MCQ 11 Mark
Value of $\sum^{\infty}_{\text{k}=1}\sum^{\text{k}}_{\text{r}=0}\frac{1}{3^{\text{k}}}\big({^\text{k}}\text{C}_{\text{r}}\big)$ is:
- ✓
$2$
- B
$\frac{2}{3}$
- C
$\frac{1}{3}$
- D
$\text{None of these}$
Answer$\sum\frac{1}{3^{\text{k}}}{^\text{k}}\text{C}_{\text{r}}$
$=\frac{1}{3^{\text{k}}}\sum{^\text{k}}\text{C}_{\text{r}}$
$=\frac{2^{\text{k}}}{3^{\text{k}}}$
This is a $G.P$
Therefore, the sum of the series will be
$\text{S}=\frac{\frac{2}{3}}{1-\frac{20}{3}}=2$
View full question & answer→MCQ 21 Mark
If the $r^{th}$ term in the expansion of $\Big(\frac{\text{x}}{3}-\frac{2}{\text{x}^{2}}\Big)^{10}$ contains $x^4$, then $r$ is equal to:
View full question & answer→MCQ 31 Mark
Sum of the coefficients of $(1 - x)^{25}$ is:
Answer$(1-x)^{25}=1-{ }^{25} C_{1 x}+{ }^{25} C_2 x^2-{ }^{25} C_3 x^3+{ }^{25} C_4 x^4-{ }^{25} C_5 x^5 \ldots-{ }^{25} C_{25} x^{25}$
Putting $x=1$, we get
$0=1-{ }^{25} C_1+{ }^{25} C_2-{ }^{25} C_3+{ }^{25} C_4-{ }^{25} C_5 \ldots-{ }^{25} C_{25}$
Hence, sum of coefficients is $0.$
View full question & answer→MCQ 41 Mark
The number of integral terms in the expansion of $\Big(3^{\frac{1}{8}}+5^{\frac{1}{4}}\Big)^{1024}$ is:
AnswerTotal number of integral term will be
$\frac{1024}{\text{L}.\text{C}.\text{M}(4,8)}+1$
$=\frac{1024}{8}+1$
$=128+1$
$=129.$
Hence there are total $129$ integral terms.
View full question & answer→MCQ 51 Mark
The $4^{th}$ term in the expansion of $\Big(\sqrt{\text{x}}+\frac{1}{\text{x}}\Big)^{12}$ is:
AnswerCorrect option: B. $220\text{x}^{\frac{3}{2}}$
Expansion is $\Big(\sqrt{\text{x}}+\frac{1}{\text{x}}\Big)^{12}$
$\text{T}_{\text{r}+1}=12_{\text{C}\text{r}}\big(\frac{1}{\text{x}}\big)^{\text{r}}\cdot(\sqrt{\text{x}})^{12-\text{r}}=12_{\text{C}\text{r}}\cdot\text{x}^{6-1.5\text{r}}$
$4^{th}$ term is $\text{T}_4=12_{\text{C}3}\cdot\text{x}^{6-1.5\times3}=220\text{x}^{\frac{3}{2}}$
View full question & answer→MCQ 61 Mark
The expansion $\Big(\text{x}-\frac{\text{x}^{2}}{2}\Big)^{40}$ is a polynomial of $n^{th}$ degree in $x,$ then $n =$
Answer$T_{r+1}={ }^{40} C_r x^{40-r} x^{2 r} 2^{-r}$
The power of $x = 40 + r$
Highest power of $x$ occurs when $r = 40($ last term$)$
Hence, highest power of $x$ is $80.$
Hence, the polynomial is of degree $80.$
View full question & answer→MCQ 71 Mark
The approximate value of $(7.995)^{\frac{1}{3}}$ correct to $4$ decimal places is:
- ✓
$1.9995$
- B
$1.9996$
- C
$1.9990$
- D
$1.9991$
AnswerCorrect option: A. $1.9995$
View full question & answer→MCQ 81 Mark
Sum of the coefficients of $(1 + x)^n$ is always $a:$
AnswerTo determine the sum of coefficients, we substitute $x = 1$ in the above expression.
Thus sum of coefficients $ = (1 + 1)^n = 2^n$
Hence, its a positive integer.
View full question & answer→MCQ 91 Mark
The number of terms with integral coefficients in the expansion of $\Big(7^{\frac{1}{3}}+5^{\frac{1}{2}}\cdot\text{x}\Big)^{600}$ is:
AnswerThe number of integral terms will be
$1+\frac{100}{\text{L}.\text{C}.\text{M}(2,3)}$
$=1+\frac{600}{6}$
$=1+100$
$=101$
View full question & answer→MCQ 101 Mark
The total number of terms in the expansion of $(x + a)^{100}+ (x - a)^{100}$ after simplification is:
AnswerIn the above binomial expansion, the terms at the even places will get eliminated, and we would be left with twice the sum of the terms at odd places.
Hence there will be
$\frac{\text{n}}{2}+1$
$=\frac{100}{2}+1$
$=51\text{terms}$
View full question & answer→MCQ 111 Mark
In the expansion of $\Big(\frac{3\sqrt{\text{x}}}{3}-\frac{\sqrt{3}}{\text{x}}\Big)10, x > 0,$ the constant term is:
View full question & answer→MCQ 121 Mark
If in the expansion of $\Big(\text{x}-\frac{1}{3\text{x}^{3}}\Big)^{9},$ the term independent of $x$ is:
- A
$\text{T}_{3}$
- ✓
$\text{T}_{4}$
- C
$\text{T}_{5}$
- D
AnswerCorrect option: B. $\text{T}_{4}$
Suppose $T_{r+1}$ is the term in the given expression that is independent of $x.$
Thus, we have
$\text{T}_{\text{r}+1}={^\text{9}}\text{C}_{\text{r}}\ \text{x}^{9-\text{r}}\Big(\frac{-1}{3\text{x}^{2}}\Big)^{\text{r}}$
$=(-1)^{\text{r}}\ {^\text{9}}\text{C}_{\text{r}}\frac{1}{3^{\text{r}}}\ \text{x}^{9-\text{r}-2\text{r}}$
For this term to be independent of $x,$ we must have
$9-3\text{r}=0$
$\Rightarrow\text{r}=3$
Hence, the required term is the $4^{th}$ term i.e. $\text{T}_{4}$
View full question & answer→MCQ 131 Mark
The number of terms in the expansion of $(a_1 + b_1) (a_2 + b_2) ..... (a_n + b_n)$.
- ✓
$2^n$
- B
$3^n$
- C
$3^{2n}$
- D
$2^{2n}$
AnswerEach bracket in the above expansion contains $2$ elements
Therefore
$2$ brackets will have $2^2 = 4$ elements
$3$ brackets will have $2^3 = 8$ elements
$4$ brackets will have $2^4 = 16$ elements
$:$
$:$
$n$ brackets will have $2^n$ elements
Hence, there will be $2^n$ elements
View full question & answer→MCQ 141 Mark
The number of terms with integral coefficients in the expansion of $\Big(17^{\frac{1}{3}}+35^{\frac{1}{2}}\text{x}\Big)^{600}$ is:
AnswerThe general term $T_{r+1}$ in the given expansion is given by ${^\text{600}}\text{C}_{\text{r}}(17^{\frac{1}{3}})^{600-\text{r}}\Big(35^{\frac{1}{2}}\text{x}\Big)^{\text{r}}$
$={^\text{600}}\text{C}_{\text{r}}(17^{\frac{1}{3}})^{200-\frac{\text{r}}{3}}\times\Big(35^{\frac{\text{r}}{2}}\text{x}\Big)^{\text{r}}$
Now, $T_{r+1}$ is an integer if $\frac{\text{r}}{2}$ and $\frac{\text{r}}{2}$ are integers for all $0\leq\text{r}\leq600$
Thus, we have
$r = 0, 6, 12, ....600$
Since, It is an $A.P$
So, $600 = 0 + (\text{n} - 1)6$
$\Rightarrow \text{n}=101$
Hence, there are $101$ terms with integral coefficients.
View full question & answer→MCQ 151 Mark
If the $(n + 1)$ numbers $a, b, c, d, ... a, b, c, d, ...$ be all different and each of them a prime number, then the number of different factors $($other than $1)$ of $a^m.b.c.d....$ is:
- A
$m - 2^n$
- B
$(m + 1)2^n$
- ✓
$(m + 1)2^n - 1$
- D
AnswerCorrect option: C. $(m + 1)2^n - 1$
No. of elements in $b.c.d... = n$
Choose $a^k$, where $\text{k}\in0, 0, 1, 2, .. m$ at a time $= (m + 1)$ for every $k,$ no. possible factors from $n$ elements
$ = {^nC_0} + {^nC_1} + {^nC_2} +..... +{^nC_n}$
$ = 2^n$
Total factors $=$ No. of possible powers of $a \times$ every $k,$ no. possible factors from $n$ elements
$ = (m + 1)2^n$
If factor $1$ is excluded then
$(m + 1)2n - 1$
View full question & answer→MCQ 161 Mark
$\sum\limits^{16}_{\text{r}+2}{^{16}}\text{C}_\text{r}=$
- A
$ 2^{15}-15 $
- B
$ 2^{16}-16 $
- ✓
$ 2^{16}-17 $
- D
AnswerCorrect option: C. $ 2^{16}-17 $
Consider given the binomial expression,
$\sum\limits^{16}_{\text{r}+2}{^{16}}\text{C}_\text{r}={^{16}}\text{C}_{2}+{^{16}}\text{C}_{3}+{^{16}}\text{C}_{3}+\ .....\ {^{16}}\text{C}_{16}$
$=2^{16}-17$
Hence, this is the answer.
View full question & answer→MCQ 171 Mark
If in the expansion of $(1+\text{x})^{20},$ the coefficients of $r^{th}$ and $(r + 4)$ terms are equal, then $r$ is equal to:
AnswerCoefficients of the rth and $(r + 4)^{th}$ terms in the given expansion are ${^\text{20}}\text{C}_{\text{r}-1}$ and ${^\text{20}}\text{C}_{\text{r}}.$
Here,
${^\text{20}}\text{C}_{\text{r}-1}={^\text{20}}\text{C}_{\text{r}+3}$
$\Rightarrow \text{r}-1+\text{r}+3=20$
$\Rightarrow \text{r}=2$ or $2\text{r}=18$
$\Rightarrow \text{r}=9$
View full question & answer→MCQ 181 Mark
If the number of terms in $\Big(\text{x}+1+\frac{1}{\text{x}}\Big)^\text{n}(\text{n}\in\text{I}^{+})$ is $401,$ then $n$ is greater than.
AnswerIf we substitute $2$ in place of $1$ in the above expression, we get
$\Big(\text{x}+2+\frac{1}{\text{x}}\Big)^\text{n}$
$\Big(\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)\Big)^{2\text{n}}$
Since the total number of terms is $401$
$2n + 1 = 401$
$2n = 400$
$n = 200$
However the actual question is
$\Big(\text{x}+1+\frac{1}{\text{x}}\Big)^\text{n}$
Hence $n$ is greater than $199.$
View full question & answer→MCQ 191 Mark
If $(1-\text{x}^{2})^{\text{n}}=\sum_{\text{r}=0}^{\text{n}}\text{a}_{\text{r}}\text{x}^{\text{r}}(1-\text{x})^{2\text{n}-\text{r}},$ then $a_r$ is equal to:
- A
${ }^n C_r$
- ✓
${ }^n C_r 3^r$
- C
${ }^{2 n} C_r$
- D
AnswerCorrect option: B. ${ }^n C_r 3^r$
$(1-\text{x})^{\text{n}}(1+\text{x})^{\text{n}}=\sum_{\text{r}=0}^{\text{n}}\text{a}_{\text{r}}\text{x}^{\text{r}}(1-\text{x})^{\text{n}}(1-\text{x})^{\text{n}-\text{r}}$
$\Rightarrow(1-\text{x}+2\text{x})^{\text{n}}=\sum_{\text{r}=0}^{\text{n}}\text{a}_{\text{r}}\text{x}^{\text{r}}(1-\text{x})^{\text{n}-\text{r}}$
$\Rightarrow\sum_{\text{r}=0}^{\text{n}}{^\text{n}}\text{C}_{\text{r}}(1-\text{x})^{\text{n}-\text{r}}(2\text{x})^{\text{r}}$
$=\sum_{\text{r}=0}^{\text{n}}\text{a}_{\text{r}}\text{x}^{\text{r}}(1-\text{x})^{\text{n}-\text{r}}$
Comparing general term, we get $\text{a}^{\text{r}}={^\text{n}}\text{C}_{\text{r}}2^{\text{r}}$
View full question & answer→MCQ 201 Mark
Number of irrational terms in the binomial expansion of $\Big(3^{\frac{1}{5}}+7^{\frac{1}{3}}\Big)^{100}$ is:
AnswerAs $n = 100$ hence there are $101$ terms
The genral term for expansion is given as $\text{T}_{\text{r}+1}={^{100}}\text{C}_{\text{r}}\cdot3\frac{100-\text{r}}{5}\cdot7\frac{\text{r}}{3}$
So, we need to find such values of $r$ for which $\frac{100-\text{r}}{5}$ and $\frac{\text{r}}{3}$ is a natural number.
For such value we find value of $r$ i.e $r = 0, 15, 30, 45, 60, 75, 90$
hence there are $7$ values for which it is rational.
So there are $101 - 7 = 94$ irrational terms.
View full question & answer→MCQ 211 Mark
If the sum of the binomial coefficients of the expansion $\Big(2\text{x}+\frac{1}{\text{x}}\Big)^{\text{n}}$ is equal to $256,$ then the term independent of $x$ is:
AnswerCorrect option: A. $1120$
Suppose $(r+1)^{\text {th}}$ term in the given expansion is independent of $x.$
Then, we have
$\text{T}_{\text{r}+1}={^\text{n}}\text{C}_{\text{r}}(2\text{x})^{\text{n}-\text{r}}\Big(\frac{1}{\text{x}}\Big)^{\text{r}}$
$={^\text{n}}\text{C}_{\text{r}}(2)^{\text{n}-\text{r}}\text{x}^{\text{n}-2\text{r}}$
For this term to be independent of $x,$ we must have
$\text{n}-2\text{r}=0$
$\Rightarrow \text{r}=\frac{\text{n}}{2}$
$\therefore$ Required term $={^\text{n}}\text{C}_{\frac{\text{n}}{2}}\ 2^{\text{n}-\frac{\text{n}}{2}}=\frac{\text{n!}}{\big[(\frac{\text{n}}{2})\big]}\ 2^{\frac{\text{n}}{2}}$
We know,
Sum of the given expansion $= 256$
Thus, we have
$2^{\text{n}}.1^{\text{n}}=256$
$\Rightarrow \text{n}=8$
$\therefore$ Required term $=\frac{8!}{(4)!(4)!}2^{4}=1120$
View full question & answer→MCQ 221 Mark
Which of the following is the highest value?
- A
$12^9$
- B
$10^{11}$
- C
$11^{10}$
- ✓
AnswerWe know a prime $x > 6$ can be always written as $6k + 1$ or $6k - 1,$ when $k$ is an integer.
Option $A \rightarrow 889 = 6 \times 148 + 1$
Option $B \rightarrow 997 = 6 \times 166 + 1$
Option $C \rightarrow 899 = 6 \times 133 + 1$
Option $D \rightarrow 1147 = 6 \times 191 + 1$
$\therefore$ All the options are correct.
View full question & answer→MCQ 231 Mark
$[$ AS $1]$ If $\text{A}=\frac{1}{3}\text{B}$ and $\text{B}=\frac{1}{2}\text{C},$ then $\text{A : B : C} = ..$
- ✓
$1 : 3 : 6$
- B
$2 : 3 : 6$
- C
$3 : 2 : 6$
- D
$3 : 1 : 2$
AnswerCorrect option: A. $1 : 3 : 6$
$\text{A}=\frac{\text{B}}{3}....(1)$
$\text{B}=\frac{\text{C}}{2}$
$\Rightarrow\text{C}=2\text{B}.....(2)$
From $(1)$ and $(2),$
$\text{A}:\text{B}:\text{C}=\frac{\text{B}}{3}:\text{B}:2\text{B}$
$=\frac{1}{3}:1:2$
$=1:3:6$
View full question & answer→MCQ 241 Mark
If the co $-$ efficient of $x$ in $\Big(\text{x}^2+\frac{\lambda}{\text{x}}\Big)^5$is $270$, then $ \lambda=$
AnswerNow, the co $-$ efficient of $x$ in $\Big(\text{x}^2+\frac{\lambda}{\text{x}}\Big)^5$is ${^6}\text{C}_{3}\cdot(\lambda)^3$or $10(\lambda)^3.$
According to the problem $10(\lambda)^{3}=270,$ or $(\lambda)=3.$
View full question & answer→MCQ 251 Mark
If the coefficients of $2^{nd}, 3^{rd}$ and $4^{th}$ terms in the expansion of $(1+\text{x})^{\text{n}}, \text{n}\in\text{N}$ are in $A.P.$ then $n =$
AnswerCoefficients of $2^{nd}, 3^{rd}$ and $4^{th}$ terms in the expansion of are ${^\text{n}}\text{C}_{\text{1}},{^\text{n}}\text{C}_{\text{2}}, {^\text{n}}\text{C}_{\text{3}}.$
we have,
$2\times{^\text{n}}\text{C}_{\text{2}}={^\text{n}}\text{C}_{\text{1}}+{^\text{n}}\text{C}_{\text{3}}$
Dividing both sides by $^nC_r$, we get
$2=\frac{{^\text{n}}\text{C}_{\text{1}}}{{^\text{n}}\text{C}_{\text{2}}}+\frac{{^\text{n}}\text{C}_{\text{3}}}{{^\text{n}}\text{C}_{\text{2}}}$
$\Rightarrow 2=\frac{2}{\text{n}-1}+\frac{\text{n}-2}{3}$
$\Rightarrow 6\text{n}-6=6+\text{n}^{2}+2-3\text{n}$
$\Rightarrow \text{n}^{2}-9\text{n}+14=0$
$\Rightarrow \text{n}=7$
View full question & answer→MCQ 261 Mark
If the sum of odd numbered terms and the sum of even numbered terms in the expansion of $(\text{x}+\text{a})^{\text{n}}$ are $A$ and $B$ respectively, then the value of $(\text{x}^{2}-\text{a}^{2})^{\text{n}}$ is:
- ✓
$A^2- B^2$
- B
$A^2+ B^2$
- C
$4AB$
- D
AnswerCorrect option: A. $A^2- B^2$
If $A$ and $B$ denote respectively the sums of odd terms and even terms in the expansion $(\text{x}+\text{a})^{\text{n}}.$
Then,
$(\text{x}+\text{a})^{\text{n}}=\text{A}+\text{B}\ ...(\text{i})$
$(\text{x}-\text{a})^{\text{n}}=\text{A}-\text{B}\ ...(\text{ii})$
Multplying both the equations we get,
$(\text{x}+\text{a})^{\text{n}}(\text{x}-\text{a})^{\text{n}}=\text{A}^{2}-\text{B}^{2}$
$\Rightarrow (\text{x}-\text{a})^{\text{n}}=\text{A}^{2}-\text{B}^{2}$
View full question & answer→MCQ 271 Mark
If $n$ is the positive integer, then $2^{3n}- 7n - 1$ is divisible by.
AnswerGiven: $2^{3 n}-7 n-1$.
It can also be written as $8^n-7 n-1$
Let $8^n-7 n-1=0$
So, $8^n=7 n+1$
$8^n=(1+7)^n$
By applying binomial theorem, we get
$8 n-1-7 n=49 \text { (or) } 2^{3 n}-7 n-1=49$
Hence, $2^{3 n}-7 n-1$ is divisible by $49.$
View full question & answer→MCQ 281 Mark
If $(1 + ax)^n = 1 + 8x + 24x^2 + ....$ then $a \times n$ is:
Answer$(1+a x)^n=1+8 x+24 x^2+\ldots \ldots \ldots $
$\Rightarrow{ }^n C_0+{ }^n C_1(a x)+{ }^n C_2 \cdot(a x)^2+\ldots \ldots=1+8 x+24 a x^2 \ldots \ldots \ldots $
$ \Rightarrow 1+(n a) x+{ }^n C_2 \cdot(a x)^2+\ldots \ldots=1+8 x+24 a x^2 \ldots \ldots \ldots$
Comparing coefficient of $x$ in $\text{R.H.S}$ to that in $\text{L.H.S}$.
Thus $n \times a = 8$
View full question & answer→MCQ 291 Mark
If in the expansion of $(\text{a}+\text{b})^{\text{n}}$ and $(\text{a}+\text{b})^{\text{n}}+3,$ the ratio of the coefficients of coefficients of second and third terms, and third and fourth terms respectively are equal, then $n$ is:
AnswerCoefficients of the $2^{nd}$ and $3^{rd}$ terms in $\Big(\text{a}+\text{b}\Big)^{\text{n}}$ are ${^\text{n}}\text{C}_{\text{}1}$ and ${^\text{n}}\text{C}_{\text{}2}$
Coefficients of the $2^{nd}$ and $3^{rd}$ terms in $\Big(\text{a}+\text{b}\Big)^{\text{n}+3}$ are ${^\text{n+3}}\text{C}_{\text{}2}$ and ${^\text{n+3}}\text{C}_{\text{}3}$
Thus, we have
$\frac{{^\text{n}}\text{C}_{\text{}1}}{{^\text{n}}\text{C}_{\text{}2}}=\frac{{^\text{n+3}}\text{C}_{\text{}1}}{{^\text{n+3}}\text{C}_{\text{}3}}$
$\Rightarrow \frac{2}{\text{n}-1}=\frac{3}{\text{n}+1}$
$\Rightarrow 2\text{n}+2=3\text{n}-3$
$\Rightarrow \text{n}=5$
View full question & answer→MCQ 301 Mark
If in the binomial expansion of $(1 + x)n$ where $n$ is a natural number, the coefficients of the $5^{th,} 6^{th}$ and $7^{th}$ terms are in $A.P.,$ then $n$ is equal to:
- A
$7$ or $13$
- ✓
$7$ or $14$
- C
$7$ or $15$
- D
$7$ or $17$
AnswerCorrect option: B. $7$ or $14$
View full question & answer→MCQ 311 Mark
The coefficient of $x^5$ in the expansion of $(1+\text{x})^{21}+(1+\text{x})^{22}+...+(1+\text{x})^{30}$
- A
${^\text{51}}\text{C}_{\text{5}}$
- B
${^\text{9}}\text{C}_{\text{5}}$
- ✓
${^\text{31}}\text{C}_{\text{6}}-{^\text{21}}\text{C}_{\text{6}}$
- D
${^\text{30}}\text{C}_{\text{5}}+{^\text{20}}\text{C}_{\text{5}}$
AnswerCorrect option: C. ${^\text{31}}\text{C}_{\text{6}}-{^\text{21}}\text{C}_{\text{6}}$
we have,
$(1+\text{x})^{21}+(1+\text{x})^{22}+...+(1+\text{x})^{30}$
$=(1+\text{x})^{21}\Big[\frac{(1+\text{x})^{10}-1}{(1+\text{x})+1}\Big]$
$=\frac{1}{\text{x}}\Big[(1+\text{x})^{31}-(1+\text{x})^{21}\Big]$
Coefficient of $x^5$ in the given expansion = Coefficient of $x^5$ in $=\frac{1}{\text{x}}\Big[(1+\text{x})^{31}-(1+\text{x})^{21}\Big]$
$=$ Coefficient of $x^6$ in $\Big[(1+\text{x})^{31}-(1+\text{x})^{21}\Big]$
$={^\text{31}}\text{C}_{\text{6}}-{^\text{21}}\text{C}_{\text{6}}$
View full question & answer→MCQ 321 Mark
The term without $x$ in the expansion of $(2\text{x}-\frac{1}{2\text{x}^{2}}\Big)^{12}$ is:
- A
$495$
- B
$-495$
- C
$-7920$
- ✓
$7920$
AnswerCorrect option: D. $7920$
Suppose the $(r + 1)^{th}$ term in the given expansion is independent of $x.$
Then, we have,
$\text{T}_{\text{r}+1}={^\text{12}}\text{C}_{\text{r}}(2\text{x})^{12-\text{r}}\Big(\frac{-1}{2\text{x}^{2}}\Big)^{\text{r}}$
$=(-1)^{\text{r}}\ {^\text{12}}\text{C}_{\text{r}}\ 2^{12-2\text{r}}\ \text{x}^{12-\text{r}-2\text{r}}$
For this term to be independent of $x,$ we must have:
$=12-3\text{r}=0$
$\Rightarrow\text{r}=4 $
$\therefore$ Required term,
$(-1)^{4}\ {^\text{12}}\text{C}_{\text{4}}\ 2^{12-8}$
$=\frac{12\times11\times10\times9}{4\times3\times2}\times16$
$=7920$
View full question & answer→MCQ 331 Mark
The coefficient of $x^{-3}$ in the expansion of $\Big(\text{x}-\frac{\text{m}}{\text{x}}\Big)^{11}$ is:
- A
$-924m^7$
- B
$-792m^5$
- C
$-792m^6$
- ✓
$-330m^7$
AnswerCorrect option: D. $-330m^7$
Let $x^{-3}$ occur at $(r + 1)^{th}$ term in the given expansion.
Then, we have
$\text{T}_{r+1}={^\text{11}}\text{C}_{\text{r}}\ \text{x}^{11-\text{r}}\ \Big(\frac{-\text{m}}{\text{x}}\Big)^{\text{r}}$
$=(-1)^{\text{r}}\times{^\text{11}}\text{C}_{\text{r}}\ \text{m}^{\text{r}}\ \text{x}^{11-\text{r}-\text{r}}$
For this term to contain $x^{-3}$, we must have
$=11-2\text{r}=-3$
$\Rightarrow \text{r}=7$
Required coefficient $=(-1)^{7}\ {^\text{11}}\text{C}_{\text{7}}\ \text{m}^{7}$
$=-\frac{11\times10\times9\times8}{4\times3\times2}\ \text{m}^{7}$
$=-330\text{m}^{7}$
View full question & answer→MCQ 341 Mark
How many terms are there in the expansion of $(1 + 2x + x^2)^{10}$?
AnswerNow, $(1 + 2x + x^2)^{10} = ((1 + x)^2)^{10}= (1 + x)^{20}$
Now, the number of terms in the expansion of $(1 + x)^n$ are $n + 1$.
Thus, the number of terms in the expansion of $(1 + x)^{20}$ will be $20 + 1 =21$.
View full question & answer→MCQ 351 Mark
The sum of the coefficients of all the even powers of $x$ in the expansion of $(2x^2- 3x + 1)^{11}$ is:
- A
$2.6^{10}$
- ✓
$3.6^{10}$
- C
$6^{11}$
- D
AnswerCorrect option: B. $3.6^{10}$
Given equation is $\left(2 x^2-3 x+1\right)^{11}$
$ =(2 x-1)^{11}(x-1)^{11} $
$ =(3)^{11} \cdot(2)^{11-1} $
$ =3^{11} \cdot 2^{10} $
$ =6^{10} \cdot 3$
View full question & answer→MCQ 361 Mark
The $4^{th}$ term from the end in the expansion of $\Big(\frac{\text{x}^3}{2}-\frac{2}{\text{x}^{2}}\Big)^7$ is:
- A
$35x$
- B
$70x^2$
- C
$35x^2$
- ✓
$70x$
AnswerFor the above question
$T_{r+1}= {^7C_rx^{21-5r}2^{2r-7}}$
For the fourth term, from the end $r = 4$
$T_{5+1} ={^7C_4x^12}$
$= (35)(2)x$
$= 70x$
View full question & answer→MCQ 371 Mark
The coefficient of $x^4$ in the expansion of $(1 - 2x)^5$ is equal to:
AnswerGeneral term of $(1-2 x)^5$ is given by
$ T_{r+1}={ }^5 C_r(-2 x)^r $
$ ={ }^5 C_r(-2) x^r$
For coefficient of $x^4$,
power of $x=4$
$\therefore r=4$
$\therefore$ Coefficient $pf\ \mathrm{x}^4={ }^5 \mathrm{C}_4(-2)^4$
$=5 \times 16$
$=80$
View full question & answer→MCQ 381 Mark
The number of non $-$ zero terms in the expansion of $\big(1+3\sqrt{2}\text{x}\big)^{9}+\big(1-3\sqrt{2}\text{x}\big)^{9}$ is:
AnswerIn the expansion of $\big(1+3\sqrt{2}\text{x}\big)^{9}+\big(1-3\sqrt{2}\text{x}\big)^{9} 2^{nd}, 4^{th}, 6^{th}, 8^{th}$ and $10^{th}$ terms get cancelled.
$\therefore$ Number of non $-$ zero terms in $2\Big[{^9}\text{C}_{0}+{^9}\text{C}_{2}(3\sqrt{2}\text{x})^{2}+\ ...\ +{^9}\text{C}_{8}(3\sqrt{2}\text{x})^{8}\Big]$ is $5.$
View full question & answer→MCQ 391 Mark
If $n$ is even in the expansion of $(a + b)n,$ the middle term is:
- A
$\text{n}^\text{th }\text{term}$
- B
$\big(\frac{\text{n}}{2}\big)^\text{th }\text{term}$
- C
$\big[\big(\frac{\text{n}}{2})-1\big]^\text{th }\text{term}$
- ✓
$\big[\big(\frac{\text{n}}{2})+1\big]^\text{th }\text{term}$
AnswerCorrect option: D. $\big[\big(\frac{\text{n}}{2})+1\big]^\text{th }\text{term}$
In general, if $“n”$ is the even in the expansion of $(a + b)^n$, then the number of terms will be odd. $(i.e) n + 1.$
Hence, the middle term of the expansion $(a + b)^n$ is $\big[\big(\frac{\text{n}}{2})+1\big]^\text{th }\text{term}.$
View full question & answer→MCQ 401 Mark
The sum of the co $-$ efficients of all odd degree terms in the expansion of$(\text{x}+\sqrt{\text{x}^3-1})^5+(\text{x}+\sqrt{\text{x}^3-1})^5(\text{x}>1)$ is:
AnswerSum of the coefficient of odd term is given by
$ =2\left[{ }^5 \mathrm{C}_0+{ }^5 \mathrm{C}_2+{ }^5 \mathrm{C}_4\right]$
$ =2[1+10-10+5-10+5] $
$ =2(1+5+5-10)$
$=2$
View full question & answer→MCQ 411 Mark
Constant term in the expansion of $\Big(\text{x}-\frac{1}{\text{x}}\Big)^{10}$ is:
- A
$152$
- B
$-152$
- ✓
$-252$
- D
$252$
AnswerCorrect option: C. $-252$
Suppose $(r + 1)^{th}$ term is the constant term in the given expansion.
Then, we have
$\text{T}_{\text{r}+1}={^\text{10}}\text{C}_{\text{r}}\ \text{x}^{10-\text{r}}\ \Big(\frac{\text{1}}{\text{x}}\Big)^{\text{r}}$
$={^\text{10}}\text{C}_{\text{r}} (-1)^{\text{r}}\ \text{x}^{10-\text{r}-\text{r}}$
For this term to be constant, we must have
$10-2\text{r}=0$
$\Rightarrow \text{r}=5$
Required term $={^\text{10}}\text{C}_{\text{5}}=-252$
View full question & answer→MCQ 421 Mark
${ }^{(2 n+1)} C_0-{ }^{(2 n+1)} C_1+{ }^{(2 n+1)} C_2-\ldots{ }^{2 n+1} C_{2 n}=$
AnswerIn some questions, substituting $n =$ a positive number in both the question and the answer is the fastest way to achieve the correct option.
Although there is always alternate options like writing the general term and splitting it to a format that can be solved but that takes long in a limited time paper.
Try to put $n = 1.$ In the end only option $A$ remains.
View full question & answer→MCQ 431 Mark
If $\frac{\text{T}_{2}}{\text{T}_{3}}$ in the expansion of $(\text{a}+\text{b})^{\text{n}}$ and $\frac{\text{T}_{3}}{\text{T}_{4}}$ in the expansion of $(\text{a}+\text{b})^{\text{n}+3}$ are equal, then $n =$
AnswerIn the expansion $(\text{a}+\text{b})^{\text{n}},$ we have
$\frac{\text{T}_{2}}{\text{T}_{3}}=\frac{{^\text{n}}\text{C}_{\text{1}}\text{a}^{\text{n}-1}\times\text{b}^{1}}{{^\text{n}}\text{C}_{\text{2}}\text{a}^{\text{n}-2}\times\text{b}^{2}}$
In the expansion $(\text{a}+\text{b})^{\text{n}+3},$ we have
$\frac{\text{T}_{3}}{\text{T}_{4}}=\frac{{^\text{n+3}}\text{C}_{\text{2}}\text{a}^{\text{n}+1}\times\text{b}^{2}}{{^\text{n+3}}\text{C}_{\text{3}}\text{a}^{\text{n}}\times\text{b}^{3}}$
Thus, we have
$\frac{\text{T}_{2}}{\text{T}_{3}}=\frac{\text{T}_{3}}{\text{T}_{4}}$
$\Rightarrow \frac{{^\text{n}}\text{C}_{\text{1}}\ \text{a}}{{^\text{n}}\text{C}_{\text{2}}\ \text{b}}=\frac{{^\text{n+3}}\text{C}_{\text{2}}\ \text{a}}{{^\text{n+3}}\text{C}_{\text{3}}\ \text{b}}$
$\Rightarrow \frac{2}{\text{n}-1}=\frac{3}{\text{n}+1}$
$\Rightarrow 2\text{n}+2=3\text{n}-3$
$\Rightarrow \text{n}=5$
View full question & answer→MCQ 441 Mark
The sum of the coefficients of the middle terms of $(1 + x)^{2n-1}$ is:
- A
$ { }^{2 n-1} C_n $
- B
$ { }^{2 n-1} C_{n+1} $
- C
$ { }^{2 n} C_{n-1} $
- ✓
$ { }^{2 n} C_n $
AnswerCorrect option: D. $ { }^{2 n} C_n $
Consider: $(1 + x)^{2n-1}$
Since $2n - 1$ is odd, the middle terms are $\Big(\frac{2\text{n}-1+1}{2}\Big)^{\text{th}}$ and $\Big(\frac{2\text{n}-1+1}{2}+1\Big)^{\text{th}}$ terms
Now, consider the following
$\operatorname{Tr}+1={ }^n C_r a^{n-r} b^r...(i)$
Where $T$ represents the term
Here the middle terms are the $n^{th}$ term and $(n+1)^{th}$ term.
$T_{n+1}={ }^{2 n-1} C_n 1^{n-1} x^n={ }^{2 n-1} C_n x^n$
Hence, the coefficient is $^{2 n-1} C_n ...(1)$
$T_n={ }^{2 n-1} C_{n-1} 1^n x^{n-1}={ }^{2 n-1} C_{n-1} x^{n-1}$
Hence, ${ }^{2 n-1} C_{n-1}$ is the coefficient $...(2)$
Therefore, sum of the coefficients of the middle terms is
$ { }^{2 n-1} C_n+{ }^{2 n-1} C_{n-1} $
$ ={ }^{2 n} C_n $
View full question & answer→MCQ 451 Mark
Total number of rational terms in the expansion of $(\sqrt[4]{3}+\sqrt[3]{4})^{136}$ is:
AnswerThe number of rational terms is equal to
$\frac{136}{\text{L}.\text{C}.\text{M}(4,3)}+1$
$=\frac{136}{12}+1$
$= 11.33 + 1$ but we consider the integral value of $\frac{136}{12}.$
$=11 + 1$
Hence, there are $12$ rational terms.
View full question & answer→MCQ 461 Mark
The coefficient of the $8^{th}$ term in the expansion of $(1 + x)^{10}$ is
- ✓
$120$
- B
$7$
- C
$^{10}C_8$
- D
$210$
Answer$(1+x)^{10}={ }^{10} \mathrm{C}_0+{ }^{10} \mathrm{C}_1 \mathrm{x}+{ }^{10} \mathrm{C}_2 \mathrm{x}^2+\ldots \ldots . .+{ }^{10} \mathrm{C}_7 \mathrm{x}^7+{ }^{10} \mathrm{C}_8 \mathrm{x}^8+{ }^{10} \mathrm{C}_9 \mathrm{x}^9+{ }^{10} \mathrm{C}_{10} \mathrm{x}^{10}$
So here, first term is ${ }^{10} \mathrm{C}_0$ then $8^{\text {th }}$ term will be ${ }^{10} \mathrm{C}_7{x}^7$.
$\Rightarrow$ Coefficient of the $8^{\text {th }}$ term $={ }^{10} \mathrm{C}_7$
$=\frac{10!}{7!3!}$
$=\frac{10\times9\times8\times7!}{7!\times3\times2\times1}$
$=120$
View full question & answer→MCQ 471 Mark
If $x^4$ occurs in the $r^{th}$ term in the expansion of $\big(\text{x}^4+\frac{1}{\text{x}^3}\big)15,$ then what is the value of $r?$
View full question & answer→MCQ 481 Mark
The number of terms with integral coefficient in the expansion of $\Big(17^{\frac{1}{3}}+32^{\frac{1}{2}}\Big)^{300}$ is:
AnswerThe number of rational terms will be
$1+\frac{300}{\text{L}.\text{C}.\text{M}(3,2)}$
$=1+\frac{300}{6}$
$=1+50$
$=51$ rational terms.
View full question & answer→MCQ 491 Mark
${ }^{(n+1)} C_1+{ }^{(n+1)} C_2+{ }^{(n+1)} C_3+\ldots+{ }^{(n+1)} C_n=$
- A
$2(2^n+1)$
- ✓
$2(2^n−1)$
- C
$2^{n+1}$
- D
$(2^{n+1}- 1)$
AnswerCorrect option: B. $2(2^n−1)$
As in the hint required expression $+{ }^{n+1} C_0+{ }^{n+1} C_{n+1}=2^{n+1}$
$\Rightarrow$ required. expression $=2^{n+1}-2=2\left(2^n-1\right)$
View full question & answer→MCQ 501 Mark
Using binomial theorem, the value of $(0.999)^3$ correct to $3$ decimal places is:
- A
$0.999$
- B
$0.998$
- ✓
$0.997$
- D
AnswerCorrect option: C. $0.997$
$ (0.999)^3=(1-0.001)^3 $
$ ={ }^3 C_0-{ }^3 C_1(0.001)+{ }^3 C_2(0.001)^2-{ }^3 C_3(0.001)^3 $
$ =1-0.003+3(0.000001)-(0.000000001) $
$ =0.997 $
View full question & answer→