frac{16}{81}\left(81-k^2\right)=8 \\
&h=\pm 2 \sqrt{2}
\end{aligned}
$$
Area of triangle $\mathrm{AOB}=\frac{648 \times \sqrt{2}}{2 \sqrt{2} \times 9}=36$
sq unit

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2014 (12 Apr Online)
The minimum distance between any two points P1 and P2 while considering point P1 on one circle and point P2 on the other circle for the given circles' equations

x2+y2-10x-10y+41=0

x2+y2-24x-10y+160=0 is ________

Options:

The correct answer is:
1

Solution:
Given C1(5,5), r1=3 and C2(12,5), r2=3

C1C2=7

Now, C1C2>r1+r2

Thus, P1P2min=7-6=1

Topic: Mathematics, Circle, JEE Main, JEE Main 2021 (17 Mar Shift 1)
The minimum distance between the curves y=tan⁡x, ∀x∈-π2,π2 and x-2-π42+y2=1 is

Options:
A 2-1
B 5-1
C 5+1
D 2

The correct answer is:
5-1

Solution:
The minimum distance between two curves lies along their common normal.
Let, Ph,tanh lies on y=tan⁡x
Then, the equation of normal at P
is y-tan⁡h=-1sec2hx-h
this passes through the centre of the circle, hence,
tan⁡hsec2⁡h=2+π4-h
⇒h=π4
Minimum distance = distance between π4,1 and 2+π4,0-1
=5-1

Topic: Mathematics, Application of Derivatives, JEE Main
The minimum distance of a point on the curve y=x2-4 from the origin is

Options:
A 152 units
B 192 units
C 152 units
D 192 units

The correct answer is:
152 units

Solution:
A variable point on the given parabola can be taken as α, α2-4
Distance from origin D= α2+α2-42
⇒ D2=α2+α4+16-8α2
=α4-7α2+16
dD2dα=4α3-14 α=0
⇒ 2α2α2-7=0
⇒ α=0, 72, -72
∴ Dmin 2 at α2=72
Dmin2=494-492+16
= -494+16=154
Dmin=152

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2016 (09 Apr Online)
The minimum distance of a point on the curve y=x2-4 from the origin is

Options:
A 152 units
B 192 units
C 152 units
D 192 units

The correct answer is:
152 units

Solution:
A variable point on the given parabola can be taken as Î±, Î±2-4
Distance from origin D= Î±2+Î±2-42
â D2=Î±2+Î±4+16-8Î±2
=Î±4-7Î±2+16
dD2dÎ±=4Î±3-14 Î±=0
â 2Î±2Î±2-7=0
â Î±=0, 72, -72
â´ Dmin 2 at Î±2=72
Dmin2=494-492+16
= -494+16=154
Dmin=152

Topic: Mathematics, Application of Derivatives, JEE Main
The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly :

Options:
A 1.5 eV
B 13.6 eV
C 1.9 eV
D 12.1 eV

The correct answer is:
12.1 eV

Solution:
For Balmer series, transition is from n=3,4,5,6, etc to n=2.

For transition to n=2, the minimum state to transition from is n=3.

Therefore, energy required ΔE=E3-E1=-13.69--13.6=12.1eV

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2024 (01 Feb Shift 1)
The minimum energy that must be possessed by photons in order to produce the photoelectric effect with platinum metal is:

[Given: The threshold frequency of platinum is 1.3×1015 s-1 and h=6.6×10-34Js]

Options:
A 8.58×10-19J
B 9.76×10-20J
C 3.21×10-14J
D 6.24×10-16J

The correct answer is:
8.58×10-19J

Solution:
Radiation energy = threshold energy of metal + kinetic energy of ejected electron

For minimum energy,

Radiation energy = threshold energy of metal

E=hνE=6.6×10-34×1.3×1015=8.58×10-19J

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2022 (25 Jun Shift 2)
The minimum force required to move a body up an inclined plane is three times the minimum force required to prevent it from sliding down the plane. If the coefficient of friction between the body and the inclined plane is 123, then the angle of the inclined plane is

Options:
A 60°
B 45°
C 30°
D 15°

The correct answer is:
30°

Solution:
Minimum force required to move a body up a rough inclined plane,
F1=mg(sin⁡θ+μcos⁡θ)
Minimum force required to prevent the body from sliding down the rough inclined plane,
F2=mgsin⁡θ-μcos⁡θ
According to the question,
F1=3F2
mgsin⁡θ+μcos⁡θ=3mgsin⁡θ-μmgcos⁡θ
4μcos⁡θ=2sin⁡θ
sin⁡θ=2μcos⁡θ
tan⁡θ=2μ=13
θ=30°

Topic: Physics, Laws of Motion, JEE Main
The minimum intensity of sound is zero at a point due to two sources of nearly equal frequencies, when

Options:
A Two sources are vibrating in opposite phase
B The amplitude of two sources are equal
C At the point of observation, the amplitudes of two S.H.M. produced by two sources are equal and both the S.H.M. are along the same straight line
D Both the sources are in the same phase

The correct answer is:
At the point of observation, the amplitudes of two S.H.M. produced by two sources are equal and both the S.H.M. are along the same straight line

Solution:
If two waves of nearly equal frequency superpose, they give beats if they both travel in straight line and $I_{\text {min }}=0$ if they have equal amplitudes.

Topic: Physics, Waves and Sound, JEE Main
The minimum number of \( 2 \)-input \( N A N D \) gates required to implement a \( N O R \) gate

Options:
A \( 2 \)
B \( 3 \)
C \( 4 \)
D \( 5 \)

The correct answer is:
\( 4 \)

Solution:
The minimum number of 2-input NAND gates required to implement a NOR gate is 4.

Topic: Physics, Semiconductors, JEE Main
The minimum number of elements that must be added to relation R=a,b,b,c,b,d on the set a,b,c,d, so that it is an equivalence relation is

Options:

The correct answer is:
13

Solution:
Let

A={a, b, c,d}, R : A→A

R=a,b,b,c,b,d

Reflexive:

For reflexivity, we must add a,a, b,b, c,c, d,d

Symmetric and Transitive:

For symmetric and transitivity, we must add

b,a, c,b, d,b

a,c, a,d, c,d, d,c, c,a, d,a

So, we can say that minimum 13 elements are required to make an equivalence relation.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2023 (24 Jan Shift 2)
The minimum number of elements that must be added to the relation R=(a,b),(b,c) on the set {a,b,c} so that it becomes symmetric and transitive is:

Options:
A 4
B 7
C 5
D 3

The correct answer is:
7

Solution:
Given:

A=a,b,c,d

And,

R=a,b,b,c

For symmetric relation, we must add b,a, c,b.

So, R=a,b,b,a, b,c,c,b.

For transitive relation, we must add

a,c, b,b, c,a, a,a, c,c

Hence,

R=b,a, c,b, a,c, b,b, c,a, a,a, c,c, a,b, b,c

So, number of new elements added=7

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2023 (30 Jan Shift 1)
The minimum number of moles of O2 required for complete combustion of 1 mole of propane and 2 moles of butane is..............

Options:

The correct answer is:
18

Solution:
(1) Combustion of propane.

C3H8 + 5O2g→3CO2g+4H2O1 mole 5 mole

(2) Combustion of butane

C4H10+132O2→4CO2g+5H2O1 mole 132 mole

so total mole of O2 =5+13=18

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2020 (05 Sep Shift 1)
The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96, is

Options:

The correct answer is:
8

Solution:
P2+P3â¦â¥0.96
1-P0+P1â¥0.96
1-0.96â¥nC012n+nC112n
â0.042nâ¥1+n
â4.2nâ¥n+1100
â2n+2â¥n+1100
As, 210â¥900
ân=8

Topic: Mathematics, Probability, JEE Main
The minimum number of times one has to toss a fair coin so that the probability of observing at least one head is at least 90% is:

Options:
A 2
B 4
C 5
D 3

The correct answer is:
4

Solution:
Let a fair coin is tossed n times, then
P(at least one head) =1–P(all tails)
=1–12n>90100 (given)
⇒12n<1-90100=110
⇒2n>10
⇒n≥4

Topic: Mathematics, Probability, JEE Main, JEE Main 2019 (08 Apr Shift 2)
The minimum possible distance between the points Aa,a-1 and Bb,b2+b+1 âa,bâR is D units, then the value of D2 is

Options:

The correct answer is:
2

Solution:
Given points lie on the line y=x-1 and the parabola y=x2+x+1
Thus, the minimum distance occurs along the common normal
∴ At B, dydx=1

or 2x+1=1
⇒x=0
∴B is 0,1
Hence, the shortest distance is D=0-1-12=2 units

Topic: Mathematics, Application of Derivatives, JEE Main
The minimum quantity of sodium chloride which is necessary to precipitate 10 litres of sol in two hours is $0.585 \mathrm{gm}$. The flocculation value of sodium chloride is

Options:
A 0.585
B 0.0585
C 0.1
D One

The correct answer is:
One

Solution:
Correct option is 4. One
The molar mass of $\mathrm{NaCl}$ is $58.5 \mathrm{~g} / \mathrm{mol}$
no. of moles $=\frac{\text { mass }}{\text { molar mass }}=\frac{0.585}{58.5}=10$ millimoles
Flocculation value of $\mathrm{NaCl}=\frac{\text { millimoles }}{V \text { ol of sol in litres }}=\frac{10}{10}=1$

Topic: Chemistry, Surface Chemistry, JEE Main
The minimum value of 2sinx+2cosx is :

Options:
A 2-1+12
B 2-1+2
C 21-2
D 21-12

The correct answer is:
21-12

Solution:
Using A.M≥G.M.

2sinx+2cosx2≥2sinx·2cosx

⇒2sinx+2cosx≥21+sinx+cosx2

Now -2≤sinx+cosx≤2

So minimum value of

2sinx+2cosx=21+-22=21-12

Topic: Mathematics, Functions, JEE Main, JEE Main 2020 (04 Sep Shift 2)
The minimum value of $3 \cos x+4 \sin x+5$ is

Options:
A 5
B 9
C 7
D 0

The correct answer is:
0

Solution:
The minimum value of $3 \cos x+4 \sin x$ is $-\sqrt{3^2+4^2}=-5$ Hence the minimum value of $3 \cos x+4 \sin x+5$ $=-5+5=0$

Topic: Mathematics, Trigonometric Equations, JEE Main
The minimum value of 8Z-8+2Z-4 exists, when Z is equal to (where, Z is a complex number)

Options:
A 2
B 1.5
C 0
D 1

The correct answer is:
1

Solution:
Given expression is
8Z-1+2Z-2=6Z-1+2Z-1+Z-2
Now, minimum Z-1+Z-2 is equal to 1, when Z lies on the line segment joining Z=1 and Z=2.
And, minimum Z-1 is equal to 0, when Z=1
⇒ Minimum 6Z-1+2Z-1+Z-2 is equal to 0+21=2 at Z=1

Topic: Mathematics, Complex Number, JEE Main
The minimum value of $9 \tan ^2 \theta+4 \cot ^2 \theta$ is

Options:
A 13
B 9
C 6
D 12

The correct answer is:
12

Solution:
$\begin{aligned}
& \text { A.M. } \geq \text { G.M. } \\
& \Rightarrow \frac{9 \tan ^2 \theta+4 \cot ^2 \theta}{2} \geq \sqrt{4 \cot ^2 \theta .9 \tan ^2 \theta} \\
& \Rightarrow 9 \tan ^2 \theta+4 \cot ^2 \theta \geq 12
\end{aligned}$
Therefore, the minimum value is 12 .

Topic: Mathematics, Trigonometric Equations, JEE Main
The minimum value of $27 \tan ^2 \theta+3 \cot ^2 \theta$ is

Options:
A $15$
B $18$
C $24$
D $30$

The correct answer is:
$18$

Solution:
Given trigonometrical equation is
$27 \tan ^2 \theta+3 \cot ^2 \theta$
$\begin{aligned} & \therefore \quad \frac{27 \tan ^2 \theta+3 \cot ^2 \theta}{2} \geq \sqrt{27 \tan ^2 \theta \cdot 3 \cot ^2 \theta} \\ & \Rightarrow \frac{27 \tan ^2 \theta+3 \cot ^2 \theta}{2} \geq \sqrt{81} \\ & \Rightarrow 27 \tan ^2 \theta+3 \cot ^2 \theta \geq 18\end{aligned}$
Hence, minimum value of given equation is 18

Topic: Mathematics, Trigonometric Equations, JEE Main
The minimum value of $\cos \theta+\sin \theta$ is

Options:
A 0
B $-\sqrt{2}$
C $1 / 2$
D $\sqrt{2}$

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

Solution:
$\begin{aligned}
& \text { Let } f(x)=\cos \theta+\sin \theta=\sqrt{2} \cos \left(\theta-\frac{\pi}{4}\right) \\
& \text { Since }-1 \leq \cos \left(\theta-\frac{\pi}{4}\right) \leq 1 \\
& -\sqrt{2} \leq \sqrt{2} \cos \left(\theta-\frac{\pi}{4}\right) \leq \sqrt{2}
\end{aligned}$
Since
Thus, the minimum value of $f(x)$ is $-\sqrt{2}$.

Topic: Mathematics, Trigonometric Equations, JEE Main
The minimum value of fx=aax+a1-ax, where a, x∈R and a>0, is equal to:

Options:
A a+1
B 2a
C a+1a
D 2a

The correct answer is:
2a

Solution:
A.M.≥G.M.

fx=aax+a1-ax=aax+aaax≥2a

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (25 Feb Shift 2)
The minimum value of function $f(x)=3 x^4-8 x^3+12 x^2-48 x+25$ on $[0,3]$ is equal to

Options:
A $25$
B $-39$
C $-25$
D $39$

The correct answer is:
$-39$

Solution:
$\begin{aligned} & f(x)=3 x^4-8 x^3+12 x^2-48 x+25 \\ & \therefore \quad f(x)=12 x^3-24 x^2+24 x-48 \\ & \quad=12\left[x^3-2 x^2+2 x-4\right]=12\left[(x-2)\left(x^2+2\right)\right]\end{aligned}$
For maximum and minimum value of the function $f^{\prime}(x)=0$
$\begin{aligned}& \Rightarrow x=2 \text {. Now } f^{\prime \prime}(x)=12\left[3 x^2-4 x+2\right] \\
& \therefore f^{\prime \prime}(2)=12[12-8+2]=72\gt0\end{aligned}$
Hence the function is minimum at $x=2$
Minimum value of $f(x)$ on $[0,3]$
$=\min \{f(0), f(2), f(3)\}=\min \{25,-39,16\}=-39 \text {. }$

Topic: Mathematics, Application of Derivatives, JEE Main
The minimum value of p for which the lines 3x-4y=2, 3x-4y=12, 12x+5y=7 and 12x+5y=p constitute the sides of a rhombus is

Options:
A 33
B 19
C -19
D 9

The correct answer is:
-19

Solution:
If the parallelogram is a rhombus, the distance between pair of parallel sides is equal.
Hence, 12-232+42=±p-7122+52

⇒p-7=±26

⇒p=33 or -19

Hence, the minimum value of p is -19

Topic: Mathematics, Straight Lines, JEE Main
The minimum value of the expression 3x+2y (âx,y>0), where xy2=10, occurs when the value of y is equal to

Options:
A 10
B 103
C 303
D 130

The correct answer is:
303

Solution:
3x+2y=310y2+2y=fy=30y2+y+y
Minimum value occurs when y=3013 usingAM≥GM

Topic: Mathematics, Application of Derivatives, JEE Main
The minimum value of the expression 3z-3+2z-4 is equal to (where, z is a complex number)

Options:
A 2
B 1.5
C 3
D 1

The correct answer is:
2

Solution:
Given expression is,
3z-1+2z-2
=z-1+2z-1+z-2
Now, minimum z-1+z-2 is equal to 1
when z lies on the line segment joining 1 & 2

and minimum z-1 is equals to 0, when z=1
⇒ Minimum of z-1+2z-1+z-2 is equal to 0+21=2 at z=1

Topic: Mathematics, Complex Number, JEE Main