MCQ 511 Mark
The positive integer just greater than $(1 + 0.0001)^{10000}$ is:
Answer$(1+0.0001)^{10000}$
$=\big(1+\frac{1}{10000}\big)^{10000}$
$=\big(1+\frac{1}{\text{n}}\big)^{\text{n}}=\text{n}\text{c}_{0}(1)^{\text{n}}+\text{n}\text{c}_{1}(1)^{\text{n}-1}\cdot\frac{1}{\text{n}}+\text{n}\text{c}_{2}(1)^{\text{n}-2}\frac{1}{\text{n}^{2}}+\ ....$
$=1+\text{n}\cdot\frac{1}{\text{n}}+\frac{\text{n}(\text{n}-1)}{2}\cdot\frac{1}{\text{n}^2}+\ ....$
$=2+\frac{\text{n}(\text{n}-1)}{2\text{n}^{2}}+\ .... > 2$ Integer just greater than $2$ is $3.$
View full question & answer→MCQ 521 Mark
The total number of rational terms in the expansion of $\Big(7^{\frac{1}{3}}+11^{\frac{1}{9}}\Big)^{6561}$ is:
AnswerTotal number of rational terms will be
$1+\frac{6561}{\text{L}.\text{C}.\text{M}.(3,9)}$
$=1+\frac{6561}{9}$
$=729+1$
$=730$
View full question & answer→MCQ 531 Mark
In the binomial expansion of $(a - b)n, n^3\ 5$ the sum of the 5th and 6th terms is zero. Then $\frac{\text{a}}{\text{b}}$ equals:
- A
$\text{n}-\frac{5}{6}$
- ✓
$\text{n}-\frac{4}{5}$
- C
$\frac{5}{\text{n}-{4}}$
- D
$\frac{6}{\text{n}-{5}}$
AnswerCorrect option: B. $\text{n}-\frac{4}{5}$
View full question & answer→MCQ 541 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 551 Mark
If $n$ is an integer lying between $0$ and $21,$ then the least value of $n!(21 - n)!$ is:
- A
$1!20!$
- ✓
$11!10!$
- C
$9!12!$
- D
AnswerCorrect option: B. $11!10!$
In pascals triangle middle terms has the highest value.
Therefore consider $^{21}C_n$
$=\frac{21!}{(21-\text{n})!\text{n}!}$
For $(21 - n)!n!$
to be least $^{21}C_n$ has to be maximum.
Therefore, since $21$ is odd we have two middle terms $T_{11}$ and $T_{12}$
Hence for $n = 11,$
$(21 - n)!n! = 11!(10)!$
which is less than $12!(9!)$ for $n = 12.$
View full question & answer→MCQ 561 Mark
The value of $(126)^{\frac{1}{3}}$ up to three decimal places is:
- A
$5.011$
- B
$5.012$
- ✓
$5.013$
- D
$5.014$
AnswerCorrect option: C. $5.013$
$(126)^{\frac{1}{3}}$ can also be written as the cube root of $126.$
Hence, $ (126)^{\frac{1}{3}}$is approximately equal to $5.013.$
View full question & answer→MCQ 571 Mark
The coefficients of $x^p$ and $x^q (p$ and $q$ are positive integers$)$ in the expansion of $(1 + x)^{p+q}$ are:
- ✓
- B
Equal with opposite signs
- C
- D
AnswerThe general term is
$t_{r+1}={ }^{p+q} C_r X^r$
For coefficient of $x^p, r=p$ and hence coefficient is ${ }^{p+q} C_p$
For coefficient of $x^q, r=q$ and hence coefficient is ${ }^{p+q} C_q$
${ }^{p+q} C_p={ }^{p+q} C_q$
View full question & answer→MCQ 581 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 591 Mark
What are the values of $k$ if the term independent of $x$ in the expansion of $\Big(\sqrt{\text{x}}+\frac{\text{k}}{\text{x}^{2}}\Big)^{10}$ is $405?$
- ✓
$\pm\ 3$
- B
$\pm\ 6$
- C
$\pm\ 5$
- D
$\pm\ 4$
AnswerCorrect option: A. $\pm\ 3$
View full question & answer→MCQ 601 Mark
The largest coefficient in the expansion of $(1 + x)10$ is:
AnswerCorrect option: A. $\frac{10!}{(5!)^2}$
Given: $(1 + x)10$
The greatest coefficient will always occur in the middle term.
Hence, the total number of terms in an expansion is $11. ($ i.e. $10 + 1 = 11)$
Therefore, middle term $ =\Big[\big(\frac{10}{2}\big)+1\Big]=5+1=6\text{th }\text{term}.$
So, $T6 ={^{10}C_5} \times x^5$
Therefore, the coefficient of the greatest term $={^{10}C_5}$$=\frac{10!}{(5!)^2}.$
View full question & answer→MCQ 611 Mark
The coefficient of $x^3y^4$ in $(2x + 3y^2)^5$ is:
- A
$360$
- ✓
$720$
- C
$240$
- D
$1080$
AnswerGiven: $\left(2 x+3 y^2\right)^5$
Therefore, the general form for the expression $\left(2 x+3 y^2\right)^5$ is $T_{r+1}={ }^5 C_r \times(2 x)^r \times\left(3 y^2\right)^{5-r}$
Hence, $T_{3+1}={ }^5 C_3(2 x)^3 \times\left(3 y^2\right)^{5-3}$
$ T_4={ }^5 C_3(2 x)^3 \times\left(3 y^2\right)^2 $
$ T_4={ }^5 C_3 \times 8 x^3 \times 9 y^4$
On simplification, we get
$T_4=720 x^3 y^4$
Therefore, the coefficient of $x^3 y^4$ in $\left(2 x+3 y^2\right)^2$ is $720 .$
View full question & answer→MCQ 621 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:
- A
$4(A + B)$
- B
$4(A - B)$
- C
$AB$
- ✓
$4AB$
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 631 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 641 Mark
Choose the correct answer.The two successive terms in the expansion of $(1 + x)^{24}$ whose coefficients are in the ratio $1 : 4$ are:
- A
$3^{rd}$ and $4^{th}$.
- B
$4^{th}$ and $5^{th}$
- ✓
$5^{th}$ and $6^{th}$.
- D
$6^{th}$ and $7^{th.}$
AnswerCorrect option: C. $5^{th}$ and $6^{th}$.
Let the two successive terms in the expansion of $(1 + x)^{24}$ be $(r + 1)(r + 2)^{th}$ terms.
Now, $\text{T}_{\text{r}+1}=\ ^{24}\text{C}_\text{r}\text{x}^\text{r}$ and $\text{T}_{\text{r}+2}=\ ^{24}\text{C}_{\text{r}+1}\text{x}^{\text{r}+1}$
Given that, $\frac{^{24}\text{C}_\text{r}}{^{24}\text{C}_{\text{r}+1}}=\frac{1}{4}$
$\Rightarrow\frac{\frac{(24)!}{\text{r}!(24-\text{r})!}}{\frac{(24)!}{(\text{r}+1)!(24-\text{r}-1)!}}=\frac{1}{4}$
$\Rightarrow\frac{(\text{r}+1)\text{r}!(23-\text{r})!}{\text{r}!(24-\text{r})(23-\text{r})!}=\frac{1}{4}$
$\Rightarrow\frac{\text{r}+1}{24-\text{r}}=\frac{1}{4}$
$\Rightarrow4\text{r}+4=24-\text{r}$
$\Rightarrow\text{r}=4$
$\therefore\text{T}_{4+1}=\text{T}_5$ and $\text{T}_{4+2}=\text{T}_6$
Hence $5^{th}$ and $6^{th}$ terms.
View full question & answer→MCQ 651 Mark
The number of terms with integral coefficient in the expansion of $\Big((27)^{\frac{1}{6}}+\sqrt[10]{32\text{x}}\Big)^{600}$ is:
Answer$\Big(27^{\frac{1}{6}}+32^{\frac{1}{10}}\text{x}\Big)^{600}$
$=\Big(3^{\frac{1}{2}}+2^{\frac{1}{2}}\Big)^{600}$
Total number of integral terms will be
$=\frac{600}{\text{L}.\text{C}.\text{M}(2,2)}+1$
$=\frac{600}{2}+1$
$=301$
View full question & answer→MCQ 661 Mark
The total number of terms in the expansion of $(x + a)51 - (x - a)51$ after simplification is:
View full question & answer→MCQ 671 Mark
The number of rational terms in the expansion of $(9^{\frac{1}{4}}+8^{\frac{1}{6}})^{1000}$ is:
AnswerThe general term in the expansion of $(9^{\frac{1}{4}}+8^{\frac{1}{6}})^{1000}$ is
$\text{T}_{\text{r}+1}=^{1000}\text{C}\text{r}\big(9^{\frac{1}{4}}\big)^{1000-\text{r}}+\big(8^{\frac{1}{6}}\big)^{\text{r}}$
$=^{1000}\text{C}_{\text{r}}3\frac{1000-\text{r}}{2}2\frac{1}{2}$
The above term will be rational if exponent of $3$ and $2$ are integers.
It means $\frac{1000-\text{r}}{2}$ and $\frac{\text{r}}{2}$ must be integers
The possible set of values of $r$ is $\{0, 2, 4, ...., 1000\}$
Hence, number of rational terms is $501.$
View full question & answer→MCQ 681 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 691 Mark
The total number of terms which are dependent on the value of $x$ in the expansion of $\Big(\text{x}^2-2+\frac{1}{\text{x}^{2}}\Big)^{\text{n}}$ is equal to:
Answer$\Big(\text{x}^2-2+\frac{1}{\text{x}^{2}}\Big)^{\text{n}}$
$=\Big(\big(\text{x}-\frac{1}{\text{x}}\big)^{2}\Big)^{\text{n}}$
$=\big(\text{x}-\frac{1}{\text{x}}\big)^{{2}{\text{n}}}$
$\text{T}_{\text{r}+1}={^{10}}\text{C}_{\text{r}}\text{x}^{2\text{n}-2\text{r}}$
For term independent of $x$
$2n - 2r = 0$
$n = r$
Hence there will be 1 term independent of $x.$
Since the total number of terms are $2n + 1.$
Hence the total number of term dependent on $x$ will be
Total number of terms $- ($total number of terms independent of $x)$.
$= 2n + 1 − 1$
$= 2n$
View full question & answer→MCQ 701 Mark
If the $4^{th}$ term in the binomial expansion of $(p + 1)n$ is $\frac{5}{2}$ then:
- A
$\text{n}=8, \text{p}=6$
- B
$\text{n}=8, \text{p}=\frac{1}{2}$
- ✓
$\text{n}=6, \text{p}=\frac{1}{2}$
- D
$\text{n}=6, \text{p}=6$
AnswerCorrect option: C. $\text{n}=6, \text{p}=\frac{1}{2}$
View full question & answer→MCQ 711 Mark
Find the sum of coefficient of middle terms of the expansion $\Big(3\text{x}-\frac{\text{x}^3}{6}\Big)^7:$
- ✓
$\frac{595}{48}$
- B
$-\frac{595}{48}$
- C
$-\frac{595}{24}$
- D
AnswerCorrect option: A. $\frac{595}{48}$
Total number of terms are $8.$
So, middle term will be the $4^{th}$ and $5^{th}$ term.
$\therefore\text{t}_{3+1}=^7\text{C}_3(-1)^{3}(3\text{x})^{7-3}\big(\frac{\text{x}^3}{6}\big)^3=-\frac{105\text{x}^{13}}{8}$
$\therefore\text{t}_{4+1}=^7\text{C}_4(-1)^{4}(3\text{x})^{7-4}\big(\frac{\text{x}^3}{6}\big)^4=\frac{35\text{x}^{15}}{48}$
So, $-\frac{105}{8}+\frac{35}{48}=-\frac{595}{48}$
View full question & answer→MCQ 721 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 731 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 741 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 751 Mark
In the expansion of $\big(\frac{\text{x}+2}{\text{x}^2}\big)15,$ the term independent of $x$ is:
- A
$15C_6.26$
- ✓
$15C_5.25$
- C
$15C_4.24$
- D
AnswerCorrect option: B. $15C_5.25$
View full question & answer→MCQ 761 Mark
${ }^{15} C_3+{ }^{15} C_5+\ldots .+{ }^{15} C_{15}$ will be equal to:
- A
$2^{14}$
- ✓
$2^{14}- 15$
- C
$2^{14}+ 15$
- D
$2^{14}− 1$
AnswerCorrect option: B. $2^{14}- 15$
We know
$ { }^{15} C_1+{ }^{15} C_3+{ }^{15} C_5+\ldots{ }^{15} C_{15}=2^{15-1} $
$ \therefore{ }^{15} C_3+{ }^{15} C_5+\ldots+{ }^{15} C_{15}=2^{14}-15 $
View full question & answer→MCQ 771 Mark
The coefficients of the expansions are arranged in an array. This array is called $………$
View full question & answer→MCQ 781 Mark
The number of terms in the expansion of $[(a + 4b)^3(a - 4b)^3]^2$ are:
Answer$[(a + 4b)^3(a - 4b)^3]^2$
$ = [(a + 4b)(a - 4b)]^6$
$ = [a^2- 16b^2]^6$
Hence total number of terms is $n + 1$
Here $n = 6$
Therefore, total number of terms is $7.$
View full question & answer→MCQ 791 Mark
Expand the following binomials: $(x - 3)^5$
- A
$ x^5+25 x^4+90 x^3-270 x^2+405 x-243 $
- B
$ x^5-15 x^4+90 x^3-270 x^2-405 x-243 $
- C
$ x^5-15 x^4+80 x^3-270 x^2+405 x-243 $
- ✓
$ x^5-15 x^4+90 x^3-270 x^2+405 x-243 $
AnswerCorrect option: D. $ x^5-15 x^4+90 x^3-270 x^2+405 x-243 $
$(x-3)^5={ }^5 C_0 x^5+{ }^5 C_1 x^4(-3)^1+{ }^5 C_2 x^3(-3)^2+{ }^5 C_3 x^2(-3)^3+{ }^5 C_4 x(-3)^4+{ }^5 C_5(-3)^5$
$=x^5-15 x^4+90 x^3-270 x^2+405 x-243$
View full question & answer→MCQ 801 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 811 Mark
The term independent of $x$ in the expansion of $\Big(2\text{x}+\frac{1}{3\times2}\Big)^9.$
- A
$2^{nd}$
- B
$3^{rd}$
- ✓
$4^{th}$
- D
$5^{th}$
AnswerCorrect option: C. $4^{th}$
View full question & answer→MCQ 821 Mark
The number of terms whose values depends on $x$ in the expansion of $\Big(\text{x}^{2}-2+\frac{1}{\text{x}^{2}}\Big)^{\text{n}}$ is:
Answer$\Big(\text{x}^{2}-2+\frac{1}{\text{x}^{2}}\Big)^{\text{n}}$
$=\Big[\big(\text{x}-\frac{1}{\text{x}}\big)^{2}\Big]^{\text{n}}$
$=\big(\text{x}-\frac{1}{\text{x}}\big)^{{2}\text{n}}$
Hence there will be $2n + 1$ terms.
The middle term i.e $n + 1^{th}$ term will be independent of $x.$
Hence total number of terms, dependent on $x$ will be
$2n + 1 - (1)$
$= 2n$ terms.
View full question & answer→MCQ 831 Mark
The sum of the coefficients of the first $10$ terms in the expansion of $(1 - x)^{-3}$
AnswerFor $(1 - x)^{-3}$, the sum of the first $r$ terms will be
$^{n+r-1}C_n$
Replacing $n = 3$ and $r = 10$ in the above formula, we get
$^{3+10-1}C_n$
$= ^{12}C_3$
$=\frac{12.11.10}{3!}$
$=\frac{12.11.10}{6}$
$= 2.11.10$
$= 220$
View full question & answer→MCQ 841 Mark
The coefficient of $x^3$ in $\Big(\sqrt{\text{x}^5}+\frac{3}{\sqrt{\text{x}^{3}}}\Big)^5$ is:
Answer$\text{r}=\frac{6\times\frac{5}{2}-3}{\frac{5}{2}+\frac{3}{2}}=\frac{15-3}{4}=3$
$\therefore$ Coefficient of $x^3$ is ${ }^6 \mathrm{C}_3 3^3$
$=\frac{6\times5\times4}{3\times2\times1}\cdot27$
$= 5 \times 4 \times 27$
$= 540$
View full question & answer→MCQ 851 Mark
What is the approximate value of $(1.02)8?$
- ✓
$1.171$
- B
$1.175$
- C
$1.177$
- D
$1.179$
AnswerCorrect option: A. $1.171$
View full question & answer→MCQ 861 Mark
The sum of the coefficients in the expansion of $(1 + 5x - 7x^3)^{3165}$ is:
- A
$1$
- B
$2^{3165}$
- C
$2^{3164}$
- ✓
AnswerTo get sum of coefficient put $x = 1$
Hence sum of the coefficients in the expansion of $(1 + 5x - 7x^3)^{3165}$ is,
$ = (1 + 5 - 7)^{3165}$
$= (-1)⋅(−1)^{3164}$
$= -1$
View full question & answer→MCQ 871 Mark
Number of rational term is the expansion of $(7^{\frac{1}{3}}+1^{\frac{11}{9}})^{729}$
AnswerSince $7$ and $11$ are prime numbers,
hence application of general formula for number of rational terms will be
$=1+\frac{729}{\text{L}.\text{C}.\text{M}(1,3)}$
$=1+\frac{729}{9}$
$=1+81$
$=82$ rational terms.
View full question & answer→MCQ 881 Mark
The coefficient of $x^5$ in the expansion of $(1 + x^2)^5 (1 + x)^4$ is?
Answeraccording to binomial expansion,
$ \left(1+x^2\right)^5=5 C_0 \times 1+5 C_1\left(x^2\right)+5 C_2\left(x^2\right)^2+5 C_3\left(x^2\right)^3+5 C_4\left(x^2\right)^4+5 C_5\left(x_2\right)^5 $
$ =1+5 x^2+10 x^4+10 x^6+5 x^8+x^0 \Rightarrow(1) $
$ =(1+x)^4=1+4 C_1 x+4 C_2 x^2+4 C_3 x^3+4 C_4 x^4 \Rightarrow (2)$
from $1$ and $2$ we can see that coefficient of $5$ is not there
so $x^5$ is zero.
View full question & answer→MCQ 891 Mark
If the coefficient of $(2r + 4)^{th}$ term and $(r – 2)^{th}$ term in the expansion of $(1 + x)^{18}$ are equal, then $r$ is equal to:
View full question & answer→MCQ 901 Mark
In the expansion of $\Big(\text{x}-\frac{1}{\text{x}}\Big)^6,$ the constant term is:
AnswerIn expansion of $\Big(\text{x}-\frac{1}{\text{x}}\Big)^6,$ the constant term is ${^{6}}\text{C}_{\text{k}}\text{x}^{6-\text{k}}\big(\frac{-1}{\text{x}}\big)^{\text{k}}$
In constant term, power of $x$ must be zero
$\therefore 6 - k - k = 02k = 6.k = 3$
Hence, the constant term is $- ^6C_3 = -20.$
View full question & answer→MCQ 911 Mark
If $n$ is a positive integer, then the number of terms in the expansion of $(x + a)^n$ is:
AnswerCorrect option: B. $n + 1$
In binomial expansion the terms goes from $nC_0 x^n$ to $nC _na^n$ i.e the base of $C$ goes from $0$ to $n$ and this shows that there must be $(n + 1)$ terms.
View full question & answer→MCQ 921 Mark
The sum of the coefficients in the expansion of $(x + 2y + z)^{10}$ is:
- A
$2^{10}$
- ✓
$4^{10}$
- C
$3^{10}$
- D
AnswerCorrect option: B. $4^{10}$
Given expression is $(x + 2y + z)^{10}$ Substituting $x = y = z = 1,$ we get the sum of the coefficients as
$(1 + 2 + 1)^{10}$
$= 4^{10}$.
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Choose the correct answer.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:
Answer$(1+\text{x})^\text{n}=\ ^\text{n}\text{C}_0+\ ^\text{n}\text{C}_1\text{x}+\ ^\text{n}\text{C}_2\text{x}^2+\ ^\text{n}\text{C}_3\text{x}^3+...+\ ^\text{n}\text{C}_\text{n}\text{x}^\text{n}$
So, coefficients of $2^{\text {nd }}, 3^{\text {rd }}$ and $4^{th}$ terms are ${ }^n C_1,{ }^n C_2$ and ${ }^n C_3$, respectively.
Given that, ${ }^n C_1,{ }^n C_2$ and ${ }^n C_3$, are in $A.P.$
$\therefore2\ ^\text{n}\text{C}_2=\ ^\text{n}\text{C}_1+\ ^\text{n}\text{C}_3$
$\Rightarrow2\Big[\frac{\text{n}!}{(\text{n}-2)!2!}\Big]=\text{n}+\frac{\text{n}!}{3!(\text{n}-3)!}$
$\Rightarrow2\Big[\frac{\text{n}(\text{n}-1)}{2!}\Big]=\text{n}+\frac{\text{n}(\text{n}-1)(\text{n}-2)}{3!}$
$\Rightarrow(\text{n}-1)=1+\frac{(\text{n}-1)(\text{n}-2)}{6}$
$\Rightarrow6\text{n}-6=6+\text{n}^2-3\text{n}+2$
$\Rightarrow\text{n}^2-9\text{n}+14=0$
$\Rightarrow(\text{n}-7)(\text{n}-2)=0$
$\therefore\text{n}=2$ or $\text{n}=7$
Since $n = 2$ is not possible, so $n = 7.$
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The coefficient of $x^8y^{10}$ 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 ${^\text{18}}\text{C}_{\text{8}}$
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If $ \text{z}=\Big(\frac{\sqrt{3}}{2}+\frac{\text{i}}{2}\Big)5+\Big(\frac{\sqrt{3}}{2}-\frac{\text{i}}{2}\Big)5,$ then:
- A
$\text{Re (z) = 0}$
- ✓
$\text{Im (z) = 0}$
- C
$\text{Re (z) > 0, Im (z) > 0}$
- D
$\text{Re (z) > 0, Im (z) < 0}$
AnswerCorrect option: B. $\text{Im (z) = 0}$
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The number of terms in the expression of $(x + y)^{n-1}$ is $2018$ then $n.$
AnswerCorrect option: A. $2018$
Here, the no. of terms in binomial expansion of $(x + y)^{n-1}$ is $(n - 1) + 1$
i.e; one more the exponent,
$\Rightarrow (n - 1) + 1 = 2018$
$\Rightarrow n = 2018$
View full question & answer→MCQ 971 Mark
The coefficient of $x^{18}$ in the product $(1 + x)(1 - x)^{10} (1 + x + x^2)^9$ is?
Answer$(1+x)(1-x)^{10}\left(1+x+x^2\right)^9 $
$\left(1-x^2\right)\left(1-x^3\right)^9 $
${ }^9 C_6=84 $
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If coefficient of $x^{100}$ in $1 + (1 + x) (1 + x)^2+$ ..... $+ (1 + x)^n ($if $n ≥ 100)$ is $\text{C}^{201}_{101}$ then the value of $n$ equals.
Answer${ }^n C_r+{ }^n C_{(r+1)}={ }^{(n+1)} C_{(r+1)}$
coefficient of $x^{100}$ is ${ }^{100} \mathrm{C}_{100}+{ }^{101} \mathrm{C}_{100}+{ }^{102} \mathrm{C}_{100}+\ldots \ldots .+{ }^{\mathrm{n}} \mathrm{C}_{100}$
Which is equal to ${ }^{(n+1)} C_{101}$.
Therefore, $n + 1 = 201$
Which implies $n = 200$
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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 1001 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 $= {^\text{14}}\text{C}_{\text{7}}\ \text{a}^{14-7}\ \text{b}^{7}=\frac{14!}{(7)!}\ \text{a}^{7}\ \text{b}^{7}$
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