datavorous/entrance-exam-dataset huggingface dataset:
The coefficient of linear expansion of brass and steel rods are $\alpha_1$ and $\alpha_2$ respectively. Lengths of brass and steel rods are $l_1$ and $l_2$ respectively. If $\left(l_2-l_1\right)$ is maintained same at all temperatures, which one of the following relations is correct?

Options:
A $\alpha_1 l_2=\alpha_2 l_1$
B $\alpha_1^2 l_2=\alpha_2^2 l_1$
C $\alpha_{12} l_2^2=\alpha_2 l_1^2$
D $l_1 \alpha_1=l_2 \alpha_2$

The correct answer is:
$l_1 \alpha_1=l_2 \alpha_2$

Solution:
The correct option is (D).
Concept: Length of a metalic rod at a given temperature is given by,
$l=l_0(1+\alpha \Delta T)$
where, $l_0$ is the initial length, $\alpha$ the coefficient of linear expansion and $\Delta T$ the temperature change.
At any given temperature the lengths of the brass and steel rods can be written as, $l_b=l_1\left(1+\alpha_1 \Delta T\right)$ and $l_s=l_2\left(1+\alpha_2 \Delta T\right)$ respectively.
Therefore, at difference of length is given by,
$\begin{aligned} & l_b-l_s=l_1\left(1+\alpha_1 \Delta T\right)-l_2\left(1+\alpha_2 \Delta T\right) \\ & \Rightarrow l_b-l_s=\left(l_1-l_2\right)+\left(\alpha_1 l_1-l_2 \alpha_2\right) \Delta T\end{aligned}$
Since, $l_b-l_{\mathrm{s}}=l_1-l_2$ as the difference of length remains constant.
Therefore, $\left(\alpha_1 l_1-l_2 \alpha_2\right)=0$
So, $\alpha_1 l_1=l_2 \alpha_2$ is the answer.
Since, $l_b-l_s=l_1-l_2$ as the difference of length remains constant.
Therefore, $\left(\alpha_1 l_1-l_2 \alpha_2\right)=0$ So, $\alpha_1 l_1=l_2 \alpha_2$ is the answer.

Topic: Physics, Thermal Properties of Matter, MHT CET, MHT CET 2022 (05 Aug Shift 1)
The coefficient of linear expansion of crystal in one direction is Î± 1 and that in the other two directions perpendicular to it is Î± 2. The coefficient of cubical expansion is

Options:
A α 1 + α 2
B 2 α 1 + α 2
C α 1 + 2 α 2
D None of these

The correct answer is:
α 1 + 2 α 2

Solution:
V = V 0 1 + γ Δ θ
or L 3 = L 0 1 + α 1 Δ θ L 0 2 1 + α 2 Δ θ 2
⇒ L 3 = L 0 3 1 + α 1 Δ θ 1 + α 2 Δ θ 2
Since L 0 3 = V 0 , hence
1 + γ Δ θ = 1 + α 1 Δ θ 1 + α 2 Δ θ 2
≅ 1 + α 1 Δ θ 1 + 2 α 2 Δ θ
≅ 1 + α 1 Δ θ + 2 α 2 Δ θ
∴ γ = α 1 + 2 α 2

Topic: Physics, Thermal Properties of Matter, JEE Main
The coefficient of linear expansions of brass and steel are Î± 1 and Î± 2 respectively. When we take a brass rod of length â 1 and a steel rod of length â 2 at 0 oC, then the difference in their lengths â 2 - â 1 will remain the same at all temperatures if

Options:
A Î± 1 â 1 = Î± 2 â 2
B α 1 ℓ 2 = α 2 ℓ 1
C α 1 2 ℓ 2 = α 2 2 ℓ 1
D α 1 ℓ 2 2 = α 2 ℓ 1 2

The correct answer is:
α 1 ℓ 1 = α 2 ℓ 2

Solution:
Let ΔT be increase in the temperature of brass wire. Then the length of brass wire.
ℓ 1 ′ = ℓ 1 ( 1+ α 1  ΔT )
ℓ 2 ′ = ℓ 2 ( 1+ α 2  ΔT )
ℓ 2 ′ − ℓ 1 ′ =( ℓ 1 − ℓ 2 )+( ℓ 1 α 1 − ℓ 2 α 2 )ΔT
( ℓ 2 ′ − ℓ 1 ′ ) is constant at all temperature
ℓ 1 α 1 = ℓ 2 α 2

Topic: Physics, Thermal Properties of Matter, NEET
The coefficient of linear expansions of brass and steel are Î±1 and Î±2 respectively. When we take a brass rod of length l1 and steel rod of length l2 at 0o C , then the difference in their lengths l2-l1 will remain the same at all temperatures if

Options:
A α1l1=α2l2
B α1l2=α2l1
C α12l2=α22l1
D α1l22=α2l12

The correct answer is:
α1l1=α2l2

Solution:
Let ∆T be increase in the temperature of brass wire. Then the length of brass wire,
l′=l11+α1∆T
Similarly, the length of steel wire when the temperature is increased by ∆T
l′′=l21+α2∆T
According to question, l′-l′′=l1-l2
l1+l1α∆T-l2+l2α2∆T=l1-l2
l1α1=l2α2 ,
As ∆T≠0

Topic: Physics, Thermal Properties of Matter, NEET
The coefficient of middle term in the expansion of $(1+x)^{10}$ is

Options:
A $\frac{10!}{5!6!}$
B $\frac{10!}{(5!)^2}$
C $\frac{10!}{5!7!}$
D None of these

The correct answer is:
$\frac{10!}{(5!)^2}$

Solution:
Here, index $=10$, an even number, therefore, middle term is
$\left(\frac{10+2}{2}\right)^{t h}$ i.e. $6^{\text {th }}$ term.
Also, $T_6={ }^{10} C_5 x^5 \Rightarrow$ Coefficient of middle term $={ }^{10} C_5$.

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of restitution between a snooker ball and the side cushion is 13. If the ball hits the cushion and then rebounds at right angles to its original direction, then what is the angle made with the side cushion by the direction of motion before and after impact?

Options:
A tan -1 1 3
B tan-13
C cot -1 3
D cot -1 1 3

The correct answer is:
tan-13

Solution:
coefficient of restitution = 1 3
u cosθ=v sinθ ... (1)
v cosθ=u3sinθ ......(2)

From (1) and (2)
tan2θ=3
θ=tan-13

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
The coefficient of restitution $e$ for a perfectly elastic collision is

Options:
A 1
B 0
C $\infty$
D -1

The correct answer is:
1

Solution:
$(\mathrm{O} . \mathrm{R} / \mathrm{e})=\frac{\text { Relative speed after collision }}{\text { Relative speed before collision }}$
For an elastic collision coefficient of restitution is 1.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
The coefficient of restitution $e$ for a perfectly inelastic collision is

Options:
A 1
B 0
C $\infty$
D -1

The correct answer is:
0

Solution:
The Coefficient of Restitution is a measure of the "bounciness" of a collision between two objects: how much of the kinetic energy remains for the objects to rebound from one another vs. how much is lost as heat, or work done deforming the objects.
The coefficient, $\mathbf{e}$ is defined as the ratio of relative speeds after and before an impact, taken along the line of the impact:
$\mathrm{e}=\frac{\text { Speed of separation }}{\text { Speed of approach }}$
The bodies stick together in perfectly inelastic collision and move together hence velocity of separation is zero.
(i) For perfectly elastic collision $\mathrm{e}=1$
(ii) For perfectly inelastic collision $\mathrm{e}=\mathrm{o}$
(iii) For other collision $\mathrm{o} \lt \mathrm{e} \lt 1$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
The coefficient of self inductance of a solenoid is $0.18 \mathrm{mH}$. If a crode of soft iron of relative permeability 900 is inserted, then the coefficient of self inductance will become nearly

Options:
A $5.4 \mathrm{mH}$
B $162 \mathrm{mH}$
C $0.006 \mathrm{mH}$
D $0.0002 \mathrm{mH}$

The correct answer is:
$162 \mathrm{mH}$

Solution:
$L=\mu n i$ $\Rightarrow \frac{L_2}{L_1}=\frac{\mu}{\mu_0}$ ( $n$ and $i$ are same)
$\Rightarrow L_2=\mu_r L_1=900 \times 0.18=162 \mathrm{mH}$

Topic: Physics, Electromagnetic Induction, JEE Main
The coefficient of static friction between a wooden block of mass 0.5 kg and a vertical rough wall is 0.2. The magnitude of the horizontal force that should be applied on the block to keep it adhere to the wall will be______N. g=10 m s-2

Options:

The correct answer is:
25

Solution:
F.B.D. of the block is shown in the diagram

Since block is at rest therefore

fr-mg=0 ...1

F-N=0 ...2

fr≤μN

In limiting case

fr=μN=μF …3

Using eq. 1 and 3

∴ μF=mg

⇒F=0.5×100.2=25 N

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2021 (24 Feb Shift 1)
The coefficient of static friction between two blocks is 0.5 and the table is smooth. The maximum horizontal force that can be applied to move the blocks together is _______N (take g=10 m s-2

Options:

The correct answer is:
15

Solution:
For Fmax with relative rest, fs between

2 kg&1 kg must be maximum.

Acommon ≤amax

F3≤102

F≤15

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2021 (26 Aug Shift 2)
The coefficient of t4 in the expansion of 1-t61-t3 is

Options:
A 10
B 14
C 15
D 12

The correct answer is:
15

Solution:
1-t61-t3=1-t631-t-3
=1-t18-3t6+3t12 1-t-3
∴ Co-efficient of t4=1× 3+4-1C4= 6C4=15

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2019 (09 Jan Shift 2)
The coefficient of the 2m+1th and 4m+5th terms in the expansion of 1+x100 are equal, then the value of m2 is equal to

Options:

The correct answer is:
8

Solution:
T2m+1=C1002mx2m
T4m+5=C1004m+4x4m+4
Coefficients of T2m+1 and T4m+5 are equal

⇒C1002m=C1004m+4
⇒2m+4m+4=100
⇒6m=96⇒m=16
Hence, m2=8

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of the middle term in the binomial expansion in powers of $x$ of $(1+\alpha x)^4$ and of $(1-\alpha x)^6$ is the same if $\alpha$ equals

Options:
A
$-\frac{5}{3}$

B
$\frac{3}{5}$

C
$\frac{-3}{10}$

D
$\frac{10}{3}$

The correct answer is:
$\frac{-3}{10}$

Solution:
Coefficient of Middle term in $(1+\alpha \mathrm{x})^4=\mathrm{t}_3={ }^4 \mathrm{C}_2 \cdot \alpha^2$
Coefficient of Middle term in $(1-\alpha \mathrm{x})^6=\mathrm{t}_4={ }^6 \mathrm{C}_3(-\alpha)^3$
$$
{ }^4 \mathrm{C}_2 \alpha^2=-{ }^6 \mathrm{C}_3 \cdot \alpha^3 \Rightarrow-6=20 \alpha \Rightarrow \alpha=\frac{-3}{10}
$$

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2004
The coefficient of the term independent of $x$ in the expansion of $\left(1+x+2 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$ is

Options:
A $\frac{1}{3}$
B $\frac{19}{54}$
C $\frac{17}{54}$
D $\frac{1}{4}$

The correct answer is:
$\frac{17}{54}$

Solution:
The general term in the expansion of $\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$ is $T_{r+1}={ }^9 C_r\left(\frac{3}{2} x^2\right)^{9-r}\left(-\frac{1}{3 x}\right)^r$ $={ }^9 C_r\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right) x^{18-3 r}...(i)$
Now, the coefficient of the term independent of $x$ in the expansion of $\left(1+x+2 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9...(ii)$
$=$ Sum of the coefficient of the terms $x^0, x^{-1}$ and $x^{-3}$ in $\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$.
For $x^0$ in (i) above, $18-3 r=0 \Rightarrow r=6$. For $x^{-1}$ in (i) above, there exists no value of $r$ and hence no such term exists. For $x^{-3}$ in (i) $18-3 r=-3 \Rightarrow r=7$
$\therefore$ For term independent of $x$, in (ii) the coefficient
$\begin{aligned}& =1 \times{ }^9 C_6(-1)^6\left(\frac{3}{2}\right)^{9-6}\left(\frac{1}{3}\right)^6+2 \times{ }^9 C_7(-1)^7\left(\frac{3}{2}\right)^{9-7}\left(\frac{1}{3}\right)^7 \\
& =\frac{9.8 .7}{1.2 \cdot 3} \cdot \frac{3^3}{2^3} \cdot \frac{1}{3^6}+2 \frac{9.8}{1.2}(-1) \frac{3^2}{2^2} \cdot \frac{1}{3^7}=\frac{7}{18}-\frac{2}{27}=\frac{17}{54}\end{aligned}$

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of the term independent of x in the expansion of 1+x+2x332x2-13x9 is

Options:
A 13
B 1954
C 1754
D 14

The correct answer is:
1754

Solution:
Coefficient of the term independent of x is the coefficient of x0 in 32x2-13x9+x32x2-13x9+2x332x2-13x9
Now, in 32x2-13x9, Tr+1=9Cr32x29-r-13xr=9Cr329-r-13rx18-3r
i Coefficient of x0⇒18-3r=0⇒r=6
Coefficient of x0 is 9C6323-136
ii Coefficient of x-1⇒18-3r=-1⇒r=193 (not possible)
iii Coefficient of x-3⇒18-3r=-3⇒r=7
Coefficient of x-3 is 9C7322-137
Hence, the required coefficient is
1×9C6323-136+1×0+2×9C7322-137
=9×8×73×2×+181273+2×9×82214-124327
=718-227=1754

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of the term independent of x in x3+3x210 is

Options:
A 53
B 45
C 6
D 12

The correct answer is:
53

Solution:
Given,x3+3x210
General term, Tr+1= 10Crx31210-r3x2r
â¹Tr+1= 10Cr1310-r23rx1210-r-2r
For the term independent of x, put
1210-r-2r=0
â¹ r=2
â´T2+1=T3= 10C2138232
=45Ã1Ã381=53

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of thermal expansion of a rod is temperature-dependent and is given by the formula Î±=aT, where a is 3.465Ã10-5 Â°C-2 and T in oC. If the length of the rod is l at temperature 0oC, then the temperature at which the length will be 2l is [Given, ln2=0.693]

Options:
A 12 oC
B 20 oC
C 200 oC
D 100 oC

The correct answer is:
200 oC

Solution:
As; dl=Î±ldT
â´â«l2ldll=a â«0TTdT
ln2=aT22
â´T=ln4a12=2Ã0.6933.465Ã10-512=25Ã105=200 oC

Topic: Physics, Thermal Properties of Matter, JEE Main
The coefficient of variation and standard deviation of an ungrouped data are 60 and 21 respectively. If 15 is added to every observation of the data, then the coefficient of variation of the new data is

Options:
A 30
B 42
C 40
D 20

The correct answer is:
42

Solution:
According to the given information,
Coefficient of variance $(\mathrm{CV})=\frac{\sigma}{\mu} \times 100$
Where $\sigma$ is standard deviation and $\mu$ is mean of an ungrouped data.
$\because \quad \frac{\sigma}{\mu} \times 100=60$ and $\sigma=21$
So, $\quad \mu=\frac{21 \times 10}{6}=35$
After adding 15 to each observation of the data, the new mean $\mu^{\prime}=35+15=50$ but
$\Sigma\left|\mu^{\prime}-x_i^{\prime}\right|=\Sigma\left|\mu-x_i\right|$, [where $x_i^{\prime}=x_i+15$ ]
Now, $\sigma=\sqrt{\frac{\sum\left(\mu-x_i\right)^2}{n}}=21$
$\Rightarrow \Sigma\left(\mu-x_i\right)^2=(21)^2 \times n$
So, new standard deviation
$$
\sigma^{\prime}=\sqrt{\frac{\sum\left(\mu^{\prime}-x_i^{\prime}\right)^2}{n}}=\sqrt{\frac{(21)^2 \times n}{n}}=21
$$
$\therefore$ New coefficient of variance
$$
=\frac{\sigma^{\prime}}{\mu^{\prime}} \times 100=\frac{21}{50} \times 100=21 \times 2=42
$$
Hence, option (b) is correct.

Topic: Mathematics, Statistics, JEE Main
The coefficient of volume expansion of glycerine is 49Ã10-5 K-1. The fractional change in its density for a 30 Â°C rise in temperature will be

Options:
A 1.45×10-2
B 2×10-4
C 3.5×10-3
D 2.5×10-2

The correct answer is:
1.45×10-2

Solution:
Let V0 be the initial volume of glycerine, i.e, at 0 °C (dry). If Vt be its volume at 30 °C
Then Vt=V01+v∆t
= V01+49×10-5×30
Vt=V01+0.01470=1.0147070 V0
⇒ V0Vt=11.01470
Let ρo and ρt be the initial and final densities of glycertine then initial density, ρ0=mv0 and final density, ρt=mVt
Where, m=mass of glycertine
∆ρρ0= fractional change in the density
ρt-ρ0ρ0=m1Vt-1V0mV0=V0Vt-1
∆ρρ0=11.01470-1=-0.0145
Here, negative sign shows that density decreases with rise in temperature
∆ρρ0=0.0145=1.45×10-2
∆ρρ0=1.45×10-2

Topic: Physics, Thermal Properties of Matter, JEE Main
The coefficient of x⁡ 1 0 1 2 in the expansion of 1+xn+x25310, (where n≤22 is any positive integer), is

Options:
A C4253
B C410
C 4n
D 1

The correct answer is:
C410

Solution:
Given 1+xn+x25310
=1+x253+xn10

Using the binomial expansion a+bn=C0nanb0+C1nan-1b+C2nan-2b2+...+Cnna0bn,
=10C01+x25310xn0+10C11+x2539xn1+10C21+x2538xn2+...+10C101+x2530xn10
As 2 5 3 = 2 3 × 1 1 and 1012=253×4, also n≤22
⇒ Coefficient of x1012 will come only from the first term, i.e. in

10C01+x25310xn0=1+x25310

The general term in the expansion of 1+an is Tr+1=Crnar

Hence, the general term in the expansion of 1+x25310 is Tr+1=Cr10x253r=Cr10x253r

Since, 1012=253×4, hence r=4

Thus, the required coefficient is=10C4.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2014 (19 Apr Online)
The coefficient of x2 in the expansion of 1-x+2x2x+1x10 is

Options:
A 210
B 714
C 504
D 240

The correct answer is:
714

Solution:
The required coefficient is the coefficient of x12 in
1-x+2x2x2+110
=1-x+2x21+10C1x2+10C2x4+.......+10C9x18+x20
Coefficient of x12 is 1×10C6+2×10C5=10×93×8×74×3×2+2×10×93×82×7×65×4×3×2
=210+504
=714

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of x2 in the expansion of the product 2-x21+2x+3x26+1-4x26 is

Options:
A 107
B 108
C 155
D 106

The correct answer is:
106

Solution:
We have, (1+2x+3x2)6=∑r=06 6Cr2x+3x2r
=6C0+6C12x+3x2+6C22x+3x22+ …+6C62x+3x26
∴ Coefficient of x2=18+60=78
Again, coefficient of x2 in 1-4x26=-24
Constant term in 1+2x+3x26+1-4x26 is 1+1=2.
Thus, the required coefficient of x2=2 (coefficient of x2) in 1+2x+3x26+1-4x26-constant term in 1+2x+3x26+1-4x26
=278-24-2
=106.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2018 (16 Apr Online)
The coefficient of $x^3 y^4 z^5$ in the expansion of $(x y+y z+x z)^6$ is

Options:
A 70
B 60
C 50
D None of these

The correct answer is:
60

Solution:
We have
$\begin{aligned}
(x y+y z+z x)^6 & =\sum_{r+s+t=6} \frac{6 !}{r ! s ! t !}(x y)^r(y z)^s(z x)^t \\
& =\sum_{r+s+t=6} \frac{6 !}{r ! s ! t !} x^{r+t} y^{r+s} z^{s+t}
\end{aligned}$
If the general term in the above expansion. contains $x^3 y^4 z^5$, then
$r+t=3, r+s=4 \text { and } s+t=5$
Also, $r+s+t=6$
Solving these equations, we get
$r=1, s=3, t=2$
$\therefore$ Coefficient of
$x^3 y^4 z^5=\frac{6 !}{1 ! 3 ! 2 !}=\frac{6 !}{2 ! 3 !}=60$

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of x4 in the expansion 1-x81+x12 is equal to

Options:
A 1
B -48
C 28
D -19

The correct answer is:
-19

Solution:
1-x81+x12=1-x281+x4
=1-8x2+28x4….1+4x+6x2+4x3+x4
Coefficient of x4=1-48+28=-19

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of x4 in the expansion of 1+5x+9x2+13x3+17x4+.â¦â¦1+x211 is equal to

Options:
A C211+4C111+3
B C211+3C111+4
C 3C211+4C111+3
D 171

The correct answer is:
171

Solution:
Coefficient of x4 in
1+5x+9x2+13x3+17x4+.…….1+11x2+C211x4+.……
Coefficient of x4 is
=C112+9×11+17×1
=11×102+99+17
=55+99+17
=171

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of x4 in the expansion of 1-x-2x28 is

Options:
A 144
B –144
C 154
D –154

The correct answer is:
–154

Solution:
1-x1+2x8
=1-8C1x1+2x+8C2x21+2x2
-8C3x31+2x3+8C4x41+2x4.........
The coefficient of x4 is
8C2⋅22-8C3⋅2⋅3+8C4
=8×72×22-8×7×63×2.3.2+8×7×6×54×3×2
=112-336+70
=–154

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of x4 in the expansion of 1+x+x26 is

Options:
A 72
B 90
C 96
D 112

The correct answer is:
90

Solution:
1+x1+x6=1+6C1x1+x+6C2x21+x2
+6C3x3(1+x)3+6C4x4(1+x)4+....
Coefficient of x4 is
6C2+6C3.3+6C4
=6×52+6×5×43×2×3+6×52
=15+60+15
=90

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of x4 in the expansion of 1+x+x210 is ________

Options:

The correct answer is:
615

Solution:
General term 10!α!β!γ!xβ+2γ
For coefficient of x4⇒β+2γ=4
γ=0,β=4,α=6⇒10!6!4!0!=210
γ=1,β=2,α=7⇒10!7!2!1!=360
γ=2,β=0,α=8⇒10!8!0!2!=45
Total =615

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2020 (09 Jan Shift 1)
The coefficient of x4 in the expansion of 1+x+x2+x36 in powers of x, is …..

Options:

The correct answer is:
120

Solution:
1+x+x2+x36=1+x61+x26

Coefficient of x4=C46+C26·C16+C26=15+15·6+15=120.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2020 (05 Sep Shift 2)
The coefficient of $\mathrm{x}^{4}$ in the expansion of $\left(1+x+x^{2}+x^{3}\right)^{11},$ is

Options:
A 440
B 770
C 990
D 1001

The correct answer is:
990

Solution:
We have coefficient of $x^{4}$ in $\left(1+x+x^{2}+x^{3}\right)^{11}$

$$

\begin{array}{l}

=\text { coefficient of } \mathrm{x}^{4} \text { in }\left(1+\mathrm{x}^{2}\right)^{11}(1+\mathrm{x})^{11} \\

=\text { coefficient of } \mathrm{x}^{4} \text { in }(1+\mathrm{x})^{11}+\text { coefficient } \\

\text { of } \mathrm{x}^{2} \text { in } \\

\text { 11. }(1+\mathrm{x})^{11}+\text { constant term is }{ }^{11} \mathrm{C}_{2} \cdot(1+\mathrm{x})^{11} \\

={ }^{11} \mathrm{C}_{4}+11 .{ }^{11} \mathrm{C}_{2}+{ }^{11} \mathrm{C}_{2}=990

\end{array}

$$

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of x-5 in the binomial expansion of x+1x23-x13+1 -x-1x-x1210 where x≠0,1 is

Options:
A -1
B 4
C 1
D -4

The correct answer is:
1

Solution:
x13+1 x23-x13+1x23-x13+1-x-1 x+1x x-110
=x13+1-1-1x1210 =x13-x-1210
tr+1=10Crx1310-rx−12r ( −1 ) r
tr+1=10Crx103-5r6-1r
∵ 103-5r6=-5
5r6=506 ⇒r=10
Hence, 10C10-110=1

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2017 (09 Apr Online)
The coefficient of x5 in the expansion of 1+x1!+x22!+x33!+x44!+x55!2 is

Options:
A 215
B 415
C 130
D 245

The correct answer is:
415

Solution:
Coefficient of x5 in
1+x1!+x22!+x33!+x44!+x55!1+x1!+x22!+x33!+x44!+x55!
=1×15!+11!×14!+12!×13!+13!×13!×12!+14!×11!+15!×1
=15!5!1!5!+5!1!4!+5!2!3!+5!3!2!+5!4!1!+5!5!0!
=15!5C0+5C1+5C2+5C3+5C4+5C5
=255!=32120=415.

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of x5 in the expansion of 2x3-13x25 is

Options:
A 809
B 9
C 8
D 263

The correct answer is:
809

Solution:
General term of the binomial expansion of 2x3-13x25 is

Tr+1=5Cr(2x3)5-r -13x2r

⇒Tr+1=Cr525-r-13rx15-5r

For coefficient of x5, we must have

15-5r=5⇒r=2

So, required coefficient is

=C25(2)319=10×8×19=809

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (13 Apr Shift 2)
The coefficient of $x^5$ in the expansion of $\left(x^2-x-2\right)^5$ is

Options:
A $-83$
B $-82$
C $-81$
D $0$

The correct answer is:
$-81$

Solution:
$\begin{aligned} & \left(x^2-x-2\right)^5=(x-2)^5(1+x)^5 \\
& =\left[{ }^5 C_0 x^5-{ }^5 C_1 x^4 \times 2+\ldots\right] \quad\left[{ }^5 C_0+{ }^5 C_1 x+\ldots\right]\end{aligned}$
Collecting the coefficient of $x^5: 1-5.5 .2+10.10 .4-10.10 .8+5.5 .16-32$ $=1-50+400-800+400-32=-81$.

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of $x^5$ in the expansion of $(x+3)^6$ is

Options:
A $18$
B $6$
C $12$
D $10$

The correct answer is:
$18$

Solution:
$T_{r+1}={ }^6 C_r \times{ }^{6-r} 3^r$
This contains $x^5$ if $6-r=5 \Rightarrow r=1$
Coefficient of $x^5={ }^6 C_1 3^1=18$.

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of x6 in 1+x6+1+x7+…+1+x15 is

Options:
A C169
B C165-C165
C C166-1
D C166

The correct answer is:
C169

Solution:
Required coefficient is C66+C67+C68+…+C614+C615
=C77+C76+C86+…+C6+14C615
=C78+C68+C69+…+C614+C615
=C79+C69+…+C614+C615
Adding in the same order =C716=C916

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of x6 in the expansion of 1-x81+x12 is equal to

Options:
A 168
B -8
C 28
D 104

The correct answer is:
104

Solution:
1-x81+x12=1-x281+x4
=1-8x2+28x4-56x6+…1+4x+6x2+4x3+x4
Hence, the coefficient of x6=-8+168-56=160=104

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of x6 in the expansion of 1+x+x26 is

Options:
A 131
B 141
C 151
D 167

The correct answer is:
141

Solution:
Coefficient of x6 in 1+x1+x6
=1+C1x61+x+C2x621+x2+C36x31+x3+C46x41+x4+C56x51+x5+C66x61+x6
⇒ The coefficient of x6 is
C36ÃC33+C46ÃC24+C56ÃC15+C66ÃC06
=6Ã5Ã43Ã2Ã1+6Ã52Ã4Ã32+6Ã5+1Ã1
=20+90+30+1=141

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of x-6, in the expansion of 4x5+52x29, is

Options:

The correct answer is:
5040

Solution:
Given,

4x5+52x29

Now rth term of the expression is given by,

Tr+1=Cr94x59-r52x2r

⇒Tr+1=Cr9459-r52rx9-r-2r

Now equating 9-r-2r=-6⇒r=5

So, coefficient of x-6 will be,

=C59459-5525

=9×8×7×64×3×2×1×454525

=5040

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (31 Jan Shift 2)
The coefficient of x7 in ax2+1bx11 and the coefficient of x-7 in ax-1bx211 are equal, then the value of ab is equal to

Options:

The correct answer is:
1

Solution:
For ax2+1bx11

Tr+1=Cr11ax211-r1bxr=Cr11a11-rb-rx22-3r

For x7, 22-3r=7

⇒3r=15⇒r=5
Coefficient of x7=C115a6b-5
For ax-1bx211

Tr+1=Cr11ax11-r-1r1bx2r
=C11ra11-rb-r-1rx11-3r
For x-7, 11-3r=-7

⇒r=6
⇒ coefficient of x-7=C116a5b-6
According to the question,
C611a5b-6=C511a6b-5⇒ab=1.

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of $x^7$ in the expansion of $\left(1-x-x^2+x^3\right)^6$ is

Options:
A
$-132$

B
$-144$

C
$132$

D
$144$

The correct answer is:
$-144$

Solution:
$$
\begin{aligned}
& {\left[1-\mathrm{x}-\mathrm{x}^2(1-\mathrm{x})\right]^6=(1-\mathrm{x})^6\left(1-\mathrm{x}^2\right)^6} \\
& =\left[{ }^6 \mathrm{C}_0-{ }^6 \mathrm{C}_1 \mathrm{x}+{ }^6 \mathrm{C}_2 \mathrm{x}^2-{ }^6 \mathrm{C}_3 \mathrm{x}^3+{ }^6 \mathrm{C}_4 \mathrm{x}^4-{ }^6 \mathrm{C}_5 \mathrm{x}^5+{ }^6 \mathrm{C}_6 \mathrm{x}^6\right] \times\left[{ }^6 \mathrm{C}_0-{ }^6 \mathrm{C}_1 \mathrm{x}^2+{ }^6 \mathrm{C}_2 \mathrm{x}^4-{ }^6 \mathrm{C}_3 \mathrm{x}^6+\ldots .\right]
\end{aligned}
$$
Coefficient of $\mathrm{x}^7={ }^6 \mathrm{C}_1{ }^6 \mathrm{C}_3-{ }^6 \mathrm{C}_3{ }^6 \mathrm{C}_2+{ }^6 \mathrm{C}_5{ }^6 \mathrm{C}_1=120-300+36=-144$

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2011
The coefficient of x7 in the expression 1+x10+x1+x9+x21+x8+....+x10, is

Options:
A 210
B 330
C 120
D 420

The correct answer is:
330

Solution:
As the given series is in G.P., so its sum is

1+x101-x1+x111-x1+x

1+x101+x11-x111+x11×11+x=1+x11-x11

Coefficient of x7, is 11C7=11C4=330

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2020 (07 Jan Shift 2)
The coefficient of x8 in the expansion of 1+x22!+x44!+x66!+x88!2 is

Options:
A 1315
B 2315
C 1105
D 1210

The correct answer is:
1315

Solution:
Coefficient of x8 in
1+x22!+x44!+x66!+x88!1+x22!+x44!+x66!+x88!
=1×18!+12!×16!+14!×14!+16!×12!+18!×1
=10!8!+12!6!+14!4!+16!2!+18!0!
=18!8!0!8!+8!2!6!+8!4!4!+8!6!2!+8!8!0!
=18!8C0+8C2+8C4+8C6+8C8
=28-18!=278!=1315

Topic: Mathematics, Binomial Theorem, JEE Main