Binomial Theorem — Page 12 | JEE Mathematics

Q111

The coefficient of x18 in the product (1 + x) (1 – x)10 (1 + x + x2)9 is :

A 126

B - 84

C - 126

D 84

Correct Answer

Option D

Solution

Coefficient of x18 in (1 + x) (1 - x)10 (1 + x + x2)9 $\Rightarrow$ Coefficient of x18 in {(1 - x) (1 - x2) (1 + x + x2)}9 $\Rightarrow$ Coefficient of x18 in (1 - x2) (1 - x3)9 $\Rightarrow$ 9C6 - 0 = 84

Q112

The sum of the real values of x for which the middle term in the binomial expansion of

{\left( {{{{x^3}} \over 3} + {3 \over x}} \right)^8}

equals 5670 is :

A 0

B 8

C 6

D 4

Correct Answer

Option A

Solution

{T_5} = {}^8{C_4}{{{x^{12}}} \over {81}} \times {{81} \over {{x^4}}} = 5670

\Rightarrow 70{x^8} = 5670

\Rightarrow x = \pm \sqrt 3

Q113

If the coefficient of

{x^7}

in

{\left[ {a{x^2} + \left( {{1 \over {bx}}} \right)} \right]^{11}}

equals the coefficient of

{x^{ - 7}}

in

{\left[ {ax - \left( {{1 \over {b{x^2}}}} \right)} \right]^{11}}

, then

a

and

b

satisfy the relation

A

a - b = 1

B

a + b = 1

C

{a \over b} = 1

D

ab = 1

Correct Answer

Option D

Solution

General term of

{\left[ {a{x^2} + \left( {{1 \over {bx}}} \right)} \right]^{11}}

is Tr+1. Tr+1 =

{}^{11}{C_r}{\left( {a{x^2}} \right)^{11 - r}}{\left( {{1 \over {bx}}} \right)^r}

=

{}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( b \right)^{ - r}}{\left( x \right)^{22 - 3r}}

For the coefficient of x7, $\Rightarrow$ 22 - 3r = 7 $\Rightarrow$ r = 5 So coefficient of x7 =

{}^{11}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}}

......(1) Now General term of

{\left[ {ax - \left( {{1 \over {b{x^2}}}} \right)} \right]^{11}}

is Tr+1. Tr+1 =

{}^{11}{C_r}{\left( {a{x}} \right)^{11 - r}}{\left( { - {1 \over {bx}}} \right)^r}

=

{}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( { - 1} \right)^r}{\left( b \right)^{ - r}}{\left( x \right)^{11 - r}}{\left( x \right)^{ - 2r}}

For the coefficient of x-7, 11 - 3r = -7 $\Rightarrow$ r = 6 $\therefore$ Coefficient of x-7 =

{}^{11}{C_6}{\left( a \right)^5}{\left( { - 1} \right)^6}{\left( b \right)^{ - 6}}

According to question, Coefficient of x7 = Coefficient of x-7 $\Rightarrow$

{}^{11}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}}

=

{}^{11}{C_6}{\left( a \right)^5}{\left( { - 1} \right)^6}{\left( b \right)^{ - 6}}

$\Rightarrow$

ab = 1

Q114

The sum of the series

{}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .....\, - \,.....\, + {}^{20}{C_{10}}

is

A

0

B

{}^{20}{C_{10}}

C

- {}^{20}{C_{10}}

D

{1 \over 2}{}^{20}{C_{10}}

Correct Answer

Option D

Solution

We know

{}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... + {}^{20}{C_{10}} - {}^{20}{C_{11}}+ ...... + {}^{20}{C_{20}} = 0

$\Rightarrow$

({}^{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}})

(As

{}^{20}{C_{11}} = {}^{20}{C_9}

) $\Rightarrow$

2({}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}})

+{}^{20}{C_{10}}

= 0 $\Rightarrow$

{}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}}

=

- {1 \over 2}{}^{20}{C_{10}}

Adding

{}^{20}{C_{10}}

both sides, $\Rightarrow$

{}^{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}}

$\Rightarrow$

{}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}}

+ {}^{20}{C_{10}}

=

{1 \over 2}{}^{20}{C_{10}}

Q115

The value of

\,{}^{50}{C_4} + \sum\limits_{r = 1}^6 {^{56 - r}} {C_3}

is

A

{}^{55}{C_4}

B

{}^{55}{C_3}

C

{}^{56}{C_3}

D

{}^{56}{C_4}

Correct Answer

Option D

Solution

Given,

{}^{50}{C_4} + \sum\limits_{n = 1}^6 {{}^{56 - r}{C_3}}

$\Rightarrow$

{}^{50}{C_4}

+

{}^{55}{C_3}

+

{}^{54}{C_3}

+

{}^{53}{C_3}

+

{}^{52}{C_3}

+

{}^{51}{C_3}

+

{}^{50}{C_3}

Arrange those this way $\Rightarrow$

{}^{50}{C_4}

+

{}^{50}{C_3}

+

{}^{51}{C_3}

+

{}^{52}{C_3}

+

{}^{53}{C_3}

+

{}^{54}{C_3}

+

{}^{55}{C_3}

We know this formula [

{{}^n{C_r}}

+

{{}^n{C_{r - 1}}}

=

{{}^{n + 1}{C_r}}

] which is used to solve this problem. $\Rightarrow$

{}^{51}{C_4}

+

{}^{51}{C_3}

+

{}^{52}{C_3}

+

{}^{53}{C_3}

+

{}^{54}{C_3}

+

{}^{55}{C_3}

$\Rightarrow$

{}^{52}{C_4}

+

{}^{52}{C_3}

+

{}^{53}{C_3}

+

{}^{54}{C_3}

+

{}^{55}{C_3}

$\Rightarrow$

{}^{53}{C_4}

+

{}^{53}{C_3}

+

{}^{54}{C_3}

+

{}^{55}{C_3}

$\Rightarrow$

{}^{54}{C_4}

+

{}^{54}{C_3}

+

{}^{55}{C_3}

$\Rightarrow$

{}^{55}{C_4}

+

{}^{55}{C_3}

$\Rightarrow$

{}^{56}{C_4}

Q116

The positive value of

\lambda

for which the co-efficient of x2 in the expression x2

{\left( {\sqrt x + {\lambda \over {{x^2}}}} \right)^{10}}

is 720, is -

A 4

B

2\sqrt 2

C 3

D

\sqrt 5

Correct Answer

Option A

Solution

The general term in the expansion of the binomial expression $(a+b)^n$ is

T_{r+1}={ }^n C_r a^{n-r} b^r

Therefore, the general term in the expansion of the binomial expression $x^2\left(\sqrt{x}+\dfrac{\lambda}{x^2}\right)^{10}$ is

\begin{aligned} T_{r+1} & =x^2\left({ }^{10} C_r(\sqrt{x})^{10-r}\left(\frac{\lambda}{x^2}\right)^r\right) \\\\ & ={ }^{10} C_r x^2 \cdot x^{\frac{10-r}{2}} \lambda^r x^{-2 r} \\\\ & ={ }^{10} C_r \lambda^r x^{2+\frac{10-r}{2}-2 r} \end{aligned}

Now, for the coefficient of $x^2$ ,

\begin{aligned} 2+\frac{10-r}{2}-2 r =2 \\\\ \Rightarrow \frac{10-r}{2}-2 r =0 \\\\ \Rightarrow 10-r =4 r \Rightarrow r=2 \end{aligned}

So, the coefficient of $x^2$ is ${ }^{10} C_2 \lambda^2=720$

\begin{aligned} \Rightarrow & \frac{10 !}{2 ! 8 !} \lambda^2 =720 \\\\ \Rightarrow & \frac{10 \cdot 9 \cdot 8 !}{2 \cdot 8 !} \lambda^2 =720 \\\\ \Rightarrow & 45 \lambda^2 =720 \\\\ \Rightarrow & \lambda^2 =16 \\\\ \Rightarrow & \lambda = \pm 4 \end{aligned}

Given the problem asks for the positive value of $\lambda$ , $\lambda = 4$ . So, the correct option is (A) 4.

Q117

Fractional part of the number

\dfrac{4^{2022}}{15}

is equal to

A

\dfrac{8}{15}

B

\dfrac{4}{15}

C

\dfrac{1}{15}

D

\dfrac{14}{15}

Correct Answer

Option C

Solution

\begin{aligned} & \left\{\frac{4^{2022}}{15}\right\}=\left\{\frac{2^{4044}}{15}\right\} \\\\ & =\left\{\frac{(1+15)^{1011}}{15}\right\} \\\\ & =\frac{1}{15} \end{aligned}

Q118

Let

{s_1} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right){}^{10}} {C_j}

,

{{s_2} = \sum\limits_{j = 1}^{10} {} } j.{}^{10}{C_j}

and

{{s_3} = \sum\limits_{j = 1}^{10} {{j^2}.{}^{10}{C_j}.} }

Statement-1 :

{{S_3} = 55 \times {2^9}}

. Statement-2 :

{{S_1} = 90 \times {2^8}}

and

{{S_2} = 10 \times {2^8}}

.

A Statement - 1 is true, Statement- 2 is true; Statement - 2 is not a correct explanation for Statement - 1.

B Statement - 1 is true, Statement-2 is false.

C Statement - 1 is false, Statement-2 is true.

D Statement - 1 is true, Statement-2 is true: -Statement - 2 is a correct explanation for Statement - 1.

Correct Answer

Option B

Solution

Note :

\sum\limits_{r = 0}^n {r.{}^n{C_r}}

=

= n{.2^{n - 1}}

\sum\limits_{r = 0}^n {{r^2}.{}^n{C_r}} = n\left( {n + 1} \right){2^{n - 2}}

Given that,

{s_1} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right){}^{10}} {C_j}

=

\sum\limits_{j = 1}^{10} {{j^2}.{}^{10}} {C_j} - \sum\limits_{j = 1}^{10} {j.{}^{10}} {C_j}

= 10 $\times$ 11 $\times$

{2^{10 - 2}}

- 10 $\times$

{2^{10 - 1}}

= 10 $\times$

{2^{8}}

(11 - 2) = 10 $\times$ 9 $\times$

{2^{8}}

= 90 $\times$

{2^{8}}

{{s_2} = \sum\limits_{j = 1}^{10} {} } j.{}^{10}{C_j}

= 10 $\times$

{2^{10-1}}

= 10 $\times$

{2^{9}}

{{s_3} = \sum\limits_{j = 1}^{10} {{j^2}.{}^{10}{C_j}.} }

= 10 $\times$ 11 $\times$

{2^{10-2}}

=

{{110} \over 2} \times {2^9}

= 55 $\times$

{2^9}

Q119

The term independent of

x

in expansion of

{\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}

is

A 4

B 120

C 210

D 310

Correct Answer

Option C

Solution

{\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}

=

{\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \right)^{10}}

=

{\left( {{{\left( {{x^{1/3}} + 1} \right)\left( {{x^{2/3}} - {x^{1/3}} + 1} \right)} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)} \over {\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}

=

{\left( {\left( {{x^{1/3}} + 1} \right) - {{\left( {\sqrt x + 1} \right)} \over {\sqrt x }}} \right)^{10}}

=

{\left( {\left( {{x^{1/3}} + 1} \right) - \left( {1 + {1 \over {\sqrt x }}} \right)} \right)^{10}}

=

{\left( {{x^{1/3}} - {1 \over {{x^{1/2}}}}} \right)^{10}}

[Note: For

{\left( {{x^\alpha } \pm {1 \over {{x^\beta }}}} \right)^n}

the

\left( {r + 1} \right)

th term with power m of x is

r = {{n\alpha - m} \over {\alpha + \beta }}

] Here

\alpha = {1 \over 3}

,

\beta = {1 \over 2}

and m = 0 then

r = {{10 \times {1 \over 3} - 0} \over {{1 \over 3} + {1 \over 2}}}

=

{{10} \over 3} \times {6 \over 5}

= 4 $\therefore$ T5 is the term independent of x. $\therefore$ T5 =

{}^{10}{C_4}

= 210

Q120

For an integer

n \geq 2

, if the arithmetic mean of all coefficients in the binomial expansion of

(x+y)^{2 n-3}

is 16 , then the distance of the point

\mathrm{P}\left(2 n-1, n^2-4 n\right)

from the line

x+y=8

is

A

\sqrt{2}

B

2 \sqrt{2}

C

5 \sqrt{2}

D

3 \sqrt{2}

Correct Answer

Option D

Solution

\begin{aligned} & \text{ Mean }=\frac{{ }^{2 n-3} C_0+{ }^{2 n-3} C_1+{ }^{2 n-3} C_2+\cdots{ }^{2 n-3} C_{2 n-3}}{2 n-2}=16 \\ &=2^{2 n-3}=16(2 n-2) \\ &=2^{2 n-3}=2^5(n-1) \\ & \Rightarrow n=5 \\ & \therefore \quad P(9,5) \\ & d=\left|\frac{9+5-8}{\sqrt{2}}\right|=\frac{6}{\sqrt{2}}=3 \sqrt{2} \end{aligned}