datavorous/entrance-exam-dataset huggingface dataset:
Pick out incorrect statement

Options:
A In a electrolysis experiment, α -amino acids migrate at the isoelectric point towards electrodes
B p-aminobenzenesulphonic acid as a dipolar ion; while p-aminobenzoic acid does not
C Sulphanilic acid is soluble in base, but not in acid
D H3N+CH2COOH is more acidic than RCH2COOH

The correct answer is:
In a electrolysis experiment, α -amino acids migrate at the isoelectric point towards electrodes

Solution:
A. The pH at which Anion = Cation is called isoelectric point. At isoelectric point, α -amino acids do not migrate when electric field is applied.
B. -SO3H is strongly acidic and donates H+ to weakly basic arylamino group. ArCOOH is not acidic enough to transfer H+ to the arylamino group.
C. In p-H3N+C6H4SO3,H3N+ is acid enough to transfer H+ to bases to give the soluble anion, p-H2N-C6H4SO3.-SO3 is too feebly basic and cannot accept H+ from acids.
D. α-H3N+ group increases acidity, because of its electron-withdrawing inductive effect.

Topic: Chemistry, Biomolecules, JEE Main
Pick out the complex compound in which the central metal atom obeys EAN rule strictly

Options:
A $\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$
B $\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]$
C $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_3$
D $\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right] \mathrm{SO}_4$

The correct answer is:
$\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$

Solution:
In complex $\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$ the $\mathrm{Fe}$ obey EAN rule strictly.

Topic: Chemistry, Coordination Compounds, JEE Main
Pick out the complex compound in which the central metal atom obeys $\mathrm{EAN}$ rule strictly.

Options:
A $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$
B $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$
C $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3}$
D $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{SO}_{4}$

The correct answer is:
$\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$

Solution:
In $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$, the central metal atom obeys EAN rule.
EAN $=Z-X+Y=26-2+12=36$
where,
$Z$ = atomic number
$X=$ number of electrons lost, while forming metal ion.
$Y=$ number of electron donated by ligands.

Topic: Chemistry, Coordination Compounds, JEE Main
Pick out the correct set of statements for HI.
1. It reduces iodic acid to I2
2. It reduces H2SO4 to SO2
3. It decolourises acidified KMnO4
4. It liberates I2 with CuSO4 solution

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

The correct answer is:
1, 2, 3, 4

Solution:
Here all statements are correct here regarding HI as follows
2KMnO4+3H2SO4+10HI→K2SO4+2MnSO4+8H2O+5I2
H2SO4+2HI→SO2+I2+2H2O
HIO3+5HI→3I2+3H2O
2CuSO4+4HI→2CuI2+2H2SO4
2CuI2→Cu2I2+I2

Topic: Chemistry, d and f Block Elements, JEE Main
Pick out the correct statement from the following and choose answer from the codes given below:
I. Prop - 1,2 - diene is conjugated diene
II. Hexa - 1,3 - diene is conjugated diene
III. Buta - 1, 3 -diene is an isolated diene
IV. Prop - 1,2 -diene is a cumulative diene

Options:
A I , II
B II, IV
C I, II
D II, III

The correct answer is:
I, II

Solution:
II, IV
Conjugated diene - have attenate double single bonds
Cumulative diene - adjacent double bonds
Isolated diene - double bonds far from them each other
1) $\mathrm{H}_2 \mathrm{C}=\mathrm{C}=\mathrm{CH}_2$
Prop - 1, 2- diene
as both the double are on some C-atom so it is not conjugated diene, so statement is incorrect.
2) $\mathrm{H}_2 \mathrm{C}=\mathrm{HC}-\mathrm{HC}=\mathrm{CH}-\mathrm{CH}_2-\mathrm{CH}_3$
Hexa - 1,3 diene.
As it has alternate double & single bonds, it is a conjugated diene. So the statement is correct.
3) $\mathrm{H}_2 \mathrm{C}=\mathrm{HC}-\mathrm{HC}=\mathrm{CH}_2$
Bute $-1,3$ diene.
As it has alternative double & single bonds it is a conjugated diene \& not isolated diene so this statement is incorrect.
4) $\mathrm{H}_2 \mathrm{C}=\mathrm{C}=\mathrm{CH}_2$
Prop - 1,2 diene
As both double bonds are on same c-atom so it is a cumulative diene. So the statement is correct.

Topic: Chemistry, Hydrocarbons, JEE Main
Pick out the correct statement set for transition metals
1. They form interstitial compounds
2. 5d-elements have higher energies than 3d or 4d-elements
3. Zr and Hf have almost identical atomic and ionic radii
4. They have low melting and boiling points (or low enthalpies of atomization)

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

The correct answer is:
1, 2, 3

Solution:
Here statement 1, 2 and 3 are correct as follows

Because small size, atoms of H, C, N occupy the vacant spaces in the crystal lattice of transition metal.

This due to the greater effective nuclear charge acting on outer valence electrons, because of weak shielding of nucleus by 4f-electrons

The normal size increase from Zr to Hf is almost exactly balanced by lanthanide contraction

Transition metals have high enthalpies of atomization, because they form strong metallic bonds, due to participation of (n – 1) d-electrons along with ns-electrons in metallic bonding.

Topic: Chemistry, d and f Block Elements, JEE Main
Pick out the correct statement.

Options:
A Both glycine and p-aminobenzoic acid form dipoler ions
B Both sulphanilic acid and p-aminobenzoic acid form dipolar ions
C Glycine does not form dipolar ion, but p-aminobenzoic acid forms dipolar ion
D Both sulphanilic acid and glycine form dipolar ions, but p-aminobenzoic acid does not form dipolar ion

The correct answer is:
Both sulphanilic acid and glycine form dipolar ions, but p-aminobenzoic acid does not form dipolar ion

Solution:
The aliphatic -NH2 is sufficiently basic to accept H+ from -COOH group in glycine, NH2CH2COOH . The -COOH is not strong enough to donate H+ to the weakly basic aromatic NH2 in p-aminobenzoic acid, The -SO3H is a sufficiently strong acid to donate H+ to weakly basic aromatic NH2 .

Topic: Chemistry, Biomolecules, JEE Main
Pick out the incorrect statement

Options:
A MnO2 dissolves in conc. HCl, but does not form Mn4+ ions
B MnO2 oxidizes hot concentrated H2SO4 liberating oxygen
C K2MnO4 is formed, when MnO2 is fused KOH is oxidized by air, KNO3,PbO2 or NaBiO3
D Decomposition of acidic KMnO4 is not catalysed by sunlight

The correct answer is:
Decomposition of acidic KMnO4 is not catalysed by sunlight

Solution:
Decomposition of acidic KMnO4 is catalysed by sunlight.
4MnO4-+4H+→4MnO2+3O2+2H2O
A. On dissolution, MnO2 forms Mn2+ ions and HCl is oxidized to Cl2 .
MnO2+4HCl-→Mn2++2Cl-+Cl2+H2O
B. H2SO4 is oxidized by MnO2
2MnO2+2H2SO4→2MnSO4+O2+2H2O
C. MnO2+2KOH+12O2→K2MnO4Green+H2O

Topic: Chemistry, d and f Block Elements, JEE Main
Pick out the incorrect statement.

Options:
A Mg also burns in gases such as CO2 and SO2
B Excess of CO2 when passed in lime-water turns it milky
C MgCl2.6H2O on heating hydrolyses to form MgO (magnesia), which is refractory
D Alkaline-earth metals are denser and harder than alkali metals

The correct answer is:
Excess of CO2 when passed in lime-water turns it milky

Solution:
Here statement in option B is incorrect as when CO2 is passed in lime-water, it becomes milky due to the formation of CaCO3 but excess of CO2 leads to the formation of clear solution, due to the formation of soluble Ca(HCO3)2 .
A. Mg+CO2→2MgO+C
and 2Mg+SO2→2MgO+S
D. Inter-atomic forces (or metallic bonding) in alkaline-earth metals are stronger than in alkali metals.

Topic: Chemistry, Practical Chemistry, JEE Main
Pick out the incorrect statement.

Options:
A I2O5 is formed by heating HIO3 to 170°C
B I2O5 is stable to heat
C I2O5 is used in the estimation of CO
D I2 combines with O3 to form I4O9 . When heated above 75°C , it I4O9 decomposes to form I2O5

The correct answer is:
I2O5 is stable to heat

Solution:
Here statement B is incorrect as I2O5 is unstable to heat
It decomposes when heated to 300°C .
2I2O5→300°C2I2+5O2
Rest other statements are correct as follows
(i) 2HIO3→170°CI2O5+H2O
(iii) I2O5 oxidizes CO to CO2 quantitatively at room temperature, liberating I2 , which can be titrated with sodium thiosulphate.
I2O5+5CO→5CO2+I2
(iv) 2I2+3O3→I4O9
4I4O9→>75°C6I2O5+2I2+3O2

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Pick out the incorrect statement.

Options:
A The geometry around 'N' atom in trimethylamine is pyramidal
B The geometry around N atom in trisilylamine is planar
C The nitrogen atom in trimethylamine is sp3 -hybridized; whilst in trisilylamine it is sp2 -hybridized
D Trisilylamine has donor properties; whilst trimethylamine has no donor properties

The correct answer is:
Trisilylamine has donor properties; whilst trimethylamine has no donor properties

Solution:
In trimethylamine, 'N' atom is sp3 -hybridized.

Due to the presence of lone pair in sp3 -hybrid orbital, the geometry of (CH3)3N becomes pyramidal around nitrogen.

This lone pair of electrons can be donated to electron-deficient species, e.g., (CH3)3N→BF3 . This lone pair cannot be transferred to carbon atom, because carbon atom has no d-orbital in its valence shell.
In (SiH3)3N, the 'N' atom is sp2 -hybridized.

The orientation of three sp2 -hybrid orbitals is plane-triangular, therefore, the geometry around 'N' becomes plane-triangular. Here the lone pair of electrons is transferred to vacant d-orbitals to each of three silicon atoms forming pπ-dπ multiple bonds. Consequently, SiH33N has no donor properties. For example
(SiH3)3N+BF3→ No reaction.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Pick out the incorrect statement regarding halogens.

Options:
A Chlorine is hydrolysed by water to form hydrochloric acid and hypochlorous acid
B Bromine and iodine react with NaOH solution to form halide and halate ion
C Chlorine reacts with cold dilute NaOH solution to give sodium chloride and sodium chlorate
D Iodine forms a deep blue colour with starch solution

The correct answer is:
Chlorine reacts with cold dilute NaOH solution to give sodium chloride and sodium chlorate

Solution:
Cl2 reacts with cold dilute NaOH solution to form NaCl and NaOCl.
2NaOH+Cl2→NaCl+NaClO+H2O
Cl2 form s NaCl and NaClO3 only above 75°C
6NaOH+3Cl2→5NaCl+NaClO3+3H2O
A. Cl2aq+2H2O⇌H3O++Cl-+HOCl
B. 6NaOH+3X2→5NaX+NaXO3+3H2;X=Br,I
D. It is thought that I2 molecules reside inside the spiral structure of starch molecule giving deep blue colour.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Pick out the incorrect statement

Options:
A PH4+ ion is tetrahedral like the NH4+ ion and is obtained when PH3 is bonded to proton
B PH4I is one of the most stable salts containing the phosphonium ion. It is also more stable than ammonium salts
C PH4I is decomposed by water to form PH3
D PH3 converts silver salts in solution to silver phosphide, which subsequently reacts to give free metal

The correct answer is:
PH4I is one of the most stable salts containing the phosphonium ion. It is also more stable than ammonium salts

Solution:
Phosphonium iodide is less thermally stable than ammonium salts, because phosphine is less basic than ammonia and decomposition occurs at 60°C .
PH4I→PH3+HI
A. PH3+HI→PH4I
C. PH4I+H2O→PH3+H3O++I-
D. 6Ag++PH3+3H2O→6Ag+H3PO3+6H+

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Pick out the molecule which has zero dipole moment

Options:
A $\mathrm{NH}_3$
B $\mathrm{H}_2 \mathrm{O}$
C $\mathrm{BCl}_3$
D $\mathrm{SO}_2$

The correct answer is:
$\mathrm{BCl}_3$

Solution:
$\mathrm{BCl}_3$ has zero dipole moment because of its trigonal planar geometry.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Pick the correct statement.

Options:
A Relative lowering of vapour pressure is independent of $\mathrm{T}$.
B Osmotic pressure always depends on the nature of solute.
C Elevation of boiling point is independent of nature of the solvent.
D Lowering of freezing point is proportional to the molar concentration of solute.

The correct answer is:
Relative lowering of vapour pressure is independent of $\mathrm{T}$.

Solution:
Relative lowering of vapour pressure is $\frac{\Delta \mathrm{P}}{\mathrm{P}^{\circ}}$, which is equal to mole fraction of solute which is independent of temperature.

Topic: Chemistry, Solutions, JEE Main
Pinus is a gymnosperm because it

Options:
A bears pollen grains
B is a large tree growing in colder areas
C lacks ovary but possesses exposed ovules
D possesses vascular tissues

The correct answer is:
lacks ovary but possesses exposed ovules

Solution:
The word Gymnosperm is derived from Greek words 'gymno' meaning naked and 'sperma' meaning seed. Thus, gymnosperms are naked seed-bearing plants in which seeds are not present inside fruits, as gymnosperms are nonflowering plants in which ovules are borne directly on megasporophyll. Thus, ovules remain exposed before and after fertilization.

Topic: Biology, Plant Kingdom, NEET
Pinus is classified under Gymnosperms because

Options:
A It is a large tree
B It is pollinated by wind
C It has exposed ovules
D It produces seeds and it has narrow leaves

The correct answer is:
It has exposed ovules

Solution:
Gymngosperms are naked seed bearing plants in ovules remains exposed before and after fertilization. These are non flowering plants in which ovules develops directly on megasporophyll.

Topic: Biology, Plant Kingdom, NEET
pK a of a weak acid HA and pK b of a weak base BOH are 3.2 and 3.4 respectively. The pH of their salt AB solution at 25 o C is

Options:
A 6.9
B 7.0
C 1.0
D 7.2

The correct answer is:
6.9

Solution:
Hydrolysis of salt of weak acid and weak base
pH= 1 2 pK w + pK a â pK b
=1214+3.2-3.4
=6.9

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2017 (02 Apr)
Place the following alcohols in decreasing order of rate of dehydration with concentrated H2SO4 :

1. CH3CH2CH(OH)CH2CH2CH3

2. (CH3)2C(OH)CH2CH2CH3

3. (CH3)2C(OH)CH(CH3)2

4. CH3CH2CH(OH)CH(CH3)2

​5. CH3CH2CH2CH2CH2CH2OH

Options:
A 3 > 2 > 4 > 5 > 1
B 3 > 2 > 4 > 1 > 5
C 3 > 2 > 1 > 4 > 5
D 3 > 2 > 1 > 5 > 4

The correct answer is:
3 > 2 > 4 > 1 > 5

Solution:
The alcohols (3) and (2) are both 3o, but alcohol (3) gives a more substituted alkene. Alcohol (4) and (1) are both 2o, but alcohol (4) can give a more substituted alkene and alcohol (5) is 1o. Rate of dehydration of alcohols with concentrated H2SO4 follows the order 3o > 2o > 1o.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Plane polarised light is passed through a polaroid. On viewing through the polaroid; we find that when the polaroid is given one complete rotation about the direction of the light, one of the following is observed

Options:
A the intensity of light gradually decreases to zero and remains at zero
B the intensity of light gradually increases to a maximum and remains at maximum
C there is no change in intensity
D the intensity of light is twice maximum and twice zero

The correct answer is:
the intensity of light is twice maximum and twice zero

Solution:
During rotation →
When θ=0° Intensity goes to maximum
When θ=90° Intensity becomes zero
When θ=180° Intensity become maximum again
When θ=270° Intensity again become zero.
Where θ is the angle of rotation.

Topic: Physics, Wave Optics, NEET
Planet A has mass M and radius R. Planet B has half the mass and half the radius of Planet A. If the escape velocities from the Planets A and B are vA and vB, respectively, then vAvB=n4. The value of n is:

Options:
A 4
B 1
C 2
D 3

The correct answer is:
4

Solution:
ve=2GMR
∴vAvB=2GMR2GM/2R/2=1=n4
⇒n=4

Topic: Physics, Gravitation, JEE Main, JEE Main 2020 (09 Jan Shift 2)
Plants deficient in elemental nitrogen, often show its effect on the biosynthesis of:

Options:
A Proteins
B Nucleic acids
C Vitamins
D More than one option is correct

The correct answer is:
More than one option is correct

Solution:
Nitrogen is the essential nutrient element required by plants in the greatest amount. It is absorbed mainly as NO3–, though some are also taken up as NO2– or NH4+. Nitrogen is required by all parts of a plant, particularly the meristematic tissues and the metabolically active cells. Nitrogen is one of the major constituents of proteins, nucleic acids, vitamins and hormones.

Topic: Biology, Mineral Nutrition, NEET
Plants which produce characteristic pneumatophores and show vivipary belong to:

Options:
A Mesophytes
B Halophytes
C Psammophytes
D Hydrophytes

The correct answer is:
Halophytes

Solution:
This is a salt-tolerant plant group that grows in soil or waters of high salinity, coming into contact with saline water through its roots or by salt spray, such as in saline semi-deserts, mangrove swamps, marshes and sloughs and seashores.

Halophytes growing in saline soils or near the delta of a river, show:

Vivipary which is in-situ seed germination. The seeds, while remaining attached in the mother plant, germinate and then shaded off.

Pneumatophores or the breathing roots are for gaseous exchange.

Topic: Biology, Plant Kingdom, NEET, NEET 2017
Plants with ovaries having only one or a few ovules, are generally pollinated by

Options:
A bees
B butterflies
C birds
D wind

The correct answer is:
wind

Solution:
The transfer of pollen grains from anther to stigma is called pollination. Anemophily is a mode of pollination accomplished through wind, while entomophily is the mode of pollination accomplished by insects like moth, butterflies, wasps, bees, etc. The plants with ovaries having only one or a few ovules are generally pollinated by wind but sometimes by insects.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2012 (Mains)
Plasma membrane helps in

Options:
A Transportation of only water in and out of cell
B Protein synthesis
C Osmoregulation
D Nucleic acid synthesis

The correct answer is:
Osmoregulation

Solution:
All cells are enclosed by a thin, film-like liable membrane called the plasma membrane or plasmalemma. The main function of the plasma membrane is to regulate the flow of materials into and out of the cell (osmoregulation). The membrane is selectively permeable.

Topic: Biology, Cell - The Unit of Life, NEET
Plasma protein which helps in osmotic balance is

Options:
A Albumin
B Globulin
C Fibrinogen
D Heparin

The correct answer is:
Albumin

Solution:
Blood consists of plasma and corpuscles. Plasma forms 55 percent of total blood. It contains three types of plasma proteins. These are Albumin, Globulin and Fibrinogen. Percentage of Albumin is maximum of all the plasma proteins. Its main function is maintenance of Osmotic balance. Globulin is plasma protein responsible for immunity. Fibrinogen acts as clotting factor.

Topic: Biology, Body Fluids and Circulation, NEET
Plasmodesmata are :

Options:
A Connection between adjacent cells
B Lignified cemented layers between cells
C Locomotory structures
D Membranes connecting the nucleus with plasmalemma

The correct answer is:
Connection between adjacent cells

Solution:
No Solution Available

Topic: Biology, Transport in Plants, NEET, NEET 2009 (Mains)
Plasmodesmata connections help in

Options:
A cytoplasmic streaming
B synchronous mitotic divisions
C locomotion of unicellular organism
D movement of substances between cells.

The correct answer is:
movement of substances between cells.

Solution:
A characteristic of most plant cells is the presence of cytoplasmic material that establish continuity between adjacent cells. These bridges are called plasmodesmata. These permit the free circulation of fluid, which is essential to the maintenance of plant cell toxicity.

Topic: Biology, Cell - The Unit of Life, AIIMS, AIIMS 2003
Plates of a parallel plate capacitor, having a potential difference 100 V applied across them, carry a surface charge density of 50 nC cm-2. Spacing between the plates is

Options:
A 329 Î¼m
B 259 Î¼m
C 177 Î¼m
D 125 Î¼m

The correct answer is:
177 Î¼m

Solution:
E=σε0
Potential difference V=Ed
∴d=VE=10050×10-7×8.85×10-12=177×10-6 m=177 μm

Topic: Physics, Electrostatics, JEE Main
Platinum, palladium, iridium etc., are called noble metals because

Options:
A Alfred Noble discovered them
B They are inert towards many common reagents
C They are shining lustrous and pleasing to look at
D They are found in active state

The correct answer is:
They are inert towards many common reagents

Solution:
They are inert towards many common reagents.

Topic: Chemistry, d and f Block Elements, JEE Main
Plot of log (x/m) against log P is a straight line inclined at an angle 45o. When the pressure is 0.5 atm and Freundich parameter k is 10, the amount of solute adsorbed per gram of adsorbent will be (log5 = 0.699)

Options:
A 1 g
B 2 g
C 3 g
D 5 g

The correct answer is:
5 g

Solution:
log x/m â=âlogâkâ+â1/nâlogâP
1/nâ=âtanâ 45 o â=â1;âlogâkâ=âlogâ10â=â1
x/m =  kp 1 n  = 10* 0.5 1 = 5
When m = 1 g, x = 5 g

Topic: Chemistry, Chemical Equilibrium, JEE Main
Plot of log x/m against log p is a straight line inclined at an angle of 45° . When the pressure is 0.5 atm and Freundlich parameter, k is 10, the amount of soluble adsorbed per gram of adsorbent will be (log 5 = 0.6990)

Options:

The correct answer is:
5

Solution:
According to Freundlich equation
xm=kp1n
log⁡xm=log⁡k+1nlog⁡p
So plot of log x/m vs log p is linear wit slope = 1/n and intercept = log k.
Thus 1n=tan⁡θ=tan⁡45°=1
or n = 1
At p = 0.5 atm and k = 10
xm=10×(0.5)1=5
so x = 5 g

Topic: Chemistry, Surface Chemistry, JEE Main
Plot the curve for $|y|=\left|e^{-x}\right|$.

Options:
A

B

C

D

The correct answer is:

Solution:
Here; curve for $y=e^{-x}$ is shown as;
(i) $y=e^{-x} \longrightarrow y=\left|e^{-x}\right|$
(ii) $y=\left|e^{-x}\right| \longrightarrow|y|=\left|e^{-x}\right|$.

Topic: Mathematics, Functions, JEE Main
Plot the curve $y=-\log (-x)$

Options:
A

B

C

D

The correct answer is:

Solution:
As $y=\log x$ is known;
(i) Take image about $y$-axis, for $y=\log (-x)$.
(ii) Take image of $y=\log (-x)$ about $x$-axis, for $y=-\log (-x)$.

Topic: Mathematics, Functions, JEE Main
Plot $y=2 \sin x$.

Options:
A

B

C

D

The correct answer is:

Solution:
We know; $y=\sin x$ and $f(x) \rightarrow a f(x)$
$\Rightarrow$ Stretch the graph of $f(x)$ ' $a$ ' times along $y$-axis.
$\therefore y=2 \sin x \Rightarrow$ stretch the graph of $\sin x, (2)$ times along $y$-axis.

Topic: Mathematics, Functions, JEE Main
Plot $y=e^x+1 ; \quad y=e^x-1$, with the help of $y=e^x$.

Options:
A

B

C

D

The correct answer is:

Solution:
We know; $y=e^x \quad$ (exponential function) could be plotted as;
$y=e^x+1$, is shifted upwards by 1 unit, shown as
Also $y=e^x-1$, is shifted downwards by 1 unit, shown as

Topic: Mathematics, Functions, JEE Main
Plot $y=\sin ^{-1}(2 x)$.

Options:
A

B

C

D

The correct answer is:

Solution:
Here; $y=\sin ^{-1}(2 x)$, is to shrink (or contract) the graph of $\sin ^{-1} x, (2)$ times along $x$-axis.

Topic: Mathematics, Functions, JEE Main
Poaching of tiger is a burning issue in today's world. What implication would this activity have on the functioning of the ecosystem of which the tigers are an integral part?

Options:

None

Solution:
Tiger represents an integral part of the food web and helps in maintaining the stability of the ecosystem. As a carnivore, it keeps a check on the unlimited growth of herbivory and removes sick or old animals from the population as well. It also acts as an indicator of the forest's health.
Saving the tiger is saving the forest. Tiger are canivores of the food chain and cannot live in places where trees or herbivores, have disappeared. Restoring ecosystem can help in restoration of tigers.

Topic: Biology, Ecosystem
Point $(1,2)$ relative to the circle $x^{2}+y^{2}+4 x-2 y-4=0$ is a/ an

Options:
A exterior point
B interior point, but not centre
C boundary point
D centre

The correct answer is:
exterior point

Solution:
We put the co-ordinates of the given point in the given equation of circle $x^{2}+y^{2}+4 x-2 y-4=0$
$\operatorname{At}(1,2)$
$(1)^{2}+(2)^{2}+4(1)-2(2)-4$
$=1+4+4-4-4=1>0$
$\Rightarrow$ Point $(1,2)$ lies out side the circle i.e, an exterior point.

Topic: Mathematics, Circle, JEE Main