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In the expansion of $\left ( \frac{x}{cos\theta }+\frac{1}{x\sin \theta } \right )^{16}$ , if $L_{1}$ is the least value of the term independent of $x$ when $\frac{\pi }{8}\leq \theta \leq \frac{\pi }{4}$ and $L_{2}$ is the least value of the term independent of $x$ when $\frac{\pi }{16}\leq \theta \leq \frac{\pi }{8}$ , then the ratio $L_{2}:L_{1}$ is equal to :
Option: 1 $16:1$
Option: 2 $8:1$
Option: 3 $1:8$
Option: 4 $1:16$

General Term of Binomial Expansion $\left(T_{r+1}\right)^{\mathrm{th}} \text { term is called as general term in }(x+y)^{n}\;\text{and general term is given by}$

$\mathrm{T}_{\mathrm{r}+1}=^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{\;x}^{\mathrm{n}-\mathrm{r}} \cdot \mathrm{y}^{\mathrm{r}}$

Term independent of x: It means term containing x⁰,

Now,

$\\\mathrm{T}_{\mathrm{r}+1}=^{16} \mathrm{C}_{\mathrm{r}}\left(\frac{\mathrm{x}}{\cos \theta}\right)^{16-\mathrm{r}}\left(\frac{1}{\mathrm{x} \sin \theta}\right)^{\mathrm{r}}\\\text{for r = 8 term is free from 'x' }\\\begin{aligned} &\mathrm{T}_{9}=^{16} \mathrm{C}_{8} \frac{1}{\sin ^{8} \theta \cos ^{8} \theta}\\ &\mathrm{T}_{9}=^{16} \mathrm{C}_{8} \frac{2^{8}}{(\sin 2 \theta)^{8}}\\ &\text { in } \theta \in\left[\frac{\pi}{8}, \frac{\pi}{4}\right], L_{1}=^{16} \mathrm{C}_{8} 2^{8} \end{aligned}$

$\\\because \text{Min value of L}_1\;\text{at }\theta=\pi/4\\\text { in } \theta \in\left[\frac{\pi}{16}, \frac{\pi}{8}\right], L_{2}=16 \mathrm{C}_{8} \frac{2^{8}}{\left(\frac{1}{\sqrt{2}}\right)^{8}}=^{16} \mathrm{C}_{8} \cdot 2^{8} \cdot 2^{4}\\\\\because \text{Min value of L}_2\;\text{at }\theta=\pi/8\\\frac{L_{2}}{L_{1}}=\frac{16 \mathrm{C}_{8} \cdot 2^{8} 2^{4}}{^{16} \mathrm{C}_{8} \cdot 2^{8}}=16$

Correct option 1

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Posted by

avinash.dongre

If a, b and c are the greatest values of $^{19}C_{p},^{20}C_{q},^{21}C_{r}$ respectively, then:

Option: 1 $\frac{a}{11}=\frac{b}{22}=\frac{c}{42}$
Option: 2 $\frac{a}{10}=\frac{b}{11}=\frac{c}{42}$
Option: 3 $\frac{a}{11}=\frac{b}{22}=\frac{c}{21}$
Option: 4 $\frac{a}{10}=\frac{b}{11}=\frac{c}{21}$

Binomial Coefficient of the middle term is greatest.

Now,

$\\^nC_{r}\;\text{ is max at middle term}\\\begin{array}{l}{a=^{19} C_{p}=^{19} C_{10}=^{19} C_{9}} \\ {b=^{20} C_{q}=^{20} C_{10}} \\ {c=^{21} C_{r}=^{21} C_{10}=^{21} C_{11}}\end{array}$

$\frac{a}{^{19}C_9}=\frac{b}{ ^{20} \mathrm{C}_{10}}=\frac{c}{^{21} \mathrm{C}_{11}}$

$\frac{a}{^{19}C_9}=\frac{b}{\frac{20}{10} \cdot ^{19} \mathrm{C}_9}=\frac{c}{\frac{21}{11} \cdot \frac{20}{10} ^{19} \mathrm{C}_{9}}$

$\\\frac{a}{1}=\frac{b}{2}=\frac{c}{42 / 11}\\\frac{a}{11}=\frac{b}{22}=\frac{c}{42}$

Correct Option (1)

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Posted by

Kuldeep Maurya

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If the sum of the coefficients of all even powers of x in the product $(1+x+x^{2}+....+x^{2n})(1-x+x^{2}-x^{3}+....+x^{2m})$ is 61, then n is equal to _________.
Option: 1 30
Option: 260
Option: 315
Option: 4 45

$\text { Let } (1+x+x^{2}+....+x^{2n})(1-x+x^{2}-x^{3}+....+x^{2n})=a_{0}+a_{1} x+a_{2} x^{2}+\ldots \ldots$

$\\Put \,\, x=1 \\ (2 n+1).1 =a_{0} + a_{1} + a_{2}+ \ldots \ldots \\ Put\, x =-1 \\ 1.(2 n+1)= a_{0} - a_{1} + a_{2}+\ldots \ldots \\ From (i)+(ii) \\ 4 n+2 =2 (a_{0}+a_{2}+\ldots ) \\ So,\,\, 2 n+1=61 \\ \Rightarrow n=30$

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Posted by

Kuldeep Maurya

If $\alpha$ and $\beta$ be the coefficients of $x^{4}$ and $x^{2}$ respectively in the expansion of $\left ( x+\sqrt{x^{2}-1} \right )^{6}+\left ( x-\sqrt{x^{2}-1} \right )^{6},$ then :
Option: 1 $\alpha +\beta =30$
Option: 2 $\alpha -\beta =-132$
Option: 3 $\alpha +\beta =60$
Option: 4 $\alpha -\beta =60$

We know

$\mathrm{(x+y)^{n}+(x-y)^{n}=2\left[^{n} C_{0}\; x^{n}\; y^{0}+^{n} C_{2} \;x^{n-2}\; y^{2}+^{n} C_{4}\; x^{n-4} \;y^{4}+\ldots .\right]}$

Now,

$\left ( x+\sqrt{x^{2}-1} \right )^{6}+\left ( x-\sqrt{x^{2}-1} \right )^{6}$

$={2\left[^{6} \mathrm{C}_{0} \mathrm{x}^{6}+^{6} \mathrm{C}_{2} \mathrm{x}^{4}\left(\mathrm{x}^{2}-1\right)+^{6} \mathrm{C}_{4} \mathrm{x}^{2}\left(\mathrm{x}^{2}-1\right)^{2}+^{6} \mathrm{C}_{6}\left(\mathrm{x}^{2}-1\right)^{3}\right]} \\ {\quad=2\left[\mathrm{x}^{6}+15\left(\mathrm{x}^{6}-\mathrm{x}^{4}\right)+15 \mathrm{x}^{2}\left(\mathrm{x}^{4}-2 \mathrm{x}^{2}+1\right)+\left(\mathrm{x}^{6}-1-3 \mathrm{x}^{4}+3 \mathrm{x}^{2}\right)\right]} \\ {\quad=2\left(32 \mathrm{x}^{6}-48 \mathrm{x}^{4}+18 \mathrm{x}^{2}-1\right)} \\ {\quad\alpha=-96 \text { and } \beta=36} \\ {\therefore \quad \alpha-\beta=-132}$ Correct Option (2)

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Posted by

vishal kumar

JEE Main high-scoring chapters and topics

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The number of ordered pairs (r,k) for which $6\cdot ^{35}C_{r}=(k^{2}-3)\cdot ^{36}C_{r+1},$ where k is an integer, is :
Option: 1 $4$
Option: 2 6
Option: 3 2
Option: 4 3

As we have learnt

$^{n} C_{r}=\frac{n}{r} \cdot^{n-1} C_{r-1}$

Now,

$^{36}C_{r+1} (k^{2}-3)={^{35}C_{r}}\times {6}$

$\frac{36}{r+1}. ^{35}C_{r} (k^{2}-3)={^{35}C_{r}}\times {6}$

${k^{2}-3=\frac{r+1}{6} \Rightarrow k^{2}=3+\frac{r+1}{6}}$

r should be less than or equal to 35

Hence for k to be an integer, r can be 5 and 35

For r=5 we get k = -2, 2

For r=35 we get k=-3,3

We get 4 ordered pair (5,-2), (5,2), (35,-3), (35, 3)

Correct Option (1)

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Posted by

Ritika Jonwal

The coefficient of $x^{7}$ in the expression $(1+x)^{10}+x(1+x)^{9}+x^{2}(1+x)^{8}+....+x^{10}$ is :
Option: 1 $420$
Option: 2 $330$
Option: 3 $210$
Option: 4 $120$

Binomial Theorem

$(x+y)^{n}=^{n} C_{0} x^{n}+^{n} C_{1} x^{n-1} y+^{n} C_{2} x^{n-2} y^{2}+\cdots+^{n} C_{n} y^{n}\;\;where,\;n\in \mathbb{N}$

Now,

Given series S is a GP, with a = (1+x)¹⁰ , r = x/(1+x), n = 11

So, S = $\frac{(1+x)^{10} \left ( \frac{x}{1+x}^{11}-1 \right )}{(\frac{x}{1+x})-1}$

= (1+x)¹¹- x¹¹

Hence coefficient of x⁷ is ¹¹C₇= 330

Correct option (2)

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Posted by

Ritika Jonwal

NEET 2024 Most scoring concepts

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$\sum_{k=0}^{20}\left({ }^{20} \mathrm{C}_{\mathrm{k}}\right)^{2}$ is equal to :
Option: 1 ${ }^{40} \mathrm{C}_{21}$
Option: 2 ${ }^{41} \mathrm{C}_{20}$
Option: 3 ${ }^{40} C_{20}$
Option: 4 ${ }^{40} C_{19}$

$s= \sum_{k= 0}^{20}\left (\, ^{20}C_{k} \right )^{2}= \, ^{20}C_{0}\: ^{2}+ \, ^{20}C_{1}\: ^{2}+ \, ^{20}C_{2}\: ^{2}+\cdots ^{20}C_{20}\, ^{2}$
it is same as coefficient of $x^{20}$ in the expansion of $\left ( 1+x \right )^{20}\left ( x+1 \right )^{20}$
$= coeff\cdot of\, x^{20}\pi\left ( 1+x \right )^{40}$
$= \, ^{40}C_{20}$
option (3)

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Posted by

Kuldeep Maurya

If the sum of the coefficients in the expansion of $(x+y)^{\mathrm{n}}\: \: is \: \: 4096$ , then the greatest coefficient in the expansion is__________

Sum of coefficients $= 2^{n}=4096$
$\Rightarrow n= 12$

$greatest\: coefficient= \, ^{12}C_{6}$ = 924

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Posted by

Kuldeep Maurya

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If the coefficient of $a^{7} b^{8}$ in the expansion of $(a+2 b+4 a b)^{10} \text { is } K \cdot 2^{16}$ , then $\mathrm{K}$ is equal to _____________.

General Term $= \frac{10!}{p!q!r!}(a)^{p}(2p)^{q}(4ab)^{r}\\$

$=\frac{10!}{p!q!r!}2^{q+2r}a^{p+r}b^{q+r}\\$

$where\; p+q+r=10\\$ .......(1)

$p+r=7\\$ .......(2)

$and\; q+r=8\\$ ........(3)

$(2)+(3)-(1)\Rightarrow r=5, So\; p=2,q=3\\$

$Coefficient=\frac{10!}{5!3!2!}\times2^{3+2\times5}\\$

$=\frac{10\times9\times8\times7\times6}{3\times2\times2}\times2^{13}=315\times2^{16}\\$

$\Rightarrow k=315$

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Posted by

Kuldeep Maurya

If $\left(\frac{3^{6}}{4^{4}}\right) \mathrm{k} ,$ is the term, independent of $x$ in the binomial expansion of $\left(\frac{x}{4}-\frac{12}{x^{2}}\right)^{12}$ , then $\mathrm{k}$ is equal to_________.

General Term $T_{r+1}= ^{12}C_{r}\left ( \frac{x}{4} \right )^{12-r}\left ( \frac{-12}{x^{2}} \right )^{r}$
Term independent of x
$12-r-2r= 0\Rightarrow r= 4$
$T_{5}= \frac{12!}{4!8!}\times \left ( \frac{1}{4} \right )^{8}\times \left ( -12 \right )^{4}$
$= \frac{12\times 11\times 10\times 9}{4\times 3\times 2\times 1}\times \frac{12\times 12\times 12\times 12}{4\times 4\times 4\times4\times4\times4\times4\times4}$
$= 55\times \frac{3^{6}}{4^{4}}$
$\Rightarrow k= 55$

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Get Answers to all your Questions

All Questions

In the expansion of , if is the least value of the term independent of when and is the least value of the term independent of when , then the ratio is equal to : Option: 1 Option: 2 Option: 3 Option: 4

Posted by

avinash.dongre

If a, b and c are the greatest values of respectively, then: Option: 1 Option: 2 Option: 3 Option: 4

Posted by

Kuldeep Maurya

Crack CUET with india's "Best Teachers"

If the sum of the coefficients of all even powers of x in the product is 61, then n is equal to _________. Option: 1 30 Option: 260 Option: 315 Option: 4 45

Posted by

Kuldeep Maurya

If and be the coefficients of and respectively in the expansion of then : Option: 1 Option: 2 Option: 3 Option: 4

Posted by

vishal kumar

JEE Main high-scoring chapters and topics

The number of ordered pairs (r,k) for which where k is an integer, is : Option: 1 Option: 2 6 Option: 3 2 Option: 4 3

If the sum of the coefficients of all even powers of x in the product $(1+x+x^{2}+....+x^{2n})(1-x+x^{2}-x^{3}+....+x^{2m})$ is 61, then n is equal to _________.
Option: 1 30
Option: 260
Option: 315
Option: 4 45

The number of ordered pairs (r,k) for which $6\cdot ^{35}C_{r}=(k^{2}-3)\cdot ^{36}C_{r+1},$ where k is an integer, is :
Option: 1 $4$
Option: 2 6
Option: 3 2
Option: 4 3

The coefficient of $x^{7}$ in the expression $(1+x)^{10}+x(1+x)^{9}+x^{2}(1+x)^{8}+....+x^{10}$ is :
Option: 1 $420$
Option: 2 $330$
Option: 3 $210$
Option: 4 $120$

$\sum_{k=0}^{20}\left({ }^{20} \mathrm{C}_{\mathrm{k}}\right)^{2}$ is equal to :
Option: 1 ${ }^{40} \mathrm{C}_{21}$
Option: 2 ${ }^{41} \mathrm{C}_{20}$
Option: 3 ${ }^{40} C_{20}$
Option: 4 ${ }^{40} C_{19}$

If the sum of the coefficients in the expansion of $(x+y)^{\mathrm{n}}\: \: is \: \: 4096$ , then the greatest coefficient in the expansion is__________

If the coefficient of $a^{7} b^{8}$ in the expansion of $(a+2 b+4 a b)^{10} \text { is } K \cdot 2^{16}$ , then $\mathrm{K}$ is equal to _____________.

If $\left(\frac{3^{6}}{4^{4}}\right) \mathrm{k} ,$ is the term, independent of $x$ in the binomial expansion of $\left(\frac{x}{4}-\frac{12}{x^{2}}\right)^{12}$ , then $\mathrm{k}$ is equal to_________.