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stringclasses 178
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stringlengths 26
9.64k
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stringlengths 2
1.63k
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stringclasses 5
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stringclasses 293
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|---|---|---|---|---|---|---|---|---|---|---|
0TpOHs7adCRBwUIj
|
maths
|
binomial-theorem
|
general-term
|
The coefficient of $${x^n}$$ in expansion of $$\left( {1 + x} \right){\left( {1 - x} \right)^n}$$ is
|
[{"identifier": "A", "content": "$${\\left( { - 1} \\right)^{n - 1}}n$$ "}, {"identifier": "B", "content": "$${\\left( { - 1} \\right)^n}\\left( {1 - n} \\right)$$ "}, {"identifier": "C", "content": "$${\\left( { - 1} \\right)^{n - 1}}{\\left( {n - 1} \\right)^2}$$ "}, {"identifier": "D", "content": "$$\\left( {n - 1} \\right)$$ "}]
|
["B"]
| null |
Given $$\left( {1 + x} \right){\left( {1 - x} \right)^n}$$
<br><br>= $${\left( {1 - x} \right)^n}$$ + $$x{\left( {1 - x} \right)^n}$$
<br><br>General term of $${\left( {1 - x} \right)^n}$$ = $${}^n{C_r}.{\left( { - 1} \right)^r}.{x^r}$$
<br><br>$$\therefore$$ Term containing $${x^n}$$ in $${\left( {1 - x} \right)^n}$$ = $${}^n{C_n}.{\left( { - 1} \right)^n}.{x^n}$$
<br><br>So coefficient of $${x^n}$$ in $${\left( {1 - x} \right)^n}$$ = $${}^n{C_n}.{\left( { - 1} \right)^n}$$
<br><br>General term of $$x{\left( {1 - x} \right)^n}$$ = $${}^n{C_r}.{\left( { - 1} \right)^r}.{x^{r + 1}}$$
<br><br>$$\therefore$$ Term containing $${x^n}$$ in $$x{\left( {1 - x} \right)^n}$$ = $${}^n{C_{n - 1}}.{\left( { - 1} \right)^{n - 1}}.{x^n}$$
<br><br>So coefficient of $${x^n}$$ in $$x{\left( {1 - x} \right)^n}$$ = $${}^n{C_{n - 1}}.{\left( { - 1} \right)^{n - 1}}$$
<br><br>$$\therefore$$ coefficient of $${x^n}$$ in $${\left( {1 - x} \right)^n}$$ + $$x{\left( {1 - x} \right)^n}$$
<br><br>= $${}^n{C_n}.{\left( { - 1} \right)^n}$$ + $${}^n{C_{n - 1}}.{\left( { - 1} \right)^{n - 1}}$$
<br><br>= $${\left( { - 1} \right)^{n - 1}}\left[ {{}^n{C_{n - 1}} - {}^n{C_n}} \right]$$
<br><br>= $${\left( { - 1} \right)^{n - 1}}\left[ {n - 1} \right]$$
<br><br>= $${\left( { - 1} \right)^n}\left[ {1 - n} \right]$$
|
mcq
|
aieee-2004
|
rTXLqH9tIJS0LPKw
|
maths
|
binomial-theorem
|
general-term
|
If the coefficient of $${x^7}$$ in $${\left[ {a{x^2} + \left( {{1 \over {bx}}} \right)} \right]^{11}}$$ equals the coefficient of $${x^{ - 7}}$$ in $${\left[ {ax - \left( {{1 \over {b{x^2}}}} \right)} \right]^{11}}$$, then $$a$$ and $$b$$ satisfy the relation
|
[{"identifier": "A", "content": "$$a - b = 1$$ "}, {"identifier": "B", "content": "$$a + b = 1$$"}, {"identifier": "C", "content": "$${a \\over b} = 1$$ "}, {"identifier": "D", "content": "$$ab = 1$$ "}]
|
["D"]
| null |
General term of $${\left[ {a{x^2} + \left( {{1 \over {bx}}} \right)} \right]^{11}}$$ is T<sub>r+1</sub>.
<br><br>T<sub>r+1</sub> = $${}^{11}{C_r}{\left( {a{x^2}} \right)^{11 - r}}{\left( {{1 \over {bx}}} \right)^r}$$
<br><br>= $${}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( b \right)^{ - r}}{\left( x \right)^{22 - 3r}}$$
<br><br>For the coefficient of x<sup>7</sup>,
<br><br>$$ \Rightarrow $$ 22 - 3r = 7
<br><br>$$ \Rightarrow $$ r = 5
<br><br>So coefficient of x<sup>7</sup> = $${}^{11}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}}$$ ......(1)
<br><br>Now General term of $${\left[ {ax - \left( {{1 \over {b{x^2}}}} \right)} \right]^{11}}$$ is T<sub>r+1</sub>.
<br><br>T<sub>r+1</sub> = $${}^{11}{C_r}{\left( {a{x}} \right)^{11 - r}}{\left( { - {1 \over {bx}}} \right)^r}$$
<br><br>= $${}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( { - 1} \right)^r}{\left( b \right)^{ - r}}{\left( x \right)^{11 - r}}{\left( x \right)^{ - 2r}}$$
<br><br>For the coefficient of x<sup>-7</sup>,
<br><br>11 - 3r = -7
<br><br>$$ \Rightarrow $$ r = 6
<br><br>$$\therefore$$ Coefficient of x<sup>-7</sup> = $${}^{11}{C_6}{\left( a \right)^5}{\left( { - 1} \right)^6}{\left( b \right)^{ - 6}}$$
<br><br>According to question,
<br><br>Coefficient of x<sup>7</sup> = Coefficient of x<sup>-7</sup>
<br><br>$$ \Rightarrow $$ $${}^{11}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}}$$ = $${}^{11}{C_6}{\left( a \right)^5}{\left( { - 1} \right)^6}{\left( b \right)^{ - 6}}$$
<br><br>$$ \Rightarrow $$ $$ab = 1$$
|
mcq
|
aieee-2005
|
cpKuiseKzcfD4PcY
|
maths
|
binomial-theorem
|
general-term
|
If the coefficients of r<sup>th</sup>, (r+1)<sup>th</sup>, and (r + 2)<sup>th</sup> terms in the binomial expansion of $${{\rm{(1 + y )}}^m}$$ are in A.P., then m and r satisfy the equation
|
[{"identifier": "A", "content": "$${m^2} - m(4r - 1) + 4\\,{r^2} - 2 = 0$$ "}, {"identifier": "B", "content": "$${m^2} - m(4r + 1) + 4\\,{r^2} + 2 = 0$$ "}, {"identifier": "C", "content": "$${m^2} - m(4r + 1) + 4\\,{r^2} - 2 = 0$$ "}, {"identifier": "D", "content": "$${m^2} - m(4r - 1) + 4\\,{r^2} + 2 = 0$$ "}]
|
["C"]
| null |
Let r = 2
<br><br>$$\therefore$$ 2nd, 3rd and 4th terms are in AP.
<br><br>2nd term = T<sub>2</sub> = $${}^m{C_1}.y$$
<br><br>Coefficient of T<sub>2</sub> = $${}^m{C_1}$$
<br><br>3rd term = T<sub>3</sub> = $${}^m{C_2}.{y^2}$$
<br><br>Coefficient of T<sub>3</sub> = $${}^m{C_2}$$
<br><br>4th term = T<sub>4</sub> = $${}^m{C_3}.{y^3}$$
<br><br>Coefficient of T<sub>2</sub> = $${}^m{C_3}$$
<br><br>$$\therefore$$ 2.$${}^m{C_2}$$ = $${}^m{C_1}$$ + $${}^m{C_3}$$
<br><br>$$ \Rightarrow $$ $$2.{{m\left( {m - 1} \right)} \over {1.2}}$$ = $${m \over 1}$$ + $${{m\left( {m - 1} \right)\left( {m - 2} \right)} \over {1.2.3}}$$
<br><br>$$ \Rightarrow $$ 6m<sup>2</sup> - 6m = 6m +m(m<sup>2</sup> - 3m + 2)
<br><br>$$ \Rightarrow $$ 6m<sup>2</sup> - 6m = 6m + m<sup>3</sup> - 3m<sup>2</sup> + 2m
<br><br>$$ \Rightarrow $$ 6m - 6 = 6 + m<sup>2</sup> - 3m + 2
<br><br>$$ \Rightarrow $$ m<sup>2</sup> - 9m + 14 = 0
<br><br>Now put r = 2 at each option and find answer.
<br><br>In option C, $${m^2} - m(4r + 1) + 4\,{r^2} - 2 = 0$$ putting r = 2 we get
<br><br>m<sup>2</sup> - 9m + 14 = 0. So Option C is correct.
|
mcq
|
aieee-2005
|
nV7NCqS5hYaVuBUA
|
maths
|
binomial-theorem
|
general-term
|
In the binomial expansion of $${\left( {a - b} \right)^n},\,\,\,n \ge 5,$$ the sum of $${5^{th}}$$ and $${6^{th}}$$ terms is zero, then $$a/b$$ equals
|
[{"identifier": "A", "content": "$${{n - 5} \\over 6}$$ "}, {"identifier": "B", "content": "$${{n - 4} \\over 5}$$ "}, {"identifier": "C", "content": "$${5 \\over {n - 4}}$$ "}, {"identifier": "D", "content": "$${6 \\over {n - 5}}$$ "}]
|
["B"]
| null |
According to the question,
<br><br>t<sub>5</sub> + t<sub>6</sub> = 0
<br><br>$$\therefore$$ $${}^n{C_4}.{a^{n - 4}}.{b^4}$$ + $$\left( { - {}^n{C_5}.{a^{n - 5}}.{b^5}} \right)$$ = 0
<br><br>By solving we get,
<br><br>$${a \over b} = {{n - 4} \over 5}$$
|
mcq
|
aieee-2007
|
3Hpkd6AMmJo5duCX
|
maths
|
binomial-theorem
|
general-term
|
The term independent of $$x$$ in expansion of
<br/> $${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$$ is
|
[{"identifier": "A", "content": "4 "}, {"identifier": "B", "content": "120"}, {"identifier": "C", "content": "210"}, {"identifier": "D", "content": "310"}]
|
["C"]
| null |
$${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$$
<br><br>= $${\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \right)^{10}}$$
<br><br>= $${\left( {{{\left( {{x^{1/3}} + 1} \right)\left( {{x^{2/3}} - {x^{1/3}} + 1} \right)} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)} \over {\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}$$
<br><br>= $${\left( {\left( {{x^{1/3}} + 1} \right) - {{\left( {\sqrt x + 1} \right)} \over {\sqrt x }}} \right)^{10}}$$
<br><br>= $${\left( {\left( {{x^{1/3}} + 1} \right) - \left( {1 + {1 \over {\sqrt x }}} \right)} \right)^{10}}$$
<br><br>= $${\left( {{x^{1/3}} - {1 \over {{x^{1/2}}}}} \right)^{10}}$$
<br><br>[<b>Note:</b>
<br><br>For $${\left( {{x^\alpha } \pm {1 \over {{x^\beta }}}} \right)^n}$$ the $$\left( {r + 1} \right)$$<sup>th</sup> term with power m of x is
<br><br>$$r = {{n\alpha - m} \over {\alpha + \beta }}$$]
<br><br>Here $$\alpha = {1 \over 3}$$, $$\beta = {1 \over 2}$$ and m = 0
<br><br>then $$r = {{10 \times {1 \over 3} - 0} \over {{1 \over 3} + {1 \over 2}}}$$ = $${{10} \over 3} \times {6 \over 5}$$ = 4
<br><br>$$\therefore$$ T<sub>5</sub> is the term independent of x.
<br><br>$$\therefore$$ T<sub>5</sub> = $${}^{10}{C_4}$$ = 210
|
mcq
|
jee-main-2013-offline
|
QUwbxq162QAKcHvSXAx82
|
maths
|
binomial-theorem
|
general-term
|
For x $$ \in $$ <b>R</b>, x $$ \ne $$ -1,
<br/><br/>if (1 + x)<sup>2016</sup> + x(1 + x)<sup>2015</sup> + x<sup>2</sup>(1 + x)<sup>2014</sup> + . . . . + x<sup>2016</sup> =
<br/><br/>$$\sum\limits_{i = 0}^{2016} {{a_i}} \,{x^i},\,\,$$ then a<sub>17</sub> is equal to :
|
[{"identifier": "A", "content": "$${{2017!} \\over {17!\\,\\,\\,2000!}}$$"}, {"identifier": "B", "content": "$${{2016!} \\over {17!\\,\\,\\,1999!}}$$"}, {"identifier": "C", "content": "$${{2017!} \\over {2000!}}$$"}, {"identifier": "D", "content": "$${{2016!} \\over {16!}}$$"}]
|
["A"]
| null |
Assume,
<br><br>P = (1 + x)<sup>2016</sup> + x(1 + x)<sup>2015</sup> + . . . . .+ x<sup>2015</sup> . (1 + x) + x<sup>2016</sup> . . . . .(1)
<br><br>Multiply this with $$\left( {{x \over {1 + x}}} \right),$$
<br><br>$$\left( {{x \over {1 + x}}} \right)P = $$ x(1 + x)<sup>2015</sup> + x<sup>2</sup>(1 + x)<sup>2014</sup> +
<br><br> . . . . . . + x<sup>2016</sup> + $${{{x^{2017}}} \over {1 + x}}$$ . . . . . (2)
<br><br>Performing (1) $$-$$ (2), we get
<br><br>$${P \over {1 + x}} = $$ (1 + x)<sup>2016</sup> $$-$$ $${{{x^{2017}}} \over {1 + x}}$$
<br><br>$$ \Rightarrow $$ P = (1 + x)<sup>2017</sup> $$-$$ x<sup>2017</sup>
<br><br>$$ \therefore $$ a<sub>17</sub> = coefficient of x<sup>17</sup> $$=$$ <sup>2017</sup>C<sub>17</sub> $$=$$ $${{2017!} \over {17!\,\,2000!}}$$
|
mcq
|
jee-main-2016-online-9th-april-morning-slot
|
XA8U3jV98E5XEVrRGyjEj
|
maths
|
binomial-theorem
|
general-term
|
If the coefficients of x<sup>−2</sup> and x<sup>−4</sup> in the expansion of $${\left( {{x^{{1 \over 3}}} + {1 \over {2{x^{{1 \over 3}}}}}} \right)^{18}},\left( {x > 0} \right),$$ are m and n respectively, then $${m \over n}$$ is equal to :
|
[{"identifier": "A", "content": "182"}, {"identifier": "B", "content": "$${4 \\over 5}$$"}, {"identifier": "C", "content": "$${5 \\over 4}$$"}, {"identifier": "D", "content": "27"}]
|
["A"]
| null |
T<sub>r+1</sub> = <sup>18</sup>C<sub>r</sub> $${\left( {{x^{{1 \over 3}}}} \right)^{18 - r}}$$ . $${\left( {{1 \over {2{x^{{1 \over 3}}}}}} \right)^r}$$
<br><br>= <sup>18</sup>C<sub>r</sub> $${\left( {{1 \over 2}} \right)^r}\,\,.\,\,{x^{{{18 - 2r} \over 3}}}$$
<br><br>For coefficient of x<sup>$$-$$2</sup>,
<br><br>$${{18 - 2r} \over 3}$$ = $$-$$2
<br><br>$$ \Rightarrow $$ r = 12
<br><br>$$ \therefore $$ Coefficient of x<sup>$$-$$2</sup> is (m) = <sup>18</sup>C<sub>12</sub> $${\left( {{1 \over 2}} \right)^{12}}$$
<br><br>For coefficient of x<sup>$$-$$4</sup>,
<br><br>$${{18 - 2r} \over 3}$$ = $$-$$ 4
<br><br>$$ \Rightarrow $$ r = 15
<br><br>$$ \therefore $$ Coefficient of x<sup>$$-$$4</sup> is (n) = <sup>18</sup>C<sub>15</sub> $$\left( {{1 \over {2}}} \right)$$<sup>15</sup>
<br><br>$$ \therefore $$ $${m \over n} = {{^{18}{C_{12}}{{\left( {{1 \over 2}} \right)}^{12}}} \over {^{18}{C_{15}}{{\left( {{1 \over 2}} \right)}^{15}}}}$$
<br><br>= $${{{}^{18}{C_6} \times {{\left( 2 \right)}^3}} \over {{}^{18}{C_3}}}$$
<br><br>= 182
|
mcq
|
jee-main-2016-online-10th-april-morning-slot
|
MegnCy5y6WH4JxsjLHqWr
|
maths
|
binomial-theorem
|
general-term
|
The coefficient of x<sup>−5</sup> in the binomial expansion of
<br/><br/>$${\left( {{{x + 1} \over {{x^{{2 \over 3}}} - {x^{{1 \over 3}}} + 1}} - {{x - 1} \over {x - {x^{{1 \over 2}}}}}} \right)^{10}},$$ where x $$ \ne $$ 0, 1, is :
|
[{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "$$-$$ 4"}, {"identifier": "D", "content": "$$-$$ 1"}]
|
["A"]
| null |
$${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$$
<br><br>= $${\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \right)^{10}}$$
<br><br>= $${\left( {{{\left( {{x^{1/3}} + 1} \right)\left( {{x^{2/3}} - {x^{1/3}} + 1} \right)} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)} \over {\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}$$
<br><br>= $${\left( {\left( {{x^{1/3}} + 1} \right) - {{\left( {\sqrt x + 1} \right)} \over {\sqrt x }}} \right)^{10}}$$
<br><br>= $${\left( {\left( {{x^{1/3}} + 1} \right) - \left( {1 + {1 \over {\sqrt x }}} \right)} \right)^{10}}$$
<br><br>= $${\left( {{x^{1/3}} - {1 \over {{x^{1/2}}}}} \right)^{10}}$$
<br><br>[<b>Note:</b>
<br><br>For $${\left( {{x^\alpha } \pm {1 \over {{x^\beta }}}} \right)^n}$$ the $$\left( {r + 1} \right)$$<sup>th</sup> term with power m of x is
<br><br>$$r = {{n\alpha - m} \over {\alpha + \beta }}$$]
<br><br>Here $$\alpha = {1 \over 3}$$, $$\beta = {1 \over 2}$$ and m = -5
<br><br>then $$r = {{10 \times {1 \over 3} - (-5)} \over {{1 \over 3} + {1 \over 2}}}$$ = $${{25} \over 3} \times {6 \over 5}$$ = 10
<br><br>$$\therefore$$ T<sub>11</sub> is the term with x<sup>-5</sup>.
<br><br>$$\therefore$$ T<sub>11</sub> = $${}^{10}{C_{10}}$$ = 1
|
mcq
|
jee-main-2017-online-9th-april-morning-slot
|
SVFbx781jSVe3AbPjftul
|
maths
|
binomial-theorem
|
general-term
|
The total number of irrational terms in the binomial expansion of (7<sup>1/5</sup> – 3<sup>1/10</sup>)<sup>60</sup> is :
|
[{"identifier": "A", "content": "54 "}, {"identifier": "B", "content": "55"}, {"identifier": "C", "content": "49"}, {"identifier": "D", "content": "48"}]
|
["A"]
| null |
General term T<sub>r+1</sub> = <sup>60</sup><sup></sup>C<sub>r</sub>, $${7^{{{60 - r} \over 5}}}{3^{{r \over {10}}}}$$
<br><br>$$ \therefore $$ for rational term, r = 0, 10, 20, 30, 40, 50, 60
<br><br>$$ \Rightarrow $$ no of rational terms = 7
<br><br>$$ \therefore $$ number of irrational terms = 54
|
mcq
|
jee-main-2019-online-12th-january-evening-slot
|
toC7gwkJsULkHFwpDn3rsa0w2w9jxb4rh40
|
maths
|
binomial-theorem
|
general-term
|
The term independent of x in the expansion of
<br/>$$\left( {{1 \over {60}} - {{{x^8}} \over {81}}} \right).{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}$$ is equal to :
|
[{"identifier": "A", "content": "36"}, {"identifier": "B", "content": "- 108"}, {"identifier": "C", "content": "- 36"}, {"identifier": "D", "content": "- 72"}]
|
["C"]
| null |
Given expression = $$\left( {{1 \over {60}} - {{{x^8}} \over {81}}} \right).{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}$$
<br><br>= $${1 \over {60}}{\left( {2{x^3} - {3 \over {{x^2}}}} \right)^6} - {{{x^8}} \over {81}}{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}$$
<br><br>So its general term is
<br><br>T<sub>r + 1</sub> = $${1 \over {60}}{}^6{C_r}{\left( {2{x^2}} \right)^{6 - r}}{\left( { - {3 \over {{x^2}}}} \right)^r} - {{{x^8}} \over {81}}{}^6{C_r}{\left( {2{x^2}} \right)^{6 - r}}{\left( { - {3 \over {{x^2}}}} \right)^r}$$
<br><br>= $${1 \over {60}}{}^6{C_r}{\left( 2 \right)^{6 - r}}{\left( { - 3} \right)^r}{x^{12 - 4r}} - {1 \over {81}}{}^6{C_r}{\left( 2 \right)^{6 - r}}{\left( { - 3} \right)^r}{x^{20 - 4r}}$$ .....(i)
<br><br>For this term to be independent of x, put r = 3 in
1<sup>st</sup> part and r = 5 in 2<sup>nd</sup> part.
<br><br>So from (i) the term independent of<br> x = $${1 \over {60}} \times {2^3} \times {\left( { - 3} \right)^3} \times {}^6{C_3} + \left( { - {1 \over {81}}} \right)(2){( - 3)^5} \times {}^6{C_5}$$<br>
= -72 + 36 = <b>-36</b>
|
mcq
|
jee-main-2019-online-12th-april-evening-slot
|
PTfHNLUxmzZnbpCIRr3rsa0w2w9jx5ztehd
|
maths
|
binomial-theorem
|
general-term
|
The coefficient of x<sup>18</sup> in the product
<br/>(1 + x) (1 – x)<sup>10</sup> (1 + x + x<sup>2</sup>)<sup>9</sup>
is :
|
[{"identifier": "A", "content": "126"}, {"identifier": "B", "content": "- 84"}, {"identifier": "C", "content": "- 126"}, {"identifier": "D", "content": "84"}]
|
["D"]
| null |
Coefficient of x<sup>18</sup> in (1 + x) (1 - x)<sup>10</sup> (1 + x + x<sup>2</sup>)<sup>9</sup><br><br>
$$ \Rightarrow $$ Coefficient of x<sup>18</sup> in {(1 - x) (1 - x<sup>2</sup>) (1 + x + x<sup>2</sup>)}<sup>9</sup><br><br>
$$ \Rightarrow $$ Coefficient of x<sup>18</sup> in (1 - x<sup>2</sup>) (1 - x<sup>3</sup>)<sup>9</sup><br><br>
$$ \Rightarrow $$ <sup>9</sup>C<sub>6</sub> - 0 = 84
|
mcq
|
jee-main-2019-online-12th-april-morning-slot
|
yO8G2rHmbJfQTsDYPk3rsa0w2w9jx23emrp
|
maths
|
binomial-theorem
|
general-term
|
The smallest natural number n, such that the coefficient of x in the expansion of $${\left( {{x^2} + {1 \over {{x^3}}}} \right)^n}$$ is <sup>n</sup>C<sub>23</sub>, is :
|
[{"identifier": "A", "content": "23"}, {"identifier": "B", "content": "58"}, {"identifier": "C", "content": "38"}, {"identifier": "D", "content": "35"}]
|
["C"]
| null |
General term<br><br>
$${T_{r + 1}} = {}^n{C_r}{x^{2n - 2r}}.{x^{ - 3r}}$$<br><br>
$$ \therefore $$ $$2n - 5r = 1 \Rightarrow 2n = 5r + 1$$<br><br>
$$ \therefore $$ $$r = {{2n - 1} \over 5}$$<br><br>
$$ \Rightarrow $$ Coefficient of x = $${}^n{C_{\left( {{{2n - 1} \over 5}} \right)}} = {}^n{C_{23}}$$<br><br>
$$ \Rightarrow $$ $${{2n - 1} \over 5} = 23\,\,or\,\,n - \left( {{{2n - 1} \over 5}} \right) = 23$$<br><br>
$$ \Rightarrow $$ 2n - 1 = 115 $$ \Rightarrow $$ n = 58<br><br>
and n = 38<br><br>
$$ \therefore $$ smallest n = 38
|
mcq
|
jee-main-2019-online-10th-april-evening-slot
|
3TQ0nlDrlHmOpomuWHpzD
|
maths
|
binomial-theorem
|
general-term
|
If the fourth term in the binomial expansion of $${\left( {{2 \over x} + {x^{{{\log }_8}x}}} \right)^6}$$
(x > 0) is 20 × 8<sup>7</sup>, then a value of
x is :
|
[{"identifier": "A", "content": "8<sup>\u20132</sup>"}, {"identifier": "B", "content": "8<sup>2</sup>"}, {"identifier": "C", "content": "8<sup>3</sup>"}, {"identifier": "D", "content": "8"}]
|
["B"]
| null |
$${\left( {{2 \over x} + {x^{{{\log }_8}x}}} \right)^6}$$
<br><br>Given T<sub>4</sub> = 20 × 8<sup>7</sup>
<br><br>$$ \Rightarrow $$ $${}^6{C_3}{\left( {{2 \over x}} \right)^3}{\left( {{x^{{{\log }_8}x}}} \right)^3}$$ = 20 × 8<sup>7</sup>
<br><br>$$ \Rightarrow $$ 20$$ \times $$$${8 \over {{x^3}}} \times {x^{3{{\log }_8}x}}$$ = 20 × 8<sup>7</sup>
<br><br>$$ \Rightarrow $$ $${x^{3{{\log }_8}x - 3}}$$ = 8<sup>6</sup>
<br><br>Taking $${{{\log }_8}}$$ both side
<br><br>$$ \Rightarrow $$ ($${3{{\log }_8}x - 3}$$) $$ \times $$ $${{{\log }_8}x}$$ = 6
<br><br>$$ \Rightarrow $$ $$3{\left( {{{\log }_8}x} \right)^2}$$ - 3$${{{\log }_8}x}$$ = 6
<br><br>$$ \Rightarrow $$ $${\left( {{{\log }_8}x} \right)^2}$$ - $${{{\log }_8}x}$$ = 2
<br><br>$$ \Rightarrow $$ ($${{{\log }_8}x}$$ - 2)($${{{\log }_8}x}$$ + 1) = 0
<br><br>$$ \Rightarrow $$ $${{{\log }_8}x}$$ = 2 or $${{{\log }_8}x}$$ = -1
<br><br>$$ \Rightarrow $$ x = 8<sup>2</sup> or x = $${1 \over 8}$$
|
mcq
|
jee-main-2019-online-9th-april-morning-slot
|
en2LblqYlH2raE9M7kcBX
|
maths
|
binomial-theorem
|
general-term
|
If the fourth term in the binomial expansion of<br/>
$${\left( {\sqrt {{x^{\left( {{1 \over {1 + {{\log }_{10}}x}}} \right)}}} + {x^{{1 \over {12}}}}} \right)^6}$$ is equal to 200, and x > 1,
then the value of x is :
|
[{"identifier": "A", "content": "100"}, {"identifier": "B", "content": "10<sup>3</sup>"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "10<sup>4</sup>"}]
|
["C"]
| null |
Fourth term (T<sub>4</sub>)
<br><br>= $${}^6{C_3}{\left( {\sqrt {{x^{\left( {{1 \over {1 + {{\log }_{10}}x}}} \right)}}} } \right)^3}{\left( {{x^{{1 \over {12}}}}} \right)^3}$$
<br><br>= $$20{\left( {{x^{\left( {{1 \over {1 + {{\log }_{10}}x}}} \right)}}} \right)^{{3 \over 2}}}\left( {{x^{{1 \over 4}}}} \right)$$
<br><br>= $$20 \times {x^{\left( {{1 \over {1 + {{\log }_{10}}x}}} \right){3 \over 2}}} \times {x^{{1 \over 4}}}$$
<br><br>= $$20 \times {x^{\left( {{3 \over {2\left( {1 + {{\log }_{10}}x} \right)}} + {1 \over 4}} \right)}}$$
<br><br>Given, T<sub>4</sub> = 200
<br><br>$$ \therefore $$ $$20 \times {x^{\left( {{3 \over {2\left( {1 + {{\log }_{10}}x} \right)}} + {1 \over 4}} \right)}}$$ = 200
<br><br>$$ \Rightarrow $$ $${x^{\left( {{3 \over {2\left( {1 + {{\log }_{10}}x} \right)}} + {1 \over 4}} \right)}}$$ = 10
<br><br>Taking log<sub>10</sub> on both sides
<br><br>$$\left( {{3 \over {2\left( {1 + {{\log }_{10}}x} \right)}} + {1 \over 4}} \right){\log _{10}}x$$ = 1
<br><br>put log<sub>10</sub> x = t
<br><br>$$\left( {{3 \over {2\left( {1 + t} \right)}} + {1 \over 4}} \right)t$$ = 1
<br><br>$$ \Rightarrow $$ $$\left( {{{\left( {1 + t} \right) + 6} \over {4\left( {1 + t} \right)}}} \right) \times t$$ = 1
<br><br>$$ \Rightarrow $$ t<sup>2</sup> + 7t = 4 + 4t
<br><br>$$ \Rightarrow $$ t<sup>2</sup> + 3t - 4 = 0
<br><br>$$ \Rightarrow $$ (t + 4)(t - 1) = 0
<br><br>$$ \Rightarrow $$ t = 1 or t = - 4
<br><br>$$ \therefore $$ log<sub>10</sub> x = 1
<br><br>$$ \Rightarrow $$ x = 10
<br><br>or log<sub>10</sub> x = - 4
<br><br>$$ \Rightarrow $$ x = 10<sup>-4</sup>
<br><br>But as x > 1 so x $$ \ne $$ 10<sup>-4</sup>
<br><br>$$ \therefore $$ x = 10
|
mcq
|
jee-main-2019-online-8th-april-evening-slot
|
8ItKCdunQp0EBnpdGh1Tw
|
maths
|
binomial-theorem
|
general-term
|
A ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of $${\left( {{2^{1/3}} + {1 \over {2{{\left( 3 \right)}^{1/3}}}}} \right)^{10}}$$ is :
|
[{"identifier": "A", "content": "1 : 2(6)<sup>1/3</sup>"}, {"identifier": "B", "content": "1 : 4(6)<sup>1/3</sup>"}, {"identifier": "C", "content": "2(36)<sup>1/3</sup> : 1"}, {"identifier": "D", "content": "4(36)<sup>1/3</sup> : 1"}]
|
["D"]
| null |
$${{{T_5}} \over {T_5^1}} = {{{}^{10}{C_4}{{\left( {{2^{1/3}}} \right)}^{10 - 4}}{{\left( {{1 \over {2{{\left( 3 \right)}^{1/3}}}}} \right)}^4}} \over {{}^{10}{C_4}{{\left( {{1 \over {2\left( {{3^{1/3}}} \right)}}} \right)}^{10 - 4}}{{\left( {{2^{1/3}}} \right)}^4}}} = 4.{\left( {36} \right)^{1/3}}$$
|
mcq
|
jee-main-2019-online-12th-january-morning-slot
|
xJn72qJEiHbMt3IWxBsj5
|
maths
|
binomial-theorem
|
general-term
|
The positive value of $$\lambda $$ for which the co-efficient of x<sup>2</sup>
in the expression x<sup>2</sup> $${\left( {\sqrt x + {\lambda \over {{x^2}}}} \right)^{10}}$$ is 720, is -
|
[{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "$$2\\sqrt 2 $$"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "$$\\sqrt 5 $$"}]
|
["A"]
| null |
The general term in the expansion of the binomial expression $(a+b)^n$ is
<br/><br/>$$
T_{r+1}={ }^n C_r a^{n-r} b^r
$$
<br/><br/>Therefore, the general term in the expansion of the binomial expression <br/><br/>$x^2\left(\sqrt{x}+\frac{\lambda}{x^2}\right)^{10}$ is
<br/><br/>$$
\begin{aligned}
T_{r+1} & =x^2\left({ }^{10} C_r(\sqrt{x})^{10-r}\left(\frac{\lambda}{x^2}\right)^r\right) \\\\
& ={ }^{10} C_r x^2 \cdot x^{\frac{10-r}{2}} \lambda^r x^{-2 r} \\\\
& ={ }^{10} C_r \lambda^r x^{2+\frac{10-r}{2}-2 r}
\end{aligned}
$$
<br/><br/>Now, for the coefficient of $x^2$,
<br/><br/>$$
\begin{aligned}
2+\frac{10-r}{2}-2 r =2 \\\\
\Rightarrow \frac{10-r}{2}-2 r =0 \\\\
\Rightarrow 10-r =4 r \Rightarrow r=2
\end{aligned}
$$
<br/><br/>So, the coefficient of $x^2$ is ${ }^{10} C_2 \lambda^2=720$
<br/><br/>$$
\begin{aligned}
\Rightarrow & \frac{10 !}{2 ! 8 !} \lambda^2 =720 \\\\
\Rightarrow & \frac{10 \cdot 9 \cdot 8 !}{2 \cdot 8 !} \lambda^2 =720 \\\\
\Rightarrow & 45 \lambda^2 =720 \\\\
\Rightarrow & \lambda^2 =16 \\\\
\Rightarrow & \lambda = \pm 4
\end{aligned}
$$
<br/><br/>Given the problem asks for the positive value of $\lambda$, $\lambda = 4$.
<br/><br/>So, the correct option is (A) 4.
|
mcq
|
jee-main-2019-online-10th-january-evening-slot
|
p3tVtM7wc6QwB2WNUeyvG
|
maths
|
binomial-theorem
|
general-term
|
If the third term in the binomial expansion <br/>of $${\left( {1 + {x^{{{\log }_2}x}}} \right)^5}$$ equals 2560, then a possible value of x is -
|
[{"identifier": "A", "content": "$$2\\sqrt 2 $$"}, {"identifier": "B", "content": "$$4\\sqrt 2 $$"}, {"identifier": "C", "content": "$${1 \\over 8}$$"}, {"identifier": "D", "content": "$${1 \\over 4}$$"}]
|
["D"]
| null |
$${\left( {1 + {x^{{{\log }_2}x}}} \right)^5}$$
<br><br>$${T_3} = {}^5{C_2}.{\left( {{x^{{{\log }_2}x}}} \right)^2} = 2560$$
<br><br>$$ \Rightarrow \,\,10.{x^{2{{\log }_2}x}} = 2560$$
<br><br>$$ \Rightarrow \,\,{x^{2\log 2x}} = 256$$
<br><br>$$ \Rightarrow \,\,2{({\log _2}x)^2} = {\log _2}256$$
<br><br>$$ \Rightarrow 2{({\log _2}x)^2} = 8$$
<br><br>$$ \Rightarrow \,\,{({\log _2}x)^2} = 4$$
<br><br>$$ \Rightarrow \,\,{\log _2}x = 2$$ or $$-$$ 2
<br><br>$$x = 4$$ or $${1 \over 4}$$
|
mcq
|
jee-main-2019-online-10th-january-morning-slot
|
n2K8x8uBYrxmQfciha7k9k2k5itwjym
|
maths
|
binomial-theorem
|
general-term
|
The coefficient of x<sup>4</sup> is the expansion of
(1 + x + x<sup>2</sup>)<sup>10</sup> is _____.
|
[]
| null |
615
|
(1 + x + x<sup>2</sup>)<sup>10</sup>
<br><br>= <sup>10</sup>C<sub>0</sub>(1 + x)<sup>10</sup> + <sup>10</sup>C<sub>1</sub>(1 + x)<sup>9</sup>.x<sup>2</sup> + <sup>10</sup>C<sub>2</sub>(1 + x)<sup>8</sup>.x<sup>4</sup>+ .....
<br><br>Coefficient of x<sup>4</sup>
<br><br>= <sup>10</sup>C<sub>0</sub>.<sup>10</sup>C<sub>4</sub> + <sup>10</sup>C<sub>1</sub>.<sup>9</sup>C<sub>2</sub> + <sup>10</sup>C<sub>2</sub>.<sup>8</sup>C<sub>0</sub>
<br><br>= 210 + 360 + 45
<br><br>= 615
|
integer
|
jee-main-2020-online-9th-january-morning-slot
|
mfa9gcoLNQtl4I0BUejgy2xukg0d0dyo
|
maths
|
binomial-theorem
|
general-term
|
If the constant term in the binomial expansion
of <br/>$${\left( {\sqrt x - {k \over {{x^2}}}} \right)^{10}}$$ is 405, then |k| equals :
|
[{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2"}]
|
["A"]
| null |
$${\left( {\sqrt x - {k \over {{x^2}}}} \right)^{10}}$$
<br><br>r<sup>th</sup> term of the expansion,
<br><br>T<sub>r+1</sub> = <sup>10</sup>C<sub>r</sub>$${\left( {\sqrt x } \right)^{10 - r}}{\left( {{{ - k} \over {{x^2}}}} \right)^r}$$
<br><br>= <sup>10</sup>C<sub>r</sub>.$${x^{{{10 - r} \over 2}}}.{\left( { - k} \right)^r}.{x^{ - 2r}}$$
<br><br>= <sup>10</sup>C<sub>r</sub>.$${x^{{{10 - 5r} \over 2}}}.{\left( { - k} \right)^r}$$
<br><br>If it is constant term then <br>$${{{10 - 5r} \over 2}}$$ = 0
<br>$$ \Rightarrow $$ r = 2
<br><br>T<sub>3</sub> = <sup>10</sup>C<sub>2</sub>.(-k)<sup>2</sup> = 405
<br><br>$$ \Rightarrow $$ k<sup>2</sup> = $${{405} \over {45}}$$ = 9
<br><br>$$ \Rightarrow $$ k = $$ \pm $$ 3
<br><br>$$ \Rightarrow $$ |k| = 3
|
mcq
|
jee-main-2020-online-6th-september-evening-slot
|
57I50P6FMvU7vioAmejgy2xukfjjs70u
|
maths
|
binomial-theorem
|
general-term
|
The natural number m, for which the coefficient of x in the binomial expansion of<br/><br/>
$${\left( {{x^m} + {1 \over {{x^2}}}} \right)^{22}}$$ is 1540, is .............
|
[]
| null |
13
|
General term,
<br><br>$${T_{r + 1}} = {}^{22}{C_r}{({x^m})^{22 - r}}{\left( {{1 \over {{x^2}}}} \right)^r} = {}^{22}{C_r}{x^{22m - mr - 2r}}$$<br><br>$$ \because $$ $${}^{22}{C_3} = {}^{22}{C_{19}} = 1540$$<br><br>$$ \therefore $$ $$r = 3\,or\,19$$<br><br>$$22m - mr - 2r = 1$$<br><br>$$m = {{2r + 1} \over {22 - 5}}$$<br><br>When $$r = 3$$, $$m = {7 \over {19}} \notin N$$<br><br>When $$r = 19$$, $$m = {{38 + 1} \over {22 - 19}} = {{39} \over 3} = 13$$<br><br>$$ \therefore $$ $$m = 13$$
|
integer
|
jee-main-2020-online-5th-september-morning-slot
|
9JtYhkWW4qfFVXr41hjgy2xukf443l60
|
maths
|
binomial-theorem
|
general-term
|
If the term independent of x in the expansion of
<br/>$${\left( {{3 \over 2}{x^2} - {1 \over {3x}}} \right)^9}$$ is k, then 18 k is equal to :
|
[{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "11"}]
|
["C"]
| null |
General term,
<br><br>$${T_{r + 1}} = {}^9{C_r}{\left[ {{3 \over 2}{x^2}} \right]^{9 - r}}{\left( { - {1 \over {3x}}} \right)^r}$$<br><br>$${T_{r + 1}} = {}^9{C_r}{\left[ {{3 \over 2}} \right]^{9 - r}}{\left( { - {1 \over 3}} \right)^r}{x^{18 - 3r}}$$<br><br>For independent of x <br><br>18 $$ - $$ 3r = 0 $$ \Rightarrow $$ r = 6<br><br>$$ \therefore $$ $${T_7} = {}^9{C_6}{\left( {{3 \over 2}} \right)^3}{\left( { - {1 \over 3}} \right)^6} = {{21} \over {54}} = k$$<br><br>$$ \therefore $$ $$18k = {{21} \over {54}} \times 18 = 7$$
|
mcq
|
jee-main-2020-online-3rd-september-evening-slot
|
dPu5jf9EUPJ9Gi2nwLjgy2xukf0wsoo1
|
maths
|
binomial-theorem
|
general-term
|
If the number of integral terms in the expansion
<br/>of (3<sup>1/2</sup> + 5<sup>1/8</sup>)<sup>n</sup> is exactly 33, then the least value
of n is :
|
[{"identifier": "A", "content": "264"}, {"identifier": "B", "content": "256"}, {"identifier": "C", "content": "128"}, {"identifier": "D", "content": "248"}]
|
["B"]
| null |
General term of the expression,<br><br>$${T_{r + 1}} = {}^n{C_r}{\left( {{3^{{1 \over 2}}}} \right)^{n - r}}{\left( {{5^{{1 \over 8}}}} \right)^r}$$<br><br>$$ = {}^n{C_r}{\left( 3 \right)^{{{n - r} \over 2}}}{\left( 5 \right)^{{r \over 8}}}$$<br><br>We will get integral term when $${{n - r} \over 2}$$ and $${r \over 8}$$ are integer<br><br>$$ \therefore $$ <b>(1)</b> n $$-$$ r is multiple of 2<br><br>$$ \Rightarrow $$ n $$-$$ r = 0, 2, 4, ......<br><br><b>(2)</b> r is multiple of 8<br><br>$$ \Rightarrow $$ r = 0, 8, 16, .......<br><br>From this two conditions common values are = 0, 8, 16, ....... which will becomes integral terms.<br><br>Given that there are 33 integral terms.<br><br>Here first integral term at 0<sup>th</sup> position.<br><br>Second integral term at 8<sup>th</sup> position.<br><br>$$ \therefore $$ 33<sup>th</sup> integral term will be at = 0 + (33 $$-$$ 1)8 = 256<br><br>So, there should be at least 256 terms.
|
mcq
|
jee-main-2020-online-3rd-september-morning-slot
|
teMqBnHZQ2oEeeo2dBjgy2xukewmb3vi
|
maths
|
binomial-theorem
|
general-term
|
Let
$$\alpha $$ > 0,
$$\beta $$ > 0 be such that
<br/>$$\alpha $$<sup>3</sup> + $$\beta $$<sup>2</sup> = 4. If the
maximum value of the term independent of x in
<br/>the binomial expansion of
$${\left( {\alpha {x^{{1 \over 9}}} + \beta {x^{ - {1 \over 6}}}} \right)^{10}}$$
is 10k,
<br/>then k is equal to :
|
[{"identifier": "A", "content": "176"}, {"identifier": "B", "content": "336"}, {"identifier": "C", "content": "352"}, {"identifier": "D", "content": "84"}]
|
["B"]
| null |
General term
<br><br>T<sub>r + 1</sub> = <sup>10</sup>C<sub>r</sub>$${\alpha ^{10 - r}}.{\left( x \right)^{{{10 - r} \over 9}}}.{\beta ^r}{\left( x \right)^{ - {r \over 6}}}$$
<br><br>= <sup>10</sup>C<sub>r</sub>$${\alpha ^{10 - r}}{\beta ^r}.{\left( x \right)^{{{10 - r} \over 9} - {r \over 6}}}$$
<br><br>If T<sub>r + 1</sub> is independent of x
<br><br>$$ \therefore $$ $${{{10 - r} \over 9} - {r \over 6}}$$ = 0
<br><br>$$ \Rightarrow $$ r = 4
<br><br>$$ \therefore $$ T<sub>5</sub> = <sup>10</sup>C<sub>4</sub> $${\alpha ^6}{\beta ^4}$$
<br><br>Also given, $$\alpha $$<sup>3</sup> + $$\beta $$<sup>2</sup> = 4
<br><br>By AM-GM inequality
<br><br>$${{{\alpha ^3} + {\beta ^2}} \over 2} \ge {\left( {{\alpha ^3}{\beta ^2}} \right)^{{1 \over 2}}}$$
<br><br>$$ \Rightarrow $$ (2)<sup>2</sup> $$ \ge $$ $${{\alpha ^3}{\beta ^2}}$$
<br><br>$$ \Rightarrow $$ $${\alpha ^6}{\beta ^4}$$ $$ \le $$ 16
<br><br>$$ \therefore $$ 10k = <sup>10</sup>C<sub>4</sub> (16)
<br><br>$$ \Rightarrow $$ k = 336
|
mcq
|
jee-main-2020-online-2nd-september-morning-slot
|
ClPz4ye6HjzLfomsWE7k9k2k5khuqun
|
maths
|
binomial-theorem
|
general-term
|
In the expansion of $${\left( {{x \over {\cos \theta }} + {1 \over {x\sin \theta }}} \right)^{16}}$$, if $${\ell _1}$$ is
the least value of the term independent of x
when $${\pi \over 8} \le \theta \le {\pi \over 4}$$ and $${\ell _2}$$ is the least value of the
term independent of x when $${\pi \over {16}} \le \theta \le {\pi \over 8}$$, then
the ratio $${\ell _2}$$ : $${\ell _1}$$ is equal to :
|
[{"identifier": "A", "content": "8 : 1"}, {"identifier": "B", "content": "16 : 1"}, {"identifier": "C", "content": "1 : 8"}, {"identifier": "D", "content": "1 : 16"}]
|
["B"]
| null |
T<sub>r + 1</sub> = <sup>16</sup>C<sub>r</sub>$${\left( {{x \over {\cos \theta }}} \right)^{16 - r}}{\left( {{1 \over {x\sin \theta }}} \right)^r}$$
<br><br>= <sup>16</sup>C<sub>r</sub>$${\left( x \right)^{16 - 2r}} \times {1 \over {{{\left( {\cos \theta } \right)}^{16 - r}}{{\left( {\sin \theta } \right)}^r}}}$$
<br><br>term is independent of x when
<br><br>$$ \therefore $$ 16 – 2r = 0
<br><br>$$ \Rightarrow $$ r = 8
<br><br>T<sub>9</sub> = <sup>16</sup>C<sub>8</sub> $$ \times $$ $${1 \over {{{\cos }^8}\theta {{\sin }^8}\theta }}$$
<br><br>= <sup>16</sup>C<sub>8</sub> $$ \times $$ $${{{2^8}} \over {{{\left( {\sin 2\theta } \right)}^8}}}$$
<br><br>If $$\theta \in \left[ {{\pi \over 8},{\pi \over 4}} \right]$$ then $$2\theta \in \left[ {{\pi \over 4},{\pi \over 2}} \right]$$
<br><br>In the range $$\left[ {{\pi \over 4},{\pi \over 2}} \right]$$, sin 2$$\theta $$ is increasing.
<br><br>And value of T<sub>9</sub> is least when sin 2$$\theta $$ is maximum.
<br><br>And sin 2$$\theta $$ is maximum in the range $$\left[ {{\pi \over 4},{\pi \over 2}} \right]$$ <br>when 2$$\theta $$ = $${{\pi \over 2}}$$
<br><br>$$ \therefore $$ $${l_1}$$ = <sup>16</sup>C<sub>8</sub> $$ \times $$ 2<sup>8</sup>
<br><br>AgainIf $$\theta \in \left[ {{\pi \over {16}},{\pi \over 8}} \right]$$ then $$2\theta \in \left[ {{\pi \over 8},{\pi \over 4}} \right]$$
<br><br>In the range $$\left[ {{\pi \over 8},{\pi \over 4}} \right]$$, sin 2$$\theta $$ is increasing.
<br><br>And value of T<sub>9</sub> is least when sin 2$$\theta $$ is maximum.
<br><br>And sin 2$$\theta $$ is maximum in the range $$\left[ {{\pi \over 8},{\pi \over 4}} \right]$$ <br>when 2$$\theta $$ = $${{\pi \over 4}}$$
<br><br>$$ \therefore $$ $${l_2}$$ = <sup>16</sup>C<sub>8</sub> $$ \times $$ $${{{2^8}} \over {{{\left( {{1 \over {\sqrt 2 }}} \right)}^8}}}$$
<br><br> = <sup>16</sup>C<sub>8</sub> $$ \times $$ $${{2^8}{2^4}}$$
<br><br>$$ \therefore $$ $${{{l_2}} \over {{l_1}}}$$ = $${{{2^4}} \over 1}$$ = $${{16} \over 1}$$
|
mcq
|
jee-main-2020-online-9th-january-evening-slot
|
whKykbRukRrJZDlsAZ7k9k2k5hjr4v7
|
maths
|
binomial-theorem
|
general-term
|
If $$\alpha $$ and $$\beta $$ be the coefficients of x<sup>4</sup> and x<sup>2</sup>
respectively in the expansion of<br/>
$${\left( {x + \sqrt {{x^2} - 1} } \right)^6} + {\left( {x - \sqrt {{x^2} - 1} } \right)^6}$$, then
|
[{"identifier": "A", "content": "$$\\alpha + \\beta = 60$$"}, {"identifier": "B", "content": "$$\\alpha - \\beta = 60$$"}, {"identifier": "C", "content": "$$\\alpha + \\beta = -30$$"}, {"identifier": "D", "content": "$$\\alpha - \\beta = -132$$"}]
|
["D"]
| null |
(x+a)<sup>n </sup>+ (x – a)<sup>n</sup>
= 2(T<sub>1</sub>
+ T<sub>3</sub>
+ T<sub>5</sub>
+.....)
<br><br>$${\left( {x + \sqrt {{x^2} - 1} } \right)^6} + {\left( {x - \sqrt {{x^2} - 1} } \right)^6}$$
<br><br>= 2[T<sub>1</sub>
+ T<sub>3</sub>
+ T<sub>5</sub>
+ T<sub>7</sub>
]
<br><br>= 2[<sup>6</sup>C<sub>0</sub>
x<sup>6</sup>
+ <sup>6</sup>C<sub>2
</sub>
x<sup>4</sup>(x<sup>2</sup>
– 1) + <sup>6</sup>C<sub>4</sub>
x<sup>2</sup>(x<sup>2</sup>
–1)<sup>2</sup>
+ <sup>6</sup>C<sub>6</sub>
x<sup>0</sup>(x<sup>2</sup>–1)<sup>3</sup>]
<br><br>= 2[x<sup>6</sup>+ 15(x<sup>6</sup>
– x<sup>4</sup>) + 15x<sup>2</sup>
(x<sup>4</sup>
+ 1 –2x<sup>2</sup>) + (x<sup>6</sup>
– 3x<sup>4</sup>
+3x<sup>2</sup>
–1)]
<br><br>= 2[x<sup>6</sup>(2 + 15 + 15 + 1) + x<sup>4</sup>(–15 – 30 –3) + x<sup>2</sup>(15 + 3)]
<br><br>Coefficient of x<sup>4</sup> = $$\alpha $$ = -96
<br><br>And coefficient of x<sup>2</sup> = $$\beta $$ = 36
<br><br>$$ \therefore $$ $$\alpha - \beta = - 96 - 36 = -132$$
|
mcq
|
jee-main-2020-online-8th-january-evening-slot
|
Qc54t3Qr25N0Yctsjd7k9k2k5fiux85
|
maths
|
binomial-theorem
|
general-term
|
The coefficient of x<sup>7</sup>
in the expression
<br/>(1 + x)<sup>10</sup> + x(1 + x)<sup>9</sup>
+ x<sup>2</sup>(1 + x)<sup>8</sup>
+ ......+ x<sup>10</sup> is:
|
[{"identifier": "A", "content": "120"}, {"identifier": "B", "content": "330"}, {"identifier": "C", "content": "420"}, {"identifier": "D", "content": "210"}]
|
["B"]
| null |
(1 + x)<sup>10</sup> + x(1 + x)<sup>9</sup>
+ x<sup>2</sup>(1 + x)<sup>8</sup>
+ ......+ x<sup>10</sup>
<br><br>This is a G.P where
<br><br>First term, a = (1 + x)<sup>10</sup>
<br><br>common ratio, r = $${x \over {1 + x}}$$
<br><br>Number of terms = 11
<br><br>Sum of G.P
<br><br>= $${{{{\left( {1 + x} \right)}^{10}}\left( {1 - {{\left( {{x \over {1 + x}}} \right)}^{11}}} \right)} \over {1 - {x \over {1 + x}}}}$$
<br><br>= (1 + x)<sup>11</sup>
– x<sup>11</sup>
<br><br>So Coefficient of x<sup>7</sup>
is <sup>11</sup>C<sub>7</sub> = 330
|
mcq
|
jee-main-2020-online-7th-january-evening-slot
|
7j0Ea0CxBrRDVTtcVm1klug2yrr
|
maths
|
binomial-theorem
|
general-term
|
The maximum value of the term independent of 't' in the expansion <br/>of $${\left( {t{x^{{1 \over 5}}} + {{{{(1 - x)}^{{1 \over {10}}}}} \over t}} \right)^{10}}$$ where x$$\in$$(0, 1) is :
|
[{"identifier": "A", "content": "$${{10!} \\over {\\sqrt 3 {{(5!)}^2}}}$$"}, {"identifier": "B", "content": "$${{2.10!} \\over {3\\sqrt 3 {{(5!)}^2}}}$$"}, {"identifier": "C", "content": "$${{10!} \\over {3{{(5!)}^2}}}$$"}, {"identifier": "D", "content": "$${{2.10!} \\over {3{{(5!)}^2}}}$$"}]
|
["B"]
| null |
$${T_{r + 1}} = {}^{10}{C_r}{(t{x^{1/5}})^{10 - r}}{\left[ {{{{{(1 - x)}^{1/10}}} \over t}} \right]^r}$$<br><br>$$ = {}^{10}{C_r}{t^{(10 - 2r)}} \times {x^{{{10 - r} \over 5}}} \times {(1 - x)^{{r \over {10}}}}$$<br><br>$$ \Rightarrow 10 - 2r = 0 \Rightarrow r = 5$$
<br><br>$$ \therefore $$ $${T_6} = {}^{10}{C_5} \times x\sqrt {1 - x} $$<br><br>At maximum, $${{d{T_6}} \over {dx}} = {}^{10}{C_5}\left[ {\sqrt {1 - x} - {x \over {2\sqrt {1 - x} }}} \right] = 0$$<br><br>$$ \Rightarrow $$ $$ 1 - x = x/2 \Rightarrow 3x = 2 \Rightarrow x = 2/3$$<br><br>$${T_6}{|_{\max }} = {{10!} \over {5!5!}} \times {2 \over {3\sqrt 3 }}$$ = $${{2.10!} \over {3\sqrt 3 {{(5!)}^2}}}$$
|
mcq
|
jee-main-2021-online-26th-february-morning-slot
|
jHrylvcW7IqT8Wci3y1kmhx5zo4
|
maths
|
binomial-theorem
|
general-term
|
If n is the number of irrational terms in the <br/>expansion of $${\left( {{3^{1/4}} + {5^{1/8}}} \right)^{60}}$$, then (n $$-$$ 1) is divisible by :
|
[{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "26"}]
|
["D"]
| null |
$${T_{r + 1}} = {}^{60}{C_r}{\left( {{3^{1/4}}} \right)^{60 - r}}{\left( {{5^{1/8}}} \right)^r}$$<br><br>rational if $${{60 - r} \over 4},{r \over 8}$$, both are whole numbers, $$r \in \{ 0,1,2,......60\} $$<br><br>$${{60 - r} \over 4} \in W \Rightarrow r \in \{ 0,4,8,....60\} $$<br><br>and $${r \over 8} \in W \Rightarrow r \in \{ 0,8,16,.....56\} $$<br><br>$$ \therefore $$ Common terms $$r \in \{ 0,8,16,.....56\} $$<br><br>So, 8 terms are rational<br><br>Then Irrational terms = $$61 - 8 = 53 = n$$<br><br>$$ \therefore $$ $$n - 1 = 52 = 13 \times {2^2}$$<br><br>Factors 1, 2, 4, 13, 26, 52
|
mcq
|
jee-main-2021-online-16th-march-morning-shift
|
H3IeTt4KAIUgRCZ0P31kmjb2rhf
|
maths
|
binomial-theorem
|
general-term
|
If the fourth term in the expansion of $${(x + {x^{{{\log }_2}x}})^7}$$ is 4480, then the value of x where x$$\in$$N is equal to :
|
[{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "2"}]
|
["D"]
| null |
T<sub>4</sub> = $${}^7{C_3}{x^4}{({x^{{{\log }_2}x}})^3} = 4480$$<br><br>$$ \Rightarrow 35{x^4}{({x^{{{\log }_2}x}})^3} = 4480$$<br><br>$$ \Rightarrow {x^4}{({x^{{{\log }_2}x}})^3} = 128$$<br><br>take log w.r.t. base 2 we get,
<br><br>$$4{\log _2}x + 3{\log _2}({x^{{{\log }_2}x}}) = {\log _2}128$$<br><br>Let $${\log _2}x = y$$<br><br>$$4y + 3{y^2} = 7$$<br><br>$$ \Rightarrow y = 1,{{ - 7} \over 3}$$<br><br>$$ \Rightarrow {\log _2}x = 1,{{ - 7} \over 3}$$<br><br>$$ \Rightarrow $$ $$x = 2,x = {2^{ - 7/3}}$$
|
mcq
|
jee-main-2021-online-17th-march-morning-shift
|
3EvGB4JDbb7CRrEJIt1kmknn38u
|
maths
|
binomial-theorem
|
general-term
|
Let the coefficients of third, fourth and fifth terms in the expansion of $${\left( {x + {a \over {{x^2}}}} \right)^n},x \ne 0$$, be in the ratio 12 : 8 : 3. Then the term independent of x in the expansion, is equal to ___________.
|
[]
| null |
4
|
$${T_{r + 1}} = {n_{C_r}}{x^{n - r}}.{\left( {{a \over {{x^2}}}} \right)^r}$$<br><br>$$ = {}^n{C_r}{a^r}{x^{n - 3r}}$$<br><br>$${T_3} = {}^n{C_2}{a^2}{x^{n - 6}}$$, $${T_4} = {}^n{C_3}{a^3}{x^{n - 9}}$$, $${T_5} = {}^n{C_4}{a^4}{x^{n - 12}}$$<br><br>Now, $${{coefficient\,of\,{T_3}} \over {coefficient\,of\,{T_4}}} = {{{}^n{C_4}.{a^2}} \over {{}^n{C_3}.{a^3}}} = {3 \over {a(n - 2)}} = {3 \over 2}$$<br><br>$$ \Rightarrow a(n - 2) = 2$$ .......... (i)<br><br>and $${{coefficient\,of\,{T_4}} \over {coefficient\,of\,{T_5}}} = {{{}^n{C_3}.{a^3}} \over {{}^n{C_4}.{a^4}}} = {4 \over {a(n - 3)}} = {8 \over 3}$$<br><br>$$ \Rightarrow a(n - 3) = {3 \over 2}$$ ........ (ii)<br><br>by (i) and (ii) $$n = 6,\,a = {1 \over 2}$$<br><br>for term independent of 'x'<br><br>$$n - 3r = 0 \Rightarrow r = {n \over 3} \Rightarrow r = {6 \over 3} = 2$$<br><br>$${T_3} = {}^6{C_2}{\left( {{1 \over 2}} \right)^2}{x^0} = {{15} \over 4} = 3.75 \approx 4$$
|
integer
|
jee-main-2021-online-17th-march-evening-shift
|
YzDnj4t7QGMWPSQlHa1kmm40p2r
|
maths
|
binomial-theorem
|
general-term
|
The term independent of x in the expansion of <br/><br/>$${\left[ {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right]^{10}}$$, x $$\ne$$ 1, is equal to ____________.
|
[]
| null |
210
|
$${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$$
<br><br>= $${\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \right)^{10}}$$
<br><br>= $${\left( {{{\left( {{x^{1/3}} + 1} \right)\left( {{x^{2/3}} - {x^{1/3}} + 1} \right)} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)} \over {\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}$$
<br><br>= $${\left( {\left( {{x^{1/3}} + 1} \right) - {{\left( {\sqrt x + 1} \right)} \over {\sqrt x }}} \right)^{10}}$$
<br><br>= $${\left( {\left( {{x^{1/3}} + 1} \right) - \left( {1 + {1 \over {\sqrt x }}} \right)} \right)^{10}}$$
<br><br>= $${\left( {{x^{1/3}} - {1 \over {{x^{1/2}}}}} \right)^{10}}$$
<br><br>$${T_{r + 1}} = {}^{10}{C_r}{\left( {{x^{{1 \over 3}}}} \right)^{(10 - r)}}{\left( {{x^{ - {1 \over 2}}}} \right)^r}$$<br><br>For being independent of $$x:{{10 - r} \over 3} - {r \over 2} = 0 \Rightarrow r = 4$$<br><br>Term independent of $$x = {}^{10}{C_4} = 210$$
|
integer
|
jee-main-2021-online-18th-march-evening-shift
|
1krq0nwdb
|
maths
|
binomial-theorem
|
general-term
|
The number of rational terms in the binomial expansion of $${\left( {{4^{{1 \over 4}}} + {5^{{1 \over 6}}}} \right)^{120}}$$ is _______________.
|
[]
| null |
21
|
$${\left( {{4^{{1 \over 4}}} + {5^{{1 \over 6}}}} \right)^{120}}$$<br><br>$${T_{r + 1}} = {}^{120}{C_r}{({2^{1/2}})^{120 - r}}{(5)^{r/6}}$$<br><br>for rational terms r = 6$$\lambda$$ <br><br>0 $$\le$$ r $$\le$$ 120<br><br>So total no of terms are 21.
|
integer
|
jee-main-2021-online-20th-july-morning-shift
|
1krubh7ej
|
maths
|
binomial-theorem
|
general-term
|
If the constant term, in binomial expansion of $${\left( {2{x^r} + {1 \over {{x^2}}}} \right)^{10}}$$ is 180, then r is equal to __________________.
|
[]
| null |
8
|
$${\left( {2{x^r} + {1 \over {{x^2}}}} \right)^{10}}$$<br><br>General term $$ = {}^{10}{C_R}{(2{x^2})^{10 - R}}{x^{ - 2R}}$$<br><br>$$ \Rightarrow {2^{10 - R}}{}^{10}{C_R} = 180$$ ....... (1)<br><br>& (10 $$-$$ R)r $$-$$ 2R = 0<br><br>$$r = {{2R} \over {10 - R}}$$<br><br>$$r = {{2(R - 10)} \over {10 - R}} + {{20} \over {10 - R}}$$<br><br>$$ \Rightarrow r = - 2 + {{20} \over {10 - R}}$$ ....... (2)<br><br>R = 8 or 5 reject equation (1) not satisfied<br><br>At R = 8<br><br>$$ \Rightarrow {2^{10 - R}}\times{}^{10}{C_R} = 180 \Rightarrow r = 8$$
|
integer
|
jee-main-2021-online-22th-july-evening-shift
|
1krw3ge50
|
maths
|
binomial-theorem
|
general-term
|
The term independent of 'x' in the expansion of <br/>$${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$$, where x $$\ne$$ 0, 1 is equal to ______________.
|
[]
| null |
210
|
$${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$$
<br><br>= $${\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \right)^{10}}$$
<br><br>= $${\left( {{{\left( {{x^{1/3}} + 1} \right)\left( {{x^{2/3}} - {x^{1/3}} + 1} \right)} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)} \over {\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}$$
<br><br>= $${\left( {\left( {{x^{1/3}} + 1} \right) - {{\left( {\sqrt x + 1} \right)} \over {\sqrt x }}} \right)^{10}}$$
<br><br>= $${\left( {\left( {{x^{1/3}} + 1} \right) - \left( {1 + {1 \over {\sqrt x }}} \right)} \right)^{10}}$$
<br><br>= $${\left( {{x^{1/3}} - {1 \over {{x^{1/2}}}}} \right)^{10}}$$
<br><br>[<b>Note:</b>
<br><br>For $${\left( {{x^\alpha } \pm {1 \over {{x^\beta }}}} \right)^n}$$ the $$\left( {r + 1} \right)$$<sup>th</sup> term with power m of x is
<br><br>$$r = {{n\alpha - m} \over {\alpha + \beta }}$$]
<br><br>Here $$\alpha = {1 \over 3}$$, $$\beta = {1 \over 2}$$ and m = 0
<br><br>then $$r = {{10 \times {1 \over 3} - 0} \over {{1 \over 3} + {1 \over 2}}}$$ = $${{10} \over 3} \times {6 \over 5}$$ = 4
<br><br>$$\therefore$$ T<sub>5</sub> is the term independent of x.
<br><br>$$\therefore$$ T<sub>5</sub> = $${}^{10}{C_4}$$ = 210
|
integer
|
jee-main-2021-online-25th-july-morning-shift
|
1krxgk2ja
|
maths
|
binomial-theorem
|
general-term
|
A possible value of 'x', for which the ninth term in the expansion of $${\left\{ {{3^{{{\log }_3}\sqrt {{{25}^{x - 1}} + 7} }} + {3^{\left( { - {1 \over 8}} \right){{\log }_3}({5^{x - 1}} + 1)}}} \right\}^{10}}$$ in the increasing powers of $${3^{\left( { - {1 \over 8}} \right){{\log }_3}({5^{x - 1}} + 1)}}$$ is equal to 180, is :
|
[{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$$-$$1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}]
|
["D"]
| null |
$${}^{10}{C_8}({25^{(x - 1)}} + 7) \times {({5^{(x - 1)}} + 1)^{ - 1}} = 180$$<br><br>$$ \Rightarrow {{{{25}^{x - 1}} + 7} \over {{5^{(x - 1)}} + 1}} = 4$$<br><br>$$ \Rightarrow {{{t^2} + 7} \over {t + 1}} = 4$$;<br><br>$$\Rightarrow$$ t = 1, 3 = 5<sup>x $$-$$ 1</sup><br><br>$$\Rightarrow$$ x $$-$$ 1 = 0 (one of the possible value).<br><br>$$\Rightarrow$$ x = 1
|
mcq
|
jee-main-2021-online-27th-july-evening-shift
|
1krz59pdx
|
maths
|
binomial-theorem
|
general-term
|
The sum of all those terms which are rational numbers in the <br/><br/>expansion of (2<sup>1/3</sup> + 3<sup>1/4</sup>)<sup>12</sup> is :
|
[{"identifier": "A", "content": "89"}, {"identifier": "B", "content": "27"}, {"identifier": "C", "content": "35"}, {"identifier": "D", "content": "43"}]
|
["D"]
| null |
$${T_{r + 1}} = {}^{12}{C_r}{\left( {{2^{1/3}}} \right)^r}.{\left( {{3^{1/4}}} \right)^{12 - 4}}$$<br><br>T<sub>r + 1</sub> will be rational number when r = 0, 3, 6, 9, 12 & r = 0, 4, 8, 12<br><br>$$\Rightarrow$$ r = 0, 12<br><br>T<sub>1</sub> + T<sub>13</sub> = 1 $$\times$$ 3<sup>3</sup> + 1 $$\times$$ 2<sup>4</sup> $$\times$$ 1<br><br>= 24 + 16 = 43
|
mcq
|
jee-main-2021-online-25th-july-evening-shift
|
1krzlrpjk
|
maths
|
binomial-theorem
|
general-term
|
If the greatest value of the term independent of 'x' in the <br/><br/>expansion of $${\left( {x\sin \alpha + a{{\cos \alpha } \over x}} \right)^{10}}$$ is $${{10!} \over {{{(5!)}^2}}}$$, then the value of 'a' is equal to :
|
[{"identifier": "A", "content": "$$-$$1"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$-$$2"}, {"identifier": "D", "content": "2"}]
|
["D"]
| null |
$${T_{r + 1}} = {}^{10}{C_r}{(x\sin \alpha )^{10 - r}}{\left( {{{a\cos \alpha } \over x}} \right)^r}$$<br><br>r = 0, 1, 2, ......., 10<br><br>T<sub>r + 1</sub> will be independent of x when 10 $$-$$ 2r = 0 $$\Rightarrow$$ r = 5<br><br>$${T_6} = {}^{10}{C_5}{(x\sin \alpha )^5} \times {\left( {{{a\cos \alpha } \over x}} \right)^5}$$<br><br>$$ = {}^{10}{C_5} \times {a^5} \times {1 \over {{2^5}}}{(\sin 2\alpha )^5}$$<br><br>will be greatest when sin2$$\alpha$$ = 1<br><br>$$ \Rightarrow {}^{10}{C_5}{{{a^5}} \over {{2^5}}} = {}^{10}{C_5} \Rightarrow a = 2$$
|
mcq
|
jee-main-2021-online-25th-july-evening-shift
|
1krzrliu5
|
maths
|
binomial-theorem
|
general-term
|
If the co-efficient of x<sup>7</sup> and x<sup>8</sup> in the expansion of $${\left( {2 + {x \over 3}} \right)^n}$$ are equal, then the value of n is equal to _____________.
|
[]
| null |
55
|
$${}^n{C_7}{2^{n - 7}}{1 \over {{3^7}}} = {}^n{C_8}{2^{n - 8}}{1 \over {{3^8}}}$$<br><br>$$\Rightarrow$$ n $$-$$ 7 = 48 $$\Rightarrow$$ n = 55
|
integer
|
jee-main-2021-online-25th-july-evening-shift
|
1ks07cgn0
|
maths
|
binomial-theorem
|
general-term
|
If the coefficients of x<sup>7</sup> in $${\left( {{x^2} + {1 \over {bx}}} \right)^{11}}$$ and x<sup>$$-$$7</sup> in $${\left( {{x} - {1 \over {bx^2}}} \right)^{11}}$$, b $$\ne$$ 0, are equal, then the value of b is equal to :
|
[{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$$-$$1"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "$$-$$2"}]
|
["C"]
| null |
Coefficient of x<sup>7</sup> in $${\left( {{x^2} + {1 \over {bx}}} \right)^{11}}$$ :<br><br>General Term = $${}^{11}{C_r}{({x^2})^{11 - r}}.{\left( {{1 \over {bx}}} \right)^r}$$<br><br>= $${}^{11}{C_r}{x^{22 - 3r}}.{1 \over {{b^r}}}$$<br><br>$$22 - 3r = 7$$<br><br>$$r = 5$$<br><br>$$\therefore$$ Required Term = $${}^{11}{C_5}.{1 \over {{b^5}}}.{x^7}$$<br><br>Coefficient of x<sup>$$-$$7</sup> in $${\left( {x - {1 \over {b{x^2}}}} \right)^{11}}$$ :<br><br>General Term = $${}^{11}{C_r}{(x)^{11 - r}}.{\left( { - {1 \over {b{x^2}}}} \right)^r}$$<br><br>= $${}^{11}{C_r}{x^{11 - 3r}}.{{{{( - 1)}^r}} \over {{b^r}}}$$<br><br>$$11 - 3r = - 7$$ $$\therefore$$ $$r = 6$$<br><br>$$ \therefore $$ Required Term = $${}^{11}{C_6}.{1 \over {{b^6}}}{x^{ - 7}}$$
<br><br>According to the question,
<br><br>$${}^{11}{C_5}.{1 \over {{b^5}}} = {}^{11}{C_6}.{1 \over {{b^6}}}$$<br><br>Since, b $$\ne$$ 0 $$\therefore$$ b = 1
|
mcq
|
jee-main-2021-online-27th-july-morning-shift
|
1ktisyc5x
|
maths
|
binomial-theorem
|
general-term
|
If $$\left( {{{{3^6}} \over {{4^4}}}} \right)k$$ is the term, independent of x, in the binomial expansion of $${\left( {{x \over 4} - {{12} \over {{x^2}}}} \right)^{12}}$$, then k is equal to ___________.
|
[]
| null |
55
|
$${\left( {{x \over 4} - {{12} \over {{x^2}}}} \right)^{12}}$$<br><br>$${T_{r + 1}} = {( - 1)^r}\,.\,{}^{12}{C_r}{\left( {{x \over 4}} \right)^{12 - r}}{\left( {{{12} \over {{x^2}}}} \right)^r}$$<br><br>$${T_{r + 1}} = {( - 1)^r}\,.\,{}^{12}{C_r}{\left( {{1 \over 4}} \right)^{12 - r}}{\left( {12} \right)^r}\,.\,{(x)^{12 - 3r}}$$<br><br>Term independent of x $$\Rightarrow$$ 12 $$-$$ 3r = 0 $$\Rightarrow$$ r = 4<br><br>$${T_5} = {( - 1)^r}\,.\,{}^{12}{C_r}{\left( {{1 \over 4}} \right)^8}{\left( {12} \right)^4} = {{{3^6}} \over {{4^4}}}.\,k$$<br><br>$$\Rightarrow$$ k = 55
|
integer
|
jee-main-2021-online-31st-august-morning-shift
|
1l54uczwv
|
maths
|
binomial-theorem
|
general-term
|
<p>Let the coefficients of x<sup>$$-$$1</sup> and x<sup>$$-$$3</sup> in the expansion of $${\left( {2{x^{{1 \over 5}}} - {1 \over {{x^{{1 \over 5}}}}}} \right)^{15}},x > 0$$, be m and n respectively. If r is a positive integer such that $$m{n^2} = {}^{15}{C_r}\,.\,{2^r}$$, then the value of r is equal to __________.</p>
|
[]
| null |
5
|
<p>Given, Binomial expansion</p>
<p>$${\left( {2{x^{{1 \over 5}}} - {1 \over {{x^{{1 \over 5}}}}}} \right)^{15}}$$</p>
<p>$$\therefore$$ General Term</p>
<p>$${T_{r + 1}} = {}^{15}{C_r}\,.\,{\left( {2{x^{{1 \over 5}}}} \right)^{15 - r}}\,.\,{\left( { - {1 \over {{x^{{1 \over 5}}}}}} \right)^r}$$</p>
<p>$$ = {}^{15}{C_r}\,.\,{2^{15 - r}}\,.\,{x^{{1 \over 5}(15 - r - r)}}\,.\,{( - 1)^r}$$</p>
<p>For $${x^{ - 1}}$$ term;</p>
<p>$${1 \over 5}(15 - 2r) = - 1$$</p>
<p>$$ \Rightarrow 15 - 2r = - 5$$</p>
<p>$$ \Rightarrow 2r = 20$$</p>
<p>$$ \Rightarrow r = 10$$</p>
<p>m is the coefficient of $${x^{ - 1}}$$ term,</p>
<p>$$\therefore$$ $$m = {}^{15}{C_{10}}\,.\,{2^{15 - 10}}\,.\,{( - 1)^{10}}$$</p>
<p>$$ = {}^{15}{C_{10}}\,.\,{2^5}$$</p>
<p>For $${x^{ - 3}}$$ term;</p>
<p>$${1 \over 5}(15 - 2r) = - 3$$</p>
<p>$$ \Rightarrow 15 - 2r = - 15$$</p>
<p>$$ \Rightarrow 2r = 30$$</p>
<p>$$ \Rightarrow r = 15$$</p>
<p>n is the coefficient of $${x^{ - 3}}$$ term,</p>
<p>$$\therefore$$ $$n = {}^{15}{C_{15}}\,.\,{2^{15 - 15}}\,.\,{( - 1)^{15}}$$</p>
<p>$$ = 1\,.\,1\,.\, - 1$$</p>
<p>$$ = - 1$$</p>
<p>Given,</p>
<p>$$m{n^2} = {}^{15}{C_r}\,.\,{2^r}$$</p>
<p>$$ \Rightarrow {}^{15}{C_{10}}\,.\,{2^5}\,.\,{(1)^2} = {}^{15}{C_r}\,.\,{2^r}$$ [putting value of m and n]</p>
<p>$$ \Rightarrow {}^{15}{C_{15 - 10}}\,.\,{2^5} = {}^{15}{C_r}\,.\,{2^r}$$</p>
<p>$$ \Rightarrow {}^{15}{C_5}\,.\,{2^5} = {}^{15}{C_r}.\,{2^r}$$</p>
<p>Comparing both side, we get</p>
<p>$$r = 5$$.</p>
|
integer
|
jee-main-2022-online-29th-june-evening-shift
|
1l55h6lp9
|
maths
|
binomial-theorem
|
general-term
|
<p>The term independent of x in the expansion of <br/><br/>$$(1 - {x^2} + 3{x^3}){\left( {{5 \over 2}{x^3} - {1 \over {5{x^2}}}} \right)^{11}},\,x \ne 0$$ is :</p>
|
[{"identifier": "A", "content": "$${7 \\over {40}}$$"}, {"identifier": "B", "content": "$${33 \\over {200}}$$"}, {"identifier": "C", "content": "$${39 \\over {200}}$$"}, {"identifier": "D", "content": "$${11 \\over {50}}$$"}]
|
["B"]
| null |
<p>General term of Binomial expansion $${\left( {{5 \over 2}{x^3} - {1 \over {5{x^2}}}} \right)^{11}}$$ is</p>
<p>$${T_{r + 1}} = {}^{11}{C_r}\,.\,{\left( {{5 \over 2}{x^3}} \right)^{11 - r}}\,.\,{\left( { - {1 \over {5{x^2}}}} \right)^r}$$</p>
<p>$$ = {}^{11}{C_r}\,.\,{\left( {{5 \over 2}} \right)^{11 - r}}\,.\,{\left( { - {1 \over 5}} \right)^r}\,.\,{x^{33 - 5r}}$$</p>
<p>In the term,</p>
<p>$$\left( {1 - {x^2} + 3{x^3}} \right){\left( {{5 \over 2}{x^3} - {1 \over {5{x^2}}}} \right)^{11}}$$</p>
<p>Term independent of x is when</p>
<p>(1) $$33 - 5r = 0$$</p>
<p>$$ \Rightarrow r = {{33} \over 5}\,\, \notin $$ integer</p>
<p>(2) $$33 - 5r = -2$$</p>
<p>$$ \Rightarrow 5r = 35$$</p>
<p>$$ \Rightarrow r = 7\,\, \in $$ integer</p>
<p>(3) $$33 - 5r = - 3$$</p>
<p>$$ \Rightarrow 5r = 36$$</p>
<p>$$ \Rightarrow r = {{36} \over 5}\,\, \notin $$ integer</p>
<p>$$\therefore$$ Only for r = 7 independent of x term possible.</p>
<p>$$\therefore$$ Independent of x term</p>
<p>$$ = - \left( {{}^{11}{C_7}{{\left( {{5 \over 2}} \right)}^4}\,.\,{{\left( { - {1 \over 5}} \right)}^7}} \right)$$</p>
<p>$$ = - \left( {{{11\,.\,10\,.\,9\,.\,8} \over {4\,.\,3\,.\,2\,.\,1}}\,.\,{{{5^4}} \over {{2^4}}}\,.\, - {1 \over {{5^7}}}} \right)$$</p>
<p>$$ = {{11\,.\,10\,.\,3} \over {{2^4}\,.\,{5^3}}}$$</p>
<p>$$ = {{11\,.\,3} \over {{2^3}\,.\,{5^2}}}$$</p>
<p>$$ = {{33} \over {200}}$$</p>
|
mcq
|
jee-main-2022-online-28th-june-evening-shift
|
1l567u2dw
|
maths
|
binomial-theorem
|
general-term
|
<p>The number of positive integers k such that the constant term in the binomial expansion of $${\left( {2{x^3} + {3 \over {{x^k}}}} \right)^{12}}$$, x $$\ne$$ 0 is 2<sup>8</sup> . l, where l is an odd integer, is ______________.</p>
|
[]
| null |
2
|
<p>Given Binomial expression is</p>
<p>$${\left( {2{x^3} + {3 \over {{x^k}}}} \right)^{12}}$$</p>
<p>General term,</p>
<p>$${T_{r + 1}} = {}^{12}{C_r}{(2{x^3})^r}\,.\,{\left( {{3 \over {{x^k}}}} \right)^{12 - r}}$$</p>
<p>$$ = \left( {{}^{12}{C_r}\,.\,{2^r}\,.\,{3^{12 - r}}} \right)\,.\,{x^{3r - 12k + kr}}$$</p>
<p>For constant term,</p>
<p>$$3r - 12k + kr = 0$$</p>
<p>$$ \Rightarrow k(12 - r) = 3r$$</p>
<p>$$ \Rightarrow k = {{3r} \over {12 - r}}$$</p>
<p>For r = 1, $$k = {3 \over {11}}$$ (not integer)</p>
<p>For r = 2, $$k = {6 \over {10}}$$ (not integer)</p>
<p>For r = 3, $$k = {9 \over {9}}=1$$ (integer)</p>
<p>For r = 6, $$k = {18 \over {6}}=3$$ (integer)</p>
<p>For r = 8, $$k = {24 \over {4}}=6$$ (integer)</p>
<p>For r = 9, $$k = {27 \over {3}}=9$$ (integer)</p>
<p>For r = 10, $$k = {30 \over {2}}=15$$ (integer)</p>
<p>For r = 11, $$k = {33 \over {1}}=33$$ (integer)</p>
<p>So, for r = 3, 6, 8, 9, 10 and 11 k is positive integer.</p>
<p>When k = 1 then r = 3 and constant term is</p>
<p>$$ = {}^{12}{C_3}\,.\,{2^3}\,.\,{3^9}$$</p>
<p>$$ = {{12\,.\,11\,.\,10} \over {3\,.\,2\,.\,1}}\,.\,{2^3}\,.\,{3^9}$$</p>
<p>$$ = 2\,.\,11\,.\,2\,.\,5\,.\,{2^3}\,.\,{3^9}$$</p>
<p>$$ = 11\,.\,5\,.\,{2^5}\,.\,{3^9}$$</p>
<p>$$ = {2^5}\,.\,(55\,.\,{3^9})$$</p>
<p>$$ = {2^5}(l)$$</p>
<p>$$ \ne {2^8}\,.\,l$$</p>
<p>When x = 3 then r = 6 and constant term</p>
<p>$$ = {}^{12}{C_6}\,.\,{2^6}\,.\,{3^6}$$</p>
<p>$$ = {{12\,.\,11\,.\,10\,.\,9\,.\,8\,.\,7} \over {6\,.\,5\,.\,4\,.\,3\,.\,2\,.\,1}}\,.\,{2^6}\,.\,{3^6}$$</p>
<p>$$ = {2^8}\,.\,231\,.\,{3^6}$$</p>
<p>$$ = {2^8}(l)$$</p>
<p>When k = 6 then r = 8 and constant term</p>
<p>$$ = {}^{12}{C_8}\,.\,{2^8}\,.\,{3^4}$$</p>
<p>$$ = {{12\,.\,11\,.\,10\,.\,9} \over {4\,.\,3\,.\,2\,.\,1}}\,.\,{2^8}\,.\,{3^4}$$</p>
<p>$$ = {2^8}\,.\,55\,.\,{3^6}$$</p>
<p>$$ = {2^8}(l)$$</p>
<p>When x = 9 then r = 9 and constant term</p>
<p>$$ = {}^{12}{C_9}\,.\,{2^9}\,.\,{3^3}$$</p>
<p>$$ = {{12\,.\,11\,.\,10} \over {3\,.\,2\,.\,1}}\,.\,{2^9}\,.\,{3^3}$$</p>
<p>$$ = {2^{11}}\,.\,55\,.\,{3^3}$$</p>
<p>Here power of 2 is 11 which is greater than 8. So, k = 9 is not possible.</p>
<p>Similarly for k = 15 and k = 33, $${2^8}\,.\,l$$ form is not possible.</p>
<p>$$\therefore$$ k = 3 and k = 6 is accepted.</p>
<p>$$\therefore$$ For 2 positive integer value of k, $${2^8}\,.\,l$$ form of constant term possible.</p>
|
integer
|
jee-main-2022-online-28th-june-morning-shift
|
1l57p2f15
|
maths
|
binomial-theorem
|
general-term
|
<p>If the coefficient of x<sup>10</sup> in the binomial expansion of $${\left( {{{\sqrt x } \over {{5^{{1 \over 4}}}}} + {{\sqrt 5 } \over {{x^{{1 \over 3}}}}}} \right)^{60}}$$ is $${5^k}\,.\,l$$, where l, k $$\in$$ N and l is co-prime to 5, then k is equal to _____________.</p>
|
[]
| null |
5
|
<p>Given Binomial Expansion</p>
<p>$$ = {\left( {{{\sqrt x } \over {{5^{{1 \over 4}}}}} + {{\sqrt x } \over {{5^{{1 \over 3}}}}}} \right)^{60}}$$</p>
<p>$$\therefore$$ General term</p>
<p>$${T_{r + 1}} = {}^{60}{C_r}\,.\,{\left( {{{{x^{1/2}}} \over {{5^{1/4}}}}} \right)^{60 - r}}\,.\,{\left( {{{{5^{1/2}}} \over {{x^{1/3}}}}} \right)^r}$$</p>
<p>$$ = {}^{60}{C_r}\,.\,{5^{\left( {{r \over 4} - 15 + {r \over 2}} \right)}}\,.\,{x^{\left( {30 - {r \over 2} - {r \over 3}} \right)}}$$</p>
<p>$$ = {}^{60}{C_r}\,.\,{5^{\left( {{{3r - 60} \over 4}} \right)}}\,.\,{x^{\left( {{{180 - 5r} \over 6}} \right)}}$$</p>
<p>For x<sup>10</sup> term,</p>
<p>$${{180 - 5r} \over 6} = 10$$</p>
<p>$$ \Rightarrow 5r = 120$$</p>
<p>$$ \Rightarrow r = 24$$</p>
<p>$$\therefore$$ Coefficient of $${x^{10}} = {}^{60}{C_{24}}\,.\,{5^{\left( {{{3 \times 24 - 60} \over 4}} \right)}}$$</p>
<p>$$ = {}^{60}{C_{24}}\,.\,{5^3}$$</p>
<p>$$ = {{60!} \over {24!\,\,36!}}\,.\,{5^3}$$</p>
<p>It is given that,</p>
<p>$${{60!} \over {24!\,\,36!}}\,.\,{5^3} = {5^k}\,.\,l$$ ...... (1)</p>
<p>Also given that, l is coprime to 5 means l can't be multiple of 5. So we have to find all the factors of 5 in 60!, 24! and 36!</p>
<p>[Note : Formula for exponent or degree of prime number in n!.</p>
<p>Exponent of p in $$n! = \left\lceil {{n \over p}} \right\rceil + \left\lceil {{n \over {{p^2}}}} \right\rceil + \left\lceil {{n \over {{p^3}}}} \right\rceil + $$ ..... until 0 comes</p>
<p>here p is a prime number. ]</p>
<p>$$\therefore$$ Exponent of 5 in 60!</p>
<p>$$= \left\lceil {{{60} \over 5}} \right\rceil + \left\lceil {{{60} \over {{5^2}}}} \right\rceil + \left\lceil {{{60} \over {{5^3}}}} \right\rceil + $$ .....</p>
<p>$$ = 12 + 2 + 0 + $$ .....</p>
<p>$$ = 14$$</p>
<p>Exponent of 5 in 24!</p>
<p>$$ = \left\lceil {{{24} \over 5}} \right\rceil + \left\lceil {{{24} \over {{5^2}}}} \right\rceil + \left\lceil {{{24} \over {{5^3}}}} \right\rceil + $$ ......</p>
<p>$$ = 4 + 0 + 0$$ ......</p>
<p>$$ = 4$$</p>
<p>Exponent of 5 in 36!</p>
<p>$$ = \left\lceil {{{36} \over 5}} \right\rceil + \left\lceil {{{36} \over {{5^2}}}} \right\rceil + \left\lceil {{{36} \over {{5^3}}}} \right\rceil + $$ .......</p>
<p>$$ = 7 + 1 + 0$$ ......</p>
<p>$$ = 8$$</p>
<p>$$\therefore$$ From equation (1), exponent of 5 overall</p>
<p>$${{{5^{14}}} \over {{5^4}\,.\,{5^8}}}\,.\,{5^3} = {5^k}$$</p>
<p>$$ \Rightarrow {5^5} = {5^k}$$</p>
<p>$$ \Rightarrow k = 5$$</p>
|
integer
|
jee-main-2022-online-27th-june-morning-shift
|
1l59jwnfk
|
maths
|
binomial-theorem
|
general-term
|
<p>The coefficient of x<sup>101</sup> in the expression $${(5 + x)^{500}} + x{(5 + x)^{499}} + {x^2}{(5 + x)^{498}} + \,\,.....\,\, + \,\,{x^{500}}$$, x > 0, is</p>
|
[{"identifier": "A", "content": "<sup>501</sup>C<sub>101</sub> (5)<sup>399</sup>"}, {"identifier": "B", "content": "<sup>501</sup>C<sub>101</sub> (5)<sup>400</sup>"}, {"identifier": "C", "content": "<sup>501</sup>C<sub>100</sub> (5)<sup>400</sup>"}, {"identifier": "D", "content": "<sup>500</sup>C<sub>101</sub> (5)<sup>399</sup>"}]
|
["A"]
| null |
<p>Given,</p>
<p>$${(5 + x)^{500}} + x{(5 + x)^{499}} + {x^2}{(5 + x)^{498}}\,\, + $$ ...... $${x^{500}}$$</p>
<p>This is a G.P. with first term $${(5 + x)^{500}}$$</p>
<p>Common ratio $$ = {{x{{(5 + x)}^{499}}} \over {{{(5 + x)}^{500}}}} = {x \over {5 + x}}$$ and 501 terms present.</p>
<p>$$\therefore$$ Sum $$ = {{{{(5 + x)}^{500}}\left( {{{\left( {{x \over {5 + x}}} \right)}^{501}} - 1} \right)} \over {{x \over {5 + x}} - 1}}$$</p>
<p>$$ = {{{{(5 + x)}^{500}}\left( {{{{x^{501}} - {{(5 + x)}^{501}}} \over {{{(5 + x)}^{501}}}}} \right)} \over {{{x - 5 - x} \over {5 + x}}}}$$</p>
<p>$$ = {{{{{x^{501}} - {{(5 + x)}^{501}}} \over {5 + x}}} \over {{{ - 5} \over {5 + x}}}}$$</p>
<p>$$ = {1 \over 5}\left( {{{\left( {5 + x} \right)}^{501}} - {x^{501}}} \right)$$</p>
<p>Coefficient of x<sup>101</sup> in $${(5 + x)^{501}}$$ is $$ = {}^{501}{C_{101}}\,.\,{5^{400}}$$</p>
<p>$$\therefore$$ In $${1 \over 5}\left( {{{(5 + x)}^{500}} - {x^{501}}} \right)$$ coefficient of x<sup>101</sup> is $$ = {1 \over 5}\,.\,{}^{501}{C_{101}}\,.\,{5^{400}}$$</p>
<p>$$ = {}^{501}{C_{101}}\,.\,{5^{399}}$$</p>
|
mcq
|
jee-main-2022-online-25th-june-evening-shift
|
1l59l7q3l
|
maths
|
binomial-theorem
|
general-term
|
<p>If the sum of the co-efficient of all the positive even powers of x in the binomial expansion of $${\left( {2{x^3} + {3 \over x}} \right)^{10}}$$ is $${5^{10}} - \beta \,.\,{3^9}$$, then $$\beta$$ is equal to ____________.</p>
|
[]
| null |
83
|
<p>Given, Binomial Expansion</p>
<p>$${\left( {2{x^3} + {3 \over x}} \right)^{10}}$$</p>
<P>General term</p>
<p>$${T_{r + 1}} = {}^{10}{C_r}\,.\,{(2{x^3})^{10 - r}}\,.\,{\left( {{3 \over x}} \right)^r}$$</p>
<p>$$ = {}^{10}{C_r}\,.\,{2^{10 - r}}\,.\,{3^r}\,.\,{x^{30 - 3r}}\,.\,{x^{ - r}}$$</p>
<p>$$ = {}^{10}{C_r}\,.\,{2^{10 - r}}\,.\,{3^r}\,.\,{x^{30 - 4r}}$$</p>
<p>For positive even power of x, 30 $$-$$ 4r should be even and positive.</p>
<p>For r = 0, 30 $$-$$ 4 $$\times$$ 0 = 30 (even and positive)</p>
<p>For r = 1, 30 $$-$$ 4 $$\times$$ 1 = 26 (even and positive)</p>
<p>For r = 2, 30 $$-$$ 4 $$\times$$ 2 = 22 (even and positive)</p>
<p>For r = 3, 30 $$-$$ 4 $$\times$$ 3 = 18 (even and positive)</p>
<p>For r = 4, 30 $$-$$ 4 $$\times$$ 4 = 14 (even and positive)</p>
<p>For r = 5, 30 $$-$$ 4 $$\times$$ 5 = 10 (even and positive)</p>
<p>For r = 6, 30 $$-$$ 4 $$\times$$ 6 = 6 (even and positive)</p>
<p>For r = 7, 30 $$-$$ 4 $$\times$$ 7 = 2 (even and positive)</p>
<p>For r = 8, 30 $$-$$ 4 $$\times$$ 8 = $$-$$2 (even but not positive)</p>
<p>So, for r = 1, 2, 3, 4, 5, 6 and 7 we can get positive even power of x.</p>
<p>$$\therefore$$ Sum of coefficient for positive even power of x</p>
<p>$$ = {}^{10}{C_0}\,.\,{2^{10}}\,.\,{3^0} + {}^{10}{C_1}\,.\,{2^9}\,.\,{3^1} + {}^{10}{C_2}\,.\,{2^8}\,.\,{3^2} + {}^{10}{C_3}\,.\,{2^7}\,.\,{3^3} + {}^{10}{C_4}\,.\,{2^6}\,.\,{3^4} + {}^{10}{C_5}\,.\,{2^5}\,.\,{3^5} + {}^{10}{C_6}\,.\,{2^4}\,.\,{3^6} + {}^{10}{C_7}\,.\,{2^3}\,.\,{3^7}$$</p>
<p>$$ = {}^{10}{C_{10}}\,.\,{2^{10}}\,.\,{3^0} + {}^{10}{C_1}\,.\,{2^9}\,.\,{3^1}\,\, + \,\,.....\,\, + \,\,{}^{10}{C_{10}}\,.\,{2^0}\,.\,{3^{10}} - \left[ {{}^{10}{C_8}\,.\,{2^2}\,.\,{3^8} + {}^{10}{C_9}\,.\,2\,.\,{3^9} + {}^{10}{C_{10}}\,.\,{2^0}\,.\,{3^{10}}} \right]$$</p>
<p>$$ = {(2 + 3)^{10}} - \left[ {45\,.\,4\,.\,{3^8} + 10\,.\,2\,.\,{3^9} + 1\,.\,1\,.\,{3^{10}}} \right]$$</p>
<p>$$ = {5^{10}} - \left[ {60 \times {3^9} + 20\,.\,{3^9} + 3\,.\,{3^9}} \right]$$</p>
<p>$$ = {5^{10}} - \left( {60 + 20 + 3} \right){3^9}$$</p>
<p>$$ = {5^{10}} - 83\,.\,{3^9}$$</p>
<p>$$\therefore$$ $$\beta = 83$$</p>
|
integer
|
jee-main-2022-online-25th-june-evening-shift
|
1l5vzfdaf
|
maths
|
binomial-theorem
|
general-term
|
<p>For two positive real numbers a and b such that $${1 \over {{a^2}}} + {1 \over {{b^3}}} = 4$$, then minimum value of the constant term in the expansion of $${\left( {a{x^{{1 \over 8}}} + b{x^{ - {1 \over {12}}}}} \right)^{10}}$$ is :</p>
|
[{"identifier": "A", "content": "$${{105} \\over 2}$$"}, {"identifier": "B", "content": "$${{105} \\over 4}$$"}, {"identifier": "C", "content": "$${{105} \\over 8}$$"}, {"identifier": "D", "content": "$${{105} \\over 16}$$"}]
|
["C"]
| null |
<p>Given, Binomial expansion,</p>
<p>$${\left( {a{x^{{1 \over 8}}} + b{x^{ - {1 \over {12}}}}} \right)^{10}}$$</p>
<p>General term,</p>
<p>$${T_{r + 1}} = {}^{10}{C_r}\,.\,{\left( {a{x^{{1 \over 8}}}} \right)^{10 - r}}\,.\,{\left( {b{x^{ - {1 \over {12}}}}} \right)^r}$$</p>
<p>$$ = {}^{10}{C_r}\,.\,{a^{10 - r}}\,.\,{b^r}\,.\,{x^{{{10 - r} \over 8}}}\,.\,{x^{ - {r \over {12}}}}$$</p>
<p>$$ = {}^{10}{C_r}\,.\,{a^{10 - r}}\,.\,{b^r}\,.\,{x^{\left( {{{10 - r} \over 8} - {r \over {12}}} \right)}}$$</p>
<p>$$ = {}^{10}{C_r}\,.\,{a^{10 - r}}\,.\,{b^r}\,.\,{x^{{{30 - 3r - 2r} \over {24}}}}$$</p>
<p>$$ = {}^{10}{C_r}\,.\,{a^{10 - r}}\,.\,{b^r}\,.\,{x^{{{30 - 5r} \over {24}}}}$$</p>
<p>For constant term,</p>
<p>$${{30 - 5r} \over {24}} = 0$$</p>
<p>$$ \Rightarrow r = 6$$</p>
<p>$$\therefore$$ Constant term,</p>
<p>$${T_{r + 1}} = {T_{6 + 1}} = {}^{10}{C_6}\,.\,{a^4}\,.\,{b^6}$$</p>
<p>$$ = {{10!} \over {6!\,4!}}{a^4}\,.\,{b^6}$$</p>
<p>$$ = {{10 \times 9 \times 8 \times 7} \over {4 \times 3 \times 2 \times 1}}\,.\,{a^4}\,.\,{b^6}$$</p>
<p>$$ = 210{a^4}{b^6}$$</p>
<p>We know, $$GM \ge HM$$</p>
<p>For terms a<sup>2</sup> and b<sup>3</sup>,</p>
<p>$$\sqrt {{a^2}{b^3}} \ge {2 \over {{1 \over {{a^2}}} + {1 \over {{b^3}}}}}$$</p>
<p>$$ \Rightarrow \sqrt {{a^2}{b^3}} \ge {2 \over 4}$$</p>
<p>$$ \Rightarrow {a^2}{b^3} \ge {1 \over 4}$$</p>
<p>$$ \Rightarrow {({a^2}{b^3})^2} \ge {1 \over {16}}$$</p>
<p>$$\therefore$$ $${a^4}{b^6} \ge {1 \over {16}}$$</p>
<p>$$\therefore$$ Minimum value of $${a^4}{b^6} = {1 \over {16}}$$</p>
<p>$$\therefore$$ Minimum value of constant term</p>
<p>$${T_7} = 210 \times {a^4}{b^6}$$</p>
<p>$$ = 210 \times {1 \over {16}}$$</p>
<p>$$ = {{105} \over 8}$$</p>
|
mcq
|
jee-main-2022-online-30th-june-morning-shift
|
1l6dx5rjl
|
maths
|
binomial-theorem
|
general-term
|
<p>If the maximum value of the term independent of $$t$$ in the expansion of $$\left(\mathrm{t}^{2} x^{\frac{1}{5}}+\frac{(1-x)^{\frac{1}{10}}}{\mathrm{t}}\right)^{15}, x \geqslant 0$$, is $$\mathrm{K}$$, then $$8 \mathrm{~K}$$ is equal to ____________.</p>
|
[]
| null |
6006
|
<p>General term of $${\left( {{t^2}{x^{{1 \over 5}}} + {{{{(1 - x)}^{{1 \over {10}}}}} \over t}} \right)^{15}}$$ is</p>
<p>$${T_{r + 1}} = {}^{15}{C_r}\,.\,{\left( {{t^2}{x^{{1 \over 5}}}} \right)^{15 - r}}\,.\,{\left( {{{{{(1 - x)}^{{1 \over {10}}}}} \over t}} \right)^r}$$</p>
<p>$$ = {}^{15}{C_r}\,.\,{t^{30 - 2r}}\,.\,{x^{{{15 - r} \over 5}}}\,.\,{\left( {1 - x} \right)^{{r \over {10}}}}\,.\,{t^{ - r}}$$</p>
<p>$$ = {}^{15}{C_r}\,.\,{t^{30 - 3r}}\,.\,{x^{{{15 - r} \over 5}}}\,.\,{\left( {1 - x} \right)^{{r \over {10}}}}$$</p>
<p>Term will be independent of $$\mathrm{t}$$ when $$30 - 3r = 0 \Rightarrow r = 10$$</p>
<p>$$\therefore$$ $${T_{10 + 1}} = {T_{11}}$$ will be independent of $$\mathrm{t}$$</p>
<p>$$\therefore$$ $${T_{11}} = {}^{15}{C_{10}}\,.\,{x^{{{15 - 10} \over 5}}}\,.\,{\left( {1 - x} \right)^{{{10} \over {10}}}}$$</p>
<p>$$ = {}^{15}{C_{10}}\,.\,{x^1}\,.\,{\left( {1 - x} \right)^1}$$</p>
<p>$$\mathrm{T_{11}}$$ will be maximum when $$x(1 - x)$$ is maximum.</p>
<p>Let $$f(x) = x(1 - x) = x - {x^2}$$</p>
<p>$$f(x)$$ is maximum or minimum when $$f'(x) = 0$$</p>
<p>$$\therefore$$ $$f'(x) = 1 - 2x$$</p>
<p>For maximum/minimum $$f'(x) = 0$$</p>
<p>$$\therefore$$ $$1 - 2x = 0$$</p>
<p>$$ \Rightarrow x = {1 \over 2}$$</p>
<p>Now, $$f''(x) = - 2 < 0$$</p>
<p>$$\therefore$$ At $$ x = {1 \over 2}$$, $$f(x)$$ maximum</p>
<p>$$\therefore$$ Maximum value of $$\mathrm{T_{11}}$$ is</p>
<p>$$ = {}^{15}{C_{10}}\,.\,{1 \over 2}\left( {1 - {1 \over 2}} \right)$$</p>
<p>$$ = {}^{15}{C_{10}}\,.\,{1 \over 4}$$</p>
<p>Given $$K = {}^{15}{C_{10}}\,.\,{1 \over 4}$$</p>
<p>Now, $$8K = 2\left( {{}^{15}{C_{10}}} \right)$$</p>
<p>$$ = 6006$$</p>
|
integer
|
jee-main-2022-online-25th-july-morning-shift
|
1l6klfrzl
|
maths
|
binomial-theorem
|
general-term
|
<p>Let for the $$9^{\text {th }}$$ term in the binomial expansion of $$(3+6 x)^{\mathrm{n}}$$, in the increasing powers of $$6 x$$, to be the greatest for $$x=\frac{3}{2}$$, the least value of $$\mathrm{n}$$ is $$\mathrm{n}_{0}$$. If $$\mathrm{k}$$ is the ratio of the coefficient of $$x^{6}$$ to the coefficient of $$x^{3}$$, then $$\mathrm{k}+\mathrm{n}_{0}$$ is equal to :</p>
|
[]
| null |
24
|
<p>$${(3 + 6x)^n} = {3^n}{(1 + 2x)^n}$$</p>
<p>If T<sub>9</sub> is numerically greatest term</p>
<p>$$\therefore$$ $${T_8} \le {T_9} \le {T_{10}}$$</p>
<p>$${}^n{C_7}{3^{n - 7}}{(6x)^7} \le {}^n{C_8}{3^{n - 8}}{(6x)^8} \ge {}^n{C_9}{3^{n - 9}}{(6x)^9}$$</p>
<p>$$ \Rightarrow {{n!} \over {(n - 7)!7!}}9 \le {{n!} \over {(n - 8)!8!}}3\,.\,(6x) \ge {{n!} \over {(n - 9)!9!}}{(6x)^2}$$</p>
<p>$$ \Rightarrow \underbrace {{9 \over {(n - 7)(n - 8)}}}_{} \le \underbrace {{{18\left( {{3 \over 2}} \right)} \over {(n - 8)8}} \ge {{36} \over {9.8}}{9 \over 4}}_{}$$</p>
<p>$$72 \le 27(n - 7)$$ and $$27 \ge 9(n - 8)$$</p>
<p>$${{29} \over 3} \le n$$and $$n \le 11$$</p>
<p>$$\therefore$$ $${n_0} = 10$$</p>
<p>For $${(3 + 6x)^{10}}$$</p>
<p>$${T_{r + 1}} = {}^{10}{C_r}$$</p>
<p>$${3^{10 - r}}{(6x)^r}$$</p>
<p>For coeff. of x<sup>6</sup></p>
<p>$$r = 6 \Rightarrow {}^{10}{C_6}{3^4}{.6^6}$$</p>
<p>For coeff. of x<sup>3</sup></p>
<p>$$r = 3 \Rightarrow {}^{10}{C_3}{3^7}{.6^3}$$</p>
<p>$$\therefore$$ $$k = {{{}^{10}{C_6}} \over {{}^{10}{C_3}}}.{{{3^4}{{.6}^6}} \over {{3^7}{{.6}^3}}} = {{10!7!3!} \over {6!4!10!}}.8$$</p>
<p>$$ \Rightarrow k = 14$$</p>
<p>$$\therefore$$ $$k + {n_0} = 24$$</p>
|
integer
|
jee-main-2022-online-27th-july-evening-shift
|
1l6p3efud
|
maths
|
binomial-theorem
|
general-term
|
<p>Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of $$\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{\mathrm{n}}$$, in the increasing powers of $$\frac{1}{\sqrt[4]{3}}$$ be $$\sqrt[4]{6}: 1$$. If the sixth term from the beginning is $$\frac{\alpha}{\sqrt[4]{3}}$$, then $$\alpha$$ is equal to _________.</p>
|
[]
| null |
84
|
<p>Fifth term from beginning $$ = {}^n{C_4}{\left( {{2^{{1 \over 4}}}} \right)^{n - 4}}{\left( {{3^{{{ - 1} \over 4}}}} \right)^4}$$</p>
<p>Fifth term from end $$ = {(n - 5 + 1)^{th}}$$ term from begin $$ = {}^n{C_{n - 4}}{\left( {{2^{{1 \over 4}}}} \right)^3}{\left( {{3^{{{ - 1} \over 4}}}} \right)^{n - 4}}$$</p>
<p>Given $${{{}^n{C_4}{2^{{{n - 4} \over 4}}}\,.\,{3^{ - 1}}} \over {{}^n{C_{n - 3}}{2^{{4 \over 4}}}\,.\,{3^{ - \left( {{{n - 4} \over 4}} \right)}}}} = {6^{{1 \over 4}}}$$</p>
<p>$$ \Rightarrow {6^{{{n - 8} \over 4}}} = {6^{{1 \over 4}}}$$</p>
<p>$$ \Rightarrow {{n - 8} \over 4} = {1 \over 4} \Rightarrow n = 9$$</p>
<p>$${T_6} = {T_{5 + 1}} = {}^9{C_5}{\left( {{2^{{1 \over 4}}}} \right)^4}{\left( {{3^{{{ - 1} \over 4}}}} \right)^5}$$</p>
<p>$$ = {{{}^9{C_5}\,.\,2} \over {{3^{{1 \over 4}}}\,.\,3}} = {{84} \over {{3^{{1 \over 4}}}}} = {\alpha \over {{3^{{1 \over 4}}}}}$$</p>
<p>$$ \Rightarrow \alpha = 84.$$</p>
|
integer
|
jee-main-2022-online-29th-july-morning-shift
|
1ldo784qa
|
maths
|
binomial-theorem
|
general-term
|
<p>Let the sixth term in the binomial expansion of $${\left( {\sqrt {{2^{{{\log }_2}\left( {10 - {3^x}} \right)}}} + \root 5 \of {{2^{(x - 2){{\log }_2}3}}} } \right)^m}$$ in the increasing powers of $$2^{(x-2) \log _{2} 3}$$, be 21 . If the binomial coefficients of the second, third and fourth terms in the expansion are respectively the first, third and fifth terms of an A.P., then the sum of the squares of all possible values of $$x$$ is __________.</p>
|
[]
| null |
4
|
${ }^m C_1,{ }^m C_2,{ }^m C_3$ are first, third and fifth term of $A P$
<br/><br/>$$
\begin{aligned}
\therefore \quad & a={ }^m C_1 \\\\
& a+2 d={ }^m C_2 \\\\
& a+4 d={ }^m C_3 \\\\
\therefore \quad & 2{ }^m C_2-{ }^m C_3=m \\\\
\Rightarrow & m=7 \text { or } m=2 \\\\
\because & m=2 \text { is not possible } \\\\
\therefore & m=7
\end{aligned}
$$
<br/><br/>$\mathrm{T}_6={ }^{\mathrm{m}} \mathrm{C}_5\left(10-3^{\mathrm{x}}\right)^{\frac{\mathrm{m}-5}{2}} \cdot\left(3^{\mathrm{x}-2}\right)=21$
<br/><br/>Putting value of m = 7, we get
<br/><br/>$\begin{aligned} & T_{5+1}={ }^7 C_5\left(10-3^x\right)^{\frac{7-5}{2}} 3^{x-2}=21 \\\\ & \Rightarrow \frac{10.3^x-\left(3^x\right)^2}{3^2}=1\end{aligned}$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \left(3^x\right)^2-10 \cdot 3^x+9=0 \\\\
& \Rightarrow 3^x=9,1 \\\\
& \Rightarrow x=0,2
\end{aligned}
$$
<br/><br/>Sum of squares of values of x = 0<sup>2</sup> + 2<sup>2</sup> = 4
|
integer
|
jee-main-2023-online-1st-february-evening-shift
|
1ldo7goz9
|
maths
|
binomial-theorem
|
general-term
|
<p>If the term without $$x$$ in the expansion of $$\left(x^{\frac{2}{3}}+\frac{\alpha}{x^{3}}\right)^{22}$$ is 7315 , then $$|\alpha|$$ is equal to ___________.</p>
|
[]
| null |
1
|
Given expansion $\left(x^{\frac{2}{3}}+\frac{\alpha}{x^3}\right)^{22}$
<br/><br/>$$
T_{r+1}={ }^{22} C_r\left(x^{\frac{2}{3}}\right)^{22-r}\left(\frac{\alpha}{x^3}\right)^r
$$
<br/><br/>For constant term
<br/><br/>$$
\begin{aligned}
& \frac{44-2 r}{3}-3 r=0 \\\\
& \Rightarrow r=4
\end{aligned}
$$
<br/><br/>Now ${ }^{22} \mathrm{C}_4 \alpha^4=7315$
<br/><br/>$$
\begin{aligned}
& \frac{22 \times 21 \times 20 \times 19}{4 \times 3 \times 2 \times 1} \alpha^4=7315 \\\\
& \therefore \alpha^4=1 \\\\
& \therefore |\alpha|=1
\end{aligned}
$$
|
integer
|
jee-main-2023-online-1st-february-evening-shift
|
ldoaj02i
|
maths
|
binomial-theorem
|
general-term
|
The coefficient of $x^{-6}$, in the
<br/><br/>expansion of $\left(\frac{4 x}{5}+\frac{5}{2 x^{2}}\right)^{9}$, is
|
[]
| null |
5040
|
Coeff of $x^{-6}$ in $\left(\frac{4 x}{5}+\frac{5}{2 x^{2}}\right)^{9}$
<br/><br/>$$
\begin{aligned}
T_{r+1} & ={ }^{9} C_{r}\left(\frac{4 x}{5}\right)^{9-r}\left(\frac{5}{2 x^{2}}\right)^{r} \\\\
9-3 r & =-6 \\\\
r & =5
\end{aligned}
$$
<br/><br/>Coeff of $x^{-6}={ }^{9} C_{5}\left(\frac{4}{5}\right)^{4}\left(\frac{5}{2}\right)^{5}$
$$
=5040
$$
|
integer
|
jee-main-2023-online-31st-january-evening-shift
|
ldoavd66
|
maths
|
binomial-theorem
|
general-term
|
If the constant term in the binomial expansion of $\left(\frac{x^{\frac{5}{2}}}{2}-\frac{4}{x^{l}}\right)^{9}$ is $-84$ and the coefficient of $x^{-3 l}$ is
$2^{\alpha} \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to ________.
|
[]
| null |
98
|
Given binomial expansion of $\left(\frac{x^{\frac{5}{2}}}{2}-\frac{4}{x^{l}}\right)^{9}$
<br/><br/>$$
\begin{aligned}
T_{r+1} & ={ }^{9} C_{r}\left(\frac{x^{\frac{5}{2}}}{2}\right)^{9-r}\left(\frac{-4}{x^{l}}\right)^{r} \\
& ={ }^{9} C_{r} x^{\frac{45-5 r}{2}-lr} \cdot 2^{r-9} \cdot 4^{r} \cdot(-1)^{r}
\end{aligned}
$$
<br/><br/>For constant term, power of x is zero.
<br/><br/>So, $\frac{45-5 r}{2}=l r \Rightarrow 2 l r+5 r=45$
<br/><br/>Now constant term $=-84$
<br/><br/>and ${ }^{9} C_{r} \cdot 2^{3 r-9}(-1)^{r}=-84$
<br/><br/>So, $r=3$ and $l=5$
<br/><br/>Now for $x^{-15}, \frac{45-5 r}{2}-5 r=-15$
<br/><br/>$$ \Rightarrow $$ $$
45-15 r=-30
$$
<br/><br/>$$ \Rightarrow $$ $$
r=5
$$
<br/><br/>$\therefore $ Coefficient $=-{ }^{9} C_{5} 2^{6}=-63.2^{7}$
<br/><br/>$\therefore \alpha=7, \beta=-63$
<br/><br/>and $|\alpha l-\beta|=|7 \times 5+63|=98$
|
integer
|
jee-main-2023-online-31st-january-evening-shift
|
1ldptjmpy
|
maths
|
binomial-theorem
|
general-term
|
<p>Let $$\alpha>0$$, be the smallest number such that the expansion of $$\left(x^{\frac{2}{3}}+\frac{2}{x^{3}}\right)^{30}$$ has a term $$\beta x^{-\alpha}, \beta \in \mathbb{N}$$. Then $$\alpha$$ is equal to ___________.</p>
|
[]
| null |
2
|
$\mathrm{T}_{\mathrm{r}+1}={ }^{30} \mathrm{C}_{\mathrm{r}}\left(\mathrm{x}^{2 / 3}\right)^{30-\mathrm{r}}\left(\frac{2}{\mathrm{x}^{3}}\right)^{\mathrm{r}}$
<br/><br/>$={ }^{30} \mathrm{C}_{\mathrm{r}} \cdot 2^{\mathrm{r}} \cdot \mathrm{x}^{\frac{60-11 \mathrm{r}}{3}}$
<br/><br/>$\frac{60-11 \mathrm{r}}{3}<0
$
<br/><br/>$\Rightarrow 11 \mathrm{r}>60 $
<br/><br/>$\Rightarrow \mathrm{r}>\frac{60}{11} $
<br/><br/>$\Rightarrow \mathrm{r}=6$
<br/><br/>$\mathrm{T}_{7}={ }^{30} \mathrm{C}_{6} \cdot 2^{6} \mathrm{x}^{-2}$
<br/><br/>We have also observed $\beta={ }^{30} \mathrm{C}_{6}(2)^{6}$ is a natural number.
<br/><br/>$\therefore \alpha=2$
|
integer
|
jee-main-2023-online-31st-january-morning-shift
|
1ldr72ghg
|
maths
|
binomial-theorem
|
general-term
|
<p>If the coefficient of $$x^{15}$$ in the expansion of $$\left(\mathrm{a} x^{3}+\frac{1}{\mathrm{~b} x^{1 / 3}}\right)^{15}$$ is equal to the coefficient of $$x^{-15}$$ in the expansion of $$\left(a x^{1 / 3}-\frac{1}{b x^{3}}\right)^{15}$$, where $$a$$ and $$b$$ are positive real numbers, then for each such ordered pair $$(\mathrm{a}, \mathrm{b})$$ :</p>
|
[{"identifier": "A", "content": "a = 3b"}, {"identifier": "B", "content": "ab = 1"}, {"identifier": "C", "content": "ab = 3"}, {"identifier": "D", "content": "a = b"}]
|
["B"]
| null |
<p>For $$\left( {a{x^3} + {1 \over {b{x^{{1 \over 3}}}}}} \right)$$</p>
<p>$${T_{r + 1}} = {}^{15}{C_r}{(a{x^3})^{15 - r}}{\left( {{1 \over {b{x^{{1 \over 3}}}}}} \right)^1}$$</p>
<p>$$\therefore$$ $${x^{15}} \to 3(15 - r) - {r \over 3} = 15$$</p>
<p>$$ \Rightarrow 30 = {{10r} \over 3} \Rightarrow r = 9$$</p>
<p>Similarly, for $${\left( {a{x^{{1 \over 3}}} - {1 \over {b{x^3}}}} \right)^{15}}$$</p>
<p>$${T_{r + 1}} = {}^{15}{C_r}{\left( {a{x^{{1 \over 3}}}} \right)^{15 - r}}{\left( { - {1 \over {b{x^3}}}} \right)^2}$$</p>
<p>$$\therefore$$ For $${x^{ - 15}} \to {{15 - r} \over 3} - 3r = - 15 \Rightarrow r = 6$$</p>
<p>$$\therefore$$ $${}^{15}{C_9}{{{a^6}} \over {{b^9}}} = {}^{15}{C_6}{{{a^9}} \over {{b^6}}} \Rightarrow ab = 1$$</p>
|
mcq
|
jee-main-2023-online-30th-january-morning-shift
|
1ldswq8el
|
maths
|
binomial-theorem
|
general-term
|
<p>Let the coefficients of three consecutive terms in the binomial expansion of $$(1+2x)^n$$ be in the ratio 2 : 5 : 8. Then the coefficient of the term, which is in the middle of those three terms, is __________.</p>
|
[]
| null |
1120
|
$\mathrm{t}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}(2 \mathrm{x})^{\mathrm{r}}$
<br/><br/>
$$
\begin{aligned}
& \Rightarrow \frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}(2)^{\mathrm{r}-1}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}(2)^{\mathrm{r}}}=\frac{2}{5} \\\\
& \Rightarrow \frac{\frac{n !}{(r-1) !(n-r+1) !}}{\frac{n !(2)}{r !(n-r) !}}=\frac{2}{5} \\\\
& \Rightarrow \frac{\mathrm{r}}{\mathrm{n}-\mathrm{r}+1}=\frac{4}{5} \Rightarrow 5 \mathrm{r}=4 \mathrm{n}-4 \mathrm{r}+4 \\\\
& \Rightarrow 9 \mathrm{r}=4(\mathrm{n}+1) \quad\quad...(1)\\\\
& \Rightarrow \frac{{ }^{n} C_{r}(2)^{\mathrm{r}}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}(2)^{\mathrm{r}+1}}=\frac{5}{8} \\\\
& \Rightarrow \frac{\frac{n !}{r !(n-r) !}}{\frac{n !}{(r+1) !(n-r-1) !}}=\frac{5}{4} \Rightarrow \frac{r+1}{n-r}=\frac{5}{4} \\\\
& \Rightarrow 4 \mathrm{r}+4=5 \mathrm{n}-5 \mathrm{r} \Rightarrow 5 \mathrm{n}-4=9 \mathrm{r} \quad\quad...(2)
\end{aligned}
$$
<br/><br/>
From (1) and (2)
<br/><br/>
$$
\Rightarrow 4 \mathrm{n}+4=5 \mathrm{n}-4 \Rightarrow \mathrm{n}=8
$$
<br/><br/>
$(1) \Rightarrow r=4$
<br/><br/>
so, coefficient of middle term is
<br/><br/>
$$
{ }^{8} \mathrm{C}_{4} 2^{4}=16 \times \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}=16 \times 70=1120
$$
|
integer
|
jee-main-2023-online-29th-january-morning-shift
|
1ldswsczz
|
maths
|
binomial-theorem
|
general-term
|
<p>If the co-efficient of $$x^9$$ in $${\left( {\alpha {x^3} + {1 \over {\beta x}}} \right)^{11}}$$ and the co-efficient of $$x^{-9}$$ in $${\left( {\alpha x - {1 \over {\beta {x^3}}}} \right)^{11}}$$ are equal, then $$(\alpha\beta)^2$$ is equal to ___________.</p>
|
[]
| null |
1
|
Coefficient of $\mathrm{x}^{9}$ in $\left(\alpha x^{3}+\frac{1}{\beta x}\right)={ }^{11} C_{6} \cdot \frac{\alpha^{5}}{\beta^{6}}$
<br/><br/>
$\because$ Both are equal
<br/><br/>
$\therefore \frac{11}{C_{6}} \cdot \frac{\alpha^{5}}{\beta^{6}}=-\frac{11}{C_{5}} \cdot \frac{\alpha^{6}}{\beta^{5}}$
<br/><br/>
$\Rightarrow \frac{1}{\beta}=-\alpha$
<br/><br/>
$\Rightarrow \alpha \beta=-1$
<br/><br/>
$\Rightarrow(\alpha \beta)^{2}=1$
|
integer
|
jee-main-2023-online-29th-january-morning-shift
|
1ldv2styk
|
maths
|
binomial-theorem
|
general-term
|
<p>The constant term in the expansion of $${\left( {2x + {1 \over {{x^7}}} + 3{x^2}} \right)^5}$$ is ___________.</p>
|
[]
| null |
1080
|
Constant term in the expansion of
<br/><br/>
$$
\begin{aligned}
& \left(2 x+\frac{1}{x^{7}}+3 x^{2}\right)^{5} \\\\
& \frac{1}{x^{35}}\left(2 x^{8}+1+3 x^{9}\right)^{5} \\\\
& \frac{1}{x^{35}}\left(1+x^{8}(3 x+2)\right)^{5}
\end{aligned}
$$
<br/><br/>
Term independent of $x=$ coefficient of $x^{35}$ in
<br/><br/>
$$
\begin{aligned}
& ^{5} C_{4}\left(x^{8}(3 x+2)\right)^{4} \\\\
= & { }^{5} C_{4} \text { coefficient of } x^{3} \text { in }(2+3 x)^{4} \\\\
= & { }^{5} C_{4} \times{ }^{4} C_{3}(2)^{1}(3)^{3} \\\\
= & 5 \times 4 \times 2 \times 27 \\\\
= & 1080
\end{aligned}
$$
|
integer
|
jee-main-2023-online-25th-january-morning-shift
|
1ldwxmgwd
|
maths
|
binomial-theorem
|
general-term
|
<p>Let the sum of the coefficients of the first three terms in the expansion of $${\left( {x - {3 \over {{x^2}}}} \right)^n},x \ne 0.~n \in \mathbb{N}$$, be 376. Then the coefficient of $$x^4$$ is __________.</p>
|
[]
| null |
405
|
$S=1-3 n+\frac{9 n(n-1)}{2}=376$
<br/><br/>
$$
\begin{aligned}
& 3 n^{2}-5 n-250=0 \\\\
& n=10, \frac{-25}{3} \text { (Rejected) } \\\\
& T_{r+1}={ }^{n} C_{r} \cdot x^{n-r}\left(\frac{-3}{x^{2}}\right)^{r} \\\\
& ={ }^{n} C_{r} x x^{n-3 r}(-3)^{r} \\\\
& ={ }^{10} C_{r} x^{10-3 r}(-3)^{r}
\end{aligned}
$$
<br/><br/>
Here $r=2$
<br/><br/>
$$
\begin{aligned}
\text {Required coefficient } & ={ }^{10} \mathrm{C}_{2}(-3)^{2} \\\\
& =45 \times 9 \\\\
& =405
\end{aligned}
$$
|
integer
|
jee-main-2023-online-24th-january-evening-shift
|
1lgowbuor
|
maths
|
binomial-theorem
|
general-term
|
<p>The coefficient of $$x^{5}$$ in the expansion of $$\left(2 x^{3}-\frac{1}{3 x^{2}}\right)^{5}$$ is :</p>
|
[{"identifier": "A", "content": "$$\\frac{26}{3}$$"}, {"identifier": "B", "content": "$$\\frac{80}{9}$$"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "8"}]
|
["B"]
| null |
Given, $$\left(2 x^{3}-\frac{1}{3 x^{2}}\right)^{5}$$
<br/><br/>General term,
<br/><br/>$$
\begin{aligned}
& T_{r+1}={ }^5 C_r\left(2 x^3\right)^{5-r}\left(\frac{-1}{3 x^2}\right)^r={ }^5 C_r \frac{(2)^{5-r}}{(-3)^r}(x)^{15-5 r} \\\\
& \therefore 15-5 \mathrm{r}=5 \\\\
& \therefore \mathrm{r}=2 \\\\
& T_3=10\left(\frac{8}{9}\right) x^5
\end{aligned}
$$
<br/><br/>So, coefficient is $\frac{80}{9}$.
|
mcq
|
jee-main-2023-online-13th-april-evening-shift
|
1lgq11k4d
|
maths
|
binomial-theorem
|
general-term
|
<p>Let $$\alpha$$ be the constant term in the binomial expansion of $$\left(\sqrt{x}-\frac{6}{x^{\frac{3}{2}}}\right)^{n}, n \leq 15$$. If the sum of the coefficients of the remaining terms in the expansion is 649 and the coefficient of $$x^{-n}$$ is $$\lambda \alpha$$, then $$\lambda$$ is equal to _____________.</p>
|
[]
| null |
36
|
Given expression $(\sqrt{x}-\frac{6}{x^{3/2}})^n$. Here, $a = \sqrt{x}$ and $b = -\frac{6}{x^{3/2}}$.
<br/><br/>The $r$-th term of the binomial expansion of $(a+b)^n$ is given by
<br/><br/>$T_{r} = {}^n{C_r}a^{n-r}b^{r}$.
<br/><br/>Substitute $a$ and $b$ in this formula, we get:
<br/><br/>$T_{r} = {}^n{C_r}(\sqrt{x})^{n-r}\left(-\frac{6}{x^{3/2}}\right)^r = {}^n{C_r}(-6)^r x^{\frac{n-4r}{2}}$.
<br/><br/>The constant term in the binomial expansion is obtained when the power of $x$ in the terms equals zero.
<br/><br/>This happens when $\frac{n-4r}{2} = 0$, which gives $n = 4r$.
<br/><br/>$$
\begin{aligned}
& { }^n C_{\frac{n}{4}}(-6)^{\frac{n}{4}}=\alpha \\\\
& (-5)^n-{ }^n C_{\frac{n}{4}}(-6)^{n / 4}=649 \\\\
& \text { By observation (625 + 24 = 649) , we get n = 4 } \\\\
& \therefore \alpha=-24
\end{aligned}
$$
<br/><br/>Now, for coefficient of $x^{-4}$
<br/><br/>$$
\begin{aligned}
& \frac{n-4 r}{2}=-4 \\\\
& n=4 r-8 \Rightarrow r=3 \\\\
& \lambda(-24)=(-6)^3 \cdot{ }^4 C_3 \\\\
& \Rightarrow \lambda=36
\end{aligned}
$$
|
integer
|
jee-main-2023-online-13th-april-morning-shift
|
1lgsubacw
|
maths
|
binomial-theorem
|
general-term
|
<p>The sum of the coefficients of three consecutive terms in the binomial expansion of $$(1+\mathrm{x})^{\mathrm{n}+2}$$, which are in the ratio $$1: 3: 5$$, is equal to :</p>
|
[{"identifier": "A", "content": "63"}, {"identifier": "B", "content": "92"}, {"identifier": "C", "content": "25"}, {"identifier": "D", "content": "41"}]
|
["A"]
| null |
The problem asks for the sum of the coefficients of three consecutive terms in the binomial expansion of $(1+x)^{n+2}$, which are in the ratio 1 : 3 : 5.
<br/><br/>Given that the ratios of the coefficients are 1:3:5, we let the terms be $T_r$, $T_{r+1}$, and $T_{r+2}$. The coefficients of these terms are ${ }^{n+2} C_{r-1}$, ${ }^{n+2} C_{r}$, and ${ }^{n+2} C_{r+1}$, respectively.
<br/><br/>
$$
\begin{aligned}
& \frac{T_{r+1}}{T_r}=\frac{{ }^{n+2} C_r}{{ }^{n+2} C_{r-1}}=\frac{n+2-r+1}{r}=\frac{n+3-r}{r}=3 \\\\
& n-4 r+3=0 ......(1) \\\\
& \frac{T_{r+2}}{T_{r+1}}=\frac{{ }^{n+2} C_{r+1}}{{ }^{n+2} C_r}=\frac{(n+2)-(r+1)+1}{r+1}=\frac{n-r+2}{r+1}=\frac{5}{3} \\\\
& 3 n-8 r+1=0 ......(2)
\end{aligned}
$$
<br/><br/>By solving (1) and (2), we get
<br/><br/>$\Rightarrow n=5, r=2$
<br/><br/>$$
\begin{aligned}
T_r+T_{r+1}+T_{r+2} & ={ }^7 C_1+{ }^7 C_2+{ }^7 C_3 \\\\
& =7+21+35=63
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-11th-april-evening-shift
|
1lgsvbm4o
|
maths
|
binomial-theorem
|
general-term
|
<p>If the $$1011^{\text {th }}$$ term from the end in the binominal expansion of $$\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^{2022}$$ is 1024 times $$1011^{\text {th }}$$R term from the beginning, then $$|x|$$ is equal to</p>
|
[{"identifier": "A", "content": "$$\n\\frac{5}{16}\n$$"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "12"}, {"identifier": "D", "content": "15"}]
|
["A"]
| null |
$\mathrm{T}_{1011}$ from beginning $=\mathrm{T}_{1010+1}$
<br/><br/>$$
={ }^{2022} \mathrm{C}_{1010}\left(\frac{4 \mathrm{x}}{5}\right)^{1012}\left(\frac{-5}{2 \mathrm{x}}\right)^{1010}
$$
<br/><br/>$\mathrm{T}_{1011}$ from end
<br/><br/>$$
={ }^{2022} \mathrm{C}_{1010}\left(\frac{-5}{2 \mathrm{x}}\right)^{1012}\left(\frac{4 \mathrm{x}}{5}\right)^{1010}
$$
<br/><br/>$$
\begin{aligned}
& \text { Given : }{ }^{2022} \mathrm{C}_{1010}\left(\frac{-5}{2 \mathrm{x}}\right)^{1012}\left(\frac{4 \mathrm{x}}{5}\right)^{1010} \\\\
& =2^{10} \cdot{ }^{2022} \mathrm{C}_{1010}\left(\frac{-5}{2 \mathrm{x}}\right)^{1010}\left(\frac{4 \mathrm{x}}{5}\right)^{1012}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
&\Rightarrow \left(\frac{-5}{2 x}\right)^2=2^{10}\left(\frac{4 x}{5}\right)^2 \\\\
&\Rightarrow x^4=\frac{5^4}{2^{16}} \\\\
& \Rightarrow |x|=\frac{5}{16}
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-11th-april-evening-shift
|
1lguwxkf8
|
maths
|
binomial-theorem
|
general-term
|
<p>The mean of the coefficients of $$x, x^{2}, \ldots, x^{7}$$ in the binomial expansion of $$(2+x)^{9}$$ is ___________.</p>
|
[]
| null |
2736
|
We have, binomial coefficient, $(2+x)^9$
<br/><br/>$$
T_{r+1}={ }^n C_r 2^{n-r} \times x^r
$$
<br/><br/>Coefficient of $x\left(T_1\right)={ }^9 C_1 \times 2^8$
<br/><br/>Coefficient of $x^2\left(T_2\right)={ }^9 C_2 \times 2^7$
<br/><br/>Coefficient of $x^3\left(T_3\right)={ }^9 C_3 \times 2^6$
<br/> . .
<br/> . .
<br/> . .
<br/><br/>Coefficient of $x^7\left(T_7\right)={ }^9 C_7 \times 2^2$
<br/><br/>$$
\text { Mean }=\frac{{ }^9 C_1 \times 2^8+{ }^9 C_2 \times 2^7+{ }^9 C_3 \times 2^6+\ldots+{ }^9 C_7 \times 2^2}{7}
$$
<br/><br/>$$
\begin{aligned}
& { }^9 C_0 \times 2^9+{ }^9 C_1 \times 2^8+{ }^9 C_2 \times 2^7+{ }^9 C_3 \times 2^6+\ldots .+{ }^9 C_7 \times 2^2 \\
& =\frac{+{ }^9 C_8 \times 2^1+{ }^9 C_9 \times 2^0-{ }^9 C_0 \times 2^9-{ }^9 C_8 \times 2^1-{ }^9 C_9 \times 2^0}{7}
\end{aligned}
$$
<br/><br/>$$
=\frac{(1+2)^9-2^9-18-1}{7}=\frac{19152}{7}=2736
$$
|
integer
|
jee-main-2023-online-11th-april-morning-shift
|
1lguwzu1x
|
maths
|
binomial-theorem
|
general-term
|
<p>The number of integral terms in the expansion of $$\left(3^{\frac{1}{2}}+5^{\frac{1}{4}}\right)^{680}$$ is equal to ___________.</p>
|
[]
| null |
171
|
$$
\begin{aligned}
& \text { General term of the expansion }\left(3^{\frac{1}{2}}+5^{\frac{1}{4}}\right)^{680} \\\\
& \qquad={ }^{680} C_r\left(3^{1 / 2}\right)^{680-r}\left(5^{1 / 4}\right)^r={ }^{680} C_r \times 3^{\frac{680-r}{2}} \times 5^{\frac{r}{4}}
\end{aligned}
$$
<br/><br/>The term will be integral if $r$ is a multiple of 4 .
<br/><br/>$$
\begin{gathered}
\therefore r=0,4,8,12, \ldots, 680(\text { which is an } \mathrm{AP}) \\\\
680=0+(n-1) 4 \\\\
n=\frac{680}{4}+1=171
\end{gathered}
$$
|
integer
|
jee-main-2023-online-11th-april-morning-shift
|
1lgvpl71w
|
maths
|
binomial-theorem
|
general-term
|
<p>If the coefficients of $$x$$ and $$x^{2}$$ in $$(1+x)^{\mathrm{p}}(1-x)^{\mathrm{q}}$$ are 4 and $$-$$5 respectively, then $$2 p+3 q$$ is equal to :</p>
|
[{"identifier": "A", "content": "66"}, {"identifier": "B", "content": "60"}, {"identifier": "C", "content": "69"}, {"identifier": "D", "content": "63"}]
|
["D"]
| null |
We have, coefficient of $x$ in $(1+x)^p(1-x)^q=4$ and
<br/><br/>coefficient of $x^2$ in $(1+x)^p(1-x)^q=-5$
<br/><br/>$$
\begin{aligned}
& (1+x)^p(1-x)^q \\\\
& =\left(1+p x+\frac{p(p-1)}{2} x^2+\ldots\right)\left(1-q x+\frac{q(q-1)}{2} x^2+\ldots\right) \\\\
& =1+(p-q) x+\left(\frac{p(p-1)}{2}+\frac{q(q-1)}{2}-p q\right) x^2+\ldots \ldots
\end{aligned}
$$
<br/><br/>Coefficient of $x$ in $(1+x)^p(1-x)^q=-q+p$
<br/><br/>$\Rightarrow p-q=4$ ...........(i)
<br/><br/>$$
\text { Coefficient of } x^2 \text { in }(1+x)^p(1-x)^q=\frac{q(q-1)}{2}-p q+\frac{p(p-1)}{2}
$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \frac{q^2-q-2 p q+p^2-p}{2} =-5 \\\\
& \Rightarrow \frac{p^2+q^2-2 p q-(p+q)}{2} =-5 \\\\
& \Rightarrow (p-q)^2-(p+q) =-10 \\\\
& \Rightarrow (4)^2-(p+q) =-10 \quad[\because \text { From Eq. (i) }] \\\\
& \Rightarrow p+q =26 ...........(ii)
\end{aligned}
$$
<br/><br/>Form Eqs. (i) and (ii), we get $p=15, q=11$
<br/><br/>$\therefore 2 p+3 q=2 \times 15+3 \times 11=30+33=63$
|
mcq
|
jee-main-2023-online-10th-april-evening-shift
|
1lgxt15ef
|
maths
|
binomial-theorem
|
general-term
|
<p>If the coefficient of $${x^7}$$ in $${\left( {ax - {1 \over {b{x^2}}}} \right)^{13}}$$ and the coefficient of $${x^{ - 5}}$$ in $${\left( {ax + {1 \over {b{x^2}}}} \right)^{13}}$$ are equal, then $${a^4}{b^4}$$ is equal to :</p>
|
[{"identifier": "A", "content": "22"}, {"identifier": "B", "content": "33"}, {"identifier": "C", "content": "44"}, {"identifier": "D", "content": "11"}]
|
["A"]
| null |
The given expression is $\left(a x-\frac{1}{b x^2}\right)^{13}$
<br/><br/>So,
<br/><br/>$$
\begin{aligned}
T_{r+1} & ={ }^{13} C_r(a x)^{13-r}\left(-\frac{1}{b x^2}\right)^r \\\\
& ={ }^{13} C_r(a)^{13-r} x^{13-r-2 r}(-1 / b)^r \\\\
& ={ }^{13} C_r(a)^{13-r}\left(-\frac{1}{b}\right)^r x^{13-3 r}
\end{aligned}
$$
<br/><br/>For coefficient of $x^7$ in $\left(a x-\frac{1}{b x^2}\right)^{13}$
<br/><br/>$$
\begin{aligned}
& \quad 13-3 r=7 \\\\
& \Rightarrow 3 r=6 \Rightarrow r=2 \\\\
& \therefore \text { Coefficient of } x^7={ }^{13} C_2 \cdot(a)^{11} \cdot \frac{1}{b^2}
\end{aligned}
$$
<br/><br/>Again, the another expression is $\left(a x+\frac{1}{b x^2}\right)^{13}$
<br/><br/>So, $T_{n+1}={ }^{13} C_r(a x)^{13-r}\left(\frac{1}{b x^2}\right)^r={ }^{13} C_r(a)^{13-r}\left(\frac{1}{b}\right)^r x^{13-3 r}$
<br/><br/>For coefficient $x^{-5}$ in $\left(a x+\frac{1}{b x^2}\right)^{13}$
<br/><br/>$$
\begin{aligned}
&13-3 r =-5 \\\\
&\Rightarrow r =6
\end{aligned}
$$
<br/><br/>So, coefficient of $x^{-5}={ }^{13} C_6(a)^7 \frac{1}{b^6}$
<br/><br/>Now, according to the question,
<br/><br/>${ }^{13} C_2(a)^{11} \frac{1}{b^2}={ }^{13} C_6(a)^7 \frac{1}{b^6}$
<br/><br/>$$
\begin{aligned}
& \Rightarrow a^4 b^4=\frac{{ }^{13} C_6}{{ }^{13} C_2} \\\\
& \therefore a^4 b^4=22
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-10th-april-morning-shift
|
1lgyliytd
|
maths
|
binomial-theorem
|
general-term
|
<p>The absolute difference of the coefficients of $$x^{10}$$ and $$x^{7}$$ in the expansion of $$\left(2 x^{2}+\frac{1}{2 x}\right)^{11}$$ is equal to :</p>
|
[{"identifier": "A", "content": "$$11^{3}-11$$"}, {"identifier": "B", "content": "$$13^{3}-13$$"}, {"identifier": "C", "content": "$$12^{3}-12$$"}, {"identifier": "D", "content": "$$10^{3}-10$$"}]
|
["C"]
| null |
General term of $\left(2 x^2+\frac{1}{2 x}\right)^{11}$ is :
<br/><br/>$$
\begin{aligned}
& \mathrm{T}_{\mathrm{r}+1}={ }^{11} \mathrm{C}_r\left(2 x^2\right){ }^{11-r}\left(\frac{1}{2 x}\right)^r \\\\
& ={ }^{11} \mathrm{C}_r 2^{11-r} x^{22-2 r} 2^{-r} x^{-r} \\\\
& ={ }^{11} \mathrm{C}_r 2^{11-r} x^{22-3 r}
\end{aligned}
$$
<br/><br/>Now, $22-2 r=10$ and $22-3 r=7$
<br/><br/>$$
\begin{array}{ll}
\Rightarrow 3 r=12 &&& \Rightarrow 3 r=15 \\\\
\Rightarrow r=4 &&& \Rightarrow r=5
\end{array}
$$
<br/><br/>$\therefore$ Coeff. of $x^{10}={ }^{11} \mathrm{C}_4 \cdot 2^{11-8}={ }^{11} \mathrm{C}_4 \times 8$
<br/><br/>Coeff. of $x^7={ }^{11} C_5 \cdot 2^{11-10}={ }^{11} C_4 \times 2$
<br/><br/>Now, required difference
<br/><br/>$$
\begin{aligned}
& ={ }^{11} \mathrm{C}_4 \times 8-{ }^{11} \mathrm{C}_5 \times 2 \\\\
& =\frac{11 \times 10 \times 9 \times 8 \times 7 !}{4 ! \times 7 !} \times 8-\frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 ! \times 2}{5 ! 6 !}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& =\frac{11 \times 10 \times 9 \times 8 \times 8}{24}-\frac{11 \times 10 \times 9 \times 8 \times 7 \times 2}{120} \\\\
& =11 \times 10 \times 8 \times 3-11 \times 3 \times 4 \times 7 \\\\
& =11 \times 3 \times 4[20-7] \\\\
& =11 \times 12 \times 13=(12-1) \times 12 \times(12+1) \\\\
& =12\left(12^2-1\right)=12^3-12
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-8th-april-evening-shift
|
1lh00ko4t
|
maths
|
binomial-theorem
|
general-term
|
<p>Let $$[t]$$ denote the greatest integer $$\leq t$$. If the constant term in the expansion of $$\left(3 x^{2}-\frac{1}{2 x^{5}}\right)^{7}$$ is $$\alpha$$, then $$[\alpha]$$ is equal to ___________.</p>
|
[]
| null |
1275
|
Let $\mathrm{T}_{r+1}$ be the constant term.
<br/><br/>$$
\mathrm{T}_{r+1}={ }^7 \mathrm{C}_r\left(3 x^2\right)^{7-r}\left(\frac{-1}{2 x^5}\right)^r
$$
<br/><br/>For constant term, power of $x$ should be zero.
<br/><br/>$$
\begin{aligned}
& \text { i.e., } 14-2 r-5 r=0 \\\\
& \Rightarrow 14=7 r \Rightarrow r=2
\end{aligned}
$$
<br/><br/>Now, constant term $=\alpha$
<br/><br/>$$
\begin{aligned}
& \Rightarrow{ }^7 C_2(3)^5\left(\frac{-1}{2}\right)^2=\alpha \\\\
& \Rightarrow 21 \times 243 \times \frac{1}{4}=\alpha \\\\
& \Rightarrow[\alpha]=[1275.75]=1275
\end{aligned}
$$
|
integer
|
jee-main-2023-online-8th-april-morning-shift
|
1lh23fm6b
|
maths
|
binomial-theorem
|
general-term
|
<p>If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $$\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{\mathrm{n}}$$ is $$\sqrt{6}: 1$$, then the third term from the beginning is :</p>
|
[{"identifier": "A", "content": "$$30 \\sqrt{2}$$"}, {"identifier": "B", "content": "$$60 \\sqrt{3}$$"}, {"identifier": "C", "content": "$$60 \\sqrt{2}$$"}, {"identifier": "D", "content": "$$30 \\sqrt{3}$$"}]
|
["B"]
| null |
$$
\mathrm{T}_{r+1}={ }^n \mathrm{C}_r x^{n-r} a^r
$$
<br/><br/>$$
\frac{T_5}{T_5^{\prime}}=\frac{{ }^n C_4 \times\left((2)^{\frac{1}{4}}\right)^{n-4}\left(\frac{1}{3^{1 / 4}}\right)^4}{{ }^n C_4\left(\frac{1}{3^{1 / 4}}\right)^{n-4}\left(2^{1 / 4}\right)^4}=\sqrt{6}
$$
<br/><br/>$\left[\because r\right.$th term from end in the expansion of $(x+y)^n=r$th term from beginning in the expansion of $\left.(y+x)^n\right]$
<br/><br/>$$
\Rightarrow \frac{{ }^n C_4(2)^{\frac{n-4}{4}}\left(\frac{1}{3}\right)^{4 / 4}}{{ }^n C_4\left(\frac{1}{3}\right)^{\frac{n-4}{4}}(2)^{4 / 4}}=\frac{\sqrt{6}}{1}
$$
<br/><br/>$$
\begin{aligned}
&\Rightarrow (2)^{\frac{n-8}{4}}(3)^{\frac{n-8}{4}} =6^{1 / 2} \\\\
&\Rightarrow 6^{\frac{n-8}{4}} =6^{1 / 2} \\\\
&\Rightarrow \frac{n-8}{4} =\frac{1}{2} \\\\
&\Rightarrow n-8=2 \Rightarrow n =10
\end{aligned}
$$
<br/><br/>$$
\therefore T_3={ }^{10} C_2(\sqrt[4]{2})^8\left(\frac{1}{\sqrt[4]{3}}\right)^2=45(2)^{\frac{8}{4}} \frac{1}{3^{1 / 2}}
$$
<br/><br/>$$
=45(4) \times \frac{\sqrt{3}}{3}=60 \sqrt{3}
$$
|
mcq
|
jee-main-2023-online-6th-april-morning-shift
|
1lh2xsi9y
|
maths
|
binomial-theorem
|
general-term
|
<p>If the coefficient of $${x^7}$$ in $${\left( {a{x^2} + {1 \over {2bx}}} \right)^{11}}$$ and $${x^{ - 7}}$$ in $${\left( {ax - {1 \over {3b{x^2}}}} \right)^{11}}$$ are equal, then :</p>
|
[{"identifier": "A", "content": "$$243ab = 64$$"}, {"identifier": "B", "content": "$$32ab = 729$$"}, {"identifier": "C", "content": "$$64ab = 243$$"}, {"identifier": "D", "content": "$$729ab = 32$$"}]
|
["D"]
| null |
General term of $\left(a x^2+\frac{1}{2 b x}\right)^{11}$ is
<br/><br/>$$
T_{r+1}={ }^{11} C_r\left(a x^2\right)^{11-r}\left(\frac{1}{2 b x}\right)^r={ }^{11} C_r(a)^{11-r}\left(\frac{1}{2 b}\right)^r x^{22-3 r}
$$
<br/><br/>$$
\begin{array}{rlrl}
&\text { Now, } 22-3 r =7 \\\\
&\Rightarrow 15 =3 r \\\\
&\Rightarrow r =5
\end{array}
$$
<br/><br/>and general term of $\left(a x-\frac{1}{3 b x^2}\right)^{11}$ is
<br/><br/>$$
\begin{aligned}
T_{r+1} & ={ }^{11} C_r(a x)^{11-r}\left(-\frac{1}{3 b x^2}\right)^r \\\\
& ={ }^{11} C_r a^{11-r}\left(-\frac{1}{3 b}\right)^r x^{11-3 r}
\end{aligned}
$$
<br/><br/>Now, $11-3 r=-7$
<br/><br/>$$
\Rightarrow 18=3 r \Rightarrow r=6
$$
<br/><br/>$\begin{aligned} & \text { Since, coefficient of } x^7 \text { in }\left(a x^2+\frac{1}{2 b x}\right)^{11} \\\\ & =\text { Coefficient of } x^{-7} \text { in }\left(a x-\frac{1}{3 b x^2}\right)^{11} \\\\ & \Rightarrow{ }^{11} C_5(a)^6\left(\frac{1}{2 b}\right)^5={ }^{11} C_6(a)^5\left(-\frac{1}{3 b}\right)^6 \\\\ & \Rightarrow \frac{a}{32 b^5}=\frac{1}{729 b^6} \Rightarrow 729 a b=32\end{aligned}$
|
mcq
|
jee-main-2023-online-6th-april-evening-shift
|
jaoe38c1lscoeie7
|
maths
|
binomial-theorem
|
general-term
|
<p>The coefficient of $$x^{2012}$$ in the expansion of $$(1-x)^{2008}\left(1+x+x^2\right)^{2007}$$ is equal to _________.</p>
|
[]
| null |
0
|
<p>$$\begin{aligned}
& (1-x)(1-x)^{2007}\left(1+x+x^2\right)^{2007} \\
& (1-x)\left(1-x^3\right)^{2007} \\
& (1-x)\left({ }^{2007} C_0-{ }^{2007} C_1\left(x^3\right)+\ldots \ldots .\right)
\end{aligned}$$</p>
<p>General term</p>
<p>$$\begin{aligned}
& (1-x)\left((-1)^r{ }^{2007} C_r x^{3 r}\right) \\
& (-1)^{r 2007} C_r x^{3 r}-(-1)^{r 2007} C_r x^{3 r+1} \\
& 3 r=2012 \\
& r \neq \frac{2012}{3} \\
& 3 r+1=2012 \\
& 3 r=2011 \\
& r \neq \frac{2011}{3}
\end{aligned}$$</p>
<p>Hence there is no term containing $$\mathrm{x}^{2012}$$.</p>
<p>So coefficient of $$\mathrm{x}^{2012}=0$$</p>
|
integer
|
jee-main-2024-online-27th-january-evening-shift
|
lv3vef7y
|
maths
|
binomial-theorem
|
general-term
|
<p>If the term independent of $$x$$ in the expansion of $$\left(\sqrt{\mathrm{a}} x^2+\frac{1}{2 x^3}\right)^{10}$$ is 105 , then $$\mathrm{a}^2$$ is equal to :</p>
|
[{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "9"}]
|
["B"]
| null |
<p>$$\begin{aligned}
& \left(\sqrt{a} x^2+\frac{1}{2 x^3}\right)^{10} \\
& T_{r+1}={ }^{10} C_r\left(\sqrt{a} x^2\right)^{10-r}\left(\frac{1}{2 x^3}\right)^r
\end{aligned}$$</p>
<p>Independent of $$x \Rightarrow 20-2 r-3 r=0$$</p>
<p>$$r=4$$</p>
<p>Independent of $$x$$ is $${ }^{10} C_4(\sqrt{a})^6\left(\frac{1}{2}\right)^4=105$$</p>
<p>$$\begin{gathered}
\frac{210}{2 \times 8} a^3=105 \\
\Rightarrow \quad a=2 \\
a^2=4
\end{gathered}$$</p>
|
mcq
|
jee-main-2024-online-8th-april-evening-shift
|
lv7v47v9
|
maths
|
binomial-theorem
|
general-term
|
<p>If the constant term in the expansion of $$\left(1+2 x-3 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$$ is $$\mathrm{p}$$, then $$108 \mathrm{p}$$ is equal to ________.</p>
|
[]
| null |
54
|
<p>$$\text { General term of }\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$$</p>
<p>$$T_{r+1}={ }^9 C_r\left(\frac{3}{2} x^2\right)^{9-r}\left(-\frac{1}{3 x}\right)^r={ }^9 C_r(-1)^r 3^{9-2 r} 2^{r-9} x^{18-35}$$</p>
<p>Constant term in expansion of $$\left(1+2 x-3 x^3\right)$$</p>
<p>$$\begin{aligned}
& \left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9 \\
& =T_7-3 T_8={ }^9 C_6 3^{-3} \cdot 2^{-3}+3{ }^9 C_7 \cdot 3^{-5} \cdot 2^{-2} \\
& =\frac{3 \times 4 \times 7}{3^3 \cdot 2^3}+\frac{3 \times 9 \times 4}{3^5 \times 2^2}= \\
& p=\frac{42+12}{108}=\frac{54}{108} \\
& 108 p=54
\end{aligned}$$</p>
|
integer
|
jee-main-2024-online-5th-april-morning-shift
|
lv9s1zz5
|
maths
|
binomial-theorem
|
general-term
|
<p>If the constant term in the expansion of $$\left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[3]{5}}\right)^{12}, x \neq 0$$, is $$\alpha \times 2^8 \times \sqrt[5]{3}$$, then $$25 \alpha$$ is equal to :</p>
|
[{"identifier": "A", "content": "724"}, {"identifier": "B", "content": "742"}, {"identifier": "C", "content": "693"}, {"identifier": "D", "content": "639"}]
|
["C"]
| null |
<p>$$\begin{aligned}
& \left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[5]{3}}\right)^{12} \\
& T_{r+1}={ }^{12} C r\left(\frac{\sqrt[5]{3}}{x}\right)^{12-r}\left(\frac{2 x}{\sqrt[3]{5}}\right)^r
\end{aligned}$$</p>
<p>For constant term $$-12+r+r=0$$</p>
<p>$$\begin{aligned}
& \Rightarrow \quad r=6 \\
& \therefore \quad \text { Constant term }={ }^{12} C_6 \frac{(3)^{\frac{6}{5}}}{(5)^{\frac{6}{3}}}(2)^6
\end{aligned}$$</p>
<p>$$\begin{aligned}
& ={ }^{12} C_6 \times \frac{2^6}{25} \times 3.3^{\frac{1}{5}} \\
& =\frac{231}{25} \times 2^8 \cdot 3^{\frac{1}{5}} \cdot 3 \\
& =\frac{693}{25} \cdot 2^8 \sqrt[5]{3} \\
& \therefore \quad \alpha=\frac{693}{25} \\
& 25 \alpha=693
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-5th-april-evening-shift
|
lvc57np0
|
maths
|
binomial-theorem
|
general-term
|
<p>If the second, third and fourth terms in the expansion of $$(x+y)^n$$ are 135, 30 and $$\frac{10}{3}$$, respectively, then $$6\left(n^3+x^2+y\right)$$ is equal to __________.</p>
|
[]
| null |
806
|
<p>$$\begin{aligned}
& T_2={ }^n C_1 y^1 \cdot x^{n-1}=135 \\
& T_3={ }^n C_2 y^2 \cdot x^{n-2}=30 \\
& T_4={ }^n C_3 y^3 x^{n-3}=\frac{10}{3}
\end{aligned}$$</p>
<p>$$\begin{aligned}
\Rightarrow \frac{135}{30} & =\left(\frac{x}{y}\right) \frac{n \cdot 2}{n(n-1)}=\left(\frac{2}{n-1}\right)\left(\frac{x}{y}\right) \quad \text{... (i)}\\
\frac{30}{\frac{10}{3}} & =\frac{n(n-1)}{2} \frac{3!}{n(n-1)(n-2)}\left(\frac{x}{y}\right) \\
9 & =\left(\frac{3}{n-2}\right)\left(\frac{x}{y}\right)
\end{aligned}$$</p>
<p>$$\begin{aligned}
\Rightarrow & 3(n-2)=\frac{135}{60}(n-1) \Rightarrow n=5 \\
\Rightarrow & x=9 y \quad \text{.... (i)}\\
& y \cdot x^4=27 \Rightarrow \frac{x}{9} \cdot x^4=3^3 \\
\Rightarrow & x^5=3^5 \Rightarrow x=3 y=\frac{1}{3} \\
\Rightarrow & 6\left(5^3+3^2+\frac{1}{3}\right)=6\left(125+9+\frac{1}{3}\right)
\end{aligned}$$</p>
<p>$$=6(134)+2=806$$</p>
|
integer
|
jee-main-2024-online-6th-april-morning-shift
|
MdI6myzXplT0kKOP
|
maths
|
binomial-theorem
|
integral-and-fractional-part-of-a-number
|
If $$n$$ is a positive integer, then $${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$$ is :
|
[{"identifier": "A", "content": "an irrational number "}, {"identifier": "B", "content": "an odd positive integer "}, {"identifier": "C", "content": "an even positive integer "}, {"identifier": "D", "content": "a rational number other than positive integers "}]
|
["A"]
| null |
Let $${\left( {a + x} \right)^n}$$ = Odd trems(A) + Even terms(B)
<br><br>So $${\left( {a - x} \right)^n}$$ = Odd terms(A) - Even terms(B)
<br><br>$$\therefore$$ $${\left( {a + x} \right)^n} - {\left( {a - x} \right)^n}$$
<br><br>= (A + B) - (A - B)
<br><br>= 2B
<br><br>= 2[even terms]
<br><br>= 2[ T<sub>2</sub> + T<sub>4</sub> + T<sub>6</sub> + ....... ]
<br><br>So in case of $${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$$
<br><br>= 2[ T<sub>2</sub> + T<sub>4</sub> + T<sub>6</sub> + ....... ]
<br><br>= 2[ $${}^{2n}{C_1}.{\left( {\sqrt 3 } \right)^{2n - 1}} + {}^{2n}{C_3}.{\left( {\sqrt 3 } \right)^{2n - 3}} + ...$$
<br><br>Here in $${\left( {\sqrt 3 } \right)^{2n - 1}}$$, 2n - 1 is odd number. So there will be always $${\sqrt 3 }$$ in $${\left( {\sqrt 3 } \right)^{2n - 1}}$$.
<br><br>So $${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$$ will be always irrational number.
|
mcq
|
aieee-2012
|
hSlHFjRByfCzOzj87g1j0
|
maths
|
binomial-theorem
|
integral-and-fractional-part-of-a-number
|
If the fractional part of the number $$\left\{ {{{{2^{403}}} \over {15}}} \right\} is \, {k \over {15}}$$, then k is equal to :
|
[{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "14"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "1"}]
|
["A"]
| null |
$${{{2^{403}}} \over {15}}$$
<br><br>$$ = {{{2^3}\, \cdot \,{2^{400}}} \over {15}}$$
<br><br>$$ = {8 \over {15}}{\left( {16} \right)^{100}}$$
<br><br>$$ = {8 \over {15}}{\left( {15 + 1} \right)^{100}}$$
<br><br>$$ = {8 \over {15}}\left( {{}^{100}{C_0} + {}^{100}{C_1}\,15 + {}^{100}{C_2}{{\left( {15} \right)}^2} + .....{{\left( {15} \right)}^{100}}} \right)$$
<br><br>$$ = {8 \over {15}} + 8\left( {{}^{100}{C_1} + {}^{100}{C_2}\,\left( {15} \right) + ..... + {{\left( {15} \right)}^{99}}} \right)$$
<br><br>$$ = {8 \over {15}} + 8$$ (integer)
<br><br>$$ \therefore $$ Fractional part $$ = {8 \over {15}}$$
<br><br>According to the question,
<br><br>$${k \over {15}} = {8 \over {15}}$$
<br><br>$$ \Rightarrow $$ K $$=$$ 8
|
mcq
|
jee-main-2019-online-9th-january-morning-slot
|
cYkeSxybbUbNNx1ExBjgy2xukfuvxjpi
|
maths
|
binomial-theorem
|
integral-and-fractional-part-of-a-number
|
If {p} denotes the fractional part of the number p, then
<br/>$$\left\{ {{{{3^{200}}} \over 8}} \right\}$$, is equal to :
|
[{"identifier": "A", "content": "$${5 \\over 8}$$"}, {"identifier": "B", "content": "$${7 \\over 8}$$"}, {"identifier": "C", "content": "$${1 \\over 8}$$"}, {"identifier": "D", "content": "$${3 \\over 8}$$"}]
|
["C"]
| null |
$$\left\{ {{{{3^{200}}} \over 8}} \right\}$$
<br><br>= $$\left\{ {{{{{\left( {{3^2}} \right)}^{100}}} \over 8}} \right\}$$
<br><br>= $$\left\{ {{{{{\left( {1 + 8} \right)}^{100}}} \over 8}} \right\}$$
<br><br>= $$\left\{ {{{1 + {}^{100}{C_1}.8 + {}^{100}{C_2}{{.8}^2} + .... + {}^{100}{C_{100}}{{.8}^{100}}} \over 8}} \right\}$$
<br><br>= $$\left\{ {{{1 + 8K} \over 8}} \right\}$$
<br><br>= $$\left\{ {{1 \over 8} + K} \right\}$$ where K $$ \in $$ Integer
<br><br>$$ \therefore $$ Fractional part = $${{1 \over 8}}$$
|
mcq
|
jee-main-2020-online-6th-september-morning-slot
|
ldqy761e
|
maths
|
binomial-theorem
|
integral-and-fractional-part-of-a-number
|
Let $x=(8 \sqrt{3}+13)^{13}$ and $y=(7 \sqrt{2}+9)^9$. If $[t]$ denotes the greatest integer $\leq t$, then :
|
[{"identifier": "A", "content": "$[x]$ is odd but $[y]$ is even"}, {"identifier": "B", "content": "$[x]$ and $[y]$ are both odd"}, {"identifier": "C", "content": "$[x]+[y]$ is even"}, {"identifier": "D", "content": "$[x]$ is even but $[y]$ is odd"}]
|
["C"]
| null |
<p>If $${I_1} + f = {(8\sqrt 3 + 13)^{13}},f' = {(8\sqrt 3 - 13)^{13}}$$</p>
<p>$${I_1} + f - f'=$$ Even</p>
<p>$${I_1} = $$ Even</p>
<p>$${I_2} + f - f' = {(7\sqrt 2 + 9)^9} + {(7\sqrt 2 - 9)^9}$$</p>
<p>= Even</p>
<p>$${I_2} = $$ Even</p>
|
mcq
|
jee-main-2023-online-30th-january-evening-shift
|
SwstiO1vnflVEJ5e
|
maths
|
binomial-theorem
|
middle-term
|
The coefficient of the middle term in the binomial expansion in powers of $$x$$ of $${\left( {1 + \alpha x} \right)^4}$$ and $${\left( {1 - \alpha x} \right)^6}$$ is the same if $$\alpha $$ equals
|
[{"identifier": "A", "content": "$${3 \\over 5}$$ "}, {"identifier": "B", "content": "$${10 \\over 3}$$"}, {"identifier": "C", "content": "$${{ - 3} \\over {10}}$$ "}, {"identifier": "D", "content": "$${{ - 5} \\over {3}}$$"}]
|
["C"]
| null |
For $${\left( {1 + \alpha x} \right)^4}$$ the middle term $${T_{{4 \over 2} + 1}}$$ = $${}^4{C_2}.{\alpha ^2}{x^2}$$
<br><br>$$\therefore$$ Coefficient of middle term = $${}^4{C_2}.{\alpha ^2}$$
<br><br>For $${\left( {1 - \alpha x} \right)^6}$$ the middle term $${T_{{6 \over 2} + 1}}$$ = $${}^6{C_3}.-{\alpha ^3}{x^3}$$
<br><br>$$\therefore$$ Coefficient of middle term = $${}^6{C_3}.{-\alpha ^3}$$
<br><br>$$\therefore$$ According to question,
<br><br>$${}^4{C_2}.{\alpha ^2}$$ = $${}^6{C_3}.{-\alpha ^3}$$
<br><br>$$ \Rightarrow 6 = 20 \times - \alpha $$
<br><br>$$ \Rightarrow \alpha = - {3 \over {10}}$$
|
mcq
|
aieee-2004
|
AcHYxPg1DlPdQGa51uedV
|
maths
|
binomial-theorem
|
middle-term
|
The sum of the real values of x for which the middle term in the binomial expansion of $${\left( {{{{x^3}} \over 3} + {3 \over x}} \right)^8}$$ equals 5670 is :
|
[{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "4"}]
|
["A"]
| null |
$${T_5} = {}^8{C_4}{{{x^{12}}} \over {81}} \times {{81} \over {{x^4}}} = 5670$$
<br><br>$$ \Rightarrow 70{x^8} = 5670$$
<br><br>$$ \Rightarrow x = \pm \sqrt 3 $$
|
mcq
|
jee-main-2019-online-11th-january-morning-slot
|
1krw2nhf2
|
maths
|
binomial-theorem
|
middle-term
|
The ratio of the coefficient of the middle term in the expansion of (1 + x)<sup>20</sup> and the sum of the coefficients of two middle terms in expansion of (1 + x)<sup>19</sup> is _____________.
|
[]
| null |
1
|
Coeff. of middle term in (1 + x)<sup>20</sup> = $${}^{20}{C_{10}}$$ & Sum of coeff. of two middle terms in (1 + x)<sup>19</sup> = $${}^{19}{C_{9}}$$ + $${}^{19}{C_{10}}$$<br><br>So required ratio = $${{{}^{20}{C_{10}}} \over {^{19}{C_9}{ + ^{19}}{C_{10}}}} = {{^{20}{C_{10}}} \over {^{20}{C_{10}}}} = 1$$
|
integer
|
jee-main-2021-online-25th-july-morning-shift
|
1l6novn6c
|
maths
|
binomial-theorem
|
middle-term
|
<p>Let the coefficients of the middle terms in the expansion of $$\left(\frac{1}{\sqrt{6}}+\beta x\right)^{4},(1-3 \beta x)^{2}$$ and $$\left(1-\frac{\beta}{2} x\right)^{6}, \beta>0$$, respectively form the first three terms of an A.P. If d is the common difference of this A.P. , then $$50-\frac{2 d}{\beta^{2}}$$ is equal to __________.</p>
|
[]
| null |
57
|
<p>Coefficients of middle terms of given expansions are $${}^4{C_2}{1 \over 6}{\beta ^2},\,{}^2{C_1}( - 3\beta ),\,{}^6{C_3}{\left( {{{ - \beta } \over 2}} \right)^3}$$ form an A.P.</p>
<p>$$\therefore$$ $$2.2( - 3\beta ) = {\beta ^2} - {{5{\beta ^3}} \over 2}$$</p>
<p>$$ \Rightarrow - 24 = 2\beta - 5{\beta ^2}$$</p>
<p>$$ \Rightarrow 5{\beta ^2} - 2\beta - 24 = 0$$</p>
<p>$$ \Rightarrow 5{\beta ^2} - 12\beta + 10\beta - 24 = 0$$</p>
<p>$$ \Rightarrow \beta (5\beta - 12) + 2(5\beta - 12) = 0$$</p>
<p>$$\beta = {{12} \over 5}$$</p>
<p>$$d = - 6\beta - {\beta ^2}$$</p>
<p>$$\therefore$$ $$50 - {{2d} \over {{\beta ^2}}} = 50 - 2{{( - 6\beta - {\beta ^2})} \over {{\beta ^2}}} = 50 + {{12} \over \beta } + 2 = 57$$</p>
|
integer
|
jee-main-2022-online-28th-july-evening-shift
|
1ldsfiyxc
|
maths
|
binomial-theorem
|
middle-term
|
<p>Let K be the sum of the coefficients of the odd powers of $$x$$ in the expansion of $$(1+x)^{99}$$. Let $$a$$ be the middle term in the expansion of $${\left( {2 + {1 \over {\sqrt 2 }}} \right)^{200}}$$. If $${{{}^{200}{C_{99}}K} \over a} = {{{2^l}m} \over n}$$, where m and n are odd numbers, then the ordered pair $$(l,\mathrm{n})$$ is equal to</p>
|
[{"identifier": "A", "content": "(50, 101)"}, {"identifier": "B", "content": "(50, 51)"}, {"identifier": "C", "content": "(51, 101)"}, {"identifier": "D", "content": "(51, 99)"}]
|
["A"]
| null |
<p>$$K = {2^{98}}$$</p>
<p>$$a = {}^{200}{C_{100}}\,{2^{50}}$$</p>
<p>$$\therefore$$ $${{{}^{200}{C_{99}}\,.\,{2^{98}}} \over {{}^{200}{C_{100}}\,.\,{2^{50}}}} = {{{2^l}m} \over n}$$</p>
<p>$$ \Rightarrow {{100} \over {101}}\,.\,{2^{48}} = {{{2^l}m} \over n}$$</p>
<p>$$ \Rightarrow {{25} \over {101}}\,.\,{2^{50}} = {{{2^l}m} \over n}$$</p>
<p>$$\therefore$$ $$l = 50,m = 25,n = 101$$</p>
|
mcq
|
jee-main-2023-online-29th-january-evening-shift
|
mSr47psylDpWgZwl9xjgy2xukf8zzatb
|
maths
|
binomial-theorem
|
multinomial-theorem
|
Let $${\left( {2{x^2} + 3x + 4} \right)^{10}} = \sum\limits_{r = 0}^{20} {{a_r}{x^r}} $$<br/><br/>
Then $${{{a_7}} \over {{a_{13}}}}$$ is equal to ______.
|
[]
| null |
8
|
<b>Note : </b> <b>Multinomial Theorem : </b>
<br><br>The general term of $${\left( {{x_1} + {x_2} + ... + {x_n}} \right)^n}$$ the expansion is
<br><br>$${{n!} \over {{n_1}!{n_2}!...{n_n}!}}x_1^{{n_1}}x_2^{{n_2}}...x_n^{{n_n}}$$
<br><br>where n<sub>1</sub> + n<sub>2</sub> + ..... + n<sub>n</sub> = n
<br><br>Here, in $${(2{x^2} + 3x + 4)^{10}}$$ general term is <br><br>$$ = {{10!} \over {{n_1}!{n_2}!{n_3}!}}{(2{x^2})^{{n_1}}}{(3x)^{{n_2}}}{(4)^{{n_3}}}$$<br><br>$$ = {{10!} \over {{n_1}!{n_2}!{n_3}!}}{.2^{{n_1}}}{.3^{{n_2}}}{.4^{{n_3}}}.{x^{2{n_1} + {n_2}}}$$<br><br>$$ \therefore $$ Coefficient of $$ {x^{2{n_1} + {n_2}}}$$ is <br><br>$${{10!} \over {{n_1}!{n_2}!{n_3}!}}{.2^{{n_1}}}{.3^{{n_2}}}{.4^{{n_3}}}$$<br><br>where $${n_1} + {n_2} + {n_3} = 10$$<br><br> For, Coefficient of x<sup>7</sup> : <br>2n<sub>1</sub> + n<sub>2</sub> = 7<br><br>Possible values of n<sub>1</sub>, n<sub>2</sub> and n<sub>3</sub> are <br><br><table>
<thead>
<tr>
<th>$${n_1}$$</th>
<th>$${n_2}$$</th>
<th>$${n_3}$$</th>
</tr>
</thead>
<tbody>
<tr>
<td>3</td>
<td>1</td>
<td>6</td>
</tr>
<tr>
<td>2</td>
<td>3</td>
<td>5</td>
</tr>
<tr>
<td>1</td>
<td>5</td>
<td>4</td>
</tr>
<tr>
<td>0</td>
<td>7</td>
<td>3</td>
</tr>
</tbody>
</table><br><br>$$ \therefore $$ Coefficient of x<sup>7</sup><br><br>$$ = {{10!} \over {3!1!6!}}{(2)^3}{(3)^1}{(4)^6} + {{10!} \over {3!1!6!}}{(2)^2}{(3)^3}{(4)^5} + {{10!} \over {1!5!4!}}{(2)^1}{(3)^5}{(4)^4} + {{10!} \over {0!7!3!}}{(2)^0}{(3)^7}{(4)^3}$$<br><br>Coefficient of x<sup>13</sup> = a<sub>13</sub><br><br>Here 2n<sub>1</sub> + n<sub>2</sub> = 13<br><br><br>possible values of n<sub>1</sub>, n<sub>2</sub> and n<sub>3</sub> are<br><br><table>
<thead>
<tr>
<th>$${n_1}$$</th>
<th>$${n_2}$$</th>
<th>$${n_3}$$</th>
</tr>
</thead>
<tbody>
<tr>
<td>6</td>
<td>1</td>
<td>3</td>
</tr>
<tr>
<td>5</td>
<td>3</td>
<td>2</td>
</tr>
<tr>
<td>4</td>
<td>5</td>
<td>1</td>
</tr>
<tr>
<td>3</td>
<td>7</td>
<td>0</td>
</tr>
</tbody>
</table><br><br>$$ \therefore $$ Coefficient of x<sup>13</sup><br><br>$$ = {{10!} \over {6!1!3!}}{(2)^6}{(3)^1}{(4)^3} + {{10!} \over {5!3!2!}}{(2)^5}{(3)^3}{(4)^2} + {{10!} \over {4!5!1!}}{(2)^4}{(3)^5}{(4)^1} + {{10!} \over {3!7!0!}}{(2)^3}{(3)^7}{(4)^0}$$<br><br>$$ \therefore $$ $${{{a_7}} \over {{a_{13}}}} = 8$$
|
integer
|
jee-main-2020-online-4th-september-morning-slot
|
1ktkde0y0
|
maths
|
binomial-theorem
|
multinomial-theorem
|
If the coefficient of a<sup>7</sup>b<sup>8</sup> in the expansion of (a + 2b + 4ab)<sup>10</sup> is K.2<sup>16</sup>, then K is equal to _____________.
|
[]
| null |
315
|
$${{10!} \over {\alpha !\beta !\gamma !}}{a^\alpha }{(2b)^\beta }.{(4ab)^\gamma }$$<br><br>$${{10!} \over {\alpha !\beta !\gamma !}}{a^{\alpha + \gamma }}.\,{b^{\beta + \gamma }}\,.\,{2^\beta }\,.\,{4^\gamma }$$<br><br>$$\alpha + \beta + \gamma = 10$$ ..... (1)<br><br>$$\alpha + \gamma = 7$$ .... (2)<br><br>$$\beta + \gamma = 8$$ ..... (3)<br><br>$$(2) + (3) - (1) \Rightarrow \gamma = 5$$<br><br>$$\alpha = 2$$<br><br>$$\beta = 3$$<br><br>so coefficients = $${{10!} \over {2!3!5!}}{2^3}{.2^{10}}$$<br><br>$$ = {{10 \times 9 \times 8 \times 7 \times 6 \times 5} \over {2 \times 3 \times 2 \times 5!}} \times {2^{13}}$$<br><br>$$ = 315 \times {2^{16}} \Rightarrow k = 315$$
|
integer
|
jee-main-2021-online-31st-august-evening-shift
|
1l545j8gt
|
maths
|
binomial-theorem
|
multinomial-theorem
|
<p>If the constant term in the expansion of
<br/><br/>$${\left( {3{x^3} - 2{x^2} + {5 \over {{x^5}}}} \right)^{10}}$$ is 2<sup>k</sup>.l, where l is an odd integer, then the value of k is equal to:</p>
|
[{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "9"}]
|
["D"]
| null |
<b>Note : </b> <b>Multinomial Theorem : </b>
<br><br>The general term of $${\left( {{x_1} + {x_2} + ... + {x_n}} \right)^n}$$ the expansion is
<br><br>$${{n!} \over {{n_1}!{n_2}!...{n_n}!}}x_1^{{n_1}}x_2^{{n_2}}...x_n^{{n_n}}$$
<br><br>where n<sub>1</sub> + n<sub>2</sub> + ..... + n<sub>n</sub> = n
<br/><br/><p>Given,</p>
<p>$${\left( {3{x^2} - 2{x^2} + {5 \over {{x^5}}}} \right)^{10}}$$</p>
<p>$$ = {{{{(3{x^8} - 2{x^7} + 5)}^{10}}} \over {{x^{50}}}}$$</p>
<p>Now constant term in $${\left( {3{x^3} - 2{x^2} + {5 \over {{x^5}}}} \right)^{10}} = {x^{50}}$$ term in $${(3{x^8} - 2{x^7} + 5)^{10}}$$</p>
<p>General term in $${(3{x^8} - 2{x^7} + 5)^{10}}$$ is</p>
<p>$$ = {{10!} \over {{n_1}!\,{n_2}!\,{n_3}!}}{(3{x^8})^{{n_1}}}{( - 2{x^7})^{{n_2}}}{(5)^{{n_3}}}$$</p>
<p>$$ = {{10!} \over {{n_1}!\,{n_2}!\,{n_3}!}}{(3)^{{n^1}}}{( - 2)^{{n_2}}}{(5)^{{n^3}}}\,.\,{x^{8{n_1} + 7{n_2}}}$$</p>
<p>$$\therefore$$ Coefficient of $${x^{8{n_1} + 7{n_2}}}$$ is</p>
<p>$$ = {{10!} \over {{n_1}!\,{n_2}!\,{n_3}!}}{(3)^{{n_1}}}{( - 2)^{{n_2}}}{(5)^{{n_3}}}$$</p>
<p>where $${n_1} + {n_2} + {n_3} = 0$$</p>
<p>For coefficient of x<sup>50</sup> :</p>
<p>$$8{n_1} + 7{n_2} = 50$$</p>
<p>$$\therefore$$ Possible values of n<sub>1</sub>, n<sub>2</sub> and n<sub>3</sub> are</p>
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<col style="width: 93px">
</colgroup>
<thead>
<tr>
<th class="tg-baqh">n$$_1$$</th>
<th class="tg-baqh">n$$_2$$</th>
<th class="tg-baqh">n$$_3$$</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">1</td>
<td class="tg-baqh">6</td>
<td class="tg-baqh">3</td>
</tr>
</tbody>
</table></p>
<p>$$\therefore$$ Coefficient of x<sup>50</sup></p>
<p>$$ = {{10!} \over {1!\,6!\,3!}}{(3)^1}{( - 2)^6}{(5)^3}$$</p>
<p>$$ = {{10 \times 9 \times 8 \times 7} \over 6} \times 3 \times {5^3} \times {2^6}$$</p>
<p>$$ = 5 \times 3 \times 8 \times 7 \times 3 \times {5^3} \times {2^6}$$</p>
<p>$$ = 7 \times {5^4} \times {3^2} \times {2^9}$$</p>
<p>$$ = {2^k}\,.\,l$$</p>
<p>$$\therefore$$ $$l = 7 \times {5^4} \times {3^2}$$ = An odd integer</p>
<p>and $${2^k} = {2^9}$$</p>
<p>$$ \Rightarrow k = 9$$</p>
|
mcq
|
jee-main-2022-online-29th-june-morning-shift
|
lgnxjgya
|
maths
|
binomial-theorem
|
multinomial-theorem
|
Let $\left(a+b x+c x^{2}\right)^{10}=\sum\limits_{i=0}^{20} p_{i} x^{i}, a, b, c \in \mathbb{N}$.<br/><br/> If $p_{1}=20$ and $p_{2}=210$, then
$2(a+b+c)$ is equal to :
|
[{"identifier": "A", "content": "15"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "12"}]
|
["D"]
| null |
<p>We are given that $\left(a+bx+cx^2\right)^{10} = \sum_{i=0}^{20} p_i x^i$, and we are given that $p_1 = 20$ and $p_2 = 210$.<br/><br/> We need to find the value of $2(a+b+c)$.</p>
Using the multinomial theorem, we can express the expansion of $(a + bx + cx^2)^{10}$ as follows:
<br/><br/>$$
\sum\limits_{k_1+k_2+k_3=10} {{10!} \over {{k_1}!{k_2}!{k_3}!}} a^{k_1} (bx)^{k_2} (cx^2)^{k_3}
$$
<br/><br/>Now we need to find the coefficients of $x^1$ and $x^2$ in the expansion:
<br/><br/>For $x^1$ term, we have:
<br/><br/>$$
k_2 = 1, k_1 = 9, k_3 = 0
$$
<br/><br/>So,
<br/><br/>$$
p_1 = {{10!} \over {9!1!0!}} a^9 b^1 = 10a^9 b
$$
<br/><br/>For $x^2$ term, there are two possibilities:
<br/><br/>$$
k_2 = 2, k_1 = 8, k_3 = 0 \quad \text{and} \quad k_2 = 0, k_1 = 9, k_3 = 1
$$
<br/><br/>So,
<br/><br/>$$
p_2 = {{10!} \over {8!2!0!}} a^8 b^2 + {{10!} \over {9!0!1!}} a^9 c = 45a^8 b^2 + 10a^9 c
$$
<p>Now we are given $p_1 = 20$ and $p_2 = 210$. So,
$$
10a^9 b = 20 \implies a^9 b = 2
$$</p>
<p>and $$
45a^8 b^2 + 10a^9 c = 210
$$</p>
<p>Now, divide the second equation by $a^8$:
$$
45b^2 + 10ac = 210
$$</p>
<p>We know that $a^9 b = 2$. Taking the $9^{th}$ root of both sides:
$$
ab = \sqrt[9]{2}
$$</p>
<p>Now, let $k = ab = \sqrt[9]{2}$. We can rewrite the equation for $x^2$ term as:
$$
45k^2 + 10k^9 = 210
$$</p>
<p>From the equation $ab = k = \sqrt[9]{2}$, we know that $a$ and $b$ are positive integers. Thus, $k = 2$ (as both $a$ and $b$ must be factors of 2). Now we have:</p>
<p>$$
a+b = 2
$$</p>
<p>and from the equation $a^9 b = 2$, we get $a = 1, b = 2$ or vice versa. </p>
<p>Now we need to find the value of $c$. We can use the equation for the $x^2$ term again:</p>
<p>$$
45a^8 b^2 + 10a^9 c = 210
$$</p>
<p>Using $a=1$ and $b=2$, we get:</p>
<p>$$
45(1)^8 (2)^2 + 10(1)^9 c = 210 \implies 180 + 10c = 210 \implies c = 3
$$</p>
<p>So, $a=1$, $b=2$, and $c=3$. Now, we need to find the value of $2(a+b+c)$:</p>
<p>$$
2(a+b+c) = 2(1+2+3) = 2(6) = 12
$$</p>
<p>Therefore, the answer is $\boxed{12}$.</p>
|
mcq
|
jee-main-2023-online-15th-april-morning-shift
|
1lgxwe9tk
|
maths
|
binomial-theorem
|
multinomial-theorem
|
<p>The coefficient of $$x^7$$ in $${(1 - x + 2{x^3})^{10}}$$ is ___________.</p>
|
[]
| null |
960
|
Given expression is $\left(1-x+2 x^3\right)^{10}$
<br/><br/>So, general term is $\frac{10 !}{r_{1} ! r_{2} ! r_{3} !}(1)^{r_1}(-1)^{r_2} \cdot(2)^{r_3} \cdot(x)^{r_2+r_3}$
<br/><br/>Where, $r_1+r_2+r_3=10$ and $r_2+3 r_3=7$
<br/><br/>Now, for possibility,
<br/><br/>$\begin{array}{ccc}r_1 & r_2 & r_3 \\ 3 & 7 & 0 \\ 7 & 1 & 2 \\ 5 & 4 & 1\end{array}$
<br/><br/>Thus, required co-efficient
<br/><br/>$$
\begin{aligned}
& =\frac{10 !}{3 ! 7 !}(-1)^7+\frac{10 !}{5 ! 4 !}(-1)^4(2)+\frac{10 !}{7 ! 2 !}(-1)^1(2)^2 \\\\
& =-120+2520-1440 \\\\
& =2520-1560=960
\end{aligned}
$$
|
integer
|
jee-main-2023-online-10th-april-morning-shift
|
lsapwdnz
|
maths
|
binomial-theorem
|
multinomial-theorem
|
If the Coefficient of $x^{30}$ in the expansion of $\left(1+\frac{1}{x}\right)^6\left(1+x^2\right)^7\left(1-x^3\right)^8 ; x \neq 0$ is $\alpha$, then $|\alpha|$ equals ___________.
|
[]
| null |
678
|
$\begin{aligned} & \text { Coefficient of } x^{30} \text { in } \frac{(1+x)^6\left(1+x^2\right)^7\left(1-x^3\right)^8}{x^6} \\\\ & \Rightarrow \text { Coefficient of } x^{36} \text { in }(1+x)^6\left(1+x^2\right)^7\left(1-x^3\right)^8 \\\\ & \Rightarrow \text { General term }={ }^6 C_{r_1}{ }^7 C_{r_2}{ }^8 C_{r_3}(-1)^{r_3} x^{r_1+2 r_2+3 r_3} \\\\ & \Rightarrow r_1+2 r_2+3 r_3=36\end{aligned}$
<br/><br/>$$
\text { Case-I : } \begin{array}{|c|c|c|}
\hline \mathrm{r}_1 & \mathrm{r}_2 & \mathrm{r}_3 \\
\hline 0 & 6 & 8 \\
\hline 2 & 5 & 8 \\
\hline 4 & 4 & 8 \\
\hline 6 & 3 & 8 \\
\hline
\end{array}
$$
<br/><br/>$r_1+2 r_2=12 \quad\left(\right.$ Taking $\left.r_3=8\right)$
<br/><br/>$$
\begin{aligned}
&\text { Case-II :}\\\\
&\begin{array}{|l|l|l|}
\hline r_1 & r_2 & r_3 \\
\hline 1 & 7 & 7 \\
\hline 3 & 6 & 7 \\
\hline 5 & 5 & 7 \\
\hline
\end{array}
\end{aligned}
$$
<br/><br/>$r_1+2 r_2=15\left(\right.$ Taking $\left.r_3=7\right)$
<br/><br/>$$
\begin{aligned}
&\text { Case-III : }\\\\
&\begin{array}{|l|l|l|}
\hline r_1 & r_2 & r_3 \\
\hline 4 & 7 & 6 \\
\hline 6 & 6 & 6 \\
\hline
\end{array}
\end{aligned}
$$
<br/><br/>$\begin{aligned} & \text { Coefficient}=7+(15 \times 21)+(15 \times 35)+(35) \\\\ & -(6 \times 8)-(20 \times 7 \times 8)-(6 \times 21 \times 8)+(15 \times 28) \\\\ & +(7 \times 28)=-678=\alpha \\\\ & |\alpha|=678\end{aligned}$
|
integer
|
jee-main-2024-online-1st-february-morning-shift
|
jaoe38c1lse5mmmr
|
maths
|
binomial-theorem
|
multinomial-theorem
|
<p>Let $$a$$ be the sum of all coefficients in the expansion of $$\left(1-2 x+2 x^2\right)^{2023}\left(3-4 x^2+2 x^3\right)^{2024}$$ and $$b=\lim _\limits{x \rightarrow 0}\left(\frac{\int_0^x \frac{\log (1+t)}{t^{2024}+1} d t}{x^2}\right)$$. If the equation $$c x^2+d x+e=0$$ and $$2 b x^2+a x+4=0$$ have a common root, where $$c, d, e \in \mathbb{R}$$, then $$\mathrm{d}: \mathrm{c}:$$ e equals</p>
|
[{"identifier": "A", "content": "$$2: 1: 4$$\n"}, {"identifier": "B", "content": "$$1: 1: 4$$\n"}, {"identifier": "C", "content": "$$1: 2: 4$$\n"}, {"identifier": "D", "content": "$$4: 1: 4$$"}]
|
["B"]
| null |
<p>Put $$x=1$$</p>
<p>$$\therefore \mathrm{a}=1$$</p>
<p>$$\mathrm{b}=\lim _\limits{\mathrm{x} \rightarrow 0} \frac{\int_\limits0^{\mathrm{x}} \frac{\ln (1+\mathrm{t})}{1+\mathrm{t}^{2024}} \mathrm{dt}}{\mathrm{x}^2}$$</p>
<p>Using L' HOPITAL Rule</p>
<p>$$\mathrm{b}=\lim _\limits{\mathrm{x} \rightarrow 0} \frac{\ln (1+\mathrm{x})}{\left(1+\mathrm{x}^{2024}\right)} \times \frac{1}{2 \mathrm{x}}=\frac{1}{2}$$</p>
<p>Now, $$\mathrm{cx}^2+\mathrm{dx}+\mathrm{e}=0, \mathrm{x}^2+\mathrm{x}+4=0$$</p>
<p>$$(\mathrm{D}<0)$$</p>
<p>$$\therefore \frac{\mathrm{c}}{1}=\frac{\mathrm{d}}{1}=\frac{\mathrm{e}}{4}$$</p>
|
mcq
|
jee-main-2024-online-31st-january-morning-shift
|
5xKBLBPPtldLjMPl
|
maths
|
binomial-theorem
|
negative-and-fractional-index
|
If $$x$$ is positive, the first negative term in the expansion of $${\left( {1 + x} \right)^{27/5}}$$ is
|
[{"identifier": "A", "content": "6th term "}, {"identifier": "B", "content": "7th term "}, {"identifier": "C", "content": "5th term "}, {"identifier": "D", "content": "8th term."}]
|
["D"]
| null |
General term of $${\left( {1 + x} \right)^{n}}$$ is ($${T_{r + 1}}$$) = $${{n\left( {n - 1} \right).....\left( {n - r + 1} \right)} \over {1.2.3....r}}{x^r}$$
<br><br>$$\therefore$$ General term of $${\left( {1 + x} \right)^{27/5}}$$ = $${{{{27} \over 5}\left( {{{27} \over 5} - 1} \right).....\left( {{{27} \over 5} - r + 1} \right)} \over {1.2.3....r}}{x^r}$$
<br><br>For first negative term, $${\left( {{{27} \over 5} - r + 1} \right)}$$ < 0
<br><br>$$ \Rightarrow r > {{27} \over 5} + 1$$
<br><br>$$ \Rightarrow r > {{32} \over 5}$$
<br><br>$$ \Rightarrow r > 6.4$$
<br><br>$$\therefore$$ r = 7
<br><br>$${T_{7 + 1}} = {T_8}$$ means 8<sup>th</sup> term is the first negative term.
|
mcq
|
aieee-2003
|
S34Ufc1rYv3mtMqj
|
maths
|
binomial-theorem
|
negative-and-fractional-index
|
If $$x$$ is so small that $${x^3}$$ and higher powers of $$x$$ may be neglected, then $${{{{\left( {1 + x} \right)}^{{3 \over 2}}} - {{\left( {1 + {1 \over 2}x} \right)}^3}} \over {{{\left( {1 - x} \right)}^{{1 \over 2}}}}}$$ may be approximated as
|
[{"identifier": "A", "content": "$$1 - {3 \\over 8}{x^2}$$ "}, {"identifier": "B", "content": "$$3x + {3 \\over 8}{x^2}$$ "}, {"identifier": "C", "content": "$$ - {3 \\over 8}{x^2}$$ "}, {"identifier": "D", "content": "$${x \\over 2} - {3 \\over 8}{x^2}$$ "}]
|
["C"]
| null |
$${\left( {1 + x} \right)^{{3 \over 2}}}$$ = 1 + $${3 \over 2}x + {{{3 \over 2}.{1 \over 2}} \over {1.2}}{x^2} + ...$$
<br><br>= 1 + $${3 \over 2}x + {3 \over 8}{x^2}$$ (As $$x$$ is so small, so $${x^3}$$ and higher powers of $$x$$ neglected)
<br><br>$${{{{\left( {1 + x} \right)}^{{3 \over 2}}} - {{\left( {1 + {1 \over 2}x} \right)}^3}} \over {{{\left( {1 - x} \right)}^{{1 \over 2}}}}}$$
<br><br>= $${{\left( {1 + {3 \over 2}x + {3 \over 8}{x^2}} \right) - \left( {1 + {}^3{C_0}.{x \over 2} + {}^3{C_1}.{{\left( {{x \over 2}} \right)}^2}} \right)} \over {{{\left( {1 - x} \right)}^2}}}$$
<br><br>= $${{{3 \over 8}{x^2} - {3 \over 4}{x^2}} \over {{{\left( {1 - x} \right)}^2}}}$$
<br><br>= $${{x^2}\left( {{3 \over 8} - {3 \over 4}} \right){{\left( {1 - x} \right)}^{ - 2}}}$$
<br><br>= $${ - {3 \over 8}{x^2}\left( {1 - {1 \over 2}\left( { - x} \right) + ....} \right)}$$
<br><br>= $$ - {3 \over 8}{x^2} - {3 \over {16}}{x^3}$$
<br><br>[ As x<sup>3</sup> is so small we can ignore $$-{3 \over {16}}{x^3}$$]
<br><br>= $$ - {3 \over 8}{x^2}$$
|
mcq
|
aieee-2005
|
hI198LyRc6DYt2by
|
maths
|
binomial-theorem
|
negative-and-fractional-index
|
If the expansion in powers of $$x$$ of the function $${1 \over {\left( {1 - ax} \right)\left( {1 - bx} \right)}}$$ is $${a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3}.....$$ then $${a_n}$$ is
|
[{"identifier": "A", "content": "$${{{b^n} - {a^n}} \\over {b - a}}$$ "}, {"identifier": "B", "content": "$${{{a^n} - {b^n}} \\over {b - a}}$$ "}, {"identifier": "C", "content": "$${{{a^{n + 1}} - {b^{n + 1}}} \\over {b - a}}$$ "}, {"identifier": "D", "content": "$${{{b^{n + 1}} - {a^{n + 1}}} \\over {b - a}}$$ "}]
|
["D"]
| null |
$${1 \over {\left( {1 - ax} \right)\left( {1 - bx} \right)}}$$
<br><br>= $${\left( {1 - ax} \right)^{ - 1}}{\left( {1 - bx} \right)^{ - 1}}$$
<br><br>= $$\left[ {1 + \left( { - 1} \right)\left( { - ax} \right) + {{\left( { - 1} \right)\left( { - 2} \right)} \over {1.2}}{{\left( { - ax} \right)}^2} + ...} \right]$$ -
<br> $$\left[ {1 + \left( { - 1} \right)\left( { - bx} \right) + {{\left( { - 1} \right)\left( { - 2} \right)} \over {1.2}}{{\left( { - bx} \right)}^2} + ...} \right]$$
<br><br>= $$\left[ {1 + ax + {a^2}{x^2} + ... + {a^{n - 1}}{x^{n - 1}} + {a^n}{x^n}}+.... \right]$$ -
<br> $$\left[ {1 + bx + {b^2}{x^2} + ... + {b^{n - 1}}{x^{n - 1}} + {b^n}{x^n}}+.... \right]$$
<br><br>Coefficient of x<sup>n</sup> =
<br><br>$${a^n} + {a^{n - 1}}b + {a^{n - 2}}{b^2} + .... + {b^n}$$
<br><br>= $${a^n}\left[ {1 + {b \over a} + {{{b^2}} \over {{a^2}}} + ..... + {{{b^n}} \over {{a^n}}}} \right]$$
<br><br>= $${a^n}\left[ {{{{{\left( {{b \over a}} \right)}^{n + 1}} - 1} \over {{b \over a} - 1}}} \right]$$
<br><br>= $${a^n}\left[ {{{{b^{n + 1}} - {a^{n + 1}}} \over {{a^{n + 1}}\left( {{{b - a} \over a}} \right)}}} \right]$$
<br><br>= $${{{{b^{n + 1}} - {a^{n + 1}}} \over {b - a}}}$$
|
mcq
|
aieee-2006
|
1krvyf364
|
maths
|
binomial-theorem
|
negative-and-fractional-index
|
If b is very small as compared to the value of a, so that the cube and other higher powers of $${b \over a}$$ can be neglected in the identity $${1 \over {a - b}} + {1 \over {a - 2b}} + {1 \over {a - 3b}} + ..... + {1 \over {a - nb}} = \alpha n + \beta {n^2} + \gamma {n^3}$$, then the value of $$\gamma$$ is :
|
[{"identifier": "A", "content": "$${{{a^2} + b} \\over {3{a^3}}}$$"}, {"identifier": "B", "content": "$${{a + b} \\over {3{a^2}}}$$"}, {"identifier": "C", "content": "$${{{b^2}} \\over {3{a^3}}}$$"}, {"identifier": "D", "content": "$${{a + {b^2}} \\over {3{a^3}}}$$"}]
|
["C"]
| null |
$${(a - b)^{ - 1}} + {(a - 2b)^{ - 1}} + .... + {(a - nb)^{ - 1}}$$<br><br>$$ = {1 \over a}\sum\limits_{r = 1}^n {{{\left( {1 - {{rb} \over a}} \right)}^{ - 1}}} $$<br><br>$$ = {1 \over a}\sum\limits_{r = 1}^n {\left\{ {\left( {1 + {{rb} \over a} + {{{r^2}{b^2}} \over {{a^2}}}} \right) + (terms\,to\,be\,neglected)} \right\}} $$<br><br>$$ = {1 \over a}\left[ {n + {{n(n + 1)} \over 2}.{b \over a} + {{n(n + 1)(2n + 1)} \over 6}.{{{b^2}} \over {{a^2}}}} \right]$$<br><br>$$ = {1 \over a}\left[ {{n^3}\left( {{{{b^2}} \over {3{a^2}}}} \right) + .....} \right]$$<br><br>So, $$\gamma = {{{b^2}} \over {3{a^3}}}$$
|
mcq
|
jee-main-2021-online-25th-july-morning-shift
|
LV5BeeNJAp7oiwUq
|
maths
|
binomial-theorem
|
problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient
|
If the sum of the coefficients in the expansion of $$\,{\left( {a + b} \right)^n}$$ is 4096, then the greatest coefficient in the expansion is
|
[{"identifier": "A", "content": "1594 "}, {"identifier": "B", "content": "792 "}, {"identifier": "C", "content": "924 "}, {"identifier": "D", "content": "2924"}]
|
["C"]
| null |
We know, $$\,{\left( {a + b} \right)^n}$$ = $${}^n{C_0}.{a^n} + {}^n{C_1}.{a^{n - 1}}.b + ... + {}^n{C_n}.{b^n}$$
<br><br>Remember to find sum of coefficient of binomial expansion we ave to put 1 in place of all the variable.
<br><br>So put $$a$$ = b = 1
<br><br>$$\therefore$$ 2<sup>n</sup> = $${}^n{C_0} + {}^n{C_1} + {}^n{C_2}... + {}^n{C_n}$$
<br><br>According to question, 2<sup>n</sup> = 4096 = 2<sup>12</sup>
<br><br>$$ \Rightarrow n = 12$$
<br><br>So $$\,{\left( {a + b} \right)^n}$$ = $$\,{\left( {a + b} \right)^{12}}$$
<br><br>Here n = 12 is even so formula for greatest term is
<br>$${T_{{n \over 2} + 1}} = {}^n{C_{{n \over 2}}}.{a^{{n \over 2}}}.{b^{{n \over 2}}}$$
<br><br>For n = 12, greatest term $${T_{6 + 1}} = {}^{12}{C_6}.{a^6}.{b^6}$$
<br><br>$$\therefore$$ Coefficient of the greatest term = $${}^{12}{C_6}$$ = $${{12!} \over {6!6!}}$$ = 924
|
mcq
|
aieee-2002
|
Dv471cd1hiN4I75F
|
maths
|
binomial-theorem
|
problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient
|
$$r$$ and $$n$$ are positive integers $$\,r > 1,\,n > 2$$ and coefficient of $$\,{\left( {r + 2} \right)^{th}}$$ term and $$3{r^{th}}$$ term in the expansion of $${\left( {1 + x} \right)^{2n}}$$ are equal, then $$n$$ equals
|
[{"identifier": "A", "content": "$$3r$$"}, {"identifier": "B", "content": "$$3r + 1$$ "}, {"identifier": "C", "content": "$$2r$$ "}, {"identifier": "D", "content": "$$2r + 1$$"}]
|
["C"]
| null |
$$\,{\left( {r + 2} \right)^{th}}$$ term = $${}^{2n}{C_{r+1}}{\left( x \right)^r}$$
<br><br>And coefficient of $$\,{\left( {r + 2} \right)^{th}}$$ = $${}^{2n}{C_{r+1}}$$
<br><br>$$3{r^{th}}$$ term = $${}^{2n}{C_{3r - 1}}{\left( x \right)^{3r - 1}}$$
<br><br>And coefficient of $$3{r^{th}}$$ term = $${}^{2n}{C_{3r - 1}}$$
<br><br>According to the question,
<br>$${}^{2n}{C_{r+1}}$$ = $${}^{2n}{C_{3r - 1}}$$
<br><br>$$ \Rightarrow \left( {r + 1} \right) + \left( {3r - 1} \right) = 2n$$
<br><br>[As if $${}^n{C_p} = {}^n{C_q}$$ then p + q = n]
<br><br>$$ \Rightarrow 4r = 2n$$
<br><br>$$ \Rightarrow n = 2r$$
|
mcq
|
aieee-2002
|
9o9lRt7vfpS59rnX
|
maths
|
binomial-theorem
|
problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient
|
If $${S_n} = \sum\limits_{r = 0}^n {{1 \over {{}^n{C_r}}}} \,\,and\,\,{t_n} = \sum\limits_{r = 0}^n {{r \over {{}^n{C_r}}},\,} $$then $${{{t_{ n}}} \over {{S_n}}}$$ is equal to
|
[{"identifier": "A", "content": "$${{2n - 1} \\over 2}$$ "}, {"identifier": "B", "content": "$${1 \\over 2}n - 1$$ "}, {"identifier": "C", "content": "n - 1"}, {"identifier": "D", "content": "$${1 \\over 2}n$$ "}]
|
["D"]
| null |
$${S_n} = \sum\limits_{r = 0}^n {{1 \over {{}^n{C_r}}}}$$
<br><br>=$${1 \over {{}^n{C_0}}} + {1 \over {{}^n{C_1}}} + .... + {1 \over {{}^n{C_n}}}$$
<br><br>$${t_n} = \sum\limits_{r = 0}^n {{r \over {{}^n{C_r}}}}$$
<br><br>= $${0 \over {{}^n{C_0}}} + {1 \over {{}^n{C_1}}} + {2 \over {{}^n{C_2}}}.... + {n \over {{}^n{C_n}}}$$.........(1)
<br><br>We can write $${t_n}$$ by rearranging like this,
<br><br>$${t_n}$$ = $${n \over {{}^n{C_n}}} + {{n - 1} \over {{}^n{C_{n - 1}}}} + ... + {1 \over {{}^n{C_1}}} + {0 \over {{}^n{C_0}}}$$
<br><br>= $${n \over {{}^n{C_0}}} + {{n - 1} \over {{}^n{C_1}}} + ... + {1 \over {{}^n{C_{n - 1}}}} + {0 \over {{}^n{C_n}}}$$.........(2)
<br><br>[as $${{}^n{C_0}}$$ = $${{}^n{C_n}}$$, $${{}^n{C_1}}$$ = $${{}^n{C_{n - 1}}}$$......]
<br><br>By adding (1) and (2) we get,
<br><br>$$2{t_n}$$ = $${n \over {{}^n{C_0}}} + {n \over {{}^n{C_1}}} + ... + {n \over {{}^n{C_{n - 1}}}} + {n \over {{}^n{C_n}}}$$
<br><br>= $$n\left[ {{1 \over {{}^n{C_0}}} + {1 \over {{}^n{C_1}}} + ... + {1 \over {{}^n{C_{n - 1}}}} + {1 \over {{}^n{C_n}}}} \right]$$
<br><br>= n$${S_n}$$
<br><br>$$\therefore$$ $${{{t_n}} \over {{S_n}}} = {n \over 2}$$
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mcq
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aieee-2004
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