Binomial Theorem — Page 14 | JEE Mathematics

Q131

The value of

\sum\limits_{r = 0}^6 {\left( {{}^6{C_r}\,.\,{}^6{C_{6 - r}}} \right)}

is equal to :

A 924

B 1024

C 1124

D 1324

Correct Answer

Option A

Solution

Given,

\sum\limits_{r = 0}^6 {{}^6{C_r}{}^6{C_{6 - r}}}

=

{}^6{C_0}.{}^6{C_6} + {}^6{C_1}.{}^6{C_5} + ... + {}^6{C_6}.{}^6{C_0}

Now,

\begin{aligned}& = \left( {{}^6{C_0} + {}^6{C_1}x + {}^6{C_2}{x^2} + ... + {}^6{C_6}{x^6}} \right) \\ & \left( {{}^6{C_0} + {}^6{C_1}x + {}^6{C_2}{x^2} + ... + {}^6{C_6}{x^6}} \right)\end{aligned}

Comparing coefficient of x6 both sides

{}^6{C_0}.{}^6{C_6} + {}^6{C_1}.{}^6{C_5} + ... + {}^6{C_6}.{}^6{C_0}

=

{}^{12}{C_6}

= 924

Q132

The greatest positive integer k, for which 49k + 1 is a factor of the sum 49125 + 49124 + ..... + 492 + 49 + 1, is:

A 32

B 60

C 63

D 65

Correct Answer

Option C

Solution

1 + 49 + 492 + ..... + 49125 sum of G.P. =

{{1.\left( {{{49}^{126}} - 1} \right)} \over {49 - 1}}

=

{{\left( {{{49}^{63}} + 1} \right)\left( {{{49}^{63}} - 1} \right)} \over {48}}

Also 4963 - 1 = (1 + 48)63 - 1 = [63C0 $\times$ 1 + 63C1 $\times$ 48 + 63C2 $\times$ (48)2 + ....

] - 1 = [1 + 48 $\lambda$ ] - 1 = 48 $\lambda$ So

{{\left( {{{49}^{63}} - 1} \right)} \over {48}}

= integer $\therefore$ 4963 + 1 is a factor. So k = 63.

Q133

Statement - 1 :

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

Statement - 2 :

\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}}.}

A Statement - 1 is false, Statement - 2 is true

B Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1

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

D Statement - 1 is true, Statement - 2 is false

Correct Answer

Option B

Solution

Check Statement - 1

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

=

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

+

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

=

\sum\limits_{r = 1}^n {r.{n \over r}{}^{n - 1}{C_{r - 1}}}

+ {2^n}

=

n\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}}

+ {2^n}

=

n \times {2^{n - 1}}

+ {2^n}

=

{2^{n - 1}}\left[ {n + 2} \right]

Check Statement 2 :

\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}{x^r}}

=

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

+

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

=

n\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}.{x^r}}

+ {\left( {1 + x} \right)^n}

=

nx\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}.{x^{r - 1}}} + {\left( {1 + x} \right)^n}

=

nx{\left( {1 + x} \right)^{n - 1}} + {\left( {1 + x} \right)^n}

Substitude x = 1 in the statement 2 and we get,

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

So Option B is correct.

Q134

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 :

A 88

B 64

C 100

D 80

Correct Answer

Option D

Solution

{(1 - y)^m}{(1 + y)^n} = 1 + {a_1}y + {a_2}{y^2} + .... + {a_{m + n}}{y^{m + n}}

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} + ....

\Rightarrow n - m = 10

..... (i)

\Rightarrow {}^m{C_2} + {}^n{C_2} - mn = 10

...... (ii)

{{m(m - 1)} \over 2} + {{n(n - 1)} \over 2} - mn = 10

\Rightarrow {{{m^2} - m} \over 2} + {{(10 + m)(9 + m)} \over 2} - m(10 + m) = 10

\Rightarrow {m^2} - m + {m^2} + 19m + 90 - 2({m^2} + 10m) = 20

\Rightarrow 18m + 90 - 20m = 20

\Rightarrow 2m = 70

\Rightarrow m = 35

&

n = 45

m + n = 80

Q135

If the term independent of x in the expansion of

{\left( {{3 \over 2}{x^2} - {1 \over {3x}}} \right)^9}

is k, then 18 k is equal to :

A 5

B 9

C 7

D 11

Correct Answer

Option C

Solution

General term,

{T_{r + 1}} = {}^9{C_r}{\left[ {{3 \over 2}{x^2}} \right]^{9 - r}}{\left( { - {1 \over {3x}}} \right)^r}

{T_{r + 1}} = {}^9{C_r}{\left[ {{3 \over 2}} \right]^{9 - r}}{\left( { - {1 \over 3}} \right)^r}{x^{18 - 3r}}

For independent of x 18 $-$ 3r = 0 $\Rightarrow$ r = 6 $\therefore$

{T_7} = {}^9{C_6}{\left( {{3 \over 2}} \right)^3}{\left( { - {1 \over 3}} \right)^6} = {{21} \over {54}} = k

$\therefore$

18k = {{21} \over {54}} \times 18 = 7

Q136

The coefficient of the middle term in the binomial expansion in powers of

x

of

{\left( {1 + \alpha x} \right)^4}

and

{\left( {1 - \alpha x} \right)^6}

is the same if

\alpha

equals

A

{3 \over 5}

B

{10 \over 3}

C

{{ - 3} \over {10}}

D

{{ - 5} \over {3}}

Correct Answer

Option C

Solution

For

{\left( {1 + \alpha x} \right)^4}

the middle term

{T_{{4 \over 2} + 1}}

=

{}^4{C_2}.{\alpha ^2}{x^2}

$\therefore$ Coefficient of middle term =

{}^4{C_2}.{\alpha ^2}

For

{\left( {1 - \alpha x} \right)^6}

the middle term

{T_{{6 \over 2} + 1}}

=

{}^6{C_3}.-{\alpha ^3}{x^3}

$\therefore$ Coefficient of middle term =

{}^6{C_3}.{-\alpha ^3}

$\therefore$ According to question,

{}^4{C_2}.{\alpha ^2}

=

{}^6{C_3}.{-\alpha ^3}

\Rightarrow 6 = 20 \times - \alpha

\Rightarrow \alpha = - {3 \over {10}}

Q137

If the fourth term in the binomial expansion of

{\left( {\sqrt {{x^{\left( {{1 \over {1 + {{\log }_{10}}x}}} \right)}}} + {x^{{1 \over {12}}}}} \right)^6}

is equal to 200, and x > 1, then the value of x is :

A 100

B 103

C 10

D 104

Correct Answer

Option C

Solution

Fourth term (T4) =

{}^6{C_3}{\left( {\sqrt {{x^{\left( {{1 \over {1 + {{\log }_{10}}x}}} \right)}}} } \right)^3}{\left( {{x^{{1 \over {12}}}}} \right)^3}

=

20{\left( {{x^{\left( {{1 \over {1 + {{\log }_{10}}x}}} \right)}}} \right)^{{3 \over 2}}}\left( {{x^{{1 \over 4}}}} \right)

=

20 \times {x^{\left( {{1 \over {1 + {{\log }_{10}}x}}} \right){3 \over 2}}} \times {x^{{1 \over 4}}}

=

20 \times {x^{\left( {{3 \over {2\left( {1 + {{\log }_{10}}x} \right)}} + {1 \over 4}} \right)}}

Given, T4 = 200 $\therefore$

20 \times {x^{\left( {{3 \over {2\left( {1 + {{\log }_{10}}x} \right)}} + {1 \over 4}} \right)}}

= 200 $\Rightarrow$

{x^{\left( {{3 \over {2\left( {1 + {{\log }_{10}}x} \right)}} + {1 \over 4}} \right)}}

= 10 Taking log10 on both sides

\left( {{3 \over {2\left( {1 + {{\log }_{10}}x} \right)}} + {1 \over 4}} \right){\log _{10}}x

= 1 put log10 x = t

\left( {{3 \over {2\left( {1 + t} \right)}} + {1 \over 4}} \right)t

= 1 $\Rightarrow$

\left( {{{\left( {1 + t} \right) + 6} \over {4\left( {1 + t} \right)}}} \right) \times t

= 1 $\Rightarrow$ t2 + 7t = 4 + 4t $\Rightarrow$ t2 + 3t - 4 = 0 $\Rightarrow$ (t + 4)(t - 1) = 0 $\Rightarrow$ t = 1 or t = - 4 $\therefore$ log10 x = 1 $\Rightarrow$ x = 10 or log10 x = - 4 $\Rightarrow$ x = 10-4 But as x > 1 so x $\ne$ 10-4 $\therefore$ x = 10

Q138

In the expansion of

{\left( {{x \over {\cos \theta }} + {1 \over {x\sin \theta }}} \right)^{16}}

, if

{\ell _1}

is the least value of the term independent of x when

{\pi \over 8} \le \theta \le {\pi \over 4}

and

{\ell _2}

is the least value of the term independent of x when

{\pi \over {16}} \le \theta \le {\pi \over 8}

, then the ratio

{\ell _2}

:

{\ell _1}

is equal to :

A 8 : 1

B 16 : 1

C 1 : 8

D 1 : 16

Correct Answer

Option B

Solution

Tr + 1 = 16Cr

{\left( {{x \over {\cos \theta }}} \right)^{16 - r}}{\left( {{1 \over {x\sin \theta }}} \right)^r}

= 16Cr

{\left( x \right)^{16 - 2r}} \times {1 \over {{{\left( {\cos \theta } \right)}^{16 - r}}{{\left( {\sin \theta } \right)}^r}}}

term is independent of x when $\therefore$ 16 – 2r = 0 $\Rightarrow$ r = 8 T9 = 16C8 $\times$

{1 \over {{{\cos }^8}\theta {{\sin }^8}\theta }}

= 16C8 $\times$

{{{2^8}} \over {{{\left( {\sin 2\theta } \right)}^8}}}

If

\theta \in \left[ {{\pi \over 8},{\pi \over 4}} \right]

then

2\theta \in \left[ {{\pi \over 4},{\pi \over 2}} \right]

In the range

\left[ {{\pi \over 4},{\pi \over 2}} \right]

, sin 2 $\theta$ is increasing.

And value of T9 is least when sin 2 $\theta$ is maximum.

And sin 2 $\theta$ is maximum in the range

\left[ {{\pi \over 4},{\pi \over 2}} \right]

when 2 $\theta$ =

{{\pi \over 2}}

$\therefore$

{l_1}

= 16C8 $\times$ 28 AgainIf

\theta \in \left[ {{\pi \over {16}},{\pi \over 8}} \right]

then

2\theta \in \left[ {{\pi \over 8},{\pi \over 4}} \right]

In the range

\left[ {{\pi \over 8},{\pi \over 4}} \right]

, sin 2 $\theta$ is increasing.

And value of T9 is least when sin 2 $\theta$ is maximum.

And sin 2 $\theta$ is maximum in the range

\left[ {{\pi \over 8},{\pi \over 4}} \right]

when 2 $\theta$ =

{{\pi \over 4}}

$\therefore$

{l_2}

= 16C8 $\times$

{{{2^8}} \over {{{\left( {{1 \over {\sqrt 2 }}} \right)}^8}}}

= 16C8 $\times$

{{2^8}{2^4}}

$\therefore$

{{{l_2}} \over {{l_1}}}

=

{{{2^4}} \over 1}

=

{{16} \over 1}

Q139

If

\alpha

and

\beta

be the coefficients of x4 and x2 respectively in the expansion of

{\left( {x + \sqrt {{x^2} - 1} } \right)^6} + {\left( {x - \sqrt {{x^2} - 1} } \right)^6}

, then

A

\alpha + \beta = 60

B

\alpha - \beta = 60

C

\alpha + \beta = -30

D

\alpha - \beta = -132

Correct Answer

Option D

Solution

(x+a)n + (x – a)n = 2(T1 + T3 + T5 +.....)

{\left( {x + \sqrt {{x^2} - 1} } \right)^6} + {\left( {x - \sqrt {{x^2} - 1} } \right)^6}

= 2[T1 + T3 + T5 + T7 ] = 2[6C0 x6 + 6C2 x4(x2 – 1) + 6C4 x2(x2 –1)2 + 6C6 x0(x2–1)3] = 2[x6+ 15(x6 – x4) + 15x2 (x4 + 1 –2x2) + (x6 – 3x4 +3x2 –1)] = 2[x6(2 + 15 + 15 + 1) + x4(–15 – 30 –3) + x2(15 + 3)] Coefficient of x4 = $\alpha$ = -96 And coefficient of x2 = $\beta$ = 36 $\therefore$

\alpha - \beta = - 96 - 36 = -132

Q140

If the fourth term in the expansion of

{(x + {x^{{{\log }_2}x}})^7}

is 4480, then the value of x where x

\in

N is equal to :

A 3

B 1

C 4

D 2

Correct Answer

Option D

Solution

T4 =

{}^7{C_3}{x^4}{({x^{{{\log }_2}x}})^3} = 4480

\Rightarrow 35{x^4}{({x^{{{\log }_2}x}})^3} = 4480

\Rightarrow {x^4}{({x^{{{\log }_2}x}})^3} = 128

take log w.r.t. base 2 we get,

4{\log _2}x + 3{\log _2}({x^{{{\log }_2}x}}) = {\log _2}128

Let

{\log _2}x = y

4y + 3{y^2} = 7

\Rightarrow y = 1,{{ - 7} \over 3}

\Rightarrow {\log _2}x = 1,{{ - 7} \over 3}

$\Rightarrow$

x = 2,x = {2^{ - 7/3}}