datavorous/entrance-exam-dataset huggingface dataset:
The coefficient of x8 in the expansion of 1+x+x2+x34 is

Options:
A 30
B 31
C 32
D 36

The correct answer is:
31

Solution:
We have,
1+x+x2+x34=1+x1+x24
=1+x41+x24
=1+4x+6x2+4x3+x41+4x2+6x4+4x6+x8
Therefore, the coefficient of x8
=1×1+6×4+1×6=31

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of x9 in the expansion of 1+x+x2+x331-x6 is

Options:
A –7
B 7
C 9
D –9

The correct answer is:
–9

Solution:
Since, 1+x+x2+x3=1-x41-x⇒1+x+x2+x31-x=1-x4
⇒1+x+x2+x331-x31-x3
=1-x431-x3
=1-3x4+3x8-x121-3x+3x2-x3
Coefficient of x9 in this product is =3–3=–9

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of x9 in the expansion of x3+12log2x3211 is equal to

Options:
A -5
B 330
C 520
D 5+log23

The correct answer is:
330

Solution:
log2x32=log212x32=3212log2x=log2x3
â2log2x32=2log2x3=x3
We consider the expansion of x3+1x311.
tr+1=11Crx311-r1x3r=11Crx33-3r-3r
=11Crx33-6r
For the coefficient of x9, we get 33-6r=9
â6r=24âr=4.
Thus, the coefficient of x9 is 11C4=11Ã10Ã93Ã84Ã3Ã2=330.

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of $x^{10}$ in the expansion of $(1+x)^2$ $\left(1+x^2\right)^3\left(1+x^3\right)^4$ is equal to

Options:
A
52

B
44

C
50

D
56

The correct answer is:
52

Solution:
$\because(1+x)^2=1+2 x+x^2$,
$$
\left(1+x^2\right)^3=1+3 x^2+3 x^4+x^6,
$$
and $\left(1+x^3\right)^4=1+4 x^3+6 x^6+4 x^9+x^{12}$
So, the possible combinations for $x^{10}$ are:
$$
x \cdot x^9, x \cdot x^6 \cdot x^3, x^2 \cdot x^2 \cdot x^6, x^4 \cdot x^6
$$
Corresponding coefficients are $2 \times 4,2 \times 1 \times$ $4,1 \times 3 \times 6,3 \times 6$ or $8,8,18,18$.
$\therefore$ Sum of the coefficient is
$$
8+8+18+18=52
$$
Therefore, the coefficient of $x^{10}$ in the expansion of $(1+x)^2\left(1+x^2\right)^3\left(1+x^3\right)^4$ is 52 .

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
The coefficient of x10 in the expansion of 1+x15+ 1+x16+ 1+x17+...+ 1+x30 is

Options:
A 31C10-15C10
B 31C11-15C11
C 30C10-15C10
D Â 31C10-Â 14C11

The correct answer is:
31C11-15C11

Solution:
Coefficient of x10 in the given expansion
= 15C10+ 16C10+ 17C10+...30C10
= 10C10+ 11C10 +...+ 14C10+ 15C10+...+ 30C10- 10C10+ 11C10+...+ 14C10
= 31C11- 15C11

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of x13 in the expansion of 1-x51+x+x2+x34 is

Options:
A 24
B 12
C 6
D 4

The correct answer is:
4

Solution:
Let, E=1-x51+x1+x24
=1-x1-x41+x41+x24
=1-x1-x241+x24
=1-x1-x44
=1-x1-4x4+6x8-4x12+x16
Coefficient of x13 is -1-4=4

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of x18 in the expansion of x4-1x315 is ____________

Options:

The correct answer is:
5005

Solution:
Given,

Binomial expression x4-1x315

Now general term is given by,

Tr+1=Cr15x415-r-1x3r

⇒Tr+1=Cr15x60-7r(-1)r

For coefficient of x18, equate 60-7r=18

⇒r=6

Coefficient of x18=-16·C615=5005

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (06 Apr Shift 1)
The coefficient of x18 in the product 1+x1-x101+x+x29 is

Options:
A 84
B -84
C -126
D 126

The correct answer is:
84

Solution:
1+x1-x101+x+x29
=1+x1-x1-x91+x+x29
=1-x21-x1+x+x29
=1-x21-x39
So, Coefficient of x18 in 1+x1-x101+x+x29
= Coefficient of x18 in 1-x21-x39
= Coefficient of x18 in 1-x2C0 9+C1 9-x3+C2-x32+… 9
=C6 9-16=C3 9=9.8.71.2.3=84

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2019 (12 Apr Shift 1)
The coefficient of $x^{20}$ in the expansion of $\left(1+3 x+3 x^2+x^3\right)^{20}$ is

Options:
A ${ }^{60} C_{40}$
B ${ }^{30} C_{20}$
C ${ }^{15} \mathrm{C}_2$
D None of these

The correct answer is:
${ }^{60} C_{40}$

Solution:
Given expansion is $\left(1+3 x+3 x^2+x^3\right)^{20}$
$$
\begin{aligned}
& =\left((1+x)^3\right)^{20} \\
& =(1+x)^{60}
\end{aligned}
$$
In the expansion of $(a+b)^n,(r+1)$ th term is given by
$$
t_{r+1}={ }^n C_r a^r \cdot b^{n-r}
$$
or $\quad t_{r+1}={ }^n C_r a^{n-r} \cdot b^r$
$\therefore 21$ st term in Eq. (i) will contain the coefficient of $x^{20}$ and it is given by
$$
t_{21}={ }^{60} C_{20} x^{20}, 1^{40}={ }^{60} C_{20} x^{20}
$$
$\therefore$ Coefficient of $x^{20}$ in $(1+x)^{60}$ is ${ }^{60} C_{20}$ or ${ }^{60} C_{40}$.

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of x20 in the expansion of 1+x240x2+2+1x2-5 is

Options:
A 30C10
B 30C15
C 30C25
D 30C20

The correct answer is:
30C25

Solution:
1+x240x2+1x2+2⋅x⋅1x-5
=1+x240x+1x2-5
=1+x240x2+1x-10
=1+x2401+x2-10x-10=x101+x230
Coefficient of x20 in x101+x230= coefficient of x10 in 1+x230
1+x230=∑r=030Cr30x2r
2r=10⇒r=5
So, the required coefficient is 30C5=30C25

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of $x^{20}$ in the expansion of
$\left(1+x^{2}\right)^{40} \cdot\left(x^{2}+2+\frac{1}{x^{2}}\right)^{-5}$ is

Options:
A ${ }^{30} \mathrm{C}_{10}$
B ${ }^{30} \mathrm{C}_{25}$
C 1
D None of these

The correct answer is:
${ }^{30} \mathrm{C}_{25}$

Solution:
Expression $=\left(1+x^{2}\right)^{40} \cdot\left(x+\frac{1}{x}\right)^{-10}$
$=\left(1+x^{2}\right)^{30} \cdot x^{10}$
The coefficient of $x^{20}$ in $x^{10}\left(1+x^{2}\right)^{30}$ $=$ the coefficient of $\mathrm{x}^{10}$ in $\left(1+\mathrm{x}^{2}\right)^{30}$
$={ }^{30} \mathrm{C}_{5}={ }^{30} \mathrm{C}_{30-5}={ }^{30} \mathrm{C}_{25}$

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of x24 in 1+x2121+x121+x24 is

Options:
A C126+3
B C126+1
C C126
D C126+2

The correct answer is:
C126+2

Solution:
1+x2121+x12+x24+x36
=1+x12+x24+x361+C112x2+C212x4+…+C1112x22+C1212x24
Coefficient of x24 in the product is,
C1212+C126+1=1+C126+1
=C126+2

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of $x^{32}$ in the expansion of $\left(x^4-\frac{1}{x^3}\right)^{15}$ is

Options:
A ${ }^{15} C_5$
B ${ }^{15} C_6$
C ${ }^{15} C_4$
D ${ }^{15} C_7$

The correct answer is:
${ }^{15} C_4$

Solution:
Let $T_{r+1}$ term containing $x^{32}$.
Therefore ${ }^{15} C_r x^{4 r}\left(\frac{-1}{x^3}\right)^{15-r}$
$\Rightarrow x^{4 r} x^{-45+3 r}=x^{32} \Rightarrow 7 r=77 \Rightarrow r=11$
Hence coefficient of $x^{32}$ is ${ }^{15} C_{11}$ or ${ }^{15} C_4$

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of x48 in the expansion of 1+x4121+x241+x48 is

Options:
A C612+3
B C126
C C126+1
D C126+2

The correct answer is:
C126+2

Solution:
Given expansion is 1+x24+x48+x72C120+C121x4+…+x48
Coefficient of x48=1+C126+1=C126+2

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of x50 in 1+x411-x+x240 is

Options:

The correct answer is:
0

Solution:
Coefficient of x50 in
1+x1+x401-x+x240
=1+x1+x1-x+x240
=1+x1-x340
=1+x1-40C1x3+40C2x6+....+40C16x48-40C17x51+....+40C40x120
Coefficient of x50=0
As there is no term containing x50 and x49 in the second bracket

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of $x^{50}$ in the binomial expansion of $(1+x)^{1000}+x(1+x)^{999}+x^2(1+x)^{998}+\ldots$ $+x^{1000}$ is:

Options:
A
$\frac{(1000) !}{(50)(! 95 \varphi !}$

B
$\frac{(1000) !}{(49)(! 95) !}$

C
$\frac{(1001) !}{(51)(! 95 \phi !}$

D
$\frac{(1001) !}{(50)(! 95) !}$

The correct answer is:
$\frac{(1001) !}{(50)(! 95) !}$

Solution:
Let given expansion be
$$
\begin{aligned}
&\mathrm{S}=(1+x)^{1000}+x(1+x)^{999}+x^2 \\
&(1+x)^{998}+\ldots+\ldots+x^{1000}
\end{aligned}
$$
Put $1+x=t$
$$
\mathrm{S}=t^{1000}+x t^{999}+x^2(t)^{998}+\ldots+x^{1000}
$$
This is a G.P with common ratio $\frac{x}{t}$
$$
\begin{aligned}
\mathrm{S} &=\frac{t^{1000}\left[1-\left(\frac{x}{t}\right)^{1001}\right]}{1-\frac{x}{t}} \\
&=\frac{(1+x)^{1000}\left[1-\left(\frac{x}{1+x}\right)^{1001}\right]}{1-\frac{x}{1+x}} \\
=& \frac{(1+x)^{1001}\left[(1+x)^{1001}-x^{1001}\right]}{(1+x)^{1001}}
\end{aligned}
$$
$$
=\left[(1+x)^{1001}-x^{1001}\right]
$$
Now coeff of $x^{50}$ in above expansion is equal to coeff of $x^{50}$ in $(1+x)^{1001}$ which is ${ }^{1001} C_{50}=\frac{(1001) !}{50 !(951) !}$

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2014 (11 Apr Online)
The coefficient of $\mathrm{x}^{53}$ in the following expansions $\sum_{m=0}^{100} 100 C_{m}(x-3)^{100-m} .2^{m}$ is

Options:
A ${ }^{100} \mathrm{C}_{47}$
B ${ }^{100} \mathrm{C}_{53}$
C $-{ }^{100} \mathrm{C}_{53}$
D $-{ }^{100} \frac{53}{\mathrm{C}}_{100}$

The correct answer is:
$-{ }^{100} \mathrm{C}_{53}$

Solution:
$\sum_{\mathrm{m}=0}^{100} 100 \mathrm{C}_{\mathrm{m}}(\mathrm{x}-3)^{100-\mathrm{m}} \cdot 2^{\mathrm{m}}$
Above expansion can be rewritten as
$$
\begin{array}{l}
{[(x-3)+2]^{100}=(x-1)^{100}=(1-x)^{100}} \\
\therefore x^{53} \text { will occur in } T_{54} \\
\Rightarrow T_{54}={ }^{100} C_{53}(-x)^{53} \\
\therefore \text { Required coefficient is }-{ }^{100} C_{53}
\end{array}
$$

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of $x^{70}$ in $x^2(1+x)^{98}+x^3(1+x)^{97}+x^4(1+x)^{96}+\ldots+x^{54}(1+x)^{46}$ is ${ }^{99} \mathrm{C}_{\mathrm{p}}-{ }^{46} \mathrm{C}_{\mathrm{q}}$. Then a possible value of $p+q$ is :

Options:
A 55
B 83
C 61
D 68

The correct answer is:
83

Solution:
$\begin{aligned}
& x^2(1+x)^{98}+x^3\left(1+x^{97}\right)+x^4(1+x)^{96}+\ldots \ldots \\
& x^{54}(1+x)^{46}
\end{aligned}$
Coeff. of $\mathrm{x}^{70}:{ }^{98} \mathrm{C}_{68}+{ }^{97} \mathrm{C}_{67}+{ }^{96} \mathrm{C}_{66}+\ldots \ldots \ldots$
$\begin{aligned}
& { }^{47} \mathrm{C}_{17}+{ }^{46} \mathrm{C}_{16} \\
& ={ }^{46} \mathrm{C}_{30}+{ }^{47} \mathrm{C}_{30}+\ldots \ldots \ldots . .{ }^{98} \mathrm{C}_{30} \\
& =\left({ }^{46} \mathrm{C}_{31}+{ }^{46} \mathrm{C}_{30}\right)+{ }^{47} \mathrm{C}_{30}+\ldots \ldots \ldots .{ }^{98} \mathrm{C}_{30}-{ }^{46} \mathrm{C}_{31} \\
& ={ }^{47} \mathrm{C}_{31}+{ }^{47} \mathrm{C}_{30}+\ldots \ldots \ldots . .{ }^{98} \mathrm{C}_{30}-{ }^{46} \mathrm{C}_{31} \\
& \ldots \ldots \\
& ={ }^{99} \mathrm{C}_{31}-{ }^{46} \mathrm{C}_{31}={ }^{99} \mathrm{C}_{\mathrm{p}}-{ }^{46} \mathrm{C}_{\mathrm{q}}
\end{aligned}$
Possible values of $(p+q)$ are $62,83,99,46$
$\Rightarrow p+q=83$

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2024 (09 Apr Shift 1)
The coefficient of x101 in the expression
5+x500+x5+x499+x25+x498+……+x500,x>0 is

Options:
A 501C101×5399
B 501C101×5400
C 501C100×5400
D 500C101×5399

The correct answer is:
501C101×5399

Solution:
Given,

5+x500+x5+x499+x25+x498+……+x500

Using sum of the G.P in the given expression we get,

⇒5+x500x5+x501-1x5+x-1=x501-5+x501-5

=155+x501-x5015

Now coefficient of x101 in 155+x501 will be 15C1015015501-101=5399C101501

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2022 (25 Jun Shift 2)
The coefficient of x256 in the expansion of 1-x101x2+x+1100 is:

Options:
A C16100
B C15100
C C16-100
D C15-100

The correct answer is:
C15100

Solution:
We have,

1-x101x2+x+1100

=1-x100x2+x+11001-x

=1-xx2+x+1100·1-x

=1-x3100·1-x

=1-x3100⏟No term of x256-x1-x3100⏟we find cofficient of x255

Now,

1-x3100=C0100-C1100x3+C2100x6-...-C85100-x255

⇒x1-x3100=C0100x-C1100x4+C2100x7-...+C85100x256

Required coefficient

=C85100=C15100

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (20 Jul Shift 1)
The coefficient of x301 in 1+x500+x1+x499+x21+x498+…..+x500 is:

Options:
A C302501
B C301500
C C300500
D C200501

The correct answer is:
C200501

Solution:
We have,

1+x500+x1+x499+x21+x498+…..+x500

This is a geometric progression.

=1+x500·1-x1+x5011-x1+x

=1+x5001+x501-x5011+x501·1+x

=(1+x)501-x501

Coefficient of x301 in (1+x)501-x501 is given by

C301501=C200501

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (30 Jan Shift 1)
The coefficient of $x^{2012}$ in $\frac{1+x}{\left(1+x^{2}\right)(1-x)}$ is -

Options:
A 2010
B 2011
C 2012
D None of these

The correct answer is:
None of these

Solution:
Coeff. Of $x^{2012}$
$\frac{(1+x)^{2}}{\left(1+x^{2}\right)\left(1-x^{2}\right)}$
$=(1+x)^{2}\left(1-x^{4}\right)^{-1}$
$=\left(1+2 x+x^{2}\right)\left(1-x^{4}\right)^{-1}$
Coeff. Of $x^{2012}+2$ Coeff of $x^{2011}+$ Coeff of $x^{2010}$ in the expansion of $\left(1-x^{4}\right)^{-1}$
$x^{2011}$ and $x^{2010}$ not possible in $\left(1-x^{4}\right)^{-1}$
$=$ only coeff. Of $x^{2012}$ in the expassion of $\left(1-x^{4}\right)^{-1}$
${ }^{1+503-1} C_{503}=1$

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of x2012 in the expansion of 1-x20081+x+x22007 is equal to _____.

Options:

The correct answer is:
0

Solution:
Let, y=1-x20081+x+x22007

⇒y=1-x1-x20071+x+x22007

⇒y=1-x1-x1+x+x22007

⇒y=1-x1-x32007

⇒y=1-xC02007-C12007x3+......

General term in the expansion is given by,

Tr+1=1-xCr2007 x3r -1r

⇒Tr+1=2007Cr x3r -1r- 2007Cr x3r+1 -1r

For coefficient of x12,

3r=2012

But, r≠20123

Also, 3r+1=2012

⇒3r=2011

But, r≠20113

Hence, there is no term containing x2012.

So coefficient of x2012=0

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2024 (27 Jan Shift 2)
The coefficient of xn-2 in the polynomial x-1x-2x-3………x-n is

Options:
A nn2+23n+124
B nn2-13n+224
C nn2+13n+424
D nn2-23n-224

The correct answer is:
nn2-13n+224

Solution:
Let, E=x-α1x-α2x-α3…x-αn where α1=1,α2=2 etc.
=xn-∑α1xn-1+∑α1α2xn-2+...
Hence, the coefficient of xn-2= the sum of all the products of the first ‘n’ natural numbers taken two at a time
1+2+3+.……+n2-12+22+32+.…….+n22=nn2-13n+224

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of $x^n$ in expansion of $(1+x)(1-x)^n$ is

Options:
A
$(n-1)$

B
$(-1)^n(1-n)$

C
$(-1)^{n-1}(n-1)^2$

D
$(-1)^{n-1} n$

The correct answer is:
$(-1)^n(1-n)$

Solution:
Coefficient of $x^n$ in $(1+x)(1-x)^n=(1+x)\left({ }^n C_0-{ }^n C_1 x+\ldots \ldots .+(-1)^{n-1}{ }^n C_{n-1} x^{n-1}+(-1)^n\right.$
$\left.{ }^n C_n x^n\right)$
$=(-1)^n{ }^n C_n+(-1)^{n-1}{ }^n C_{n-1}=(-1)^n(1-n)$

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2004
The coefficient of $\mathrm{x}^{\mathrm{n}}$ in the expansion of $\log _{a}(1+x)$ is

Options:
A $\frac{(-1)^{\mathrm{n}-1}}{\mathrm{n}}$
B $\frac{(-1)^{\mathrm{n}-1}}{\mathrm{n}} \log _{\mathrm{a}} \mathrm{e}$
C $\frac{(-1)^{\mathrm{n}-1}}{\mathrm{n}} \log _{\mathrm{e}} \mathrm{a}$
D $\frac{(-1)^{\mathrm{n}}}{\mathrm{n}} \log _{\mathrm{a}} \mathrm{e}$

The correct answer is:
$\frac{(-1)^{\mathrm{n}-1}}{\mathrm{n}} \log _{\mathrm{a}} \mathrm{e}$

Solution:
$$
\begin{array}{l}
\log _{\mathrm{a}}(1+\mathrm{x})=\log _{\mathrm{e}}(1+\mathrm{x}) \log _{\mathrm{a}} \mathrm{e} \\
=\log _{\mathrm{a}} \mathrm{e}\left[\sum_{\mathrm{n}=1}^{\infty}(-1)^{\mathrm{n}-1} \frac{\mathrm{x}^{\mathrm{n}}}{\mathrm{n}}\right]
\end{array}
$$
So, the coefficient of $x^{n}$ in $\log _{a}(1+x)$ is
$$
\frac{(-1)^{\mathrm{n}-1}}{\mathrm{n}} \log _{\mathrm{a}} \mathrm{e}
$$

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of x n in the polynomial ( x+   2n+1 C 0 )( x+   2n+1 C 1 )( x+   2n+1 C 2 ).....( x+   2n+1 C n ) is

Options:
A 2 n+1
B 2 2n+1 −1
C 2 2n −1
D 2 2n

The correct answer is:
2 2n

Solution:
Given Expression is, ( x+ â 2n+1 C 0 )( x+ â 2n+1 C 1 )( x+ â 2n+1 C 2 ).....( x+ â 2n+1 C n )
If P is coefficient of x n then,
P= â 2n+1 C 0 + â 2n+1 C 1 + â 2n+1 C 2 +.....+ â 2n+1 C n -------(1)
âP= â 2n+1 C 2n+1 + â 2n+1 C 2n + â 2n+1 C 2nâ1 +.....+ â 2n+1 C n+1 ------(2) ( âµ â n C r = n C nâr )
adding (1) and (2)
2P=( 2n+1 C 0 + 2n+1 C 1 +....... + 2n+1 C 2n+1 )
2P= 2 2n+1
â´âââP= 2 2n

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of \( x^{n} \) in the expansion of \( (1-\mathrm{x})^{-1} \) is \( (0 < |\mathrm{x}| < 1) \)

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

The correct answer is:
\( 1 \)

Solution:
Expanding using binomial expansion for negative index,

⇒1-x-1=1+x+x2+…+xn+…∞
From the expansion, we get the coefficient of xn
∴The coefficient of xn=1

Topic: Mathematics, Binomial Theorem, JEE Main
The coefficient of \(x^n\) in the expansion of \(\frac{e^{7 x}+e^x}{e^{3 x}}\) is

Options:
A \(\frac{4^{\mathrm{n}-1}-(-2)^{\mathrm{n}-1}}{\lfloor\mathrm{n}}\)
B \(\frac{4^{n-1}-2^{n-1}}{\lfloor\text { n }}\)
C \(\frac{4^{\mathrm{n}}-2^{\mathrm{n}}}{\lfloor\mathrm{n}}\)
D \(\frac{4^{\mathrm{n}}+(-2)^{\mathrm{n}}}{\underline{n}}\)

The correct answer is:
\(\frac{4^{\mathrm{n}}+(-2)^{\mathrm{n}}}{\underline{n}}\)

Solution:
Hints: \(\frac{e^{7 x}+e^x}{e^{3 x}}=e^{4 x}+e^{-2 x}\)
Co-efficient of \(x^n\)
\(\frac{(4)^{\mathrm{n}}}{\mathrm{n} !}+\frac{(2)^{\mathrm{n}}}{\mathrm{n} !}(-1)^{\mathrm{n}}=\frac{4^{\mathrm{n}}+(-2)^{\mathrm{n}}}{\mathrm{n} !}\)

Topic: Mathematics, Sequences and Series, JEE Main
The coefficients a, b and c of the quadratic equation, ax2+bx+c=0 are obtained by throwing a dice three times. The probability that this equation has equal roots is:

Options:
A 172
B 136
C 154
D 5216

The correct answer is:
5216

Solution:
ax2+bx+c=0

a, b, c∈{1, 2, 3, 4, 5, 6}

n(s)=6×6×6=216

For equal root D=0

⇒b2=4ac

Case I: ac=1

a,b,c=1,2,1

Case II: ac=4

a,b,c=1,4,4

or 4,4,1

or 2,4,2

Case-III: ac=9

a,b,c=3,6,3

Required probability =5216

Topic: Mathematics, Probability, JEE Main, JEE Main 2021 (25 Feb Shift 1)
The coefficients $a, b, c$ in the quadratic equation $a x^2+b x+c=0$ are from the set $\{1,2,3,4,5,6\}$. If the probability of this equation having one real root bigger than the other is $p$, then 216 p equals :

Options:
A 57
B 76
C 38
D 19

The correct answer is:
38

Solution:
$\begin{aligned}
& \mathrm{D}>0 \\
& \mathrm{~b}^2>4 \mathrm{ac} \\
& \mathrm{b}=3:(\mathrm{a}, \mathrm{c})=(1,1)(1,2)(2,1) \\
& \mathrm{b}=4:(\mathrm{a}, \mathrm{c})=(1,1)(1,2)(2,1)(1,3)(3,1) \\
& \mathrm{b}=5:(\mathrm{a}, \mathrm{c})=(1,1)(1,2)(2,1)(1,3)(3,1)(1,4)(4,1) \\
&(1,5)(5,1)(1,6)(6,1)(2,3)(3,2)(2,2) \\
& \mathrm{b}=6:(\mathrm{a}, \mathrm{c})=(1,1)(1,2)(2,1)(1,3)(3,1)(1,4)(4,1) \\
&(1,5)(5,1)(1,6)(6,1)(2,3)(3,2)(2,4)(4,2)(2,2)
\end{aligned}$
fav. cases $=38$
Prob. : $\frac{38}{6 \times 6 \times 6}$

Topic: Mathematics, Probability, JEE Main, JEE Main 2024 (05 Apr Shift 2)
The coefficients of apparent expansion of a liquid when determined using two different vessels $A$ and $B$ are $y_1$ and $y_2$, respectively. If the coefficient of linear expansion of the vessel $A$ is $\alpha$, the coefficient of linear expansion of the vessel $B$ is

Options:
A $\frac{a \gamma_1 \gamma_2}{\gamma_1+\gamma_2}$
B $\frac{\gamma_1-\gamma_2}{2 \alpha}$
C $\frac{\gamma_1-\gamma_2+\alpha}{3}$
D $\frac{\gamma_1-\gamma_2}{3}+\alpha$

The correct answer is:
$\frac{\gamma_1-\gamma_2}{3}+\alpha$

Solution:
Coefficient of apparent expansion of liquid for vessel $(A)=\gamma_1$
Coefficient of apparent expansion of liquid for vessel $(B)=\gamma_2$
For vessel $A$
$$
\begin{aligned}
& \gamma_{\text {real }}=\gamma_{\text {app }}+\gamma_{\text {vessel }} \\
& \text { For vessel } \quad \gamma_{\text {vessel }}=3 \alpha_A=3 \alpha \\
& \because \quad \gamma_{\text {real }}=\gamma_1+3 \alpha \\
& \text { For vessel } B, \quad \gamma_{\text {real }}=\gamma_2+3 \alpha_B \\
& \therefore \quad \gamma_1+3 \alpha=\gamma_2+3 \alpha_B \\
& \alpha_B=\frac{\gamma_1-\gamma_2+3 \alpha}{3}=\frac{\gamma_1-\gamma_2}{3}+\alpha \\
&
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, JEE Main
The coefficients of $x^p$ and $x^q$ in the expansion of $(1+x)^{p+q}$ are

Options:
A
equal

B
equal with opposite signs

C
reciprocals of each other

D
none of these

The correct answer is:
equal

Solution:
We have $t_{p+1}={ }^{p+q} C_p x^p$ and $t_{q+1}={ }^{p+q} C_q x^q \quad{ }^{p+q} C_p={ }^{p+q} C_q$

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2002
The coercivity of a small magnet, where the ferromagnet gets demagnetised is 3×103 A/m. The current required to be passed in a solenoid of length 10 cm and number of turns 100, so that the magnet gets demagnetised when inside the solenoid is

Options:
A 30 mA
B 60 mA
C 3 A
D 6 A

The correct answer is:
3 A

Solution:
For solenoid, the magnetic field needed to be magnetised the magnet.

B=μ0nI

where, n=100,l=10 cm=10100 m=0.1 m

⇒3×103=1000.1×I

⇒I=3 A

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2014 (06 Apr)
The cofactor of the element ' 4 ' in the determinant $\left|\begin{array}{llll}1 & 3 & 5 & 1 \\ 2 & 3 & 4 & 2 \\ 8 & 0 & 1 & 1 \\ 0 & 2 & 1 & 1\end{array}\right|$ is

Options:
A 4
B 10
C -10
D -4

The correct answer is:
10

Solution:
The cofactor of element 4 , in the $2^{\text {nd }}$ row and $3^{\text {rd }}$ column is
$\begin{aligned} & =(-1)^{2+3}\left|\begin{array}{lll}1 & 3 & 1 \\ 8 & 0 & 1 \\ 0 & 2 & 1\end{array}\right|=-\{1(-2)-3(8-0)+1.16\} \\ & =10 .\end{aligned}$

Topic: Mathematics, Determinants, JEE Main
The coil of area $0.1 \mathrm{~m}^2$ has 500 turns. After placing the coil in a magnetic field of strength $4 \times 10^{-4} \mathrm{wb} / \mathrm{m}^2$, if rotated through $90^{\circ}$ in $0.1 \mathrm{~s}$, the average emf induced in the coil is

Options:
A $0.012 \mathrm{~V}$
B $0.05 \mathrm{~V}$
C $0.1 \mathrm{~V}$
D $0.2 \mathrm{~V}$

The correct answer is:
$0.2 \mathrm{~V}$

Solution:
$e=\frac{-N B A\left(\cos \theta_2-\cos \theta_1\right)}{\Delta t}$
$=-\frac{500 \times 4 \times 10^{-4} \times 0.1(\cos 90-\cos 0)}{0.1}=0.2 \mathrm{~V}$

Topic: Physics, Electromagnetic Induction, JEE Main
The coil of choke in a circuit

Options:
A Increases the current
B Decreases the current
C Does not change the current
D Has high resistance to dc circuit

The correct answer is:
Decreases the current

Solution:
In a circuit with a.c,source,choke coil which is essentially an inductor with high reactance, is used to decrease the current without loss of energy.No heat is generated so no loss of energy. When we use resistance to reduce current, there is loss of electrical energy in the form of heat generated.

Topic: Physics, Alternating Current, JEE Main
The coils of a step-down transformer have 500 and 5000 turns. In the primary coil an \(\mathrm{AC}\) current of \(4 \mathrm{~A}\) at \(2200 \mathrm{~V}\) is sent. The value of the current and potential difference in the secondary coil is

Options:
A \(20 \mathrm{~A}, 220 \mathrm{~V}\)
B \(0.4 \mathrm{~A}, 22000 \mathrm{~V}\)
C \(40 \mathrm{~A}, 220 \mathrm{~V}\)
D \(40 \mathrm{~A}, 22000 \mathrm{~V}\)

The correct answer is:
\(40 \mathrm{~A}, 220 \mathrm{~V}\)

Solution:
For a step-down transformer,
\(\begin{aligned}
N_P & =5000, N_S=500 \\
I_P & =4 \mathrm{~A}, V_P=2200 \mathrm{~V}
\end{aligned}\)
For a transformer, we know that
\(\begin{aligned}
& \frac{N_S}{N_P}=\frac{V_S}{V_P}=\frac{I_P}{I_S} \\
& \text {Using, } \quad \frac{N_S}{N_P}=\frac{V_S}{V_P} \\
& \Rightarrow \quad V_S=\frac{N_S V_P}{N_P}=\frac{500 \times 2200}{5000}=220 \mathrm{~V} \\
& \text {Again, using } \frac{V_S}{V_P}=\frac{I_P}{I_S} \\
& I_S=\frac{I_P \times V_P}{V_S}=\frac{4 \times 2200}{220}=40 \mathrm{~A} \\
\end{aligned}\)

Topic: Physics, Alternating Current, JEE Main
The cold junction of a thermocouple is at $0^{\circ} \mathrm{C}$. The thermo e.m.f produced in the thermocouple is given by an equation $E=16 T-0.04 T^2$, where $T$ is the temperature of the hot junction. The temperature of inversion and neutral temperature of the thermocouple are

Options:
A $200^{\circ} \mathrm{C} ; 400^{\circ} \mathrm{C}$
B $400^{\circ} \mathrm{C} ; 200^{\circ} \mathrm{C}$
C $200^{\circ} \mathrm{C} ; 300^{\circ} \mathrm{C}$
D $300^{\circ} \mathrm{C} ; 200^{\circ} \mathrm{C}$

The correct answer is:
$400^{\circ} \mathrm{C} ; 200^{\circ} \mathrm{C}$

Solution:
Thermo emf $E=16 T-0.04 T^2$
At inversion temperature $\left(T i\right), E=0$
$$
\therefore \quad \begin{aligned}
0 & =16 T_i-0.04 T_i^2 \\
16 T i & =0.04 T_i^2 \\
T i & =\frac{16}{0.04}=400^{\circ} \mathrm{C}
\end{aligned}
$$
Neutral temperature $T_n=\frac{T i+T}{2}$
$T=$ Temperature of cold end
$$
T_n=\frac{400+0}{2}=200^{\circ} \mathrm{C}
$$

Topic: Physics, Thermal Properties of Matter, JEE Main