MCQ Questions for Class 11 Engineering Maths Binomial Theorem Quiz 15

Co-efficient of

$x^{n-1}$ in the expansion of,

$(x+3)^{n}+(x+3)^{n-1}(x+2)+(x+3)^{n-2}(x+2)^{2}+...+(x+2)^{n}$ is:

0%
$^{n+1}C_{2}(3)$
0%
$^{n-1}C_{2}(5)$
0%
$^{n+1}C_{2}(5)$
0%
$^{n}C_{2}(5)$

In the expansion of

$(x+y+z)^{10}$

0%
the coefficient of $x^{2}y^{2}z^{3}$ is 0
0%
the coefficient of $x^{8}yz$ is 90
0%
the number of terms is 66
0%
none of these

If in the expansion of

$\left( 2 ^ { x } + \dfrac { 1 } { 4 ^ { x } } \right) ^ { n } , T _ { 3 } = 7 T _ { 2 }$ and sum of the binomial coefficients of second and third terms is

$36 ,$ then the value of

$x$ is -

0%
$- 1 / 3$
0%
$- 1 / 2$
0%
$1 / 3$
0%
$1 / 2$

The first three terms in the expansion of

$(1 - ax)^{n}$ , where

$n$ is positive integer are,

$1 - 4x + 7x^{2}$ , then the value of

$a$

0%
$4$
0%
$\dfrac {1}{4}$
0%
$8$
0%
$\dfrac {1}{2}$

The value of x for which the sixth term in the expansion of

$\left [ 2^{\log_{2}{\sqrt{9^{x-1}+7}}}+\dfrac{1}{2^{\frac{1}{5}\log_{2}{(3^{x-1}+1)}}} \right ]^{7}$ is equal to 84 is

0%
4
0%
1 or 2
0%
0 or 1
0%
3.

Explanation

Here we have,

$\left [ 2^{\log_{2}{\sqrt{9^{x-1}+7}}}+\dfrac{1}{2^{\frac{1}{5}\log_{2}{(x-1)}+1}} \right ]^{7}=\left[ \sqrt{9^{x-1}+7} +(3^{x-1}+1)^{(-1/5)}\right]^7$

The sixth term in the expansion is given by,

$T_6=\ ^7C_5 ( \sqrt{9^{x-1}+7})^2\cdot \left\{(3^{x-1}+1)^{(-1/5)}\right\}^5\\ \Rightarrow 84=21 \cdot(9^{x-1}+7)\cdot(3^{x-1}+1)^{-1}\\ \Rightarrow \dfrac{84}{21}=\dfrac{9^{x-1}+7}{3^{x-1}+1}\\\Rightarrow 4\cdot (3^{x-1}+1)=(3^{x-1})^2+7\\ \Rightarrow (3^{x-1})^2-4\times 3^{x-1}+3=0\\ \Rightarrow (3^{x-1}-1)(3^{x-1}-3)=0$

$\Rightarrow 3^{x-1}=1$ or

$3$

$\Rightarrow 3^{x-1}=3^0$ or

$3^1$

$\Rightarrow x-1=0$ or

$1$

$\Rightarrow x=1, 2$

Hence, option (B) is correct.

In the expansion of

$\left( 1+ \right) ^{ 50 },$ the sum of the coefficient of odd powers of x is _______.

0%
226
0%
249
0%
250
0%
251

Find the middle term(s) in the expansion of :

$\left(2ax-\dfrac{b}{x^{2}}\right)^{12}$

0%
$\dfrac {59136\ a^6 b^6} {x^6}$
0%
$\dfrac {59163\ a^5 b^5} {x^5}$
0%
$\dfrac {59631\ a^7 b^7} {x^7}$
0%
None of these

Explanation

Given that to find middle term in expansion of

$\left(20x-\dfrac{b}{x^{2}}\right)^{12}$

$n=12\Rightarrow$ even

for even number middle term

$=\left(\dfrac{n}{2}+1\right)th$

middle term

$=\left(\dfrac{12}{2}+1\right)th\Rightarrow 7$

We know that

$T_{r+1}=\ ^{n}C_{r}(a)^{n-r}b^{r}$

$T_{7}=T_{6+1}=\ ^{12}C_{6}(29 x)^{12-6}\left(\dfrac{-b}{x^{2}}\right)^{6}$

$=\dfrac{12!}{6!6!}2^{6}a^{6}x^{6}\dfrac{b^{6}}{x^{12}}$

$=\dfrac{12\times 11\times 10\times 9\times 8\times 7\times 6!}{6\times 5\times 4\times 3\times 2\times 1\times 6!}2^{6}\dfrac{a^{6}b^{6}}{x^{6}}$

$=\dfrac{59136 a^{6}b^{6}}{x^{6}}$

The constant term in the expansion of

$(x^{2}-\frac{1}{x^{2}})^{16}$ is

0%
$C^{16}_{8}$
0%
$C^{16}_{7}$
0%
$C^{16}_{9}$
0%
$C^{16}_{10}$

Find the middle term(s) in the expansion of :

$\left(2x-\dfrac{x^{2}}{4}\right)^{9}$

0%
$\dfrac {63}{5}x^{13}, -\dfrac {62}{31}x^{15}$
0%
$\dfrac {63}{4}x^{13}, -\dfrac {63}{32}x^{14}$
0%
$\dfrac {61}{4}x^{11}, \dfrac {61}{33}x^{13}$
0%
None of these

Explanation

Given expansion term is

$\left(2x-\dfrac{x^{2}}{4}\right)^{9}$

$n=9=$ odd

middle term

$=\left(\dfrac{9+1}{2}\right)^{th}+\left(\dfrac{9+3}{2}\right)^{th}$

$=5^{th}$ or

$6^{th}$ term

$T_{5}=T_{4+1}=\ ^{9}C_{4}(2x)^{5}\left(\dfrac{-x^{2}}{4}\right)^{4}$

$=\dfrac{9!}{4!5!}2^{5}\times x^{5}(-1)^{4}\times \dfrac{x^{8}}{4^{4}}$

$=\dfrac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 5!}\times \dfrac{2^{5}.x^{13}.1}{2^{5}}$

$T_{5}=\dfrac{63}{4}x^{13}$

$T_{6}=T_{5+1}=\ ^{9}C_{5}(2x)^{4}\left(\dfrac{-x^{2}}{4}\right)^{5}$

$=\dfrac{9!}{4!5!}\times 2^{4}x^{4}(-1)^{5}\dfrac{x^{10}}{4^{5}}$

$=\dfrac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 5!}2^{4}\dfrac{x^{14}}{2^{10}}$

$T_{6}=\dfrac{-63}{32}x^{14}$

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

$(1+\alpha x)^{4}$ and of

$(1-\alpha x)^{6}$ is the same if

$\alpha$ equals :

0%
$\frac{3}{5}$
0%
$\frac{10}{3}$
0%
$\frac{-3}{10}$
0%
$\frac{-5}{3}$

Explanation

$\textbf{Step 1: Calculating coefficient of individual term}$

$\text{The middle term of the binomial expansion }(1+\alpha x)^4$

$\text{ will be the }(\dfrac42+1)=\text{ 3rd term}$

$\text{The middle term of the binomial expansion }(1-\alpha x)^6 \text{will be the }(\dfrac62+1)=\text{ 4th term}$

$\text{Coefficient of3rd term of }(1+\alpha x)^4=^4C_2\alpha^2$

$\text{Coefficient of 4th term of }(1-\alpha x)^6=-^6C_3\alpha^3$

$\textbf{Step 2: Calculating }\mathbf{\alpha}$

$\text{Equating both the coefficients}$

$\implies ^4C_2\alpha^2=-^6C_3\alpha^3$

$\implies 6\alpha^2=-20\alpha^3$

$\implies \alpha=-\dfrac6{20}=-\dfrac3{10}$

$\mathbf{\text{Hence, The value of }\alpha =-\dfrac3{10}}$

Find the coefficient of

$x^{7}$ in the expansion of

$\left(x-\dfrac{1}{x^{2}}\right)^{40}$

0%
$-^{40}C_{11}$
0%
$-^{40}C_{10}$
0%
$-^{40}C_{12}$
0%
None of these

Find the middle term in the expansion of

$\left(\dfrac{2}{3}x-\dfrac{3}{2x}\right)^{20}$

0%
$^{20}C_{10}x^{10}y^{10}$
0%
$^{20}C_{11}x^{11}y^{11}$
0%
$^{20}C_{9}x^{11}y^{10}$
0%
None of these

Explanation

Given that,

$\left(\dfrac{2x}{3}-\dfrac{3y}{2}\right)^{20}$

$n=20\Rightarrow$ even

For even values of

$n$ , middle term is

$\left(\dfrac{n}{2}+1\right)th$

$\Rightarrow$ middle term

$=\left(\dfrac{20}{2}+1\right)th=11th$

Now use the formula

$T_{r+1}=\ ^{n}C_{r}(a)^{n-r}(b)^{r}$

Here,

$r=10, a=\dfrac{2x}{3},b=\dfrac{3y}{2}$

$T_{11}=T_{10+1}=\ ^{20}C_{10}\left(\dfrac{2x}{3}\right)^{20-10}\left(\dfrac{3y}{2}\right)^{10}$

$T_{11}=\ ^{20}C_{10}\left(\dfrac{2x}{3}\right)^{10}\left(\dfrac{3y}{x}\right)^{10}$

$T_{11}=\ ^{20}C_{10}x^{10}y^{10}$

Find the middle term(s) in the expansion of :

$\left(\dfrac{p}{x}+\dfrac{x}{p}\right)^{9}$

0%
$\dfrac {126x}{p}$ , $\dfrac {126p}{x}$
0%
$\dfrac {126}{p}$ , $126x$
0%
$\dfrac {160x^2}{p}$ , $\dfrac {162x}{p^2}$
0%
None of these

Explanation

Given that to find middle term in expansion of

$\left(\dfrac{P}{x}+\dfrac{x}{P}\right)^{9}$

$n=9\Rightarrow$ odd

For off values there are two middle terms

First middle term

$=\left(\dfrac{n+1}{2}\right)^{th}$ term

$\Rightarrow \left(\dfrac{9+1}{2}\right)^{th}$ term

$=5^{th}$ term

second middle term

$=\left(\dfrac{n+1}{2}+1\right)^{th}$ term

$=\left(\dfrac{9+1}{2}+1\right)^{th}$ term

$=6^{th}$ term

We know that

$T_{r+1}=\ ^{9}C_{r}\left(\dfrac{P}{x}\right)^{9-r}\left(\dfrac{x}{P}\right)^{r}$

$T_{r+1}=T_{5}\Rightarrow r+1=5\Rightarrow r=4$

$T_{4+1}=\ ^{9}C_{4}\left(\dfrac{P}{x}\right)^{5}\left(\dfrac{x}{P}\right)^{4}$

$=\dfrac{9!}{4!5!}\dfrac{P^{5}}{x^{5}}\Rightarrow 126 \dfrac{P}{x}$

$T_{r+1}=T_{0}\Rightarrow r+1=6\Rightarrow r=5$

$T_{5+1}=\ ^{9}C_{5}\left(\dfrac{P}{x}\right)^{4}\left(\dfrac{x}{P}\right)^{5}$

$=\dfrac{9!}{4!5!}\dfrac{P^{4}}{x^{4}}\dfrac{r^{5}}{P^{5}}$

$=126\dfrac{x}{P}$

Find the middle term(s) in the expansion of :

$\left(3x-\dfrac{x^{3}}{6}\right)^{9}$

0%
$\dfrac {189}{8}x^{15},- \dfrac {21}{16} x^{17}$
0%
$\dfrac {189}{8}x^{17},- \dfrac {21}{16} x^{19}$
0%
$\dfrac {189}{7}x^{15},- \dfrac {23}{13} x^{19}$
0%
None of these

Explanation

Given that to find middle term in expansion of

$\left(3x-\dfrac{x^{3}}{6}\right)^{9}$

$n=9\Rightarrow$ odd

For odd values, there are two middle terms.

First middle term

$=\dfrac{n+1}{2}^{th}$ term :

$\Rightarrow \left(\dfrac{9+1}{2}\right)^{th}$ term

$=5^{th}$ term

Second middle term

$=\left(\dfrac{n+1}{2}+1\right)^{th}$ term;

$\Rightarrow \left(\dfrac{9+1}{2}+1\right)^{th}$ term

$=6^{th}$ term

We know that

$T_{r+1}=\ ^{9}C_{r}(3x)^{r}\left(\dfrac{-x^{3}}{6}\right)^{r}$

$=\ ^{9}C_{r}(3^{r})x^{r}(-1) x^{3r}\left(\dfrac{1}{6}\right)^{r}$

$T_{r+1}=T_{5}\Rightarrow r+1=5\Rightarrow r=4$

$T_{4+1}=\ ^{9}C_{4}(3x)^{9-4}\left(\dfrac{-x^{3}}{6}\right)^{4}$

$T_{5+1}=\ ^{9}C_{5}(3x)^{9-5}\left(\dfrac{-x^{3}}{6}\right)^{5}$

$=\dfrac{9!}{5!4!}3^{5}x^{5}\dfrac{x^{12}}{6^{4}}$

$=\dfrac{9!}{4!5!}3^{4}x^{4}\dfrac{x^{15}}{6^{5}}$

$=126\dfrac{3^{5}x^{12}\times x^{5}}{64}$

$=\dfrac{-21}{16}x^{19}$

$=\dfrac{189}{7}x^{17}$

Find the middle term in the expansion of :

$\left(x^{2}-\dfrac{2}{x}\right)^{10}$

0%
$-8604 \ x^7$
0%
$-8064 \ x^5$
0%
$-804 \ x^4$
0%
None of these

Explanation

Given to find middle term in the expansion of

$\left(x^{2}-\dfrac{2}{x}\right)^{10}$

$n=10\Rightarrow$ even

for even numbers middle term is

$\left(\dfrac{n}{2}+1\right)th$

middle term

$=\left(\dfrac{10}{2}+1\right)th=6^{th}$

We know that

$T_{r+1}=\ ^{n}C_{r}(a)^{n-r}(b)^{r}$

$T_{6}=T_{5+1}=\ ^{10}C_{5}(x^{2})^{5}\left(\dfrac{-2}{x}\right)^{5}$

$=\dfrac{-10!}{5!5!}x^{10}\dfrac{2^{5}}{x^{5}}$

$T_{6}=-8064 x^{5}$

In the expansion of

$(1 + ax + bx^2)(1-3x)^{15}$ , if coefficient of

$x^2$ is

$0$ then order pair

$(a, b)$ is equal to

0%
$(28, 325)$
0%
$(18, 315)$
0%
$(28, 315)$
0%
$(18, 325)$

Explanation

Coefficient of

$x^2$ in

$(1 + ax + bx^2 )(1-3x)^{15}$

$=\ ^{15}C_2(-3)^2+a. \ ^{15}C_1(-3)+ b. \ ^{15}C_0 = 0$

$\dfrac{15.14}{2}. 9 - 3.a.15+b = 0$

$15 \times 63 - 45a + b = 0$ ....(1)
Coefficient of

$x^3$ in

$(1 + ax + bx^2) (1 - 3x)^{15}$

$\ ^{15}C_3(-3)^3 + a . \ ^{15}C_2(-3)^2 + b.\ ^{15}C_1(-3) = 0$

$= \dfrac{15.14.13}{3\times 2} . 3^2 - a.3. \dfrac{15.14}{2} + 15.b = 0$

$7.13.3 - 21a + b = 0$ ...(2)
by using (1) - (2)

$672 - 24a = 0 \Rightarrow a = 28$
Hence

$b = 315$

The number of terms in the expansion of

$\left\{ ( 2x +3y)^9 + ( 2x - 3y)^9 \right\}$ is

0%
10
0%
8
0%
4
0%
5

Explanation

$\textbf{Step 1: Find number of terms of expansion}$

$\text{Given equation is -}$

$\left\{ ( 2x +3y)^9 + ( 2x - 3y)^9 \right\}$

$\text{We know that for expansion of (x + y)$^n$ + (x – y)$^n$ $\\$$\Rightarrow$ If n is odd then, number of terms = $\dfrac{n+1}{2} $}$

$\text{Comparing with standard equation,n=9}$

$\therefore \text{Numer of terms=}$

$\dfrac{9+1}{2}$

$=\dfrac{10}{2}$

$=5$

$\textbf{Therefore,the expansion of}$

$\mathbf{ \left\{ ( 2x +3y)^9 + ( 2x - 3y)^9 \right\}}$

$\textbf{has 5 terms.}$

The number of terms in the expansion of

$\left\{( x+a)^{16} + (x -a)^{16} \right\}$ is

0%
7
0%
8
0%
9
0%
17

Explanation

$\textbf{Step 1: Find number of terms of expansion}$

$\text{Given equation is -}$

$({x} + {16})^{16} +( {x} - {16})^{16}$

$\text{We know that for expansion of (x + y)$^n$ + (x – y)$^n -$}$

$\text{Comparing with standard equation,n=16}$

$\Rightarrow \text{If n is even then, Number of terms = $\dfrac{n}{2} $+ 1}$

$\therefore \text{Numer of terms=}$

$\dfrac{16}{2} +1$

$=8+1$

$=9$

$\textbf{Therefore,the expansion of $\mathbf{ ({x} + {a})^{16} +( {x} - {a})^{16} }$ has 9 terms.}$

The numbers of terms in the expansion of

$( 3x +y)^{10}$ is

0%
9
0%
11
0%
10
0%
8

Explanation

$\textbf{Step 1: Expand the Expression to find number of terms}$

$\text{Given-Equation is } (3x+y)^{10}$

$\text{Binomial Theorem is Given by-}$

$(b+a)^n=b^n+C_{1}^n(b)^{n-1}a+C_{2}^n(b)^{n-2}a^2+C_{3}^n(b)^{n-3}a^3+..............+C_{n}^n(a)^n$

$\text{Comparing with standard formula, n=10, a=3x and b=y}$

$(3x+y)^{10}=(3x)^{10}+C_{1}^{10}(3x)^{10-1}(y)+C_{2}^{10}(3x)^{10-2}(y)^2+C_{3}^{10}(2x)^{10-3}(3y)^3+.....+C_{10}^{10}(y)^{10}$

$\text{From expansion, total number of terms=n+1}$

$\therefore \text{Total number of terms= 10+1}$

$=11$

$\textbf{Therefore, total number of terms are 11}$

$\{ C_{ 0 }+3C_{ 1 }+5C_{ 2 }+...+(2n+1)C_{ n }\} =?$

0%
$(n+1)2^{ n }$
0%
$(n+1)2^{ n-1 }$
0%
$(n-1)(n+2)$
0%
none of these

Explanation

$\textbf{Step 1: Find the value of given expression}$

$\text{Let X=} C_{ 0 }+3C_{ 1 }+5C_{ 2 }+...+(2n+1)C_{ n }$

$\Rightarrow X=(C_{ 0 }+C_{ 1 }+C_{ 2 }+...+C_{ n })+2(C_{ 1 }+2C_{ 2 }+3C_{ 3 }+...+nC_{ n })$

$\text{In binomial expansion,sum of coefficient= } 2^n$

$\therefore (C_{ 0 }+C_{ 1 }+C_{ 2 }+...+C_{ n }) = 2^n$

$(1)$

$\Rightarrow X=(C_{ 0 }+C_{ 1 }+C_{ 2 }+...+C_{ n })+2(C_{ 1 }+2C_{ 2 }+3C_{ 3 }+...+nC_{ n })$

$=2^n +2\times\bigg(n+2\dfrac{n(n-1)}{2}+3\dfrac{n(n-1)(n-2)}{3\times 2}+........+n.1\bigg)$

$\textbf{[From equation 1]}$

$=2^n +n\bigg(1+(n-1)+\dfrac{(n-1)(n-2)}{ 2}+........+1\bigg)$

$=2^n+2n[ ^{n-1}C_{ 0 }+^{n-1}C_{ 1 }+3C_{ 2 }+......+^{n-1}C_{ n-1 }]$

$=2^{ n }+(2n).2^{ n-1 }$

$[\because \mathbf{ C_{ 0 }+C_{ 1 }+C_{ 2 }+...+C_{ n-1 } = 2^{n-1}}]$

$=2^{ n }+(n).2^{ n-1+1 }$

$=2^{ n }+(n).2^{ n }$

$=(n+1).2^{ n }$

$\textbf{Hence, value of given expression is }$

$\mathbf{(n+1).2^{ n } }$

The sum of coefficients of the two middle terms in the expansion of

$(1 + x)^{2n - 1}$ is equal to

0%
$^{2n -1}C_n$
0%
$^{2n -1}C_{n+1}$
0%
$^{2n}C_{n-1}$
0%
$^{2n}C_n$

The cofficient of

$x^{32}$ in the expansion of

$( x^4 - \frac {1}{x^3})^{15}$ is

0%
273
0%
546
0%
1365
0%
1092

Explanation

$\textbf{Step 1: Find value of general term for expansion}$

$\text{Given equation is}$

$\bigg ( x^4 -\dfrac {1}{x^3}\bigg)^{15}$

$\text{General term of expansion $ (p+q)^n$ is given by-}$

$T_n= T_{r +1 }=^nC_r p^{ (n-r)}.q^r$

$\text{Comparing given equation with standard form, p=x$^4$and q=-1/x$^3$ and n=15}$

$\Rightarrow T_n=T_{r+1}$

$= ^{15}C_r (x^4)^{(15-r)} .\bigg (\dfrac {-1}{x^3}\bigg)^r$

$= (-1)^r. ^{15}C_r. x^{(60-4r)} .\bigg (\dfrac {1}{x^3}\bigg)^r$

$=(-1)^r . ^{15}C_r x^{(60 -4r -3r)}$

$=(-1)^r . ^{15}C_r x^{(60 -7r)}$

$\textbf{Step 2: Find value of coefficient of term x}^{32}$

$\text{To obtain term of x}^{32}$

$\Rightarrow 60-7r=32$

$\Rightarrow$

$7r=28$

$\therefore$

$r=4$

$\text{For r=4,5th term is given by-}$

$T_5=T_{4+1} = (-1)^4.^{15}C_4 x^{(60-7 \times 4)} = ^{15}C_4 x^{32} .$

$\therefore$

$\text{Coefficient of } x^{32}$

$=^{15}C_4$

$= \dfrac { 15 \times 14 \times 13 \times 12 }{ 4 \times 3 \times 2\times 1}$

$= 1365$

$\textbf{Therefore,value of coefficient of term x$^{32}$ is 1365}$

The number of terms in the expansion of

$\displaystyle{(\sqrt{3} + ^4 \sqrt{5})^{124}}$ which are integers, is equal to

0%
nil
0%
30
0%
31
0%
32

The coefficient of

$x^3$ in the expansion of

$(1+2x)^6(1-x)^7$ is

0%
43
0%
-43
0%
63
0%
-63

Explanation

$\textbf{Step 1: Expand the Expression }$

$\text{Binomial Theorem is Given by-}$

$(1+p)^n=1+^nC_{1}(p)^1+^nC_{2}(p)^2+^nC_{3}(p)^3+..............+^nC_{n}(p)^n$

$\text{For expansion of} (1+2x)^6:$

$\text{By Comparing with Formula p=2x and n=6}$

$(1+2x)^6 =\{ 1+^{ 6 }C_{ 1 }.2x+^{ 6 }C_{ 2 }.(2x)^{ 2 }+^{ 6 }C_{ 3 }.(2x)^{ 3 }+...\}$

$\text{For expansion of} (1-x)^7 :$

$\text{By Comparing with Formula p=x and n=7}$

$(1-x)^7 = \{ 1-^{ 7 }C_{ 1 }.x+^{ 7 }C_{ 2 }.x^{ 2 }-^{ 7 }C_{ 3 }.x^{ 3 }+..........\}$

$\textbf{Step 2: Find the coefficient of } \mathbf{ x^3}$

$\text{Let P}= (1+2x)^6(1-x)^7$

$\Rightarrow P =\{ 1+^{ 6 }C_{ 1 }.2x+^{ 6 }C_{ 2 }.(2x)^{ 2 }+^{ 6 }C_{ 3 }.(2x)^{ 3 }+...\} \times \{ 1-^{ 7 }C_{ 1 }.x+^{ 7 }C_{ 2 }.x^{ 2 }-^{ 7 }C_{ 3 }.x^{ 3 }+..\}$

$\therefore \text{Coefficient of }x^3 = -^{7}C_3 + ^{7}C_2 \times 2 ^{6}C_1 + 4\times ^{6}C_2 \times (- ^{7}C_1 )+8\times ^{7}C_3$

$=1\times(-35)+(12\times21)+60 \times (-7) + (160 \times 1)$

$= (-35 +252-420+160)$

$=-43$

$\textbf{Therefore, coefficient of } \mathbf{ x^3 } \textbf{is equal to -43}$

The number of terms in the expansion of

$( 1 + \sqrt [3]{2}x)^9 + ( 1 - \sqrt [3]{2}x )^9$ is

0%
5
0%
7
0%
9
0%
10

Explanation

$\textbf{Step 1: Find number of terms of expansion}$

$\text{Given equation is -}$

$( 1 + \sqrt [3]{2x})^9 + ( 1 + \sqrt [3]{2x} )^9$

$\text{We know that for expansion of (x + y)$^n$ + (x – y)$^n$}$

$\text{Comparing with standard equation,n=9}$

$\Rightarrow \text{ If n is even then, Number of terms = }$

$\dfrac{n+1}{2}$

$\therefore \text{Numer of terms=}$

$\dfrac{9+1}{2}$

$=\dfrac{10}{2}$

$=5$

$\textbf{Therefore,the expansion of}$

$\mathbf{ ( 1 + \sqrt [3]{2x})^9 + ( 1 + \sqrt [3]{2x} )^9 }$

$\textbf{has 5 terms.}$

If p and q are positive integers, then the coefficients of

$x^p$ and

$x^q$ in the expansion of

$( 1+ x)^{p+q}$ are

0%
euqal
0%
equal with opposite signs
0%
reciprocal to each other
0%
none of these

Explanation

$\textbf{Step 1: Find value of (p+1)th and (q+1)th term from expansion}$

$\text{General term of expansion $ (1+x)^n$ is given by-}$

$T_n= T_{r +1 }=^nC_r x^{ (n-r)}$

$\text{Comparing given equation with standard form, n=p+q}$

$\Rightarrow$

$T_{p+1} = ^{(p+q)}C_p . x^{p+1-1}$

$= ^{(p+q)}C_p . x^{p}$

$\therefore \text{Coefficient of x$^{p}$ is $ = ^{(p+q)}C_p $}$

$(1)$

$\Rightarrow T_{q+1} = ^{(p+q)}C_q. x^{q+1-1}$

$= ^{(p+q)}C_p . x^{q}$

$\therefore \text{Coefficient of x$^{q}$ is $ = ^{(p+q)}C_q $}$

$(2)$

$\Rightarrow ^{(p+q)}C_p = ^{(p+q)}C_{(p+q)-p}$

$\mathbf{[^{n}C_r = ^{n}C_{n-r}]}$

$=^{(p+q)}C_q$

$\therefore \text{Coefficient of x$^p$ and x$^q$ are same}$

$\textbf{[From equation (1) and (2)]}$

$\textbf{Therefore, coefficient of both terms are equal.}$

The total number of terms in the expansion of

$(a + y)^{100 } + ( a -y)^{100 }$ is

0%
50
0%
51
0%
101
0%
102

Explanation

$\textbf{Step 1: Find number of terms of expansion}$

$\text{Given equation is -}$

$({a} + {y})^{100} +( {a} - {y})^{100}$

$\text{We know that for expansion of (x + y)$^n$ + (x – y)$^n -$ $\\$$\Rightarrow$ If n is even then, Number of terms = $\dfrac{n}{2} $+ 1}$

$\text{Comparing with standard equation,n=100}$

$\therefore \text{Numer of terms=}$

$\dfrac{100}{2} +1$

$=50+1$

$=51$

$\textbf{Therefore,the expansion of $\mathbf{ ({a} + {y})^{100} +( {a} - {y})^{100} }$ has 51 terms.}$

If the coefficients of the second, third and fourth terms in the expansion of

$(1+x)^{2n}$ are in AP , then

0%
$n^2-3n+8=0$
0%
$2n^2 - 7n +5 = 0$
0%
$2n^2-9n+7 =0$
0%
none of these

Explanation

$\textbf{Step 1: Expand the Expression }$

$\text{Binomial Theorem is Given by-}$

$(1+x)^p=1+C_{1}^p(x)^1+C_{2}^p(x)^2+C_{3}^p(x)^3+..............+C_{p}^p(x)^p$

$\text{By Comparing with Formula p=2n}$

$\Rightarrow (1+x)^{-2n}$

$=C_{1}^{2n}(x)^1+C_{2}^{2n}(x)^2+C_{3}^{2n}(x)^3+..............+C_{2n}^{2n}(x)^{2n}$

$\textbf{Step 2: Find value of 2nd, 3rd and 4th term from expansion}$

$\text{General term of expansion$(1+x)^{2n}$ is given by-}$

$T_n= T_{r +1 }=^{2n}C_r x^{ r}$

$\Rightarrow T_2=T_{1+1}=^{2n}C_1x^1$

$\therefore \text{Coefficient of}$

$T_2 =^{2n}C_1x$

$=2n$

$\Rightarrow T_3=T_{2+1}=^{2n}C_2x^2$

$\therefore \text{Coefficient of}$

$T_3$

$=^{2n}C_2x^2$

$=\dfrac { 2n(2n-1) }{ 2 }$

$\Rightarrow T_4=T_{3+1} =^{2n}C_3x^3$

$\therefore \text{Coefficient of}$

$T_4$

$=^{2n}C_2x^3$

$=\dfrac { 2n(2n-1)(2n-2) }{ 6 }$

$\textbf{Step 3: Find condition for three terms to be in AP}$

$\text{Given-Coefficient of $T_2,T_3$ and $T_4$ are in AP }$

$\text{Condition for three terms a,b and c to be in AP- 2b=a+c }$

$\therefore 2 T_3=T_2 + T_4$

$2\bigg(\dfrac { 2n(2n-1) }{ 2 }\bigg) =2n+\dfrac { 2n(2n-1)(2n-2) }{ 6 }$

$\Rightarrow$

$3(2n-1)=3+2n^2-3n+1$

$\Rightarrow$

$6n-3=4+2n^2-3n$

$\therefore$

$2n^2 - 9n+7=0$ .

$\textbf{Therefore,option c is the correct answer}$

The term independent of x in the expansion of

$(x -\frac {1}{x})^{12}$ is

0%
924
0%
462
0%
231
0%
693

Explanation

$\textbf{Step 1: Find value of general term}$

$\text{Given equation is}$

$\bigg ( x -\dfrac {1}{x}\bigg)^{12}$

$\text{General term of expansion $ (p+q)^n$ is given by-}$

$T_n= T_{r +1 }=^nC_r p^{ (n-r)}.q^r$

$\text{Comparing given equation with standard form, p=x and q=-1/x and n=12}$

$\Rightarrow T_n=T_{r+1} = ^{12}C_r. (x)^{(12-r)} .\bigg( \dfrac {-1}{x}\bigg)^r$

$= (-1)^r. ^{12}C_r. (x)^{(12-r)} .x^{-r}$

$= (-1)^r. ^{12}C_r. (x)^{(12-r-r)}$

$= (-1)^r. ^{12}C_r. (x)^{(12-2r)}$

$\textbf{Step 2: Find value of coefficient of term independent of x}$

$\text{To obtain term independent of x-:Coefficient of x will be 0}$

$\Rightarrow 12-2r=0$

$\Rightarrow$

$2r=12$

$\therefore$

$r=6$

$\text{For r=6,7th term is given by-}$

$T_7=T_{6+1} = (-1)^6.^{12}C_6 x^{(12-2x6)} = ^{12}C_6 x^0 .$

$\therefore$

$\text{Coefficient of } x^0$

$=^{12}C_6$

$= \dfrac { 12\times11\times10 \times 9 \times 8 \times 7}{6\times5 \times 4 \times 3 \times 2 \times 1}$

$= 924$

$\textbf{Therefore,value of coefficient of term independent of x will be 924}$

The sum of possible values of

$x$ is

0%
$1$
0%
$3$
0%
$4$
0%
$None\ of \these$

The middle term in the expansion of

$(x/2+2)^{8}$ is

$1120$ ; then

$x\in R$ is equal to

0%
$-2$
0%
$3$
0%
$-3$
0%
$2$

If the coefficients of

$r^{th}$ and

$(r+1)^{th}$ terms in the expansion of

$(3+7x)^{29}$ are equal, then

$r$ equals

0%
$15$
0%
$21$
0%
$14$
0%
none of these

Explanation

Given expansion is

$(3+7x)^{29}$

General term

$T_{r+1}=^{29}C_r.3^{29-r}.(7x)^{r}$

For

$r^{th}$ term

Let put

$r=r-1$

$\implies T_{r-1+1}=^{29}C_{r-1}3^{29-r+1}.(7x)^{r-1}~~~~-(1)$

AS per question coefficient are equal

$\implies ^{29}C_r.3^{29-r}.(7)^{r}=^{29}C_{r-1}.3^{30-r}.(7)^{r-1}$

On solving the above expression we get

$\dfrac{29-r+1}{r}\times7=3$

$\implies 210-7r=3r\\\implies 10r=210\\\implies r=21$

The middle term in the expansion of

$(a x)^n$ is

0%
$56a^3 x^5$
0%
$-56a^3 x^5$
0%
$70a^4 x^4$
0%
$-70a^4 x^4$

CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 15 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 15 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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