MCQ 11 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 21 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 31 Mark
If in the expansion of $(1+\text{x})^{20},$ the coefficients of rth 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 41 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 $(r + 1)^{th}$ 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 51 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 $^nC_2,$ 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 61 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 71 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 2nd and 3rd 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 2nd and 3rd 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 81 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 91 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 101 Mark
The coefficient of $x^{-3}$ in the expansion of $\Big(\text{x}-\frac{\text{m}}{\text{x}}\Big)^{11}$ is:
- A
$-924\ m^7$
- B
$-792\ m^5$
- C
$-792\ m^6$
- ✓
$-330\ m^7$
AnswerCorrect option: D. $-330\ m^7$
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 111 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 121 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 131 Mark
In the expansion of $\Big(\text{x}^{2}-\frac{1}{3\text{x}}\Big)^{9},$ the term without $x$ is equal to:
- A
$\frac{28}{81}$
- B
$\frac{-28}{243}$
- ✓
$\frac{28}{243}$
- D
AnswerCorrect option: C. $\frac{28}{243}$
Suppose the $(r + 1)^{th}$ term in the given expansion is independent of $x.$
Then, we have
$\text{T}_{\text{r}+1}={^\text{9}}\text{C}_{\text{r}}(\text{x}^{2})^{9-\text{r}}\Big(\frac{-1}{3\text{x}}\Big)^{\text{r}}$
$=(-1)^{\text{r}}\ {^\text{9}}\text{C}_{\text{r}}\frac{1}{3\text{r}}\ \text{x}^{18-2\text{r}-\text{r}}$
For this term to be independent of $x,$ we must have
$18-3\text{r}=0$
$\Rightarrow \text{r}=6$
$\therefore$ Required term $=(-1)^{6}\ {^\text{9}}\text{C}_{\text{6}}\ \frac{1}{3^{6}}$
$=\frac{9\times8\times7}{3\times2}\times\frac{1}{3^{6}}$
$=\frac{28}{243}$
View full question & answer→MCQ 141 Mark
The coefficient of $x^4$ in $\Big(\frac{\text{x}}{2}-\frac{3}{\text{x}^{2}}\Big)$ is:
- ✓
$\frac{405}{256}$
- B
$\frac{504}{259}$
- C
$\frac{450}{263}$
- D
AnswerCorrect option: A. $\frac{405}{256}$
Suppose $x^4$ occurs at the $(r + 1)^{th}$ term in the given expansion.
Then, we have
$\text{T}_{\text{r}+1}={^\text{10}}\text{C}_{\text{r}}\Big(\frac{\text{x}}{2}\Big)^{10-\text{r}}\Big(\frac{-3}{2\text{x}^{2}}\Big)$
$=(-1)^{\text{r}}\ {^\text{10}}\text{C}_{\text{r}}\frac{3^{\text{r}}}{2^{10-\text{r}}}\text{x}^{10-\text{r}-2\text{r}}$
For this term to contain $x^4,$ we must have
$10-3\text{r}=4$
$\Rightarrow \text{r}=2$
$\therefore$ Required coefficient $={^\text{10}}\text{C}_{\text{2}}\frac{3^{2}}{2^{8}}$
$=\frac{10\times9\times9}{2\times2^{8}}$
$=\frac{405}{256}$
View full question & answer→MCQ 151 Mark
If A and B are the sums of odd and even terms respectively in the expansion of $(\text{x}+\text{a})^{\text{n}},$ then $(\text{x}+\text{a})^{\text{2n}}-(\text{x}-\text{a})^{2\text{n}}$ is equal to:
AnswerIf A and B denote respectively the sums of odd terms and even terms in the expansion $(\text{x}+\text{a})^{\text{n}}$
Then, $(\text{x}+\text{a})^{\text{n}}=\text{A}+\text{B}\ ...(\text{i})$
$(\text{x}-\text{a})^{\text{n}}=\text{A}-\text{B}\ ...(\text{ii})$
Squaring and subtraction equation (ii) from(i) we get,
$ (\text{x}+\text{a})^{2\text{n}}-(\text{x}-\text{a})^{2\text{n}}=(\text{A}+\text{B})^{2}-(\text{A}-\text{B})^{2}$
$\Rightarrow (\text{x}+\text{a})^{2\text{n}}-(\text{x}-\text{a})^{2\text{n}}=4\text{AB}$
View full question & answer→MCQ 161 Mark
If the fifth term of the expansion $\Big(\text{a}^{\frac{2}{3}}+\text{a}^{-1}\Big)^{\text{n}}$ does not contain 'a'. Then n is equal to:
Answer$\text{T}_{5}=\text{T}_{4+1}$
$={^\text{n}}\text{C}_{\text{4}}\big(\text{a}^{\frac{2}{3}}\big)^{\text{n-4}}(\text{a}^{-1})^{4}$
$={^\text{n}}\text{C}_{\text{4}}\ \text{a}^{\big(\frac{2\text{n}-8}{3}-4\big)}$
For this term to be independent of a, we must have
$\frac{2\text{n}-8}{3}-4=0$
$\Rightarrow 2\text{n}-20=0$
$\Rightarrow \text{n}=10$
View full question & answer→MCQ 171 Mark
The total number of terms in the expansion of $(\text{x}+\text{a})^{100}+(\text{x}-\text{a})^{100}$ after simplification is:
AnswerHere, n i.e. 100 is even.
$\therefore$ Total number of terms in the expansion $=\frac{\text{n}}{2}+1=\frac{100}{2}+1=51$
View full question & answer→MCQ 181 Mark
The middle term in the expansion of $\Big(\frac{2\text{x}^{2}}{3}+\frac{3}{2\text{x}^{2}}\Big)^{10}$ is:
AnswerHence, $n,$ i.e., $10,$ is an even number.
$\therefore$ Middle term $=\Big(\frac{10}{2}+1\Big)^{\text{th}}$ term $= 6^{th}$ term
Thus, we have
$\text{T}_{6}=\text{T}_{5+1}$
$={^\text{10}}\text{C}_{\text{5}}\Big(\frac{2\text{x}^{2}}{3}\Big)^{10-5}\Big(\frac{3}{2\text{x}^{2}}\Big)^{5}$
$=\frac{10\times9\times8\times7\times6}{5\times4\times3\times2}\times\frac{2^{5}}{3^{5}}\times\frac{3^{5}}{2^{5}}$
$=252$
View full question & answer→MCQ 191 Mark
If $r^{th}$ term is the middle term in the expansion of $\Big(\text{x}^{2}-\frac{1}{2\text{x}}\Big)^{20},$ then $(r + 3)^{th}$ term is:
- A
${^\text{20}}\text{C}_{\text{14}}\ \Big(\frac{\text{x}}{2^{14}}\Big)$
- B
${^\text{20}}\text{C}_{\text{12}}\ \text{x}^{2}\ 2^{-12}$
- ✓
$-{^\text{20}}\text{C}_{\text{7}}\ \text{x}\ 2^{-13}$
- D
AnswerCorrect option: C. $-{^\text{20}}\text{C}_{\text{7}}\ \text{x}\ 2^{-13}$
Here, $n$ is even,
So, The middle term in the given expansion is $\Big(\frac{20}{2}+1\Big)^{\text{th}}=11^{\text{th}}$
Therefore, $(r + 3)^{th}$ term is the $14^{th}$ term
$\text{T}_{14}={^\text{20}}\text{C}_{\text{13}}(\text{x}^{2})^{20-13}\ \Big(\frac{-1}{2\text{x}}\Big)$
$=(-1)^{13}\ {^\text{20}}\text{C}_{\text{13}}\ \frac{\text{x}^{14-3}}{2^{13}}$
$=-{^\text{20}}\text{C}_{\text{7}}\ \text{x}\ 2^{-13}$
View full question & answer→MCQ 201 Mark
If in the expansion of $\Big(\text{x}^{4}-\frac{1}{\text{x}^{3}}\Big)^{15},\text{x}^{-17}$ occurs in $r^{th}$ term, then
- A
$r = 10$
- B
$r = 11$
- ✓
$r = 12$
- D
$r = 13$
AnswerCorrect option: C. $r = 12$
Here,
$\text{T}^{\text{r}}={^\text{15}}\text{C}_{\text{r}-1}(\text{x}^{4})^{15-\text{r}+1}\Big(\frac{-1}{\text{x}^{3}}\Big)^{\text{r}-1}$
$=(-1)^{\text{r}}\times{^\text{15}}\text{C}_{\text{r}-1}\ \text{x}^{64-4\text{r}-3\text{r}+3}$
For this term to contain $x^{-17},$ we must have
$67-7\text{r}=-17$
$\Rightarrow \text{r}=12 $
View full question & answer→MCQ 211 Mark
The coefficient of $x^8 y^{10}$ in the expansion of $\text{(x + y)}^{18}$ is:
AnswerCorrect option: A. ${^\text{18}}\text{C}_{\text{8}}$
Suppose $(r + 1)^{th}$ term in the given expansion is independent of $x.$
Then, we have
$\text{T}_{\text{r}+1}={^\text{18}}\text{C}_{\text{r}}(\text{x})^{18-\text{r}}\ \text{y}^{\text{r}}$
For this term to be independent of $x,$ we must have
$\text{r}=10$
Hence, the required coefficient is ${^\text{18}}\text{C}_{\text{10}}$ or ${^\text{18}}\text{C}_{\text{8}}$
View full question & answer→MCQ 221 Mark
The middle term in the expansion of $\Big(\frac{2\text{x}}{3}=\frac{3}{2\text{x}^{2}}\Big)^{2\text{n}}$ is:
- A
${^\text{2n}}\text{C}_{\text{n}}$
- ✓
$(-1)\ {^\text{2n}}\text{C}_{\text{n}}\ \text{x}^{-\text{n}}$
- C
${^\text{2n}}\text{C}_{\text{n}}\ \text{x}^{-\text{n}}$
- D
AnswerCorrect option: B. $(-1)\ {^\text{2n}}\text{C}_{\text{n}}\ \text{x}^{-\text{n}}$
Here, n is even,
Middle term in the given expansion $=\Big(\frac{2\text{n}}{2}+1\Big)^{\text{th}}=(\text{n}+1)$
$={^\text{2n}}\text{C}_{\text{n}}\Big(\frac{2\text{x}}{3}\Big)^{2\text{n}-\text{n}}\ \Big(\frac{-3}{2\text{x}^{2}}\Big)^{\text{n}}$
$(-1)\ {^\text{2n}}\text{C}_{\text{n}}\ \text{x}^{-\text{n}}$
View full question & answer→MCQ 231 Mark
The coefficient of the term independent of $x$ in the expansion of $\Big(\text{ax}+\frac{\text{b}}{\text{x}}\Big)^{14}$ is:
- A
$14!\ \text{a}^{7}\ \text{b}^{7}$
- B
$\frac{14!}{7!}\ \text{a}^{7}\ \text{b}^{7}$
- ✓
$\frac{14!}{(7!)^{2}}\ \text{a}^{7}\ \text{b}^{7}$
- D
$\frac{14!}{(7!)^{3}}\ \text{a}^{7}\ \text{b}^{7}$
AnswerCorrect option: C. $\frac{14!}{(7!)^{2}}\ \text{a}^{7}\ \text{b}^{7}$
Suppose $(r + 1)^{th}$ term in the given expansion is independent of $x.$
Then, we have
$\text{T}_{\text{r}+1}={^\text{14}}\text{C}_{\text{r}}(\text{ax})^{14-\text{x}}\ \Big(\frac{\text{b}}{\text{a}}\Big)^{\text{r}}$
$={^\text{14}}\text{C}_{\text{r}}\ \text{a}^{14-\text{r}}\ \text{b}^{\text{r}}\ \text{x}^{14-2\text{r}}$
For this term to be independent of $x,$ we must have
$=14-2\text{r}=0$
$\Rightarrow \text{r}=7$
Required term $= {^\text{14}}\text{C}_{\text{7}}\ \text{a}^{14-7}\ \text{b}^{7}=\frac{14!}{(7)!}\ \text{a}^{7}\ \text{b}^{7}$
View full question & answer→MCQ 241 Mark
If the coefficients of the $(n + 1)^{th}$ term and the $(n + 3)^{th}$ term in the expansion of $(1+\text{x})^{20}$ are equal, then the value of $n$ is:
AnswerCoefficient of $(r + 1)^{th}$ term $=$ Coefficient of $(n + 3)^{th}$
Then, we have
${^\text{20}}\text{C}_{\text{n}}={^\text{20}}\text{C}_{\text{n}+2}$
$\Rightarrow 2\text{n}+2=20$
$\Rightarrow \text{n}=9$
View full question & answer→MCQ 251 Mark
If an the expansion of $(1+\text{x})^{15},$ the coefficients of $(2\text{r}+3)^{\text{th}}$ and $(\text{r}-1)^{\text{th}}$ terms are equal, then the value of r is:
AnswerCoefficients of $(2\text{r}+3)^{\text{th}}$ and $(\text{r}-1)^{\text{th}}$ terms in the given expansion are ${^\text{15}}\text{C}_{\text{2r}+2}$ and ${^\text{15}}\text{C}_{\text{2r}-2}.$
Then, we have
${^\text{15}}\text{C}_{\text{2r}+2}={^\text{15}}\text{C}_{\text{r}-2}$
$\Rightarrow 2\text{r}+2=\text{r}-2$ or $2\text{r}+2+\text{r}-2=15$
$\Rightarrow \text{r}=-4$ or $\text{r}=5$
Neglecting the negative value, We have
$\text{r}=5$
View full question & answer→MCQ 261 Mark
If the coefficient of $x$ in $\Big(\text{x}^{2}+\frac{\lambda}{\text{x}}\Big)^{5}$ is $270,$ then $\lambda=$
AnswerThe coefficient of $x$ in the given expansion where $x$ occurs at the $(r + 1)^{th}$ term.
We have,
${^\text{15}}\text{C}_{\text{r}}(\text{x}^{2})^{5-\text{r}}\ \Big(\frac{\lambda}{\text{x}}\Big)^{\text{r}}$
$={^\text{15}}\text{C}_{\text{r}}\ \lambda^{\text{r}}\ \text{x}^{10-2\text{r}-\text{r}}$
For it to contain $x,$ we must have
$10-3\text{r}=1$
$\Rightarrow \text{r}=3$
Coefficient of $x$ in the given expansion,
$={^\text{15}}\text{C}_{\text{3}}\ \lambda^{\text{3}}=10\lambda^{3}$
Now, we have
$10\lambda^{3}=270$
$\Rightarrow \lambda^{3}=27$
$\Rightarrow \lambda=3$
View full question & answer→MCQ 271 Mark
The coefficient of $\frac{1}{\text{x}}$ in the expansion of $(1+\text{x})^{\text{n}}+\Big(1+\frac{1}{\text{x}}\Big)^{\text{n}}$ is:
- A
$\frac{\text{n}!}{\big[(\text{n}-1)!(\text{n}+1)!\big]}$
- ✓
$\frac{(\text{2n})!}{\big[(\text{n}-1)!(\text{n}+1)!\big]}$
- C
$\frac{(\text{2n})!}{\big[(\text{2n}-1)!(\text{2n}+1)!\big]}$
- D
AnswerCorrect option: B. $\frac{(\text{2n})!}{\big[(\text{n}-1)!(\text{n}+1)!\big]}$
Coefficient of $\frac{1}{\text{x}}$ in the given expansion = Coefficient of 1 in $(1+\text{x})^{\text{n}}$ × Coefficient of $\frac{1}{\text{x}}$
$={^\text{n}}\text{C}_{\text{0}}\times{^\text{n}}\text{C}_{\text{1}}+{^\text{n}}\text{C}_{\text{1}}\times{^\text{n}}\text{C}_{\text{2}}$
$=\text{n}+\text{n}\times\frac{\text{n}!}{2(\text{n}-2)!}$
$=\text{n}+\text{n}\frac{\text{n}(\text{n}-1)}{2}$
View full question & answer→MCQ 281 Mark
The coefficient of $x^{−17}$ in the expansion of $\Big(\text{x}^{4}-\frac{1}{\text{x}^{3}}\Big)^{15}$ is:
- A
$1365$
- ✓
$-1365$
- C
$3003$
- D
$-3003$
AnswerCorrect option: B. $-1365$
Suppose the $(r + 1)^{th}$ term in the given expansion contains the coefficient of $x^{−17}.$
Then, we have
$\text{T}_{\text{r}+1}={^\text{15}}\text{C}_{\text{r}}(\text{x}^{4})^{15-\text{r}}\Big(\frac{-1}{\text{x}^{3}}\Big)^{\text{r}}$
$\Rightarrow (1)^{\text{r}}\ {^\text{15}}\text{C}_{\text{r}-1}\ \text{x}^{60-4\text{r}-3\text{r}}$
For this term to contain $x^{-17},$ we mst have
$60-7\text{r}=-17$
$\Rightarrow 7\text{r}=77$
$\Rightarrow \text{r}=11$
$\therefore$ Required coefficient $=(-1)^{11}\ {^\text{15}}\text{C}_{\text{11}}=-\frac{15\times14\times13\times12}{4\times3\times2}=-1365$
View full question & answer→MCQ 291 Mark
In the expansion of $\Big(\frac{1}{2}\text{x}^{\frac{1}{3}}+\text{x}^{\frac{-1}{5}}\Big)^{8},$ the term independent of $x$ is:
- A
$\text{T}_{5}$
- ✓
$\text{T}_{6}$
- C
$\text{T}_{7}$
- D
$\text{T}_{8}$
AnswerCorrect option: B. $\text{T}_{6}$
Suppose the $(r + 1)^{th}$ term in the given expansion is independent of $x.$
Thus, we have
$\text{T}_{\text{r}+1}={^\text{8}}\text{C}_{\text{r}}\Big(\frac{1}{2}\text{x}^{\frac{1}{3}}\Big)^{8-\text{r}}\Big(\text{x}^{\frac{-1}{5}}\Big)^{\text{r}}$
$={^\text{8}}\text{C}_{\text{r}}\frac{1}{2^{8-\text{r}}}\ \text{x}^{\frac{8-\text{r}}{3}-\frac{\text{r}}{5}}$
For this term to be independent of $x,$ we must have
$\frac{8-\text{r}}{3}-\frac{\text{r}}{5}=0$
$\Rightarrow 40-5\text{r}-3\text{r}=0$
$\Rightarrow \text{r}=5$
Hence, the required term is the 6th term, i.e. $\text{T}_{6}$
View full question & answer→MCQ 301 Mark
If the coefficients of $x^2$ and $x^3$ in the expansion of $(3+\text{ax})^{9}$ are the same, then the value of $a$ is:
- A
$-\frac{7}{9}$
- B
$-\frac{9}{7}$
- C
$\frac{7}{9}$
- ✓
$\frac{9}{7}$
AnswerCorrect option: D. $\frac{9}{7}$
Coefficients of $x^2 = $ Coefficients of $x^3$
${^\text{9}}\text{C}_{\text{2}}\times3^{9-2}\text{a}^{2}={^\text{9}}\text{C}_{\text{3}}\times3^{9-3}\ \text{a}^{3}$
$\Rightarrow \text{a}=\frac{{^\text{9}}\text{C}_{\text{2}}}{{^\text{9}}\text{C}_{\text{3}}}\times3$
$=\frac{9!\times3!\times6!\times3}{2!\times7!\times9!}$
$=\frac{9}{7}$
View full question & answer→MCQ 311 Mark
The number of irrational terms in the expansion of $\Big(4^{\frac{1}{5}}+7^{\frac{1}{10}}\Big)^{45}$ is:
AnswerThe general term $T_{r+1}$ in the given expansion is given by ${^\text{45}}\text{C}_{\text{r}}\Big(4^{\frac{1}{5}}\Big)^{45-\text{r}}\Big(7^{\frac{1}{10}}\Big)^{\text{r}}$
For $T_{r+1}$ to be an integer, we must have $\frac{\text{r}}{5}$ and $\frac{\text{r}}{10}$ as integers i.e. $0\leq\text{r}\leq45$
$\therefore \text{r}=0,10,20,30,40$
Hence, there are $5$ rational and $41,$ i.e. $46 - 5,$ irrational terms.
View full question & answer→MCQ 321 Mark
If rth term in the expansion of $\Big(2\text{x}^{2}-\frac{1}{\text{x}}\Big)^{12}$ is without x, then r is equal to:
Answerrth term in the given expansion is ${^\text{20}}\text{C}_{\text{r}-1}\Big(2\text{x}^{2}\Big)^{12-\text{r}+1}\Big(\frac{-1}{\text{x}}\Big)^{\text{r}-1}$
$=(-1)^{\text{r}-1}\ {^\text{20}}\text{C}_{\text{r}-1}\ 2^{13-\text{r}}\ \text{x}^{26-2\text{r}-\text{r}+1}$
For this term to be independent of x, we must have,
$27-3\text{r}=0$
$\Rightarrow \text{r}=9$
Hence, the term in the expansion is independent.
View full question & answer→MCQ 331 Mark
If in the expansion of $(1+\text{y})^{\text{n}},$ the coefficients of $5^{th}, 6^{th}$ and $7^{th}$ terms are in $A.P.,$ then $n$ is equal to:
- A
$7, 11$
- ✓
$7, 14$
- C
$8, 16$
- D
AnswerCorrect option: B. $7, 14$
Coefficients of the $5^{th}, 6^{th}$ and $7^{th}$ terms in the given expansion are ${^\text{n}}\text{C}_{\text{4}},{^\text{n}}\text{C}_{\text{5}}$ and ${^\text{n}}\text{C}_{\text{6}}.$
These coefficients are in $AP.$
Thus, we have
$2\ {^\text{n}}\text{C}_{\text{5}}={^\text{n}}\text{C}_{\text{4}}+{^\text{n}}\text{C}_{\text{6}}$
On dividing both sides by ${^\text{n}}\text{C}_{\text{5}},$ we get
$2=\frac{{^\text{n}}\text{C}_{\text{4}}}{{^\text{n}}\text{C}_{\text{5}}}+\frac{{^\text{n}}\text{C}_{\text{6}}}{{^\text{n}}\text{C}_{\text{5}}}$
$\Rightarrow 2=\frac{5}{\text{n}-4}+\frac{\text{n}-5}{6}$
$\Rightarrow 12\text{n}-48=30+\text{n}^{2}-4\text{n}-5\text{n}+20$
$\Rightarrow \text{n}^{2}-21\text{n}+98=0$
$\Rightarrow (\text{n}-14)(\text{n}-7)=0$
$\Rightarrow \text{n}=7,14$
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