Mathematics Class XI

Question 1 of 20

1. Question
1 point(s)
$\sum\limits_{r = 0}^n {{}^n{C_r}{4^r}} = $
- ${4^n}$
- ${5^n}$
- ${4^{ – n}}$
- ${5^{ – n}}$
Correct

Incorrect

Hint

${\left( {1 + x} \right)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}{x^r}}$ putting x=4
${\left( {1 + 4} \right)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}{4^r}}$
$\Rightarrow \sum\limits_{r = 0}^n {{}^n{C_r}{4^r}} = {5^n}$
Question 2 of 20

2. Question
1 point(s)
${(1.1)^{10000}}$ is ….${2^{10}}$
- greater than
- smaller than
- Equal to
- None of these
Correct

Incorrect

Hint

${(1.1)^{10000}} = {(1 + 0.1)^{10000}}$ $=^{10000}{C_0} + {}^{10000}{C_1} \times (0.1){ + ^{10000}}{C_2} \times {(0.1)^2}+ \text {other the terms }$
$= 1 + 10000 \times \frac{1}{{10}} + 10000 \times \frac{1}{{10}} +$ other the terms
=1+1000+100+other the terms
=1101+other the terms
$> {2^{10}} = 1024$
Question 3 of 20

3. Question
1 point(s)
${{{10}^{th}}}$ term in expansion of ${(2{x^2} + \frac{1}{{{x^2}}})^{25}}$ is….
- ${2^{16}}\frac{{(25!)}}{{(9!)(16!)}}{x^{14}}$
- ${2^{15}}\frac{{(25!)}}{{(8!)(17!)}}{x^{14}}$
- ${2^{15}}\frac{{(25!)}}{{(10!)(15!)}}{x^{15}}$
- None of these
Correct

Incorrect

Hint

${{T_{10}} = {T_{9 + 1}} = {}^{25}{C_9}{{(2{x^2})}^{25 - 9}}{{(\frac{1}{{{x^2}}})}^9}}$
$= {}^{25}{C_9}{(2{x^2})^{16}}{(\frac{1}{{{x^2}}})^9}$
$= \frac{{(25!)}}{{(9!)(16!)}}{2^{16}}.{x^{32}}.\frac{1}{{{x^{18}}}}$
$= {2^{16}}\frac{{(25!)}}{{(9!)(16!)}}{x^{14}}$
Question 4 of 20

4. Question
1 point(s)
If coefficient of ${x^7}$ in ${(a{x^2} + \frac{a}{{bx}})^{11}}$ equals the cofficient of ${x^{ – 7}}$ in ${(a{x^2} – \frac{1}{{bx}})^{11}}$ then
- $\dfrac{a}{b} = 1$
- $ab = 1$
- $a – b = 1$
- $a + b = 1$
Correct

Incorrect

Hint

${T_{r + 1}} = {}^n{C_r}{(a)^{n - r}}{b^r} = {}^{11}{C_r}{(a{x^2})^{11 - r}}{(\frac{a}{{bx}})^r}$
$= {}^{11}{C_r}{a^{11 - r}}\frac{{{a^r}}}{{{b^r}}}{(x)^{11 - 2r}}$
${x^7} = {x^{11 - 2r}}\Rightarrow =11-2r =7 \Rightarrow r=2$
$={T_3} = {}^{11}{C_2}{a^{11 - 2}}(\frac{{{a^2}}}{{{b^2}}}{(x)^7}) = {}^{11}{C_2}\frac{{{a^{11}}}}{{{b^2}}}{x^7}$
${T_{p + 1}} = {}^{11}{C_p}{\left( {ax} \right)^{11 - p}}{\left( { - \frac{1}{{b{x^2}}}} \right)^p} = {}^{11}{C_p}{a^{11 - p}}{\left( { - \frac{1}{b}} \right)^p}{\left( x \right)^{11 - 3p}}$
$\therefore {x^{ - 7}} = {x^{11 - 3p}} = 11 - 3p = - 7 = p = 6{}^{11}{C_2}\frac{{{a^{11}}}}{{{b^2}}}{x^7}$
$\therefore$ ${T_7} = {}^{11}{C_6}{(ax)^5}{( - \frac{1}{{bx}})^{11}} =$ ${}^{11}{C_6}\frac{{{a^5}}}{{{b^6}}}{x^7}$
$\therefore {}^{11}{C_2}\frac{{{a^{11}}}}{{{b^2}}} =$ ${}^{11}{C_6}\frac{{{a^5}}}{{{b^6}}} =$
$\dfrac{{10 \times 11}}{2}\frac{{{a^{11}}}}{{{b^2}}} = 462\frac{{{a^5}}}{{{b^6}}}$
$\begin{array}{l} = 55{a^6} = \frac{{462}}{{{b^4}}} = {a^6}{b^4} = \frac{{462}}{{55}} = 8.4\end{array}$
$\Rightarrow {a^2}{\left( {ab} \right)^4} = 8.4 \times 1 \Rightarrow {a^2} = 8.4\& {\left( {ab} \right)^4} = 1$
$\Rightarrow ab = 1$
Question 5 of 20

5. Question
1 point(s)
If ${(r + 1)^{th}}$ term in expansion of ${(2{x^2} + \frac{5}{{{x^2}}})^{10}}$ is constant then r =
- 6
- 7
- 5
- 8
Correct

Incorrect

Hint

${T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}$ $= {}^{10}{C_r}{(2x)^{10 - r}}{(\frac{5}{{{x^2}}})^r}$
$= {}^{10}{C_r}{2^{10 - r}}{.5^r}.{x^{20 - 2r - 2r}}$
$\therefore$ ${x^{20 - 4r}} = {x^0}$
20-4r=0
=4r=20
r=5
Question 6 of 20

6. Question
1 point(s)
If ${r^{{\text{th}}}}$ term in expansion of ${\left( {x + \frac{1}{{2x}}} \right)^{12}}$ is constant then r =
- 5
- 6
- 7
- 9
Correct

Incorrect

Hint

${T_r} = {T_{(r - 1) + 1}} = {}^n{C_{r - 1}}{a^{n - r + 1}}{b^{r - 1}}$
$= {}^{12}{C_{r - 1}}{x^{12 - r + 1}}{(\frac{1}{{2x}})^{r - 1}}$
$= {}^{12}{C_{r - 1}}{(\frac{1}{2})^{r - 1}}{(x)^{13 - r - }}^{r + 1}$
${}^{12}{C_{r - 1}}{(\frac{1}{2})^{r - 1}}{(x)^{14 - 2r}}$
$\therefore{x^{14 - 2r}} = {x^0} =14-2r=0$
2r=14
r=7
Question 7 of 20

7. Question
1 point(s)
Co-efficient of ${x^{ – 3}}$ in expansion of ${\left( {x – \dfrac{a}{x}} \right)^{11}}$ is
- $ – 792{a^2}$
- $-923{a^7}$
- $-792{a^6}$
- $-330{a^7}$
Correct

Incorrect

Hint

${T_n}{\left( {x - \frac{a}{x}} \right)^n}$ ,Here n=11
${{T_{r + 1}}{}^n{C_r}{\alpha ^{n - r}}{\beta ^r}}$ $= {}^{11}{C_r}{(x)^{11 - r}}{( - \dfrac{a}{x})^r}$
$= {}^{11}{C_r}{( - a)^r}{x^{11 - r - r}}$
$= {x^{11 - 2r}} = {x^{ - 3}}$
11-2r=-3
2r=14
r=7
$\therefore$ ${T_{7 + 1}} = {T_8} = {}^{11}{C_7}{( - a)^7}{x^{11 - 14}}$
coefficient of ${T_8}\frac{{(11!)}}{{(7!)(4!)}}{( - a)^7}$
$= - 330{a^7}$
Question 8 of 20

8. Question
1 point(s)
The ${{n^ \text {th}}}$ term in expansion of ${\left( {{x^3} – \frac{1}{{{x^3}}}} \right)^{12}}$ is constant term, then find r
- 4
- 8
- 7
- 6
Correct

Incorrect

Hint

${T_r} = {T_{(r - 1) + 1}}{ = {}^n{C_{r - 1}}{a^{n - r}}{b^r}}$
$= {}^{11}{C_{r - 1}}{\left( {{x^3}} \right)^{12 - r}}{\left( { - \frac{1}{{{x^3}}}} \right)^r}$
$= {}^{11}{C_{r - 1}}{\left( { - 1} \right)^{11}}{x^{36 - 3r - 3r}}$
$\therefore {x^{36 - 6r}} = {x^0} \Rightarrow 36 - 6r = 0$
$\Rightarrow 6r = 36 \Rightarrow r = 6$
Question 9 of 20

9. Question
1 point(s)
Find the coefficient of $ x^{-5}$ in expansion of ${\left( {{x^3} – \frac{2}{{{x^2}}}} \right)^7}$.
- 510
- 530
- 560
- 660
Correct

Incorrect

Hint

${T_{r + 1}} = {}^n{C_r}{a^{n - r}}{B^r} = {}^7{C_r}{\left( {{x^3}} \right)^{5 - r}}{\left( { - \frac{2}{{{x^2}}}} \right)^r}$
$= {}^7{C_r}{\left( { - 2} \right)^r}{\left( x \right)^{15 - 3r - 2r}}$
$\therefore {x^{15 - 5r}} = {x^{ - 5}}$
$\Rightarrow 15 - 5r = - 5 \Rightarrow 5r = 20 \Rightarrow r = 4$
$\therefore {\text{Coefficient of }}{T_5} = {}^7{C_4}{\left( { - 2} \right)^4}$
$= \frac{{7!}}{{\left( {3!} \right)\left( {4!} \right)}} \cdot 16 = 35 \times 16 = 560$
Question 10 of 20

10. Question
1 point(s)
Coefficient of $ x^{11}$ in the expansion of ${\left( {1 + {x^2}} \right)^4}{\left( {1 + {x^3}} \right)^7}{\left( {1 + {x^2}} \right)^{12}}$ is
- 1051
- 1106
- 1113
- 1120
Correct

Incorrect

Hint

${(1 + {x^m})^n} = \sum\limits_{r = 0}^n { = {}^n{C_r}{{({x^m})}^n}}$ $= \sum\limits_{r = 0}^n { = {}^n{C_r}({x^m}^n)}$
Coefficient of ${x^{11}}$ in the expansion of
${\left( {1 + {x^2}} \right)^4}{\left( {1 + {x^3}} \right)^7}{\left( {1 + {x^4}} \right)^{12}}$ is
$({}^4{C_0}.{}^7{C_1}.{}^{12}{C_2})$ + $({}^4{C_1}{}^7{C_3}.{}^{12}{C_0})$ + $({}^4{C_2}{}^7{C_3}.{}^{12}{C_1})$ + $({}^4{C_2}{}^7{C_3}{}^{12}{C_1})$
= $(1\cdot7 \cdot 66)+(4\cdot 35\cdot 1)+(6\cdot 7\cdot 12)+(1\cdot 7)$
462+140+504+7=1113
Question 11 of 20

11. Question
1 point(s)
For r=0,1,…,10. Let ${A_r},{B_r}$ and ${C_r}$ denote respectively, the coeffiecent of ${x^r}$ in expansion of ${\left( {1 + x} \right)^{10}},{\left( {1 + x} \right)^{20}}$ and ${\left( {1 + x} \right)^{30}}$. Then $\sum\limits_{r = 1}^n {{A_r}({B_{10}}{B_r} – {C_{10}}{A_r})} $ is equal to
- ${{B_{10}} – {C_{10}}}$
- ${{A_{10}}({B^2}_{10} – {C_{10}}{A_{10}})}$
- 0
- ${{C_{10}} – {B_{10}}}$
Correct

Incorrect

Hint

$\sum\limits_{r - 1}^{10} {{A_r}{B_r}}$ = coeffecient of ${x^20}$ in the expansion of $\{ {\left( {1 + x} \right)^{10}}{\left( {x + 1} \right)^{20}}\} - 1$ $={{C_{20}} - 1 = {C_{10}} - 1}$ and $\sum\limits_{r - 1}^{10} {{{({A_r})}^2}}$ = coffiecient of ${x^{10}}$ in $\{ {\left( {1 + x} \right)^{10}}{\left( {x + 1} \right)^{20}}\} - 1$
${ = {b_{10}} - 1}$
We have , ${{A_r}({B_{10}}{B_r}) - {C_{10}}{A_r}}$ ${ = {B_{10}}{A_r}{B_r} - {C_{10}}{{({A_r})}^2}}$
${ = ({B_{10}}({C_{10}} - 1) - {C_{10}}({B_{10}} - 1)}$
${ = {C_{10}} - {B_{10}}}$
Question 12 of 20

12. Question
1 point(s)
The Expression ${\{ x + {\left( {{x^3} – 1} \right)^{1/2}}\} ^5} + {\{ x – {\left( {{x^3} – 1} \right)^{1/2}}\} ^5}$ is a polynomial of degree.
- 5
- 6
- 7
- 8
Correct

Incorrect

Hint

${\{ x + {\left( {{x^3} - 1} \right)^{1/2}}\} ^5} + {\{ x - {\left( {{x^3} - 1} \right)^{1/2}}\} ^5}$
$=2\{ {x^5} + 10{x^3}({x^3} - 1) + 5x{\left( {{x^3} - 1} \right)^2}\}$
The degree of the polynomial is 7.
Since degree of $5x{\left( {{x^3} - 1} \right)^2}$ is 7.
Question 13 of 20

13. Question
1 point(s)
The sum $\sum\limits_{i = 1}^m {\left( \begin{array}10\\i\end{array} \right)} \left( \begin{array}20\\m – i\end{array} \right)$, where $ \dfrac pq =0 $ of p>q is maximum when m is
- 5
- 10
- 15
- 20
Correct

Incorrect

Hint

$\sum\limits_{i = 1}^m {\left( \begin{array}{l}10\\i\end{array} \right)} \left( \begin{array}{l}20\\m - i\end{array} \right)$ is the coefficient of $x^m$ in the expansion of ${\left( {1 + x} \right)^{10}}{\left( {x + 1} \right)^{20}}$ .
$\Rightarrow \sum\limits_{i = 1}^m {\left( \begin{array}{l}10\\i\end{array} \right)} \left( \begin{array}{l}20\\m - i\end{array} \right)=$ Coefficient of $x^m$ in expansion of ${\left( {x + 1} \right)^{30}}$ .
$\Rightarrow \sum\limits_{i = 1}^m {\left( \begin{array}{l}10\\i\end{array} \right)} \left( \begin{array}{l}20\\m - i\end{array} \right) = {}^{30}{C_m}$
${}^{30}{C_m}$ is maximum when $15 \cdot \dfrac {30}{2}$
Question 14 of 20

14. Question
1 point(s)
In expansion of ${\left( {a – b} \right)^n},n \ge 5,$ the sum of the ${5^\text {th}}$ and ${6^\text {th}}$ term is zero, then $\dfrac{a}{b}$ equals to
- $\dfrac{{n – 5}}{6}$
- $\dfrac{{n – 4}}{5}$
- $\dfrac{5}{{n – 4}}$
- $\dfrac{6}{{n – 5}}$
Correct

Incorrect

Hint

${T_5} = {T_{4 + 1}} = {}^n{C_4}{(a)^{n - 4}}{( - b)^5}$ $ = {}^n{C_4}{(a)^{n – 4}}{b^4}$
${T_6} = {T_{5 + 1}} = {}^n{C_5}{a^{n - 5}}{( - b)^5}$ $= - {}^n{C_5}{a^{n - 5}}{b^5}$
$\therefore$ ${T_5} + {T_6} = 0$
${}^n{C_4}{a^{n - 4}}{b^4} - {}^n{C_5}{a^{n - 5}}{b^5} = 0$
$= \frac{{{a^{n - 4}}{b^4}}}{{{a^{n - 5}}{b^5}}} = \frac{{{}^n{C_5}}}{{{}^n{C_4}}}$
$\frac{a}{b} = (\frac{{n!}}{{(n - 5!)5!}})(\frac{{4!\left( {n - 4} \right)}}{{n!}}) = \frac{{n - 4}}{5}$
Question 15 of 20

15. Question
1 point(s)
In the expansion of ${\left( {1 + x} \right)^m}{\left( {1 – x} \right)^n}$, the coefficient of x and $ x^2 $ are 3 and -6 respectively, then m equals to
- 6
- 9
- 12
- 24
Correct

Incorrect

Hint

${\left( {1 + x} \right)^m}{\left( {1 - x} \right)^n} = \left[ {1 + mx + \dfrac{{m\left( {m - 1} \right)}}{2}{x^2} + \cdots } \right]\left[ {1 - nx + \dfrac{{n\left( {n - 1} \right)}}{2}{x^2} + \cdots } \right]$
$= 1 + \left( {m - n} \right)x + \left[ {\dfrac{{m\left( {m - 1} \right)}}{2} + \dfrac{{n\left( {n - 1} \right)}}{2} - mn} \right]{x^2} + \cdots$
$\therefore m - n = 3$
$\dfrac{{m\left( {m - 1} \right)}}{2} + \dfrac{{n\left( {n - 1} \right)}}{2} - mn = - 6$
$\Rightarrow {m^2} - m + {n^2} - n - 2mn = - 12 \Rightarrow {\left( {m - n} \right)^2} - \left( {m + n} \right) = - 12$
$\Rightarrow m + n = 9 + 12 = 21$
$\therefore 2m = 21 + 3 = 24$
$\Rightarrow m = 12$
Question 16 of 20

16. Question
1 point(s)
If ${{C_r}}$ stands for ${}^n{C_r}$ then the sum of the series ${\frac{{2(\frac{n}{2})!(\frac{n}{2})!}}{{n!}}}$${[C_0^2 – 2C_1^2 + ….. + {{\left( { – 1} \right)}^n}\left( {n + 1} \right)C_n^2]}$ where n is an ever positive integer, is _______
- 0
- ${\left( { – 1} \right)^{n/2}}(n + 1)$
- ${\left( { – 1} \right)^n}(n + 2)$
- None of these
Correct

Incorrect

Hint

n=2k,k $\in$ N
$f\left( k \right) = (\frac{{2\left( {k!} \right)\left( {k!} \right)}}{{(2k)!}})$ ${[C_0^2 - 2C_1^2 + 3C_2^2 + .... + {{\left( { - 1} \right)}^{2k}}{}^{(2k + 1)}C_{2k}^2]}$
$f\left( k \right) = (\frac{{2\left( {k!} \right)\left( {k!} \right)}}{{(2k)!}})$ ${[{}^{\left( {2k + 1} \right)}C_0^2 – 2C_1^2 + …. + C_{2k}^2]}$
$f\left( k \right) = (\frac{{2\left( {k!} \right)\left( {k!} \right)}}{{(2k)!}})$ ${[\left( {2k + 1} \right)C_0^2 - 2kC_1^2 + .... + C_{2k}^2]}$
$2f\left( k \right) = \frac{2}{{{}^{2k}{C_k}}}2\left( {k + 1} \right){}^{2k}{C_k}$
$\therefore$ $f\left( k \right) = {\left( { - 1} \right)^{n/2}}\left( {\frac{n}{2} + 1} \right).2$
${ = {{\left( { - 1} \right)}^{n/2}}\left( {n + 2} \right){}^{(2k + 1)}}$
Question 17 of 20

17. Question
1 point(s)
The coefficient of $ x^4 $ in ${\left( {\frac{x}{2} – \frac{3}{{{x^2}}}} \right)^{10}}$ is
- $\dfrac{{405}}{{256}}$
- $\dfrac{{504}}{{259}}$
- $\dfrac{{450}}{{263}}$
- $\dfrac{{540}}{{273}}$
Correct

Incorrect

Hint

${T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}$
$= {}^{10}{C_r}{\frac{x}{2}^{10 - r}}{\left( { - \fra c{3}{{{x^2}}}} \right)^r}$
$= {}^{10}{C_r}\frac{{ - {3^r}}}{{{2^{10 - r}}}}{x^{10 - r - 2r}}$
$\therefore$ ${x^{10 - r - 2r}} = {x^4} \Rightarrow 10 - 3x = 4 \Rightarrow r = 2$
cofficient of ${{x^4}}$ $= {}^{10}{C_2}\frac{{{{( - 3)}^2}}}{{{2^8}}}$
$\dfrac{{9 \times 10}}{2} \times \dfrac{9}{{256}} = \dfrac{{405}}{{256}}$
Question 18 of 20

18. Question
1 point(s)
The cofficient of ${\left( {3r} \right)^{th}}$ and ${(r + 2)^{th}}$ terms in the expansion of ${\left( {1 + x} \right)^{2n}}$ are equal, then (where r>1, n>1)
- n=2r
- n=2r+1
- n=3r
- n=3r-1
Correct

Incorrect

Hint

${T_{r + 1}} = {}^{2n}{C_{r - 1}}{x^r}$
${T_r} = {}^{2n}{C_{r - 1}}{x^{r - 1}}$
${T_{3r}} = {}^{2n}{C_{3r - 1}}{x^{3r - 1}}$
${T_{r + 2}} = {}^{2n}{C_{r + 1}}{x^{r + 1}}$
$\therefore$ ${}^{2n}{C_{3r - 1}}$ $= {}^{2n}{C_{r + 1}}$
$\Rightarrow$ 3r-1=r+1 or 2n=(3r-1)+(r+1)
$\Rightarrow$ 2r=2 or 2n=4r
$\Rightarrow$ r=1 or n=2r
But r>1. $\therefore n=2r$
Question 19 of 20

19. Question
1 point(s)
For a position integer n, Let $ a(n) = 1 + \frac{1}{2} + \frac{1}{3} + …… + \frac{1}{{{2^n} – 1}}$, then
- $a(100)$<100
- $a(100)>100$
- $a(200) \le 100$
- None of these
Correct

Incorrect

Hint

$a(n) = 1 + \frac{1}{2} + \frac{1}{3} + ...... + \frac{1}{{{2^n} - 1}}$
$\begin{array}{l}1 + (\frac{1}{2} + \frac{1}{3}) + (\frac{1}{4} + ...... + \frac{1}{7}) + ...... + (\frac{1}{{{2^n} - 1}} + .... + \frac{1}{{{2^n} - 1}})\\\therefore m - n = 3a(100) \le 100\end{array}$
$< 1 + \frac{2}{2} + \frac{4}{4} + ......... + \frac{{{2^n} - 1}}{{{2^n} - 1}} = n$ $\therefore a(100)<100$
Question 20 of 20

20. Question
1 point(s)
Let $x = {({}^{10}{C_1})^2} + 2{({}^{10}{C_2})^2} + …. + {}^{10}{({}^{10}{C_{10}})^2}$, where ${}^{10}{C_r}, r \in \{ 1,2,…,10\} $ denote binomial cofficients. Then, the value of $\dfrac{1}{{1430}}x$ is ____________
- 554
- 336
- 547
- 646
Correct

Incorrect

Hint

$x = \sum\limits_{r = 1}^{10} r {({}^{10}{C_r})^2}$ $\sum\limits_{r = 1}^{10} r{}^{10}{C_r}{}^{10}{C_r}$
$=\sum\limits_{r=1}^{10}{}^9{C_{r - 1}}{}^{10}{C_{10 - r}} = 10{}^{19}{C_9}$
$\therefore$ $\frac{x}{{1430}} = \frac{{10{}^{19}{C_9}}}{{1430}}$ $= \frac{{{}^{19}{C_9}}}{{143}} = 646$

Mathematics Class XI

Practice Questions 2-Binomial Theorem-Class XI

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