Binomial Theorem — Page 8 | JEE Mathematics

Q71

If the coefficients of

x

and

x^{2}

in

(1+x)^{\mathrm{p}}(1-x)^{\mathrm{q}}

are 4 and

-

5 respectively, then

2 p+3 q

is equal to :

A 66

B 60

C 69

D 63

Correct Answer

Option D

Solution

We have, coefficient of $x$ in $(1+x)^p(1-x)^q=4$ and coefficient of $x^2$ in $(1+x)^p(1-x)^q=-5$

\begin{aligned} & (1+x)^p(1-x)^q \\\\ & =\left(1+p x+\frac{p(p-1)}{2} x^2+\ldots\right)\left(1-q x+\frac{q(q-1)}{2} x^2+\ldots\right) \\\\ & =1+(p-q) x+\left(\frac{p(p-1)}{2}+\frac{q(q-1)}{2}-p q\right) x^2+\ldots \ldots \end{aligned}

Coefficient of $x$ in $(1+x)^p(1-x)^q=-q+p$ $\Rightarrow p-q=4$ ...........(i)

\text{ Coefficient of } x^2 \text{ in }(1+x)^p(1-x)^q=\frac{q(q-1)}{2}-p q+\frac{p(p-1)}{2}

\begin{aligned} & \Rightarrow \frac{q^2-q-2 p q+p^2-p}{2} =-5 \\\\ & \Rightarrow \frac{p^2+q^2-2 p q-(p+q)}{2} =-5 \\\\ & \Rightarrow (p-q)^2-(p+q) =-10 \\\\ & \Rightarrow (4)^2-(p+q) =-10 \quad[\because \text{ From Eq. (i) }] \\\\ & \Rightarrow p+q =26 ...........(ii) \end{aligned}

Form Eqs. (i) and (ii), we get $p=15, q=11$ $\therefore 2 p+3 q=2 \times 15+3 \times 11=30+33=63$

Q72

If the coefficient of

{x^7}

in

{\left( {ax - {1 \over {b{x^2}}}} \right)^{13}}

and the coefficient of

{x^{ - 5}}

in

{\left( {ax + {1 \over {b{x^2}}}} \right)^{13}}

are equal, then

{a^4}{b^4}

is equal to :

A 22

B 33

C 44

D 11

Correct Answer

Option A

Solution

The given expression is $\left(a x-\dfrac{1}{b x^2}\right)^{13}$ So,

\begin{aligned} T_{r+1} & ={ }^{13} C_r(a x)^{13-r}\left(-\frac{1}{b x^2}\right)^r \\\\ & ={ }^{13} C_r(a)^{13-r} x^{13-r-2 r}(-1 / b)^r \\\\ & ={ }^{13} C_r(a)^{13-r}\left(-\frac{1}{b}\right)^r x^{13-3 r} \end{aligned}

For coefficient of $x^7$ in $\left(a x-\dfrac{1}{b x^2}\right)^{13}$

\begin{aligned} & \quad 13-3 r=7 \\\\ & \Rightarrow 3 r=6 \Rightarrow r=2 \\\\ & \therefore \text{ Coefficient of } x^7={ }^{13} C_2 \cdot(a)^{11} \cdot \frac{1}{b^2} \end{aligned}

Again, the another expression is $\left(a x+\dfrac{1}{b x^2}\right)^{13}$ So, $T_{n+1}={ }^{13} C_r(a x)^{13-r}\left(\dfrac{1}{b x^2}\right)^r={ }^{13} C_r(a)^{13-r}\left(\dfrac{1}{b}\right)^r x^{13-3 r}$ For coefficient $x^{-5}$ in $\left(a x+\dfrac{1}{b x^2}\right)^{13}$

\begin{aligned} &13-3 r =-5 \\\\ &\Rightarrow r =6 \end{aligned}

So, coefficient of $x^{-5}={ }^{13} C_6(a)^7 \dfrac{1}{b^6}$ Now, according to the question, ${ }^{13} C_2(a)^{11} \dfrac{1}{b^2}={ }^{13} C_6(a)^7 \dfrac{1}{b^6}$

\begin{aligned} & \Rightarrow a^4 b^4=\frac{{ }^{13} C_6}{{ }^{13} C_2} \\\\ & \therefore a^4 b^4=22 \end{aligned}

Q73

If the coefficients of three consecutive terms in the expansion of

(1+x)^{n}

are in the ratio

1: 5: 20

, then the coefficient of the fourth term is

A 3654

B 1827

C 5481

D 2436

Correct Answer

Option A

Solution

\begin{aligned} & \text{ Given: }{ }^n \mathrm{C}_{r-1}:{ }^n \mathrm{C}_r:{ }^n \mathrm{C}_{r+1} \\\\ & =1: 5: 20 \\\\ & \Rightarrow \frac{n !}{(r-1) !(n-r+1) !} \times \frac{r !(n-r) !}{n !}=\frac{1}{5} \\\\ & \Rightarrow \frac{r}{(n-r+1)}=\frac{1}{5} \\\\ & \Rightarrow 5 r=n-r+1 \\\\ & \Rightarrow n=6 r-1 ........(i) \end{aligned}

\begin{aligned} & \text{ Also, } \frac{n}{r !(n-r) !} \times \frac{(r+1) !(n-r-1) !}{n !}=\frac{5}{20}=\frac{1}{20} \\\\ & \Rightarrow \frac{(r+1)}{(n-r)}=\frac{1}{4} \\\\ & \Rightarrow 4 r+4=n-r \\\\ & \Rightarrow n=5 r+4 ..........(ii) \end{aligned}

From (i) and (ii), we get

\begin{aligned} & 6 r-1=5 r+4 \\\\ & \Rightarrow r=5 \\\\ & \text{ So, } n=5(5)+4=29 \\\\ & \text{ So, coefficient of } 4{ }^{\text{th }} \text{ terms }={ }^n \mathrm{C}_3={ }^{29} \mathrm{C}_3 \\\\ & =\frac{29 !}{3 ! 26 !}=\frac{29 \times 28 \times 27}{3 \times 2}=3654 \end{aligned}

Q74

The coefficien of x10 in the expansion of (1 + x)2(1 + x2)3(1 + x3)4 is equal to :

A 52

B 56

C 50

D 44

Correct Answer

Option A

Solution

$\because$

\,\,\,

(1 + x)2 = 1 + 2x + x2, (1 + x2)3 = 1 + 3x2 + 3x4 + x6 and (1 + x3)4 = 1 + 4x3 + 6x6 + 4x9 + x12 So, the possible combination for x10 are : x . x9, x . x6. x3, x2 . x2 . x6 , x4 . x6 Corresponding coefficients are 2 $\times$ 4, 2 $\times$ 1 $\times$ 4, 1 $\times$ 3 $\times$ 6, 3 $\times$ 6 or 8, 8, 18, 18.

\therefore\,\,\,

Sum of the coefficient is 8 + 8 + 18 + 18 = 52 Therefore, the coefficient of x10 in the expansion of (1 + x)2 (1 + x2)3 (1 + x3)4 is 52.

Q75

If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of

\left(\sqrt[4]{2}+\dfrac{1}{\sqrt[4]{3}}\right)^{\mathrm{n}}

is

\sqrt{6}: 1

, then the third term from the beginning is :

A

30 \sqrt{2}

B

60 \sqrt{3}

C

60 \sqrt{2}

D

30 \sqrt{3}

Correct Answer

Option B

Solution

\mathrm{T}_{r+1}={ }^n \mathrm{C}_r x^{n-r} a^r

\frac{T_5}{T_5^{\prime}}=\frac{{ }^n C_4 \times\left((2)^{\frac{1}{4}}\right)^{n-4}\left(\frac{1}{3^{1 / 4}}\right)^4}{{ }^n C_4\left(\frac{1}{3^{1 / 4}}\right)^{n-4}\left(2^{1 / 4}\right)^4}=\sqrt{6}

$\left[\because r\right.$ th term from end in the expansion of $(x+y)^n=r$ th term from beginning in the expansion of $\left.(y+x)^n\right]$

\Rightarrow \frac{{ }^n C_4(2)^{\frac{n-4}{4}}\left(\frac{1}{3}\right)^{4 / 4}}{{ }^n C_4\left(\frac{1}{3}\right)^{\frac{n-4}{4}}(2)^{4 / 4}}=\frac{\sqrt{6}}{1}

\begin{aligned} &\Rightarrow (2)^{\frac{n-8}{4}}(3)^{\frac{n-8}{4}} =6^{1 / 2} \\\\ &\Rightarrow 6^{\frac{n-8}{4}} =6^{1 / 2} \\\\ &\Rightarrow \frac{n-8}{4} =\frac{1}{2} \\\\ &\Rightarrow n-8=2 \Rightarrow n =10 \end{aligned}

\therefore T_3={ }^{10} C_2(\sqrt[4]{2})^8\left(\frac{1}{\sqrt[4]{3}}\right)^2=45(2)^{\frac{8}{4}} \frac{1}{3^{1 / 2}}

=45(4) \times \frac{\sqrt{3}}{3}=60 \sqrt{3}

Q76

If the coefficient of

{x^7}

in

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

and

{x^{ - 7}}

in

{\left( {ax - {1 \over {3b{x^2}}}} \right)^{11}}

are equal, then :

A

243ab = 64

B

32ab = 729

C

64ab = 243

D

729ab = 32

Correct Answer

Option D

Solution

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

T_{r+1}={ }^{11} C_r\left(a x^2\right)^{11-r}\left(\frac{1}{2 b x}\right)^r={ }^{11} C_r(a)^{11-r}\left(\frac{1}{2 b}\right)^r x^{22-3 r}

\begin{array}{rlrl} &\text{ Now, } 22-3 r =7 \\\\ &\Rightarrow 15 =3 r \\\\ &\Rightarrow r =5 \end{array}

and general term of $\left(a x-\dfrac{1}{3 b x^2}\right)^{11}$ is

\begin{aligned} T_{r+1} & ={ }^{11} C_r(a x)^{11-r}\left(-\frac{1}{3 b x^2}\right)^r \\\\ & ={ }^{11} C_r a^{11-r}\left(-\frac{1}{3 b}\right)^r x^{11-3 r} \end{aligned}

Now, $11-3 r=-7$

\Rightarrow 18=3 r \Rightarrow r=6

$\begin{aligned} & \text{ Since, coefficient of } x^7 \text{ in }\left(a x^2+\dfrac{1}{2 b x}\right)^{11} \\\\ & =\text{ Coefficient of } x^{-7} \text{ in }\left(a x-\dfrac{1}{3 b x^2}\right)^{11} \\\\ & \Rightarrow{ }^{11} C_5(a)^6\left(\dfrac{1}{2 b}\right)^5={ }^{11} C_6(a)^5\left(-\dfrac{1}{3 b}\right)^6 \\\\ & \Rightarrow \dfrac{a}{32 b^5}=\dfrac{1}{729 b^6} \Rightarrow 729 a b=32\end{aligned}$

Q77

Among the statements : (S1) :

2023^{2022}-1999^{2022}

is divisible by 8 (S2) :

13(13)^{n}-12 n-13

is divisible by 144 for infinitely many

n \in \mathbb{N}

A both (S1) and (S2) are incorrect

B only (S1) is correct

C only (S2) is correct

D both (S1) and (S2) are correct

Correct Answer

Option D

Solution

We have, $S_1$ : $(2023)^{2022}-(1999)^{2022}$

\begin{aligned} & =(1999+24)^{2022}-(1999)^{2022}={ }^{2022} C_0(1999)^{2022}(24)^0 \\\\ & +{ }^{2022} C_1(1999)^{2021}(24)^1+{ }^{2022} C_2(1999)^{2020}(24)^2 \\\\ & +\ldots-(1999)^{2022} \\\\ & ={ }^{2022} C_1(1999)^{2021}(24)+{ }^{2022} C_2(1999)^{2022}(24)^2 \\\\ & =24\left({ }^{2022} C_1(1999)^{2021}+{ }^{2022} C_2(1999)^{2022}(24)+\ldots+\ldots\right) \\\\ & \Rightarrow S_1 \text{ is divisible by } 24 \end{aligned}

\begin{aligned} & \text{ Now, } S_2: 13(13)^n-12 n-13 \\\\ & \text{ Here, } 13^n=(1+12)^n \\\\ & \quad=1+{ }^n C_1 12+{ }^n C_2(12)^2+{ }^n C_3(12)^3 \\\\ & \begin{aligned} \therefore S_2: & 13\left(1+{ }^n C_1(12)+{ }^n C_2(12)^2+{ }^n C_3(12)^3+\ldots\right)-12 n-13 \\\\ \quad= & 13+156 n+13\left({ }^n C_2(12)^2+{ }^n C_3(12)^3+\ldots\right)-12 n-13 \\\\ \quad= & 144 \times 13\left({ }^n C_2+{ }^n C_3(12)+\ldots\right) \end{aligned} \end{aligned}

$\Rightarrow S_2$ is divisible by 144 for infinitely many $n \in N$

Q78

{ }^{n-1} C_r=\left(k^2-8\right){ }^n C_{r+1}

if and only if :

A

2 \sqrt{2}<\mathrm{k}<2 \sqrt{3}

B

2 \sqrt{2}<\mathrm{k} \leq 3

C

2 \sqrt{3}<\mathrm{k}<3 \sqrt{3}

D

2 \sqrt{3}<\mathrm{k} \leq 3 \sqrt{2}

Correct Answer

Option B

Solution

{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}}=(\mathrm{k}^2-8){ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}

\underbrace{\mathrm{r}+1 \geq 0, \quad \mathrm{r} \geq 0}_{\mathrm{r} \geq 0}

\begin{aligned} & \frac{{ }^{n-1} C_r}{{ }^n C_{r+1}}=k^2-8 \\ & \frac{r+1}{n}=k^2-8 \\ & \Rightarrow k^2-8>0 \\ & (k-2 \sqrt{2})(k+2 \sqrt{2})>0 \end{aligned}

\mathrm{k} \in(-\infty,-2 \sqrt{2}) \cup(2 \sqrt{2}, \infty)\quad \text{.... (I)}

\begin{aligned} \therefore & \mathrm{n} \geq \mathrm{r}+1, \frac{\mathrm{r}+1}{\mathrm{n}} \leq 1 \\ \Rightarrow & \mathrm{k}^2-8 \leq 1 \\ & \mathrm{k}^2-9 \leq 0 \\ & -3 \leq \mathrm{k} \leq 3 \quad \text{.... (II)} \end{aligned}

From equation (I) and (II) we get

\mathrm{k} \in[-3,-2 \sqrt{2}) \cup(2 \sqrt{2}, 3]

Q79

If A denotes the sum of all the coefficients in the expansion of

\left(1-3 x+10 x^2\right)^{\mathrm{n}}

and B denotes the sum of all the coefficients in the expansion of

\left(1+x^2\right)^n

, then :

A

\mathrm{B}=\mathrm{A}^3

B

3 \mathrm{A}=\mathrm{B}

C

A=3 B

D

\mathrm{A}=\mathrm{B}^3

Correct Answer

Option D

Solution

Sum of coefficients in the expansion of

\left(1-3 \mathrm{x}+10 \mathrm{x}^2\right)^{\mathrm{n}}=\mathrm{A}

then

A=(1-3+10)^n=8^n

(put

x=1

) and sum of coefficients in the expansion of

\begin{aligned} & \left(1+x^2\right)^n=B \\ & \text{ then } B=(1+1)^n=2^n \\ & A=B^3 \end{aligned}

Q80

Let

a

be the sum of all coefficients in the expansion of

\left(1-2 x+2 x^2\right)^{2023}\left(3-4 x^2+2 x^3\right)^{2024}

and

b=\lim \limits_{x \rightarrow 0}\left(\dfrac{\int_0^x \dfrac{\log (1+t)}{t^{2024}+1} d t}{x^2}\right)

. If the equation

c x^2+d x+e=0

and

2 b x^2+a x+4=0

have a common root, where

c, d, e \in \mathbb{R}

, then

\mathrm{d}: \mathrm{c}:

e equals

A

2: 1: 4

B

1: 1: 4

C

1: 2: 4

D

4: 1: 4

Correct Answer

Option B

Solution

Put

x=1

\therefore \mathrm{a}=1

\mathrm{b}=\lim \limits_{\mathrm{x} \rightarrow 0} \frac{\int\limits_0^{\mathrm{x}} \frac{\ln (1+\mathrm{t})}{1+\mathrm{t}^{2024}} \mathrm{dt}}{\mathrm{x}^2}

Using L' HOPITAL Rule

\mathrm{b}=\lim \limits_{\mathrm{x} \rightarrow 0} \frac{\ln (1+\mathrm{x})}{\left(1+\mathrm{x}^{2024}\right)} \times \frac{1}{2 \mathrm{x}}=\frac{1}{2}

Now,

\mathrm{cx}^2+\mathrm{dx}+\mathrm{e}=0, \mathrm{x}^2+\mathrm{x}+4=0

(\mathrm{D}

\therefore \frac{\mathrm{c}}{1}=\frac{\mathrm{d}}{1}=\frac{\mathrm{e}}{4}$$