datavorous/entrance-exam-dataset huggingface dataset:
The Structural formula of an amino acid, isoleucine is

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Biomolecules, JEE Main
The structural formula of methyl aminomethane is

Options:
A $\left(\mathrm{CH}_3\right)_2 \mathrm{CHNH}_2$
B $\left(\mathrm{CH}_3\right)_3 \mathrm{~N}$
C $\left(\mathrm{CH}_3\right)_2 \mathrm{NH}$
D $\mathrm{CH}_3 \mathrm{NH}_2$

The correct answer is:
$\left(\mathrm{CH}_3\right)_2 \mathrm{NH}$

Solution:
Methylamino methane has 2 methane groups attached to $\mathrm{N}$ directly. Hence option 3 is correct

Topic: Chemistry, Amines, JEE Main
The structure of A in the given reaction is

​​​​​​​

Options:
A

B

C

D

The correct answer is:

Solution:
Alpha hydrogen in aldehydes are slightly acidic and in presence of alkali they release H+ions and carbanion.

Ethyl bromide undergoes nucleophilic substitution reaction with the carbanion generated, where bromide nucleophile is replaced by the carbanion to form ketone by SN2 mechansim

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2022 (27 Jul Shift 2)
The structure of B3N3H6 is as follows :

How many derivative structures of B 3 N 3 H 4 X 2 can be derived from the basic structure, by the replacement of two hydrogen atoms ?

Options:
A 2
B 3
C 4
D 5

The correct answer is:
4

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The structure of chloramphenicol given below

If number of chiral carbon atoms in this structure is X and number of sp2 hybridised carbon atoms is Y. The sum of X + Y is equal to?

Options:

The correct answer is:
9

Solution:
It has two chiral C-atoms and 7sp2 hybridised C-atoms so X = 2, Y = 7
Hence X+Y=2+7=9

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
The structure of D-(+)-glucose is

The structure of L-(-)-glucose is

Options:
A
B
C
D

The correct answer is:

Solution:
The structures of D(+)-glucose and L(-)-glucose are given below from these it is clear that they differ in configuration at second, third, fourth and fifth C atoms.

D-Glucose L-Glucose

Topic: Chemistry, Biomolecules, JEE Main
The structure of diborane $\left(\mathrm{B}_2 \mathrm{H}_6\right)$ contains

Options:
A
four 2c-2e bonds and two 3c-2e bonds

B
two 2c-2e bonds and four 3c-2e bonds

C
two 2c-2e bonds and two 3c-3e bonds

D
four 2c-2e bonds and four 3c-2e bonds

The correct answer is:
four 2c-2e bonds and two 3c-2e bonds

Solution:

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
The structure of diborane $\left(\mathrm{B}_2 \mathrm{H}_6\right)$ contains

Options:
A
four 2c-2e bonds and two 3c-2e bonds

B
two 2c-2e bonds and four 3c-2e bonds

C
two 2c-2e bonds and two 3c-3e bonds

D
four 2c-2e bonds and four 3c-2e bonds

The correct answer is:
four 2c-2e bonds and two 3c-2e bonds

Solution:

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2005
The structure of $\mathrm{H}_2 \mathrm{O}_2$ is

Options:
A

B

C

D

The correct answer is:

Solution:
The four atoms never lie on the same plane. At most, any $\mathrm{H}-\mathrm{O}-\mathrm{O}$ could lie on a plane. This type of structure is called open book structure.

Topic: Chemistry, Hydrogen, JEE Main
The structure of $\mathrm{H}_2 \mathrm{O}_2$ is

Options:
A Planar
B Linear
C Spherical
D Non-planar

The correct answer is:
Non-planar

Solution:
The structure of $\mathrm{H}_2 \mathrm{O}_2$ is open book shaped, and thus non-planar.

Topic: Chemistry, Hydrogen, JEE Main
The structure of $\mathrm{IF}_7$ is :

Options:
A
trigonal bipyramid

B
octahedral

C
pentagonal bipyramid

D
square pyramid

The correct answer is:
pentagonal bipyramid

Solution:
In $\mathrm{IF}_7, \mathrm{I}$ undergoes $\mathrm{sp}^3 \mathrm{~d}^3$ hybridisation

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2011
The structure of Nylon-6 is:

Options:
A
B
C
D

The correct answer is:

Solution:
Nylon 6 is obtained by heating caprolactam with water at a high temperature.

Topic: Chemistry, Polymers, JEE Main, JEE Main 2019 (08 Apr Shift 2)
The structure of Neoprene is :

Options:
A
B
C
D

The correct answer is:

Solution:
The structure of the neoprene polymer is as shown below.
It is obtained from chloroprene which is 2-chloro buta-1,3-diene. It is an addition homopolymer with rubber like structure and properties.

Neoprene is a polymer of chloroprene which is chemically 2- chlorobuta - 1, 3 -diene (isoprene). Neoprene exhibits good chemical stability and maintains flexibility over a wide temperature range.

Topic: Chemistry, Polymers, JEE Main, JEE Main 2021 (26 Feb Shift 1)
The structure of orthophosphoric acid is

Options:
A

B

C

D

The correct answer is:

Solution:
The structure of orthophosphoric acid is

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The structure of PCl5 in the solid state is :

Options:
A tetrahedral PC4+ and octahedral PCl6-
B square planar PCl4+ and octahedral PCl6-
C square pyramidal
D trigonal bipyramidal

The correct answer is:
tetrahedral PC4+ and octahedral PCl6-

Solution:
2PCl5s→PCl4+Tetrahedral PCl6-Octahedral

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2020 (05 Sep Shift 1)
The structure of protein that is unaffected by heating is :

Options:
A secondary structure
B tertiary structure
C primary structure
D Quaternary structure

The correct answer is:
primary structure

Solution:
Primary structure of proteins: Proteins may have one or more polypeptide chains. Each polypeptide in a protein has amino acids linked with each other in a specific sequence and it is this sequence of amino acids that is said to be the primary structure of that protein.

Secondary structure of proteins: The secondary structure of protein refers to the shape in which a long polypeptide chain can exist.

Tertiary structure of proteins: The tertiary structure of proteins represents overall folding of the polypeptide chains i.e., further folding of the secondary structure.

Quaternary structure of proteins: Some proteins are composed of two or more polypeptide chains referred to as subunits. The spatial arrangement of these subunits with respect to each other is known as quaternary structure.

During denaturation Quaternary, secondary and tertiary structures are destroyed, but primary structure remains intact.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2022 (29 Jun Shift 2)
The structure of the compound that gives a tribromo derivative on treatment with bromine water is

Options:
A

B

C

D

The correct answer is:

Solution:
No Solution Available

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2006
The structure of the major product formed in the following reaction is

Options:
A

B

C

D

The correct answer is:

Solution:
No Solution Available

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2006
The structure of the protein that is absolutely necessary for its biological activity is

Options:
A Primary structure
B Secondary structure
C Tertiary structure
D Quaternary structure

The correct answer is:
Tertiary structure

Solution:
In tertiary structure, the polypeptide chain having secondary structure is twisted and folded to produce a stable, three dimensional structure, which may be maintained by three types of bonds e.g. ionic bond, hydrogen bond and disulphide bond. In a large protein like hemoglobin, or in the case of an enzyme, the molecule undergoes further folding and coiling to attain functional conformation and is termed as the tertiary structure. The coils and folds of the protein molecule are so arranged as to hide non-polar amino acid side chains inside and expose the polar side chains and is absolutely necessary for many biological activity. Quaternary structure is most complex structure.

Topic: Chemistry, Biomolecules, NEET
The structure of which of the following chloro species can be explained on the basis of $\mathrm{dsp}^2$ hybridization?

Options:
A
$\mathrm{PdCl}_4^{2-}$

B
$\mathrm{FeCl}_4^{2-}$

C
$\mathrm{CoCl}_4^{2-}$

D
$\mathrm{NiCl}_4^{2-}$

The correct answer is:
$\mathrm{PdCl}_4^{2-}$

Solution:
$\left[\mathrm{PdCl}_4\right]^{2-}$ is $d s p^2$ hybridized and square planar in shape.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2013 (25 Apr Online)
The structure of X is:

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Amines, JEE Main, JEE Main 2021 (16 Mar Shift 2)
The structure of $\mathrm{XeF}_{6}$ is experimentally determined to be distorted octahedron. Its structure according to VSEPR theory is

Options:
A octahedron
B trigonal bipyramid
C pentagonal bipyramid
D tetragonal bipyramid

The correct answer is:
pentagonal bipyramid

Solution:
In $\mathrm{XeF}_{6}$. Xe, the central atom $\mathrm{F}$ contains, 8 valence electrons. Out of which 6 are utilised with fluorine in bonding (ie., it contains six bond pairs of electrons) while one pair remains as lone pair. Thus, the total pairs $=6+1=7$ Hence, its shape is pentagonal bipyramid according to VSEPR theory.

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2014
The structure shown below is of which well-known drug molecule?

Options:
A Codeine
B Morphine
C Ranitidine
D Cimetidine

The correct answer is:
Cimetidine

Solution:
The given structure is of Cimetidine. Cimetidine is an antihistamine drug which is used to treat stomach acidity and peptic ulcers. It is a sulphur-containing derivative of imidazole.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2022 (24 Jun Shift 2)
The structures of A and B formed in the following reaction are : Ph=-C6H5

Options:
A
B
C
D

The correct answer is:

Solution:
Friedel-Craft acylation reaction in benzene takes place. Aluminium chloride is Lewis acid which helps in generating electrophile as shown in first step of reaction.

This electrophile get substituted in benzene and product (A) is obtained. In next step, Clemmensen's reduction occur which converts aldehyde or ketone into hydrocarbon.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2021 (31 Aug Shift 2)
The structures of beryllium chloride in the solid state and vapour phase, respectively, are:

Options:
A chain and chain
B dimeric and chain
C chain and dimeric
D dimeric and dimeric

The correct answer is:
chain and dimeric

Solution:
BeCl2 exists as (BeCl2)n polymeric chain in solid form, while BeCl2 exists as dimer in vapour phase because Be has vacant orbital and it is electron deficient.
In vapour state
In solid state

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2019 (09 Apr Shift 2)
The structures of major products A,B and C in the following reaction are sequence.

Options:
A

B

C

D

The correct answer is:

Solution:
The structures of major product A, B and C is given by

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2023 (01 Feb Shift 2)
The structures that remain separate with forming a layer is

Options:
A Epithelial cells
B Muscle cells
C Nervous cells
D Glands

The correct answer is:
Nervous cells

Solution:
Nerve cells or neurons are the structural and functional units of the nervous system. These cells do not form any layer and are structurally separate by getting associated with the accessory cells.

Topic: Biology, Structural Organisation in Animals, NEET
The substance not likely to contain $\mathrm{CaCO}_3$ is

Options:
A
calcined gypsum

B
sea shells

C
dolomite

D
a marble statue

The correct answer is:
calcined gypsum

Solution:
Gypsum is $\mathrm{CaSO}_4 \cdot 2 \mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2003
The substance used as froth stabilisers in forth-floatation process is

Options:
A
Potassium ethyl xanthate

B
Aniline

C
Sodium cyanide

D
Copper sulphate

The correct answer is:
Aniline

Solution:
Froth stabilizers cresols, aniline are added to stabilize the froth and enhance the nonwettability of the mineral particles.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2012 (26 May Online)
The substance used in the thermite process of reducing metal ores is

Options:
A Aluminium
B Thorium
C Heated Pt gauge
D Carbon

The correct answer is:
Aluminium

Solution:
A mixture of Al powder and metallic oxide (Cr2O3, Mn3O4 etc) is called thermite.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
The substance x belonging to IA group gives a pale violet colour in flame test, x is

Options:
A NaCl
B CsCl
C KCl
D None of these

The correct answer is:
CsCl

Solution:
Na - Golden yellow colour
Li - Red
K - Lilac (pink) colour
Cs - Violet colour

Topic: Chemistry, s Block Elements, JEE Main
The sum $1 \times 1!+2 \times 2!+\ldots+50 \times 50$ ! equals

Options:
A $51 !$
B $51 !-1$
C $51 !+1$
D $2 \times 51 !$

The correct answer is:
$51 !-1$

Solution:
$1 \times 1!+2 \times 2!+\ldots+50 \times 50!$
$=(2-1) 1!+(3-1) 2!+(4-1) 3 !$
$+\ldots+(51-1) 50!$
$=(2 !-1!)+(3 !-2!)+(4 !-3 !)+(51!-50!)$
$+\ldots+(511-501)$
$=511-11=511-1$

Topic: Mathematics, Sequences and Series, JEE Main
The sum 12-2.32+3.52-4.72+5.92-…..+15.292 is _____ .

Options:

The correct answer is:
6952

Solution:
Given,

S=12-2·32+3·52............+15·292

Now rewriting the above expression we get,

S=12+2·32+3·52............+15·292⏟S1-22·32+4·72............+14·272⏟S2

Now solving S1=12+2·32+3·52............+15·292

⇒S1=∑r=115r·2r-12

⇒S1=∑r=115r·4r2+1-4r

⇒S1=∑r=1154r3+r-4r2

⇒S1=4∑r=115r3+∑r=115r-4∑r=115r2
⇒S1=415×1622+15×162-415×16×316

⇒S1=41202+120-4960

⇒S1=52760

Now solving S2=2·32+4·72............+14·272

⇒S2=∑r=172r4r-12

⇒S2=∑r=172r16r2+1-8r

⇒S2=32∑r=17r3+2∑r=17r-16∑r=17r2

⇒S2=327×822+27×82-167×8×156

⇒S2=32×784+56-16×140=22904

Hence, S=S1-2S2=52760-2×22904=6952

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (31 Jan Shift 2)
The sum $\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+\ldots$. upto 11-terms is:

Options:
A
$\frac{7}{2}$

B
$\frac{11}{4}$

C
$\frac{11}{2}$

D
$\frac{60}{11}$

The correct answer is:
$\frac{11}{2}$

Solution:
Given sum is
$$
\begin{aligned}
& \frac{3}{12}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+\ldots . . \\
& n \text {th term }=\mathrm{T}_n \\
& =\frac{2 n+1}{\frac{n(n+1)(2 n+1)}{6}}=\frac{6}{n(n+1)} \\
& \text { or } \mathrm{T}_n=6\left[\frac{1}{n}-\frac{1}{n+1}\right] \\
& \therefore \quad \mathrm{S}_n= \\
& \sum \mathrm{T}_n=6 \sum \frac{1}{n}-6 \sum \frac{1}{n+1}=\frac{6 n}{n}-\frac{6}{n+1} \\
& =6-\frac{6}{n+1}=\frac{6 n}{n+1} \\
& S_{11}=\frac{6 \times 11}{11+1}=\frac{66}{12}=\frac{33}{6}=\frac{11}{2} \\
&
\end{aligned}
$$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2013 (22 Apr Online)
The sum 1+2·3+3·32+……..+10·39 is equal to

Options:
A 2·312+104
B 19·310+14
C 5·310-2
D 9·310+12

The correct answer is:
19·310+14

Solution:
Let S=1·30+2·31+3·32+…..+10.39 ....(i)

3S=1.31+2.32……………….+9×39+10×310 .....(ii)

Now subtracting equation (ii) from (i) we get,

-2 S=1·30+31+32…..39-10.310

S=5×310-310-14

S=20.310-310+14=19.310+14

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2022 (25 Jun Shift 2)
The sum 3×1312+5×(13+23)12+22+7×(13+23+33)12+22+32+..... upto 10th term is

Options:
A 660
B 600
C 620
D 680

The correct answer is:
660

Solution:
S= 31312+513+2312+22+713+23+3312+22+32+…10 terms
Here the general terms is
Tr=2r+113+23+….+r312+22+….+r2+2r+1rr+122rr+12r+16
=32 r(r+1)
⇒S=∑r=11032rr+1=32∑r=110r2+∑r=110r
=3210.11.216+10.112=32385+55=660

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2019 (10 Apr Shift 1)
The sum and product of the mean and variance of a binomial distribution are 82.5 and 1350 respectively. They the number of trials in the binomial distribution is

Options:

The correct answer is:
96

Solution:
Given, sum of mean and variance is 82.5

Product of roots is 1350

Now, Mean and variance are the roots of

x2-sum of rootsx+product of roots=0

⇒x2-82.5x+1350=0

⇒x=60 or 22.5

So, mean =np=60 and variance =npq=22.5

⇒ q=22.560=38

So, p=58 and n=6058=96

Topic: Mathematics, Probability, JEE Main, JEE Main 2022 (29 Jul Shift 2)
The sum ∑k=1201+2+3+…+k is ___________.

Options:

The correct answer is:
1540

Solution:
∑k=120k(k+1)2
=12∑k=120k2+k
=122021416+20212
=12420×416+20×212
=122870+210
=1540

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2020 (08 Jan Shift 1)
The sum ∑k=120k12k is equal to

Options:
A 1 –11220
B 2 –21220
C 2–3217
D 2 –11219

The correct answer is:
2 –11219

Solution:
S= ∑k=120k2k=12+222+323+424+…..+20220
12 S= ∑k=120k2k+1=122+223+324+…+19220+2022
By subtraction,
12 S=12+122+123+ …….+1220–20221
=121-12201-12-20221 (Using sum of n terms of G.P.)
=1–1220 –20221 = 1 –11220
⇒S=2-11219

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2019 (08 Apr Shift 2)
The sum, ∑n=17nn+12n+14, is equal to

Options:

The correct answer is:
504

Solution:
Given,

∑n=17nn+12n+14

=14∑n=172n3+3n2+n

Now using ∑1+2+3+...+n=nn+12, ∑12+22+32+...+n2=nn+12n+16 and ∑13+23+33+...+n3=nn+122, we can write
=1427·822+37·8·156+7·82
142×49×16+28×15+28
141568+420+28==504.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2020 (08 Jan Shift 2)
The sum ∑n=12134n-14n+3 is equal to

Options:
A 787
B 729
C 1487
D 2129

The correct answer is:
729

Solution:
∑n=12134n-14n+3=34∑n=12114n-1-14n+3

=3413-17+17-111+….+183-187

=3413-187=3484261=729

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2022 (25 Jul Shift 2)
The sum of $1^3+2^3+3^3+4^3+\ldots .+15^3$, is

Options:
A $22000$
B $10,000$
C $14,400$
D $15,000$

The correct answer is:
$14,400$

Solution:
Sum of cubes of ' $n$ ' natural number
$=\frac{n^2(n+1)^2}{4}=\frac{15^2(16)^2}{4}=14,400$

Topic: Mathematics, Sequences and Series, JEE Main
The sum of 3 numbers in geometric progression is 38 and their product is 1728 . The middle number is

Options:
A $12$
B $8$
C $18$
D $6$

The correct answer is:
$12$

Solution:
Let the three numbers be $\frac{a}{r}, a$ and $a r$, then
$a^3=1728 \Rightarrow a=12$
Hence the middle term is 12.

Topic: Mathematics, Sequences and Series, JEE Main
The sum of 10 terms of the series 312×22+522×32+732×42+… is :

Options:
A 143144
B 99100
C 1
D 120121

The correct answer is:
120121

Solution:
Given that

312×22+522×32+732×42+…

Here 3,5,7,----∈A.P

First term a = 3 & Common difference d =2, tr = a + r-1d = 3 + r-12 = 2r+1

=∑r=110(2r+1)r2·(r+1)2

=∑r=110r2+2r+1-r2r2·(r+1)2

=∑r=110(r+1)2-r2r2·(r+1)2

=∑r=1101r2-1(r+1)2

=112-122+122-132+132-142+…..+1102-1112=1-1121=120121

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (31 Aug Shift 1)
The sum of 50 terms of the series 3+7+13+21+31+43+. . . . . is equal to S50, then the value of S5012500 is

Options:

The correct answer is:
3.54

Solution:
Let, S=3+7+13+21+31+43+.....+Tn
S= 3+7+13+21+31+.....+Tn-1+Tn
Subtracting both the equations, we get,
0=3+4+6+8+10+........+Tn-Tn-1-Tn
⇒Tn=3+4+6+8+10+........n-1terms=3+n-128+n-22
=3+n-14+n-2
=3+n-1n+2
=3+n2+n-2
=n2+n+1
Sn=ΣTn=Σn2+Σn+Σ1
=nn+12n+16+nn+12+n
If n=50, S50=50×51×1016+50×512+50
=25×17×101+25×51+50=44250

Topic: Mathematics, Sequences and Series, JEE Main
The sum of 100 observations and the sum of their squares are 400 & 2475, respectively. Later on, three observations 3, 4 & 5 were found to be incorrect. If the incorrect observations are omitted, then the variance of the remaining observations is

Options:
A 8.25
B 8.50
C 9.00
D 8.00

The correct answer is:
9.00

Solution:
Given ∑i=110xi=400 and
∑i=1100xi2=2475
Variance of 97 observations=σ2=∑xi2N-∑xiN2

∴ 3+4+5=12, 32+42+52=50
=242597-388972
=242597-16
=2425-155297=87397
=9

Topic: Mathematics, Statistics, JEE Main, JEE Main 2017 (09 Apr Online)
The sum of all 3-digit numbers less than or equal to 500, that are formed without using the digit 1 and they all are multiple of 11, is ______.

Options:

The correct answer is:
7744

Solution:
209,220,231,........,495 series in A.P.

Clearly, a=209, d=220-209=11 and l=495

Where a is the first term, d is the common difference and l is the last term.

⇒l=a+n-1d=495

where n is the number of terms

⇒l=209+n-111=495

⇒n-111=286

⇒n-1=26

⇒n=27

Sum=2722×209+26×11=27209+143=27×352=9504

Numbers containing 1 at unit place =231, 341, 451

Numbers containing 1 at 10th place =319, 418

Hence, required sum of all 3-digit numbers less than or equal to 500=9501-231+341+451+319+418=7744

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2021 (26 Aug Shift 2)
The sum of all integral values of k(k≠0) for which the equation 2x-1-1x-2=2k in x has no real roots, is_____.

Options:

The correct answer is:
66

Solution:
We have,

2x-1-1x-2=2k

⇒2x-4-x+1(x-1)(x-2)=2k

⇒x-3x2-3x+2=2k

⇒kx-3k=2x2-6x+4

⇒2x2-(6+k)x+3k+4=0

For no real roots D<0

⇒(6+k)2-8(3k+4)<0

⇒k2+12k+36-24k-32<0

⇒(k-6)2-32<0

⇒k-6<32

⇒6-32<k<6+32

Integral value of k=1,2,3,4, 5,6,7,8,9,10,11

Sum=11×122=66

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2021 (26 Aug Shift 1)
The sum of all natural numbers between 1 and 100 which are multiples of 3 is

Options:
A $1680$
B $1683$
C $1681$
D $1682$

The correct answer is:
$1683$

Solution:
We get series $3,6,9,12, \ldots \ldots . .99$.
Here, $n=\frac{99}{3}=33, a=3, d=3$, therefore
$S=\frac{33}{2}\{2 \times 3+(33-1) 3\}$
$=\frac{33}{2} \times 102=33 \times 51=1683$

Topic: Mathematics, Sequences and Series, JEE Main
The sum of all odd numbers between 1 and 1000 which are divisible by 3 is

Options:
A 83667
B 90000
C 83660
D None of these

The correct answer is:
83667

Solution:
Sum of odd numbers between 1 and 1000 , which is divisible by $3=3+9+15+21+27+\ldots+999=\mathrm{S}$ (let)

$\therefore$ Let $\mathrm{n}$ be the number of terms in series and $\mathrm{a}$ is first term.

$\therefore l=a+(n-1) d$,

where $l$ is last term and $d$ is is common difference.

$\begin{array}{l}

999=3+(n-1) \times 6 \\

n-1=\frac{999-3}{6}=\frac{996}{6} \\

\Rightarrow n-1=166 \\

\Rightarrow n=167 \\

\therefore \mathrm{S}=\frac{n}{2}[2 a+(n-1) d] \\

=\frac{167}{2}[2 \times 3+(167-1) \times 6] \\

=\frac{167}{2}[1002]=167 \times 501=83667

\end{array}$

Topic: Mathematics, Sequences and Series, JEE Main
The sum of all possible values of $\theta \in[-\pi, 2 \pi]$, for which $\frac{1+i \cos \theta}{1-2 i \cos \theta}$ is purely imaginary, is equal

Options:
A $3 \pi$
B $2 \pi$
C $5 \pi$
D $4 \pi$

The correct answer is:
$3 \pi$

Solution:
$\begin{aligned} & Z=\frac{1+i \cos \theta}{1-2 i \cos \theta} \\ & Z=-\bar{Z} \Rightarrow \frac{1+i \cos \theta}{1-2 i \cos \theta}=-\left(\frac{\overline{1+i \cos \theta}}{1-2 i \cos \theta}\right) \\ & (1+i \cos \theta)(\overline{1-2 i \cos \theta})=-(1-2 i \cos \theta)(\overline{1+i \cos \theta}) \\ & (1+i \cos \theta)(1+2 i \cos \theta)=-(1-2 i \cos \theta)(1-i \cos \theta) \\ & 1+3 i \cos \theta-2 \cos ^2 \theta=-\left(1-3 i \cos \theta-2 \cos ^2 \theta\right) \\ & 2-4 \cos ^2 \theta=0 \\ & \Rightarrow \cos ^2 \theta=\frac{1}{2} \Rightarrow \theta=-\frac{\pi}{4},-\frac{3 \pi}{4}, \frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4} \\ & \text { sum }=3 \pi\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2024 (08 Apr Shift 2)
The sum of all rational terms in the expansion of $\left(2^{\frac{1}{5}}+5^{\frac{1}{3}}\right)^{15}$ is equal to :

Options:
A 3133
B 931
C 6131
D 633

The correct answer is:
3133

Solution:
$\begin{aligned}
& \mathrm{T}_{\mathrm{r}+1}={ }^{15} \mathrm{C}_{\mathrm{r}}\left(5^{\frac{1}{3}}\right)^{\mathrm{r}}\left(\frac{1}{2^5}\right)^{15-\mathrm{r}} \\
& ={ }^{15} \mathrm{C}_{\mathrm{r}} 5^{\frac{\mathrm{r}}{3}} \cdot 2^{\frac{15-\mathrm{r}}{5}} \\
& \mathrm{R}=3 \lambda, 15 \mu \\
& \Rightarrow \mathrm{r}=0,15
\end{aligned}$
2 rational terms
$\begin{aligned}
& \Rightarrow{ }^{15} \mathrm{C}_0 2^3+{ }^{15} \mathrm{C}_{15}(5)^5 \\
& =8+3125=3133
\end{aligned}$

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2024 (04 Apr Shift 1)