Binomial Theorem — Page 9 | JEE Mathematics

Q81

If the fractional part of the number

\left\{ {{{{2^{403}}} \over {15}}} \right\} is \, {k \over {15}}

, then k is equal to :

A 8

B 14

C 6

D 1

Correct Answer

Option A

Solution

{{{2^{403}}} \over {15}}

= {{{2^3}\, \cdot \,{2^{400}}} \over {15}}

= {8 \over {15}}{\left( {16} \right)^{100}}

= {8 \over {15}}{\left( {15 + 1} \right)^{100}}

= {8 \over {15}}\left( {{}^{100}{C_0} + {}^{100}{C_1}\,15 + {}^{100}{C_2}{{\left( {15} \right)}^2} + .....{{\left( {15} \right)}^{100}}} \right)

= {8 \over {15}} + 8\left( {{}^{100}{C_1} + {}^{100}{C_2}\,\left( {15} \right) + ..... + {{\left( {15} \right)}^{99}}} \right)

= {8 \over {15}} + 8

(integer) $\therefore$ Fractional part

= {8 \over {15}}

According to the question,

{k \over {15}} = {8 \over {15}}

$\Rightarrow$ K $=$ 8

Q82

The sum of the coefficient of

x^{2 / 3}

and

x^{-2 / 5}

in the binomial expansion of

\left(x^{2 / 3}+\dfrac{1}{2} x^{-2 / 5}\right)^9

is

A 19/4

B 69/16

C 63/16

D 21/4

Correct Answer

Option D

Solution

\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}

For coefficient of

x^{2 / 3}

\begin{aligned} & \Rightarrow 6-\frac{16 r}{15}=\frac{2}{3} \\ & \Rightarrow 90-16 r=10 \\ & \Rightarrow r=5 \end{aligned}

For coefficient of

x^{-2 / 5}

\begin{aligned} & \Rightarrow 6-\frac{16 r}{15}=\frac{-2}{5}\\ & \Rightarrow 90-16 r=-6 \\ & \Rightarrow r=6 \end{aligned}

Sum of coefficient of

x^{2 / 3}

&

x^{-2 / 5}

\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}

Q83

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 :

A 61

B 83

C 55

D 68

Correct Answer

Option B

Solution

x^2(1+x)^{98}+x^3(1+x)^{97}+\ldots+x^{54}(1+x)^{46}

It is a G.P. with first term

=x^2(1+x)^{98}

and common ratio

=\frac{x}{1+x}

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)

=x^2(1+x)^{98}\left((1+x)-x^{53}(1+x)^{-52}\right)

\begin{aligned} & ={ }^{99} \mathrm{C}_{68}-{ }^{46} \mathrm{C}_{15} \\ & \Rightarrow p=68, q=15 \\ & \Rightarrow p+q=83 \end{aligned}

Q84

The sum of all rational terms in the expansion of

\left(2^{\dfrac{1}{5}}+5^{\dfrac{1}{3}}\right)^{15}

is equal to :

A 633

B 6131

C 3133

D 931

Correct Answer

Option C

Solution

\begin{aligned} & T_{r+1}={ }^{15} \mathrm{C}_r\left(2^{1 / 5}\right)^{15-r}\left(5^{1 / 3}\right)^r \\ & ={ }^{15} C_r 5^{r / 3} 2^{\left(3-\frac{r}{5}\right)} \end{aligned}

For rational terms,

\frac{r}{3}

and

\frac{r}{5}

must be integer 3 and 5 divide

r \Rightarrow 15

divides

r \Rightarrow r=0

and

r=15

{ }^{15} C_0 5^0 2^3+{ }^{15} C_{15} 5^5 2^{(0)}

\begin{aligned} & =8+3125 \\ & =3133 \end{aligned}

Q85

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:

A 28

B 21

C 7

D 14

Correct Answer

Option D

Solution

\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}

Q86

If the constant term in the expansion of

\left(\dfrac{\sqrt[5]{3}}{x}+\dfrac{2 x}{\sqrt[3]{5}}\right)^{12}, x \neq 0

, is

\alpha \times 2^8 \times \sqrt[5]{3}

, then

25 \alpha

is equal to :

A 724

B 742

C 693

D 639

Correct Answer

Option C

Solution

\begin{aligned} & \left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[5]{3}}\right)^{12} \\ & T_{r+1}={ }^{12} C r\left(\frac{\sqrt[5]{3}}{x}\right)^{12-r}\left(\frac{2 x}{\sqrt[3]{5}}\right)^r \end{aligned}

For constant term

-12+r+r=0

\begin{aligned} & \Rightarrow \quad r=6 \\ & \therefore \quad \text{ Constant term }={ }^{12} C_6 \frac{(3)^{\frac{6}{5}}}{(5)^{\frac{6}{3}}}(2)^6 \end{aligned}

\begin{aligned} & ={ }^{12} C_6 \times \frac{2^6}{25} \times 3.3^{\frac{1}{5}} \\ & =\frac{231}{25} \times 2^8 \cdot 3^{\frac{1}{5}} \cdot 3 \\ & =\frac{693}{25} \cdot 2^8 \sqrt[5]{3} \\ & \therefore \quad \alpha=\frac{693}{25} \\ & 25 \alpha=693 \end{aligned}

Q87

The least value of n for which the number of integral terms in the Binomial expansion of

(\sqrt[3]{7}+\sqrt[12]{11})^n

is 183, is :

A 2184

B 2172

C 2196

D 2148

Correct Answer

Option A

Solution

\begin{aligned} & \text{ General term }={ }^n C_r\left(7^{1 / 3}\right)^{n-r}\left(11^{1 / 12}\right)^r \\ & ={ }^n C_r(7)^{\frac{n-r}{3}}(11)^{r / 12} \end{aligned}

For integral terms, $r$ must be multiple of 12

\therefore \mathrm{r}=12 \mathrm{k}, \mathrm{k} \in \mathrm{~W}

Total values of $\mathrm{r}=183$ Hence $\max r=12(182)$

=2184

Min value of $n=2184$

Q88

The remainder, when

7^{103}

is divided by 23, is equal to:

A 9

B 6

C 14

D 17

Correct Answer

Option C

Solution

\begin{aligned} & 7^{103}=7\left(7^{102}\right)=7(343)^{34}=7(345-2)^{34} \\ & 7^{103}=23 \mathrm{~K}_1+7.2^{34} \\ & \text{ Now } 7.2^{34}=7 \cdot 2^2 \cdot 2^{32} \\ & =28 \cdot(256)^4 \\ & =28(253+3)^4 \\ & \therefore 28 \times 81 \Rightarrow(23+5)(69+12) \\ & 23 \mathrm{~K}_2+60 \\ & \therefore \text{ Remainder }=14 \end{aligned}

Q89

Let

\alpha, \beta, \gamma

and

\delta

be the coefficients of

x^7, x^5, x^3

and

x

respectively in the expansion of

\begin{aligned} & \left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5, x>1 \text{. If } u \text{ and } v \text{ satisfy the equations } \\\\ & \alpha u+\beta v=18, \\\\ & \gamma u+\delta v=20, \end{aligned}

then

\mathrm{u+v}

equals :

A 4

B 3

C 5

D 8

Correct Answer

Option C

Solution

To find the sum of $u$ and $v$ , we first need to expand the expression: $\left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5$ Using the Binomial Theorem, the expansion yields: $= 2\left({}^5C_0 \cdot x^5 + {}^5C_2 \cdot x^3(x^3-1) + {}^5C_4 \cdot x(x^3-1)^2\right)$ Simplifying this, we obtain: $= 2\left(5x^7 + 10x^6 + x^5 - 10x^4 - 10x^3 + 5x\right)$ From this expansion, we can identify the coefficients: The coefficient of $x^7$ is $\alpha = 10$ The coefficient of $x^5$ is $\beta = 2$ The coefficient of $x^3$ is $\gamma = -20$ The coefficient of $x$ is $\delta = 10$ Given the equations: $\alpha u + \beta v = 18$ $\gamma u + \delta v = 20$ Substituting in the coefficients: $10u + 2v = 18$ $-20u + 10v = 20$ By solving these equations, we find: From $10u + 2v = 18$ , simplify to $5u + v = 9$ .

From $-20u + 10v = 20$ , simplify to $-2u + v = 2$ .

Solving these linear equations simultaneously, we find: Subtracting equation 2 from equation 1: $5u + v = 9$ $- (-2u + v = 2)$ This yields: $7u = 7 \quad \Rightarrow \quad u = 1$ Substitute $u = 1$ back into $5u + v = 9$ : $5(1) + v = 9 \quad \Rightarrow \quad v = 4$ Thus, the sum $u + v$ is: $u + v = 1 + 4 = 5$

Q90

The total number of irrational terms in the binomial expansion of (71/5 – 31/10)60 is :

A 54

B 55

C 49

D 48

Correct Answer

Option A

Solution

General term Tr+1 = 60Cr,

{7^{{{60 - r} \over 5}}}{3^{{r \over {10}}}}

$\therefore$ for rational term, r = 0, 10, 20, 30, 40, 50, 60 $\Rightarrow$ no of rational terms = 7 $\therefore$ number of irrational terms = 54