Binomial Theorem and its Simple Applications jee mains questions | Binomial Theorem and its Simple Applications Previous Year Question

Q 21 :
If the co-efficient of $x^{9}$ in ${(α x^{3} + \frac{1}{β x})}^{11}$ and the co-efficient of $x^{- 9}$ in ${(α x - \frac{1}{β x^{3}})}^{11}$ are equal, then ${(α β)}^{2}$ is equal to ________ . [2023]

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(1)

${(r + 1)}^{th} term in$ ${(α x^{3} + \frac{1}{β x})}^{11} = {}^{11}C_{r} α^{11 - r} \cdot β^{- r} \cdot x^{33 - 4 r}$

$Here, 33 - 4 r = 9 \Rightarrow r = 6$

$∴$ $Coefficient of x^{9} = {}^{11}C_{6} α^{5} β^{- 6}$

${(r + 1)}^{th} term in$ ${(α x - \frac{1}{β x^{3}})}^{11} = {}^{11}C_{r} α^{11 - r} {(- \frac{1}{β})}^{r} \cdot x^{11 - 4 r}$

$Here, 11 - 4 r = - 9 \Rightarrow r = 5$

$∴$ $Coefficient of x^{- 9} = - {}^{11}C_{5} α^{6} β^{- 5}$

$Now, {}^{11}C_{6} \frac{α^{5}}{β^{6}} = - {}^{11}C_{5} \frac{α^{6}}{β^{5}} [∵ Given] \Rightarrow \frac{1}{β} = - α \Rightarrow {(α β)}^{2} = 1$

Q 22 :
Let $α > 0$ , be the smallest number such that the expansion of ${(x^{\frac{2}{3}} + \frac{2}{x^{3}})}^{30}$ has a term $β x^{- α}, β \in N .$ Then $α$ is equal to ________ . [2023]

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(2)

$Here, T_{r + 1} = {}^{30}C_{r} {(x^{2 / 3})}^{30 - r} {(\frac{2}{x^{3}})}^{r}$

$= {}^{30}C_{r} x^{(20 - \frac{2 r}{3} - 3 r)} {(2)}^{r} = {}^{30}C_{r} {(2)}^{r} x^{(\frac{60 - 11 r}{3})}; 0 \leq r \leq 30$

$For β x^{- α}, α > 0, r = 6, and T_{7} = {}^{30}C_{6} {(2)}^{6} x^{- 2} \Rightarrow α = 2$

Q 23 :
The coefficient of $x^{- 6}$ , in the expansion of ${(\frac{4 x}{5} + \frac{5}{2 x^{2}})}^{9}$ , is __________ . [2023]

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(5040)

$T_{r + 1} = {}^{9}C_{r} {(\frac{4 x}{5})}^{9 - r} {(\frac{5}{2 x^{2}})}^{r}$

$= {}^{9}C_{r} {(\frac{4}{5})}^{9 - r} {(\frac{5}{2})}^{r} \cdot x^{9 - r - 2 r} \Rightarrow x^{9 - r - 2 r} = x^{- 6}$

$\Rightarrow 9 - r - 2 r = - 6 \Rightarrow 3 r = 15 \Rightarrow r = 5$

$So, T_{6} = {}^{9}C_{5} {(\frac{4}{5})}^{4} {(\frac{5}{2})}^{5} x^{- 6}$

$= \frac{9!}{5! 4!} (\frac{4 \times 4 \times 4 \times 4}{5 \times 5 \times 5 \times 5}) (\frac{5 \times 5 \times 5 \times 5 \times 5}{2 \times 2 \times 2 \times 2 \times 2}) x^{- 6}$

$= \frac{9 \cdot 8 \cdot 7 \cdot 6}{4 \cdot 3 \cdot 2} (8) (5) x^{- 6} = 126 \times 40 \cdot x^{- 6} = 5040$

Q 24 :
If the constant term in the binomial expansion of ${(\frac{x^{\frac{5}{2}}}{2} - \frac{4}{x^{l}})}^{9}$ is $- 84$ and the coefficient of $x^{- 3 l}$ is $2^{α} β$ , where $β < 0$ is an odd number, then $| α l - β |$ is equal to __________ . [2023]

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(98)

$T_{r + 1} = {}^{9}C_{r} {\frac{(x^{5 / 2})}{2^{9 - r}}}^{9 - r} {(\frac{- 4}{x^{l}})}^{r}$

$= {(- 1)}^{r} \frac{{}^{9}C_{r}}{2^{9 - r}} 4^{r} x^{\frac{45}{2} - \frac{5 r}{2} - l r} \Rightarrow 45 - 5 r - 2 l r = 0$

$r = \frac{45}{5 + 2 l} \dots (i)$

$Now, according to the question, {(- 1)}^{r} \frac{{}^{9}C_{r}}{2^{9 - r}} 4^{r} = - 84$

$\Rightarrow {(- 1)}^{r} {}^{9}C_{r} 2^{3 r - 9} = - 21 \times 4$

$Only natural value of r possible if 3 r - 9 = 0 \Rightarrow r = 3 and {}^{9}C_{3} = 84$

$From equation (i): l = 5$

$For coefficient of x^{- 3 l} : \frac{45}{2} - \frac{5 r}{2} - l r = - 3 l \Rightarrow \frac{45}{2} - \frac{5 r}{2} - 5 r = - 3 l \Rightarrow r = 5$

$∴$ $The coefficient of x^{- 3 l} is {}^{9}C_{5} (- 1) \frac{4^{5}}{2^{4}} = 2^{α} \times β$

$\Rightarrow α = 7, β = - 63$

$∴$ $Value of$ $| α l - β |$ $= 7 \times 5 + 63 = 35 + 63 = 98$

Q 25 :
The sum of all rational terms in the expansion of ${(2^{\frac{1}{5}} + 5^{\frac{1}{3}})}^{15}$ is equal to [2024]

3133
633
931
6131

1

2

3

4

(1)

We have, ${(2^{1 / 5} + 5^{1 / 3})}^{15}$

$∴$ $T_{r + 1} = {}^{15}{C_{r}} {(2^{1 / 5})}^{15 - r} {(5^{1 / 3})}^{r} = {}^{15}{C_{r}} 5^{r / 3} 2^{(3 - \frac{r}{5})}$

For rational terms, $\frac{r}{3}$ and $\frac{r}{5}$ must be an integer

$∴$ 3 and 5 divide $r \Rightarrow 15$ divides $r \Rightarrow r = 0$ and $r = 15$

Hence, ${}^{15}{C_{0}} 5^{0} 2^{3} + {}^{15}{C_{15}} 5^{5} 2^{0}$

$∴$ Required sum = 8 + 3125 = 3133

Q 26 :
If the constant term in the expansion of ${(\frac{\sqrt[5]{3}}{x} + \frac{2 x}{\sqrt[3]{5}})}^{12}, x \neq 0,$ is $α \times 2^{8} \times \sqrt[5]{3},$ then $25 α$ is equal to: [2024]

639
742
724
693

1

2

3

4

(4)

$T_{r + 1} = {}^{12}{C_{r}} {(\frac{3^{1 / 5}}{x})}^{12 - r} {(\frac{2 x}{5^{1 / 3}})}^{r}$

$= {}^{12}C_{r} (\frac{3^{\frac{12 - r}{5}}}{5^{r / 3}}) \cdot 2^{r} \cdot x^{2 r - 12}$

For constant term, $2 r - 12 = 0$

$\Rightarrow r = 6$

$∴$ Constant term = $\frac{{}^{12}{C_{6} \cdot} 3^{6 / 5} \cdot 2^{6}}{5^{2}}$

$= 924 \times \frac{3^{1} 3^{1 / 5} \cdot 2^{6}}{25} = \frac{11 \times 9 \times 7 \times 2^{8} \times 3^{1 / 5}}{25}$

$\Rightarrow α = \frac{11 \times 9 \times 7}{25} \Rightarrow 25 α = 11 \times 9 \times 7 = 693$

Q 27 :
If the term independent of $x$ in the expansion of ${(\sqrt{a} x^{2} + \frac{1}{2 x^{3}})}^{10}$ is 105, then $a^{2}$ is equal to : [2024]

2
6
4
9

1

2

3

4

(3)

We have, ${(\sqrt{a} x^{2} + \frac{1}{2 x^{3}})}^{10}$

$T_{r + 1} = {}^{10}C_{r} {(\sqrt{a} x^{2})}^{10 - r} {(\frac{1}{2 x^{3}})}^{r}$

$= {}^{10}C_{r} {(\sqrt{a})}^{10 - r} {(\frac{1}{2})}^{r} x^{20 - 2 r - 3 r}$

For the term to be independent of $x,$ we have $20 - 2 r - 3 r = 0$

$\Rightarrow r = 4$

$∴$ Required term $= T_{5} = {}^{10}C_{4} {(\sqrt{a})}^{6} {(\frac{1}{2})}^{4} = 105 (given)$

$\Rightarrow \frac{210}{16} a^{3} = 105 \Rightarrow a^{3} = 8 \Rightarrow a = 2$

Hence, $a^{2} = 4$

Q 28 :
The sum of the coefficient of $x^{\frac{2}{3}}$ and $x^{\frac{- 2}{5}}$ in the binomial expansion of ${(x^{\frac{2}{3}} + \frac{1}{2} x^{\frac{- 2}{5}})}^{9}$ is [2024]

$\frac{69}{16}$
$\frac{63}{16}$
$\frac{21}{4}$
$\frac{19}{4}$

1

2

3

4

(3)

$T_{r + 1} = {}^{9}{C_{r}} {(\frac{x^{- 2 / 5}}{2})}^{r} {(x^{2 / 3})}^{9 - r}$

$= {}^{9}{C_{r}} \frac{1}{2^{r}} x^{6 - \frac{2}{3} r - \frac{2}{5} r} = {}^{9}{C_{r}} \frac{1}{2^{r}} x^{6 - \frac{16 r}{15}}$

For coefficient of $x^{2 / 3}, 6 - \frac{16 r}{15} = \frac{2}{3}$

$\Rightarrow 90 - 16 r = 10$

$\Rightarrow r = 5$

For coefficient of $x^{- 2 / 5}, 6 - \frac{16 r}{15} = \frac{- 2}{5}$

$\Rightarrow 90 - 16 r = - 6$

$\Rightarrow r = 6$

Required sum $= {}^{9}{C_{5}} \frac{1}{2^{5}} + {}^{9}{C_{6}} \frac{1}{2^{6}} = \frac{126}{2^{5}} + \frac{84}{2^{6}}$

$= \frac{336}{2^{6}} = \frac{21}{4}$

Q 29 :
Let $m$ and $n$ be the coefficients of seventh and thirteenth terms respectively in the expansion of ${(\frac{1}{3} x^{\frac{1}{3}} + \frac{1}{2 x^{\frac{2}{3}}})}^{18} .$ Then ${(\frac{n}{m})}^{\frac{1}{3}}$ is: [2024]

$\frac{1}{9}$
$\frac{1}{4}$
$\frac{4}{9}$
$\frac{9}{4}$

1

2

3

4

(4)

$T_{r + 1} = {}^{18}{C_{r}} {(\frac{1}{3} x^{1 / 3})}^{18 - r} \cdot {(\frac{1}{2} x^{- 2 / 3})}^{r}$

$∴$ Coefficient of $7^{th}$ term $= {}^{18}{C_{6}} {(\frac{1}{3})}^{12} \cdot {(\frac{1}{2})}^{6}$

and coefficient of $13^{th}$ term $= {}^{18}{C_{12}} {(\frac{1}{3})}^{6} \cdot {(\frac{1}{2})}^{12}$

So, ${(\frac{n}{m})}^{1 / 3} = {[\frac{{}^{18}{C_{12}} {(\frac{1}{3})}^{6} \cdot {(\frac{1}{2})}^{12}}{{}^{18}{C_{6}} {(\frac{1}{3})}^{12} \cdot {(\frac{1}{2})}^{6}}]}^{1 / 3} = \frac{9}{4}$

Q 30 :
If the constant term in the expansion of $(1 + 2 x - 3 x^{3}) {(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}$ is $p,$ then $108 p$ is equal to _______. [2024]

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(54)

General term of $I = {(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}$ is

$T_{r + 1} = {}^{9}{C_{r}} {(\frac{3}{2} x^{2})}^{9 - r} {(\frac{- 1}{3 x})}^{r}$

$= {}^{9}{C_{r}} {(\frac{3}{2})}^{9 - r} \frac{x^{18 - 2 r}}{x^{r}} {(\frac{- 1}{3})}^{r}$

$= {}^{9}{C_{r}} {(\frac{3}{2})}^{9 - r} r^{18 - 3 r} {(\frac{- 1}{3})}^{r}$

Now, $(1 + 2 x - 3 x^{3}) {(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}$ will have the constant term $= 1 \times coeff of x^{0} in I + 2 \times coeff of \frac{1}{x} in I - 3 \times coeff of x^{- 3} in I$

$= 1 \times {}^{9}{C_{6}} {(\frac{3}{2})}^{3} {(\frac{- 1}{3})}^{6} + 2 \times 0 - 3 \times {}^{9}{C_{7}} {(\frac{3}{2})}^{2} {(\frac{- 1}{3})}^{7}$

$= \frac{84}{2^{3} \times 3^{3}} + \frac{3 \times 36 \times 1}{2^{2} \times 3^{5}} = \frac{7}{2 \times 3^{2}} + \frac{2}{2 \times 3^{2}}$

$= \frac{9}{2 \times 3^{2}} = \frac{1}{2}$

$∴$ $p = \frac{1}{2} .$ So $108 p = \frac{1}{2} \times 108 = 54$

Topic Question Set

Q 21 :
If the co-efficient of $x^{9}$ in ${(α x^{3} + \frac{1}{β x})}^{11}$ and the co-efficient of $x^{- 9}$ in ${(α x - \frac{1}{β x^{3}})}^{11}$ are equal, then ${(α β)}^{2}$ is equal to ________ . [2023]

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Q 22 :
Let $α > 0$ , be the smallest number such that the expansion of ${(x^{\frac{2}{3}} + \frac{2}{x^{3}})}^{30}$ has a term $β x^{- α}, β \in N .$ Then $α$ is equal to ________ . [2023]

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Q 23 :
The coefficient of $x^{- 6}$ , in the expansion of ${(\frac{4 x}{5} + \frac{5}{2 x^{2}})}^{9}$ , is __________ . [2023]

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Q 24 :
If the constant term in the binomial expansion of ${(\frac{x^{\frac{5}{2}}}{2} - \frac{4}{x^{l}})}^{9}$ is $- 84$ and the coefficient of $x^{- 3 l}$ is $2^{α} β$ , where $β < 0$ is an odd number, then $| α l - β |$ is equal to __________ . [2023]

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Q 25 :
The sum of all rational terms in the expansion of ${(2^{\frac{1}{5}} + 5^{\frac{1}{3}})}^{15}$ is equal to [2024]

Q 26 :
If the constant term in the expansion of ${(\frac{\sqrt[5]{3}}{x} + \frac{2 x}{\sqrt[3]{5}})}^{12}, x \neq 0,$ is $α \times 2^{8} \times \sqrt[5]{3},$ then $25 α$ is equal to: [2024]

Q 27 :
If the term independent of $x$ in the expansion of ${(\sqrt{a} x^{2} + \frac{1}{2 x^{3}})}^{10}$ is 105, then $a^{2}$ is equal to : [2024]

Q 28 :
The sum of the coefficient of $x^{\frac{2}{3}}$ and $x^{\frac{- 2}{5}}$ in the binomial expansion of ${(x^{\frac{2}{3}} + \frac{1}{2} x^{\frac{- 2}{5}})}^{9}$ is [2024]

Q 29 :
Let $m$ and $n$ be the coefficients of seventh and thirteenth terms respectively in the expansion of ${(\frac{1}{3} x^{\frac{1}{3}} + \frac{1}{2 x^{\frac{2}{3}}})}^{18} .$ Then ${(\frac{n}{m})}^{\frac{1}{3}}$ is: [2024]

Q 30 :
If the constant term in the expansion of $(1 + 2 x - 3 x^{3}) {(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}$ is $p,$ then $108 p$ is equal to _______. [2024]

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Topic Question Set

Q 21 : If the co-efficient of x9 in (αx3+1βx)11 and the co-efficient of x-9 in (αx-1βx3)11 are equal, then (αβ)2 is equal to ________ . [2023]

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Q 22 : Let α>0, be the smallest number such that the expansion of (x23+2x3)30 has a term βx-α,β∈N. Then α is equal to ________ . [2023]

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Q 23 : The coefficient of x-6, in the expansion of (4x5+52x2)9, is __________ . [2023]

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Q 24 : If the constant term in the binomial expansion of (x522-4xl)9 is -84 and the coefficient of x-3l is 2αβ, where β<0 is an odd number, then |αl-β| is equal to __________ . [2023]

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Q 25 : The sum of all rational terms in the expansion of (215+513)15 is equal to [2024]

Q 26 : If the constant term in the expansion of (35x+2x53)12, x≠0, is α×28×35, then 25α is equal to: [2024]

Q 27 : If the term independent of x in the expansion of (ax2+12x3)10 is 105, then a2 is equal to : [2024]

Q 28 : The sum of the coefficient of x23 and x-25 in the binomial expansion of (x23+12x-25)9 is [2024]

Q 29 : Let m and n be the coefficients of seventh and thirteenth terms respectively in the expansion of (13x13+12x23)18. Then (nm)13 is: [2024]

Q 30 : If the constant term in the expansion of (1+2x-3x3)(32x2-13x)9 is p, then 108p is equal to _______. [2024]

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Q 21 :
If the co-efficient of $x^{9}$ in ${(α x^{3} + \frac{1}{β x})}^{11}$ and the co-efficient of $x^{- 9}$ in ${(α x - \frac{1}{β x^{3}})}^{11}$ are equal, then ${(α β)}^{2}$ is equal to ________ . [2023]

Q 22 :
Let $α > 0$ , be the smallest number such that the expansion of ${(x^{\frac{2}{3}} + \frac{2}{x^{3}})}^{30}$ has a term $β x^{- α}, β \in N .$ Then $α$ is equal to ________ . [2023]

Q 23 :
The coefficient of $x^{- 6}$ , in the expansion of ${(\frac{4 x}{5} + \frac{5}{2 x^{2}})}^{9}$ , is __________ . [2023]

Q 24 :
If the constant term in the binomial expansion of ${(\frac{x^{\frac{5}{2}}}{2} - \frac{4}{x^{l}})}^{9}$ is $- 84$ and the coefficient of $x^{- 3 l}$ is $2^{α} β$ , where $β < 0$ is an odd number, then $| α l - β |$ is equal to __________ . [2023]

Q 25 :
The sum of all rational terms in the expansion of ${(2^{\frac{1}{5}} + 5^{\frac{1}{3}})}^{15}$ is equal to [2024]

Q 26 :
If the constant term in the expansion of ${(\frac{\sqrt[5]{3}}{x} + \frac{2 x}{\sqrt[3]{5}})}^{12}, x \neq 0,$ is $α \times 2^{8} \times \sqrt[5]{3},$ then $25 α$ is equal to: [2024]

Q 27 :
If the term independent of $x$ in the expansion of ${(\sqrt{a} x^{2} + \frac{1}{2 x^{3}})}^{10}$ is 105, then $a^{2}$ is equal to : [2024]

Q 28 :
The sum of the coefficient of $x^{\frac{2}{3}}$ and $x^{\frac{- 2}{5}}$ in the binomial expansion of ${(x^{\frac{2}{3}} + \frac{1}{2} x^{\frac{- 2}{5}})}^{9}$ is [2024]

Q 29 :
Let $m$ and $n$ be the coefficients of seventh and thirteenth terms respectively in the expansion of ${(\frac{1}{3} x^{\frac{1}{3}} + \frac{1}{2 x^{\frac{2}{3}}})}^{18} .$ Then ${(\frac{n}{m})}^{\frac{1}{3}}$ is: [2024]

Q 30 :
If the constant term in the expansion of $(1 + 2 x - 3 x^{3}) {(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}$ is $p,$ then $108 p$ is equal to _______. [2024]