Let $m$ and $n$ be the coefficients of seventh and thirteenth terms respectively in the expansion of $\left(\frac{1}{3} x^{\frac{1}{3}}+\frac{1}{2 x^{\frac{2}{3}}}\right)^{18}$. Then $\left(\frac{\mat

$\begin{aligned} & \mathrm{t}_7={ }^{18} \mathrm{C}_6\left(\frac{\mathrm{x}^{\frac{1}{3}}}{3}\right)^{12}\left(\frac{\mathrm{x}^{\frac{-2}{3}}}{2}\right)^6={ }^{18} \mathrm{C}_6 \frac{1}{(3)^{12}} \cdot \frac{1}{2^6} \\\\ & \mathrm{t}_{13}={ }^{18} \mathrm{C}_{12}\left(\frac{\mathrm{x}^{\frac{1}{3}}}{3}\right)^6\left(\frac{\mathrm{x}^{\frac{-2}{3}}}{2}\right)^{12}={ }^{18} \mathrm{C}_{12} \frac{1}{(3)^6} \cdot \frac{1}{2^{12}} \cdot \mathrm{x}^{-6}\end{aligned}$ $$ \therefore $$ $m={ }^{18} C_6\

Binomial Theorem — Page 15 | JEE Mathematics

Q141

The value of r for which 20Cr 20C0 + 20Cr

-

1 20C1 + 20Cr

-

2 20C2 + . . . . .+ 20C0 20Cr is maximum, is

A 20

B 15

C 10

D 11

Correct Answer

Option A

Solution

Given sum = coefficient of xr in the expansion of (1 + x)20(1 + x)20, Which is equal to 40Cr It is maximum when r = 20

Q142

The number of integral terms in the expansion of

{\left( {\sqrt 3 + \sqrt[8]5 } \right)^{256}}

is

A 35

B 32

C 33

D 34

Correct Answer

Option C

Solution

General term =

{}^{256}{C_r}.{\left( {\sqrt 3 } \right)^{256 - r}}.{\left( {\sqrt[8]5 } \right)^r}

=

{}^{256}{C_r}.{\left( 3 \right)^{{{256 - r} \over 2}}}.{\left( 5 \right)^{{r \over 8}}}

When

{{{256 - r} \over 2}}

is integer then

{\left( 3 \right)^{{{256 - r} \over 2}}}

is integer. And when

{{r \over 8}}

is integer then

{\left( 5 \right)^{{r \over 8}}}

is integer. Entire general term will be integer when

{{{256 - r} \over 2}}

and

{{r \over 8}}

both are integer.

{{{256 - r} \over 2}}

is integer when r = 0, 2, 4, 6, ......, 256

{{r \over 8}}

is integer when r = 0, 8, 16 ,......., 256 Now both

{{{256 - r} \over 2}}

and

{{r \over 8}}

will be integer when r = 0, 8, 16, ...., 256 (This is an AP) $\therefore$ 256 = 0 + (n - 1)8 using formula of AP, tn = a + (n - 1)d $\therefore$ n =

{{256} \over 8} + 1

= 32 + 1 = 33

Q143

Let

m

and

n

be the coefficients of seventh and thirteenth terms respectively in the expansion of

\left(\dfrac{1}{3} x^{\dfrac{1}{3}}+\dfrac{1}{2 x^{\dfrac{2}{3}}}\right)^{18}

. Then

\left(\dfrac{\mathrm{n}}{\mathrm{m}}\right)^{\dfrac{1}{3}}

is :

A

\dfrac{1}{9}

B

\dfrac{1}{4}

C

\dfrac{4}{9}

D

\dfrac{9}{4}

Correct Answer

Option D

Solution

$\begin{aligned} & \mathrm{t}_7={ }^{18} \mathrm{C}_6\left(\dfrac{\mathrm{x}^{\dfrac{1}{3}}}{3}\right)^{12}\left(\dfrac{\mathrm{x}^{\dfrac{-2}{3}}}{2}\right)^6={ }^{18} \mathrm{C}_6 \dfrac{1}{(3)^{12}} \cdot \dfrac{1}{2^6} \\\\ & \mathrm{t}_{13}={ }^{18} \mathrm{C}_{12}\left(\dfrac{\mathrm{x}^{\dfrac{1}{3}}}{3}\right)^6\left(\dfrac{\mathrm{x}^{\dfrac{-2}{3}}}{2}\right)^{12}={ }^{18} \mathrm{C}_{12} \dfrac{1}{(3)^6} \cdot \dfrac{1}{2^{12}} \cdot \mathrm{x}^{-6}\end{aligned}$ $\therefore$ $m={ }^{18} C_6\left(\dfrac{1}{3}\right)^{12}\left(\dfrac{1}{2}\right)^6$ $n={ }^{18} C_{12}\left(\dfrac{1}{3}\right)^6\left(\dfrac{1}{2}\right)^{12}$ $\begin{aligned}\left(\dfrac{m}{n}\right)^{\dfrac{1}{3}} & =\left(\dfrac{{ }^{18} C_6\left(\dfrac{1}{3}\right)^{12}\left(\dfrac{1}{2}\right)^6}{{ }^{18} C_{12}\left(\dfrac{1}{3}\right)^6\left(\dfrac{1}{2}\right)^{12}}\right)^{\dfrac{1}{3}} \\\\ & =\left(\dfrac{\left(\dfrac{1}{3}\right)^6}{\left(\dfrac{1}{2}\right)^6}\right)^{\dfrac{1}{3}}=\left(\left(\dfrac{2}{3}\right)^6\right)^{\dfrac{1}{3}}=\dfrac{4}{9}\end{aligned}$ $\therefore\left(\dfrac{n}{m}\right)^{\dfrac{1}{3}}=\dfrac{9}{4}$

Q144

The absolute difference of the coefficients of

x^{10}

and

x^{7}

in the expansion of

\left(2 x^{2}+\dfrac{1}{2 x}\right)^{11}

is equal to :

A

11^{3}-11

B

13^{3}-13

C

12^{3}-12

D

10^{3}-10

Correct Answer

Option C

Solution

General term of $\left(2 x^2+\dfrac{1}{2 x}\right)^{11}$ is :

\begin{aligned} & \mathrm{T}_{\mathrm{r}+1}={ }^{11} \mathrm{C}_r\left(2 x^2\right){ }^{11-r}\left(\frac{1}{2 x}\right)^r \\\\ & ={ }^{11} \mathrm{C}_r 2^{11-r} x^{22-2 r} 2^{-r} x^{-r} \\\\ & ={ }^{11} \mathrm{C}_r 2^{11-r} x^{22-3 r} \end{aligned}

Now, $22-2 r=10$ and $22-3 r=7$

\begin{array}{ll} \Rightarrow 3 r=12 &&& \Rightarrow 3 r=15 \\\\ \Rightarrow r=4 &&& \Rightarrow r=5 \end{array}

$\therefore$ Coeff. of $x^{10}={ }^{11} \mathrm{C}_4 \cdot 2^{11-8}={ }^{11} \mathrm{C}_4 \times 8$ Coeff. of $x^7={ }^{11} C_5 \cdot 2^{11-10}={ }^{11} C_4 \times 2$ Now, required difference

\begin{aligned} & ={ }^{11} \mathrm{C}_4 \times 8-{ }^{11} \mathrm{C}_5 \times 2 \\\\ & =\frac{11 \times 10 \times 9 \times 8 \times 7 !}{4 ! \times 7 !} \times 8-\frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 ! \times 2}{5 ! 6 !} \end{aligned}

\begin{aligned} & =\frac{11 \times 10 \times 9 \times 8 \times 8}{24}-\frac{11 \times 10 \times 9 \times 8 \times 7 \times 2}{120} \\\\ & =11 \times 10 \times 8 \times 3-11 \times 3 \times 4 \times 7 \\\\ & =11 \times 3 \times 4[20-7] \\\\ & =11 \times 12 \times 13=(12-1) \times 12 \times(12+1) \\\\ & =12\left(12^2-1\right)=12^3-12 \end{aligned}

Q145

Let

x=(8 \sqrt{3}+13)^{13}

and

y=(7 \sqrt{2}+9)^9

. If

[t]

denotes the greatest integer

\leq t

, then :

A

[x]

is odd but

[y]

is even

B

[x]

and

[y]

are both odd

C

[x]+[y]

is even

D

[x]

is even but

[y]

is odd

Correct Answer

Option C

Solution

If

{I_1} + f = {(8\sqrt 3 + 13)^{13}},f' = {(8\sqrt 3 - 13)^{13}}

{I_1} + f - f'=

Even

{I_1} =

Even

{I_2} + f - f' = {(7\sqrt 2 + 9)^9} + {(7\sqrt 2 - 9)^9}

= Even

{I_2} =

Even

Q146

If n is the degree of the polynomial,

{\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}

and m is the coefficient of xn in it, then the ordered pair (n, m) is equal to :

A (24, (10)8)

B (8, 5(10)4)

C (12, (20)4)

D (12, 8(10)4)

Correct Answer

Option C

Solution

Given,

{\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}

=

{\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}

+

{\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}

=

{\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}

= (

{\sqrt {5{x^3} + 1} }

+

{\sqrt {5{x^3} - 1} }

)8 + (

{\sqrt {5{x^3} - 1} }

-

{\sqrt {5{x^3} - 1} }

)8 = 2

\left[ {{}^8{C_0}} \right.{(\sqrt {5{x^3} + 1} )^8}

+ +

{}^8{C_2}{(\sqrt {5{x^3} + 1} )^6}\sqrt {5{x^3} - 1}

+

{}^8{C_4}{(\sqrt {5{x^3} + 1} )^4}{(5{x^3} - 1)^2}

+

{}^8{C_6}{(\sqrt {5{x^3} + 1} )^2}{(5{x^3} - 1)^3}

+

\left. {{}^8{C_8}{{(5{x^3} - 1)}^4}} \right]

= 2

[{}^8{C_0}{(5{x^3} + 1)^4}

+

{}^8{C_2}{(5{x^3} + 1)^3}(5{x^3} - 1)

+

{}^8{C_4}(5{x^3} + 1)^2{(5{x^3} - 1)^2}

+

{}^8{C_6}(5{x^3} + 1){(5{x^3} - 1)^3}

+

{}^8{C_8}{(5{x^3} - 1)^4}]

Here maximum power of x is 12 $\therefore$ Degree of polynomial = 12 Coefficient of x12 = 2 [8C0 54 + 8C2 $\times$ 53 $\times$ 5 + 8C4 $\times$ 52 $\times$ 52 + 8C6 $\times$ 5 $\times$ 53 + 8C8 $\times$ 54] = 160000 = (20)4

Q147

If the coefficients of x−2 and x−4 in the expansion of

{\left( {{x^{{1 \over 3}}} + {1 \over {2{x^{{1 \over 3}}}}}} \right)^{18}},\left( {x > 0} \right),

are m and n respectively, then

{m \over n}

is equal to :

A 182

B

{4 \over 5}

C

{5 \over 4}

D 27

Correct Answer

Option A

Solution

Tr+1 = 18Cr

{\left( {{x^{{1 \over 3}}}} \right)^{18 - r}}

.

{\left( {{1 \over {2{x^{{1 \over 3}}}}}} \right)^r}

= 18Cr

{\left( {{1 \over 2}} \right)^r}\,\,.\,\,{x^{{{18 - 2r} \over 3}}}

For coefficient of x $-$ 2,

{{18 - 2r} \over 3}

= $-$ 2 $\Rightarrow$ r = 12 $\therefore$ Coefficient of x $-$ 2 is (m) = 18C12

{\left( {{1 \over 2}} \right)^{12}}

For coefficient of x $-$ 4,

{{18 - 2r} \over 3}

= $-$ 4 $\Rightarrow$ r = 15 $\therefore$ Coefficient of x $-$ 4 is (n) = 18C15

\left( {{1 \over {2}}} \right)

15 $\therefore$

{m \over n} = {{^{18}{C_{12}}{{\left( {{1 \over 2}} \right)}^{12}}} \over {^{18}{C_{15}}{{\left( {{1 \over 2}} \right)}^{15}}}}

=

{{{}^{18}{C_6} \times {{\left( 2 \right)}^3}} \over {{}^{18}{C_3}}}

= 182

Q148

If the expansion in powers of

x

of the function

{1 \over {\left( {1 - ax} \right)\left( {1 - bx} \right)}}

is

{a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3}.....

then

{a_n}

is

A

{{{b^n} - {a^n}} \over {b - a}}

B

{{{a^n} - {b^n}} \over {b - a}}

C

{{{a^{n + 1}} - {b^{n + 1}}} \over {b - a}}

D

{{{b^{n + 1}} - {a^{n + 1}}} \over {b - a}}

Correct Answer

Option D

Solution

{1 \over {\left( {1 - ax} \right)\left( {1 - bx} \right)}}

=

{\left( {1 - ax} \right)^{ - 1}}{\left( {1 - bx} \right)^{ - 1}}

=

\left[ {1 + \left( { - 1} \right)\left( { - ax} \right) + {{\left( { - 1} \right)\left( { - 2} \right)} \over {1.2}}{{\left( { - ax} \right)}^2} + ...} \right]

-

\left[ {1 + \left( { - 1} \right)\left( { - bx} \right) + {{\left( { - 1} \right)\left( { - 2} \right)} \over {1.2}}{{\left( { - bx} \right)}^2} + ...} \right]

=

\left[ {1 + ax + {a^2}{x^2} + ... + {a^{n - 1}}{x^{n - 1}} + {a^n}{x^n}}+.... \right]

-

\left[ {1 + bx + {b^2}{x^2} + ... + {b^{n - 1}}{x^{n - 1}} + {b^n}{x^n}}+.... \right]

Coefficient of xn =

{a^n} + {a^{n - 1}}b + {a^{n - 2}}{b^2} + .... + {b^n}

=

{a^n}\left[ {1 + {b \over a} + {{{b^2}} \over {{a^2}}} + ..... + {{{b^n}} \over {{a^n}}}} \right]

=

{a^n}\left[ {{{{{\left( {{b \over a}} \right)}^{n + 1}} - 1} \over {{b \over a} - 1}}} \right]

=

{a^n}\left[ {{{{b^{n + 1}} - {a^{n + 1}}} \over {{a^{n + 1}}\left( {{{b - a} \over a}} \right)}}} \right]

=

{{{{b^{n + 1}} - {a^{n + 1}}} \over {b - a}}}

Q149

The coefficient of x2 in the expansion of the product (2

-

x2) .((1 + 2x + 3x2)6 + (1

-

4x2)6) is :

A 107

B 106

C 108

D 155

Correct Answer

Option B

Solution

Given, (2 $-$ x2) . (1 + 2x + 3x2) 6 + (1 $-$ 4x2)6) Let, a = ((1 + 2x + 3x2)6 + (1 $-$ 4x2)6)

\therefore\,\,\,\,

Given statement becomes, (2 $-$ x2) . (a) = 2a $-$ x2 (a) Here coefficients of x2 is = 2 (coefficient of x2 in a ) $-$ 1 (constant in a) (1 + 2x + 3x2) 6 = 6C0 + 6C1 (2x + 3x2) + 6C2 (2x + 3x2)2 + . . . . . .+ (2x + 3x2)6 (1 $-$ 4x2)6 = 6C0 $-$ 6C1 (4x2) + 6C2 (4x2)2 + . . . . .+ (4x2)6 Coefficient of x2 in (1 + 2x + 3x2)6 = 6C1 $\times$ 3 + 6C2 $\times$ 4 = 18 + 60 Coefficient of x2 in (1 $-$ 4x2)6 = $-$ 6C1 $\times$ 4 = $-$ 24 Coefficient of x2 in ((1 + 2x + 3x2)6 + (1 $-$ 4x2)6) = 60 + 18 $-$ 24 = 54 Constant term in (1 + 2x + 3x2)6 = 6C0 = 1 Constant term in (1 $-$ 4x)6 = 6C0 = 1

\therefore\,\,\,\,

Constant term in ((1 + 2x + 3x2)6 + (1 $-$ 4x)6) = 1 + 1 = 2

\therefore\,\,\,\,

Coefficient of x2 in (2 $-$ x2) ((1 + 2x + 3x2)6 + (1 $-$ 4x2)6) = 2 (54) $-$ 1 (2) = 108 $-$ 2 = 106

Q150

If the third term in the binomial expansion of

{\left( {1 + {x^{{{\log }_2}x}}} \right)^5}

equals 2560, then a possible value of x is -

A

2\sqrt 2

B

4\sqrt 2

C

{1 \over 8}

D

{1 \over 4}

Correct Answer

Option D

Solution

{\left( {1 + {x^{{{\log }_2}x}}} \right)^5}

{T_3} = {}^5{C_2}.{\left( {{x^{{{\log }_2}x}}} \right)^2} = 2560

\Rightarrow \,\,10.{x^{2{{\log }_2}x}} = 2560

\Rightarrow \,\,{x^{2\log 2x}} = 256

\Rightarrow \,\,2{({\log _2}x)^2} = {\log _2}256

\Rightarrow 2{({\log _2}x)^2} = 8

\Rightarrow \,\,{({\log _2}x)^2} = 4

\Rightarrow \,\,{\log _2}x = 2

or $-$ 2

x = 4

or

{1 \over 4}