datavorous/entrance-exam-dataset huggingface dataset:
Explain why
(a) To keep a piece of paper horizontal, you should blow over, not under, it.
(b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers.
(c) The size of a needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection.
(d) A fluid flowing out of a small hole in a vessel results in a backward thurst on the vessel.
(e) a spinning cricket ball in air does not follow a parabolic trajectory.

Options:

None

Solution:
(a) When we blow over the piece of paper, the velocity of air increases. As a result, the pressure on it decreases in accordance with the Bernoulli's theorem whereas the pressure below remains the same (atmospheric pressure). Thus, the paper remains horizontal.
(b) By doing so the area of outlet of water jet is reduced, so velocity of water increases according to equation of continuity $a v=a$ constant.
(c) For a constant height, the Bernoulli's theorem is expressed as
$$
\mathrm{P}+\frac{1}{2} \rho \mathrm{v}^2=\text { constant }
$$
In this equation, the pressure $P$ occur with a single power whereas the velocity occurs with a square power. Therefore, the velocity has more effect compared to the pressure. It is for this reason that needle of the springe controls flow rate better than the thumb pressure exerted by the doctor.
(d) This is because of principle of conservation of momentum. While the flowing fluid carries forward momentum, the vessel gets a backward momentum.
(e) A spinning cricket ball would have followed a parabolic trajectory has there been no air. But because of air the Magnus effect take place. Due to the Magnus effect the spinning cricket ball deviates from its parabolic trajectory.

Topic: Physics, Mechanical Properties of Fluids
Explain why does conductivity of germanium crystals increase on doping with galium?

Options:

None

Solution:
On doping germanium with galium some of the positions of lattice of germanium are occupied by galium. Galium atom has only three valence electrons. Therefore, fourth valency of nearby germanium atom is not satisfied and the place remains vacant.
This place is deficient of electrons and is therefore called electron hole or electron vacancy. Electron from neighbouring atom comes and fills the gap, thereby creating a hole in its original position.
Under the influence of electric field electrons move towards positively charged plates through these holes and conduct electricity. The holes appear to move towards negatively charged plates.

Topic: Chemistry, Solid State
Explain why meiosis and gametogenesis are always interlinked?

Options:

None

Solution:
In sexual reproducing organisms, meiosis occurs during gametogenesis to reduce the diploid number of chromosomes $(2 n)$ to haploid number of chromosomes (n) in the gametes. Thus, gametes are formed as a result of meiosis so that their chromosome number remains constant.

Topic: Biology, Reproduction in Organisms
Explain why the reactance offered by an inductor increases with increasing frequency of an alternating voltage.

Options:

None

Solution:
When current in an inductor change the direction of induced emf. The inductor opposes flow of current through it by developing a back emf according to Lenz's law. The induced voltage has a polarity so as to maintain the current at its present value. If the current is decreasing, the polarity of the induced emf will be so as to increase the current and vice-versa.
Thus, the induced emf is proportional to the rate of change of current. So inductor provide greater reactance to the flow of current through it if the rate of change is faster, i.e., if the frequency is higher. So the reactance of an inductor, is directly proportional to the frequency. Mathematically, the reactance offered by the inductor is
$$
X_L=(2 \pi \nu) L=\omega L
$$

Topic: Physics, Alternating Current
Expression for an electric field is given by E→=4000 x2 i^ V m-1. The electric flux through the cube of side 20 cm when placed in electric field (as shown in the figure) is ______ V cm.

Options:

The correct answer is:
640

Solution:
Electric flux is expressed as ϕ=E→·A→=EAcosθ, here, A is area of the surface and θ is the angle between the electric field lines and the normal to A.

The flux will be only from DEFG surface as on the surface OABC field is zero and for rest of the surface, area vector is perpendicular to field.

So, ϕ=EA

=4000×(0.2)2×(0.2)2

=4000×16×10-4 V m

=640 V cm

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (31 Jan Shift 1)
Expression for time in terms of G (universal gravitational constant), h (Planck constant) and c (speed of light) is proportional to:

Options:
A Ghc3
B hc5G
C Ghc5
D c3Gh

The correct answer is:
Ghc5

Solution:
t=Gα. hβ cr
T=M-1L3T-2αML2T-1βLT-1γ
=M-α+βL3α+2β+γ T-2α-β-γ
- α+β=0
3α+2β+γ=0
-2α-β-γ=1
∴α=12;β=12;γ=-52

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2019 (09 Jan Shift 2)
Extraction of copper by smelting uses silica as an additive to remove.

Options:
A Cu2O
B FeS
C FeO
D Cu2S

The correct answer is:
FeO

Solution:
Smelting is a form of extractive metallurgy to produce a metal from its ore. Smelting uses heat and a chemical reducing agent to decompose the ore, driving off other elements as gasses or slag and leaving just the metal behind. The reducing agent is commonly a source of carbon such as coke, charcoal, and coal.

FeO is gangue and SiO2 is flux to form slag FeSiO3.
FeO+SiO2→FeSiO3

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2016 (10 Apr Online)
Extraction of gold and silver involves leaching with \( \mathrm{CN}^{-} \)ion. Silver or gold is later recovered by:

Options:
A Liquation
B Distillation
C Zone refining
D Displacement with \( \mathrm{Zn} \)

The correct answer is:
Displacement with \( \mathrm{Zn} \)

Solution:
Zinc metal can displace Ag and Au as Zn being more reactive than Ag and Au. The following reactions are involved in the extraction of gold or silver from their native ores.
From Native ore,
4 Ag + 8 NaCN + 2 H 2 O + O 2 → Leaching 4 Na Ag CN 2 + 4 NaOH Soluble Sodium dicyanoargentate (I)
2 Na Ag CN 2 + Zn → Displacement Na 2 Zn CN 4 + 2 Ag ↓

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Extraction of zinc from zinc blende is achieved by

Options:
A
roasting followed by self-reduction

B
electrolytic reduction

C
roasting followed by reduction with carbon

D
roasting followed by reduction with another metal

The correct answer is:
roasting followed by reduction with carbon

Solution:
Extraction of $\mathrm{Zn}$ from zinc blends is achieved by roasting followed by reduction with carbon. Reactions involved are :
$$
\begin{aligned}
& 2 \mathrm{ZnS}+3 \mathrm{O}_2 \stackrel{\text { heat }}{\longrightarrow} 2 \mathrm{ZnO}+2 \mathrm{SO}_2 \text { (Roasting) } \\
& \mathrm{ZnO}+\mathrm{C}(\text { Coke }) \stackrel{\text { heat }}{\longrightarrow} \mathrm{Zn}+\mathrm{CO} \text { (Re duction) }
\end{aligned}
$$

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2012 (19 May Online)
Extrusion of second polar body from egg nucleus occurs

Options:
A after entry of sperm, before completion of fertilization
B after completion of fertilization
C before the entry of sperm
D has no relation with sperm entry

The correct answer is:
after entry of sperm, before completion of fertilization

Solution:
In oogenesis the primary oocyte forms secondary oocyte just before ovulation. This secondary oocyte as formed starts Meiosis 2 but this is arrested at metaphase 2 and is ovulated at this stage. It enters the oviduct and remains viable for one day and then sheds in the form of menstruation alongwith the endometrial lining. Thus secondary oocyte sheds in the arrested phase only. When in the oviduct if the sperm enters there are chances of fertilization. The sperm swims the vagina, cervix and uterus are enters the oviduct where it approaches the ovum membrane. As the sperm touches the membrane the MAF (Metaphase arrested factor) disappears and thus Meiosis 2 completes following this, syngamy i.e. fusion of nuclei takes place and thus fertilization is completed.

Topic: Biology, Human Reproduction, NEET
$f:[-\infty, 0] \rightarrow[0, \infty)$ is defined as $f(x)=x^2$. The domain and range of its inverse is

Options:
A Domain $\left(f^{-1}\right)=[0, \infty)$, Range of $\left(f^{-1}\right\}=(-\infty, 0]$
B Domain of $\left(f^{-1}\right\}=[0, \infty)$, Range of $\left(f^{-1}\right)=(-\infty, \infty]$
C Dornain of $\left(f^{-1}\right\}=[0, \infty)$, Range of $\left(f^{-1}\right\}=(0, \infty)$
D $f^{-1}$ does not exist

The correct answer is:
Domain $\left(f^{-1}\right)=[0, \infty)$, Range of $\left(f^{-1}\right\}=(-\infty, 0]$

Solution:
We have a function $f:(-\infty, 0] \rightarrow[0, \infty)$ defined as $f(x)=x^2$.
The graph of above function is
Since, each line parallel to $x$ axis cuts the above curve at maximum one point, therefore $f$ is one one. Also from the graph it is clear that range $f=[0, \infty)$
$\therefore f$ is onto.
Thus, $f$ is invertible.
Hence, $f^{-1}:[0, \infty) \rightarrow(-\infty, 0]$
$\Rightarrow$ Domain $\left\langle f^{-1}\right)=[0, \infty)$ and Range of $\left(f^{-1}\right)=(-\infty, 0]$

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2017
f:-1,1→R be a function defined by fx=x2cosπx for x≠00 for x=0 The set of points where f is not differentiable is

Options:
A {x∈[-1,1]:x≠0}
B {x∈-1,1:x=0 or x=22n+1,n∈Z}
C x∈-1,1:x=22n+1,n∈Z
D -1,1

The correct answer is:
x∈-1,1:x=22n+1,n∈Z

Solution:
We have,

fx=x2cosπx for x≠00 for x=0

fx is not differentiable at cosπx=0

cosπx=0⇒πx=2n+1π2

⇒x=22n+1

∵fx is not differentiable at

x∈-1,1:x=22n+1,n∈Z

Topic: Mathematics, Continuity and Differentiability, JEE Main
$f:[1,3] \rightarrow R$ is a function defined as $f(x)=x^3+a x^2+b x$. If $f(\mathrm{l})-f(3)=0$ and $f^{\prime}\left(\frac{2 \sqrt{3}+1}{\sqrt{3}}\right)=0$. Then, $a-b$ is equal to

Options:
A 5
B -17
C $4 \sqrt{3}$
D $-2 \sqrt{3}$

The correct answer is:
-17

Solution:
Given, $f(x)=x^3+a x^2+b x$
Also, $f(1)-f(3)=0$
$$
\begin{aligned}
\Rightarrow & f(1) =f(3) \Rightarrow 1+a+b=27+9 a+3 b \\
\Rightarrow & 8 a+2 b =-26
\end{aligned}
$$
Now, $\quad f^{\prime}(x)=3 x^2+2 a x+b$
$$
\begin{aligned}
& \therefore f^{\prime}\left(\frac{2 \sqrt{3}+1}{\sqrt{3}}\right)=3\left(\frac{2 \sqrt{3}+1}{\sqrt{3}}\right)^2+2 a\left(\frac{2 \sqrt{3}+1}{\sqrt{3}}\right)+b=0 \\
& \Rightarrow 12+1+4 \sqrt{3}+\frac{4 a \sqrt{3}+2 a}{\sqrt{3}}+b=0 \\
& \Rightarrow 13 \sqrt{3}+12+4 a \sqrt{3}+2 a+b \sqrt{3}=0 \\
& \Rightarrow 13 \sqrt{3}+12+4 a \sqrt{3}+2 a+(-13-4 a) \sqrt{3}=0
\end{aligned}
$$
\{from Eq. (i)\}
$$
\Rightarrow \quad a=\frac{-12}{2}=-6
$$
$\begin{array}{ll}\therefore & b=-13-4 a=-13+24=11 \\ \therefore & a-b=-6-11=-17\end{array}$

Topic: Mathematics, Functions, JEE Main
$f:[-2,2] \rightarrow[-2,2]$ and $g:[-2,2] \rightarrow[0,4]$ are two functions defined as $f(x)=\left\{\begin{array}{l}-2,-2 \leq x \leq 0 \\ x^2-2,0 \leq x \leq 2\end{array}\right.$ and $g(x)=|f(x)|+f(|x|)$, then

Options:
A f and g are injective mappings
B f and g are surjective mappings
C f is bijective mapping and g is injective mapping
D f is not bijective mapping and g is surjective mapping

The correct answer is:
f and g are surjective mappings

Solution:
Given, $f(x)=\left\{\begin{array}{cc}-2, & -2 \leq x \leq 0 \\ x^2-2, & 0 \leq x \leq 2\end{array}\right.$
$\begin{gathered}\therefore \quad|f(x)|=\left\{\begin{array}{cc}2, & -2 \leq x \leq 0 \\ 2-x^2, & 0 \leq x \leq \sqrt{2} \\ x^2-2, & \sqrt{2} < x \leq 2\end{array}\right. \\ \therefore g(x)=|f(x)|+f(|x|)=\left\{\begin{array}{cc}x^2, & -2 \leq x \leq 0 \\ 0, & 0 \leq x \leq \sqrt{2} \\ 2\left(x^2-2\right), & \sqrt{2} < x \leq 2\end{array}\right.\end{gathered}$
Here, $g(x)$ is not one-one but onto because range $=[0,4]=$ codomain .

Topic: Mathematics, Functions, JEE Main
$f:[-2,2] \rightarrow[-2,2]$ and $g:[-2,2] \rightarrow[0,4]$ are two functions defined as $f(x)=\left\{\begin{array}{l}-2,-2 \leq x \leq 0 \\ x^2-2,0 \leq x \leq 2\end{array}\right.$ and $g(x)=|f(x)|+f(|x|)$, then

Options:
A f and g are injective mappings
B f and g are surjective mappings
C f is bijective mapping and g is injective mapping
D f is not bijective mapping and g is surjective mapping

The correct answer is:
f and g are surjective mappings

Solution:
Given, $f(x)=\left\{\begin{array}{cc}-2, & -2 \leq x \leq 0 \\ x^2-2, & 0 \leq x \leq 2\end{array}\right.$
$\begin{gathered}\therefore \quad|f(x)|=\left\{\begin{array}{cc}2, & -2 \leq x \leq 0 \\ 2-x^2, & 0 \leq x \leq \sqrt{2} \\ x^2-2, & \sqrt{2} < x \leq 2\end{array}\right. \\ \therefore g(x)=|f(x)|+f(|x|)=\left\{\begin{array}{cc}x^2, & -2 \leq x \leq 0 \\ 0, & 0 \leq x \leq \sqrt{2} \\ 2\left(x^2-2\right), & \sqrt{2} < x \leq 2\end{array}\right.\end{gathered}$
Here, $g(x)$ is not one-one but onto because range $=[0,4]=$ codomain .

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2021 (06 Aug Shift 1)
FeCl3 on reaction with K4[FeCN)6 in aqueous solution gives blue colour. These are separated by a semipermeable membrane AB as shown. Due to osmosis there is

Options:
A blue colour formation in side X
B blue colour formation in side Y
C blue colour formation in both of the side X and Y
D no blue colour formation

The correct answer is:
no blue colour formation

Solution:
There will be no blue colour formation because in the process of osmosis, solvent molecules migrate across the membrane and not solute molecules.

Topic: Chemistry, Solutions, JEE Main
[Fe(H2O)6]2+ and [Fe(CN)6]4- differ in

Options:
A geometry, magnetic moment
B geometry, hybridization
C magnetic moment, colour
D hybridisation, number of d-electrons

The correct answer is:
magnetic moment, colour

Solution:
[Fe(H2O)6]2+ and [Fe(CN)6]4- differ in magnetic moment and colour.
Fe2+ in [Fe(H2O)6]2+
Colour: Pale green μ=4.9 B.M.; octahedral
Fe2+ in [Fe(CN)6]4-
Colour: Yellow; μ=0 ; octahedral.

Topic: Chemistry, Coordination Compounds, JEE Main
F=Î±Î²e-xÎ±kt
k= Boltzmann constant
t= temperature
x= distance
The dimensions of Î² is

Options:
A M L T-4
B M-2 L-2 T4
C M L T-2
D M2 L T-1

The correct answer is:
M-2 L-2 T4

Solution:
F=αβe-xαKT
xα=ML2T-2
α=M-1L-1T-2
MLT-2=M-1L-1T2β-1
β=M-2 L-2 T4

Topic: Physics, Units and Dimensions, JEE Main
$\Delta_{\mathrm{f}} \mathrm{G}^{\circ}$ at $500 \mathrm{~K}$ for substance ' $\mathrm{S}$ ' in liquid state and gaseous state are $+100.7 \mathrm{kcal} \mathrm{mol}^{-1}$ and $+103$ $\mathrm{kcal} \mathrm{mol}^{-1}$, respectively. Vapour pressure of liquid ' $\mathrm{S}$ ' at $500 \mathrm{~K}$ is approximately equal to:
$$
\left(\mathrm{R}=2 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right) \text {. }
$$

Options:
A
$100 \mathrm{~atm}$

B
$1 \mathrm{~atm}$

C
$10 \mathrm{~atm}$

D
$0.1 \mathrm{~atm}$

The correct answer is:
$10 \mathrm{~atm}$

Solution:
$$
\begin{aligned}
&\Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=\Delta_{\mathrm{f}} \mathrm{G}^{\circ} \text { (vapour) }-\Delta_{\mathrm{f}} \mathrm{G}^{\circ} \text { (liquid) } \\
&\Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=103-100.7=2.3 \mathrm{kcal} / \mathrm{mol} \\
&=2300 \mathrm{cal} / \mathrm{mol} \\
&\Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=-\mathrm{RT} \ln \mathrm{K} \\
&2300 \mathrm{cal} / \mathrm{mol}=-2 \mathrm{cal} / \mathrm{mol} \mathrm{K} \times 500 \mathrm{~K} \times \ln \mathrm{K} \\
&\quad \ln \mathrm{K}=2.3 \\
&\mathrm{~K}=10 \mathrm{~atm}=\text { Vapour pressure of liquid ' } \mathrm{S} \text { ' }
\end{aligned}
$$
$\therefore \quad$ Vapour pressure of liquid ' $\mathrm{S}$ ' at $500 \mathrm{~K}$ is approximately equal to $10 \mathrm{~atm}$.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
f,g:R→R be two real valued function defined as fx=-x+3,x<0ex,x≥0 and gx=x2+k1x,x<04x+k2,x≥0, where k1 and k2 are real constants. If gof is differentiable at x=0, then gof-4+gof4 is equal to

Options:
A 4e4+1
B 22e4+1
C 4e4
D 22e4-1

The correct answer is:
22e4-1

Solution:
Here fx=x+3;x<-3-x+3;-3≤x<0ex;x≥0

and gx=x2+k1x;x<04x+k2;x≥0

Now gfx=fx2+k1fx;fx<04fx+k2;fx≥0

i.e. gfx=x+32+k1x+3;x<-3x+32-k1x+3;-3≤x<04ex+k2;x≥0

For continuity at x=0,

gof0=gf0-=gf0+

i.e. 4+k2=9-3k1=4+k2

⇒3k1+k2=5 ⋯i

Now differentiating, we get

gfx'=2x+3+k1;x<-32x+3-k1;-3≤x<04ex;x≥0

At x=0,

6-k1=4

⇒k1=2 …ii

∴ k1=2, k2=-1 (from i)

So gofx=x+32+2x+3;x<-3x+32-2x+3;-3≤x<04ex-1;x≥0

Hence, gof-4+gof4=4e4-2=22e4-1

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2022 (26 Jun Shift 1)
f is defined in [-5, 5] as f(x) = x if x is rational and = -x is irrational. Then

Options:
A
f(x) is continuous at every x, except x = 0

B
f(x) is discontinuous at every x, except x = 0

C
f(x) is continuous everywhere

D
f(x) is discontinuous everywhere

The correct answer is:
f(x) is discontinuous at every x, except x = 0

Solution:
No Solution Available

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2002
' $F$ ' is the force between the two identical charged particles placed at a distance ' $\mathrm{Y}$ ' from each other. If the distance between the charges is reduced to half the previous distance, then force between them becomes

Options:
A $\frac{F}{4}$
B $4 \mathrm{~F}$
C $2 \mathrm{~F}$
D $\frac{\mathrm{F}}{2}$

The correct answer is:
$4 \mathrm{~F}$

Solution:
$$
\begin{aligned}
& \mathrm{F}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}^2}{\mathrm{r}^2} \\
& \therefore \frac{\mathrm{F}_2}{\mathrm{~F}_1}=\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^2=(2)^2=4 \\
& \therefore \mathrm{F}_2=4 \mathrm{~F}_1
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main
$\boldsymbol{f}: \boldsymbol{R} \rightarrow \boldsymbol{R}$ and $g:[0, \infty) \rightarrow R$ defined by $f(x)=x^2$ and $g(x)=\sqrt{x}$. Which one of the following is not true?

Options:
A $(f \circ g)(-4)=4$
B $(f \circ g)(2)=2$
C $(g \circ f)(-2)=2$
D $(g \circ f)(4)=4$

The correct answer is:
$(f \circ g)(-4)=4$

Solution:
$f o g(x)=f(g(x))$
As domain of $g(x) \rightarrow[0, \infty)$, so domain of $f \circ g(\lambda)$ also $[0, \infty)$
$\therefore \quad f(g(x))=(\sqrt{x})^2$
$\begin{aligned} & g \circ f(x)=g(f(x)) \\ & \qquad g \circ f(x)=\sqrt{x^2}=|x|\end{aligned}$
Hence, $f o g(-4)=(\sqrt{-4})^2$ is not define.

Topic: Mathematics, Functions, KCET, KCET 2023
f:R→R, fx=λx+sin⁡x is onto if λ is an element of the set P and fx is one-one if λ is an element of the set Q, then (given, λ is a real number)

Options:
A P=Q
B P⊂Q
C P-Q=0
D Q⊂P

The correct answer is:
Q⊂P

Solution:
f'x=λ+cos⁡x
If fx is one-one, then λ≥-cos⁡x or λ≤-cos⁡x ∀x∈R⇒λ∈-∞,-1∪1,∞
Further, fx has range R for λ∈R-0
Hence, Q⊂P

Topic: Mathematics, Functions, JEE Main
$f: R \rightarrow R$ is a function such that $f(0)=1$ and for all $x, y \in R f(x y+1)=f(x) f(y)-f(y)$, $-x+2$, then $\frac{d f}{d x}$ at $x=e$ is

Options:
A 0
B -1
C $e$
D 1

The correct answer is:
1

Solution:
Given functional relation is,
$$
f(x y+1)=f(x) f(y)-f(y)-x+2
$$
Now, put $y=0$, then we are getting
$$
\begin{gathered}
f(\mathrm{l})=f(x) \cdot f(0)-f(0)-x+2 \\
\Rightarrow \quad f(\mathrm{l})=f(x)-1-x+2\{\because f(0)=1\}
\end{gathered}
$$
On differentiating with respect to ' $x$ '
$$
0=\frac{d f(x)}{d x}-0-1+0
$$
$$
\Rightarrow \quad \frac{d f(x)}{d x}=1
$$
So, $\frac{d f(x)}{d x}($ at $x=e)=1$

Topic: Mathematics, Differentiation, JEE Main
$f\left(x_1\right)-f\left(x_2\right)=f\left(\frac{x_1-x_2}{1-x_1 x_2}\right)$ for $x_1, x_2 \in[-1,1]$, then $f(x)$ is

Options:
A $\log \frac{(1-x)}{(1+x)}$
B $\tan ^{-1} \frac{(1-x)}{(1+x)}$
C $\log \frac{(1+x)}{(1-x)}$
D $\tan ^{-1} \frac{(1+x)}{(1-x)}$

The correct answers are:
$\log \frac{(1-x)}{(1+x)}$, $\tan ^{-1} \frac{(1-x)}{(1+x)}$, $\log \frac{(1+x)}{(1-x)}$

Solution:
When $x_1=-1$ and $x_2=1$, then
$f(-1)-f(1)=f\left[\frac{-1-1}{1+1(1)}\right]=f(-1) \Rightarrow f(1)=0$
Which is satisfied when $f(x)=\tan ^{-1}\left(\frac{1-x}{1+x}\right)$
When $x_1=x_2=0$, then
$f(0)-f(0)=f\left[\frac{0-0}{1-0}\right]=f(0) \Rightarrow f(0)=0$
When $x_1=-1$ and $x_2=0$ then
$f(-1)-f(0)=f\left(\frac{-1-0}{1-0}\right)=f(-1) \Rightarrow f(0)=0$
Which is satisfied when
$f(x)=\log \left(\frac{1-x}{1+x}\right) \text { and } f(x)=\log \left(\frac{1+x}{1-x}\right)$

Topic: Mathematics, Functions, JEE Main
$f(x)=\frac{x}{1+x}$, then $f^{-1}(x)$ is equal to

Options:
A $\frac{(1+x)}{x}$
B $\frac{1}{(1+x)}$
C $\frac{(1+x)}{(1-x)}$
D $\frac{x}{(1-x)}$

The correct answer is:
$\frac{x}{(1-x)}$

Solution:
$\begin{aligned} & f(x)=\frac{x}{1+x} . \text { Let } y=f(x) \Rightarrow x=f^{-1}(y) \\ & \quad y=\frac{x}{1+x} \Rightarrow y+y x=x \Rightarrow x=\frac{y}{1-y} \\ & \therefore \quad f^{-1}(y)=\frac{y}{1-y} \Rightarrow f^{-1}(x)=\frac{x}{1-x} .\end{aligned}$

Topic: Mathematics, Functions, JEE Main
$f(x)= \begin{cases}2 a-x & \text { in }-a < x < a \\ 3 x-2 a & \text { in } a \leq x\end{cases}$
Then, which of the following is true ?

Options:
A $f(x)$ is discontinuous at $x=a$
B $f(x)$ is not differentiable at $x=a$
C $f(x)$ is differentiable at $x \geq a$
D $f(x)$ is continuous at all $x < a$

The correct answer is:
$f(x)$ is not differentiable at $x=a$

Solution:
Given, $\quad \mathrm{f}(\mathrm{x})=\left\{\begin{array}{llc}2 \mathrm{a}-\mathrm{x} & \text { in } & -a < x < a \\ 3 \mathrm{x}-2 \mathrm{a} & \text { in } & \mathrm{a} \leq \mathrm{x}\end{array}\right.$
At $x=a$
$\mathrm{LHL}=\lim _{\mathrm{x} \rightarrow \mathrm{a}^{-}} \mathrm{f}(\mathrm{x})$
$$
=\lim _{x \rightarrow a} 2 a-x=a
$$
$$
\begin{aligned}
\mathrm{RHL} &=\lim _{\mathrm{x} \rightarrow \mathrm{a}^{+}} \mathrm{f}(\mathrm{x}) \\
&=\lim _{\mathrm{x} \rightarrow \mathrm{a}} 3 \mathrm{x}-2 \mathrm{a}=\mathrm{a}
\end{aligned}
$$
and $\quad \mathrm{f}(\mathrm{a})=3(\mathrm{a})-2 \mathrm{a}=\mathrm{a}$
$\therefore \quad \mathrm{LHL}=\mathrm{RHL}=\mathrm{f}(\mathrm{a})$
Hence, it is continuous at $\mathrm{x}=\mathrm{a}$
Again, at $x=a$
$$
\begin{aligned}
\mathrm{LHD} &=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(\mathrm{a}-\mathrm{h})-\mathrm{f}(\mathrm{a})}{-\mathrm{h}} \\
&=\lim _{\mathrm{h} \rightarrow 0} \frac{2 \mathrm{a}-(\mathrm{a}-\mathrm{h})-\mathrm{a}}{-\mathrm{h}}=\frac{\mathrm{h}}{-\mathrm{h}}=-1
\end{aligned}
$$
$$
\text { and } \begin{aligned}
\mathrm{RHD}=& \lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(\mathrm{a}+\mathrm{h})-\mathrm{f}(\mathrm{a})}{h} \\
=& \lim _{\mathrm{h} \rightarrow 0} \frac{3(\mathrm{a}+\mathrm{h})-2 \mathrm{a}-\mathrm{a}}{\mathrm{h}}=3 \\
& \mathrm{LHD} \neq \mathrm{RHD}
\end{aligned}
$$
Hence, it is not differentiable at $\mathrm{x}=\mathrm{a}$.
Alternate Solution
Let $y=f(x)= \begin{cases}2 a-x & \text { in }-a < x < a \\ 3 x-2 a & \text { in } \quad a \leq x\end{cases}$
It is clear from the graph that $\mathrm{f}(\mathrm{x})$ is continuous everywhere it is also differentiable everywhere except a point $\mathrm{A}$ at $\mathrm{x}=\mathrm{a}$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
$\int \frac{f^{\prime}(x)}{[f(x)]^2} d x=$

Options:
A $-[f(x)]^{-1}+c$
B $\log [f(x)]+c$
C $e^{f(x)}+c$
D None of these

The correct answer is:
$-[f(x)]^{-1}+c$

Solution:
Put $f(x)=t \Rightarrow f^{\prime}(x) d x=d t$, then
$\int \frac{f^{\prime}(x)}{|f(x)|^2} d x=\int \frac{1}{t^2} d t=-\frac{1}{t}+c=-\frac{1}{f(x)}+c .$

Topic: Mathematics, Indefinite Integration, JEE Main
$f(x)=\frac{2 x+1}{3 x-2}$, then $(f \circ f)(2)$ is equal to

Options:
A $1$
B $3$
C $4$
D $2$

The correct answer is:
$2$

Solution:
Here
$\begin{aligned}
& f(2)=\frac{5}{4} \\
& (f o f)(2)=f(f(2))=f\left(\frac{5}{4}\right)=\frac{2 \times \frac{5}{4}+1}{3 \times \frac{5}{4}-2}=2
\end{aligned}$

Topic: Mathematics, Functions, JEE Main
$f(x)=\left\{\begin{array}{ll}\frac{2}{5-x}, & \text { when } x \lt 3 \\ 5-x, & \text { when } x\gt3\end{array}\right.$, then

Options:
A $\lim _{x \rightarrow 3+} f(x)=0$
B $\lim _{x \rightarrow 3-} f(x)=0$
C $\lim _{x \rightarrow 3+} f(x) \neq \lim _{x \rightarrow 3-} f(x)$
D None of these

The correct answer is:
$\lim _{x \rightarrow 3+} f(x) \neq \lim _{x \rightarrow 3-} f(x)$

Solution:
$\lim _{x \rightarrow 3+} f(x)=5-3=2, \lim _{x \rightarrow 3-} f(x)=\frac{2}{5-3}=1$

Topic: Mathematics, Limits, JEE Main