Binomial Theorem Test

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Binomial Theorem Test - 12

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Weekly Quiz Competition

Question 1
1 / -0

The seventh term in the expansion of
$$ \left(4x-\dfrac{1}{2\sqrt{x}}\right)^{13} $$ is

A
$$ ^{13}C_7.2^5.x^4 $$

B
$$ ^{13}C_6.2^8.x^3 $$

C
$$ ^{13}C_6.2^8.x^4 $$

D
$$ ^{13}C_6.2^6.x^5 $$

Solution

$$(x-a)^{n}={}^{n}C_{0} x^{n}-{}^{n}C_{1}x^{n-1}a+{}^{n}C_{2} x^{n-2} a^{2}-.....$$
$$(r+1)$$th term in the expansion of $${ (x-a) }^{ n }$$ is given by

$${ T }_{ r+1 }= ({ -1 })^{ r }\ \ { }^{n}{ C }_{ r }{ x }^{ n-r }{ a }^{ r }$$

Using above formula for $$ \left(4x-\dfrac{1}{2\sqrt{x}}\right)^{13}$$, seventh term in the expansion is
$$T_{7} =T_{6+1} =({ -1 })^{ 6 }\ \ { }^{13}{ C }_{ 6 }(4x)^{13-6}\left(\dfrac{1}{2\sqrt{x}}\right)^{6}$$
$$= { }^{13}{ C }_{ 6 }(4x)^{7}\left(\dfrac{1}{2^{6}(\sqrt{x})^{6}}\right)$$
$$= { }^{13}{ C }_{ 6 } \cdot 2^{14}\cdot x^{7}\left(\dfrac{1}{2^{6}x^{3}}\right)$$
$$= { }^{13}{ C }_{ 6 } \cdot 2^{8}\cdot x^{4}$$
Question 2
1 / -0

The coefficients of $$ x^p $$ and $$ x^q $$ ($$p$$ and $$q$$ are positive integers) in the expansion of $$ (1 + x)^{p+q} $$ are

A
equal

B
equal with opposite signs

C
reciprocal to each other

D
unequal

Solution

The general term is

$$t_{r+1}=^{p+q}C_{r}x^{r}$$

For coefficient of $$x^p$$, $$r=p$$ and hence coefficient is $$^{p+q}C_{p}$$

For coefficient of $$x^q$$, $$r=q$$ and hence coefficient is $$^{p+q}C_{q}$$

$$^{p+q}C_{p}$$ = $$^{p+q}C_{q}$$
Question 3
1 / -0

In the expansion of $$ (3+ax)^9 $$ coefficients of $$ x^2 $$ and $$ x^3 $$ are equal then $$a =$$

A
$$ \dfrac{9}{7} $$

B
$$ \dfrac{7}{9} $$

C
$$ -\dfrac{9}{7} $$

D
$$ -\dfrac{7}{9} $$

Solution

It is known that $${ \left( r+1 \right) }^{ th }$$ term, $$({T}_{r+1}),$$ in the binomial expansion of $${ \left( a+b \right) }^{ n }$$ is given by $${ T }_{ r+1 }=^{ n }{ { C }_{ r } }{ a }^{ n-r }{ b }^{ r }.$$
Assuming that $${x}^{2}$$ occurs in the $${ \left( r+1 \right) }^{ th }$$ term in the expansion of $${ \left( 3+ax \right) }^{ 9 },$$ we obtain
$${ T }_{ r+1 }=^{ 9 }{ { C }_{ r } }{ \left( 3 \right) }^{ 9-r }=^{ 9 }{ { C }_{ r } }{ \left( 3 \right) }^{ 9-r }{ a }^{ r }{ x }^{ r }$$
Comparing the indices of $$x$$ in $${x}^{2}$$ and in $${T}_{r+1},$$ we obtain $$r=2.$$
Thus, the coefficients of $${x}^{2}$$ is $$\displaystyle ^{ 9 }{ { C }_{ 2 }{ \left( 3 \right) }^{ 9-2 } }{ a }^{ 2 }=\frac { 9! }{ 2!7! } { \left( 3 \right) }^{ 7 }{ a }^{ 2 }=36{ \left( 3 \right) }^{ 7 }{ a }^{ 2 }.$$
Arruming that $${x}^{3}$$ occurs in the $${ \left( k+1 \right) }^{ n }$$ term in the expansion of $${ \left( 3+ax \right) }^{ 9 },$$ we obtain
$${ T }_{ k+1 }=^{ 9 }{ { C }_{ k } }{ \left( 3 \right) }^{ 9-k }{ \left( ax \right) }^{ k }=^{ 9 }{ { C }_{ k } }{ \left( 3 \right) }^{ 9-k }{ a }^{ k }{ x }^{ k }$$
Comparing the indices of $$x$$ in $${x}^{3}$$ and $${T}_{k+1},$$ we obtain $$k=3$$
Thus, the coefficient of $${x}^{3}$$ is $$\displaystyle ^{ 9 }{ { C }_{ 3 } }{ \left( 3 \right) }^{ 9-3 }{ a }^{ 3 }=\frac { 9! }{ 3!6! } { \left( 3 \right) }^{ 6 }{ a }^{ 3 }=84{ \left( 3 \right) }^{ 6 }{ a }^{ 3 }$$
In is given that the coefficient of $${x}^{2}$$ and $${x}^{3}$$ are the same,
$$\displaystyle 84{ \left( 3 \right) }^{ 6 }{ a }^{ 3 }=36{ \left( 3 \right) }^{ 7 }{ a }^{ 2 }\Rightarrow 84a=36\times 3\Rightarrow a=\frac { 36\times 3 }{ 84 } =\frac { 104 }{ 84 } \Rightarrow a=\frac { 9 }{ 7 } $$
Thus, the required value of $$a$$ is $$\displaystyle \frac{9}{7} $$
Question 4
1 / -0

If the coefficient of $$ x^7 $$ in $$(ax^2 + \dfrac{1}{bx})^{11} $$ is equal to the coefficient of $$ x^{-7} $$ in $$ (ax - \dfrac{1}{bx^2})^{11} $$ then

A
$$ ab = -1 $$

B
$$ a = b $$

C
$$ ab = 1 $$

D
$$ a+b = 0 $$

Solution

General term $$T_{r+1}$$ in the expansion of $$(a+b)^{n}$$ is given by

$$T_{r+1}=\:^nC_{r}a^{n-r}b^r$$

Applying to $$\left(ax^2+\dfrac{1}{bx}\right)^{11}$$, we get

$$T_{r+1}={}^{11}C_{r}x^{22-3r}a^{11}(ab)^{-r}$$ ...(i)

Therefore $$22-3r=7$$ for coefficient of $$x^7$$

$$\Rightarrow r=5$$

Thus, $$(i)$$ reduces to

$$T_{6}=\:^{11}C_{5}x^{7}a^{11}(ab)^{-5}$$

Similarly applying to $$\left(ax-\dfrac{1}{bx^2}\right)^{11}$$, we get

$$T_{r+1}=(-1)^r\:^{11}C_{r}x^{11-3r}a^{11}(ab)^{-r}$$ ...(ii)

Therefore $$11-3r=-7$$ for coefficient of $$x^{-7}$$

$$\Rightarrow r=6$$

Equation $$(ii)$$ reduces to

$$T_{7}=\:^{11}C_{6}x^{-7}a^{11}{ab}^{-6}$$ ...(b)

Hence applying the given condition we get

$$\:^{11}C_{6}a^{11}(ab)^{-6}=\:^{11}C_{5}a^{11}(ab)^{-5}$$ from(a,b)

$$ab=1$$

Hence answer is C
Question 5
1 / -0

If the fourth term in the expansion of $$\left (px+\dfrac{1}{x}\right)^n $$ is $$\dfrac52$$, then $$(n, p) = $$

A
$$\left(3, \dfrac12\right)$$

B
$$\left(6, \dfrac12\right)$$

C
$$\left(5, \dfrac12\right)$$

D
$$(6, 2)$$

Solution

Given expression is $$(px+x^{-1})\:^n$$
General term $$T_{r+1}=\:^nC_{r}a^{n-r}b^r$$
Applying to the above question, we get
$$T_{3+1}=\:^nC_{3}(px)^{n-3}x^{-3}$$
$$=\:^nC_{3}(p)^{n-3}x^{n-6}$$ ...(i)
Since the fourth term is independent of $$x$$,
$$n-6=0$$
$$n=6$$
Substituting this value in (i), we get
$$\:^6C_{3}(p)^{3}=\dfrac{5}{2}$$
$$\Rightarrow 20p^3=\dfrac{5}{2}$$
$$\Rightarrow p^3=\dfrac{1}{8}$$
$$\Rightarrow p=\dfrac{1}{2}$$
Therefore $$(n,p)=\left(6,\dfrac{1}{2}\right)$$
Question 6
1 / -0

In the binomial expansion of $$ \left ( \sqrt{y} + \dfrac{1}{2\sqrt[4]{y}} \right )^8 $$, the terms in the expansion in which the power of $$y$$ is a natural number are:

A
$$ T_1, T_5 $$

B
$$ T_1, T_2, T_3, T_4 $$

C
$$ T_2, T_4 $$

D
$$ T_1, T_3, T_5 $$

Solution

$$T_{r+1}=\:^nC_{r}a^{n-r}b^r$$
Applying to the above question, we get
$$T_{r+1}=\:^nC_{r}y^{4-\frac{r}{2}}2^{-r}y^{\frac{r}{4}}$$
Power of $$y$$ $$=4-\displaystyle \frac{3r}{4}$$
For the power of $$y$$ to be natural $$4-\displaystyle \frac{3r}{4}>0$$ and $$4-\displaystyle \frac{3r}{4}$$ should not be a fraction.
Hence, $$r$$ must be a multiple of $$4$$ or $$0$$.
These conditions are met for $$r=0$$ and $$r=4$$.
For $$r=8$$ , the power of $$y$$ would be negative
Hence the terms are $$T_{0+1}$$ and $$T_{4+1}$$ i.e. $$T_{1},T_{5}$$
Question 7
1 / -0

In the expansion of $$ (2+\dfrac{x}{3})^n $$, coefficients of $$ x^7 $$ and $$ x^8 $$ are equal. Then $$ n = $$

A
$$49$$

B
$$50$$

C
$$55$$

D
$$56$$

Solution

We know, $$T_{r+1}=\:^nC_{r}a^{n-r}b^r$$
Applying to the above question, we get
$$T_{r+1}=\:^nC_{r}2^{n-r}3^{-r}x^{r}$$
Applying the given conditions, i.e. coefficient of $$x^7$$ is equal to coefficient of $$x^8$$
Therefore $$\:^nC_{7}2^{n-7}3^{-7}=\:^nC_{8}2^{n-8}3^{-8}$$
$$\Rightarrow 6\:^nC_{7}=\:^nC_{8}$$
$$\Rightarrow \dfrac{6.(n)!}{(n-7)!7!}=\dfrac{(n)!}{(n-8)!8!}$$
$$\Rightarrow 48=n-7$$
$$\Rightarrow n =55$$
Question 8
1 / -0

The middle term in the expansion of $$ \left (\displaystyle \frac{x^{\frac{3}{2}}}{\sqrt{a}} - \frac{y^{\frac{5}{2}}}{b^{\frac{3}{2}}} \right )^8 $$ is

A
$$ ^8C_4. \dfrac{x^6.y^{10}}{a^2b^6} $$

B
$$ ^8C_4. \dfrac{x^6.y^{8}}{a^2b^4} $$

C
$$ ^8C_5. \dfrac{x^5.y^{10}}{a^2b^5} $$

D
$$ ^8C_1. \dfrac{x^5.y^{10}}{a^2b^5} $$

Solution
Question 9
1 / -0

The middle term in the expansion of $$ \left ( x + \dfrac{1}{x} \right )^{10} $$ is

A
$$ ^{10}C_{4}.\dfrac{1}{x} $$

B
$$ ^{10}C_{5} $$

C
$$ ^{10}C_{5}.\dfrac{1}{x} $$

D
$$ ^{10}C_{6}.x $$

Solution

Middle term $$=\dfrac{n}{2}+1$$ $$=6^{th} $$ term
Now consider the following:
$$T_{r+1}=\:^nC_{r}a^{n-r}b^r$$...(i)
Where $$T$$ represents the term.
Here the middle term is the $$6^{th}$$ term.
Hence $$r+1=6$$
$$\therefore r=5$$
substituting in (i), we get
$$\:^{10}C_{5}x^{10-5}\frac{1}{x^5}$$
$$=\:^{10}C_{5}x^{5}x^{-5}$$
$$=\:^{10}C_{5}$$
Hence, answer is B.
Question 10
1 / -0

The middle term in the expansion of $$ (1+x)^{2n} $$ is

A
$$ ^{2n}C_{n} $$

B
$$ ^{2n}C_{n-1}.x^{n+1} $$

C
$$ ^{2n}C_{n-1}.x^{n-1} $$

D
$$ ^{2n}C_{n}.x^n $$

Solution

Middle term $$=\dfrac{N}{2}+1$$
$$=\dfrac{2n}{2}+1$$
$$=(n+1)^{th}$$ term
Now consider the following,
$$T_{r+1}=\:^nC_{r}a^{n-r}b^r$$...(i)
Where $$T$$ represents the term
Here the middle term is the $$(n+1)^{th}$$ term
Hence $$r+1=n+1$$
$$\therefore r=n$$
Substituting in (i), we get $$\:^{2n}C_{n}1^{2n-n}x^{n}$$
$$=\:^{2n}C_{n}x^{n}$$
Hence answer is D

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Re-Attempt Weekly Quiz Competition

Binomial Theorem Test - 12

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Binomial Theorem Test - 12

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Question 1

$$ ^{13}C_7.2^5.x^4 $$

$$ ^{13}C_6.2^8.x^3 $$

$$ ^{13}C_6.2^8.x^4 $$

$$ ^{13}C_6.2^6.x^5 $$

Solution

Question 2

equal

equal with opposite signs

reciprocal to each other

unequal

Solution

Question 3

$$ \dfrac{9}{7} $$

$$ \dfrac{7}{9} $$

$$ -\dfrac{9}{7} $$

$$ -\dfrac{7}{9} $$

Solution

Question 4

$$ ab = -1 $$

$$ a = b $$

$$ ab = 1 $$

$$ a+b = 0 $$

Solution

Question 5

$$\left(3, \dfrac12\right)$$

$$\left(6, \dfrac12\right)$$

$$\left(5, \dfrac12\right)$$

$$(6, 2)$$

Solution

Question 6

$$ T_1, T_5 $$

$$ T_1, T_2, T_3, T_4 $$

$$ T_2, T_4 $$

$$ T_1, T_3, T_5 $$

Solution

Question 7

$$49$$

$$50$$

$$55$$

$$56$$

Solution

Question 8

$$ ^8C_4. \dfrac{x^6.y^{10}}{a^2b^6} $$

$$ ^8C_4. \dfrac{x^6.y^{8}}{a^2b^4} $$

$$ ^8C_5. \dfrac{x^5.y^{10}}{a^2b^5} $$

$$ ^8C_1. \dfrac{x^5.y^{10}}{a^2b^5} $$

Solution

Question 9

$$ ^{10}C_{4}.\dfrac{1}{x} $$

$$ ^{10}C_{5} $$

$$ ^{10}C_{5}.\dfrac{1}{x} $$

$$ ^{10}C_{6}.x $$

Solution

Question 10

$$ ^{2n}C_{n} $$

$$ ^{2n}C_{n-1}.x^{n+1} $$

$$ ^{2n}C_{n-1}.x^{n-1} $$

$$ ^{2n}C_{n}.x^n $$

Solution

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