datavorous/entrance-exam-dataset huggingface dataset:
If \(\mathrm{a} > 0, \mathrm{~b} > 0, \mathrm{c} > 0\) are respectively the \(\mathrm{pth}, \mathrm{q}\) th, rth terms of GP., then the value of the determinant \(\left|\begin{array}{lll}\log \mathrm{a} & \mathrm{p} & 1 \\ \log \mathrm{b} & \mathrm{q} & 1 \\ \log \mathrm{c} & \mathrm{r} & 1\end{array}\right|\) is

Options:
A 0
B 1
C -1
D None of these

The correct answer is:
0

Solution:
Let \(A\) be the 1 st term and \(\mathrm{R}\) the common ratio of G.P., then
\(\begin{aligned}
& a=T_p=A R^{p-1} \\
& \therefore \log \mathrm{a}=\log \mathrm{A}+(\mathrm{p}-1) \log \mathrm{R}
\end{aligned}\)
Similarly, \(\log \mathrm{b}=\log \mathrm{A}+(\mathrm{q}-1) \log \mathrm{R}\)
and \(\log \mathrm{c}=\log \mathrm{A}+(\mathrm{r}-1) \log \mathrm{R}\)
\(\therefore \Delta=\left|\begin{array}{lll}
\log A+(p-1) \log R & p & 1 \\
\log A+(q-1) \log R & q & 1 \\
\log A+(r-1) \log R & r & 1
\end{array}\right|\)
Split into two determinants and in the first take \(\log A\) common and in the second take \(\log \mathrm{R}\) common
\(\Delta=\log A\left|\begin{array}{lll}
1 & p & 1 \\
1 & q & 1 \\
1 & r & 1
\end{array}\right|+\log R\left|\begin{array}{lll}
p-1 & p & 1 \\
q-1 & q & 1 \\
r-1 & r & 1
\end{array}\right|\)
Apply \(\mathrm{C}_1 \rightarrow \mathrm{C}_2 \rightarrow \mathrm{C}_3\) in the second
\(\Delta=0+\log R\left|\begin{array}{lll}
0 & p & 1 \\
0 & q & 1 \\
0 & r & 1
\end{array}\right|=0\)

Topic: Mathematics, Determinants, JEE Main
If \( \mathrm{A} \) and \( \mathrm{B} \) are any two events such that \( P(A)=\frac{2}{5} \) and \( P(A \cap B)=\frac{3}{20} \), then the conditional probability, \( P\left(A \mid\left(A^{\prime} \cup B^{\prime}\right)\right) \), where \( \mathrm{A}^{\prime} \) denotes the complement of \( \mathrm{A} \), is equal to :

Options:
A \( \frac{11}{20} \)
B \( \frac{5}{17} \)
C \( \frac{\frac{3}{\tilde{n}}}{1} \)
D \( \frac{1}{4} \)

The correct answer is:
\( \frac{5}{17} \)

Solution:
Given PA=25=820 & PA∩B=320
∴ PA'∪B'=PA∩B'
=1-PA∩B
=1-320=1720
PA∩A'∪B'=PA∩B'
=PA-PA∩B
=820-320=520
∴ PAA'∪B'=PA∩A'∪B'PA'∪B'
=5201720=517

Topic: Mathematics, Probability, JEE Main
If \(\mathrm{A}\) and \(\mathrm{B}\) are two matrices such that \(\mathrm{A}+\mathrm{B}\) and \(\mathrm{AB}\) are both defined, then

Options:
A A and \(B\) can be any matrices
B A, B are square matrices not necessarily of the same order
C A, B are square matrices of the same order
D Number of columns of \(A=\) number of rows of \(B\)

The correct answer is:
A, B are square matrices of the same order

Solution:
Hints : Addition is defined if order of \(A\) is equal to order of \(B\)
A B
\(\mathrm{nxm} \mathrm{nxm}\) is defined if \(\mathrm{m}=\mathrm{n}\)
\(\Rightarrow A, B\) are square matrices of same order

Topic: Mathematics, Matrices, JEE Main
If \( \mathrm{A} \) is an invertible matrix of order \( 3 \) and \( \mathrm{B} \) is another matrix of the same order as of \( \mathrm{A} \), such that
\( |B|=2, A^{T}|A| B=A|B| B^{T} . \) If \( \left|A B^{-1} \operatorname{adj}\left(A^{T} B\right)^{-1}\right|=K \), then the value of \( 4 K \) is equal to

Options:

The correct answer is:
0.25

Solution:
∵ATAB=ABBT
Taking determinant on both sides, we get,
ATAB=ABBT
⇒A=2
Now, AB-1adjATB-1
|A|×1|B|×1|adj(ATB)|
A×1B×1ATB2=AB3A2=116

i.e. 4K=14=0.25

Topic: Mathematics, Matrices, JEE Main
If \( \mathrm{A}=\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & \mathrm{a} & 1\end{array}\right] \& \mathrm{~A}^{-1}=\left[\begin{array}{lll}1 / 2 & -1 / 2 & 1 / 2 \\ -4 & 3 & \mathrm{C} \\ 5 / 2 & -3 / 2 & 1 / 2\end{array}\right] \), then values of a \& \( C \) are

Options:
A \( -1,1 \)
B \( 1,2 \)
C \( 1,-1 \)
D \( -2,-1 \)

The correct answer is:
\( 1,-1 \)

Solution:
We have, A=0121233a1 & A-1= 1/2-1/21/2-43C5/2-3/21/2

Since, A.A-1 = I

⇒A.A-1=0121233a1 1/2-1/21/2-43C5/2-3/21/2

=10C+1012+2C4-4a3a-32+ac

∴10C+1012+2C4-4a3a-32+ac=100010001

Now, 4-4a=0⇒a=1

and C+1=0⇒C=-1.

Topic: Mathematics, Matrices, JEE Main
If \( \mathrm{A}=\left[\begin{array}{ll}\mathrm{i} & -\mathrm{i} \\ -\mathrm{i} & \mathrm{i}\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}1 & -1 \\ -1 & 1\end{array}\right] \), then \( \mathrm{A}^{8} \) equals

Options:
A \( 4 \mathrm{~B} \)
B 128B
C \( -128 \mathrm{~B} \)
D \( -64 \mathrm{~B} \)

The correct answer is:
128B

Solution:
We have A=iB
⇒A2= iB2= i2B2= -B2
=2-2-22 =-2B
⇒A4= -2B2= 4B2= 42B= 8B
⇒A8= A42= 8B2= 64B2= 128B

Topic: Mathematics, Matrices, JEE Main
If \( \mathrm{f} \) is a real valued function defined by \( \mathrm{f}(\mathrm{x}+\mathrm{y})+\mathrm{f}(\mathrm{x}-\mathrm{y})=\mathrm{f}(2 \mathrm{x}) \), then the value of \( \int_{\mathrm{f}(2)}^{\mathrm{f}(-2)} \mathrm{f}(\mathrm{x}) \) is equal to

Options:
A \( 0 \)
B \( 1 \)
C \( 2 \)
D \( 3 \)

The correct answer is:
\( 0 \)

Solution:
fx+y+fx-y=f2x=fx+y+x-y
Put x=0, y=-x
fx+f-x=0⇒fx=-fx⇒fx is an odd function.
⇒f2=-f-2
⇒∫f2f-2fxdx=∫f2-f2fxdx=0

Topic: Mathematics, Definite Integration, JEE Main
If \( \mathrm{I}_{\mathrm{n}}=\int(\ln \mathrm{x})^{\mathrm{n}} \mathrm{dx} \), then \( \mathrm{I}_{\mathrm{n}}+\mathrm{nI}_{\mathrm{n}-1}= \)

Options:
A \( \frac{(\ln \mathrm{x})^{\mathrm{n}}}{\mathrm{x}}+\mathrm{C} \)
B \( \mathrm{x}(\ln \mathrm{x})^{\mathrm{n}-1}+\mathrm{C} \)
C \( \mathrm{x}(\ln \mathrm{x})^{\mathrm{n}}+\mathrm{C} \)
D None of these

The correct answer is:
\( \mathrm{x}(\ln \mathrm{x})^{\mathrm{n}}+\mathrm{C} \)

Solution:
Integrate by parts by taking â« ln x n dx as I n = â« ( lnâx ) n .1.âdx I n = x ln x n - â« x n ln x n - 1 x dx
= x ln x n - n I n - 1
â I n + n I n - 1 = x ln x n

Topic: Mathematics, Indefinite Integration, JEE Main
If \( \mathrm{m} \) arithmetic means (A.Ms) and three geometric means (G.Ms) are inserted between \( 3 \) and \( 243 \) such that \( 4^{t h} \) A.M. is equal to \( 2^{n d} \) G.M., then \( m \) is equal to:

Options:

The correct answer is:
39

Solution:
Let 3, A1, A2, A3,......Am, 243 are in arithmetic progression with m arithmetic means.

Common difference d=243-3m+1=240m+1

Let 3, G1, G2, G3, 243 are in geometric progression with 3 geometric means.

Common ratio r=243313+1=8114=3

Given G2=A4

⇒332=3+4240m+1

⇒27=3+960m+1

⇒m+1=40

⇒m=39

Topic: Mathematics, Sequences and Series, JEE Main
If \( \mathrm{pK}_{\mathrm{a}} \) of a weak acid is \( 5 \), then \( \mathrm{pK}_{\mathrm{b}} \) of the conjugate base will be

Options:

The correct answer is:
9

Solution:
From the formula:

pKa + pKb = 14⇒ pKb = 14 -pKa = 14 - 5 = 9

Therefore, the pKb of the conjugate base will be 9.

Topic: Chemistry, Ionic Equilibrium, JEE Main
If \( \mathrm{S}_{1}, \mathrm{~S}_{2}, \mathrm{~S}_{3} \) and \( \mathrm{S}_{4} \) are the solubility of \( \mathrm{AgCl} \) in water, in \( 0.01 \mathrm{M} \mathrm{CaCl}_{2} \) in \( 0.01 \mathrm{M} \mathrm{NaCl} \) and in \( 0.05 \mathrm{M} \mathrm{AgNO}_{3} \) respectively at a certain temperature, the correct order of solubility is

Options:
A \( \mathrm{S}_{1}>\mathrm{S}_{2}>\mathrm{S}_{3}>\mathrm{S}_{4} \)
B \( \mathrm{S}_{1}>\mathrm{S}_{3}>\mathrm{S}_{2}>\mathrm{S}_{4} \)
C \( \mathrm{S}_{1}>\mathrm{S}_{2}=\mathrm{S}_{3}>\mathrm{S}_{4} \)
D \( \mathrm{S}_{1}>\mathrm{S}_{3}>\mathrm{S}_{4}>\mathrm{S}_{2} \)

The correct answer is:
\( \mathrm{S}_{1}>\mathrm{S}_{3}>\mathrm{S}_{2}>\mathrm{S}_{4} \)

Solution:
0.01 M CaCl2 = 0.02 M Cl-,
0.01 M NaCl = 0.01 M Cl-,
0.05 M AgNO3 = 0.05 M Ag+
[Ag+][Cl-] = Ksp (const.). Hence,
S 2 = K sp 0.02 = 50K sp , S 3 = K sp 0.01 = 100K sp
S 4 = K sp 0.05 = 20K sp , ( S 1 = K sp = maximum ) â

Topic: Chemistry, Ionic Equilibrium, JEE Main
If \( \mathrm{V} \) and \( u \) are electric potential and energy density, respectively, at a distance \( r \) from a positive point charge, then which of the following graphs is correct?

Options:
A
B
C
D

The correct answer is:

Solution:
Electric potential due to a point charge at a point is, V=kQr

Electric potential due to a point charge at a point is, E=kQr2
We know that, energy density due to electric field,
u=12ε0E2=12ε0k2Q2r4

On replacing r,
u=12ε0k2Q2V4
∴ V4∝u

Topic: Physics, Electromagnetic Induction, JEE Main
If M.D. is 12, the value of S.D. will be (where M.D. is mean deviation and S.D. is standard deviation)

Options:
A $15$
B $12$
C $24$
D None of these

The correct answer is:
$15$

Solution:
We know that Q.D. $=\frac{5}{6} \times$ M.D. $=\frac{5}{6} \times 12=10$
$\therefore$ S.D. $=\frac{3}{2} \times$ Q.D. $=\frac{3}{2} \times 10 \Rightarrow$ S.D. $=15$.
where Q.D. is Quartile Deviation, S.D. is Standard deviation and M.D. is Mean deviation.

Topic: Mathematics, Statistics, JEE Main
If microwaves, X‐rays, infrared, gamma rays, ultraviolet, radio waves and visible parts of the electromagnetic spectrum are denoted respectively by M, X, I, G, U, R and V the following is the arrangement in ascending order of the wavelength

Options:
A I, M, R, U, V, X and G
B R, M, I, V, U, X and G
C M, R, V, X, U, G and I
D G, X, U, V, I, M and R

The correct answer is:
G, X, U, V, I, M and R

Solution:
The electromagnetic (EM) spectrum is the range of all types of EM radiation. The γ‐ray has a maximum frequency or minimum wavelength; since the wavelength is inversely proportional to frequency, while radio wave has a minimum frequency and maximum wavelength. The order of wavelength will be; Gamma rays G < X-rays X < Ultra violet U < Visible rays V < Infrared rays I < Microwaves M < Radio waves R.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2014 (19 Apr Online)
If minimum possible work is done by a refrigerator in converting 100 grams of water at 0 °C to ice, how much heat (in calories) is released to the surroundings at temperature 27 °C (Latent heat of ice= 80 Cal/gram) to the nearest integer?

Options:

The correct answer is:
8791

Solution:
Q2=mL=8000

wQ2=Q1-Q2Q2=T1-T2T2

Q1-80008000=27273

Q1=8791

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2020 (03 Sep Shift 2)
If \( M_{O} \) is the mass of an oxygen isotope \( { }_{8} \mathrm{O}^{17}, M_{\mathrm{P}} \) and \( M_{\mathrm{N}} \) are the masses of a proton and a neutron respectively, the nuclear binding energy of the isotope is

Options:
A \( \left(M_{\mathrm{O}}-17 M_{\mathrm{N}}\right) c^{2} \)
B \( \left(M_{\mathrm{O}}-8 M_{P}\right) c^{2} \)
C \( \left(8 M_{\mathrm{P}}+9 M_{\mathrm{N}}-M_{O}\right) c^{2} \)
D \( M_{\mathrm{O}} c^{2} \)

The correct answer is:
\( \left(8 M_{\mathrm{P}}+9 M_{\mathrm{N}}-M_{O}\right) c^{2} \)

Solution:
Binding energy = [ZMP + (A â Z) MN â MO] c2
= [8MP + (17 â 8) MN â MO]c2 = (8 MP + 9MN â MO) c2

Topic: Physics, Nuclear Physics, JEE Main
If molality of the dilute solution is doubled, the value of molal depression constant \( \left(\mathrm{K}_{\mathrm{f}}\right) \) will be

Options:
A Doubled
B Halved
C Tripled
D Unchanged

The correct answer is:
Unchanged

Solution:
Kf (molal depression constant) is a characteristic of solvent and is independent of molality.

Topic: Chemistry, Solutions, JEE Main
If molality of the dilute solutions is doubled, the value of molal depression constant (Kf) will be

Options:
A Halved
B Tripled
C Unchanged
D Doubled

The correct answer is:
Unchanged

Solution:
If molality of the dilute solution is doubled, the value of molal depression constant (Kf) remains unchanged.As molal depression constant (Kf) is a characteristic property of the solvent. It has nothing to do with nature of solute or its concentration.

Topic: Chemistry, Solutions, NEET
If molar mass of compound B is x then find x2

Options:

The correct answer is:
99

Solution:
Molar mass of B=198=x
x2=1982=99.00

Topic: Chemistry, Amines, JEE Main
If molten NaCl contains SrCl2 as impurity, crystallization can generate

Options:
A Anionic vacancies
B Cationic vacancies
C Metal excess defects
D Metal deficiency defects

The correct answer is:
Cationic vacancies

Solution:
If molten NaCl containing a little amount of SrCl2 is crystallised, some of the sites of Na+ ions are occupied by Sr2+. Each Sr2+ replaces two Na+ ions. It occupies the site of one ion and the other site remains vacant. The cationic vacancies thus produced are equal in number to that of Sr2+ ions.

Topic: Chemistry, Solid State, AP EAMCET, AP EAMCET 2022 (04 Jul Shift 1)
If momentum of a body is increased by 20%, then its kinetic energy increases by :

Options:
A 36%
B 40%
C 44%
D 48%

The correct answer is:
44%

Solution:
A body of mass m, moving with momentum p and kinetic energy K are related as follows,
K=p22m.
As the momentum of the body is increased by 20%, then its momentum becomes,
p+20100p=1.2 p
and the mass of the body remains unchanged. So, kinetic energy is directly related to the square of momentum.

Hence, new kinetic energy K' will be,
K'=1.2 p22m

K'=1.44 K.

So, the percentage change in kinetic energy
=K'-KK×100
=1.44K-KK×100
=1.44-1×100
=44%.

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2022 (29 Jul Shift 2)
If momentum(P), area (A) and time (T) are taken to be the fundamental quantities then the dimensional formula for energy is :

Options:
A P2 AT-2
B PA-1 T-1
C PA1/2T-1
D P1/2AT-1

The correct answer is:
PA1/2T-1

Solution:
Let dimension formula of energy will be

E=AaTbPc
P=M1L1T-1
A=L2
T=T1
M1L2T-2=MCL2a+cTb-c
by comparison C=1.........1
2a+c=2......2
b-c=-2........3
c=1, a=1/2,b=-1
E=A1/2T-1P1

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2020 (02 Sep Shift 2)
If $\mathrm{N}_0$ is the original mass of the substance of half- life period $\mathrm{t}_{1 / 2}=5$ years, then the amount of substance left after 15 years is

Options:
A
$\mathrm{N}_0 / 8$

B
$\mathrm{N}_0 / 16$

C
$\mathrm{N}_0 / 2$

D
$\mathrm{N}_0 / 4$

The correct answer is:
$\mathrm{N}_0 / 8$

Solution:
Amount left $=N_0 / 2^n=N_0 / 8 \quad($ Here $n=15 / 5=3)$

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2002
If ∑ n=1 5 1nn+1n+2n+3=k3, then k is equal to:

Options:
A 55336
B 17105
C 19112
D 16

The correct answer is:
55336

Solution:
Tn=13n+3-nn(n+1)(n+2)(n+3)
=131n(n+1)(n+2)-1n+1n+2n+3
∑ n=1 5 T n = 1311·2·3-12·3·4+12·3·4-13·4·5…+15·6·7-16·7·8
=1316-16·56
=1356-1336
=553336=k3
⇒k=55336

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2015 (11 Apr Online)
If ${ }^{\mathrm{n}-1} \mathrm{C}_3+{ }^{\mathrm{n}-1} \mathrm{C}_4>{ }^{\mathrm{n}} \mathrm{C}_3$, then $\mathrm{n}$ is just greater than integer

Options:
A 5
B 6
C 4
D 7

The correct answer is:
7

Solution:
$$
\begin{aligned}
& \text { Hints: }:^{n-1} C_3+{ }^{n-1} C_4>{ }^n C_3 \\
& \qquad \Rightarrow^n C_4>^n C_3 \Rightarrow \frac{n !}{4 !(n-4) !}>\frac{n !}{3 !(n-3) !} \Rightarrow \frac{1}{4}>\frac{1}{(n-3)} \Rightarrow n-3>4 \Rightarrow n>7
\end{aligned}
$$

Topic: Mathematics, Binomial Theorem, JEE Main
If $\sum_{n=1}^k \tan ^{-1}\left(\frac{1}{n^2+3 n+3}\right)=\tan ^{-1} \alpha$, then $\alpha=$

Options:
A $\frac{k}{k+2}$
B $\frac{2 k}{2 k+1}$
C $\frac{k}{2 k+5}$
D $\frac{3 k}{4 k+5}$

The correct answer is:
$\frac{k}{2 k+5}$

Solution:
We have,
$\begin{aligned}
\sum_{n=1}^k \tan ^{-1}\left(\frac{1}{n^2+3 n+3}\right) =\tan ^{-1} \alpha \\
\sum_{n=1}^k \tan ^{-1}\left(\frac{(n+2)-(n+1)}{1+(n+2)(n+1)}\right) =\tan ^{-1} \alpha \\
\Rightarrow \quad \sum_{n=1}^k\left(\tan ^{-1}(n+2)-\tan ^{-1}(n+1)\right) =\tan ^{-1} \alpha \\
\Rightarrow \quad\left(\tan ^{-1} 3-\tan ^{-1} 2\right)+\left(\tan ^{-1} 4-\tan ^{-1} 2\right) \ldots & \tan ^{-1}(k+2)-\tan ^{-1}(k+1) \\
\tan ^{-1}(k+2)-\tan ^{-1} 2 =\tan ^{-1} \alpha \\
\tan ^{-1}\left(\frac{k+2-2}{1+(k+2)(2)}\right) =\tan ^{-1} \alpha \\
\frac{k}{1+2 k+4}=\alpha \Rightarrow \alpha =\frac{k}{2 k+5}
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If n1, n2 and n3 are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency n of the string is given by:

Options:
A 1n=1n1+1n2+1n3
B 1n=1n1+1n2+1n3
C n=n1+n2+n3
D n=n1+n2+n3

The correct answer is:
1n=1n1+1n2+1n3

Solution:
For first segment we have,

n1=12l1 Tμ

For second segment we have,
n2=12l2 Tμ

For third segment we have,
n3=12l3Tμ

For the whole string we have,
n=12l Tμ

Total length of the string is,
l=l1+l2+l3

Thus, the fundamental frequency for the whole string is,
1n=1n1+1n2+1n3

Topic: Physics, Waves and Sound, JEE Main
If n>2 and α,β,γ∈R, then the value of S=αC0-α+βC1+α+2β+22γC2-α+3β+32γC3+.... upto n+1 terms is equal to
(where, Cr denotes nCr)

Options:
A 0
B 2n-2γ
C n22n-2γ
D nγ

The correct answer is:
nγ

Solution:
S=αC0-C1+C2-C3........+β-C1+2C2-3C3+.......+γ22C2-32C3+......
=α∑k=0n-1kCk+β∑k=1n-1kkCk+γ∑k=2n-1kk2Ck
But, ∑k=0n-1kCk=0,
∑k=1n-1kkCk=ddx1-xnx=1=0
and ∑k=2n-1kk2Ck=∑k=2n-1kkk-1+kCk
=∑k=2n-1kkk-1Ck+∑k=1n-1kkCk+C1
=0+0+n=n
⇒S=α0+β0+γn=nγ

Topic: Mathematics, Binomial Theorem, JEE Main
If n+2C6n-2P2=11, then n satisfies the equation:

Options:
A n2+n-110=0
B n2+2n-80=0
C n2+3n-108=0
D n2+5n-84=0

The correct answer is:
n2+3n-108=0

Solution:
n+2C6n-2P2=11
⇒ n+2 n+1 n n-1 n-2 n-36 .5 . 4 . 3. 2. 1 n-2 n-3=11
⇒ n+2 n+1 nn-1=11.10.9.8
⇒ n=9

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2016 (10 Apr Online)
If n+2C8:n-2P4=57:16, then the value of n2 is

Options:

The correct answer is:
9.5

Solution:
n+2!8!n-6!×n-6!n-2!=5716
⇒n+2n+1nn-1=5716×8!
⇒n-1nn+1n+2=57×7×6×5×4×3
⇒n-1nn+1n+2=18×19×20×21
⇒n=19
⇒n2=9.5

Topic: Mathematics, Permutation Combination, JEE Main
If n⩾2 is a positive integer, then the sum of the series C2n+1+2C22+C23+C24+…+C2n is

Options:
A nn-12n+16
B nn+12n+16
C nn+12n+212
D n2n+13n+16

The correct answer is:
nn+12n+16

Solution:
C2n+1+2C22+C23+C24+…+C2n

C2n+1+2C33+C23+C24+…+C2n

{use Cr+1n+Crn=Crn+1}

=C2n+1+2C34+C24+C35+…+C2n

=C2n+1+2C35+C25+…+C2n

=C2n+1+2C3n+C2n

=C2n+1+2·C3n+1

=n+1n2+2·n+1nn-12.3

=nn+12n+16

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (24 Feb Shift 2)
If $n(\mathrm{~A})=1000, n(\mathrm{~B})=500$ and if $n(\mathrm{~A} \cap \mathrm{B}) \geq 1$ and $n(\mathrm{~A} \cup \mathrm{B})=p$, then

Options:
A $500 \leq p \leq 1000$
B $1001 \leq p \leq 1498$
C $1000 \leq p \leq 1498$
D $1000 \leq p \leq 1499$

The correct answer is:
$1000 \leq p \leq 1499$

Solution:
$\mathrm{n}(\mathrm{A})=1000, \mathrm{n}(\mathrm{B})=500, \mathrm{n}(\mathrm{A} \cap \mathrm{B}) \geq 1$, $n(A \cup B)=p ; n(A \cup B)=n(A)+n(B)-n(A \cap B)$ $\mathrm{p}=1000+500-\mathrm{n}(\mathrm{A} \cap \mathrm{B}) \Rightarrow 1 \leq \mathrm{n}(\mathrm{A} \cap \mathrm{B}) \leq 500$
Hence $p \leq 1499$ and $p \geq 1000 \Rightarrow 1000 \leq p \leq 1499$

Topic: Mathematics, Sets and Relations, JEE Main
If NA,NB and NC are the number of significant figures in A=0.001204 m, B=43120000 m, and C=1.200 m respectively then

Options:
A NA=NB=NC
B NA>NB>NC
C NA<NB<NC
D NA>NB<NC

The correct answer is:
NA=NB=NC

Solution:
Significant figures are calculated through certain rules:

(i) All the digits that have no zero are significant.

(ii) Zeros that are written at the right hand side of the last non-zero digit in a number with the decimal are significant, while if the decimal is not present then the zeros are not significant.

(iii) The zeros that are written before non-zero digits are not significant.

(iv) Zeros that are written between the non- zero digits are significant.

So, here number of significant figures in A, B and C are four. Thus, NA=NB=NC.

Topic: Physics, Mathematics in Physics, JEE Main
If $n$ and $r$ are two positive integers such that $n \geq r$, then ${ }^n C_{r-1}+{ }^n C_r=$

Options:
A ${ }^n C_{n-r}$
B ${ }^n C_r$
C ${ }^{n-1} C_{\text {, }}$
D ${ }^{n+1} C_r$

The correct answer is:
${ }^{n+1} C_r$

Solution:
${ }^n C_{r-1}+{ }^n C_r={ }^{n+1} C_r$ is a standard formula.

Topic: Mathematics, Permutation Combination, JEE Main
If n arithmetic means are inserted between a and 100 such that the ratio of the first mean to the last mean is 1:7 and a+n=33, then the value of n is

Options:
A 21
B 22
C 23
D 24

The correct answer is:
23

Solution:
Let the A.P. be a, A1, A2...An, 100

Here, common difference, d=100-an+1

Given A1An=17⇒a+d100-d=17 ...i

Also a+n=33

From options, when n=23, a=10 and d=9024=154

from i 10+154100-154=55385=17

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2022 (28 Jun Shift 2)
If nC4, nC5 and nC6 are in A.P., then n can be

Options:
A 9
B 14
C 12
D 11

The correct answer is:
14

Solution:
Since, nC4, nC5, nC6 are in A.P.
⇒2×nC5=nC4+nC6

⇒2×n!5!n-5!=n!4!n-4!+n!6!n-6!

⇒25!n-5!=30+n-5n-46!n-4!

⇒6×2n-4=30+n2-9n+20

⇒n2-21n+98=0

⇒n-7n-14=0

⇒n=7 or 14

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2019 (12 Jan Shift 2)
If ${ }^n C_{r-1}=36,{ }^n C_r=84$ and ${ }^n C_{r+1}=126$, then the value of $r$ is

Options:
A $1$
B $2$
C $3$
D None of these

The correct answer is:
$3$

Solution:
Here $\frac{{ }^n C_{r-1}}{{ }^n C_r}=\frac{36}{84}$ and $\frac{{ }^n C_r}{{ }^n C_{r+1}}=\frac{84}{126}$.
$\Rightarrow 3 n-10 r=-3$ and $4 n-10 r=6$
On solving, we get $n=9, r=3$.

Topic: Mathematics, Permutation Combination, JEE Main
If ${ }^n C_r$ denotes the number of combination of $n$ things taken $r$ at a time, then the expression ${ }^n C_{r+1}+{ }^n C_{r-1}+2 x^n C_r$ equals

Options:
A
${ }^{n+1} C_{r+1}$

B
${ }^{n+2} C_r$

C
${ }^{n+2} C_{r+1}$

D
${ }^{n+1} C_r$

The correct answer is:
${ }^{n+2} C_{r+1}$

Solution:
After t; velocity $=\mathrm{f} \times \mathrm{t}$
$V_{B A}=\vec{f} t+(-\vec{u})=\sqrt{f^2 t^2+u^2-2 f u t \cos \alpha}$
For max. and min.
$\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{V}_{\mathrm{BA}}^2\right)=2 \mathrm{f}^2 \mathrm{t}-2 \mathrm{fu} \cos \alpha=0 \quad$ or $\quad \mathrm{t}=\frac{\mathrm{u} \cos \alpha}{\mathrm{f}}$
Therefore, total no. of values of $\mathrm{r}=33$.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2003
If ${ }^n C_r$ denotes the number of combinations of $n$ distinct things taken $r$ at a time, then the domain of the function $g(x)={ }^{(16-x)} C_{(2 x-1)}$ is

Options:
A $\{1,2,3,4,5\}$
B $\{0,1,2,3,4\}$
C $\phi$
D $\{0\}$

The correct answer is:
$\{1,2,3,4,5\}$

Solution:
$g(x)={ }^{16-x} \mathrm{C}_{2 x-1}$
$16-x>0$ and $2 x-1 \geq 0$ and $16-x \geq 2 x-1$ $x < 16$ and $x \geq \frac{1}{2}$ and $x \leq \frac{17}{3}$
Also $x$ must be integer
$$
\therefore x \in\{1,2,3,4,5\}
$$

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2023 (14 May Shift 1)
If $n, e, \tau$ and $m$ respectively represent the density, charge relaxation time and mass of the electron, then the resistance of a wire of length $I$ and area of cross-section $A$ will be

Options:
A $\frac{m l}{n e^2 \tau A}$
B $\frac{m \tau^2 A}{n e^2 l}$
C $\frac{n e^2 \tau A}{2 m l}$
D $\frac{n e^2 A}{2 m \pi}$

The correct answer is:
$\frac{m l}{n e^2 \tau A}$

Solution:
$R=\rho \frac{I}{A}=\frac{n}{n e^2 \tau} \cdot \frac{I}{A}$

Topic: Physics, Current Electricity, JEE Main
If $n_e$ and $n_n$ are the number of electrons and holes in a semiconductor heavily doped with phosphorus, then

Options:
A $n_e \gg n_n$
B $n_e \ll n_h$
C $n_e \leq n_h$
D $n_e=n_h$

The correct answer is:
$n_e \gg n_n$

Solution:
Phosphorus is a pentavalent impurity so if $n_e$ and $n_h$
are the number of electrons and holes in a semiconductor heavily doped with phosphorus, then
$n_e \gg n_h$

Topic: Physics, Semiconductors, JEE Main
If $n$ geometric means be inserted between $a$ and $b$ then the $n^{\text {th }}$ geometric mean will be

Options:
A $a\left(\frac{b}{a}\right)^{\frac{n}{n-1}}$
B $a\left(\frac{b}{a}\right)^{\frac{n-1}{n}}$
C $a\left(\frac{b}{a}\right)^{\frac{n}{n+1}}$
D $a\left(\frac{b}{a}\right)^{\frac{1}{n}}$

The correct answer is:
$a\left(\frac{b}{a}\right)^{\frac{n}{n+1}}$

Solution:
If $n$ geometric means $g_1, g_2 \ldots \ldots g_n$ are to be inserted between two positive real numbers $a$ and $b$, then $a, g_1, g_2 \ldots \ldots . g_n, b$ are in G.P. Then
$g_1=a r, g_2=a r^2 \ldots \ldots . . . g_n=a r^n$
So, $b=a r^{n+1} \Rightarrow r=\left(\frac{b}{a}\right)^{1 /(n+1)}$
Now $n^{\text {th }}$ geometric mean $\left(g_n\right)=a r^n=a\left(\frac{b}{a}\right)^{n /(n+1)}$
Aliter : As we have the $m^{\text {th }}$ G.M. is given by
$G_m=a\left(\frac{b}{a}\right)^{\frac{m}{n+1}}$
Now replace $m$ by $n$ we get the required result.

Topic: Mathematics, Sequences and Series, JEE Main