MCQ 1011 Mark
What is the coefficient of $x^3y^4$ in $(2x + 3y^2)5$?
- A
$240$
- B
$360$
- ✓
$720$
- D
$1080$
View full question & answer→MCQ 1021 Mark
The value of $(\sqrt{5}+1)^{5}-(\sqrt{5}-1)^{5} $ is:
AnswerGiven, $(\sqrt{5}+1)^{5}-(\sqrt{5}-1)^{5} $
$=(\sqrt{5}+1)^{5}-(\sqrt{5}-1)^{5} $
The even terms will get eliminated.
Hence, we get
$2\left[{ }^5 \mathrm{C}_0+{ }^5 \mathrm{C}_2 5+{ }^5 \mathrm{C}_4 5^2\right]$
$= 2 [1 + 50 + 125]$
$= 2 [176]$
$= 352.$
View full question & answer→MCQ 1031 Mark
In any binomial expansion, the number of terms are:
- A
$\geq5$
- ✓
$\geq2$
- C
$\geq3$
- D
AnswerCorrect option: B. $\geq2$
Bi $-$ nomial, involves summation of two terms.
Let the terms be $x$ and $y.$
Therefore a binomial expansion can be of the form, $(x + y)n.$
where $\text{n}\geq1$ If $n = 1,$ we get only two terms.If $n > 1$ where $n$ is an integer, then it gives us in total $(n + 1)$ terms.
Thus, number of terms has to be $\geq2.$
View full question & answer→MCQ 1041 Mark
The number of real negative terms in the binomial expansion of $(1 + ix)^{4n-2}$, $\text{n}\in\text{N}, x > 0,$ is:
Answer$ (1+i x)^{4 n-2} $
$ =\left((1+i x)^2\right)^{2 n-1} $
$ =\left(1-x^2+2 i x\right)^{2 n-1} $
$ =\left[\left(1-x^2\right)+i(2 x)\right]^{2 n-1} $
Total number of terms will be $2n - 1 + 1 = 2n.$
Hence the number of real negative terms will therefore be
$=\frac{2\text{n}}{2}$
$=\text{n}.$
View full question & answer→MCQ 1051 Mark
Sum of the coefficients of the terms of degree $mm$ in the expansion of $(1 + x)^n(1 + y)^n(1 + z)^n$ is:
AnswerThe Coefficient of $x^m =$ Number of ways of choosing $m$ balls out of $n$ black balls, $n$ green balls and $n$ blue ball.
Hence total number of balls $= 3n.$
Required is $m.$
Hence required combination is $ { }^{3 n} C_m $
Hence the coefficient of $x^m$ in $(1 + x)^n(1 + y)^n(1 + z)^n$
$= { }^{3 n} C_m $
View full question & answer→MCQ 1061 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 1071 Mark
In the expansion of $(1 + x)^n$, the sum of coefficients of odd powers of $x$ is:
Answer$(1 + x)^n= C_0+ C_{1x}+ C_2x^2 + C_3x^3+ ... + C_n x^n$
Putting $x = 1$ and $x = 1$ and subtracting, we get.
$2^n= 2(C_1 + C_3+ C_5+ ...)$
$\therefore$ $C_1 + C_3+ C_5+ ... = 2^{n-1}$
Or the sum of the coefficients of the odd power of $x$ is $2^{n-1}$.
View full question & answer→MCQ 1081 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 1091 Mark
If the coefficients of $x^7$ and $x^8$ in $\big(2+\frac{\text{x}}{3}\big)\text{n}$ are equal, then $n$ is:
View full question & answer→MCQ 1101 Mark
The coefficient of $y$ in the expansion of $\Big(\text{y}^2+\big(\frac{\text{c}}{\text{y}}\big)\Big)^5$ is:
- A
$10c$
- B
$29c$
- ✓
$10c^3$
- D
$20c^3$
AnswerCorrect option: C. $10c^3$
Given:$\Big(\text{y}^2+\big(\frac{\text{c}}{\text{y}}\big)\Big)^5$
$\Big(\text{y}^2+\big(\frac{\text{c}}{\text{y}}\big)\Big)^{5}=\ ^{5}\text{C}_{\text{r}}\times(\text{y}^2)^{\text{r}}\times\big(\frac{\text{c}}{\text{y}}\big)^{5-\text{r}}$
$\Big(\text{y}^2+\big(\frac{\text{c}}{\text{y}}\big)\Big)^{5}={5}\text{C}{\text{r}}\times\text{y}^{2{\text{r}}}\times\big(\frac{\text{c}^{5-\text{r}}}{\text{y}^{5-\text{r}}}\big)$
On solving this, we get $r = 3.$
Hence, the coefficient of $y ={^5C_3}\times c^3= 10c^3$.
View full question & answer→MCQ 1111 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 1121 Mark
Number of rational terms in the expansion of $\Big(\sqrt{2}+\sqrt[4]{3}\Big)^{100}$ is:
AnswerThe general term for the following expression is $\text{T}_{\text{r}+1}={^{100}}\text{C}_{\text{r}}2^{50-\frac{\text{r}}{2}}\cdot3\frac{\text{r}}{4}.$
Hence we get rational terms for
$r = 0, 4, 8, 12 ....100$
$a_n = a + (n - 1).d$
$100 = 0 + (n - 1).4$
$25 = n - 1$
$n = 26$
View full question & answer→MCQ 1131 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}} \times $ 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 1141 Mark
How many terms are there in the expansion of $(4x + 7y)10 + (4x - 7y)10?$
View full question & answer→MCQ 1151 Mark
The number of terms that are integers in the binomial expansion of $\big(\sqrt{7}+\sqrt[3]{5}\big)^{35}$ is:
AnswerThe general term in the given expansion $\big(\sqrt{7}+\sqrt[3]{5}\big)^{35}$ is $^{35}\text{C}_\text{r}7\frac{35-\text{r}}{2}.5\frac{\text{r}}{3},$
If $r$ is a multiple of $3$ and $35 - r$ is a multiple of $2$ then the terms are integers,
$\therefore r = 3, 9, 15, 21, 27, 33$ which are six values.
View full question & answer→MCQ 1161 Mark
The value of $\sum\limits^\text{n}_{\text{r}=0}\text{a}_{2\text{r}-1}$ is:
- A
$9^n- 1$
- ✓
$9^n + 1$
- C
$9^n - 2$
- D
AnswerCorrect option: B. $9^n + 1$
$\left(1+4 x+4 x^2\right)^n=a_0+a_1 x+a_2 x^2+\ldots\left(a_{2 n} x^{2 n}\right)$
Substituting $x=1$ we get
$9^n=a_0+a_1+a_2+\ldots\left(a_{2 n}\right)$
Substituting $x=-1$ we get
$1=a_0-a_1+a_2-a 3 \ldots\left(a_{2 n}\right)$
Adding both we get
$2\left(a_0+a_2+a_4+\ldots a^{2 n}\right)=9^n+1$
Hence
$\sum^{\text{n}}_{\text{k}=0}\text{a}_{2\text{k}}=9^{\text{n}}+1$
View full question & answer→MCQ 1171 Mark
Find the middle term in the expansion of $\Big(\frac{2\text{x}}{3}+\frac{3}{2\text{x}}\Big)^{10}.$
AnswerThe middle term will be the $6^{th}$ term.
It will also be the only term independent of $x.$
Hence the coefficient will be
$T_{5+1} ={^{10}C_5}$
$=\frac{10!}{5!(5!)}$
$= 252$
View full question & answer→MCQ 1181 Mark
The sum of the coefficient in the expansion of $(x + y)^n$ is $4096$. The greatest coefficient in the expansion is:
Answer$(x + y)^n$, Sum of coefficient $= 4096$
When $x = y = 1,$ if $n = 12$
$\Rightarrow (1 + 1)^{12} = 2^{12} = 4096$
$\Rightarrow$ Hence, greatest coefficient
${^{\text{n}}}\text{C}_{\frac{\text{n}}{2}}={^{12}}\text{C}_{6}=\frac{12!}{6!6!}=924$
Hence, this is the answer.
View full question & answer→MCQ 1191 Mark
If the coefficients of $2^{nd}, 3^{rd}$ and the $4^{th}$ terms in the expansion of $(1 + x)n$ are in $A.P.,$ then value of $n$ is:
View full question & answer→MCQ 1201 Mark
The sum of the coefficients in the expansion of $(1 - x)^{10}$
Answer$(1-x)^{10}=1-{ }^{10} C_{1 x}+{ }^{10} C_2 x^2+\ldots{ }^{10} \mathrm{C1}_0 x^{10}$
Substituting $x=1$, we get sum of coefficients as
$ 1-{ }^{10} \mathrm{C}_1+{ }^{10} \mathrm{C}_2+\ldots{ }^{10} \mathrm{C}_{10} $
$ =(1-1)^{10}$
$= 0$
View full question & answer→MCQ 1211 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 1221 Mark
The coefficient of $x - 12$ in the expansion of $\Big(\frac{\text{x}+\text{y}}{\text{x}3}\Big)^{20}$ is:
- A
$ { }^{20} \mathrm{C}_8 $
- ✓
$ { }^{20} \mathrm{C}_8 \mathrm{y}^8 $
- C
$ { }^{20} \mathrm{C}_{12} $
- D
$ { }^{20} \mathrm{C}_{12} \mathrm {~y}{12 } $
AnswerCorrect option: B. $ { }^{20} \mathrm{C}_8 \mathrm{y}^8 $
View full question & answer→MCQ 1231 Mark
$(\sqrt{3}+1)^{5}-(\sqrt{3}+1)^{5}=$
AnswerIn the above binomial expansion, the terms at the odd position will get eliminated.
We would be left with
$2({^{5}}\text{C}_{1}(\sqrt{3})^{4}+{^5}\text{C}_{3}(\sqrt{3})^{2}+{^5}\text{C}_{5})$
$=2(5(3^2)+10(3)+1)$
$=2(45+30+1)$
$=2(76)$
$=152$
View full question & answer→MCQ 1241 Mark
The term independent of $x$ in the expansion of $\Big(9\text{x}-\frac{1}{3\sqrt{{x}}2}\Big)18, x > 0$ , is $'a\ '$ times the corresponding binomial coefficient. Then $'a\ '$ is:
- A
$3$
- B
$\frac{1}{3}$
- C
$-\frac{1}{3}$
- ✓
$\text{None of these}$
AnswerCorrect option: D. $\text{None of these}$
View full question & answer→MCQ 1251 Mark
Find the sum of the series $3 \cdot{ }^n C_0-8 \cdot{ }^n C_1+13 \cdot{ }^n C_2-18 \cdot{ }^n C_3+\ldots+(n+1)$ terms.
AnswerLet $n = 2$
Hence the above expression is reduced to
$3(1) - 8(2) + 13(1)$
$= 16 - 16 = 0$
Let $n = 3$
$3(1) - 8(3) + 13(3) - 18(1)$
$= 42 - 42 = 0$
Hence the sum of the series for $n > 1$ is $0.$
View full question & answer→MCQ 1261 Mark
In the expansion of $\Big(\sqrt[3]4+\frac{1}{\sqrt[4]{6}}\Big)^{20},$
AnswerCorrect option: B. The number of irrational terms $= 19$
$\text{T}_{\text{r}+1}={^\text{n}}\text{C}_{\text{r}}4\frac{20-\text{r}}{3}6\frac{-\text{r}}{4}$
$={^\text{n}}\text{C}_{\text{r}}2\frac{40-2\text{r}}{3}-\frac{\text{r}}{4}3-\frac{\text{r}}{4}$
$={^\text{n}}\text{C}_{\text{r}}2\frac{100-11\text{r}}{12}3-\frac{\text{r}}{4}$
There are total of $21$ terms.
Hence, we get rational terms for $r = 20, 8$
Hence there are in total $21 - 2 = 19$ irrational terms.
The middle term is at $r = 10$ which is irrational.
View full question & answer→MCQ 1271 Mark
For $2\leq\text{r}\leq\text{n},$ $\Big(\frac{\text{n}+1}{\text{r}}\Big)+\Big(\frac{\text{n}}{\text{r}-1}\Big)+\Big(\frac{\text{n}}{\text{r}-2}\Big)$is equal to$-$
- A
$\Big(\frac{\text{n}+1}{\text{r}}\Big)$
- B
$2\Big(\frac{\text{n}+1}{\text{r}-1}\Big)$
- C
$2\Big(\frac{\text{n}+2}{\text{r}}\Big)$
- ✓
$\Big(\frac{\text{n}+2}{\text{r}}\Big)$
AnswerCorrect option: D. $\Big(\frac{\text{n}+2}{\text{r}}\Big)$
$ { }^{n+1} C_r+{ }^n C_{r-1}+{ }^n C_{r-2} $
$ ={ }^{n+1} C_r+{ }^{n+1} C_{r-1} $
$ ={ }^{n+2} C_r $
View full question & answer→MCQ 1281 Mark
The number of terms in the expansion of $(1 + x)21$ is:
AnswerThe number of terms in the expansion is one more than $n$ i.e., $n + 1$
So, here $n = 21$
The number of terms in the expansion $(1 + x)21 = 21 + 1 = 22.$
View full question & answer→MCQ 1291 Mark
Choose the correct answer. The total number of terms in the expansion of $(x + a)^{100} + (x - a)^{100}$ after simplification is:
AnswerNumber of terms in the expansion of $(x + a)^{100} = 101$
Number of terms in the expansion of $(x - a)^{100} = 101$
Now $50$ terms of expansion will cancel out with negative $50$ terms of $(x - a)^{100}$
So, the remaining $51$ terms of first expansion will be added to $51$ terms of other.
Therefore, the number of terms $= 51$
Hence, the correct option is $(c).$
View full question & answer→MCQ 1301 Mark
If the middle term of $\big(\frac{1}{\text{x}}+\text{x}\sin\text{x}\big)10$ is equal to $7.\frac{7}{8},$ then value of $x$ is:
- A
$2\text{n}\sqcap+\frac{\sqcap}{6}$
- B
$\text{n}\sqcap+\frac{\sqcap}{6}$
- ✓
$\text{n}\sqcap+(-1)^\text{n}\frac{\sqcap}{6}$
- D
$\text{n}\sqcap$
AnswerCorrect option: C. $\text{n}\sqcap+(-1)^\text{n}\frac{\sqcap}{6}$
View full question & answer→MCQ 1311 Mark
The coefficient of the middle term in the expansion of $(2 + 3x)4$ is:
AnswerIf the exponent of the expression is $n,$ then the total number of terms is $n + 1.$
Hence, the total number of terms is $4 + 1 = 5.$
Hence, the middle term is the $3^{rd}$ term.
Therefore, ${T}_3={ }^4 \mathrm{C}_2 \times(2)^2 \times(3 \mathrm{x})^2$
$T_3=(6) \times(4) \times\left(9 x^2\right) $
$T_3=216 x^2$ .
Therefore, the coefficient of the middle term is $216.$
View full question & answer→MCQ 1321 Mark
In the expansion of $\Big(7^{\frac{1}{3}}+11^{\frac{1}{9}}\Big)^{5832},$ the number of terms free from radicals is:
AnswerTotal number of integral terms are
$\frac{5832}{\text{L}.\text{C}.\text{M}(3,9)}+1$
$=\frac{5832}{9}+1$
$=648+1$
$=649$
View full question & answer→MCQ 1331 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 $6^{th}$ term, i.e. $\text{T}_{6}$
View full question & answer→MCQ 1341 Mark
The largest term in the expansion of $(3 + 2x)50,$ when $\text{x}=\frac{1}{5}$ is:
- A
$6^{th}$ term
- B
$7^{th}$ term
- C
$8^{th}$ term
- ✓
View full question & answer→MCQ 1351 Mark
The coefficient of the term independent of $x$ in the expansion of $\Big(\frac{\sqrt{\text{x}}}{3}+\frac{3}{2\text{x}^2}\Big)10$ is:
- ✓
$\frac{5}{4}$
- B
$\frac{7}{4}$
- C
$\frac{9}{4}$
- D
$\text{None of these}$
AnswerCorrect option: A. $\frac{5}{4}$
View full question & answer→MCQ 1361 Mark
If sum of all the coefficients in the expansion of $\Big(\text{x}^{\frac{3}{2}}+\text{x}^{\frac{1}{3}}\Big)^{\text{n}}$ is $128$, then the coefficient of $x^5$ is:
AnswerSubstituting $x = 1,$ we get the sum of the coefficients as
$(2)^n = 128$
$\therefore n = 7$
Hence writing the general term, we get
$\text{T}_{\text{r}+1}={^{7}}\text{C}_{\text{r}}\text{x}\frac{63-11\text{r}}{6}$
Hence for the coefficient of $x^5$
$63 - 11r = 6(5)$
$63 - 11r = 30$
$33 - 11r = 0$
$\therefore r = 3$
Hence coefficient is $^7C_3 = 35$.
View full question & answer→MCQ 1371 Mark
Choose the correct answer.The coefficient of $x^n$ in the expansion of $(1 + x)^{2n}$ and $(1 + x)^{2n-1}$ are in the ratio.
- A
$1 : 2.$
- B
$1 : 3.$
- C
$3 : 1.$
- ✓
$2 : 1. $
AnswerCorrect option: D. $2 : 1. $
General Term $\text{T}_{\text{r}+1}=\ ^\text{n}\text{C}_\text{r}\text{x}^{\text{n}-\text{r}}\text{y}^\text{r}$
In the expansion of $(1 + x)^{2n}$, we get $\text{T}_{\text{r}+1}=\ ^{2\text{n}}\text{C}_\text{r}\text{x}^\text{r}$
To get the coefficient of $x^n$, put $r = n$
$\therefore$ Coefficient of $\text{x}^\text{n}=\ ^{\text{2n}}\text{C}_\text{n}$
In the expansion of $(1+\text{x})^{2\text{n}-1},$ we get $\text{T}_{\text{r}+1}=\ ^{2\text{n}-1}\text{C}_\text{r}\text{x}^\text{r}$
$\therefore$ Coefficient of $\text{x}^\text{n}\ \ \text{is}=\ ^{\text{2n}-1}\text{C}_{\text{n}-1}$
The required ratio is $\frac{^{\text{2n}}\text{C}_{\text{n}-1}}{^{\text{2n}-1}\text{C}_{\text{n}-1}}$
$=\frac{\frac{2\text{n}!}{\text{n}!(\text{n}!)}}{\frac{(2\text{n}-1)!}{(\text{n}-1)!(2\text{n}-1-\text{n}+1)!}}=\frac{\frac{2\text{n}!}{\text{n}!.\text{n}!}}{\frac{(2\text{n}-1)!}{(\text{n}-1)!(\text{n}!)}}$
$=\frac{2\text{n}!}{\text{n}!\text{n}!}\times\frac{(\text{n}-\text{n})!\cdot\text{n}!}{(2\text{n}-1)!}=\frac{2\text{n}(2\text{n}-1)!}{\text{n}!\text{n}(\text{n}-1)!}\times\frac{(\text{n}-1)!\cdot\text{n}!}{(2\text{n}-1)!}$
$=\frac{2}{1}$
$=2:1$
Hence, the correct option is $(d).$
View full question & answer→MCQ 1381 Mark
$r$ and $n$ are positive integers $r > 1, n > 2$ and coefficient of $(r+2)^{th}$ term and $3^{rd}$ term in the expansion of $(1 + x)^{2n}$ are equal, then $n$ equals:
- A
$3r$
- B
$3r + 1$
- ✓
$2r$
- D
$2r + 1$
View full question & answer→MCQ 1391 Mark
Find the coefficient of $\frac{1}{\text{y}^{2}}$ in $\Big(\text{y}+\frac{\text{c}}{\text{y}^{2}}\Big)^{10}.$
- ✓
$ 210 c^4 $
- B
$ 210 c^5 $
- C
$ 120 c^3 $
- D
AnswerCorrect option: A. $ 210 c^4 $
$\Big(\text{y}+\frac{\text{c}}{\text{y}^{2}}\Big)^{10}$
$\text{T}_{\text{r}+1}={^{10}}\text{C}_{\text{r}}\text{y}^{10-3\text{r}}\text{c}^{\text{r}}$
Hence for $y^{-2}$
$10-3 r=-2$
$12=3 r$
$r=4$
Coefficient will be
${ }^{10} \mathrm{C}_4 \mathrm{c}^4$
$=210 \mathrm{c}^4$
View full question & answer→MCQ 1401 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 1411 Mark
The coefficient of the middle term in the expansion of $(2 + 3x)4$ is:
View full question & answer→MCQ 1421 Mark
The middle term in the expansion of $\Big(\frac{\text{a}}{\text{x}}+\text{bx}\Big)^{12}$ is:
- ✓
$924 a^6 b^6 $
- B
$924 a^6 b^5 $
- C
$924 a^5 b^5 $
- D
$924 a^5 b^6 $
AnswerCorrect option: A. $924 a^6 b^6 $
The middle term will be the $7^{th}$ term.
Hence $\text{T}_{6+1}=^{12}\text{C}_6\big(\frac{\text{a}}{\text{x}}\big)^6(\text{bx})^6=924\text{a}^6\text{b}^6$
View full question & answer→MCQ 1431 Mark
In the expansion of $(1 + x)^n.(1 + y)^n.(1 + z)^n$ the sum of the coefficients of the terms of degree $r$ is:
- A
$(^nC_r)^3$
- ✓
$^{3n}C_r$
- C
$^{3 \times n}C_r$
- D
AnswerCorrect option: B. $^{3n}C_r$
The given expression contains $3n$ factors
Using combination to choose $r$ brackets out of $3n$ brackets for a term of degree $r,$ we get
$^{3n}C_r$
View full question & answer→MCQ 1441 Mark
The middle term in the expansion of $\Big(\text{x}+\frac{1}{\text{x}}\Big)^{10},$ is:
- A
$^{10}\text{C}_1\frac{1}{\text{x}}$
- ✓
$ { }^{10} \mathrm{C}_5 $
- C
$ { }^{10} \mathrm{C}_6 $
- D
$ { }^{10} \mathrm{C}_7 \mathrm{x}$
AnswerCorrect option: B. $ { }^{10} \mathrm{C}_5 $
The middle term would be the $6^{th}$ term.
Hence
$T_{5+1} = { }^{10} \mathrm{C}_5 $
View full question & answer→MCQ 1451 Mark
$(x - 1)^4+ 4(x - 1)^3+ 6(x - 1)^2+ 4(x - 1) + 1 =$
AnswerConsider the following identity
$ (a+1)^4 $
$ =\left((a+1)^2\right)^2 $
$ =\left(a^2+2 a+1\right)^2 $
$ =a^4+4 a^2+1+4 a^3+4 a+2 a^2 $
$ =a^4+4 a^3+6 a^2+4 a+1 \ldots(i)$
Comparing $i$ with the given question we get
$a=(x-1)$
Therefore
$ (x-1)^4+4(x-1)^3+6(x-1)^2+4(x-1)+1 $
$ =(x-1+1)^4$ from $(i)$
$ =x^4$
View full question & answer→MCQ 1461 Mark
If the coefficent of $x^2$ in the expansion of $(1 + x)^m$ is $6$ then $m = .........$
AnswerThe coefficient of $x^r$ in $(1+x)^n$ is ${ }^n C_r$
$\Rightarrow$ the coefficient of $x^2$ in $(1+x)^m$ is ${ }^m C_2=6$
$\Rightarrow\frac{\text{m}(\text{m}-1)}{2}=6$
$\Rightarrow\text{m}^{2}-\text{m}-12=0$
$\Rightarrow\text{m}=4,-3$
$\therefore m$ is a positive number,
so $m = 4.$
View full question & answer→MCQ 1471 Mark
After simplification, what is the number of terms in the expansion of $[(3x + y)5]4 - [(3x - y)4]5?$
View full question & answer→MCQ 1481 Mark
The number of integral terms in$ (\sqrt{3}+\sqrt[8]{2})^{64}$ is-
AnswerThe general term of expansion $(x+y)^n$ is ${ }^n C_r x^{n-r} y^r$
So the general term of $ (\sqrt{3}+\sqrt[8]{2})^{64}$ is $ {^{64}}{\text{C}}_{\text{r}}3\frac{64-\text{r}}{2}2\frac{\text{r}}{8}$
For the term to be integer, $r$ must be divided by $8$ and $64 - r$ must be divided by $2$
The possible values of $r$ are $0, 8, 16, 24, 32, 40, 48, 56, 64$ the number of integral values is $9.$
View full question & answer→MCQ 1491 Mark
The number of irrational terms in the expansion of $\Big(2^{\frac{1}{5}}+3^{\frac{1}{10}}\Big)^{55}$ is:
AnswerFor the above question $\text{T}_{\text{r}+1}={^{55}}\text{C}_{\text{r2}}11-\frac{\text{r}}{5}3\frac{\text{r}}{10}$
Hence we will have rational terms at $r = 0, 10, 20, 30, 40, 50$ respectively.
Hence there will be $6$ rational terms.
The total number of terms will be
$55 + 1$
$= 56$ terms.
Hence the number of irrational terms will be
$56 - 6$
$= 50$ terms.
View full question & answer→MCQ 1501 Mark
The coefficient of $xp$ and $xq (p$ and $q$ are positive integers$)$ in the expansion of $(1 + x) p + q$ are:
- ✓
- B
Equal with opposite signs
- C
- D
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