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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": "$$\\left( {n - 1} ... | ["B"] | null | Given $$\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}$$ ... | mcq | aieee-2004 |
rTXLqH9tIJS0LPKw | maths | binomial-theorem | general-term | If the coefficient of $${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}... | 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> = $${}^m... | mcq | aieee-2005 |
nV7NCqS5hYaVuBUA | maths | binomial-theorem | general-term | In the binomial expansion of $${\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}}}}} \... | 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 ... | 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> $${\left( {{1 \over 2}} \right)^r}\,\,.\,\,{x^{{{18 -... | mcq | jee-main-2016-online-10th-april-morning-slot |
MegnCy5y6WH4JxsjLHqWr | maths | binomial-theorem | general-term | The coefficient of x<sup>−5</sup> in the binomial expansion of
<br/><br/>$${\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}}}}} \... | 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 irration... | 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 +... | 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</su... | 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>
$$ \Rightar... | mcq | jee-main-2019-online-10th-april-evening-slot |
3TQ0nlDrlHmOpomuWHpzD | maths | binomial-theorem | general-term | If the fourth term in the binomial expansion of $${\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 $$$${8 \over {{x^3}}} \times {x^{3{{\log }_... | mcq | jee-main-2019-online-9th-april-morning-slot |
en2LblqYlH2raE9M7kcBX | maths | binomial-theorem | general-term | If the fourth term in the binomial expansion of<br/>
$${\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)$$
<br><... | mcq | jee-main-2019-online-8th-april-evening-slot |
8ItKCdunQp0EBnpdGh1Tw | maths | binomial-theorem | general-term | A ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of $${\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({ }... | 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>$$ \Rightarrow 2... | mcq | jee-main-2019-online-10th-january-morning-slot |
n2K8x8uBYrxmQfciha7k9k2k5itwjym | maths | binomial-theorem | general-term | The coefficient of x<sup>4</sup> is the expansion of
(1 + x + x<sup>2</sup>)<sup>10</sup> is _____. | [] | null | 615 | (1 + x + x<sup>2</sup>)<sup>10</sup>
<br><br>= <sup>10</sup>C<sub>0</sub>(1 + x)<sup>10</sup> + <sup>10</sup>C<sub>1</sub>(1 + x)<sup>9</sup>.x<sup>2</sup> + <sup>10</sup>C<sub>2</sub>(1 + x)<sup>8</sup>.x<sup>4</sup>+ .....
<br><br>Coefficient of x<sup>4</sup>
<br><br>= <sup>10</sup>C<sub>0</sub>.<sup>10</sup>C<sub>4<... | integer | jee-main-2020-online-9th-january-morning-slot |
mfa9gcoLNQtl4I0BUejgy2xukg0d0dyo | maths | binomial-theorem | general-term | If the constant term in the binomial expansion
of <br/>$${\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} \r... | 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>$$m = {{2r + 1} \over {22 - 5}}$$<b... | 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 $$ \Rightarrow... | mcq | jee-main-2020-online-3rd-september-evening-slot |
dPu5jf9EUPJ9Gi2nwLjgy2xukf0wsoo1 | maths | binomial-theorem | general-term | If the number of integral terms in the expansion
<br/>of (3<sup>1/2</sup> + 5<sup>1/8</sup>)<sup>n</sup> is exactly 33, then the least value
of n is : | [{"identifier": "A", "content": "264"}, {"identifier": "B", "content": "256"}, {"identifier": "C", "content": "128"}, {"identifier": "D", "content": "248"}] | ["B"] | null | General term of the expression,<br><br>$${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}... | 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 + ... | 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 $${\pi \over {16}} \le \th... | [{"identifier": "A", "content": "8 : 1"}, {"identifier": "B", "content": "16 : 1"}, {"identifier": "C", "content": "1 : 8"}, {"identifier": "D", "content": "1 : 16"}] | ["B"] | null | T<sub>r + 1</sub> = <sup>16</sup>C<sub>r</sub>$${\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}}}$$
<br><... | mcq | jee-main-2020-online-9th-january-evening-slot |
whKykbRukRrJZDlsAZ7k9k2k5hjr4v7 | maths | binomial-theorem | general-term | If $$\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}$$
<br><br>= 2[T<sub>1</sub>
+ T<sub>3</sub>
+ T<sub>5</sub>
+ T<sub>7</sub>
]
<br><br>= 2[<sup>6</sup>C<sub>0</sub>... | mcq | jee-main-2020-online-8th-january-evening-slot |
Qc54t3Qr25N0Yctsjd7k9k2k5fiux85 | maths | binomial-theorem | general-term | The coefficient of x<sup>7</sup>
in the expression
<br/>(1 + x)<sup>10</sup> + x(1 + x)<sup>9</sup>
+ x<sup>2</sup>(1 + x)<sup>8</sup>
+ ......+ x<sup>10</sup> is: | [{"identifier": "A", "content": "120"}, {"identifier": "B", "content": "330"}, {"identifier": "C", "content": "420"}, {"identifier": "D", "content": "210"}] | ["B"] | null | (1 + x)<sup>10</sup> + x(1 + x)<sup>9</sup>
+ x<sup>2</sup>(1 + x)<sup>8</sup>
+ ......+ x<sup>10</sup>
<br><br>This is a G.P where
<br><br>First term, a = (1 + x)<sup>10</sup>
<br><br>common ratio, r = $${x \over {1 + x}}$$
<br><br>Number of terms = 11
<br><br>Sum of G.P
<br><br>= $${{{{\left( {1 + x} \right)}^{10}}... | 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>$$ \therefore $$ $${T_6} = {}^{10}{C_5} \time... | mcq | jee-main-2021-online-26th-february-morning-slot |
jHrylvcW7IqT8Wci3y1kmhx5zo4 | maths | binomial-theorem | general-term | If n is the number of irrational terms in the <br/>expansion of $${\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 \Rightar... | 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 $${\log _2... | mcq | jee-main-2021-online-17th-march-morning-shift |
3EvGB4JDbb7CRrEJIt1kmknn38u | maths | binomial-theorem | general-term | Let the coefficients of third, fourth and fifth terms in the expansion of $${\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}}} = {{{... | 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>= $${\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \... | integer | jee-main-2021-online-18th-march-evening-shift |
1krq0nwdb | maths | binomial-theorem | general-term | The number of rational terms in the binomial expansion of $${\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} \... | 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>= $${\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \... | integer | jee-main-2021-online-25th-july-morning-shift |
1krxgk2ja | maths | binomial-theorem | general-term | A possible value of 'x', for which the ninth term in the expansion of $${\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)}}$$ is equal ... | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$$-$$1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}] | ["D"] | null | $${}^{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$$ x $$-$$ 1 = 0 (one of the po... | mcq | jee-main-2021-online-27th-july-evening-shift |
1krz59pdx | maths | binomial-theorem | general-term | The sum of all those terms which are rational numbers in the <br/><br/>expansion of (2<sup>1/3</sup> + 3<sup>1/4</sup>)<sup>12</sup> is : | [{"identifier": "A", "content": "89"}, {"identifier": "B", "content": "27"}, {"identifier": "C", "content": "35"}, {"identifier": "D", "content": "43"}] | ["D"] | null | $${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$$ 2<sup>4</su... | mcq | jee-main-2021-online-25th-july-evening-shift |
1krzlrpjk | maths | binomial-theorem | general-term | If the greatest value of the term independent of 'x' in the <br/><br/>expansion of $${\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}} ... | 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 = $${}^{1... | mcq | jee-main-2021-online-27th-july-morning-shift |
1ktisyc5x | maths | binomial-theorem | general-term | If $$\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)^{1... | 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 __________.... | [] | 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>$$ = {}^{15}{C_r... | integer | jee-main-2022-online-29th-june-evening-shift |
1l55h6lp9 | maths | binomial-theorem | general-term | <p>The term independent of x in the expansion of <br/><br/>$$(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... | 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... | 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>$${T_{r + 1}} = {}^{60}{C_r}\,.\,{\left( {{{{x^{1/2}}} \over {{5^{1/4}}}}} \right)^{60 - r}}\,.\,{\left( {{{{5^{1/2}}} \over {{x^{1/... | integer | jee-main-2022-online-27th-june-morning-shift |
1l59jwnfk | maths | binomial-theorem | general-term | <p>The coefficient of x<sup>101</sup> in the expression $${(5 + x)^{500}} + x{(5 + x)^{499}} + {x^2}{(5 + x)^{498}} + \,\,.....\,\, + \,\,{x^{500}}$$, x > 0, is</p> | [{"identifier": "A", "content": "<sup>501</sup>C<sub>101</sub> (5)<sup>399</sup>"}, {"identifier": "B", "content": "<sup>501</sup>C<sub>101</sub> (5)<sup>400</sup>"}, {"identifier": "C", "content": "<sup>501</sup>C<sub>100</sub> (5)<sup>400</sup>"}, {"identifier": "D", "content": "<sup>500</sup>C<sub>101</sub> (5)<sup>... | ["A"] | null | <p>Given,</p>
<p>$${(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$$ Sum $$ = {{{{(... | 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>$$ = {}^{10}{C_r}\,.\,... | integer | jee-main-2022-online-25th-june-evening-shift |
1l5vzfdaf | maths | binomial-theorem | general-term | <p>For two positive real numbers a and b such that $${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^... | 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>$$ = {}^{15}{C_r}\,.\,{t^{30 - 2r}}\,.... | integer | jee-main-2022-online-25th-july-morning-shift |
1l6klfrzl | maths | binomial-theorem | general-term | <p>Let for the $$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^... | [] | 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>$$ \Rightarrow {{n!} \over {(n - 7)!7!}}9 \le {{n!} \over {... | integer | jee-main-2022-online-27th-july-evening-shift |
1l6p3efud | maths | binomial-theorem | general-term | <p>Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of $$\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]... | [] | 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>G... | 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 ... | [] | null | 4 | ${ }^m C_1,{ }^m C_2,{ }^m C_3$ are first, third and fifth term of $A P$
<br/><br/>$$
\begin{aligned}
\therefore \quad & a={ }^m C_1 \\\\
& a+2 d={ }^m C_2 \\\\
& a+4 d={ }^m C_3 \\\\
\therefore \quad & 2{ }^m C_2-{ }^m C_3=m \\\\
\Rightarrow & m=7 \text { or } m=2 \\\\
\because & m=2 \text { is not possible } \\\\
\th... | integer | jee-main-2023-online-1st-february-evening-shift |
1ldo7goz9 | maths | binomial-theorem | general-term | <p>If the term without $$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}
$$
<br/><br/>Now ${ }^{22... | integer | jee-main-2023-online-1st-february-evening-shift |
ldoaj02i | maths | binomial-theorem | general-term | The coefficient of $x^{-6}$, in the
<br/><br/>expansion of $\left(\frac{4 x}{5}+\frac{5}{2 x^{2}}\right)^{9}$, is | [] | null | 5040 | Coeff of $x^{-6}$ in $\left(\frac{4 x}{5}+\frac{5}{2 x^{2}}\right)^{9}$
<br/><br/>$$
\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}
$$
<br/><br/>Coeff of $x^{-6}={ }^{9} C_{5}\left(\frac{4}{5}\right)^{4}\left(\frac{... | integer | jee-main-2023-online-31st-january-evening-shift |
ldoavd66 | maths | binomial-theorem | general-term | If the constant term in the binomial expansion of $\left(\frac{x^{\frac{5}{2}}}{2}-\frac{4}{x^{l}}\right)^{9}$ is $-84$ and the coefficient of $x^{-3 l}$ is
$2^{\alpha} \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to ________. | [] | null | 98 | Given binomial expansion of $\left(\frac{x^{\frac{5}{2}}}{2}-\frac{4}{x^{l}}\right)^{9}$
<br/><br/>$$
\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}
$... | 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/... | 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 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>Simila... | 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/>
$$
\begin{aligned}
& \Rightarrow \frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}(2)^{\mathrm{r}-1}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}(2)^{\mathrm{r}}}=\frac{2}{5} \\\\
& \Rightarrow \frac{\frac{n !}{(r-1) ... | integer | jee-main-2023-online-29th-january-morning-shift |
1ldswsczz | maths | binomial-theorem | general-term | <p>If the co-efficient of $$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$$ is equal to ___________.</p> | [] | null | 1 | Coefficient of $\mathrm{x}^{9}$ in $\left(\alpha x^{3}+\frac{1}{\beta x}\right)={ }^{11} C_{6} \cdot \frac{\alpha^{5}}{\beta^{6}}$
<br/><br/>
$\because$ Both are equal
<br/><br/>
$\therefore \frac{11}{C_{6}} \cdot \frac{\alpha^{5}}{\beta^{6}}=-\frac{11}{C_{5}} \cdot \frac{\alpha^{6}}{\beta^{5}}$
<br/><br/>
$\Rightarrow... | integer | jee-main-2023-online-29th-january-morning-shift |
1ldv2styk | maths | binomial-theorem | general-term | <p>The constant term in the expansion of $${\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}
$$
<br/><br/>
Term independent of $x=$ coefficient of $x^{35}$ in
<br/><br/>
$$
\b... | 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/>
H... | 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/>$$
\begin{aligned}
& T_{r+1}={ }^5 C_r\left(2 x^3\right)^{5-r}\left(\frac{-1}{3 x^2}\right)^r={ }^5 C_r \frac{(2)^{5-r}}{(-3)^r}(x)^{15-5 r} \\\\
& \therefore 15-5 \mathrm{r}=5 \\\\
& \therefore \mathrm{r}=2 \\\\
& T_3=10\left(\frac... | mcq | jee-main-2023-online-13th-april-evening-shift |
1lgq11k4d | maths | binomial-theorem | general-term | <p>Let $$\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$$ is equal to _____________.</p> | [] | null | 36 | Given expression $(\sqrt{x}-\frac{6}{x^{3/2}})^n$. Here, $a = \sqrt{x}$ and $b = -\frac{6}{x^{3/2}}$.
<br/><br/>The $r$-th term of the binomial expansion of $(a+b)^n$ is given by
<br/><br/>$T_{r} = {}^n{C_r}a^{n-r}b^{r}$.
<br/><br/>Substitute $a$ and $b$ in this formula, we get:
<br/><br/>$T_{r} = {}^n{C_r}(\sqrt{... | integer | jee-main-2023-online-13th-april-morning-shift |
1lgsubacw | maths | binomial-theorem | general-term | <p>The sum of the coefficients of three consecutive terms in the binomial expansion of $$(1+\mathrm{x})^{\mathrm{n}+2}$$, which are in the ratio $$1: 3: 5$$, is equal to :</p> | [{"identifier": "A", "content": "63"}, {"identifier": "B", "content": "92"}, {"identifier": "C", "content": "25"}, {"identifier": "D", "content": "41"}] | ["A"] | null | The problem asks for the sum of the coefficients of three consecutive terms in the binomial expansion of $(1+x)^{n+2}$, which are in the ratio 1 : 3 : 5.
<br/><br/>Given that the ratios of the coefficients are 1:3:5, we let the terms be $T_r$, $T_{r+1}$, and $T_{r+2}$. The coefficients of these terms are ${ }^{n+2} C... | mcq | jee-main-2023-online-11th-april-evening-shift |
1lgsvbm4o | maths | binomial-theorem | general-term | <p>If the $$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(\... | 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/> . ... | 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 ... | 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/>$$
\begin{aligned}
& (1+x)^p(1-x)^q \\\\
& =\left(1+p x+\frac{p(p-1)}{2} x^2+\ldots\right)\left(1-q x+\frac{q(q-1)}{2} x^2+\ldots\right) \\\\
& =1+(p-q) x+\left(\frac{p(p-1)}{2}+\frac{q(q-1)}{2}-p q\ri... | mcq | jee-main-2023-online-10th-april-evening-shift |
1lgxt15ef | maths | binomial-theorem | general-term | <p>If the coefficient of $${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}
$$
<br/><... | mcq | jee-main-2023-online-10th-april-morning-shift |
1lgyliytd | maths | binomial-theorem | general-term | <p>The absolute difference of the coefficients of $$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{align... | 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/>$$
\begin{aligned}
& \text { i.e., } 14-2 r-5 r=0 \\\\
& \Rightarrow 14=7 r \Rightarrow r=2
\end{ali... | integer | jee-main-2023-online-8th-april-morning-shift |
1lh23fm6b | maths | binomial-theorem | general-term | <p>If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $$\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}
$$
<br/><br/>$\left[\because r\right.$th term from e... | mcq | jee-main-2023-online-6th-april-morning-shift |
1lh2xsi9y | maths | binomial-theorem | general-term | <p>If the coefficient of $${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... | 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>$$\begin{aligned}
& (1-x)\left((-1)^r{ }^{2007} C_r x^{3 r}\right) \\
& (-1)^{r 2007} C_... | integer | jee-main-2024-online-27th-january-evening-shift |
lv3vef7y | maths | binomial-theorem | general-term | <p>If the term independent of $$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}\rig... | 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>$$\begin{aligned}
& \left(\frac{3}... | integer | jee-main-2024-online-5th-april-morning-shift |
lv9s1zz5 | maths | binomial-theorem | general-term | <p>If the constant term in the expansion of $$\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 \t... | 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>$$\begin{aligned}
\Rightarrow \frac{135}{30} & =\left(\frac{x}{y}\right) \frac{n \cdot 2}{n(n-1)}=\left(\frac{2}{n-1}\right)\left(\frac{x}{y}\right) \quad... | integer | jee-main-2024-online-6th-april-morning-shift |
MdI6myzXplT0kKOP | maths | binomial-theorem | integral-and-fractional-part-of-a-number | If $$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}$$
<br><br>= (A + B) - (A - B)
<br><br>= 2B
<br><br>= 2[even terms]
<br><br>= 2[ T<sub>2</sub> + T<s... | mcq | aieee-2012 |
hSlHFjRByfCzOzj87g1j0 | maths | binomial-theorem | integral-and-fractional-part-of-a-number | If the fractional part of the number $$\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} + ........ | 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>= $$\left\{ {{{1 + {}^{100}{C_1}.8 + {}^{100}{C_2}{{.8}^2} + .... + {}^{100}{C_{100}}{{.8}^{100}}} \over 8}} \r... | mcq | jee-main-2020-online-6th-september-morning-slot |
ldqy761e | maths | binomial-theorem | integral-and-fractional-part-of-a-number | Let $x=(8 \sqrt{3}+13)^{13}$ and $y=(7 \sqrt{2}+9)^9$. If $[t]$ denotes the greatest integer $\leq t$, then : | [{"identifier": "A", "content": "$[x]$ is odd but $[y]$ is even"}, {"identifier": "B", "content": "$[x]$ and $[y]$ are both odd"}, {"identifier": "C", "content": "$[x]+[y]$ is even"}, {"identifier": "D", "content": "$[x]$ is even but $[y]$ is odd"}] | ["C"] | null | <p>If $${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}$$... | mcq | aieee-2004 |
AcHYxPg1DlPdQGa51uedV | maths | binomial-theorem | middle-term | The sum of the real values of x for which the middle term in the binomial expansion of $${\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}}$$ is... | [] | null | 57 | <p>Coefficients of middle terms of given expansions are $${}^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}$$</p>... | integer | jee-main-2022-online-28th-july-evening-shift |
1ldsfiyxc | maths | binomial-theorem | middle-term | <p>Let K be the sum of the coefficients of the odd powers of $$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}$$, where m and n are odd numbers, then the ordered pair $... | [{"identifier": "A", "content": "(50, 101)"}, {"identifier": "B", "content": "(50, 51)"}, {"identifier": "C", "content": "(51, 101)"}, {"identifier": "D", "content": "(51, 99)"}] | ["A"] | null | <p>$$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}} = {{{... | 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 $$... | integer | jee-main-2020-online-4th-september-morning-slot |
1ktkde0y0 | maths | binomial-theorem | multinomial-theorem | If the coefficient of a<sup>7</sup>b<sup>8</sup> in the expansion of (a + 2b + 4ab)<sup>10</sup> is K.2<sup>16</sup>, then K is equal to _____________. | [] | null | 315 | $${{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$$ .... (2)<br><br>$$\b... | 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,... | 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/>$$
\sum\limits_{k_1+k_2+k_3=... | mcq | jee-main-2023-online-15th-april-morning-shift |
1lgxwe9tk | maths | binomial-theorem | multinomial-theorem | <p>The coefficient of $$x^7$$ in $${(1 - x + 2{x^3})^{10}}$$ is ___________.</p> | [] | null | 960 | Given expression is $\left(1-x+2 x^3\right)^{10}$
<br/><br/>So, general term is $\frac{10 !}{r_{1} ! r_{2} ! r_{3} !}(1)^{r_1}(-1)^{r_2} \cdot(2)^{r_3} \cdot(x)^{r_2+r_3}$
<br/><br/>Where, $r_1+r_2+r_3=10$ and $r_2+3 r_3=7$
<br/><br/>Now, for possibility,
<br/><br/>$\begin{array}{ccc}r_1 & r_2 & r_3 \\ 3 & 7 & 0 \\ 7 &... | integer | jee-main-2023-online-10th-april-morning-shift |
lsapwdnz | maths | binomial-theorem | multinomial-theorem | If the Coefficient of $x^{30}$ in the expansion of $\left(1+\frac{1}{x}\right)^6\left(1+x^2\right)^7\left(1-x^3\right)^8 ; x \neq 0$ is $\alpha$, then $|\alpha|$ equals ___________. | [] | null | 678 | $\begin{aligned} & \text { Coefficient of } x^{30} \text { in } \frac{(1+x)^6\left(1+x^2\right)^7\left(1-x^3\right)^8}{x^6} \\\\ & \Rightarrow \text { Coefficient of } x^{36} \text { in }(1+x)^6\left(1+x^2\right)^7\left(1-x^3\right)^8 \\\\ & \Rightarrow \text { General term }={ }^6 C_{r_1}{ }^7 C_{r_2}{ }^8 C_{r_3}(-1)... | integer | jee-main-2024-online-1st-february-morning-shift |
jaoe38c1lse5mmmr | maths | binomial-theorem | multinomial-theorem | <p>Let $$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$$ have a common root, ... | [{"identifier": "A", "content": "$$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... | 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 $${\left( {1 + x} \right)^{27/5}}$$ = $${{{{27} \over 5}\left( {{{27} \over 5} - 1} \right).....\left( {{{27} \over 5} - r... | mcq | aieee-2003 |
S34Ufc1rYv3mtMqj | maths | binomial-theorem | negative-and-fractional-index | If $$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... | 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>= $$\left[ {1 + \left( { - 1} \right)\left( { - ax} \right) + {{\left( { - 1} \right)\left( { - 2} \right)} \over {1.2}}{{\left( { - ax} \right)}^2} + ...} \right]$$ -
<... | 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>$$ = {1 \over a}\sum\limits_{r = 1}^n {\left\{ {\left( {1 + {{rb} \over a} + {{{r^2}{b^2}} \over {{a^2}}}} \right) + (terms\,to\,be\,neglected)} \ri... | mcq | jee-main-2021-online-25th-july-morning-shift |
LV5BeeNJAp7oiwUq | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If the sum of the coefficients in the expansion of $$\,{\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} + ... | 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 = $${}^{2n}{C_{3r - 1}}$... | mcq | aieee-2002 |
9o9lRt7vfpS59rnX | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If $${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... | mcq | aieee-2004 |
Subsets and Splits
Chapter Question Count Chart
Displays the number of questions in each chapter and a graphical representation, revealing which chapters have the most questions for focused study.
SQL Console for archit11/jee_math
Counts the number of occurrences of each paper ID, which could help identify duplicates but lacks deeper analytical insight.