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1cwigZ8CK7EUvRnc | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The value of $$\,{}^{50}{C_4} + \sum\limits_{r = 1}^6 {^{56 - r}} {C_3}$$ is | [{"identifier": "A", "content": "$${}^{55}{C_4}$$ "}, {"identifier": "B", "content": "$${}^{55}{C_3}$$"}, {"identifier": "C", "content": "$${}^{56}{C_3}$$"}, {"identifier": "D", "content": "$${}^{56}{C_4}$$"}] | ["D"] | null | Given, $${}^{50}{C_4} + \sum\limits_{n = 1}^6 {{}^{56 - r}{C_3}} $$
<br><br>$$ \Rightarrow $$ $${}^{50}{C_4}$$ + $${}^{55}{C_3}$$ + $${}^{54}{C_3}$$ + $${}^{53}{C_3}$$ + $${}^{52}{C_3}$$ + $${}^{51}{C_3}$$ + $${}^{50}{C_3}$$
<br><br>Arrange those this way
<br><br>$$ \Rightarrow $$ $${}^{50}{C_4}$$ + $${}^{50}{C_3}$$ + $${}^{51}{C_3}$$ + $${}^{52}{C_3}$$ + $${}^{53}{C_3}$$ + $${}^{54}{C_3}$$ + $${}^{55}{C_3}$$
<br><br>We know this formula [ $${{}^n{C_r}}$$ + $${{}^n{C_{r - 1}}}$$ = $${{}^{n + 1}{C_r}}$$ ] which is used to solve this problem.
<br><br>$$ \Rightarrow $$ $${}^{51}{C_4}$$ + $${}^{51}{C_3}$$ + $${}^{52}{C_3}$$ + $${}^{53}{C_3}$$ + $${}^{54}{C_3}$$ + $${}^{55}{C_3}$$
<br><br>$$ \Rightarrow $$ $${}^{52}{C_4}$$ + $${}^{52}{C_3}$$ + $${}^{53}{C_3}$$ + $${}^{54}{C_3}$$ + $${}^{55}{C_3}$$
<br><br>$$ \Rightarrow $$$${}^{53}{C_4}$$ + $${}^{53}{C_3}$$ + $${}^{54}{C_3}$$ + $${}^{55}{C_3}$$
<br><br>$$ \Rightarrow$$$$ {}^{54}{C_4}$$ + $${}^{54}{C_3}$$ + $${}^{55}{C_3}$$
<br><br>$$ \Rightarrow $$$${}^{55}{C_4}$$ + $${}^{55}{C_3}$$
<br><br>$$ \Rightarrow$$$$ {}^{56}{C_4}$$ | mcq | aieee-2005 |
anq50ql02KdHRW7Q | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | For natural numbers $$m$$ , $$n$$, if $${\left( {1 - y} \right)^m}{\left( {1 + y} \right)^n}\,\, = 1 + {a_1}y + {a_2}{y^2} + ..........$$ and $${a_1} = {a_2} = 10,$$ then $$\left( {m,\,n} \right)$$ is | [{"identifier": "A", "content": "$$\\left( {20,\\,45} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( {35,\\,20} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {45,\\,35} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {35,\\,45} \\right)$$"}] | ["D"] | null | $${\left( {1 - y} \right)^m}{\left( {1 + y} \right)^n}\,\,$$
<br><br>= $$\left( {{}^m{C_0} - {}^m{C_1}y + {}^m{C_2}{y^2} + ....} \right)$$ -
<br> $$\left( {{}^n{C_0} + {}^n{C_1}y + {}^n{C_2}{y^2} + ....} \right)$$
<br><br>$${a_1}$$ = Coefficient of y = $${{}^n{C_1}}$$ - $${{}^m{C_1}}$$ = 10
<br><br>$$ \Rightarrow $$ n - m = 10
<br><br>$${a_2}$$ = Coefficient of y<sup>2</sup>
<br><br>= $${}^n{C_2} + {}^n{C_1} \times {}^m{C_1} + {}^m{C_2} = 10$$
<br><br>$$ \Rightarrow {{n\left( {n - 1} \right)} \over 2} - nm + {{m\left( {m - 1} \right)} \over 2} = 10$$
<br><br>$$ \Rightarrow n\left( {n - 1} \right) - 2nm + m\left( {m - 1} \right) = 20$$
<br><br>$$ \Rightarrow$$ (m + 10)(m + 9) - 2(m + 10)m + m(m - 1) = 20
<br><br>$$ \Rightarrow$$ 90 + 19m + m<sup>2</sup> - 2m<sup>2</sup> - 20m + m<sup>2</sup> - m - 20 = 0
<br><br>$$ \Rightarrow$$ 70 - 2m = 0
<br><br>$$ \Rightarrow$$ m = 35
<br><br>$$\therefore$$ n = 10 + 35 = 45 | mcq | aieee-2006 |
zCRZuqBTgvBpoM9t | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The sum of the series $${}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .....\, - \,.....\, + {}^{20}{C_{10}}$$ is | [{"identifier": "A", "content": "$$0$$"}, {"identifier": "B", "content": "$${}^{20}{C_{10}}$$ "}, {"identifier": "C", "content": "$$ - {}^{20}{C_{10}}$$ "}, {"identifier": "D", "content": "$${1 \\over 2}{}^{20}{C_{10}}$$ "}] | ["D"] | null | We know
<br><br>$${}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... + {}^{20}{C_{10}} - {}^{20}{C_{11}}+ ...... + {}^{20}{C_{20}} = 0$$
<br><br>$$ \Rightarrow $$ $$({}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}})$$ $$+{}^{20}{C_{10}}$$ $$(-{}^{20}{C_{9}}+ {}^{20}{C_{8}}+...... + {}^{20}{C_{0}})$$
<br><br>(As $${}^{20}{C_{11}} = {}^{20}{C_9}$$)
<br><br>$$ \Rightarrow $$ $$2({}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}})$$ $$+{}^{20}{C_{10}}$$ = 0
<br><br>$$ \Rightarrow $$ $${}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}}$$ = $$ - {1 \over 2}{}^{20}{C_{10}}$$
<br><br>Adding $${}^{20}{C_{10}}$$ both sides,
<br><br>$$ \Rightarrow $$ $${}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}}$$ $$ + {}^{20}{C_{10}}$$ = $$ - {1 \over 2}{}^{20}{C_{10}}$$ $$ + {}^{20}{C_{10}}$$
<br><br>$$ \Rightarrow $$ $${}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}}$$ $$ + {}^{20}{C_{10}}$$ = $$ {1 \over 2}{}^{20}{C_{10}}$$ | mcq | aieee-2007 |
lYC68b3a4PYHi6TL | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <b>Statement - 1 :</b> $$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r} = \left( {n + 2} \right){2^{n - 1}}.} $$
<br/><b>Statement - 2 :</b> $$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}{x^r} = {{\left( {1 + x} \right)}^n} + nx{{\left( {1 + x} \right)}^{n - 1}}.} $$ | [{"identifier": "A", "content": "Statement - 1 is false, Statement - 2 is true "}, {"identifier": "B", "content": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1 "}, {"identifier": "C", "content": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1 "}, {"identifier": "D", "content": "Statement - 1 is true, Statement - 2 is false "}] | ["B"] | null | <b>Check Statement - 1</b>
<br><br>$$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}}$$
<br><br>= $$\sum\limits_{r = 0}^n {r.{}^n{C_r}} $$ + $$\sum\limits_{r = 0}^n {{}^n{C_r}} $$
<br><br>= $$\sum\limits_{r = 1}^n {r.{n \over r}{}^{n - 1}{C_{r - 1}}} $$ $$ + {2^n}$$
<br><br>= $$n\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}} $$ $$ + {2^n}$$
<br><br>= $$n \times {2^{n - 1}}$$$$ + {2^n}$$
<br><br>= $${2^{n - 1}}\left[ {n + 2} \right]$$
<br><br><b>Check Statement 2 :</b>
<br><br>$$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}{x^r}}$$
<br><br>= $$\sum\limits_{r = 0}^n r .{}^n{C_r}.{x^r}$$ + $$\sum\limits_{r = 0}^n {{}^n{C_r}.{x^r}} $$
<br><br>= $$n\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}.{x^r}} $$ $$ + {\left( {1 + x} \right)^n}$$
<br><br>= $$nx\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}.{x^{r - 1}}} + {\left( {1 + x} \right)^n}$$
<br><br>= $$nx{\left( {1 + x} \right)^{n - 1}} + {\left( {1 + x} \right)^n}$$
<br><br>Substitude x = 1 in the statement 2 and we get,
<br><br>$$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r} = \left( {n + 2} \right){2^{n - 1}}.} $$
<br><br>So Option B is correct. | mcq | aieee-2008 |
nvvLZmA4D6F3MenV | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | Let $${s_1} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right){}^{10}} {C_j}$$,
<br/><br/>$${{s_2} = \sum\limits_{j = 1}^{10} {} } j.{}^{10}{C_j}$$ and <br/><br/>$${{s_3} = \sum\limits_{j = 1}^{10} {{j^2}.{}^{10}{C_j}.} }$$
<p><b>Statement-1 :</b> $${{S_3} = 55 \times {2^9}}$$.
<br/><b>Statement-2 :</b> $${{S_1} = 90 \times {2^8}}$$ and $${{S_2} = 10 \times {2^8}}$$.</p> | [{"identifier": "A", "content": "Statement - 1 is true, Statement- 2 is true; Statement - 2 is not a correct explanation for Statement - 1. "}, {"identifier": "B", "content": "Statement - 1 is true, Statement-2 is false. "}, {"identifier": "C", "content": "Statement - 1 is false, Statement-2 is true. "}, {"identifier": "D", "content": "Statement - 1 is true, Statement-2 is true: -Statement - 2 is a correct explanation for Statement - 1."}] | ["B"] | null | <b>Note :</b>
<br><br>$$\sum\limits_{r = 0}^n {r.{}^n{C_r}} $$ = $$ = n{.2^{n - 1}}$$
<br><br>$$\sum\limits_{r = 0}^n {{r^2}.{}^n{C_r}} = n\left( {n + 1} \right){2^{n - 2}}$$
<br><br>Given that,
<br><br>$${s_1} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right){}^{10}} {C_j}$$
<br><br>=$$\sum\limits_{j = 1}^{10} {{j^2}.{}^{10}} {C_j} - \sum\limits_{j = 1}^{10} {j.{}^{10}} {C_j}$$
<br><br>= 10$$ \times $$11$$ \times $$$${2^{10 - 2}}$$ - 10$$ \times $$$${2^{10 - 1}}$$
<br><br>= 10$$ \times $$$${2^{8}}$$(11 - 2)
<br><br>= 10$$ \times $$9$$ \times $$$${2^{8}}$$
<br><br>= 90$$ \times $$$${2^{8}}$$
<br><br>$${{s_2} = \sum\limits_{j = 1}^{10} {} } j.{}^{10}{C_j}$$
<br><br>= 10$$ \times $$$${2^{10-1}}$$
<br><br>= 10$$ \times $$$${2^{9}}$$
<br><br>$${{s_3} = \sum\limits_{j = 1}^{10} {{j^2}.{}^{10}{C_j}.} }$$
<br><br>= 10$$ \times $$11$$ \times $$$${2^{10-2}}$$
<br><br>= $${{110} \over 2} \times {2^9}$$
<br><br>= 55 $$ \times $$ $$ {2^9}$$ | mcq | aieee-2010 |
1oOYpeLuxiC0SOFR | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The coefficient of $${x^7}$$ in the expansion of $${\left( {1 - x - {x^2} + {x^3}} \right)^6}$$ is | [{"identifier": "A", "content": "$$-132$$ "}, {"identifier": "B", "content": "$$-144$$ "}, {"identifier": "C", "content": "$$132$$ "}, {"identifier": "D", "content": "$$144$$"}] | ["B"] | null | Given,
<br>$${\left( {1 - x - {x^2} + {x^3}} \right)^6}$$
<br><br>= $${\left[ {\left( {1 - x} \right) - {x^2}\left( {1 - x} \right)} \right]^6}$$
<br><br>= $${\left( {1 - x} \right)^6}{\left( {1 - {x^2}} \right)^6}$$
<br><br>= $$\left( {1 + {}^6{C_1}( - x) + {}^6{C_2}{{( - x)}^2} + {}^6{C_3}{{( - x)}^3} + .......} \right)\times$$
<br> $$\left( {1 + {}^6{C_1}( - {x^2}) + {}^6{C_2}{{( - {x^2})}^2} + {}^6{C_3}{{( - {x^2})}^3} + .......} \right)$$
<br><br>$$\therefore$$ Coefficient of x<sup>7</sup> = $$ - {}^6{C_1} \times - {}^6{C_3} + \left( { - {}^6{C_3}} \right) \times {}^6{C_2} + \left( { - {}^6{C_5}} \right) \times - {}^6{C_1}$$
<br><br>= 120 - 300 + 36 = - 144 | mcq | aieee-2011 |
YZsTRt5kvoMcWFZQ | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If the coefficints of $${x^3}$$ and $${x^4}$$ in the expansion of $$\left( {1 + ax + b{x^2}} \right){\left( {1 - 2x} \right)^{18}}$$ in powers of $$x$$ are both zero, then $$\left( {a,\,b} \right)$$ is equal to: | [{"identifier": "A", "content": "$$\\left( {14,{{272} \\over 3}} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( {16,{{272} \\over 3}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {16,{{251} \\over 3}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {14,{{251} \\over 3}} \\right)$$ "}] | ["B"] | null | $$\left( {1 + ax + b{x^2}} \right){\left( {1 - 2x} \right)^{18}}$$
<br><br>= $${\left( {1 - 2x} \right)^{18}} + ax{\left( {1 - 2x} \right)^{18}} + b{x^2}{\left( {1 - 2x} \right)^{18}}$$
<br><br>= $$\left( {1 + ax + b{x^2}} \right)\left[ {{}^{18}{C_0} - {}^{18}{C_1}\left( {2x} \right) + {}^{18}{C_2}{{\left( {2x} \right)}^2} - {}^{18}{C_3}{{\left( {2x} \right)}^3} + ....} \right]$$
<br><br>Coefficient of x<sup>3</sup> is
<br>$${\left( { - 2} \right)^3}.{}^{18}{C_3} + a{\left( { - 2} \right)^2}.{}^{18}{C_2} + b\left( { - 2} \right).{}^{18}{C_1}$$ = 0
<br><br>$$ \Rightarrow 153a - 9b = 1632$$ ....... (1)
<br><br>Coefficient of x<sup>4</sup> is
<br>$${\left( { - 2} \right)^4}.{}^{18}{C_4} + a{\left( { - 2} \right)^3}.{}^{18}{C_3} + b{\left( { - 2} \right)^2}.{}^{18}{C_2}$$ = 0
<br><br>$$ \Rightarrow 3b - 32a = -240$$ ....... (2)
<br><br>Solving (1) and (2), we get $$a$$ = 16, b = $${{172} \over 3}$$ | mcq | jee-main-2014-offline |
IfImSZkdzKK4fCbz | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The sum of coefficients of integral power of $$x$$ in the binomial expansion $${\left( {1 - 2\sqrt x } \right)^{50}}$$ is : | [{"identifier": "A", "content": "$${1 \\over 2}\\left( {{3^{50}} - 1} \\right)$$ "}, {"identifier": "B", "content": "$${1 \\over 2}\\left( {{2^{50}} + 1} \\right)$$ "}, {"identifier": "C", "content": "$${1 \\over 2}\\left( {{3^{50}} + 1} \\right)$$"}, {"identifier": "D", "content": "$${1 \\over 2}\\left( {{3^{50}}} \\right)$$ "}] | ["C"] | null | $${\left( {1 - 2\sqrt x } \right)^{50}}$$
<br><br>= $${}^{50}{C_0} + {}^{50}{C_1}.\left( { - 2\sqrt x } \right) + {}^{50}{C_2}.{\left( { - 2\sqrt x } \right)^2} + ....$$
<br><br>Now we need to find out those coefficient where degree of x is integer and you can see at odd terms power of x is integer.
<br><br>Let $${\left( {1 - 2\sqrt x } \right)^{50}}$$ = Odd(A) - Even(B)
<br><br>So $${\left( {1 + 2\sqrt x } \right)^{50}}$$ = A + B
<br><br>$$\therefore$$ 2A = $${\left( {1 + 2\sqrt x } \right)^{50}}$$ + $${\left( {1 - 2\sqrt x } \right)^{50}}$$
<br><br>$$ \Rightarrow A = {1 \over 2}\left[ {{{\left( {1 + 2\sqrt x } \right)}^{50}} + {{\left( {1 - 2\sqrt x } \right)}^{50}}} \right]$$
<br><br>Now to find sum of coefficient of A, put x = 1.
<br><br>$$\therefore$$ Sum of coefficient of A = $${1 \over 2}\left[ {{{\left( {1 + 2} \right)}^{50}} + {{\left( {1 - 2} \right)}^{50}}} \right]$$
<br><br>= $${1 \over 2}\left[ {{{\left( 3 \right)}^{50}} + 1} \right]$$ | mcq | jee-main-2015-offline |
L23VRvsDDUiZd6Fw | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If the number of terms in the expansion of $${\left( {1 - {2 \over x} + {4 \over {{x^2}}}} \right)^n},\,x \ne 0,$$ is 28, then the sum of the coefficients of all the terms in this expansion, is : | [{"identifier": "A", "content": "243 "}, {"identifier": "B", "content": "729 "}, {"identifier": "C", "content": "64 "}, {"identifier": "D", "content": "2187 "}] | ["B"] | null | Total no of terms in $${\left( {1 - {2 \over x} + {4 \over {{x^2}}}} \right)^n}$$ = $${}^{n + 2}{C_2}$$ = 28
<br><br>(n+2)(n+1) = 56
<br><br>$$ \Rightarrow n = 6$$
<br><br>Sum of coefficient = (1 - 2 + 4)<sup>6</sup> = 3<sup>6</sup> = 729 | mcq | jee-main-2016-offline |
ifASjcJfTU5dfaRe | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The value of $$\left( {{}^{21}{C_1} - {}^{10}{C_1}} \right) + \left( {{}^{21}{C_2} - {}^{10}{C_2}} \right) + \left( {{}^{21}{C_3} - {}^{10}{C_3}} \right)$$
<br/>$$\left( {{}^{21}{C_4} - {}^{10}{C_4}} \right)$$$$ + .... + \left( {{}^{21}{C_{10}} - {}^{10}{C_{10}}} \right)$$ is | [{"identifier": "A", "content": "$${2^{21}} - {2^{10}}$$"}, {"identifier": "B", "content": "$${2^{20}} - {2^{9}}$$"}, {"identifier": "C", "content": "$${2^{20}} - {2^{10}}$$"}, {"identifier": "D", "content": "$${2^{21}} - {2^{11}}$$"}] | ["C"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265679/exam_images/bx4nbx1k931n5ip0ploo.webp" loading="lazy" alt="JEE Main 2017 (Offline) Mathematics - Binomial Theorem Question 184 English Explanation"> | mcq | jee-main-2017-offline |
KLIGyyeNa2wGDQBt | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The sum of the co-efficients of all odd degree terms in the expansion of
<br/><br/>$${\left( {x + \sqrt {{x^3} - 1} } \right)^5} + {\left( {x - \sqrt {{x^3} - 1} } \right)^5}$$, $$\left( {x > 1} \right)$$ is | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "-1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "1"}] | ["A"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266144/exam_images/sxwjinh9ykk33t08ossz.webp" loading="lazy" alt="JEE Main 2018 (Offline) Mathematics - Binomial Theorem Question 185 English Explanation"> | mcq | jee-main-2018-offline |
1TSVDgXFTBH4OS30sfYBP | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If n is the degree of the polynomial,
<br/><br/>$${\left[ {{2 \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }}} \right]^8} + $$ $${\left[ {{2 \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }}} \right]^8}$$
<br/><br/>and m is the coefficient of x<sup>n</sup> in it, then the ordered pair (n, m) is equal to : | [{"identifier": "A", "content": "(24, (10)<sup>8</sup>)"}, {"identifier": "B", "content": "(8, 5(10)<sup>4</sup>)"}, {"identifier": "C", "content": "(12, (20)<sup>4</sup>)"}, {"identifier": "D", "content": "(12, 8(10)<sup>4</sup>)"}] | ["C"] | null | Given, <br>
$${\left[ {{2 \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }}} \right]^8}$$ + $${\left[ {{2 \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }}} \right]^8}$$<br><br>
= $${\left[ {{2 \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }} \times {{\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} } \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }}} \right]^8}$$<br><br>
+ $${\left[ {{2 \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }} \times {{\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} } \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }}} \right]^8}$$<br><br>
= $${\left[ {{{2(\sqrt {5x^3 + 1} + \sqrt {5x^3 - 1} } \over 2}} \right]^8}$$ + $${\left[ {{{2(\sqrt {5x^3 - 1} - \sqrt {5x^3 - 1} } \over 2}} \right]^8}$$<br><br>
= ($${\sqrt {5{x^3} + 1} }$$ + $${\sqrt {5{x^3} - 1} }$$)<sup>8</sup> + ($${\sqrt {5{x^3} - 1} }$$ - $${\sqrt {5{x^3} - 1} }$$)<sup>8</sup><br><br>
= 2 $$\left[ {{}^8{C_0}} \right.{(\sqrt {5{x^3} + 1} )^8}$$ +<br><br>
+ $${}^8{C_2}{(\sqrt {5{x^3} + 1} )^6}\sqrt {5{x^3} - 1} $$ <br><br>
+ $${}^8{C_4}{(\sqrt {5{x^3} + 1} )^4}{(5{x^3} - 1)^2}$$<br><br>
+ $${}^8{C_6}{(\sqrt {5{x^3} + 1} )^2}{(5{x^3} - 1)^3}$$<br><br>
+ $$\left. {{}^8{C_8}{{(5{x^3} - 1)}^4}} \right]$$<br><br>
= 2$$[{}^8{C_0}{(5{x^3} + 1)^4}$$<br><br>
+ $${}^8{C_2}{(5{x^3} + 1)^3}(5{x^3} - 1)$$<br><br>
+ $${}^8{C_4}(5{x^3} + 1)^2{(5{x^3} - 1)^2}$$ <br><br>
+ $${}^8{C_6}(5{x^3} + 1){(5{x^3} - 1)^3}$$ <br><br>
+ $${}^8{C_8}{(5{x^3} - 1)^4}]$$ <br><br>
Here maximum power of x is 12 <br><br>
$$ \therefore $$ Degree of polynomial = 12<br><br>
Coefficient of x<sup>12</sup><br><br>
= 2 [<sup>8</sup>C<sub>0</sub> 5<sup>4</sup> + <sup>8</sup>C<sub>2</sub>
$$\times$$ 5<sup>3</sup> $$\times$$ 5 + <sup>8</sup>C<sub>4</sub>
$$\times$$ 5<sup>2</sup> $$\times$$ 5<sup>2</sup> + <sup>8</sup>C<sub>6 </sub> $$\times$$ 5 $$\times$$ 5<sup>3</sup> + <sup>8</sup>C<sub>8</sub> $$\times$$ 5<sup>4</sup>]<br><br>
= 160000 = (20)<sup>4</sup>
| mcq | jee-main-2018-online-15th-april-morning-slot |
BBlE646RfwQmNFEvkpwnD | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The coefficien of x<sup>10</sup> in the expansion of (1 + x)<sup>2</sup>(1 + x<sup>2</sup>)<sup>3</sup>(1 + x<sup>3</sup>)<sup>4</sup> is equal to : | [{"identifier": "A", "content": "52"}, {"identifier": "B", "content": "56"}, {"identifier": "C", "content": "50"}, {"identifier": "D", "content": "44"}] | ["A"] | null | $$ \because $$$$\,\,\,$$ (1 + x)<sup>2</sup> = 1 + 2x + x<sup>2</sup>,
<br><br>(1 + x<sup>2</sup>)<sup>3</sup> = 1 + 3x<sup>2</sup> + 3x<sup>4</sup> + x<sup>6</sup>
<br><br>and (1 + x<sup>3</sup>)<sup>4</sup> = 1 + 4x<sup>3</sup> + 6x<sup>6</sup> + 4x<sup>9</sup> + x<sup>12</sup>
<br><br>So, the possible combination for x<sup>10</sup> are :
<br><br>x . x<sup>9</sup>, x . x<sup>6</sup>. x<sup>3</sup>, x<sup>2</sup> . x<sup>2</sup> . x<sup>6</sup> , x<sup>4</sup> . x<sup>6</sup>
<br><br>Corresponding coefficients are 2 $$ \times $$ 4, 2$$ \times $$ 1 $$ \times $$4, 1 $$ \times $$3$$ \times $$ 6,
<br><br>3 $$ \times $$ 6 or 8, 8, 18, 18.
<br><br>$$\therefore\,\,\,$$ Sum of the coefficient is 8 + 8 + 18 + 18 = 52
<br><br>Therefore, the coefficient of x<sup>10</sup> in the expansion of
<br><br>(1 + x)<sup>2</sup> (1 + x<sup>2</sup>)<sup>3</sup> (1 + x<sup>3</sup>)<sup>4</sup> is 52. | mcq | jee-main-2018-online-15th-april-evening-slot |
vHqL1tXHdlVCD1G4cwS1m | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The coefficient of x<sup>2</sup> in the expansion of the product
<br/>(2$$-$$x<sup>2</sup>) .((1 + 2x + 3x<sup>2</sup>)<sup>6</sup> + (1 $$-$$ 4x<sup>2</sup>)<sup>6</sup>) is : | [{"identifier": "A", "content": "107"}, {"identifier": "B", "content": "106"}, {"identifier": "C", "content": "108"}, {"identifier": "D", "content": "155"}] | ["B"] | null | Given,
<br><br>(2 $$-$$ x<sup>2</sup>) . (1 + 2x + 3x<sup>2</sup>) <sup>6</sup> + (1 $$-$$ 4x<sup>2</sup>)<sup>6</sup>)
<br><br>Let, a = ((1 + 2x + 3x<sup>2</sup>)<sup>6</sup> + (1 $$-$$ 4x<sup>2</sup>)<sup>6</sup>)
<br><br>$$\therefore\,\,\,\,$$ Given statement becomes,
<br><br>(2 $$-$$ x<sup>2</sup>) . (a)
<br><br>= 2a $$-$$ x<sup>2</sup> (a)
<br><br>Here coefficients of x<sup>2</sup> is
<br><br>= 2 (coefficient of x<sup>2</sup> in a ) $$-$$ 1 (constant in a)
<br><br>(1 + 2x + 3x<sup>2</sup>) <sup>6</sup> = <sup>6</sup>C<sub>0</sub> + <sup>6</sup>C<sub>1</sub> (2x + 3x<sup>2</sup>) + <sup>6</sup>C<sub>2</sub> (2x + 3x<sup>2</sup>)<sup>2</sup>
<br><br> + . . . . . .+ (2x + 3x<sup>2</sup>)<sup>6</sup>
<br><br>(1 $$-$$ 4x<sup>2</sup>)<sup>6</sup> = <sup>6</sup>C<sub>0</sub> $$-$$ <sup>6</sup>C<sub>1</sub> (4x<sup>2</sup>) + <sup>6</sup>C<sub>2</sub> (4x<sup>2</sup>)<sup>2</sup> <br><br> + . . . . .+ (4x<sup>2</sup>)<sup>6</sup>
<br><br>Coefficient of x<sup>2</sup> in (1 + 2x + 3x<sup>2</sup>)<sup>6</sup>
<br><br>= <sup>6</sup>C<sub>1</sub> $$ \times $$ 3 + <sup>6</sup>C<sub>2</sub> $$ \times $$ 4
<br><br>= 18 + 60
<br><br>Coefficient of x<sup>2</sup> in (1 $$-$$ 4x<sup>2</sup>)<sup>6</sup>
<br><br>= $$-$$ <sup>6</sup>C<sub>1</sub> $$ \times $$ 4
<br><br>= $$-$$ 24
<br><br>Coefficient of x<sup>2</sup> in ((1 + 2x + 3x<sup>2</sup>)<sup>6</sup> + (1 $$-$$ 4x<sup>2</sup>)<sup>6</sup>)
<br><br>= 60 + 18 $$-$$ 24
<br><br>= 54
<br><br>Constant term in (1 + 2x + 3x<sup>2</sup>)<sup>6</sup> = <sup>6</sup>C<sub>0</sub> = 1
<br><br>Constant term in (1 $$-$$ 4x)<sup>6</sup> = <sup>6</sup>C<sub>0</sub> = 1
<br><br>$$\therefore\,\,\,\,$$ Constant term in ((1 + 2x + 3x<sup>2</sup>)<sup>6</sup> + (1 $$-$$ 4x)<sup>6</sup>)
<br><br>= 1 + 1 = 2
<br><br>$$\therefore\,\,\,\,$$ Coefficient of x<sup>2</sup> in (2 $$-$$ x<sup>2</sup>) ((1 + 2x + 3x<sup>2</sup>)<sup>6</sup> + (1 $$-$$ 4x<sup>2</sup>)<sup>6</sup>)
<br><br>= 2 (54) $$-$$ 1 (2)
<br><br>= 108 $$-$$ 2
<br><br>= 106 | mcq | jee-main-2018-online-16th-april-morning-slot |
bD89kihDYtpZYSOBmF3rsa0w2w9jxaybbxn | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If <sup>20</sup>C<sub>1</sub> + (2<sup>2</sup>) <sup>20</sup>C<sub>2</sub> + (3<sup>2</sup>) <sup>20</sup>C<sub>3</sub> + ..... + (20<sup>2</sup>
)
<sup>20</sup>C<sub>20</sub> = A(2<sup>$$\beta $$</sup>), then the ordered pair (A, $$\beta $$) is equal to : | [{"identifier": "A", "content": "(420, 19)"}, {"identifier": "B", "content": "(420, 18)"}, {"identifier": "C", "content": "(380, 18)"}, {"identifier": "D", "content": "(380, 19)"}] | ["B"] | null | S = $${1^2}\,{}^{20}{C_1} + {2^2}\,{}^{20}{C_2} + {3^2}\,{}^{20}{C_3} + .......... + {20^2}\,{}^{20}{C_{20}}$$<br><br>
$$ \Rightarrow $$ $$\sum\limits_{r = 1}^{20} {{r^2}\,{}^{20}{C_r}} $$ <br><br>
$$ \Rightarrow $$ $$\sum\limits_{r = 1}^{20} {r\,.\left( {r.{}^{20}{C_r}} \right)} $$<br><br>
$$ \Rightarrow $$ $$20\sum\limits_{r = 1}^{20} {r\,.} {}^{19}{C_{r - 1}}$$<br><br>
$$ \Rightarrow $$ $$20\sum\limits_{r = 1}^{20} {(r - 1 + 1)\,.} {}^{19}{C_{r - 1}}$$<br><br>
$$ \Rightarrow $$ $$20\sum\limits_{r = 1}^{20} {(r - 1)\,.} {}^{19}{C_{r - 1}} + 20\sum\limits_{r = 1}^{20} {{}^{19}{C_{r - 1}}} $$<br><br>
$$ \Rightarrow $$ $$20 \times 19\sum\limits_{r = 2}^{20} {{}^{18}{C_{r - 2}}} + 20 \times {2^{19}}$$<br><br>
$$ \Rightarrow $$ 20 $$ \times $$ 19 $$ \times $$ 2<sup>18</sup> + 20 $$ \times $$ 2<sup>19</sup><br><br>
$$ \Rightarrow $$ $$20 \times {2^{18}}\left( {19 + 2} \right) = 20 \times 21 \times {2^{18}}$$<br><br>
$$ \Rightarrow $$ $$420 \times {2^{18}}$$
| mcq | jee-main-2019-online-12th-april-evening-slot |
2O8ZKwAonWPua4lhpj3rsa0w2w9jwxtytkr | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If the coefficients of x<sup>2</sup>
and x<sup>3</sup>
are both zero, in the expansion of the expression (1 + ax + bx<sup>2</sup>
) (1 – 3x)<sup>15</sup> in
powers of x, then the ordered pair (a,b) is equal to : | [{"identifier": "A", "content": "(28, 861)"}, {"identifier": "B", "content": "(28, 315)"}, {"identifier": "C", "content": "(\u201321, 714)"}, {"identifier": "D", "content": "(\u201354, 315)"}] | ["B"] | null | (1 + ax + bx<sup>2</sup>)(1 – 3x)<sup>15</sup><br><br>
Co-eff. of x<sup>2</sup> = 1.<sup>15</sup>C<sub>2</sub>(–3)<sup>2</sup> + a.<sup>15</sup>C<sub>1</sub>(–3) + b.<sup>15</sup>C<sub>0</sub><br><br>
$$ = {{15 \times 14} \over 2} \times 9 - 15 \times 3a + b = 0$$ (given)<br><br>
$$ \Rightarrow $$ 945 – 45a + b = 0 ...(i)<br><br>
Now co-eff. of x<sup>3</sup> = 0<br><br>
$$ \Rightarrow $$ <sup>15</sup>C<sub>3</sub>(–3)<sup>3</sup> + a.<sup>15</sup>C<sub>2</sub>(–3)<sup>2</sup> + b.<sup>15</sup>C<sub>1</sub>(–3) = 0<br><br>
$$ \Rightarrow {{15 \times 14 \times 13} \over {3 \times 2}} \times ( - 3 \times 3 \times 3) + a \times {{15 \times 14 \times 9} \over 2 } – b × 3 × 15 = 0$$<br><br>
$$ \Rightarrow $$ 15 × 3[–3 × 7 × 13 + a × 7 × 3 – b] = 0<br><br>
$$ \Rightarrow $$ 21a – b = 273 ...(ii)<br><br>
From (i) and (ii)<br><br>
a = +28, b = 315 $$ \equiv $$ (a, b) $$ \equiv $$ (28, 315) | mcq | jee-main-2019-online-10th-april-morning-slot |
ULgVxIF8IPjhg5YNaS18hoxe66ijvwukqgc | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If some three consecutive in the binomial
expansion of (x + 1)<sup>n</sup> is powers of x are in the ratio
2 : 15 : 70, then the average of these three
coefficient is :- | [{"identifier": "A", "content": "625"}, {"identifier": "B", "content": "227"}, {"identifier": "C", "content": "964"}, {"identifier": "D", "content": "232"}] | ["D"] | null | Given $${}^n{C_{r - 1}}:{}^n{C_r}:{}^n{C_{r + 1}} = 2:15:70$$<br><br>
$${{{}^n{C_{r - 1}}} \over {{}^n{C_r}}} = {2 \over {15}}$$<br><br>
$$ \Rightarrow {r \over {n - r + 1}} = {2 \over {15}}$$<br><br>
$$ \Rightarrow 15r = 2n - 2r + 2$$<br><br>
$$ \Rightarrow 17r = 2n + 2$$ .... (i)<br><br>
Now $${{{}^n{C_r}} \over {{}^n{C_{r + 1}}}} = {{15} \over {70}}$$<br><br>
$$ \Rightarrow {{r + 1} \over {n - r}} = {3 \over {14}}$$<br><br>
$$ \Rightarrow $$ 14r + 14 = 3n – 3r<br><br>
$$ \Rightarrow $$ 17r = 3n – 14 ... (ii)<br><br>
Now From (i) and (ii) equation<br><br>
2n + 2 = 3n - 14 $$ \Rightarrow $$ n = 16<br><br>
By putting n = 16 in equation (i)<br><br>
$$ \Rightarrow $$ r = 2<br><br>
$$ \therefore $$ Average of coefficient = $${{{}^{16}{C_1} + {}^{16}{C_2} + {}^{16}{C_3}} \over 3} $$
<br><br>= $${{16 + 120 + 560} \over 3}$$<br><br>
= $${{696} \over 3} = 232$$
| mcq | jee-main-2019-online-9th-april-evening-slot |
OIyzlxPBzW0N8G5uCPqZ5 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The sum of the co-efficients of all even
degree terms in x in the expansion of<br/>
$${\left( {x + \sqrt {{x^3} - 1} } \right)^6}$$ + $${\left( {x - \sqrt {{x^3} - 1} } \right)^6}$$, (x > 1) is equal to: | [{"identifier": "A", "content": "32"}, {"identifier": "B", "content": "26"}, {"identifier": "C", "content": "29"}, {"identifier": "D", "content": "24"}] | ["D"] | 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>= 2A
<br><br>= 2[odd terms]
<br><br>= 2[ T<sub>1</sub> + T<sub>3</sub> + T<sub>5</sub> + ....... ]
<br><br>So in case of
<br><br>$${\left( {x + \sqrt {{x^3} - 1} } \right)^6}$$ + $${\left( {x - \sqrt {{x^3} - 1} } \right)^6}$$
<br><br>= 2[ T<sub>1</sub> + T<sub>3</sub> + T<sub>5</sub> + T<sub>5</sub> ]
<br><br>= 2[ $${}^6{C_0}{x^6} + {}^6{C_2}{x^4}\left( {{x^3} - 1} \right)$$
<br><br>$$ + {}^6{C_4}{x^2}{\left( {{x^3} - 1} \right)^2} + {}^6{C_6}{\left( {{x^3} - 1} \right)^3}$$]
<br><br>= 2[ $${x^6} + 15{x^4}\left( {{x^3} - 1} \right) + 15{x^2}{\left( {{x^3} - 1} \right)^2} + {\left( {{x^3} - 1} \right)^3}$$ ]
<br><br>= 2[ $${x^6} + 15\left( {{x^7} - {x^4}} \right)$$
<br><br>$$ + 15{x^2}\left( {{x^6} - 2{x^3} + 1} \right) + \left( {{x^9} - 3{x^6} + 3{x^3} - 1} \right)$$ ]
<br><br>$$ \therefore $$ Sum of coefficient of all even degree terms
<br><br>= 2[ 1 - 15 + 15 + 15 - 3 - 1 ] = 24 | mcq | jee-main-2019-online-8th-april-morning-slot |
sd2g3rXgKxRxJIAeLbYtU | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | Let S<sub>n</sub> = 1 + q + q<sup>2</sup> + . . . . . + q<sup>n</sup> and T<sub>n</sub> = 1 + $$\left( {{{q + 1} \over 2}} \right) + {\left( {{{q + 1} \over 2}} \right)^2}$$ + . . . . . .+ $${\left( {{{q + 1} \over 2}} \right)^n}$$ where q is a real number and q $$ \ne $$ 1. If <sup>101</sup>C<sub>1</sub> + <sup>101</sup>C<sub>2</sub> . S<sub>1</sub> + .... + <sup>101</sup>C<sub>101</sub> . S<sub>100</sub> = $$\alpha $$T<sub>100</sub> then $$\alpha $$ is equal to
| [{"identifier": "A", "content": "202"}, {"identifier": "B", "content": "200"}, {"identifier": "C", "content": "2<sup>100</sup>"}, {"identifier": "D", "content": "2<sup>99</sup>"}] | ["C"] | null | <sup>101</sup>C<sub>1</sub> + <sup>101</sup>C<sub>2</sub>S<sub>1</sub> + . . . . . . . + <sup>101</sup>C<sub>101</sub>S<sub>100</sub>
<br><br>$$=$$ $$\alpha $$T<sub>100</sub>
<br><br><sup>101</sup>C<sub>1</sub> + <sup>101</sup>C<sub>2</sub>(1 + q) + <sup>101</sup>C<sub>3</sub>(1 + q + q<sup>2</sup>) +
<br><br> . . . . . .+<sup>101</sup>C<sub>101</sub>(1 + q + . . . . . + q<sup>100</sup>)
<br><br>$$ = 2\alpha {{\left( {1 - {{\left( {{{1 + q} \over 2}} \right)}^{101}}} \right)} \over {\left( {1 - q} \right)}}$$
<br><br>$$ \Rightarrow $$ <sup>101</sup>C<sub>1</sub>(1 $$-$$ q) + <sup>101</sup>C<sub>2</sub>(1 $$-$$ q<sup>2</sup>) +
<br><br> . . . . . . + <sup>101</sup>C<sub>101</sub>(1 $$-$$ q<sup>101</sup>)
<br><br>$$ = 2\alpha \left( {1 - {{\left( {{{1 + q} \over 2}} \right)}^{101}}} \right)$$
<br><br>$$ \Rightarrow $$ (2<sup>101</sup> $$-$$ 1) $$-$$ ((1 + q)<sup>101</sup> $$-$$ 1)
<br><br>$$ = 2\alpha \left( {1 - {{\left( {{{1 + q} \over 2}} \right)}^{101}}} \right)$$
<br><br>$$ \Rightarrow $$ $${2^{101}}\left( {1 - {{\left( {{{1 + q} \over 2}} \right)}^{101}}} \right) = 2\alpha \left( {1 - {{\left( {{{1 + q} \over 2}} \right)}^{101}}} \right)$$
<br><br>$$ \Rightarrow $$ $$\alpha = {2^{100}}$$ | mcq | jee-main-2019-online-11th-january-evening-slot |
rgg4nMVTcWW8M9c4pNsE7 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | Let (x + 10)<sup>50</sup> + (x $$-$$ 10)<sup>50</sup> = a<sub>0</sub> + a<sub>1</sub>x + a<sub>2</sub>x<sup>2</sup> + . . . . + a<sub>50</sub>x<sup>50</sup>, for all x $$ \in $$ R; then $${{{a_2}} \over {{a_0}}}$$ is equal to | [{"identifier": "A", "content": "12.25 "}, {"identifier": "B", "content": "12.75"}, {"identifier": "C", "content": "12.00"}, {"identifier": "D", "content": "12.50"}] | ["A"] | null | (10 + x)<sup>50</sup> + (10 $$-$$ x)<sup>50</sup>
<br><br>$$ \Rightarrow $$ a<sub>2</sub> = 2.<sup>50</sup>C<sub>2</sub> 10<sup>48</sup>, a<sub>0</sub> = 2.10<sup>50</sup>
<br><br>$${{{a_2}} \over {{a_0}}} = {{^{50}{C_2}} \over {{{10}^2}}} = 12.25$$ | mcq | jee-main-2019-online-11th-january-evening-slot |
MVYyLR9wVR6I6zEOqNoEu | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The value of r for which <sup>20</sup>C<sub>r</sub> <sup>20</sup>C<sub>0</sub> + <sup>20</sup>C<sub>r$$-$$1</sub> <sup>20</sup>C<sub>1</sub> + <sup>20</sup>C<sub>r$$-$$2</sub> <sup>20</sup>C<sub>2</sub> + . . . . .+ <sup>20</sup>C<sub>0</sub> <sup>20</sup>C<sub>r</sub> is maximum, is | [{"identifier": "A", "content": "20"}, {"identifier": "B", "content": "15"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "11"}] | ["A"] | null | Given sum = coefficient of x<sup>r</sup> in the expansion of
<br><br>(1 + x)<sup>20</sup>(1 + x)<sup>20</sup>,
<br><br>Which is equal to <sup>40</sup>C<sub>r</sub>
<br><br>It is maximum when r = 20 | mcq | jee-main-2019-online-11th-january-morning-slot |
y3wzzzbvff6SL6QK9F80V | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If $${\sum\limits_{i = 1}^{20} {\left( {{{{}^{20}{C_{i - 1}}} \over {{}^{20}{C_i} + {}^{20}{C_{i - 1}}}}} \right)} ^3} = {k \over {21}}$$ then k is equal to | [{"identifier": "A", "content": "100"}, {"identifier": "B", "content": "200"}, {"identifier": "C", "content": "50"}, {"identifier": "D", "content": "400"}] | ["A"] | null | $${\sum\limits_{i = 1}^{20} {\left( {{{^{20}{C_{i - 1}}} \over {^{20}{C_i}{ + ^{20}}{C_{i - 1}}}}} \right)} ^3} = {k \over {21}}$$
<br><br>$$ \Rightarrow \,\,\sum\limits_{i = 1}^{20} {{{\left( {{{{}^{20}{C_{i - 1}}} \over {{}^{21}{C_i}}}} \right)}^3}} = {k \over {21}}$$
<br><br>$$ \Rightarrow \,\,\sum\limits_{i = 1}^{20} {{{\left( {{i \over {21}}} \right)}^3}} = {k \over {21}}$$
<br><br>$$ \Rightarrow \,\,{1 \over {{{\left( {21} \right)}^3}}}{\left[ {{{20\left( {21} \right)} \over 2}} \right]^2} = {k \over {21}}$$
<br><br>$$ \Rightarrow 100 = k$$ | mcq | jee-main-2019-online-10th-january-morning-slot |
CZ6nSlHjDI3ZaVDxF8B92 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The coefficient of t<sup>4</sup> in the expansion of $${\left( {{{1 - {t^6}} \over {1 - t}}} \right)^3}$$ is : | [{"identifier": "A", "content": "14"}, {"identifier": "B", "content": "15"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "12"}] | ["B"] | null | $${\left( {{{1 - {t^6}} \over {1 - t}}} \right)^3}$$
<br><br>= (1 $$-$$ t<sup>6</sup>)<sup>3</sup> (1 $$-$$ t)<sup>$$-$$3</sup>
<br><br>= (1 $$-$$ <sup>3</sup>C<sub>1</sub>t<sup>6</sup> + <sup>3</sup>C<sub>2</sub>t<sup>12</sup> $$-$$ <sup>3</sup>C<sub>3</sub>t<sup>18</sup>) $$ \times $$ (1 $$-$$ t)<sup>$$-$$3</sup>
<br><br>coefficient of t<sup>4</sup> is 1 $$ \times $$ coefficient of t<sup>4</sup> in (1 $$-$$ t)<sup>$$-$$3</sup>
<br><br>= 1 $$ \times $$ <sup>3+4$$-$$1</sup>C<sub>4</sub> (By multinomial theorem)
<br><br>= <sup>6</sup>C<sub>4</sub> = 15 | mcq | jee-main-2019-online-9th-january-evening-slot |
U7i5ScdA9N27m362FQs2z | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The sum of the series
<br/><br/>2.<sup>20</sup>C<sub>0</sub>
+ 5.<sup>20</sup>C<sub>1</sub> + 8.<sup>20</sup>C<sub>2</sub> + 11.<sup>20</sup>C<sub>3</sub> + ... +62.<sup>20</sup>C<sub>20</sub> is equal to :
| [{"identifier": "A", "content": "2<sup>25</sup>"}, {"identifier": "B", "content": "2<sup>24</sup>"}, {"identifier": "C", "content": "2<sup>26</sup>"}, {"identifier": "D", "content": "2<sup>23</sup>"}] | ["A"] | null | Here general term = (3r + 2)<sup>20</sup>C<sub>r</sub>
<br><br>$$ \therefore $$ Sum of the series = $$\sum\limits_{r = 0}^{20} {\left( {3r + 2} \right)} {}^{20}{C_r}$$
<br><br>= $$3\sum\limits_{r = 0}^{20} {r.} {}^{20}{C_r} + 2\sum\limits_{r = 0}^{20} {{}^{20}{C_r}} $$
<br><br>= 3 $$ \times $$ 20$$ \times $$2<sup>20 - 1</sup> + 2$$ \times $$2<sup>20</sup>
<br><br>= 60$$ \times $$2<sup>19</sup> + 2<sup>21</sup>
<br><br>= 2<sup>21</sup> [15 + 1]
<br><br>= 2<sup>21</sup> $$ \times $$ 16
<br><br>= 2<sup>25</sup> | mcq | jee-main-2019-online-8th-april-morning-slot |
Cq5FMsIGax6YUnb7Jrjgy2xukfajwxje | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If for some positive integer n, the coefficients<br/> of three consecutive terms in the binomial<br/> expansion of (1 + x)<sup>n + 5</sup> are in the ratio<br/> 5 : 10 : 14, then the largest coefficient in this expansion is : | [{"identifier": "A", "content": "330"}, {"identifier": "B", "content": "792 "}, {"identifier": "C", "content": "252"}, {"identifier": "D", "content": "462"}] | ["D"] | null | Consider the three consecutive coefficients as <br><br>$$^{n + 5}{C_r},{\,^{n + 5}}{C_{r + 1}},{\,^{n + 5}}{C_{r + 2}}$$<br><br>$$ \because $$ $${{^{n + 5}{C_r}} \over {^{n + 5}{C_{r + 1}}}} = {1 \over 2}$$<br><br>$$ \Rightarrow {{r + 1} \over {n + 5 - r}} = {1 \over 2} \Rightarrow 3r = n + 3$$ ...(i)<br><br>and $${{^{n + 5}{C_{r + 1}}} \over {^{n + 5}{C_{r + 2}}}} = {5 \over 7}$$<br><br>$$ \Rightarrow $$ $$ \Rightarrow {{r + 2} \over {n + 4 - r}} = {5 \over 7} \Rightarrow 12r = 5n + 6$$ ...(ii)<br><br>From (i) and (ii) n = 6<br><br>Largest coefficient in the expansion is $${^{11}{C_6}}$$ = 462 | mcq | jee-main-2020-online-4th-september-evening-slot |
cqGCAjsv0eMJjeXRlCjgy2xukfqgcp3c | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The coefficient of x<sup>4</sup>
in the expansion of
<br/>(1 + x + x<sup>2</sup>
+ x<sup>3</sup>)<sup>6</sup>
in powers of x, is ______. | [] | null | 120 | (1 + x + x<sup>2</sup>
+ x<sup>3</sup>)<sup>6</sup>
<br><br>= ((1 + x) (1 + x<sup>2</sup>))<sup>6</sup>
<br><br>= (1 + x)<sup>6</sup>(1 + x<sup>2</sup>)<sup>6</sup>
<br><br>= $$\sum\limits_{r = 0}^6 {{}^6{C_r}.{x^r}} $$ $$\sum\limits_{r = 0}^6 {{}^6{C_r}.{x^{2r}}} $$
<br><br>Coefficient of x<sup>4</sup> = <sup>6</sup>C<sub>0</sub>
<sup>6</sup>C<sub>2</sub> + <sup>6</sup>C<sub>2</sub>
<sup>6</sup>C<sub>1</sub> + <sup>6</sup>C<sub>4</sub>
<sup>6</sup>C<sub>0</sub> = 120 | integer | jee-main-2020-online-5th-september-evening-slot |
n7goxh7onsO8yp6qOojgy2xukf7fj0by | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The value of $$\sum\limits_{r = 0}^{20} {{}^{50 - r}{C_6}} $$ is equal to: | [{"identifier": "A", "content": "$${}^{50}{C_6} - {}^{30}{C_6}$$"}, {"identifier": "B", "content": "$${}^{51}{C_7} - {}^{30}{C_7}$$"}, {"identifier": "C", "content": "$${}^{50}{C_7} - {}^{30}{C_7}$$"}, {"identifier": "D", "content": "$${}^{51}{C_7} + {}^{30}{C_7}$$"}] | ["B"] | null | $$\sum\limits_{r = 0}^{20} {} {}^{50 - r}{C_6} = {}^{50}{C_6} + {}^{49}{C_6} + {}^{48}{C_6} + .... + {}^{30}{C_6}$$<br><br>$$ = {}^{50}{C_6} + {}^{49}{C_6} + .... + {}^{31}{C_6} + ({}^{30}{C_6} + {}^{30}{C_7}) - {}^{30}{C_7}$$<br><br>$$ = {}^{50}{C_6} + {}^{49}{C_6} + .... + ({}^{31}{C_6} + {}^{31}{C_7}) - {}^{30}{C_7}$$<br><br>$$ = {}^{50}{C_6} + {}^{50}{C_7} - {}^{30}{C_7}$$<br><br>$$ = {}^{51}{C_7} - {}^{30}{C_7}$$<br><br>[As $${{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}}$$] | mcq | jee-main-2020-online-4th-september-morning-slot |
Hs8F4tUpgblToosyEojgy2xukezfp3c8 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | For a positive integer n,
<br/>$${\left( {1 + {1 \over x}} \right)^n}$$ is expanded
<br/>in increasing powers of x. If three consecutive
<br/>coefficients in this expansion are in the ratio,
<br/>2 : 5 : 12, then n is equal to________. | [] | null | 118 | Let, three consecutive coefficients are<br><br>
$${}^n{C_{r - 1}},{}^n{C_r},{}^n{C_{r + 1}}$$<br><br>
$${}^n{C_{r - 1}}:{}^n{C_r}:{}^n{C_{r + 1}} = 2:5:12$$<br><br>
Now, $${{{}^n{C_{r - 1}}} \over {{}^n{C_r}}} = {2 \over 5}$$<br><br>
$$ \Rightarrow 7r = 2n + 2$$ ...(i)<br><br>
$${{{}^n{C_r}} \over {{}^n{C_{r + 1}}}} = {5 \over {12}}$$<br><br>
$$ \Rightarrow 7r = 5n - 12$$ ...(ii)<br><br>
On solving (i) and (ii) we get n = 118
| integer | jee-main-2020-online-2nd-september-evening-slot |
ktfWIoaLo6BVZrCGuO7k9k2k5ki9hvm | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If C<sub>r</sub> $$ \equiv $$ <sup>25</sup>C<sub>r</sub> and<br/>
C<sub>0</sub> + 5.C<sub>1</sub> + 9.C<sub>2</sub> + .... + (101).C<sub>25</sub> = 2<sup>25</sup>.k, then k is equal to _____. | [] | null | 51 | S = 1.<sup>25</sup>C<sub>0</sub> + 5.<sup>25</sup>C<sub>1</sub> + 9.<sup>25</sup>C<sub>2</sub> + .... + (101)<sup>25</sup>C<sub>25</sub>
<br><br>S = (101).<sup>25</sup>C<sub>25</sub> + (97).<sup>25</sup>C<sub>24</sub> + .......... + (1).<sup>25</sup>C<sub>0</sub>
<br>_________________________________________
<br><br>2S = 102{<sup>25</sup>C<sub>0</sub> + <sup>25</sup>C<sub>1</sub> + ......+ <sup>25</sup>C<sub>25</sub>}
<br><br>$$ \Rightarrow $$ S = 51 $$ \times $$ 2<sup>25</sup> = k.2<sup>25</sup>
<br><br>$$ \Rightarrow $$ k = 51 | integer | jee-main-2020-online-9th-january-evening-slot |
dwSgKUsyRIPv9cmnaZ7k9k2k5e4ld1s | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If the sum of the coefficients of all even powers of x in the product<br/> (1 + x + x<sup>2</sup>
+ ....+ x<sup>2n</sup>)(1 - x + x<sup>2</sup> - x<sup>3</sup> + ...... + x<sup>2n</sup>) is 61, then n is equal to _______. | [] | null | 30 | (1 + x + x<sup>2</sup>
+ ....+ x<sup>2n</sup>)(1 - x + x<sup>2</sup> - x<sup>3</sup> + ...... + x<sup>2n</sup>)
<br><br>= a<sub>0</sub> + a<sub>1</sub>x + a<sub>2</sub>x<sup>2</sup>
+ …..
<br><br>put x = 1
<br><br> (2n + 1)$$ \times $$1 = a<sub>0</sub> + a<sub>1</sub> + a<sub>2</sub> + …… (1)
<br><br>put x = –1
<br><br> 1$$ \times $$(2n + 1) = a<sub>0</sub> – a<sub>1</sub> + a<sub>2</sub>+ …….. (2)
<br><br>Adding (1) and (2)
<br><br>4n + 2 = 2(a<sub>0</sub> + a<sub>2</sub> + ….. )
<br><br>$$ \Rightarrow $$ 4n + 2 = 2 $$ \times $$ 61
<br><br>$$ \Rightarrow $$ n = 30 | integer | jee-main-2020-online-7th-january-morning-slot |
9WFnigdEpxBY7O7zuf1klrfi0p3 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The value of
<br/>-<sup>15</sup>C<sub>1</sub> + 2.<sup>15</sup>C<sub>2</sub> – 3.<sup>15</sup>C<sub>3</sub> + ... - 15.<sup>15</sup>C<sub>15</sub> + <sup>14</sup>C<sub>1</sub> + <sup>14</sup>C<sub>3</sub> + <sup>14</sup>C<sub>5</sub> + ...+ <sup>14</sup>C<sub>11</sub> is : | [{"identifier": "A", "content": "2<sup>13</sup> - 13 "}, {"identifier": "B", "content": "2<sup>16</sup> - 1"}, {"identifier": "C", "content": "2<sup>14</sup>"}, {"identifier": "D", "content": "2<sup>13</sup> - 14"}] | ["D"] | null | $$ - {}^{15}{C_1} + 2.{}^{15}{C_2} - 3.{}^{15}{C_3} + ....\,. - 15.{}^{15}{C_{15}}$$<br><br>$$ = \sum\limits_{r = 1}^{15} {{{( - 1)}^r}.r.{}^{15}{C_r}} $$<br><br>$$ = \sum\limits_{r = 1}^{15} {{{( - 1)}^2}.r.{{15} \over r}} .{}^{14}{C_{r - 1}}$$<br><br>$$ = 15\sum\limits_{r = 1}^{15} {{{( - 1)}^2}.{}^{14}{C_{r - 1}}} $$<br><br>$$ = 15\left( { - {}^{14}{C_0} + {}^{14}{C_1} - {}^{14}{C_2} + .... - {}^{14}{C_{14}}} \right)$$<br><br>$$ = 15(0) = 0$$<br><br>We know,<br><br>$$ \Rightarrow $$2<sup>14 - 1</sup>$$ = {}^{14}{C_1} + {}^{14}{C_3} + {}^{14}{C_5} .... + {}^{14}{C_{13}}$$
<br><br>$$ \Rightarrow $$2<sup>13</sup>$$ = {}^{14}{C_1} + {}^{14}{C_3} + {}^{14}{C_5} .... + {}^{14}{C_{13}}$$
<br><br>Also let, S = <sup>14</sup>C<sub>1</sub> + <sup>14</sup>C<sub>3</sub> + <sup>14</sup>C<sub>5</sub> + ...+ <sup>14</sup>C<sub>11</sub>
<br><br>$$ \Rightarrow $$ S + <sup>14</sup>C<sub>13</sub> = <sup>14</sup>C<sub>1</sub> + <sup>14</sup>C<sub>3</sub> + <sup>14</sup>C<sub>5</sub> + ...+ <sup>14</sup>C<sub>11</sub> + <sup>14</sup>C<sub>13</sub>
<br><br>$$ \Rightarrow $$ S + <sup>14</sup>C<sub>13</sub> = 2<sup>13</sup>
<br><br>$$ \Rightarrow $$ S + 14 = 2<sup>13</sup>
<br><br>$$ \Rightarrow $$ S = 2<sup>13</sup> - 14
<br><br><b>Other Method :</b><br><br>We know, $${(1 - x)^{15}} = {}^{15}{C_0} - {}^{15}{C_1}x + {}^{15}{C_2}{x^2} - ..... - {}^{15}{C_{15}}{x^{15}}$$<br><br>Differentiating both sides with respect to x, <br><br>$$15{(1 - x)^{14}}( - 1) = - {}^{15}{C_1} + 2{}^{15}{C_2}x - 3{}^{15}{C_3}{x^2} + ....... - 15{}^{15}{C_{15}}{x^{14}}$$<br><br>Put $$x = 1$$<br><br>$$ \Rightarrow 0 = - {}^{15}{C_1} + 2{}^{15}{C_2} - 3{}^{15}{C_3} + .... - 15{}^{15}{C_{15}}$$
<br><br>We know,<br><br>$$ \Rightarrow $$2<sup>14 - 1</sup>$$ = {}^{14}{C_1} + {}^{14}{C_3} + {}^{14}{C_5} .... + {}^{14}{C_{13}}$$
<br><br>$$ \Rightarrow $$2<sup>13</sup>$$ = {}^{14}{C_1} + {}^{14}{C_3} + {}^{14}{C_5} .... + {}^{14}{C_{13}}$$
<br><br>Also let, S = <sup>14</sup>C<sub>1</sub> + <sup>14</sup>C<sub>3</sub> + <sup>14</sup>C<sub>5</sub> + ...+ <sup>14</sup>C<sub>11</sub>
<br><br>$$ \Rightarrow $$ S + <sup>14</sup>C<sub>13</sub> = <sup>14</sup>C<sub>1</sub> + <sup>14</sup>C<sub>3</sub> + <sup>14</sup>C<sub>5</sub> + ...+ <sup>14</sup>C<sub>11</sub> + <sup>14</sup>C<sub>13</sub>
<br><br>$$ \Rightarrow $$ S + <sup>14</sup>C<sub>13</sub> = 2<sup>13</sup>
<br><br>$$ \Rightarrow $$ S + 14 = 2<sup>13</sup>
<br><br>$$ \Rightarrow $$ S = 2<sup>13</sup> - 14 | mcq | jee-main-2021-online-24th-february-morning-slot |
0xkP9OO8ZCBXcnBDiN1klrlf9sl | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If $$n \ge 2$$ is a positive integer, then the sum of the series $${}^{n + 1}{C_2} + 2\left( {{}^2{C_2} + {}^3{C_2} + {}^4{C_2} + ... + {}^n{C_2}} \right)$$ is : | [{"identifier": "A", "content": "$${{n(2n + 1)(3n + 1)} \\over 6}$$"}, {"identifier": "B", "content": "$${{n(n + 1)(2n + 1)} \\over 6}$$"}, {"identifier": "C", "content": "$${{n{{(n + 1)}^2}(n + 2)} \\over {12}}$$"}, {"identifier": "D", "content": "$${{n(n - 1)(2n + 1)} \\over 6}$$"}] | ["B"] | null | $${}^{n + 1}{C_2} + 2\left( {{}^2{C_2} + {}^3{C_2} + {}^4{C_2} + ........ + {}^n{C_2}} \right)$$<br><br>$${}^{n + 1}{C_2} + 2\left( {{}^3{C_2} + {}^3{C_2} + {}^4{C_2} + ........ + {}^n{C_2}} \right)$$<br><br>use $$\left\{ {{}^n{C_{r + 1}} + {}^n{C_r} = {}^{n + 1}{C_r}} \right\}$$<br><br>$$ = {}^{n + 1}{C_2} + 2\left( {{}^4{C_3} + {}^4{C_2} + {}^5{C_3} + ........ + {}^n{C_2}} \right)$$<br><br>$$ = {}^{n + 1}{C_2} + 2\left( {{}^5{C_3} + {}^5{C_2} + ........ + {}^n{C_2}} \right)$$<br><br>$$\eqalign{
& . \cr
& . \cr
& . \cr
& . \cr
& . \cr} $$<br><br>$$ = {}^{n + 1}{C_2} + 2\left( {{}^n{C_3} + {}^n{C_2}} \right)$$<br><br>$$ = {}^{n + 1}{C_2} + 2.{}^{n + 1}{C_3}$$<br><br>$$ = {{(n + 1)n} \over 2} + 2.{{(n + 1)(n)(n - 1)} \over {2.3}}$$<br><br>$$ = {{n(n + 1)(2n + 1)} \over 6}$$ | mcq | jee-main-2021-online-24th-february-evening-slot |
vof8AROKUVRxEr6t151klrmvqnc | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | For integers n and r, let $$\left( {\matrix{
n \cr
r \cr
} } \right) = \left\{ {\matrix{
{{}^n{C_r},} & {if\,n \ge r \ge 0} \cr
{0,} & {otherwise} \cr
} } \right.$$ The maximum value of k for which the sum $$\sum\limits_{i = 0}^k {\left( {\matrix{
{10} \cr
i \cr
} } \right)\left( {\matrix{
{15} \cr
{k - i} \cr
} } \right)} + \sum\limits_{i = 0}^{k + 1} {\left( {\matrix{
{12} \cr
i \cr
} } \right)\left( {\matrix{
{13} \cr
{k + 1 - i} \cr
} } \right)} $$ exists, is equal to _________. | [] | null | 12 | As k is unbounded so maximum value is not defined.
<br><br>Question will be <b>BONUS</b>. | integer | jee-main-2021-online-24th-february-evening-slot |
tijFbckj71imHUIQj71kluhqs3z | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | Let m, n$$\in$$N and gcd (2, n) = 1. If $$30\left( {\matrix{
{30} \cr
0 \cr
} } \right) + 29\left( {\matrix{
{30} \cr
1 \cr
} } \right) + ...... + 2\left( {\matrix{
{30} \cr
{28} \cr
} } \right) + 1\left( {\matrix{
{30} \cr
{29} \cr
} } \right) = n{.2^m}$$, then n + m is equal to __________.<br/><br/>(Here $$\left( {\matrix{
n \cr
k \cr
} } \right) = {}^n{C_k}$$) | [] | null | 45 | $$30({}^{30}{C_0}) + 29({}^{30}{C_1}) + .... + 2({}^{30}{C_{28}}) + 1({}^{30}{C_{29}})$$<br><br>$$ = 30({}^{30}{C_{30}}) + 29({}^{30}{C_{29}}) + ...... + 2({}^{30}{C_2}) + 1({}^{30}{C_1})$$<br><br>$$ = \sum\limits_{r = 1}^{30} {r({}^{30}{C_r})} $$<br><br>$$ = \sum\limits_{r = 1}^{30} {r\left( {{{30} \over r}} \right)({}^{29}{C_{r - 1}}} )$$<br><br>$$ = 30\sum\limits_{r = 1}^{30} {{}^{29}{C_{r - 1}}} $$<br><br>$$ = 30({}^{29}{C_0} + {}^{29}{C_1} + {}^{29}{C_2} + ..... + {}^{29}{C_{29}})$$<br><br>$$ = 30({2^{29}}) = 15{(2)^{30}} = n{(2)^m}$$<br><br>$$ \therefore $$ n = 15, m = 30<br><br>$$ \Rightarrow $$ n + m = 45 | integer | jee-main-2021-online-26th-february-morning-slot |
k0jFMPAB7RIVNOTskF1kmhxhsc1 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | Let [ x ] denote greatest integer less than or equal to x. If for n$$\in$$N, <br/><br/>$${(1 - x + {x^3})^n} = \sum\limits_{j = 0}^{3n} {{a_j}{x^j}} $$, <br/><br/>then $$\sum\limits_{j = 0}^{\left[ {{{3n} \over 2}} \right]} {{a_{2j}} + 4} \sum\limits_{j = 0}^{\left[ {{{3n - 1} \over 2}} \right]} {{a_{2j}} + 1} $$ is equal to : | [{"identifier": "A", "content": "2<sup>n $$-$$ 1</sup>"}, {"identifier": "B", "content": "n"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}] | ["C"] | null | $${(1 - x + {x^3})^n} = \sum\limits_{j = 0}^{3n} {{a_j}{x^j}} $$<br><br>$$(1 - x + {x^3}) = {a_0} + {a_1}x + {a_2}{x^2} + ...... + {a_{3n}}{x^{3n}}$$<br><br>Put x = 1<br><br>$$1 = {a_0} + {a_1} + {a_2} + {a_3} + {a_4} + ........ + {a_{3n}}$$ ...... (1)<br><br>Put x = $$-$$1<br><br>$$1 = {a_0} - {a_1} + {a_2} - {a_3} + {a_4} + ........( - 1){}^{3n}{a_{3n}}$$ ..... (2)<br><br>Add (1) + (2)<br><br>$$ \Rightarrow {a_0} + {a_2} + {a_4} + {a_6} + ...... = 1$$<br><br>Sub (1) $$-$$ (2)<br><br>$$ \Rightarrow {a_1} + {a_3} + {a_5} + {a_7} + ...... = 0$$<br><br>Now, $$\sum\limits_{j = 0}^{\left[ {{{3n} \over 2}} \right]} {{a_{2j}}} + 4\sum\limits_{j = 0}^{\left[ {{{3n - 1} \over 2}} \right]} {{a_{2j }}} + 1 $$<br><br>$$ = ({a_0} + {a_2} + {a_4} + ......) + 4({a_1} + {a_3} + .....)$$<br><br>$$ = 1 + 4 \times 0$$ + 1<br><br>$$ = 1 + 1 = 2$$ | mcq | jee-main-2021-online-16th-march-morning-shift |
UU75IzxYFunjb8MhY91kmizvwyk | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | Let n be a positive integer. Let <br/><br/>$$A = \sum\limits_{k = 0}^n {{{( - 1)}^k}{}^n{C_k}\left[ {{{\left( {{1 \over 2}} \right)}^k} + {{\left( {{3 \over 4}} \right)}^k} + {{\left( {{7 \over 8}} \right)}^k} + {{\left( {{{15} \over {16}}} \right)}^k} + {{\left( {{{31} \over {32}}} \right)}^k}} \right]} $$. If <br/><br/>$$63A = 1 - {1 \over {{2^{30}}}}$$, then n is equal to _____________. | [] | null | 6 | $$A = \sum {{{( - 1)}^k}{}^n{C_k}{{\left( {{1 \over 2}} \right)}^k}} + \sum {{{( - 1)}^k}{}^n{C_k}{{\left( {{3 \over 4}} \right)}^k}} + .....$$<br><br>$$ = {\left( {1 - {1 \over 2}} \right)^n} + {\left( {1 - {3 \over 4}} \right)^n} + ..... + {\left( {1 - {{31} \over {32}}} \right)^n}$$<br><br>$$ = {\left( {{1 \over 2}} \right)^n} + {\left( {{1 \over 2}} \right)^{2n}} + {\left( {{1 \over 2}} \right)^{3n}} + ..... + {\left( {{1 \over 2}} \right)^{5n}}$$<br><br>$$ = {\left( {{1 \over 2}} \right)^n}\left( {{{1 - {{\left( {{1 \over 2}} \right)}^{5n}}} \over {1 - {{\left( {{1 \over 2}} \right)}^n}}}} \right) = {{{2^{5n}} - 1} \over {{2^{5n}}({2^n} - 1)}}$$
<br><br>$$ \therefore $$ $$63A = {{63\left( {{2^{5n}} - 1} \right)} \over {{2^{5n}}\left( {{2^n} - 1} \right)}}$$ = $${{63} \over {\left( {{2^n} - 1} \right)}}\left( {1 - {1 \over {{2^{5n}}}}} \right)$$
<br><br>Given, $$63A = 1 - {1 \over {{2^{30}}}}$$
<br><br>$$ \therefore $$ $${{63} \over {\left( {{2^n} - 1} \right)}}\left( {1 - {1 \over {{2^{5n}}}}} \right)$$ = $$1 - {1 \over {{2^{30}}}}$$
<br><br>For n = 6, L.H.S = R.H.S
<br><br>$$ \therefore $$ n = 6 | integer | jee-main-2021-online-16th-march-evening-shift |
VOvssU7YRUG7n4l6u61kmknei42 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The value of $$\sum\limits_{r = 0}^6 {\left( {{}^6{C_r}\,.\,{}^6{C_{6 - r}}} \right)} $$ is equal to : | [{"identifier": "A", "content": "924"}, {"identifier": "B", "content": "1024"}, {"identifier": "C", "content": "1124"}, {"identifier": "D", "content": "1324"}] | ["A"] | null | Given,<br><br>$$\sum\limits_{r = 0}^6 {{}^6{C_r}{}^6{C_{6 - r}}} $$
<br><br>= $${}^6{C_0}.{}^6{C_6} + {}^6{C_1}.{}^6{C_5} + ... + {}^6{C_6}.{}^6{C_0}$$
<br><br>Now,
<br><br>$$\eqalign{
& = \left( {{}^6{C_0} + {}^6{C_1}x + {}^6{C_2}{x^2} + ... + {}^6{C_6}{x^6}} \right) \cr
& \left( {{}^6{C_0} + {}^6{C_1}x + {}^6{C_2}{x^2} + ... + {}^6{C_6}{x^6}} \right) \cr} $$
<br><br>Comparing coefficient of x<sup>6</sup> both sides
<br><br>$${}^6{C_0}.{}^6{C_6} + {}^6{C_1}.{}^6{C_5} + ... + {}^6{C_6}.{}^6{C_0}$$
<br><br>= $${}^{12}{C_6}$$ = 924 | mcq | jee-main-2021-online-17th-march-evening-shift |
pMbhKHUGvmcFAk55Hp1kmli04du | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | Let (1 + x + 2x<sup>2</sup>)<sup>20</sup> = a<sub>0</sub> + a<sub>1</sub>x + a<sub>2</sub>x<sup>2</sup> + .... + a<sub>40</sub>x<sup>40</sup>. Then a<sub>1</sub> + a<sub>3</sub> + a<sub>5</sub> + ..... + a<sub>37</sub> is equal to | [{"identifier": "A", "content": "2<sup>20</sup>(2<sup>20</sup> $$-$$ 21)"}, {"identifier": "B", "content": "2<sup>19</sup>(2<sup>20</sup> $$-$$ 21)"}, {"identifier": "C", "content": "2<sup>19</sup>(2<sup>20</sup> $$+$$ 21)"}, {"identifier": "D", "content": "2<sup>20</sup>(2<sup>20</sup> $$+$$ 21)"}] | ["B"] | null | $${(1 + x + 2{x^2})^{20}} = {a_0} + {a_1}x + {a_2}{x^2} + .... + {a_{40}}{x^{40}}$$<br><br>Put x = 1<br><br>$$ \Rightarrow {4^{20}} = {a_0} + {a_1} + ....... + {a_{40}}$$ ..... (i)<br><br>Put x = $$-$$1<br><br>$$ \Rightarrow {2^{20}} = {a_0} - {a_1} + ....... + - {a_{39}} + {a_{40}}$$ ..... (ii)<br><br>by (i) $$-$$ (ii) we get,<br><br>$${4^{20}} - {2^{20}} = 2({a_1} + {a_3} + ...... + {a_{37}} + {a_{39}})$$<br><br>$$ \Rightarrow {a_1} + {a_3} + ...... + {a_{37}} = {2^{39}} - {2^{19}} - {a_{39}}$$ ..... (iii)<br><br>a<sub>39</sub> = coeff. x<sup>39</sup> in (1 + x + 2x<sup>2</sup>)<sup>20</sup><br><br>$$ = {{20!} \over {0!1!9!}}{(1)^0}{(1)^1}{(2)^{19}}$$<br><br>$$ = {20.2^{19}}$$<br><br>$$ \therefore $$ $${a_1} + {a_3} + ...... + {a_{37}} = {2^{39}} - {2^{19}}.21$$<br><br>$$ \Rightarrow {2^{19}}({2^{20}} - 21)$$ | mcq | jee-main-2021-online-18th-march-morning-shift |
n2yplWu9PmN2Y3682V1kmm4cd5o | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | Let $${}^n{C_r}$$ denote the binomial coefficient of x<sup>r</sup> in the expansion of (1 + x)<sup>n</sup>.
If $$\sum\limits_{k = 0}^{10} {({2^2} + 3k)} {}^{10}{C_k} = \alpha {.3^{10}} + \beta {.2^{10}},\alpha ,\beta \in R$$, then $$\alpha$$ + $$\beta$$ is equal to ___________. | [] | null | 19 | $$\sum\limits_{k = 0}^{10} {({2^2} + 3k){}^{10}{C_k}} $$<br><br>$$ = 4\sum\limits_{k = 0}^{10} {{}^{10}{C_k}} + 3\sum\limits_{k = 0}^{10} {k.{}^{10}{C_k}} $$<br><br>$$ = 4({2^{10}}) + 3\sum\limits_{k = 0}^{10} {k.{{10} \over k}.{}^9{C_{k - 1}}} $$<br><br>= $$4({2^{10}}) + 3.10({2^9})$$<br><br>$$ = 4({2^{10}}) + {3.5.2^{10}}$$<br><br>$$ = {2^{10}}(19)$$<br><br>According to question,<br><br>$$19({2^{10}}) = \alpha {.3^{10}} + \beta {.2^{10}}$$<br><br>$$ \therefore $$ $$\alpha = 0,\beta = 19$$<br><br>$$ \Rightarrow \alpha + \beta = 19$$ | integer | jee-main-2021-online-18th-march-evening-shift |
1krpw3j81 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The coefficient of x<sup>256</sup> in the expansion of <br/><br/>(1 $$-$$ x)<sup>101</sup> (x<sup>2</sup> + x + 1)<sup>100</sup> is : | [{"identifier": "A", "content": "$${}^{100}{C_{16}}$$"}, {"identifier": "B", "content": "$${}^{100}{C_{15}}$$"}, {"identifier": "C", "content": "$$-$$ $${}^{100}{C_{16}}$$"}, {"identifier": "D", "content": "$$-$$ $${}^{100}{C_{15}}$$"}] | ["B"] | null | $${(1 - x)^{101}}{({x^2} + x + 1)^{100}}$$<br/><br/>Coefficient of $${x^{256}} = {[(1 - x)(1 + x + {x^2})]^{100}}(1 - x) = {(1 - {x^3})^{100}}(1 - x)$$<br/><br/>$$ \Rightarrow ({}^{100}{C_0} - {}^{100}{C_1}{x^3} + {}^{100}{C_2}{x^6} - {}^{100}{C_3}{x^9}...)(1 - x)$$<br/><br/>$$\sum {{{( - 1)}^r}{}^{100}{C_r}{x^{3r}}(1 - x)} $$<br/><br/>$$ \Rightarrow 3r = 256$$ or $$255 \Rightarrow r = {{256} \over 3}$$ (Reject)<br/><br/>r = 85<br/><br/>Coefficient = $${}^{100}{C_{85}} = {}^{100}{C_{15}}$$ | mcq | jee-main-2021-online-20th-july-morning-shift |
1krrogpc3 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | For the natural numbers m, n, if $${(1 - y)^m}{(1 + y)^n} = 1 + {a_1}y + {a_2}{y^2} + .... + {a_{m + n}}{y^{m + n}}$$ and $${a_1} = {a_2} = 10$$, then the value of (m + n) is equal to : | [{"identifier": "A", "content": "88"}, {"identifier": "B", "content": "64"}, {"identifier": "C", "content": "100"}, {"identifier": "D", "content": "80"}] | ["D"] | null | $${(1 - y)^m}{(1 + y)^n} = 1 + {a_1}y + {a_2}{y^2} + .... + {a_{m + n}}{y^{m + n}}$$<br><br>Given, ($${a_1} = {a_2} = 10$$)$$(1 - my + {}^m{C_2}{y^2} + .....)(1 + ny + {}^n{C_2}{y^2} + .....) = 1 + {a_1}y + {a_2}{y^2} + ....$$<br><br>$$ \Rightarrow n - m = 10$$ ..... (i)<br><br>$$ \Rightarrow {}^m{C_2} + {}^n{C_2} - mn = 10$$...... (ii)<br><br>$${{m(m - 1)} \over 2} + {{n(n - 1)} \over 2} - mn = 10$$<br><br>$$ \Rightarrow {{{m^2} - m} \over 2} + {{(10 + m)(9 + m)} \over 2} - m(10 + m) = 10$$<br><br>$$ \Rightarrow {m^2} - m + {m^2} + 19m + 90 - 2({m^2} + 10m) = 20$$<br><br>$$ \Rightarrow 18m + 90 - 20m = 20$$<br><br>$$ \Rightarrow 2m = 70$$<br><br>$$ \Rightarrow m = 35$$ & $$n = 45$$<br><br>$$m + n = 80$$ | mcq | jee-main-2021-online-20th-july-evening-shift |
1kruc0v6x | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | The number of elements in the set {n $$\in$$ {1, 2, 3, ......., 100} | (11)<sup>n</sup> > (10)<sup>n</sup> + (9)<sup>n</sup>} is ______________. | [] | null | 96 | $${11^n} > {10^n} + {9^n}$$<br><br>$$ \Rightarrow {11^n} - {9^n} > {10^n}$$<br><br>$$ \Rightarrow {(10 + 1)^n} - {(10 - 1)^n} > {10^n}$$<br><br>$$ \Rightarrow 2\{ {}^n{C_1}{.10^{n - 1}} + {}^n{C_3}{10^{n - 10}} + {}^n{C_5}{10^{n - 5}} + .....\} > {10^n}$$
<br><br>$$ \Rightarrow $$ $${1 \over 5}\left[ {{}^n{C_1}{{10}^n} + {}^n{C_3}{{10}^{n - 2}} + {}^n{C_5}{{10}^{n - 4}} + .....} \right] > {10^n}$$
<br><br>$$ \Rightarrow $$ $${1 \over 5}\left[ {{}^n{C_1} + {}^n{C_3}{{10}^{ - 2}} + {}^n{C_5}{{10}^{ - 4}} + .....} \right] > 1$$
<br><br>Clearly the above inequality is true for n $$ \ge $$ 5
<br><br>For n = 4, we have $${1 \over 5}\left[ {4 + {4 \over {{{10}^2}}}} \right] = {4 \over 5}\left( {{{101} \over {100}}} \right) < 1$$
<br><br>$$\Rightarrow$$ Inequality does not hold good for n = 1, 2, 3, 4
<br><br>So, required number of elements ={5, 6, 7, ......., 100} = 96 | integer | jee-main-2021-online-22th-july-evening-shift |
1krzqmv94 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | Let n$$\in$$N and [x] denote the greatest integer less than or equal to x. If the sum of (n + 1) terms $${}^n{C_0},3.{}^n{C_1},5.{}^n{C_2},7.{}^n{C_3},.....$$ is equal to 2<sup>100</sup> . 101, then $$2\left[ {{{n - 1} \over 2}} \right]$$ is equal to _______________. | [] | null | 98 | 1. $${}^n{C_0} + 3.{}^n{C_1} + 5.{}^n{C_2} + ... + (2n + 1).{}^n{C_n}$$<br><br>$${T_r} = (2r + 1){}^n{C_r}$$<br><br>$$S = \sum {{T_r}} $$<br><br>$$S = \sum {(2r + 1){}^n{C_r}} = \sum {2r{}^n{C_r} + \sum {{}^n{C_r}} } $$<br><br>$$S = 2(n{.2^{n - 1}}) + {2^n} = {2^n}(n + 1)$$<br><br>$${2^n}(n + 1) = {2^{100}}.101 \Rightarrow n = 100$$<br><br>$$2\left[ {{{n - 1} \over 2}} \right] = 2\left[ {{{99} \over 2}} \right] = 98$$ | integer | jee-main-2021-online-25th-july-evening-shift |
1ktbck01t | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If $${{}^{20}{C_r}}$$ is the co-efficient of x<sup>r</sup> in the expansion of (1 + x)<sup>20</sup>, then the value of $$\sum\limits_{r = 0}^{20} {{r^2}.{}^{20}{C_r}} $$ is equal to : | [{"identifier": "A", "content": "$$420 \\times {2^{19}}$$"}, {"identifier": "B", "content": "$$380 \\times {2^{19}}$$"}, {"identifier": "C", "content": "$$380 \\times {2^{18}}$$"}, {"identifier": "D", "content": "$$420 \\times {2^{18}}$$"}] | ["D"] | null | $$\sum\limits_{r = 0}^{20} {{r^2}.{}^{20}{C_r}} $$<br><br>$$\sum {(4(r - 1) + r).{}^{20}{C_r}} $$<br><br>$$\sum {r(r - 1).{{20 \times 19} \over {r(r - 1)}}} .{}^{18}{C_r} + r.{{20} \over r}.\sum {{}^{19}{C_{r - 1}}} $$<br><br>$$ \Rightarrow 20 \times {19.2^{18}} + {20.2^{19}}$$<br><br>$$ \Rightarrow 420 \times {2^{18}}$$ | mcq | jee-main-2021-online-26th-august-morning-shift |
1ktd4dk0v | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | Let $$\left( {\matrix{
n \cr
k \cr
} } \right)$$ denotes $${}^n{C_k}$$ and $$\left[ {\matrix{
n \cr
k \cr
} } \right] = \left\{ {\matrix{
{\left( {\matrix{
n \cr
k \cr
} } \right),} & {if\,0 \le k \le n} \cr
{0,} & {otherwise} \cr
} } \right.$$<br/><br/>If $${A_k} = \sum\limits_{i = 0}^9 {\left( {\matrix{
9 \cr
i \cr
} } \right)\left[ {\matrix{
{12} \cr
{12 - k + i} \cr
} } \right] + } \sum\limits_{i = 0}^8 {\left( {\matrix{
8 \cr
i \cr
} } \right)\left[ {\matrix{
{13} \cr
{13 - k + i} \cr
} } \right]} $$ and A<sub>4</sub> $$-$$ A<sub>3</sub> = 190 p, then p is equal to : | [] | null | 49 | $${A_k} = \sum\limits_{i = 0}^9 {{}^9{C_i}} {}^{12}{C_{k - i}} + \sum\limits_{i = 0}^8 {{}^8{C_i}} {}^{13}{C_{k - i}}$$<br><br>$${A_k} = {}^{21}{C_k} + {}^{21}{C_k} = 2.{}^{21}{C_k}$$<br><br>$${A_4} - {A_3} = 2\left( {{}^{21}{C_4} - {}^{21}{C_3}} \right) = 2(5985 - 1330)$$<br><br>$$190p = 2(5985 - 1330) \Rightarrow p = 49$$ | integer | jee-main-2021-online-26th-august-evening-shift |
1ktelqz9h | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | $$\sum\limits_{k = 0}^{20} {{{\left( {{}^{20}{C_k}} \right)}^2}} $$ is equal to : | [{"identifier": "A", "content": "$${}^{40}{C_{21}}$$"}, {"identifier": "B", "content": "$${}^{40}{C_{19}}$$"}, {"identifier": "C", "content": "$${}^{40}{C_{20}}$$"}, {"identifier": "D", "content": "$${}^{41}{C_{20}}$$"}] | ["C"] | null | $$\sum\limits_{k = 0}^{20} {{{\left( {{}^{20}{C_k}} \right)}^2}} $$
<br><br>= $${\left( {{}^{20}{C_0}} \right)^2} + {\left( {{}^{20}{C_1}} \right)^2} + {\left( {{}^{20}{C_2}} \right)^2} + .... + {\left( {{}^{20}{C_{20}}} \right)^2}$$
<br><br>= <sup>40</sup>C<sub>20</sub>
<br><br><b>Using the formula :</b>
<br><br>$${\left( {{}^n{C_0}} \right)^2} + {\left( {{}^n{C_1}} \right)^2} + {\left( {{}^n{C_2}} \right)^2} + .... + {\left( {{}^n{C_n}} \right)^2} = {}^{2n}{C_n}$$ | mcq | jee-main-2021-online-27th-august-morning-shift |
1ktobm399 | 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 (x + y)<sup>n</sup> is 4096, then the greatest coefficient in the expansion is _____________. | [] | null | 924 | (x + y)<sup>n</sup> $$\Rightarrow$$ 2<sup>n</sup> = 4096<br><br>2<sup>10</sup> = 1024 $$\times$$ 2<br><br>$$\Rightarrow$$ 2<sup>n</sup> = 2<sup>12</sup><br><br>2<sup>11</sup> = 2048<br><br>n = 12<br><br>2<sup>12</sup> = 4096<br><br>$${}^{12}{C_6}={{12 \times 11 \times 10 \times 9 \times 8 \times 7} \over {6 \times 5 \times 4 \times 3 \times 2 \times 1}}$$<br><br>$$ = 11 \times 3 \times 4 \times 7$$<br><br>$$ = 924$$ | integer | jee-main-2021-online-1st-september-evening-shift |
1l54tl51c | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>Let n $$\ge$$ 5 be an integer. If 9<sup>n</sup> $$-$$ 8n $$-$$ 1 = 64$$\alpha$$ and 6<sup>n</sup> $$-$$ 5n $$-$$ 1 = 25$$\beta$$, then $$\alpha$$ $$-$$ $$\beta$$ is equal to</p> | [{"identifier": "A", "content": "1 + <sup>n</sup>C<sub>2</sub> (8 $$-$$ 5) + <sup>n</sup>C<sub>3</sub> (8<sup>2</sup> $$-$$ 5<sup>2</sup>) + ...... + <sup>n</sup>C<sub>n</sub> (8<sup>n $$-$$ 1</sup> $$-$$ 5<sup>n $$-$$ 1</sup>)"}, {"identifier": "B", "content": "1 + <sup>n</sup>C<sub>3</sub> (8 $$-$$ 5) + <sup>n</sup>C<sub>4</sub> (8<sup>2</sup> $$-$$ 5<sup>2</sup>) + ...... + <sup>n</sup>C<sub>n</sub> (8<sup>n $$-$$ 2</sup> $$-$$ 5<sup>n $$-$$ 2</sup>)"}, {"identifier": "C", "content": "<sup>n</sup>C<sub>3</sub> (8 $$-$$ 5) + <sup>n</sup>C<sub>4</sub> (8<sup>2</sup> $$-$$ 5<sup>2</sup>) + ...... + <sup>n</sup>C<sub>n</sub> (8<sup>n $$-$$ 2</sup> $$-$$ 5<sup>n $$-$$ 2</sup>)"}, {"identifier": "D", "content": "<sup>n</sup>C<sub>4</sub> (8 $$-$$ 5) + <sup>n</sup>C<sub>5</sub> (8<sup>2</sup> $$-$$ 5<sup>2</sup>) + ...... + <sup>n</sup>C<sub>n</sub> (8<sup>n $$-$$ 3</sup> $$-$$ 5<sup>n $$-$$ 3</sup>)"}] | ["C"] | null | <p>Given,</p>
<p>$${9^n} - 8n - 1 = 64\alpha $$</p>
<p>$$ \Rightarrow \alpha = {{{{(1 + 8)}^n} - 8n - 1} \over {64}}$$</p>
<p>$$ = {{\left( {{}^n{C_0}\,.\,1 + {}^n{C_1}\,.\,{8^1} + {}^n{C_2}\,.\,{8^2}\,\, + \,\,.....\,\, + \,\,{}^n{C_n}\,.\,{8^n}} \right) - 8n - 1} \over {{8^2}}}$$</p>
<p>$$ = {{1 + 8n + {}^n{C_2}\,.\,{8^2}\,\, + \,\,....\,\, + \,\,{}^n{C_n}\,.\,{8^n} - 8n - 1} \over {{8^2}}}$$</p>
<p>$$ = {{{}^n{C_2}\,.\,{8^2} + {}^n{C_3}\,.\,{8^3}\,\, + \,\,.....\,\, + \,\,{}^n{C_n}\,.\,{8^n}} \over {{8^2}}}$$</p>
<p>$$ = {}^n{C_2} + {}^n{C_3}\,.\,8 + {}^n{C_4}\,.\,{8^2}\,\, + \,\,.....\,\,{}^n{C_n}\,.\,{8^{n - 2}}$$</p>
<p>Also given,</p>
<p>$${6^n} - 5n - 1 = 25\beta $$</p>
<p>$$ \Rightarrow \beta = {{{{(1 + 5)}^n} - 5n - 1} \over {25}}$$</p>
<p>$$ = {{{}^n{C_0}\,.\,1 + {}^n{C_1}\,.\,5 + {}^n{C_2}\,.\,{5^2}\,\, + \,\,.....\,\, + \,\,{}^n{C_n}\,.\,{5^n} - 5n - 1} \over {{5^2}}}$$</p>
<p>$$ = {{1 + 5n + {}^n{C_2}\,.\,{5^2} + {}^n{C_3}\,.\,{5^3}\,\, + \,\,....\,\, + \,\,{}^n{C_n}\,.\,{5^2} - 5n - 1} \over {{5^2}}}$$</p>
<p>$$ = {{{}^n{C_2}\,.\,{5^2} + {}^n{C_3}\,.\,{5^3} + {}^n{C_4}\,.\,{5^4}\,\, + \,\,....\,\, + \,\,{}^n{C_n}\,.\,{5^n}} \over {{5^2}}}$$</p>
<p>$$ = {}^n{C_2} + {}^n{C_3}\,.\,5 + {}^n{C_4}\,.\,{5^2}\,\, + \,\,.....\,\, + \,\,{}^n{C_n}\,.\,{5^{n - 2}}$$</p>
<p>$$\therefore$$ $$\alpha - \beta $$</p>
<p>$$ = \left( {{}^n{C_2} + {}^n{C_3}\,.\,8 + {}^n{C_4}\,.\,{8^2}\,\, + \,\,....\,\, + \,\,{}^n{C_n}\,.\,{8^{n - 2}}} \right) - \left( {{}^n{C_2} + {}^n{C_3}\,.\,5 + {}^n{C_4}\,.\,{5^2}\,\, + \,\,....\,\, + \,\,{}^n{C_n}\,.\,{5^{n - 2}}} \right)$$</p>
<p>$$ = {}^n{C_3}\,.\,(8 - 5) + {}^n{C_4}\,.\,({8^2} - {5^2})\,\, + \,\,....\,\, + \,\,{}^n{C_n}({8^{n - 2}} - {5^{n - 2}})$$</p> | mcq | jee-main-2022-online-29th-june-evening-shift |
1l5662wlg | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>If <br/><br/>$$\sum\limits_{k = 1}^{31} {\left( {{}^{31}{C_k}} \right)\left( {{}^{31}{C_{k - 1}}} \right) - \sum\limits_{k = 1}^{30} {\left( {{}^{30}{C_k}} \right)\left( {{}^{30}{C_{k - 1}}} \right) = {{\alpha (60!)} \over {(30!)(31!)}}} } $$, <br/><br/>where $$\alpha$$ $$\in$$ R, then the value of 16$$\alpha$$ is equal to</p> | [{"identifier": "A", "content": "1411"}, {"identifier": "B", "content": "1320"}, {"identifier": "C", "content": "1615"}, {"identifier": "D", "content": "1855"}] | ["A"] | null | <p>Given,</p>
<p>$$\sum\limits_{k = 1}^{31} {\left( {{}^{31}{C_k}} \right)\left( {{}^{31}{C_{k - 1}}} \right) - \sum\limits_{k = 1}^{30} {\left( {{}^{30}{C_k}} \right)\left( {{}^{30}{C_{k - 1}}} \right) = {{\alpha (60!)} \over {(30!)(31!)}}} } $$</p>
<p>Now,</p>
<p>$$\sum\limits_{k = 1}^{31} {\left( {{}^{31}{C_k}} \right)\left( {{}^{31}{C_{k - 1}}} \right)} $$</p>
<p>$$ = \left( {{}^{31}{C_1}\,.\,{}^{31}{C_0} + {}^{31}{C_2}\,.\,{}^{31}{C_1} + {}^{31}{C_3}\,.\,{}^{31}{C_2} + \,\,......\,\, + \,\,{}^{31}{C_{31}}\,.\,{}^{31}{C_{30}}} \right)$$</p>
<p>$$ = \left( {{}^{31}{C_0}\,.\,{}^{31}{C_{31 - 1}} + {}^{31}{C_1}\,.\,{}^{31}{C_{31 - 2}} + \,\,.....\,\, + \,\,{}^{31}{C_{30}}\,.\,{}^{31}{C_{31 - 31}}} \right)$$</p>
<p>[using $${}^n{C_r} = {}^n{C_{n - r}}$$]</p>
<p>$$ = \left( {{}^{31}{C_0}\,.\,{}^{31}{C_{30}} + {}^{31}{C_1}\,.\,{}^{31}{C_{29}} + \,\,.....\,\, + \,\,{}^{31}{C_{30}}\,.\,{}^{31}{C_0}} \right)$$</p>
<p>$$ = {}^{62}{C_{30}}$$</p>
<p>Now, $$\sum\limits_{k = 1}^{30} {{}^{30}{C_k}\,.\,{}^{30}{C_{k - 1}}} $$</p>
<p>$$ = \left( {{}^{30}{C_1}\,.\,{}^{30}{C_0} + {}^{30}{C_2}\,.\,{}^{30}{C_1} + \,\,.....\,\, + \,\,{}^{30}{C_{30}}\,.\,{}^{30}{C_{29}}} \right)$$</p>
<p>$$ = \left( {{}^{30}{C_0}\,.\,{}^{30}{C_{29}} + {}^{30}{C_1}\,.\,{}^{30}{C_{28}} + \,\,....\,\, + \,\,{}^{30}{C_{29}}\,.\,{}^{30}{C_0}} \right)$$</p>
<p>$$ = {}^{60}{C_{29}}$$</p>
<p>$$\therefore$$ $${}^{60}{C_{30}} - {}^{60}{C_{29}} = {{\alpha (60!)} \over {30!\,31!}}$$</p>
<p>$$ \Rightarrow {{62\,.\,61\,.\,60!} \over {30!\,32!}} - {{60!} \over {29!\,31!}} = {{\alpha (60!)} \over {30!\,31!}}$$</p>
<p>$$ \Rightarrow {{62\,.\,61\,.\,60!} \over {30!\,32!}} - {{60!} \over {{{30!} \over {30}}\,.\,31!}} = {{\alpha (60!)} \over {30!\,31!}}$$</p>
<p>$$ \Rightarrow {{60!} \over {30!\,31!}}\left( {{{62\,.\,61} \over {32}} - 30} \right) = {{\alpha (60!)} \over {30!\,31!}}$$</p>
<p>$$ \Rightarrow \alpha = {{62\,.\,61} \over {32}} - 30$$</p>
<p>$$ \Rightarrow 16\alpha = {{62\,.\,61 - 30 \times 32} \over 2}$$</p>
<p>$$ \Rightarrow 16\alpha = {{2822} \over 2} = 1411$$</p> | mcq | jee-main-2022-online-28th-june-morning-shift |
1l56ry0nx | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>If the sum of the coefficients of all the positive powers of x, in the Binomial expansion of $${\left( {{x^n} + {2 \over {{x^5}}}} \right)^7}$$ is 939, then the sum of all the possible integral values of n is _________.</p> | [] | null | 57 | <p>Given, Binomial expression is</p>
<p>$$ = {\left( {{x^n} + {2 \over {{x^5}}}} \right)^7}$$</p>
<p>$$\therefore$$ General term</p>
<p>$${T_{r + 1}} = {}^7{C_r}\,.\,{({x^n})^{7 - r}}\,.\,{\left( {{2 \over {{x^5}}}} \right)^r}$$</p>
<p>$$ = {}^7{C_r}\,.\,{x^{7n - nr - 5r}}\,.\,{2^r}$$</p>
<p>For positive power of x,</p>
<p>$$7n - nr - 5r > 0$$</p>
<p>$$ \Rightarrow 7n > r(n + 5)$$</p>
<p>$$ \Rightarrow r < {{7n} \over {n + 5}}$$</p>
<p>As r represent term of binomial expression so r is always integer.</p>
<p>Given that sum of coefficient is 939.</p>
<p>When $$r = 0$$,</p>
<p>sum of coefficient $$ = {}^7{C_0}\,.\,{2^0} = 1$$</p>
<p>when $$r = 1$$,</p>
<p>sum of coefficient $$ = {}^7{C_0}\,.\,{2^0} + {}^7{C_1}\,.\,{2^1} = 1 + 14 = 15$$</p>
<p>when $$r = 2$$,</p>
<p>sum of coefficient</p>
<p>$$ = {}^7{C_0}\,.\,{2^0} + {}^7{C_1}\,.\,{2^1} + {}^7{C_2}\,.\,{2^2}$$</p>
<p>$$ = 1 + 14 + 84$$</p>
<p>$$ = 99$$</p>
<p>when $$r = 3$$,</p>
<p>sum of coefficient</p>
<p>$$ = {}^7{C_0}\,.\,{2^0} + {}^7{C_1}\,.\,{2^1} + {}^7{C_2}\,.\,{2^2} + {}^7{C_3}\,.\,{2^3}$$</p>
<p>$$ = 1 + 14 + 84 + 280$$</p>
<p>$$ = 379$$</p>
<p>when $$r = 4$$,</p>
<p>sum of coefficient</p>
<p>$$ = {}^7{C_0}\,.\,{2^0} + {}^7{C_1}\,.\,{2^1} + {}^7{C_2}\,.\,{2^2} + {}^7{C_3}\,.\,{2^3} + {}^7{C_4}\,.\,{2^4}$$</p>
<p>$$ = 1 + 14 + 84 + 280 + 560$$</p>
<p>$$ = 939$$</p>
<p>$$\therefore$$ For r = 4 sum of coefficient = 939</p>
<p>To get value of r = 4, value of $${{7n} \over {n + 5}}$$ should be between 4 and 5.</p>
<p>$$\therefore$$ $$4 < {{7n} \over {n + 5}} < 5$$</p>
<p>$$ \Rightarrow 4n + 20 < 7n < 5n + 25$$</p>
<p>$$\therefore$$ $$4n + 20 < 7n$$</p>
<p>$$ \Rightarrow 3n > 20$$</p>
<p>$$ \Rightarrow n > {{20} \over 3}$$</p>
<p>$$ \Rightarrow n > 6.66$$</p>
<p>and</p>
<p>$$7n < 5n + 25$$</p>
<p>$$ \Rightarrow 2n < 25$$</p>
<p>$$ \Rightarrow n < 12.5$$</p>
<p>$$\therefore$$ $$6.66 < n < 12.5$$</p>
<p>$$\therefore$$ Possible integer values of $$n = 7,8,9,10,11,12$$</p>
<p>$$\therefore$$ Sum of values of $$n = 7 + 8 + 9 + 10 + 11 + 12$$</p>
<p>$$ = 57$$</p> | integer | jee-main-2022-online-27th-june-evening-shift |
1l58h25fw | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If $$\left( {{}^{40}{C_0}} \right) + \left( {{}^{41}{C_1}} \right) + \left( {{}^{42}{C_2}} \right) + \,\,.....\,\, + \,\,\left( {{}^{60}{C_{20}}} \right) = {m \over n}{}^{60}{C_{20}}$$ m and n are coprime, then m + n is equal to ___________. | [] | null | 102 | <p>Here property used is</p>
<p>$${}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}$$</p>
<p>Given, $${}^{40}{C_0} + {}^{41}{C_1} + {}^{42}{C_2} + \,\,....\,\, + \,\,{}^{60}{C_{20}} = {m \over n}{}^{60}{C_{20}}$$</p>
<p>As $${}^{40}{C_0} = {}^{41}{C_0} = 1$$</p>
<p>So, we replace $${}^{40}{C_0}$$ with $${}^{41}{C_0}$$.</p>
<p>$$ \Rightarrow {}^{41}{C_0} + {}^{41}{C_1} + {}^{42}{C_2} + \,\,.....\,\,\, + \,{}^{60}{C_{20}} = {m \over n}\,.\,{}^{60}{C_{20}}$$</p>
<p>$$ \Rightarrow {}^{42}{C_1} + {}^{42}{C_2} + \,\,.....\,\, + \,\,{}^{60}{C_{20}} = {m \over n}\,.\,{}^{60}{C_{20}}$$</p>
<p>$$ \Rightarrow {}^{43}{C_2} + {}^{43}{C_3} + \,\,.....\,\, + \,\,{}^{60}{C_{20}} = {m \over n}\,.\,{}^{60}{C_{20}}$$</p>
<p>$$ \Rightarrow {}^{44}{C_3} + {}^{44}{C_4} + \,\,.....\,\, + \,\,{}^{60}{C_{20}} = {m \over n}\,.\,{}^{60}{C_{20}}$$</p>
<p>$$ \Rightarrow {}^{45}{C_4} + {}^{45}{C_5} + \,\,.....\,\, + \,\,{}^{60}{C_{20}} = {m \over n}\,.\,{}^{60}{C_{20}}$$</p>
<p>$$ \vdots $$</p>
<p>$$ \Rightarrow {}^{60}{C_{19}} + {}^{60}{C_{20}} = {m \over n}\,.\,{}^{60}{C_{20}}$$</p>
<p>$$ \Rightarrow {}^{61}{C_{20}} = {m \over n}\,.\,{}^{60}{C_{20}}$$</p>
<p>$$ \Rightarrow {{61!} \over {20!\,41!}} = {m \over n}\,.\,{{60!} \over {20!\,40!}}$$</p>
<p>$$ \Rightarrow {{61} \over {41}} = {m \over n}$$</p>
<p>$$\therefore$$ m = 61 and n = 41</p>
<p>$$\therefore$$ m + n = 61 + 41 = 102</p> | integer | jee-main-2022-online-26th-june-evening-shift |
1l5ajlo7t | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>Let C<sub>r</sub> denote the binomial coefficient of x<sup>r</sup> in the expansion of $${(1 + x)^{10}}$$. If for $$\alpha$$, $$\beta$$ $$\in$$ R, $${C_1} + 3.2{C_2} + 5.3{C_3} + $$ ....... upto 10 terms $$ = {{\alpha \times {2^{11}}} \over {{2^\beta } - 1}}\left( {{C_0} + {{{C_1}} \over 2} + {{{C_2}} \over 3} + \,\,.....\,\,upto\,10\,terms} \right)$$ then the value of $$\alpha$$ + $$\beta$$ is equal to ___________.</p> | [] | null | 286 | <p>Given,</p>
<p>$${C_1} + 3\,.\,2{C_2} + 5\,.\,3{C_3} + $$ ...... upto 10 terms</p>
<p>$$ = {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}$$ ($${C_0} + {{{C_1}} \over 2} + {{{C_2}} \over 3}$$ + ..... upto 10 terms)</p>
<p>Now,</p>
<p>L.H.S. :-</p>
<p>$${C_1} + 3\,.\,2{C_2} + 5\,.\,3{C_3} + $$ ...... upto 10 terms</p>
<p>$$ = 1\,.\,1{C_1} + 3\,.\,2{C_2} + 5\,.\,3{C_3} + $$ ..... upto 10 terms</p>
<p>$$ = \sum\limits_{r = 1}^{10} {r\,.\,(2r - 1){}^{10}{C_r}} $$</p>
<p>$$ = \sum\limits_{r = 1}^{10} {(2{r^2} - r)\,.\,{}^{10}{C_r}} $$</p>
<p>$$ = 2\,.\,\sum\limits_{r = 1}^{10} {{r^2}\,.\,{}^{10}{C_r} - \sum\limits_{r = 1}^{10} {r\,.\,{}^n{C_r}} } $$</p>
<p>[We know, $$\sum\limits_{r = 1}^n {r\,.\,{}^n{C_r} = n\,.\,{2^{n - 1}}} $$</p>
<p>and $$\sum\limits_{r = 1}^n {{r^2}\,.\,{}^n{C_r} = \sum\limits_{r = 1}^n {(r\,.\,{}^n{C_r})\,.\,r} } $$</p>
<p>$$ = \sum\limits_{r = 1}^n {\left( {r\,.\,{n \over r}\,.\,{}^{n - 1}{C_{r - 1}}} \right)\,.\,r} $$</p>
<p>$$ = \sum\limits_{r = 1}^n {\left( {n\,.\,{}^{n - 1}{C_{r - 1}}} \right)\,.\,r} $$</p>
<p>$$ = n\sum\limits_{r = 1}^n {(r - 1 + 1){}^{n - 1}{C_{r - 1}}} $$</p>
<p>$$ = n\,.\,\sum\limits_{r = 1}^n {(r - 1)\,.\,{}^{n - 1}{C_{r - 1}} + n\,.\,\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}} } $$</p>
<p>$$ = n\,.\,(n - 1)\,.\,{2^{n - 2}} + n\,.\,{2^{n - 1}}$$]</p>
<p>$$ = 2\left( {n(n - 1){2^{n - 2}} + n\,.\,{2^{n - 1}}} \right) - n\,.\,{2^{n - 1}}$$</p>
<p>Put $$n = 10$$</p>
<p>$$ = 2\left( {10\,.\,9\,.\,{2^8} + 10\,.\,{2^9}} \right) - 10\,.\,{2^9}$$</p>
<p>$$ = 45\,.\,{2^{10}} + 10\,.\,{2^{10}} - 5\,.\,{2^{10}}$$</p>
<p>$$ = {2^{10}}(45 + 10 - 5)$$</p>
<p>$$ = {2^{10}}\,.\,(50)$$</p>
<p>$$ = 25\,.\,{2^{11}}$$</p>
<p>R.H.S. :-</p>
<p>$${{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}$$ ($${C_0} + {{{C_1}} \over 2} + {{{C_2}} \over 3} + $$ ..... upto 10 terms)</p>
<p>$${{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}$$ ($${{{C_0}} \over 1} + {{{C_1}} \over 2} + {{{C_2}} \over 3} + $$ ..... upto 10 terms)</p>
<p>$${{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {\sum\limits_{r = 0}^n {{{{}^n{C_r}} \over {r + 1}}} } \right)$$</p>
<p>$$ = {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {\sum\limits_{r = 0}^n {{{{}^{n + 1}{C_{r + 1}}} \over {n + 1}}} } \right)$$</p>
<p>$$ = {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {{{{}^{n + 1}{C_1} + {}^{n + 1}{C_2} + \,\,....\,\, + \,\,{}^{n + 1}{C_{n + 1}}} \over {n + 1}}} \right)$$</p>
<p>$$ = {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {{{{}^{n + 1}{C_0} + {}^{n + 1}{C_2} + \,\,....\,\, + \,\,{}^{n + 1}{C_{n + 1}} - {}^{n + 1}{C_0}} \over {n + 1}}} \right)$$</p>
<p>$$ = {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {{{{2^{n + 1}} - 1} \over {n + 1}}} \right)$$</p>
<p>Putting value of $$n = 10$$, we get</p>
<p>$$ = {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {{{{2^{11}} - 1} \over {11}}} \right)$$</p>
<p>Using L.H.S. = R.H.S.</p>
<p>$$ \Rightarrow 25\,.\,{2^{11}} = {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {{{{2^{11}} - 1} \over {11}}} \right)$$</p>
<p>$$ \Rightarrow 25\,.\,{2^{11}} = {2^{11}}\left( {{\alpha \over {11}}} \right)\left( {{{{2^{11}} - 1} \over {{2^\beta } - 1}}} \right)$$</p>
<p>By comparing both sides,</p>
<p>$${\alpha \over {11}} = 25 \Rightarrow \alpha = 275$$</p>
<p>and $${{{2^{11}} - 1} \over {{2^\beta } - 1}} = 1$$</p>
<p>$$ \Rightarrow {2^{11}} = {2^\beta }$$</p>
<p>$$ \Rightarrow \beta = 11$$</p>
<p>$$\therefore$$ $$\alpha + \beta = 275 + 11 = 286$$</p> | integer | jee-main-2022-online-25th-june-morning-shift |
1l6gjfwtl | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>If the coefficients of $$x$$ and $$x^{2}$$ in the expansion of $$(1+x)^{\mathrm{p}}(1-x)^{\mathrm{q}}, \mathrm{p}, \mathrm{q} \leq 15$$, are $$-3$$ and $$-5$$ respectively, then the coefficient of $$x^{3}$$ is equal to _____________.</p> | [] | null | 23 | <p>Coefficient of x in $${(1 + x)^p}{(1 - x)^q}$$</p>
<p>$$ - {}^p{C_0}\,{}^q{C_1} + {}^p{C_1}\,{}^q{C_0} = - 3 \Rightarrow p - q = - 3$$</p>
<p>Coefficient of x<sup>2</sup> in $${(1 + x)^p}{(1 - x)^q}$$</p>
<p>$${}^p{C_0}\,{}^q{C_2} - {}^p{C_1}\,{}^q{C_1} + {}^p{C_2}\,{}^q{C_0} = - 5$$</p>
<p>$${{q(q - 1)} \over 2} - pq + {{p(q - 1)} \over 2} = - 5$$</p>
<p>$${{{q^2} - q} \over 2} - (q - 3)q + {{(q - 3)(q - 4)} \over 2} = - 5$$</p>
<p>$$ \Rightarrow q = 11,\,p = 8$$</p>
<p>Coefficient of x<sup>3</sup> in $${(1 + x)^8}{(1 - x)^{11}}$$</p>
<p>$$ = - {}^{11}{C_3} + {}^8{C_1}\,{}^{11}{C_2} - {}^8{C_2}\,{}^{11}{C_1} + {}^8{C_3} = 23$$</p> | integer | jee-main-2022-online-26th-july-morning-shift |
1l6hxu8ex | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>$$\sum\limits_{\matrix{
{i,j = 0} \cr
{i \ne j} \cr
} }^n {{}^n{C_i}\,{}^n{C_j}} $$ is equal to</p> | [{"identifier": "A", "content": "$$2^{2 n}-{ }^{2 n} C_{n}$$"}, {"identifier": "B", "content": "$${2^{2n - 1}} - {}^{2n - 1}{C_{n - 1}}$$"}, {"identifier": "C", "content": "$$2^{2 n}-\\frac{1}{2}{ }^{2 n} C_{n}$$"}, {"identifier": "D", "content": "$${2^{2n - 1}} + {}^{2n - 1}{C_n}$$"}] | ["A"] | null | <p>$$\sum\limits_{i,\,j = 0\,\,i \ne j}^n {{}^n{C_i}\,{}^n{C_j} = \sum\limits_{i,\,j = 0}^n {{}^n{C_i}\,{}^n{C_j} - \sum\limits_{i = j}^n {{}^n{C_i}\,{}^n{C_j}} } } $$</p>
<p>$$ = \sum\limits_{j = 0}^n {{}^n{C_i}\,\sum\limits_{j = 0}^n {{}^n{C_j} - \sum\limits_{i = 0}^n {{}^n{C_i}\,{C_i}} } } $$</p>
<p>$$ = {2^n}\,.\,{2^n} - {}^{2n}{C_n}$$</p>
<p>$$ = {2^{2n}} - {}^{2n}{C_n}$$</p> | mcq | jee-main-2022-online-26th-july-evening-shift |
1l6np70mw | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>If $$1 + (2 + {}^{49}{C_1} + {}^{49}{C_2} + \,\,...\,\, + \,\,{}^{49}{C_{49}})({}^{50}{C_2} + {}^{50}{C_4} + \,\,...\,\, + \,\,{}^{50}{C_{50}})$$ is equal to $$2^{\mathrm{n}} \cdot \mathrm{m}$$, where $$\mathrm{m}$$ is odd, then $$\mathrm{n}+\mathrm{m}$$ is equal to __________.</p> | [] | null | 99 | <p>$$l = 1 + (1 + {}^{49}{C_0} + {}^{49}{C_1}\, + \,....\, + \,{}^{49}{C_{49}})({}^{50}{C_2} + {}^{50}{C_4}\, + \,....\, + \,{}^{50}{C_{50}})$$</p>
<p>As $${}^{49}{C_0} + {}^{49}{C_1}\, + \,.....\, + \,{}^{49}{C_{49}} = {2^{49}}$$</p>
<p>and $${}^{50}{C_0} + {}^{50}{C_2}\, + \,....\, + \,{}^{50}{C_{50}} = {2^{49}}$$</p>
<p>$$ \Rightarrow {}^{50}{C_2} + {}^{50}{C_4}\, + \,....\, + \,{}^{50}{C_{50}} = {2^{49}} - 1$$</p>
<p>$$\therefore$$ $$l = 1 + ({2^{49}} + 1)({2^{49}} - 1)$$</p>
<p>$$ = {2^{98}}$$</p>
<p>$$\therefore$$ $$m = 1$$ and $$n = 98$$</p>
<p>$$m + n = 99$$</p> | integer | jee-main-2022-online-28th-july-evening-shift |
1l6re5ytr | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>$$\sum\limits_{r=1}^{20}\left(r^{2}+1\right)(r !)$$ is equal to</p> | [{"identifier": "A", "content": "$$22 !-21 !$$"}, {"identifier": "B", "content": "$$22 !-2(21 !)$$"}, {"identifier": "C", "content": "$$21 !-2(20 !)$$"}, {"identifier": "D", "content": "$$21 !-20$$ !"}] | ["B"] | null | <p>Given,</p>
<p>$$\sum\limits_{r = 1}^{20} {({r^2} + 1)(r!)} $$</p>
<p>Let, $$f(r) = ({r^2} + 1)(r!)$$</p>
<p>$$ = ({r^2})(r!) + r!$$</p>
<p>$$ = r(r\,r!) + r!$$</p>
<p>$$ = r[(r + 1 - 1)r!] + r!$$</p>
<p>$$ = r[(r + 1)r! - r!] + r!$$</p>
<p>$$ = r[(r + 1)! - (r!)] + r!$$</p>
<p>$$ = r(r + 1)! - r(r!) + r!$$</p>
<p>$$ = (r + 2 - 2)(r + 1)! - r(r!) + r!$$</p>
<p>$$ = (r + 2)(r + 1)! - 2(r + 1)! - [(r + 1 - 1)(r!)] + r!$$</p>
<p>$$ = (r + 2)! - 2(r + 1)! - (r + 1)! + r! + r!$$</p>
<p>$$ = (r + 2)! - 3(r + 1)! + 2r!$$</p>
<p>$$ = [(r + 2)! - (r + 1)!] - 2[(r + 1)! - r!]$$</p>
<p>$$\therefore$$ $$\sum\limits_{r = 1}^{20} {f(r)} $$</p>
<p>$$ = \sum\limits_{r = 1}^{20} {[(r + 2)! - (r + 1)!] - 2\sum\limits_{r = 1}^{20} {[(r + 1)! - r!]} } $$</p>
<p>$$ = [(22! + 21! + 20!\, + \,.....\, + \,4! + 3!) - (21! + 20! + 19!\, + \,....\, + \,3! + 2!] - 2[(21! + 20!\, + \,.....\, + \,3! + 2!) - (20! + 19!\, + \,.....\,1!)]$$</p>
<p>$$ = [(22!) - (2!)] - 2[(21)! - (1!)]$$</p>
<p>$$ = 22! - 2! - 2\,.\,(21)! + 2\,.\,1!$$</p>
<p>$$ = 22! - 2\,.\,(21)!$$</p> | mcq | jee-main-2022-online-29th-july-evening-shift |
1l6rfnnhx | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>$$
\text { If } \sum\limits_{k=1}^{10} K^{2}\left(10_{C_{K}}\right)^{2}=22000 L \text {, then } L \text { is equal to }$$ ________.</p> | [] | null | 221 | <p>Given,</p>
<p>$$\sum\limits_{k = 1}^{10} {{k^2}{{\left( {{}^{10}{C_k}} \right)}^2} = 2200\,L} $$</p>
<p>$$ \Rightarrow \sum\limits_{k = 1}^{10} {{{\left( {k\,.\,{}^{10}{C_k}} \right)}^2} = 22000\,L} $$</p>
<p>$$ \Rightarrow \sum\limits_{k = 1}^{10} {{{\left( {k\,.\,{{10} \over k}\,.\,{}^9{C_{k - 1}}} \right)}^2} = 22000\,L} $$</p>
<p>$$ \Rightarrow \sum\limits_{k = 1}^{10} {{{\left( {10\,.\,{}^9{C_{k - 1}}} \right)}^2} = 22000\,L} $$</p>
<p>$$ \Rightarrow 100\,.\,\sum\limits_{k = 1}^{10} {{{\left( {{}^9{C_{k - 1}}} \right)}^2} = 22000\,L} $$</p>
<p>$$ \Rightarrow 100\left( {{{\left( {{}^9{C_0}} \right)}^2} + {{\left( {{}^9{C_1}} \right)}^2}\, + \,....\, + \,{{\left( {{}^9{C_9}} \right)}^2}} \right) = 22000\,L$$</p>
<p>$$ \Rightarrow 100\left( {{}^{18}{C_9}} \right) = 22000\,L$$</p>
<p>[Note : $${\left( {{}^n{C_1}} \right)^2} + {\left( {{}^n{C_2}} \right)^2}\, + \,....\, + \,{\left( {{}^n{C_n}} \right)^2} = {}^{2n}{C_n}$$]</p>
<p>$$ \Rightarrow 100 \times {{18!} \over {9!\,9!}} = 22000\,L$$</p>
<p>$$ \Rightarrow L = 221$$</p> | integer | jee-main-2022-online-29th-july-evening-shift |
1ldr759il | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>The coefficient of $${x^{301}}$$ in $${(1 + x)^{500}} + x{(1 + x)^{499}} + {x^2}{(1 + x)^{498}}\, + \,...\, + \,{x^{500}}$$ is :</p> | [{"identifier": "A", "content": "$${}^{500}{C_{300}}$$"}, {"identifier": "B", "content": "$${}^{501}{C_{200}}$$"}, {"identifier": "C", "content": "$${}^{500}{C_{301}}$$"}, {"identifier": "D", "content": "$${}^{501}{C_{302}}$$"}] | ["B"] | null | <p>The coefficient of $${x^{301}}$$ in
<br/><br/>$${(1 + x)^{500}} + x{(1 + x)^{499}} + {x^2}{(1 + x)^{498}}\, + \,...\, + \,{x^{500}}$$</p>
<p>$${}^{500}{C_{301}} + {}^{499}{C_{300}} + {}^{498}{C_{299}}\, + \,...\, + \,{}^{199}{C_0}$$</p>
<p>$$ = {}^{500}{C_{199}} + {}^{499}{C_{199}} + {}^{498}{C_{199}}\, + \,...\, + \,{}^{199}{C_{199}}$$</p>
<p>$$ = {}^{501}{C_{200}}$$</p> | mcq | jee-main-2023-online-30th-january-morning-shift |
1ldv1hnql | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>If $$a_r$$ is the coefficient of $$x^{10-r}$$ in the Binomial expansion of $$(1 + x)^{10}$$, then $$\sum\limits_{r = 1}^{10} {{r^3}{{\left( {{{{a_r}} \over {{a_{r - 1}}}}} \right)}^2}} $$ is equal to </p> | [{"identifier": "A", "content": "3025"}, {"identifier": "B", "content": "4895"}, {"identifier": "C", "content": "5445"}, {"identifier": "D", "content": "1210"}] | ["D"] | null | $$
\begin{aligned}
& \mathrm{a}_{\mathrm{r}}={ }^{10} \mathrm{C}_{10-\mathrm{r}}={ }^{10} \mathrm{C}_{\mathrm{r}} \\\\
& \Rightarrow \sum_{\mathrm{r}=1}^{10} \mathrm{r}^3\left(\frac{{ }^{10} \mathrm{C}_{\mathrm{r}}}{{ }^{10} \mathrm{C}_{\mathrm{r}-1}}\right)^2=\sum_{\mathrm{r}=1}^{10} \mathrm{r}^3\left(\frac{11-\mathrm{r}}{\mathrm{r}}\right)^2=\sum_{\mathrm{r}=1}^{10} \mathrm{r}(11-\mathrm{r})^2 \\\\
& =\sum_{\mathrm{r}=1}^{10}\left(121 \mathrm{r}+\mathrm{r}^3-22 \mathrm{r}^2\right)=1210
\end{aligned}
$$ | mcq | jee-main-2023-online-25th-january-morning-shift |
1ldwxb38k | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>If $${({}^{30}{C_1})^2} + 2{({}^{30}{C_2})^2} + 3{({}^{30}{C_3})^2}\, + \,...\, + \,30{({}^{30}{C_{30}})^2} = {{\alpha 60!} \over {{{(30!)}^2}}}$$ then $$\alpha$$ is equal to :</p> | [{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "15"}, {"identifier": "D", "content": "60"}] | ["C"] | null | $$
\begin{aligned}
& \mathrm{S}=0 \cdot\left({ }^{30} \mathrm{C}_0\right)^2+1 \cdot\left({ }^{30} \mathrm{C}_1\right)^2+2 \cdot\left({ }^{30} \mathrm{C}_2\right)^2+\ldots \ldots+30 \cdot\left({ }^{30} \mathrm{C}_{30}\right)^2 \\\\
& \mathrm{S}=30 \cdot\left({ }^{30} \mathrm{C}_0\right)^2+29 \cdot\left({ }^{30} \mathrm{C}_1\right)^2+28 \cdot\left({ }^{30} \mathrm{C}_2\right)^2+\ldots . \cdot+0 \cdot\left({ }^{30} \mathrm{C}_0\right)^2 \\\\
& 2 \mathrm{~S}=30 \cdot\left({ }^{30} \mathrm{C}_0{ }^2+{ }^{30} \mathrm{C}_1{ }^2+\ldots \ldots . \cdot+{ }^{30} \mathrm{C}_{30}{ }^2\right) \\\\
& \mathrm{S}=15 \cdot{ }^{60} \mathrm{C}_{30}=15 \cdot \frac{60 !}{(30 !)^2} \\\\
& \frac{15 \cdot 10 !}{(30 !)^2}=\frac{\alpha \cdot 60 !}{(30 !)^2} \\\\
& \Rightarrow \alpha=15
\end{aligned}
$$
<br/><br/><b>Other Method :</b>
<br/><br/><p>Given,</p>
<p>$$1\,.\,{\left( {{}^{30}{C_1}} \right)^2} + 2\,.\,{\left( {{}^{30}{C_2}} \right)^2} + 3\,.\,{\left( {{}^{30}{C_3}} \right)^2}\, + \,.....\, + \,30\,.\,{\left( {{}^{30}{C_{30}}} \right)^2}$$</p>
<p>$$ = \sum\limits_{r = 1}^{30} {r\,.\,{{\left( {{}^{30}{C_r}} \right)}^2}} $$</p>
<p>$$ = \sum\limits_{r = 1}^{30} {r\,.\,{}^{30}{C_r}\,.\,{}^{30}{C_r}} $$</p>
<p>$$ = \sum\limits_{r = 1}^{30} {r\,.\,{{30} \over r}\,.\,{}^{29}{C_{r - 1}}\,.\,{}^{30}{C_r}} $$</p>
<p>$$ = 30\sum\limits_{r = 1}^{30} {{}^{29}{C_{r - 1}}\,.\,{}^{30}{C_r}} $$</p>
<p>$$ = 30 \times $$ (Coefficient of $${x^{29}}$$ in $${(1 + x)^{59}}$$)</p>
<p>$$ = 30 \times \left( {{}^{59}{C_{29}}} \right)$$</p>
<p>$$ = 30 \times {{60} \over {30}} \times {}^{59}{C_{29}} \times {{30} \over {60}}$$</p>
<p>$$ = 30 \times {}^{60}{C_{30}} \times {{30} \over {60}}$$</p>
<p>$$ = 15 \times {{60!} \over {30!\,30!}}$$</p>
<p>$$\therefore$$ $$\alpha = 15$$</p> | mcq | jee-main-2023-online-24th-january-evening-shift |
1ldyaytnw | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>The value of $$\sum\limits_{r = 0}^{22} {{}^{22}{C_r}{}^{23}{C_r}} $$ is</p> | [{"identifier": "A", "content": "$${}^{44}{C_{23}}$$"}, {"identifier": "B", "content": "$${}^{45}{C_{23}}$$"}, {"identifier": "C", "content": "$${}^{44}{C_{22}}$$"}, {"identifier": "D", "content": "$${}^{45}{C_{24}}$$"}] | ["B"] | null | <p>$$\sum\limits_{r = 0}^{22} {{}^{22}{C_r}\,.\,{}^{23}{C_r}} $$</p>
<p>$$ = \sum\limits_{r = 0}^{22} {{}^{22}{C_r}\,{}^{23}{C_{23 - r}}} $$ [using $${}^n{C_r} = {}^n{C_{n - r}}$$]</p>
<p>$$ = {}^{22}{C_0}{}^{23}{C_{23}} + {}^{22}{C_1}{}^{23}{C_{22}}\, + \,...\, + \,{}^{22}{C_{21}}{}^{23}{C_2} + {}^{22}{C_{22}}{}^{23}{C_1}$$</p>
<p>We know,</p>
<p>$${(1 + x)^{22}} = {}^{22}{C_0} + {}^{22}{C_1}x + {}^{22}{C_2}{x^2} + {}^{22}{C_3}{x^3}\, + \,...\, + \,{}^{22}{C_{22}}{x^{22}}$$</p>
<p>and</p>
<p>$${(1 + x)^{23}} = {}^{23}{C_0} + {}^{23}{C_1}x + {}^{23}{C_2}{x^2}\, + \,...\, + \,{}^{23}{C_{22}}{x^{22}} + {}^{23}{C_{23}}{x^{23}}$$</p>
<p>Now coefficient of $${x^{23}}$$ in $${(1 + x)^{22}}{(1 + x)^{23}}$$ or $${(1 + x)^{45}}$$</p>
<p>$$ = {}^{22}{C_0}{}^{23}{C_{23}} + {}^{22}{C_1}{}^{23}{C_{22}}\, + \,...\, + \,{}^{22}{C_{21}}\,.\,{}^{23}{C_2} + {}^{22}{C_{22}}\,.\,{}^{23}{C_1}$$</p>
<p>$$ = \sum\limits_{r = 0}^{22} {{}^{22}{C_r}\,.\,{}^{23}{C_{23 - r}}} $$</p>
<p>$$\therefore$$ Coefficient of $${x^{23}}$$ in $${(1 + x)^{45}} = {}^{45}{C_{23}}$$</p> | mcq | jee-main-2023-online-24th-january-morning-shift |
1ldyc6w0d | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>Suppose $$\sum\limits_{r = 0}^{2023} {{r^2}{}~^{2023}{C_r} = 2023 \times \alpha \times {2^{2022}}} $$. Then the value of $$\alpha$$ is ___________</p> | [] | null | 1012 | <p>Concept :</p>
<p>(1) $${}^n{C_r} = {n \over r}\,.\,{}^{n - 1}{C_{r - 1}}$$</p>
<p>Given,</p>
<p>$$\sum\limits_{r = 0}^{2023} {{r^2}\,.\,{}^{2023}{C_r}} $$</p>
<p>$$ = \sum\limits_{r = 0}^{2023} {{r^2}\,.\,{{2023} \over r}\,.{}^{2022}{C_{r - 1}}} $$</p>
<p>$$ = 2023\sum\limits_{r = 0}^{2023} {{r}\,.\,{}^{2022}{C_{r - 1}}} $$</p>
<p>$$ = 2023\sum\limits_{r = 0}^{2023} {[(r - 1) + 1]\,.\,{}^{2022}{C_{r - 1}}} $$</p>
<p>$$ = 2023\left[ {\sum\limits_{r = 0}^{2023} {(r - 1)\,.\,{}^{2022}{C_{r - 1}} + \sum\limits_{r = 0}^{2023} {{}^{2022}{C_{r - 1}}} } } \right]$$</p>
<p>$$ = 2023\left[ {\sum\limits_{r = 0}^{2023} {(r - 1)\,.\,{{2022} \over {(r - 1)}}\,.\,{}^{2021}{C_{r - 2}} + {2^{2022}}} } \right]$$</p>
<p>$$ = 2023\left[ {2022\sum\limits_{r = 0}^{2023} {{}^{2021}{C_{r - 2}} + {2^{2022}}} } \right]$$</p>
<p>$$ = 2023\left[ {2022\,.\,{2^{2021}} + {2^{2022}}} \right]$$</p>
<p>$$ = 2023\,.\,{2^{2021}}\left[ {2022 + 2} \right]$$</p>
<p>$$ = 2023\,.\,{2^{2021}}\,.\,2024$$</p>
<p>$$ = 2023\,.\,{{{2^{2022}}} \over {{2}}}\,.\,2024$$</p>
<p>$$ = 2023\,.\,{2^{2022}}\,.\,{1012}$$</p>
<p>$$\therefore$$ $$\alpha = {1012}$$</p> | integer | jee-main-2023-online-24th-january-morning-shift |
1lgrecwt6 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>If $$\frac{1}{n+1}{ }^{n} \mathrm{C}_{n}+\frac{1}{n}{ }^{n} \mathrm{C}_{n-1}+\ldots+\frac{1}{2}{ }^{n} \mathrm{C}_{1}+{ }^{n} \mathrm{C}_{0}=\frac{1023}{10}$$ then $$n$$ is equal to :</p> | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "8"}] | ["A"] | null | $$\frac{1}{n+1}{ }^{n} \mathrm{C}_{n}+\frac{1}{n}{ }^{n} \mathrm{C}_{n-1}+\ldots+\frac{1}{2}{ }^{n} \mathrm{C}_{1}+{ }^{n} \mathrm{C}_{0}=\frac{1023}{10}$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \sum_{r=0}^n \frac{1}{r+1}{ }^n C_r=\frac{1023}{10} \\\\
& \quad\left( \because{ }^{n+1} C_{r+1}=\frac{n+1}{r+1}{ }^n C_r\right) \\\\
& \Rightarrow \sum_{r=0}^n \frac{1}{n+1}{ }^{n+1} C_{r+1}=\frac{1023}{10} \\\\
& \Rightarrow \frac{1}{n+1}\left[{ }^{n+1} C_1+{ }^{n+1} C_2+\ldots+{ }^{n+1} C_{n+1}\right]=\frac{1023}{10} \\\\
& \Rightarrow \frac{2^{n+1}-1}{n+1}=\frac{1023}{10}=\frac{2^{10}-1}{10} \\\\
& \Rightarrow n+1=10 \\\\
& \Rightarrow n=9
\end{aligned}
$$ | mcq | jee-main-2023-online-12th-april-morning-shift |
1lgrg83q5 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>The sum, of the coefficients of the first 50 terms in the binomial expansion of $$(1-x)^{100}$$, is equal to</p> | [{"identifier": "A", "content": "$${ }^{99} \\mathrm{C}_{49}$$"}, {"identifier": "B", "content": "$${ }^{101} \\mathrm{C}_{50}$$"}, {"identifier": "C", "content": "$$-{ }^{99} \\mathrm{C}_{49}$$"}, {"identifier": "D", "content": "$$-{ }^{101} \\mathrm{C}_{50}$$"}] | ["C"] | null | $$
\begin{aligned}
& \left({ }^{100} C_0-{ }^{100} C_1+{ }^{100} C_2-\ldots . .{ }^{100} C_{49}\right)+{ }^{100} C_{50} \\\\
& +\left(-{ }^{100} C_{51}+{ }^{100} C_{52}-\ldots .+{ }^{100} C_{100}\right)=0 \\\\
& \lambda_1+{ }^{100} C_{50}+\lambda_2=0 \\\\
& \lambda_1=-\frac{1}{2}{ }^{100} C_{50} \quad\left(\because \lambda_1=\lambda_2\right) \\\\
& =-{ }^{99} C_{49}
\end{aligned}
$$ | mcq | jee-main-2023-online-12th-april-morning-shift |
1lgzzpzqm | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>If the coefficients of three consecutive terms in the expansion of $$(1+x)^{n}$$ are in the ratio $$1: 5: 20$$, then the coefficient of the fourth term is</p> | [{"identifier": "A", "content": "3654"}, {"identifier": "B", "content": "1827"}, {"identifier": "C", "content": "5481"}, {"identifier": "D", "content": "2436"}] | ["A"] | null | $$
\begin{aligned}
& \text { Given: }{ }^n \mathrm{C}_{r-1}:{ }^n \mathrm{C}_r:{ }^n \mathrm{C}_{r+1} \\\\
& =1: 5: 20 \\\\
& \Rightarrow \frac{n !}{(r-1) !(n-r+1) !} \times \frac{r !(n-r) !}{n !}=\frac{1}{5} \\\\
& \Rightarrow \frac{r}{(n-r+1)}=\frac{1}{5} \\\\
& \Rightarrow 5 r=n-r+1 \\\\
& \Rightarrow n=6 r-1 ........(i)
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \text { Also, } \frac{n}{r !(n-r) !} \times \frac{(r+1) !(n-r-1) !}{n !}=\frac{5}{20}=\frac{1}{20} \\\\
& \Rightarrow \frac{(r+1)}{(n-r)}=\frac{1}{4} \\\\
& \Rightarrow 4 r+4=n-r \\\\
& \Rightarrow n=5 r+4 ..........(ii)
\end{aligned}
$$
<br/><br/>From (i) and (ii), we get
<br/><br/>$$
\begin{aligned}
& 6 r-1=5 r+4 \\\\
& \Rightarrow r=5 \\\\
& \text { So, } n=5(5)+4=29 \\\\
& \text { So, coefficient of } 4{ }^{\text {th }} \text { terms }={ }^n \mathrm{C}_3={ }^{29} \mathrm{C}_3 \\\\
& =\frac{29 !}{3 ! 26 !}=\frac{29 \times 28 \times 27}{3 \times 2}=3654
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-morning-shift |
1lh238a1o | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>If $${ }^{2 n} C_{3}:{ }^{n} C_{3}=10: 1$$, then the ratio $$\left(n^{2}+3 n\right):\left(n^{2}-3 n+4\right)$$ is :</p> | [{"identifier": "A", "content": "$$27: 11$$"}, {"identifier": "B", "content": "$$2: 1$$"}, {"identifier": "C", "content": "$$35: 16$$"}, {"identifier": "D", "content": "$$65: 37$$"}] | ["B"] | null | $$
\begin{aligned}
& \text {We have, }{ }^{2 n} C_3:{ }^n C_3=10: 1 \\\\
& \Rightarrow \frac{{ }^{2 n} C_3}{{ }^n C_3}=\frac{10}{1} \\\\
& \Rightarrow \frac{(2 n) !}{3 !(2 n-3) !} \times \frac{3 !(n-3) !}{n !}=\frac{10}{1} \\\\
& \Rightarrow \frac{(2 n)(2 n-1)(2 n-2)}{(n)(n-1)(n-2)}=\frac{10}{1} \\\\
& \Rightarrow 4 n^2-6 n+2=5\left(n^2-3 n+2\right) \\\\
& \Rightarrow n^2-9 n+8=0 \\\\
& \Rightarrow n^2-8 n-n+8=0 \\\\
& \Rightarrow n(n-8)-1(n-8)=0 \\\\
& \Rightarrow (n-8)(n-1)=0 \\\\
& \Rightarrow n=8(n=1 \text { not valid })
\end{aligned}
$$
<br/><br/>$$
\therefore \frac{n^2+3 n}{n^2-3 n+4}=\frac{88}{44}=\frac{2}{1}=2: 1
$$ | mcq | jee-main-2023-online-6th-april-morning-shift |
1lh241fo2 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>The coefficient of $$x^{18}$$ in the expansion of $$\left(x^{4}-\frac{1}{x^{3}}\right)^{15}$$ is __________.</p> | [] | null | 5005 | $\begin{aligned} T_{r+1}= & { }^{15} C_r\left(x^4\right)^{15-r}\left(-\frac{1}{x^3}\right)^r={ }^{15} C_r(-1)^r x^{60-4 r-3 r} \\\\ = & { }^{15} C_r(-1)^r x^{60-7 r}\end{aligned}$
<br/><br/>$\begin{aligned} \therefore 60-7 r =18 \\\\ \Rightarrow 7 r =42 \\\\ \Rightarrow r =6\end{aligned}$
<br/><br/>$\therefore$ The coefficient of $x^{18}$
<br/><br/>$$
={ }^{15} C_6(-1)^6=\frac{15 \times 14 \times 13 \times 12 \times 11 \times 10}{6 \times 5 \times 4 \times 3 \times 2 \times 1}=5005
$$ | integer | jee-main-2023-online-6th-april-morning-shift |
lsbl0a9f | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | ${ }^{n-1} C_r=\left(k^2-8\right){ }^n C_{r+1}$ if and only if : | [{"identifier": "A", "content": "$2 \\sqrt{2}<\\mathrm{k}<2 \\sqrt{3}$"}, {"identifier": "B", "content": "$2 \\sqrt{2}<\\mathrm{k} \\leq 3$"}, {"identifier": "C", "content": "$2 \\sqrt{3}<\\mathrm{k}<3 \\sqrt{3}$"}, {"identifier": "D", "content": "$2 \\sqrt{3}<\\mathrm{k} \\leq 3 \\sqrt{2}$"}] | ["B"] | null | <p>$${ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}}=(\mathrm{k}^2-8){ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}$$</p>
<p>$$\underbrace{\mathrm{r}+1 \geq 0, \quad \mathrm{r} \geq 0}_{\mathrm{r} \geq 0}$$</p>
<p>$$\begin{aligned}
& \frac{{ }^{n-1} C_r}{{ }^n C_{r+1}}=k^2-8 \\
& \frac{r+1}{n}=k^2-8 \\
& \Rightarrow k^2-8>0 \\
& (k-2 \sqrt{2})(k+2 \sqrt{2})>0
\end{aligned}$$</p>
<p>$$\mathrm{k} \in(-\infty,-2 \sqrt{2}) \cup(2 \sqrt{2}, \infty)\quad \text{.... (I)}$$</p>
<p>$$\begin{aligned}
\therefore & \mathrm{n} \geq \mathrm{r}+1, \frac{\mathrm{r}+1}{\mathrm{n}} \leq 1 \\
\Rightarrow & \mathrm{k}^2-8 \leq 1 \\
& \mathrm{k}^2-9 \leq 0 \\
& -3 \leq \mathrm{k} \leq 3 \quad \text{.... (II)}
\end{aligned}$$</p>
<p>From equation (I) and (II) we get</p>
<p>$$\mathrm{k} \in[-3,-2 \sqrt{2}) \cup(2 \sqrt{2}, 3]$$</p> | mcq | jee-main-2024-online-27th-january-morning-shift |
lsbl8y41 | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | If A denotes the sum of all the coefficients in the expansion of $\left(1-3 x+10 x^2\right)^{\mathrm{n}}$ and B denotes the sum of all the coefficients in the expansion of $\left(1+x^2\right)^n$, then : | [{"identifier": "A", "content": "$\\mathrm{B}=\\mathrm{A}^3$"}, {"identifier": "B", "content": "$3 \\mathrm{A}=\\mathrm{B}$"}, {"identifier": "C", "content": "$A=3 B$"}, {"identifier": "D", "content": "$\\mathrm{A}=\\mathrm{B}^3$"}] | ["D"] | null | <p>Sum of coefficients in the expansion of $$\left(1-3 \mathrm{x}+10 \mathrm{x}^2\right)^{\mathrm{n}}=\mathrm{A}$$</p>
<p>then $$A=(1-3+10)^n=8^n$$ (put $$x=1$$)<?p>
<p>and sum of coefficients in the expansion of</p>
<p>$$\begin{aligned}
& \left(1+x^2\right)^n=B \\
& \text { then } B=(1+1)^n=2^n \\
& A=B^3
\end{aligned}$$</p> | mcq | jee-main-2024-online-27th-january-morning-shift |
jaoe38c1lsd577xu | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>Let the coefficient of $$x^r$$ in the expansion of $$(x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3}(x+2)^2+\ldots \ldots \ldots .+(x+2)^{n-1}$$ be $$\alpha_r$$. If $$\sum_\limits{r=0}^n \alpha_r=\beta^n-\gamma^n, \beta, \gamma \in \mathbb{N}$$, then the value of $$\beta^2+\gamma^2$$ equals _________.</p> | [] | null | 25 | <p>$$\begin{aligned}
& (x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3} \\
& (x+2)^2+\ldots \ldots .+(x+2)^{n-1} \\
& \sum \alpha_r=4^{n-1}+4^{n-2} \times 3+4^{n-3} \times 3^2 \ldots \ldots+3^{n-1} \\
& =4^{n-1}\left[1+\frac{3}{4}+\left(\frac{3}{4}\right)^2 \ldots .+\left(\frac{3}{4}\right)^{n-1}\right] \\
& =4^{n-1} \times \frac{1-\left(\frac{3}{4}\right)^n}{1-\frac{3}{4}} \\
& =4^n-3^n=\beta^n-\gamma^n \\
& \beta=4, \gamma=3 \\
& \beta^2+\gamma^2=16+9=25
\end{aligned}$$</p> | integer | jee-main-2024-online-31st-january-evening-shift |
jaoe38c1lse5rrmq | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>In the expansion of $$(1+x)\left(1-x^2\right)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5, x \neq 0$$, the sum of the coefficients of $x^3$ and $$x^{-13}$$ is equal to __________.</p> | [] | null | 118 | <p>$$\begin{aligned}
& (1+x)\left(1-x^2\right)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5 \\
& =(1+x)\left(1-x^2\right)\left(\left(1+\frac{1}{x}\right)^3\right)^5 \\
& =\frac{(1+x)^2(1-x)(1+x)^{15}}{x^{15}} \\
& =\frac{(1+x)^{17}-x(1+x)^{17}}{x^{15}}
\end{aligned}$$</p>
<p>$$=\operatorname{coeff}\left(\mathrm{x}^3\right)$$ in the expansion $$\approx \operatorname{coeff}\left(\mathrm{x}^{18}\right)$$ in</p>
<p>$$\begin{aligned}
& (1+x)^{17}-x(1+x)^{17} \\
& =0-1 \\
& =-1
\end{aligned}$$</p>
<p>$$\operatorname{coeff}\left(\mathrm{x}^{-13}\right)$$ in the expansion $$\approx \operatorname{coeff}\left(\mathrm{x}^2\right)$$ in</p>
<p>$$\begin{aligned}
& (1+x)^{17}-x(1+x)^{17} \\
& =\left(\begin{array}{c}
17 \\
2
\end{array}\right)-\left(\begin{array}{c}
17 \\
1
\end{array}\right) \\
& =17 \times 8-17 \\
& =17 \times 7 \\
& =119
\end{aligned}$$</p>
<p>Hence Answer $$=119-1=118$$</p> | integer | jee-main-2024-online-31st-january-morning-shift |
jaoe38c1lsf0mkur | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>$$\text { If } \frac{{ }^{11} C_1}{2}+\frac{{ }^{11} C_2}{3}+\ldots+\frac{{ }^{11} C_9}{10}=\frac{n}{m} \text { with } \operatorname{gcd}(n, m)=1 \text {, then } n+m \text { is equal to }$$ _______.</p> | [] | null | 2041 | <p>$$\begin{aligned}
& \sum_{\mathrm{r}=1}^9 \frac{{ }^{11} \mathrm{C}_{\mathrm{r}}}{\mathrm{r}+1} \\
& =\frac{1}{12} \sum_{\mathrm{r}=1}^9{ }^{12} \mathrm{C}_{\mathrm{r}+1} \\
& =\frac{1}{12}\left[2^{12}-26\right]=\frac{2035}{6} \\
& \therefore \mathrm{m}+\mathrm{n}=2041
\end{aligned}$$</p> | integer | jee-main-2024-online-29th-january-morning-shift |
1lsg4gw8t | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>Suppose $$2-p, p, 2-\alpha, \alpha$$ are the coefficients of four consecutive terms in the expansion of $$(1+x)^n$$. Then the value of $$p^2-\alpha^2+6 \alpha+2 p$$ equals</p> | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "10"}] | [] | null | <p>$$2-p, p, 2-\alpha, \alpha$$</p>
<p>Binomial coefficients are</p>
<p>$$\begin{aligned}
& { }^n C_r,{ }^n C_{r+1},{ }^n C_{r+2},{ }^n C_{r+3} \text { respectively } \\
\Rightarrow \quad & { }^n C_r+{ }^n C_{r+1}=2 \\
\Rightarrow \quad & { }^{n+1} C_{r+1}=2 \quad \ldots . .(1)
\end{aligned}$$</p>
<p>Also, $${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+2}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+3}=2$$</p>
<p>$$\Rightarrow \quad{ }^{n+1} C_{r+3}=2$$ $$\quad\text{..... (2)}$$</p>
<p>From (1) and (2)</p>
<p>$$\begin{aligned}
& { }^{n+1} C_{r+1}={ }^{n+1} C_{r+3} \\
& \Rightarrow \quad 2 \mathrm{r}+4=\mathrm{n}+1 \\
& \mathrm{n}=2 \mathrm{r}+3 \\
& { }^{2 \mathrm{r}+4} \mathrm{C}_{\mathrm{r}+1}=2 \\
\end{aligned}$$</p>
<p>Data Inconsistent</p> | mcq | jee-main-2024-online-30th-january-evening-shift |
1lsg5bloa | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>Let $$\alpha=\sum_\limits{k=0}^n\left(\frac{\left({ }^n C_k\right)^2}{k+1}\right)$$ and $$\beta=\sum_\limits{k=0}^{n-1}\left(\frac{{ }^n C_k{ }^n C_{k+1}}{k+2}\right)$$ If $$5 \alpha=6 \beta$$, then $$n$$ equals _______.</p> | [] | null | 10 | <p>$$\begin{aligned}
\alpha= & \sum_{k=0}^n \frac{{ }^n C_k \cdot{ }^n C_k}{k+1} \cdot \frac{n+1}{n+1} \\
& =\frac{1}{n+1} \sum_{k=0}^n{ }^{n+1} C_{k+1} \cdot{ }^n C_{n-k} \\
\alpha & =\frac{1}{n+1} \cdot{ }^{2 n+1} C_{n+1} \\
\beta & =\sum_{k=0}^{n-1} C_k \cdot \frac{{ }^n C_{k+1}}{k+2} \frac{n+1}{n+1} \\
& \frac{1}{n+1} \sum_{k=0}^{n-1}{ }^n C_{n-k} \cdot{ }^{n+1} C_{k+2} \\
& =\frac{1}{n+1} \cdot{ }^{2 n+1} C_{n+2} \\
\frac{\beta}{\alpha} & =\frac{2 n+1}{2 n+1} C_{n+2} \\
\frac{\beta}{\alpha} & =\frac{2 n+1-(n+2)+1}{n+2}=\frac{5}{6} \\
n & =10
\end{aligned}$$</p> | integer | jee-main-2024-online-30th-january-evening-shift |
luxwdx5b | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>The sum of the coefficient of $$x^{2 / 3}$$ and $$x^{-2 / 5}$$ in the binomial expansion of $$\left(x^{2 / 3}+\frac{1}{2} x^{-2 / 5}\right)^9$$ is</p> | [{"identifier": "A", "content": "19/4"}, {"identifier": "B", "content": "69/16"}, {"identifier": "C", "content": "63/16"}, {"identifier": "D", "content": "21/4"}] | ["D"] | null | <p>$$
\begin{aligned}
& T_{r+1}={ }^9 C_r\left(\frac{x^{-2 / 5}}{2}\right)^r\left(x^{2 / 3}\right)^{9-r} \\
& ={ }^9 C_r \frac{1}{2^r} x^{\frac{2}{3}(9-r)+\left(\frac{-2 r}{5}\right)} \\
& ={ }^9 C_r \cdot \frac{1}{2^r} \cdot x^{6-\frac{16 r}{15}}
\end{aligned}
$$</p>
<p>For coefficient of $$x^{2 / 3}$$</p>
<p>$$\begin{aligned}
& \Rightarrow 6-\frac{16 r}{15}=\frac{2}{3} \\
& \Rightarrow 90-16 r=10 \\
& \Rightarrow r=5
\end{aligned}$$</p>
<p>For coefficient of $$x^{-2 / 5}$$</p>
<p>$$\begin{aligned}
& \Rightarrow 6-\frac{16 r}{15}=\frac{-2}{5}\\
& \Rightarrow 90-16 r=-6 \\
& \Rightarrow r=6
\end{aligned}$$</p>
<p>Sum of coefficient of $$x^{2 / 3}$$ & $$x^{-2 / 5}$$</p>
<p>$$\begin{aligned}
& ={ }^9 C_5 \cdot \frac{1}{2^5}+{ }^9 C_6 \cdot \frac{1}{2^6} \\
& =\frac{9!}{5!4!}\left(\frac{1}{2^5}\right)+\frac{9!}{6!3!}\left(\frac{1}{2^6}\right)=\frac{21}{4}
\end{aligned}$$</p> | mcq | jee-main-2024-online-9th-april-evening-shift |
luy6z5lw | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>The coefficient of $$x^{70}$$ in $$x^2(1+x)^{98}+x^3(1+x)^{97}+x^4(1+x)^{96}+\ldots+x^{54}(1+x)^{46}$$ is $${ }^{99} \mathrm{C}_{\mathrm{p}}-{ }^{46} \mathrm{C}_{\mathrm{q}}$$. Then a possible value of $$\mathrm{p}+\mathrm{q}$$ is :</p> | [{"identifier": "A", "content": "61"}, {"identifier": "B", "content": "83"}, {"identifier": "C", "content": "55"}, {"identifier": "D", "content": "68"}] | ["B"] | null | <p>$$x^2(1+x)^{98}+x^3(1+x)^{97}+\ldots+x^{54}(1+x)^{46}$$</p>
<p>It is a G.P. with first term $$=x^2(1+x)^{98}$$</p>
<p>and common ratio $$=\frac{x}{1+x}$$</p>
<p>sum of these term $$=x^2(1+x)^{98}\left(\frac{\left(\frac{x}{1+x}\right)^{53}-1}{\frac{x}{1+x}-1}\right)$$</p>
<p>$$=x^2(1+x)^{98}\left((1+x)-x^{53}(1+x)^{-52}\right)$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw38sytv/376c0654-e73d-4b44-b3e8-2fec8771109f/52a01a30-1035-11ef-b980-477f779c8c59/file-1lw38sytw.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw38sytv/376c0654-e73d-4b44-b3e8-2fec8771109f/52a01a30-1035-11ef-b980-477f779c8c59/file-1lw38sytw.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 9th April Morning Shift Mathematics - Binomial Theorem Question 9 English Explanation"></p>
<p>$$\begin{aligned}
& ={ }^{99} \mathrm{C}_{68}-{ }^{46} \mathrm{C}_{15} \\
& \Rightarrow p=68, q=15 \\
& \Rightarrow p+q=83
\end{aligned}$$</p> | mcq | jee-main-2024-online-9th-april-morning-shift |
lv0vxdwu | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>Let $$a=1+\frac{{ }^2 \mathrm{C}_2}{3 !}+\frac{{ }^3 \mathrm{C}_2}{4 !}+\frac{{ }^4 \mathrm{C}_2}{5 !}+...., \mathrm{b}=1+\frac{{ }^1 \mathrm{C}_0+{ }^1 \mathrm{C}_1}{1 !}+\frac{{ }^2 \mathrm{C}_0+{ }^2 \mathrm{C}_1+{ }^2 \mathrm{C}_2}{2 !}+\frac{{ }^3 \mathrm{C}_0+{ }^3 \mathrm{C}_1+{ }^3 \mathrm{C}_2+{ }^3 \mathrm{C}_3}{3 !}+....$$ Then $$\frac{2 b}{a^2}$$ is equal to _________.</p> | [] | null | 8 | <p>$$\begin{aligned}
& a=1+\frac{{ }^2 C_2}{3!}+\frac{{ }^3 C_2}{4!}+\frac{{ }^4 C_2}{5!}+\ldots \\
& b=1+\frac{{ }^1 C_0+{ }^1 C_1}{1!}+\frac{{ }^2 C_0+{ }^2 C_1+{ }^2 C_2}{2!}+\ldots \\
& b=1+\frac{2}{1!}+\frac{2^2}{2!}+\frac{2}{3!}+\ldots=e^2
\end{aligned}$$</p>
<p>Using $$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x}{3!}+\ldots$$</p>
<p>$$\begin{aligned}
a= & 1+\sum_{r=2}^{\infty} \frac{{ }^r C_2}{(r+1)!}=1+\sum_{r=2} \frac{r(r-1)}{2(r+1)!} \\
& =1+\frac{1}{2} \sum_{r=2}^{\infty} \frac{(r+1) r-2 r}{(r+1)!} \\
& =1+\frac{1}{2} \sum_{r=2}^{\infty} \frac{1}{(r-1)!}-\frac{1}{2} \sum_{r=2} \frac{2 r}{(r+1)!} \\
& =1+\frac{1}{2}\left(\frac{1}{1!}+\frac{1}{2!}+\ldots\right)-\sum_{r=2}^{\infty} \frac{(r+1)-1}{(r+1)!} \\
& =1+\frac{1}{2}(e-1)-\sum_{r=2}^{\infty} \frac{1}{r!}+\sum_{r=2} \frac{1}{(r+1)!} \\
& =1+\frac{1}{2}(e-1)-\left(e-\frac{1}{1!}-\frac{1}{0!}\right)+\left(e-\frac{1}{1!}-\frac{1}{0!}-\frac{1}{2!}\right) \\
& =1+\frac{e}{2}-\frac{1}{2}-e+2+e-2-\frac{1}{2}=\frac{e}{2} \\
\Rightarrow & \frac{2 b}{a^2}=\frac{2}{\frac{e^2}{4}} 8
\end{aligned}$$</p> | integer | jee-main-2024-online-4th-april-morning-shift |
lv2eqvbz | maths | binomial-theorem | problems-based-on-binomial-co-efficient-and-collection-of-binomial-co-efficient | <p>If the coefficients of $$x^4, x^5$$ and $$x^6$$ in the expansion of $$(1+x)^n$$ are in the arithmetic progression, then the maximum value of $$n$$ is:</p> | [{"identifier": "A", "content": "28"}, {"identifier": "B", "content": "21"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "14"}] | ["D"] | null | <p>$$\begin{aligned}
& (1+x)^n={ }^n C_0+{ }^n C_1 x^1+{ }^n C_2 x^2+\ldots{ }^n C_n x^n \\
& { }^n C_4,{ }^n C_5 \&{ }^n C_6 \text { are in A.P. } \\
& { }^n C_5-{ }^n C_4={ }^n C_6-{ }^n C_5 \\
& \Rightarrow \frac{n!}{5!(n-5)!}-\frac{n!}{4!(n-4)!}=\frac{n!}{6!(n-6)!}-\frac{n!}{5!(n-5)!} \\
& \Rightarrow 30(n-9)(n-6)=5(n-4)(n-11) \\
& \Rightarrow 30 n^2-450 n+1620=5 n^2 \\
& \Rightarrow \frac{1}{n-5}\left[\frac{n-4-5}{5(n-4)}\right]=\frac{1}{5}\left[\frac{n-5-6}{6(n-5)}\right] \\
& \Rightarrow \frac{n-9}{5(n-4)}=\frac{1}{5}\left[\frac{n-11}{6}\right] \\
& \Rightarrow n^2-21 n+98=0 \\
& n_{\max }=14
\end{aligned}$$</p> | mcq | jee-main-2024-online-4th-april-evening-shift |
xQSkrUiNigxHOp8P | maths | circle | basic-theorems-of-a-circle | The equation of a circle with origin as a center and passing through an equilateral triangle whose median is of length $$3$$$$a$$ is : | [{"identifier": "A", "content": "$${x^2}\\, + \\,{y^2} = 9{a^2}$$ "}, {"identifier": "B", "content": "$${x^2}\\, + \\,{y^2} = 16{a^2}$$"}, {"identifier": "C", "content": "$${x^2}\\, + \\,{y^2} = 4{a^2}$$ "}, {"identifier": "D", "content": "$${x^2}\\, + \\,{y^2} = {a^2}$$"}] | ["C"] | null | Let $$ABC$$ be an equilateral triangle, whose median is $$AD.$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267493/exam_images/qh2bpsmsygtrhhpn1feb.webp" loading="lazy" alt="AIEEE 2002 Mathematics - Circle Question 157 English Explanation">
<br><br>Given $$AD=3a.$$
<br><br>In $$\Delta ABD,\,\,A{B^2} = A{D^2} + B{D^2};$$
<br><br>$$ \Rightarrow {x^2} = 9{a^2} + \left( {{x^2}/4} \right)\,\,$$
<br><br>where $$AB = BC = AC = x.$$
<br><br>$${3 \over 4}{x^2} = 9{a^2} \Rightarrow {x^2} = 12{a^2}.$$
<br><br>In $$\,\,\,\Delta OBD,O{B^2} = O{D^2} + B{D^2}$$
<br><br>$$ \Rightarrow {r^2} = {\left( {3a - r} \right)^2} + {{{x^2}} \over 4}$$
<br><br>$$ \Rightarrow {r^2} = 9{a^2} - 6ar + {r^2} + 3{a^2};$$
<br><br>$$ \Rightarrow 6ar = 12{a^2}$$
<br><br>$$ \Rightarrow r = 2a$$
<br><br>So equation of circle is $${x^2} + {y^2} = 4{a^2}$$ | mcq | aieee-2002 |
erkEpvhy5MhZbgZh | maths | circle | basic-theorems-of-a-circle | The point diametrically opposite to the point $$P(1, 0)$$ on the circle $${x^2} + {y^2} + 2x + 4y - 3 = 0$$ is : | [{"identifier": "A", "content": "$$(3, -4)$$"}, {"identifier": "B", "content": "$$(-3, 4)$$ "}, {"identifier": "C", "content": "$$(-3, -4)$$"}, {"identifier": "D", "content": "$$(3, 4)$$"}] | ["C"] | null | The given circle is $${x^2} + {y^2} + 2x + 4y - 3 = 0$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266849/exam_images/earxtruibjeyrajhopbc.webp" loading="lazy" alt="AIEEE 2008 Mathematics - Circle Question 144 English Explanation">
<br><br>Center $$(-1,-2)$$
<br><br>Let $$Q$$ $$\left( {\alpha ,\beta } \right)$$ be the point diametrically opposite to the point $$P(1,0),$$
<br><br>then $${{1 + \alpha } \over 2} = - 1$$ and $${{0 + \beta } \over 2} = - 2$$
<br><br>$$ \Rightarrow \alpha = - 3,\beta = - 4,$$ So, $$Q$$ is $$\left( { - 3, - 4} \right)$$ | mcq | aieee-2008 |
ExtzD08AoBipFNWA | maths | circle | basic-theorems-of-a-circle | Three distinct points A, B and C are given in the 2 -dimensional coordinates plane such that the ratio of the distance of any one of them from the point $$(1, 0)$$ to the distance from the point $$(-1, 0)$$ is equal to $${1 \over 3}$$. Then the circumcentre of the triangle ABC is at the point : | [{"identifier": "A", "content": "$$\\left( {{5 \\over 4},0} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( {{5 \\over 2},0} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {{5 \\over 3},0} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {0,0} \\right)$$ "}] | ["A"] | null | Given that
<br><br>$$P\left( {1,0} \right),Q\left( { - 1,0} \right)$$
<br><br>and $${{AP} \over {AQ}} = {{BP} \over {BQ}} = {{CP} \over {CQ}} = {1 \over 3}$$
<br><br>$$ \Rightarrow 3AP = AQ$$
<br><br>$$\,\,\,\,\,\,$$ Let $$A = (x,y)$$ then $$3AP = AQ \Rightarrow 9A{P^2} = A{Q^2}$$
<br><br>$$ \Rightarrow 9{\left( {x - 1} \right)^2} + 9{y^2} = {\left( {x + 1} \right)^2} + y{}^2$$
<br><br>$$ \Rightarrow 9{x^2} - 18x + 9 + 9{y^2} = {x^2} + 2x + 1 + {y^2}$$
<br><br>$$ \Rightarrow 8{x^2} - 20x + 8{y^2} + 8 = 0$$
<br><br>$$ \Rightarrow {x^2} + {y^2} - {5 \over 3}x + 1 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>$$\therefore$$ A lies on the circle given by eq. $$(1).$$ As $$B$$ and $$C$$
<br><br>also follow the same condition, - they must lie on the same circle.
<br><br>$$\therefore$$ Center of circumcircle of $$\Delta ABC$$
<br><br>$$=$$ Center of circle given by $$\left( 1 \right) = \left( {{5 \over 4},0} \right)$$ | mcq | aieee-2009 |
53q58LReyCXabFV0 | maths | circle | basic-theorems-of-a-circle | Locus of the image of the point $$(2, 3)$$ in the line $$\left( {2x - 3y + 4} \right) + k\left( {x - 2y + 3} \right) = 0,\,k \in R,$$ is a : | [{"identifier": "A", "content": "circle of radius $$\\sqrt 2 $$."}, {"identifier": "B", "content": "circle of radius $$\\sqrt 3 $$."}, {"identifier": "C", "content": "straight line parallel to $$x$$-axis "}, {"identifier": "D", "content": "straight line parallel to $$y$$-axis "}] | ["A"] | null | Intersection point of $$2x - 3y + 4 = 0$$
<br><br>and $$x-2y+3=0$$ is $$(1, 2)$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l91npsqa/a856b112-8560-4eba-878d-d29d808c6c0a/1868e920-47fd-11ed-8757-0f869593f41f/file-1l91npsqb.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l91npsqa/a856b112-8560-4eba-878d-d29d808c6c0a/1868e920-47fd-11ed-8757-0f869593f41f/file-1l91npsqb.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2015 (Offline) Mathematics - Circle Question 136 English Explanation"><br>Since, $$P$$ is the fixed point for given family of lines
<br><br>So, $$PB=PA$$
<br><br>$${\left( {\alpha - 1} \right)^2} + {\left( {\beta - 2} \right)^2} = {\left( {2 - 1} \right)^2} + {\left( {3 - 2} \right)^2}$$
<br><br>$${\left( {\alpha - 1} \right)^2} + {\left( {\beta - 2} \right)^2} = 1 + 1 = 2$$
<br><br>$${\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} = {\left( {\sqrt 2 } \right)^2}$$
<br><br>$${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}$$
<br><br>Therefore, given locus is a circle with center $$(1, 2)$$ and radius $$\sqrt 2 .$$ | mcq | jee-main-2015-offline |
Aqw7ajbxD5RLmqJW1LohR | maths | circle | basic-theorems-of-a-circle | The two adjacent sides of a cyclic quadrilateral are 2 and 5 and the angle between them is 60<sup>o</sup>. If the area of the quadrilateral is $$4\sqrt 3 $$, then the perimeter
of the quadrilateral is :
| [{"identifier": "A", "content": "12.5 "}, {"identifier": "B", "content": "13.2"}, {"identifier": "C", "content": "12"}, {"identifier": "D", "content": "13"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265487/exam_images/fnyfpg1ibpegkks9nf7r.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2017 (Online) 9th April Morning Slot Mathematics - Circle Question 124 English Explanation">
<br><br>Here; cos$$\theta $$ = $${{{a^2} + {b^2} - {c^2}} \over {2ab}}$$
<br><br>and $$\theta $$ = 60<sup>o</sup>
<br><br>$$ \Rightarrow $$ cos 60<sup>o</sup> = $${{4 + 25 - {c^2}} \over {2.2.5}}$$
<br><br>$$ \Rightarrow $$ 10 = 29 $$-$$ c<sup>2</sup>
<br><br>$$ \Rightarrow $$ c<sup>2</sup> = 19
<br><br>$$ \Rightarrow $$ <b>c = $$\sqrt {19} $$</b>
<br><br>also; cos$$\theta $$ = $${{{a^2} + {b^2} - {c^2}} \over {2ab}}$$
<br><br>and $$\theta $$ = 120<sup>o</sup>
<br><br>$$ \Rightarrow $$ $$-$$ $${1 \over 2}$$ = $${{{a^2} + {b^2} - 19} \over {2ab}}$$
<br><br>$$ \Rightarrow $$ a<sup>2</sup> + b<sup>2</sup> $$-$$ 19 = $$-$$ ab
<br><br>$$ \Rightarrow $$ a<sup>2</sup> + b<sup>2</sup> + ab = 19
<br><br>$$ \therefore $$ Area = $${1 \over 2} \times 2 \times 5$$ sin 60 + $${1 \over 2}$$ ab sin 120<sup>o</sup> = 4$$\sqrt 3 $$
<br><br>$$ \Rightarrow $$ $${{5\sqrt 3 } \over 2} + {{ab\sqrt 3 } \over 4}$$ = $$4\sqrt 3 $$
<br><br>$$ \Rightarrow $$ $${{ab} \over 4}$$ = 4 $$-$$ $${5 \over 2}$$ = $${3 \over 2}$$
<br><br>$$ \Rightarrow $$ <b>ab = 6</b>
<br><br>$$ \therefore $$ a<sup>2</sup> + b<sup>2</sup> = 13
<br><br>$$ \Rightarrow $$ a = 2, b = 3
<br><br>Perimeter = Sum of all sides
<br><br>= 2 + 5 + 2 + 3 = 12 | mcq | jee-main-2017-online-9th-april-morning-slot |
ZOFJXZMIfDQ8A2XV9I18hoxe66ijvww9skq | maths | circle | basic-theorems-of-a-circle | A rectangle is inscribed in a circle with a diameter
lying along the line 3y = x + 7. If the two adjacent
vertices of the rectangle are (–8, 5) and (6, 5), then
the area of the rectangle (in sq. units) is : | [{"identifier": "A", "content": "72"}, {"identifier": "B", "content": "84"}, {"identifier": "C", "content": "56"}, {"identifier": "D", "content": "98"}] | ["B"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263678/exam_images/ch1ypnxfax2wczso5gcp.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266115/exam_images/vwts7pqrwjuyf7wdy8m4.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266384/exam_images/sluvcuupdddpwc5u5sek.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th April Evening Slot Mathematics - Circle Question 107 English Explanation"></picture>
<br>Distance beetween point A and B is = 14 unit
<br><br>As y coordinate of point A(-8, 5) and B(6, 5) are same. So line AB is parallel to the x axis.
<br><br>Point O is the center of the circle and OP is the perpendicular to the line AB. And P is the mis point of line AB.
<br><br>$$ \therefore $$ Point P = $$\left( {{{ - 8 + 6} \over 2},{{5 + 5} \over 2}} \right)$$ = (-1, 5)
<br><br>As OP is perpendicular to the line AB and AB is parallel to the x axis then OP is parallel to the y axis. So x coordinate of line OP is constant.
<br><br>$$ \therefore $$ x coordinate of O = -1
<br><br>And point O is lies on the line 3y = x + 7
<br><br>$$ \therefore $$ 3y = -1 + 7
<br><br>$$ \Rightarrow $$ y = 2
<br><br>$$ \therefore $$ Center O = (-1, 2)
<br><br>$$ \therefore $$ OP = 3
<br><br>Then BC = 2OP = 6
<br><br>$$ \therefore $$ Area of rectangle = (AB)(BC)
<br><br>= 14 $$ \times $$ 6
<br><br>= 84
| mcq | jee-main-2019-online-9th-april-evening-slot |
EfmKOioHmsuvP1wBRvjgy2xukf0zcypw | maths | circle | basic-theorems-of-a-circle | The diameter of the circle, whose centre lies on
the line x + y = 2 in the first quadrant and which
touches both the lines x = 3 and y = 2, is
_______ . | [] | null | 3 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263616/exam_images/ssi5vdlnvmrlmfb1iuud.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 3rd September Morning Slot Mathematics - Circle Question 96 English Explanation">
<br>$$ \because $$ center lies on x + y = 2 and in 1<sup>st</sup> quadrant center = ($$\alpha $$, 2 $$-$$ $$\alpha $$)<br><br>where $$\alpha $$ > 0 and 2 $$-$$ $$\alpha $$ > 0 $$ \Rightarrow $$ 0 < $$\alpha $$ < 2<br><br>$$ \because $$ circle touches x = 3 and y = 2<br><br>$$ \therefore $$ $${{\left| {\alpha - 3} \right|} \over 1} = r$$<br><br>and $${{\left| {2 - (2 - \alpha )} \right|} \over 1} = r$$<br><br>$$ \Rightarrow \,|\alpha |\, = r$$<br><br>$$ \therefore $$ $$|\alpha - 3|\, = \,|\alpha |$$<br><br>$$ \Rightarrow $$ $${\alpha ^2} - 6\alpha + 9 = {\alpha ^2}$$<br><br>$$ \Rightarrow \alpha = {3 \over 2}$$<br><br>$$ \therefore $$ $$r = {3 \over 2}$$<br><br>$$ \Rightarrow $$ 2r = 3 = diameter. | integer | jee-main-2020-online-3rd-september-morning-slot |
xYtfGaltEQHT5L76yWjgy2xukfal1q5d | maths | circle | basic-theorems-of-a-circle | Let PQ be a diameter of the circle x<sup>2</sup> + y<sup>2</sup> = 9. If $$\alpha $$ and $$\beta $$ are the lengths of the perpendiculars from P and Q on the straight line,<br/> x + y = 2 respectively, then the maximum value of $$\alpha\beta $$ is _____. | [] | null | 7 | Let $$P(3\cos \theta ,\,3\sin \theta )$$<br><br>$$Q( - 3\cos \theta ,\, - 3\sin \theta )$$<br><br>$$\alpha = \left| {{{3\cos \theta + 3\sin \theta - 2} \over {\sqrt 2 }}} \right|$$<br><br>$$\beta = \left| {{{ - 3\cos \theta - 3\sin \theta - 2} \over {\sqrt 2 }}} \right|$$<br><br>$$\alpha \beta = \left| {{{{{\left( {3\cos \theta + 3\sin \theta } \right)}^2} - 4} \over 2}} \right|$$<br><br>$$ = \left| {{{5 + 9\sin 2\theta } \over 2}} \right|$$<br><br>$$\alpha {\beta _{\max }}$$$$ = {{5 + 9} \over 2} = 7$$ (when sin2$$\theta $$ = 1) | integer | jee-main-2020-online-4th-september-evening-slot |
0m8EX9yisbE45d5O431klrmflld | maths | circle | basic-theorems-of-a-circle | Let a point P be such that its distance from the point (5, 0) is thrice the distance of P from the point ($$-$$5, 0). If the locus of the point P is a circle of radius r, then 4r<sup>2</sup> is equal to ________ | [] | null | 56 | Let P(h, k)<br><br>Given<br><br>PA = 3PB<br><br>PA<sup>2</sup> = 9PB<sup>2</sup><br><br>$$ \Rightarrow $$ (h $$-$$ 5)<sup>2</sup> + k<sup>2</sup> = 9[(h + 5)<sup>2</sup> + k<sup>2</sup>]<br><br>$$ \Rightarrow $$ 8h<sup>2</sup> + 8k<sup>2</sup> + 100h + 200 = 0<br><br>$$ \therefore $$ Locus<br><br>$${x^2} + {y^2} + \left( {{{25} \over 2}} \right)x + 25 = 0$$<br><br>$$ \therefore $$ $$c \equiv \left( {{{ - 25} \over 4},0} \right)$$<br><br>$$ \therefore $$ $${r^2} = {\left( {{{ - 25} \over 4}} \right)^2} - 25$$<br><br>$$ = {{625} \over {16}} - 25$$<br><br>$$ = {{225} \over {16}}$$<br><br>$$ \therefore $$ $$4{r^2} = 4 \times {{225} \over {16}} = {{225} \over 4} = 56.25$$<br><br>After Round of 4r<sup>2</sup> = 56 | integer | jee-main-2021-online-24th-february-evening-slot |
OavNaE5a4W4jEIN4sj1klugtqbo | maths | circle | basic-theorems-of-a-circle | In the circle given below, let OA = 1 unit, OB = 13 unit and PQ $$ \bot $$ OB. Then, the area of the triangle PQB (in square units) is :<br/><br/><img src="data:image/png;base64,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"/> | [{"identifier": "A", "content": "24$$\\sqrt 2 $$"}, {"identifier": "B", "content": "24$$\\sqrt 3 $$"}, {"identifier": "C", "content": "26$$\\sqrt 2 $$"}, {"identifier": "D", "content": "26$$\\sqrt 3 $$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266436/exam_images/ms5tyuzkyy8vnwsy0d9h.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th February Morning Shift Mathematics - Circle Question 89 English Explanation"><br><br>Let PA = AQ = $$\lambda$$<br><br>OA . AB = AP . AQ<br><br>$$ \Rightarrow $$ 1.12 = $$\lambda$$ . $$\lambda$$<br><br>$$ \Rightarrow $$ $$\lambda$$ = 2$$\sqrt 3 $$<br><br>Area $$\Delta$$PQB = $${1 \over 2}$$ $$\times$$ 2$$\lambda$$ $$\times$$ AB<br><br>$$\Delta$$ = $${1 \over 2}$$ . 4$$\sqrt 3 $$ $$\times$$ 12<br><br>= 24$$\sqrt 3 $$ | mcq | jee-main-2021-online-26th-february-morning-slot |
1ks09eg6q | maths | circle | basic-theorems-of-a-circle | Let P and Q be two distinct points on a circle which has center at C(2, 3) and which passes through origin O. If OC is perpendicular to both the line segments CP and CQ, then the set {P, Q} is equal to : | [{"identifier": "A", "content": "{(4, 0), (0, 6)}"}, {"identifier": "B", "content": "$$\\{ (2 + 2\\sqrt 2 ,3 - \\sqrt 5 ),(2 - 2\\sqrt 2 ,3 + \\sqrt 5 )\\} $$"}, {"identifier": "C", "content": "$$\\{ (2 + 2\\sqrt 2 ,3 + \\sqrt 5 ),(2 - 2\\sqrt 2 ,3 - \\sqrt 5 )\\} $$"}, {"identifier": "D", "content": "{($$-$$1, 5), (5, 1)}"}] | ["D"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265883/exam_images/j4eb4u3olf4aobpzkvfj.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263397/exam_images/a0quemgl8enbtkiviozm.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264845/exam_images/vn5mptmseenvrcsn84v1.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Morning Shift Mathematics - Circle Question 74 English Explanation"></picture> <br><br>$$\tan \theta = - {2 \over 3}$$<br><br>Using symmetric from of line<br><br>$$P,Q:\left( {2 \pm \sqrt {13} \cos \theta ,3 \pm \sqrt {13} \sin \theta } \right)$$<br><br>$$\left( {2 \pm \sqrt {13} .\left( { - {3 \over {\sqrt {13} }}} \right),3 \pm \sqrt {13} \left( {{2 \over {\sqrt {13} }}} \right)} \right)$$<br><br>($$-$$1, 5) & (5, 1) | mcq | jee-main-2021-online-27th-july-morning-shift |
1ks0au11w | maths | circle | basic-theorems-of-a-circle | Let $$A = \{ (x,y) \in R \times R|2{x^2} + 2{y^2} - 2x - 2y = 1\} $$, $$B = \{ (x,y) \in R \times R|4{x^2} + 4{y^2} - 16y + 7 = 0\} $$ and $$C = \{ (x,y) \in R \times R|{x^2} + {y^2} - 4x - 2y + 5 \le {r^2}\} $$.<br/><br/>Then the minimum value of |r| such that $$A \cup B \subseteq C$$ is equal to | [{"identifier": "A", "content": "$${{3 + \\sqrt {10} } \\over 2}$$"}, {"identifier": "B", "content": "$${{2 + \\sqrt {10} } \\over 2}$$"}, {"identifier": "C", "content": "$${{3 + 2\\sqrt 5 } \\over 2}$$"}, {"identifier": "D", "content": "$$1 + \\sqrt 5 $$"}] | ["C"] | null | $${S_1}:{x^2} + {y^2} - x - y - {1 \over 2} = 0;{C_1}\left( {{1 \over 2},{1 \over 2}} \right)$$<br><br>$${r_1} = \sqrt {{1 \over 4} + {1 \over 4} + {1 \over 2}} = 1$$<br><br>$${S_2}:{x^2} + {y^2} - 4y + {7 \over 4} = 0;{C_2}:(0,2)$$<br><br>$${r_2} = \sqrt {4 - {7 \over 4}} = {3 \over 2}$$<br><br>$${S_3} = {x^2} + {y^2} - 4x - 2y + 5 - {r^2} = 0$$<br><br>C<sub>3</sub> (2, 1)<br><br>$${r_3} = \sqrt {4 + 1 - 5 + {r^2}} = |r|$$<br><br> <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263911/exam_images/cfj3zwxthsujcfnh8i2z.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Morning Shift Mathematics - Circle Question 72 English Explanation"> <br><br>$${C_1}{C_3} = \sqrt {{5 \over 2}} $$<br><br>$$\sqrt {{5 \over 2}} \le |r - 1| \Rightarrow \left. \matrix{
r \le 1 + \sqrt {{5 \over 2}} \hfill \cr
r \ge {3 \over 2} + \sqrt 5 \hfill \cr} \right\}$$<br><br>$${C_2}{C_3} = \sqrt 5 \le \left| {r - {3 \over 2}} \right|$$<br><br>$$\left. \matrix{
r - {3 \over 2} \ge \sqrt 5 \hfill \cr
r - {3 \over 2} \le - \sqrt 5 \hfill \cr} \right\}$$ | mcq | jee-main-2021-online-27th-july-morning-shift |
1ktbiovly | maths | circle | basic-theorems-of-a-circle | The locus of a point, which moves such that the sum of squares of its distances from the points (0, 0), (1, 0), (0, 1), (1, 1) is 18 units, is a circle of diameter d. Then d<sup>2</sup> is equal to _____________. | [] | null | 16 | Let point P(x, y)
<br><br>A(0, 0), B(1, 0), C(0, 1), D(1, 1)
<br><br>(PA)<sup>2</sup> + (PB)<sup>2</sup> + (PC)<sup>2</sup> + (PD)<sup>2</sup> = 18
<br><br>$${x^2} + {y^2} + {x^2} + {(y - 1)^2} + {(x - 1)^2} + {y^2} + {(x - 1)^2} + {(y - 1)^2}$$ = 18<br><br>$$ \Rightarrow 4({x^2} + {y^2}) - 4y - 4x = 14$$<br><br>$$ \Rightarrow {x^2} + {y^2} - x - y - {7 \over 2} = 0$$<br><br>$$d = 2\sqrt {{1 \over 4} + {1 \over 4} + {7 \over 2}} $$<br><br>$$ \Rightarrow {d^2} = 16$$ | integer | jee-main-2021-online-26th-august-morning-shift |
1ktg2remf | maths | circle | basic-theorems-of-a-circle | Let Z be the set of all integers,<br/><br/>$$A = \{ (x,y) \in Z \times Z:{(x - 2)^2} + {y^2} \le 4\} $$<br/><br/>$$B = \{ (x,y) \in Z \times Z:{x^2} + {y^2} \le 4\} $$<br/><br/>$$C = \{ (x,y) \in Z \times Z:{(x - 2)^2} + {(y - 2)^2} \le 4\} $$<br/><br/>If the total number of relation from A $$\cap$$ B to A $$\cap$$ C is 2<sup>p</sup>, then the value of p is : | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "49"}, {"identifier": "D", "content": "9"}] | ["B"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265832/exam_images/dtgcpzs2ufin73ryqlsn.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264735/exam_images/vsklxgqplszmejwhrtui.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267186/exam_images/huf8jytaaw2g3husjhna.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th August Evening Shift Mathematics - Circle Question 67 English Explanation"></picture> <br><br>(x $$-$$ 2)<sup>2</sup> + y<sup>2</sup> $$\le$$ 4<br><br>x<sup>2</sup> + y<sup>2</sup> $$\le$$ 4<br><br>No. of points common in C<sub>1</sub> & C<sub>2</sub> is 5.<br><br>(0, 0), (1, 0), (2, 0), (1, 1), (1, $$-$$1)<br><br>Similarly in C<sub>2</sub> & C<sub>3</sub> is 5.<br><br>No. of relations = 2<sup>$$ 5 \times 5$$</sup> = 2<sup>25</sup>. | mcq | jee-main-2021-online-27th-august-evening-shift |
1l54bapov | maths | circle | basic-theorems-of-a-circle | <p>Let a triangle ABC be inscribed in the circle $${x^2} - \sqrt 2 (x + y) + {y^2} = 0$$ such that $$\angle BAC = {\pi \over 2}$$. If the length of side AB is $$\sqrt 2 $$, then the area of the $$\Delta$$ABC is equal to :</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$$\\left( {\\sqrt 6 + \\sqrt 3 } \\right)/2$$"}, {"identifier": "C", "content": "$$\\left( {3 + \\sqrt 3 } \\right)/4$$"}, {"identifier": "D", "content": "$$\\left( {\\sqrt 6 + 2\\sqrt 3 } \\right)/4$$"}] | ["A"] | null | <p>Note:</p>
<p>For equation of circle $${x^2} + {y^2} + 2gx + 2fy + c = 0$$, center is $$( - g,\, - f)$$ and radius $$r = \sqrt {{g^2} + {f^2} - c} $$</p>
<p>Given,</p>
<p>equation of circle is</p>
<p>$${x^2} - \sqrt 2 (x + y) + {y^2} = 0$$</p>
<p>$$ \Rightarrow {x^2} + {y^2} - \sqrt 2 x - \sqrt 2 y = 0$$</p>
<p>$$ \Rightarrow {x^2} + {y^2} + 2\left( { - {1 \over {\sqrt 2 }}} \right)x + 2\left( { - {1 \over {\sqrt 2 }}} \right) = 0$$</p>
<p>$$\therefore$$ $$g = - {1 \over {\sqrt 2 }}$$ and $$f = - {1 \over {\sqrt 2 }}$$</p>
<p>$$\therefore$$ Center $$ = ( - g,\, - f) = \left( {{1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }}} \right)$$</p>
<p>And Radius $$ = r = \sqrt {{{\left( { - {1 \over {\sqrt 2 }}} \right)}^2} + {{\left( { - {1 \over {\sqrt 2 }}} \right)}^2} - 0} $$</p>
<p>$$ = \sqrt {{1 \over 2} + {1 \over 2}} = \sqrt 1 = 1$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5sczyp3/e6de313f-31b7-4de8-8ad3-c03114d47130/7ab55770-077b-11ed-94f6-83604aa63acb/file-1l5sczyp4.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5sczyp3/e6de313f-31b7-4de8-8ad3-c03114d47130/7ab55770-077b-11ed-94f6-83604aa63acb/file-1l5sczyp4.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th June Evening Shift Mathematics - Circle Question 62 English Explanation"></p>
<p>As AB and AC makes an angle 90$$^\circ$$ then line BC passes through the center of circle and BC is the diameter of the circle.</p>
<p>$$\therefore$$ Length of BC = 2r = 2 $$\times$$ 1 = 2</p>
<p>$$\therefore$$ AC<sup>2</sup> = BC<sup>2</sup> $$-$$ AB<sup>2</sup></p>
<p>= 2<sup>2</sup> $$-$$ ($$\sqrt2$$)<sup>2</sup></p>
<p>= 2</p>
<p>$$\Rightarrow$$ AC = $$\sqrt2$$</p>
<p>$$\therefore$$ Area of right angle triangle ABC</p>
<p>= $${1 \over 2}$$ $$\times$$ AC $$\times$$ AB</p>
<p>= $${1 \over 2}$$ $$\times$$ $$\sqrt2$$ $$\sqrt2$$</p>
<p>= 1 square unit.</p> | mcq | jee-main-2022-online-29th-june-evening-shift |
1l57p6r2o | maths | circle | basic-theorems-of-a-circle | <p>A rectangle R with end points of one of its sides as (1, 2) and (3, 6) is inscribed in a circle. If the equation of a diameter of the circle is 2x $$-$$ y + 4 = 0, then the area of R is ____________.</p> | [] | null | 16 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5qa5m7e/7e601531-e009-4eeb-99e6-ea8138a03412/cb3a40a0-0656-11ed-903e-c9687588b3f3/file-1l5qa5m7f.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5qa5m7e/7e601531-e009-4eeb-99e6-ea8138a03412/cb3a40a0-0656-11ed-903e-c9687588b3f3/file-1l5qa5m7f.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 27th June Morning Shift Mathematics - Circle Question 56 English Explanation"></p>
<p>As slope of line joining (1, 2) and (3, 6) is 2 given diameter is parallel to side</p>
<p>$$\therefore$$ $$a = \sqrt {{{(3 - 1)}^2} + {{(6 - 2)}^2}} = \sqrt {20} $$</p>
<p>and $$b/2 = {4 \over {\sqrt 5 }} \Rightarrow b = {8 \over {\sqrt 5 }}$$</p>
<p>Area $$ = ab = 2\sqrt 5 \,.\,{8 \over {\sqrt 5 }} = 16$$.</p> | integer | jee-main-2022-online-27th-june-morning-shift |
1l6rg5oz4 | maths | circle | basic-theorems-of-a-circle | <p>$$\text { Let } S=\left\{(x, y) \in \mathbb{N} \times \mathbb{N}: 9(x-3)^{2}+16(y-4)^{2} \leq 144\right\}$$ and
$$T=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}:(x-7)^{2}+(y-4)^{2} \leq 36\right\}$$. Then $$n(S \cap T)$$ is equal to __________.</p> | [] | null | 27 | $S=\left\{(x, y) \in \mathbb{N} \times \mathbb{N}: \frac{(x-3)^{2}}{16}+\frac{(y-4)^{2}}{9} \leq 1\right\}$ <br><br>represents all the integral points inside <br><br>and on the ellipse $\frac{(x-3)^{2}}{16}+\frac{(y-4)^{2}}{9}=1$, in first quadrant.
<br><br>
and $T=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}:(x-7)^{2}+(y-4)^{2} \leq 36\right\}$ represents all the points on <br><br>and inside the circle $(x-7)^{2}+(y-4)^{2}=36$<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lc8cegbu/4182b4ab-8c51-4d29-951d-0c7f9f73f6e7/fb3108a0-870e-11ed-9cef-67b5ed764fcf/file-1lc8cegbv.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lc8cegbu/4182b4ab-8c51-4d29-951d-0c7f9f73f6e7/fb3108a0-870e-11ed-9cef-67b5ed764fcf/file-1lc8cegbv.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th July Evening Shift Mathematics - Circle Question 40 English Explanation"><br>
$\therefore \quad n(S \cap T)=\{(3,1),(2,2),(3,2),(4,2),(5,2)$, $(2,3), \ldots(6,5)\}$
<br><br>
Total number of points $=27$ | integer | jee-main-2022-online-29th-july-evening-shift |
ldr0ffpu | maths | circle | basic-theorems-of-a-circle | Let $P\left(a_1, b_1\right)$ and $Q\left(a_2, b_2\right)$ be two distinct points on a circle with center $C(\sqrt{2}, \sqrt{3})$. Let $\mathrm{O}$ be the origin and $\mathrm{OC}$ be perpendicular to both $\mathrm{CP}$ and $\mathrm{CQ}$. If the area of the triangle $\mathrm{OCP}$ is $\frac{\sqrt{35}}{2}$, then $a_1^2+a_2^2+b_1^2+b_2^2$ is equal to : | [] | null | 24 | <p>$$OC \,\bot \,CP$$ and $$OC \,\bot \, CQ$$</p>
<p>$$\Rightarrow PCQ$$ is a straight line</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leol1wpz/d1fe7720-2e72-4e7b-baef-e1168e337218/c9a57f70-b795-11ed-b103-ed967fad3dff/file-1leol1wq0.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1leol1wpz/d1fe7720-2e72-4e7b-baef-e1168e337218/c9a57f70-b795-11ed-b103-ed967fad3dff/file-1leol1wq0.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 30th January Evening Shift Mathematics - Circle Question 37 English Explanation"></p>
<p>$$OC = \sqrt {{{(\sqrt 2 )}^2} + {{(\sqrt 3 )}^2}} = \sqrt 5 $$</p>
<p>Let $$CP = CQ = I$$</p>
<p>$$[OCP] = {1 \over 2} \times OC \times I = {{\sqrt {35} } \over 2}$$</p>
<p>$$I = \sqrt 7 $$</p>
<p>$$OP = OQ = \sqrt {{{(OC)}^2} + {I^2}} = \sqrt {5 + 7} = \sqrt {12} $$</p>
<p>$$a_1^2 + a_2^2 + b_1^2 + b_2^2 = \left( {a_1^2 + b_2^2} \right) + \left( {a_2^2 + b_2^2} \right)$$</p>
<p>$$O{P^2} + O{Q^2} = 12 + 12 = 24$$</p> | integer | jee-main-2023-online-30th-january-evening-shift |
1ldv1zshn | maths | circle | basic-theorems-of-a-circle | <p>The points of intersection of the line $$ax + by = 0,(a \ne b)$$ and the circle $${x^2} + {y^2} - 2x = 0$$ are $$A(\alpha ,0)$$ and $$B(1,\beta )$$. The image of the circle with AB as a diameter in the line $$x + y + 2 = 0$$ is :</p> | [{"identifier": "A", "content": "$${x^2} + {y^2} + 5x + 5y + 12 = 0$$"}, {"identifier": "B", "content": "$${x^2} + {y^2} + 3x + 5y + 8 = 0$$"}, {"identifier": "C", "content": "$${x^2} + {y^2} - 5x - 5y + 12 = 0$$"}, {"identifier": "D", "content": "$${x^2} + {y^2} + 3x + 3y + 4 = 0$$"}] | ["A"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lebxhw5x/9e3df15a-0f74-4149-87bf-67c2379ca244/4344ed40-b0a0-11ed-a0da-1fff956b892c/file-1lebxhw5y.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lebxhw5x/9e3df15a-0f74-4149-87bf-67c2379ca244/4344ed40-b0a0-11ed-a0da-1fff956b892c/file-1lebxhw5y.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 25th January Morning Shift Mathematics - Circle Question 32 English Explanation"><br>
As $A$ and $B$ satisfy both line and circle we have $\alpha=0 \Rightarrow A(0,0)$ and $\beta=1$ i.e. $B(1,1)$
<br><br>
Centre of circle as $A B$ diameter is $\left(\frac{1}{2}, \frac{1}{2}\right)$ and radius $=\frac{1}{\sqrt{2}}$
<br><br>
$\therefore$ For image of $\left(\frac{1}{2} ; \frac{1}{2}\right)$ in $x+y+z$ we get $\frac{x-\frac{1}{2}}{1}=\frac{y-\frac{1}{2}}{1}=\frac{-2(3)}{2}$
<br><br>
$\Rightarrow$ Image $\left(-\frac{5}{2},-\frac{5}{2}\right)$<br><br>$\therefore $ Equation of required circle
<br><br>
$\left(x+\frac{5}{2}\right)^{2}+\left(y+\frac{5}{2}\right)^{2}=\frac{1}{2}$
<br><br>
$\Rightarrow x^{2}+y^{2}+5 x+5 y+\frac{50}{4}-\frac{1}{2}=0$
<br><br>
$\Rightarrow x^{2}+y^{2}+5 x+5 y+12=0$ | mcq | jee-main-2023-online-25th-january-morning-shift |
1lh00nw5i | maths | circle | basic-theorems-of-a-circle | <p>Consider a circle $$C_{1}: x^{2}+y^{2}-4 x-2 y=\alpha-5$$. Let its mirror image in the line $$y=2 x+1$$ be another circle $$C_{2}: 5 x^{2}+5 y^{2}-10 f x-10 g y+36=0$$. Let $$r$$ be the radius of $$C_{2}$$. Then $$\alpha+r$$ is equal to _________.</p> | [] | null | 2 | We have,
<br/><br/>$$
\begin{aligned}
& C_1: x^2+y^2-4 x-2 y=\alpha-5 \\\\
& C_1:(x-2)^2+(y-1)^5-5=\alpha-5 \\\\
& C_1:(x-2)^2+(y-1)^2=(\sqrt{\alpha})^2
\end{aligned}
$$
<br/><br/>So, centre and radius of $C_1$ are $(2,1)$ and $\sqrt{\alpha}$ respectively
<br/><br/>Now, image of $(2,1)$ along the line $y=2 x+1$ is,
<br/><br/>$$
\frac{x-2}{2}=\frac{y-1}{-1}=\frac{-2(4-1+1)}{2^2+(-1)^2}
$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \frac{x-2}{2}=\frac{y-1}{-1}=\frac{-8}{5} \\\\
& \Rightarrow x=\frac{-6}{5} \text { and } y=\frac{13}{5}
\end{aligned}
$$
<br/><br/>Now, $\left(\frac{-6}{5}, \frac{13}{5}\right)$ will be the centre of $C_2$
<br/><br/>$$
\therefore f=\frac{6}{5} \text { and } g=\frac{-13}{5}
$$
<br/><br/>Now, radius of $\mathrm{C}_2=r=\sqrt{f^2+g^2-\frac{36}{5}}$
<br/><br/>$$
\begin{aligned}
& \Rightarrow r=\sqrt{\frac{36}{25}+\frac{169}{25}-\frac{36}{5}}=1 \\\\
& \because r=1 \text { so, } \alpha=1 \\\\
& \therefore \alpha+r=1+1=2
\end{aligned}
$$
<br/><br/><b>Concept :</b>
<br/><br/>Image of a point $\left(x_1, y_1\right)$ w.r.t. $a x+b y+c=0$ is $(x, y)$, then
<br/><br/>$$
\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{-2\left(a x_1+b y_1+c\right)}{\left(a^2+b^2\right)}
$$ | integer | jee-main-2023-online-8th-april-morning-shift |
1lh23ruwz | maths | circle | basic-theorems-of-a-circle | <p>Let the point $$(p, p+1)$$ lie inside the region $$E=\left\{(x, y): 3-x \leq y \leq \sqrt{9-x^{2}}, 0 \leq x \leq 3\right\}$$. If the set of all values of $$\mathrm{p}$$ is the interval $$(a, b)$$, then $$b^{2}+b-a^{2}$$ is equal to ___________.</p> | [] | null | 3 | Given region,
<br/><br/>$$
E=\left\{(x, y): 3-x \leq y \leq \sqrt{9-x^2}, 0 \leq x \leq 3\right\}
$$
<br/><br/>Since, point $(p, p+1)$ lie on line $y=x+1$
<br/><br/>$\therefore$ Point of intersection of $y=x+1$ and $y=3-x$
<br/><br/>i.e.,
$x+1=3-x$
<br/><br/>$\Rightarrow$ $2 x=2 \Rightarrow x=1$
<br/><br/>and $y=2$
<br/><br/>and point of intersection of $y=x+1$ and
<br/><br/>$$
y=\sqrt{9-x^2}
$$
<br/><br/>i.e., $(x+1)^2=9-x^2$
<br/><br/>$\begin{array}{lc}
&\Rightarrow x^2+1+2 x=9-x^2 \\\\
&\Rightarrow 2 x^2+2 x-8=0 \\\\
&\Rightarrow x^2+x-4=0 \\\\
&\Rightarrow x=\frac{-1 \pm \sqrt{1+4(1)(4)}}{2} \\\\
&\Rightarrow x=\frac{-1 \pm \sqrt{17}}{2}\end{array}$
<br/><br/>$\begin{array}{lll}\Rightarrow & x=\frac{-1+\sqrt{17}}{2}, & \text { (Since, } x \in[0,3]) \\\\ \therefore & p \in \left(1, \frac{-1+\sqrt{17}}{2}\right)\end{array}$
<br/><br/>$\Rightarrow a=1, b=\frac{-1+\sqrt{17}}{2}$
<br/><br/>$\begin{aligned} & \therefore b^2+b-a \\\\ &= \frac{1+17-2 \sqrt{17}}{4}+\frac{(-1+\sqrt{17})}{2}-1 \\\\ &= \frac{18-2 \sqrt{17}-2+2 \sqrt{17}-4}{4} \\\\ &= \frac{12}{4}=3\end{aligned}$ | integer | jee-main-2023-online-6th-april-morning-shift |
luy6z5cm | maths | circle | basic-theorems-of-a-circle | <p>Let a circle passing through $$(2,0)$$ have its centre at the point $$(\mathrm{h}, \mathrm{k})$$. Let $$(x_{\mathrm{c}}, y_{\mathrm{c}})$$ be the point of intersection of the lines $$3 x+5 y=1$$ and $$(2+\mathrm{c}) x+5 \mathrm{c}^2 y=1$$. If $$\mathrm{h}=\lim _\limits{\mathrm{c} \rightarrow 1} x_{\mathrm{c}}$$ and $$\mathrm{k}=\lim _\limits{\mathrm{c} \rightarrow 1} y_{\mathrm{c}}$$, then the equation of the circle is :</p> | [{"identifier": "A", "content": "$$5 x^2+5 y^2-4 x-2 y-12=0$$\n"}, {"identifier": "B", "content": "$$25 x^2+25 y^2-20 x+2 y-60=0$$\n"}, {"identifier": "C", "content": "$$25 x^2+25 y^2-2 x+2 y-60=0$$\n"}, {"identifier": "D", "content": "$$5 x^2+5 y^2-4 x+2 y-12=0$$"}] | ["B"] | null | <p>$$\begin{aligned}
& 3 x+5 y=1 \\
& (2+c) x+5 c^2 y=1 \\
& 3 c^2 x+5 c^2 y=c^2
\end{aligned}$$</p>
<p>Subtracting</p>
<p>$$\begin{aligned}
& \left(2+c-3 c^2\right) x=1-c^2 \\
& x_c=\frac{1-c^2}{2+c-3 c^2}=\frac{(1-c)(1+c)}{(1-c)(3 c+2)}=\frac{c+1}{3 c+2} \\
& y=\frac{1-3 x}{5}=\frac{1-3\left(\frac{c+1}{3 c+2}\right)}{5} \\
& =\frac{3 c+2-3 c-3}{5(3 c+2)} \\
& y_c=\frac{-1}{5(3 c+2)} \\
& \lim _{c \rightarrow 1} x_c=\frac{2}{5}=h \\
& \lim _{c \rightarrow 1} y_c=\frac{-1}{25}=k
\end{aligned}$$</p>
<p>Equation of circle is</p>
<p>$$25 x^2+25 y^2-20 x+2 y-60=0$$</p> | mcq | jee-main-2024-online-9th-april-morning-shift |
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