MCQ Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 8

$a+ ib= \sum_{k=1}^{101} i^k$ , then

$(a, b)$ equals

0%
$(0, 1)$
0%
$(1, 0)$
0%
$(0,- 1)$
0%
$(1, 1)$

Explanation

Given:-

$a + ib = \Sigma^{101}_{k = 1}{{i}^{k}} \longrightarrow \left( 1 \right)$

Solution:-

$\Sigma^{101}_{k = 1}{{i}^{k}} = i + {i}^{2} + {i}^{3} + ............... + {i}^{101}$

As te above series is in the form of G.P. with common ration

$(r)=i$ , first term

$\left( a \right) = i$ and no. of terms

$\left( n \right) = 101$

$\therefore \; {S}_{101} = \cfrac{ i \left( 1 - {i}^{101} \right)}{\left( 1 - i \right)} \; \left\{ \because {S}_{n} = \cfrac{a \left( 1 - {r}^{n} \right)}{\left( 1 - r \right)}\; \text{ for } r < 1 \right\}$

$\Rightarrow \; {S}_{n} = \cfrac{i \left( 1 - {i}^{4 \times 25}.i \right)}{\left( 1 - i \right)}$

$\Rightarrow \; {S}_{n} = \cfrac{ i \left( 1 - i \right)}{\left( 1 - i \right)} \; \left\{ \because {i}^{4n} = 1 \right\}$

$\Rightarrow \; {S}_{n} = i$

$\therefore \; \Sigma^{101}_{k = 1}{{i}^{k}} = i \longrightarrow \left( 2 \right)$

Now, from

${eq}^{n} \left( 1 \right) \& \left( 2 \right)$ , we have,

$a + ib = i$

$\Rightarrow \; a + ib = 0 + i.1$

On comparing real and imaginary parts, we have

$a = 0 \; \& \; b = 1$

Hence,

$\left( a, b \right) = \left( 0, 1 \right)$ .

$z = -3- i,$ find

$|z|$ .

0%
$\sqrt{10}$
0%
$\sqrt{9}$
0%
$\sqrt{8}$
0%
$\sqrt{7}$

Explanation

$Re(z)=-3\\Im(z)=-1\\ \bar{z}=-3+i\\|z|=3^2+1^2=10$

The real part of

$(1 - \cos\theta + 2i \sin\theta)^{-1}$ is:

0%
$\dfrac{2\sin \theta}{(1-\cos\theta)^2+4\sin^2 \theta}$
0%
$\dfrac{1+\cos \theta}{(1-\cos\theta)^2+4\sin^2 \theta}$
0%
$\cfrac { \left( 1-\cos { \theta } \right) -2i\sin { \theta } }{ { \left( 1-\cos { \theta } \right) }^{ 2 }+4{ i }^{ 2 }\sin ^{ 2 }{ \theta } }$
0%
$\dfrac{1-\cos \theta}{(1-\cos\theta)^2+4\sin^2 \theta}$

Explanation

$\Rightarrow \cfrac { 1 }{ \left( 1-\cos { \theta } \right) +2i\sin { \theta } } =\cfrac { \left( 1-\cos { \theta } \right) -2i\sin { \theta } }{ { \left( 1-\cos { \theta } \right) }^{ 2 }-4{ i }^{ 2 }\sin ^{ 2 }{ \theta } }$

$=\cfrac { \left( 1-\cos { \theta } \right) -2i\sin { \theta } }{ { \left( 1-\cos { \theta } \right) }^{ 2 }+4{ i }^{ 2 }\sin ^{ 2 }{ \theta } }$

Which of the following pair of numbers are

$\text{relatively prime}$ :-

0%
36 and 54
0%
52 and 78
0%
54 and 114
0%
59 and 61

Explanation

The question must be which of the following are relatively primes?

$(a)$ HCF

$\left(36,54\right)$

Factors of

$36=2\times 3\times 2\times 3$

Factors of

$54=2\times 3\times 3\times 3$

Common factor

$=2\times 3\times 3=18$

Hence HCF

$=18$

$\therefore\,36$ and

$54$ are not relatively primes

$(b)$ HCF $\left(52,78\right)$

Factors of $52=2\times 2\times 13$

Factors of $78=2\times 3\times 13$

Common factor $=2\times 13=26$

Hence HCF $=26$

$\therefore\,52$ and $78$ are not relatively primes

$(c)$ HCF $\left(54,114\right)$

Factors of $54=2\times 3\times 3\times 3$

Factors of $114=2\times 3\times 19$

Common factor $=2\times 3=6$

Hence HCF $=6$

$\therefore\,54$ and $114$ are not relatively primes

$(d)$ HCF

$\left(59,61\right)$

Factors of

$59=1\times 59$

Factors of

$61=1\times 61$

Common factor

$=1$

Hence HCF

$=1$

$\therefore\,59$ and

$61$ are relatively primes

$61-59=2$ they are also twin primes.

$\dfrac{{{{\left( {1 + i} \right)}^3}}}{{2 + i}}$ is equal to

0%
$\dfrac{2}{5} - \dfrac{6}{5}i$
0%
$0$
0%
$- \dfrac{1}{5} + \dfrac{6}{5}i$
0%
$- \dfrac{2}{5} + \dfrac{6}{5}i$

Explanation

$\Rightarrow \cfrac { { \left( 1+i \right) }^{ 3 } }{ 2+i } =\cfrac { { \left( 1+i \right) }^{ 2 }\left( 1+i \right) }{ 2+i } =\cfrac { \left( 1+{ i }^{ 2 }+2i \right) \left( 1+i \right) }{ \left( 2+i \right) } =\cfrac { 2i\left( 1+i \right) }{ 2+i }$

$\Rightarrow \cfrac { 2\left( i-1 \right) \left( 2-i \right) }{ \left( 2+i \right) \left( 2-i \right) } =\cfrac { 2\left( 2i-{ i }^{ 2 }-2+i \right) }{ 4-{ i }^{ 2 } } =\cfrac { -2 }{ 5 } +\cfrac { 6 }{ 5 } i$

Whether the statement is given true or false

Statement : The product of fifth roots of unity is 1.

0%
True
0%
False

The modulus of the complex number

$z=\dfrac{1-i}{3-4i}$ is

0%
$\dfrac{5}{\sqrt{2}}$
0%
$\dfrac{\sqrt{2}}{5}$
0%
$\sqrt{\dfrac{2}{5}}$
0%
none of these

Explanation

Now,

$z=\dfrac{1-i}{3-4i}$

or,

$z=\dfrac{(1-i)(3+4i)}{3^2+4^2}$

or,

$z=\dfrac{7+i}{25}$ .

Then

$|z|=\sqrt{\left(\dfrac{7}{25}\right)^2+\left(\dfrac{1}{25}\right)^2}=\dfrac{\sqrt{50}}{25}=\dfrac{\sqrt{2}}{5}$ .

Let z be a complex number such that

$z\in c\ R$ and

$\dfrac{1+z+z^2}{1-z+z^2}\in R$ , then

$|z|=3$ .

0%
True
0%
False

$z=\dfrac{1+i}{\sqrt{2}}$ , then the value of

$z^{1929}$ is

0%
$1+i$
0%
$-1$
0%
$\dfrac{1+i}{2}$
0%
$\dfrac{1+i}{\sqrt{2}}$

Explanation

Let

$z=\dfrac{1+i}{\sqrt{2}}$

$\Rightarrow |z|=\sqrt{\dfrac{1}{2}+\dfrac{1}{2}}=1$

$\Rightarrow Arg=tan^{-1}(1)=\dfrac{\pi}{4}$

$\therefore z^{1929}=(1\cdot e^{\dfrac{\pi i}{4}})^{1929}$

$=e^{1929 \times \dfrac{\pi i}{4}}$

$=e^{i\left(482\pi+\dfrac{\pi}{4}\right)}$

$=e^{i\dfrac{\pi}{4}}$

$=\dfrac{1}{\sqrt{2}}+i\dfrac{1}{\sqrt{2}}$

$\therefore z^{1929}=\dfrac{1+i}{\sqrt{2}}$

$|Z|=2,|z_{2}|=3,|z_{3}=4|$ and

$|z_{1}+z_{2}+z_{3}|=5$ then

$|4z_{2}z_{3}+9z_{3}z_{1}+16z_{1}z_{2}|=$

0%
$20$
0%
$24$
0%
$48$
0%
$120$

Explanation

Solution:- (D) 120

$\left| {z}_{1} \right| = 2, \left| {z}_{2} \right| = 3, \left| {z}_{3} \right| = 4, \left| {z}_{1} + {z}_{2} + {z}_{3} \right| = 5$

$\left| 4 {z}_{2} {z}_{3} + 9 {z}_{3} {z}_{1} + 16 {z}_{1} {z}_{2} \right| = \cfrac{\left| {z}_{1} \right| \left| {z}_{2} \right| \left| {z}_{3} \right|}{\left| {z}_{1} \right| \left| {z}_{2} \right| \left| {z}_{3} \right|} \left| 4 {z}_{2} {z}_{3} + 9 {z}_{3} {z}_{1} + 16 {z}_{1} {z}_{2} \right|$

$= \left| {z}_{1} \right| \left| {z}_{2} \right| \left| {z}_{3} \right| \left| \cfrac{4}{{z}_{1}} + \cfrac{9}{{z}_{2}} + \cfrac{16}{{z}_{3}} \right|$

$= \left( 2 \times 3 \times 4 \right) \left| \cfrac{4 \bar{{z}_{1}}}{{\left| {z}_{1} \right|}^{2}} + \cfrac{9 \bar{{z}_{2}}}{{\left| {z}_{2} \right|}^{2}} + \cfrac{16 \bar{{z}_{3}}}{{\left| {z}_{3} \right|}^{2}}\right|$

$= 24 \left| \cfrac{4 \bar{{z}_{1}}}{{2}^{2}} + \cfrac{9 \bar{{z}_{2}}}{{3}^{2}} + \cfrac{16 \bar{{z}_{3}}}{{4}^{2}} \right|$

$= 24 \left| \cfrac{4 \bar{{z}_{1}}}{4} + \cfrac{9 \bar{{z}_{2}}}{9} + \cfrac{16 \bar{{z}_{3}}}{16} \right|$

$= 24 \left| \bar{{z}_{1}} + \bar{{z}_{2}} + \bar{{z}_{3}} \right|$

$= 24 \left| \overline{{z}_{1} + {z}_{2} + {z}_{3}} \right|$

$= 24 \times 5$

$= 120$

Hence,

$\left| 4 {z}_{2} {z}_{3} + 9 {z}_{3} {z}_{1} + 16 {z}_{1} {z}_{2} \right| = 120$ .

$\dfrac{x - 3}{3 + i} + \dfrac{y - 3}{3 - i} = i$ where

$x , y \in R$ then

0%
$x = 2$ & $y = -8$
0%
$x = -2$ & $y = 8$
0%
$x = -2$ & $y = -6$
0%
$x = 2$ & $y = 8$

Explanation

$\dfrac{x-3}{3+i}+\dfrac{y-3}{3-i}=i$

$\dfrac{(x-3)(3-i)}{10}+\dfrac{(y-3)(3+i)}{10}=i$

$3(x-3)+3(y-3)-i((x-3+(3-y))=10i$

$3(x+y-6)-i(x-y)=10i$

compare real and imaginary point

$x+y=6$

$y-x=10$

$\Rightarrow y=8, x=-2$

The locus of

$z$ such that

$\left| {\dfrac{{z + i}}{{z - 1}}} \right| = 2$

0%
straight line
0%
circle with radius $2$
0%
circle with radius $\frac{{2\sqrt 2 }}{3}$
0%
none of these

Explanation

$\left|\dfrac{z+i}{z-1}\right|=2$

Put,

$z = x + iy$

$\therefore\,\left|\dfrac{x + iy+i}{x + iy-1}\right|=2$

$\Rightarrow\,\left|\dfrac{x + i\left(y+1\right)}{\left(x-1\right)+iy}\right|=2$

$\Rightarrow\,\left|x + i\left(y+1\right)\right|=2\left|\left(x-1\right)+iy\right|$

$\Rightarrow\,\sqrt{{x}^{2}+{\left(y+1\right)}^{2}}=2\sqrt{{\left(x-1\right)}^{2}+{y}^{2}}$

Squaring both sides, we get

$\Rightarrow\,{x}^{2}+{\left(y+1\right)}^{2}=4{\left(x-1\right)}^{2}+4{y}^{2}$

$\Rightarrow\,{x}^{2}-{\left(2x-2\right)}^{2}={\left(2y\right)}^{2}-{\left(y+1\right)}^{2}$

$\Rightarrow\,\left(x-2x+2\right)\left(x+2x-2\right)=\left(2y-y-1\right)\left(2y+y+1\right)$

$\Rightarrow\,\left(-x+2\right)\left(3x-2\right)=\left(y-1\right)\left(3y+1\right)$

$\Rightarrow\,-3{x}^{2}+2x+6x-4=3{y}^{2}+y-3y-1$

$\Rightarrow\,-3{x}^{2}+8x-4=3{y}^{2}-2y-1$

$\Rightarrow\,-3{x}^{2}+8x-4-3{y}^{2}+2y+1=0$

$\Rightarrow\,3{x}^{2}+3{y}^{2}-8x-2y+3=0$

$\Rightarrow\,{x}^{2}+{y}^{2}-\dfrac{8}{3}x-\dfrac{2}{3}y+1=0$ is of the form

${x}^{2}+{y}^{2}+2gx+2fy+c=0$

where

$g=\dfrac{4}{3},\,f=\dfrac{1}{3}$ and

$c=1$

Radius

$=r=\sqrt{{\left(\dfrac{4}{3}\right)}^{2}+{\left(\dfrac{1}{3}\right)}^{2}+1}=\sqrt{\dfrac{16}{9}+\dfrac{1}{9}-1}=\sqrt{\dfrac{17-9}{9}}=\dfrac{\sqrt{8}}{3}=\dfrac{2\sqrt{2}}{3}$

The locus of

$z$ is a circle with radius

$\dfrac{2\sqrt{2}}{3}$

$n\in N,\ { \left( \dfrac { 1+i }{ \sqrt { 2 } } \right) }^{ 8n }+{ \left( \dfrac { 1-i }{ \sqrt { 2 } } \right) }^{ 8n }=$

0%
$0$
0%
$1$
0%
$2$
0%
$-2$

Explanation

We have,

${{\left( \dfrac{1+i}{\sqrt{2}} \right)}^{8n}}+{{\left( \dfrac{1-i}{\sqrt{2}} \right)}^{8n}}$

${{\left[ {{\left( \dfrac{1+i}{\sqrt{2}} \right)}^{2}} \right]}^{4n}}+{{\left[ {{\left( \dfrac{1-i}{\sqrt{2}} \right)}^{2}} \right]}^{4n}}$

$={{\left[ \dfrac{{{1}^{2}}+{{i}^{2}}+2i}{2} \right]}^{4n}}+{{\left[ \dfrac{{{1}^{2}}+{{i}^{2}}-2i}{2} \right]}^{4n}}$

$={{\left[ \dfrac{1-1+2i}{2} \right]}^{4n}}+{{\left[ \dfrac{1-1-2i}{2} \right]}^{4n}}$

$={{\left( {{i}^{4}} \right)}^{n}}+{{\left( {{\left( -i \right)}^{4}} \right)}^{n}}$

$={{1}^{n}}+{{1}^{n}}$

$=1+1$

$=2$

Hence, this is the answer.

$\left(\dfrac{1 + i}{1 - i}\right)^4 + \left(\dfrac{1 - i}{1 + i}\right)^4 =$

0%
$0$
0%
$1$
0%
$2$
0%
$4$

Explanation

$\left( \dfrac{1+i}{1-i} \right)^{4}+ \left( \dfrac{1-i}{1+i} \right)^{4}$

$= \left( \dfrac{(1+i)^{2}}{1-i^{2}} \right)^{4}+ \left( \dfrac{(1-i)^{2}}{1-i^{2}} \right)^{4}$

$=\dfrac{1}{2^{4}} (1+i^{2} +2 i)^{4}+ \dfrac{1}{2^{4}} (1+i^{2}-2i)^{4}$

$=\dfrac{1}{2^{4}} (1-1+ 2i)^{4}+ \dfrac{1}{2^{4}} (1-1-2i)^{4}$

$=\dfrac{2^{4}}{2^{4}} (i^{4})+ \dfrac{2^{4}}{2^{4}} (i)^{4}$

$= 1+1 =2$

The complex number

$z$ satisfies

$z+|z|=2+8i$ . The value of

$|z|$ is

0%
$10$
0%
$13$
0%
$17$
0%
$23$

Explanation

$Assume,z=a+ib,where \ i=\sqrt { -1 }$

$Then,z+|z|=a+ib+\sqrt { { a }^{ 2 }+{ b }^{ 2 } } =2+8i$

$Thus,b=8.$

$And\quad a+\sqrt { { a }^{ 2 }+64 } =2$

${ a }^{ 2 }+64={ (2−a) }^{ 2 }$

${ a }^{ 2 }+64=4−4a+{ a }^{ 2 }$

$a=−15$

$Thus\quad z=−15+8i$

$Thus|z|=\sqrt { { (-15) }^{ 2 }+{ 8 }^{ 2 } } =17$

$|z_1+z_2|=|z_1|+|z_2|$ where

$z_1$ and

$z_2$ are different non - zero complex number, then ?

0%
$Re\left(\dfrac{z_1}{z_2}\right)=0$
0%
$Im\left(\dfrac{z_1}{z_2}\right)=0$
0%
$z_1+z_2=0$
0%
None

Explanation

$let\quad { z }_{ 1 }=a+ib\quad \quad and\quad \quad { z }_{ 2 }=c+id\\ \Rightarrow \quad \left| \quad { z }_{ 1 }+\quad { z }_{ 2 } \right| \quad \quad \quad \quad =\quad \quad \quad \left| \quad { z }_{ 1 } \right| +\left| \quad { z }_{ 2 } \right| \\ \Rightarrow \quad \quad \left| a+ib+c+id \right| \quad =\quad \quad \left| a+ib \right| +\left| c+id \right| \\ \Rightarrow \left| a+c+i\left( b+d \right) \right| \quad \quad \quad =\quad \quad \quad \left| a+ib \right| +\left| c+id \right| \\ \Rightarrow \sqrt { \left( a+c \right) ^{ 2 }+\left( b+d \right) ^{ 2 } } =\quad \quad \sqrt { { a }^{ 2 }+{ b }^{ 2 } } +\sqrt { { c }^{ 2 }+d^{ 2 } } \\ \quad \quad \quad \quad squaring\quad both\quad sides\\ \Rightarrow \left( a+c \right) ^{ 2 }+\left( b+d \right) ^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+d^{ 2 }+2\sqrt { { a }^{ 2 }+{ b }^{ 2 } } .\sqrt { { c }^{ 2 }+d^{ 2 } } \\ \Rightarrow 2ac+2bd\quad \quad =\quad \quad 2\sqrt { { a }^{ 2 }+{ b }^{ 2 } } .\sqrt { { c }^{ 2 }+d^{ 2 } } \\ \Rightarrow ac+bd\quad \quad =\quad \quad \sqrt { { a }^{ 2 }+{ b }^{ 2 } } .\sqrt { { c }^{ 2 }+d^{ 2 } } \\ \quad \quad \quad again\quad squaring\quad both\quad sides\\ \Rightarrow { \left( ac \right) }^{ 2 }+{ \left( bd \right) }^{ 2 }+2abcd\quad =\quad \left( { a }^{ 2 }+{ b }^{ 2 } \right) \left( { c }^{ 2 }+d^{ 2 } \right) \\ \Rightarrow { \left( ac \right) }^{ 2 }+{ \left( bd \right) }^{ 2 }+2abcd\quad =\quad { \left( ac \right) }^{ 2 }+{ \left( ad \right) }^{ 2 }+{ \left( bc \right) }^{ 2 }+{ \left( bd \right) }^{ 2 }\\ \Rightarrow { \left( ad \right) }^{ 2 }+{ \left( bc \right) }^{ 2 }-2abcd\quad =\quad 0\\ \Rightarrow \left( ad-bc \right) ^{ 2 }\quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 0\\ \Rightarrow ad-bc\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 0\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ad\quad \quad =\quad \quad \quad bc\quad \quad \quad \quad \longrightarrow \left( 1 \right) \\ Now\quad \quad \dfrac { { z }_{ 1 } }{ { z }_{ 2 } } \quad =\quad \dfrac { a+ib }{ c+id } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \dfrac { \left( a+ib \right) \left( c-id \right) }{ \left( c+id \right) \left( c-id \right) } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \dfrac { ac-iad+ibc-{ i }^{ 2 }bd }{ { c }^{ 2 }+{ d }^{ 2 } } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \dfrac { ac+bd+i\left( bc-ad \right) }{ { c }^{ 2 }+{ d }^{ 2 } } \quad \quad \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \dfrac { ac+bd }{ { c }^{ 2 }+{ d }^{ 2 } } \quad +\quad 0\quad \quad \quad using\quad equation\quad \left( 1 \right) \\ Imaginary\quad part\quad is\quad zero\quad as\quad there\quad is\quad only\quad real\quad part\\ \therefore \quad Im\left( \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right) =\quad 0$

$z$ be a complex number and

$\,\,\left| {\,z + 3\,} \right| \leqslant \,\,8$ then the value of

$\left| {\,z - 2\,} \right|$ lies in

0%
$[-2,13]$
0%
$[0,13]$
0%
$[2,13]$
0%
$[-13,2]$

Explanation

$|z+3|\le 8$

$-8\le z+3\le 8$

$-4\le z \le 5$

$-13\le z-2 \le 3$

$|z-2|\in [0, 13]$ . because it can attain value from

$13$ to

$0$ which will be taken as absolute values

$B$ is correct.

The number of prime numbers between

$1 \ and \ 10$ is

0%
$12$
0%
$4$
0%
$3$
0%
$2$

Explanation

The prime numbers between

$1$ to

$10$ is,

$2,3,5,7,$

There are four prime numbers in between

$1$ to

$10$ .

$z_1=3+4i$ and

$Im(z_1z_2)=0$ Find

$z_2$

0%
$z_2=3-4i$
0%
$z_2=3+4i$
0%
$z_2=3\pm 4i$
0%
None of these

Explanation

Let

$z_2=a+ib$

$z_1z_2=(3+4i)(a+ib)$

$=(3a-4b)+i(4a+3b)$

$Im(z_1z_2)=0$

$\Rightarrow 4a+3b=0$

$\Rightarrow 4a=-3b$

$\dfrac ab=\dfrac 3{-4}$

$\implies a=3$

$b=-4$

$z_2=3-4i$

Modulus of

$\dfrac{\cos \theta - i\sin \theta}{\sin \theta - i \cos \theta}$ is

0%
$0$
0%
$2\theta$
0%
$\pi - 2\theta$
0%
None of these

Explanation

$\cfrac { \cos { \theta } -i\sin { \theta } }{ \sin { \theta } -i\cos { \theta } }$

$\cfrac { \cos { \theta } -i\sin { \theta } }{ \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } } \left( \sin { \theta } +i\cos { \theta } \right) \Rightarrow \left( \cos { \theta } -i\sin { \theta } \right) \left( \sin { \theta } +i\cos { \theta } \right)$

$\Rightarrow \cos { \theta } .\sin { \theta } +\sin { \theta } .\cos { \theta } +i\cos ^{ 2 }{ \theta } -i\sin ^{ 2 }{ \theta } \Rightarrow 2\sin { \theta } .\cos { \theta } +i\left( \cos ^{ 2 }{ \theta } -\sin ^{ 2 }{ \theta } \right)$

$2\sin { \theta } .\cos { \theta } =\sin { 2\theta } ;\cos ^{ 2 }{ \theta } -\sin ^{ 2 }{ \theta } =\cos { 2\theta } \Rightarrow \sin { 2\theta } +i\cos { 2\theta }$

modulus

$=\sqrt { \sin ^{ 2 }{ 2\theta } +\cos ^{ 2 }{ 2\theta } } =\sqrt { 1 } =1$

none of these

The value of $\sum\limits_{n = 1}^{13} {\left( {{i^n} + {i^{n + 1}}} \right)}$ , where $i = \sqrt { - 1}$ equals:

0%
$i$
0%
$i - 1$
0%
$- i$
0%
0

$\left(\dfrac{1+\cos \dfrac{\pi}{8}+i\sin \dfrac{\pi}{8}}{1+\cos \dfrac{\pi}{8}-i\sin \dfrac{\pi}{8}}\right)^{8}=$ ?

0%
$1+i$
0%
$1-i$
0%
$1$
0%
$-1$

$3+2\ i\ \sin \theta$ will be real, if

$\theta=$

0%
$2n \pi$
0%
$n \pi +\pi/2$
0%
$n\pi$
0%
$none\ of\ these$

Explanation

$3+2i\sin \theta$ will be real if imaginary term is zero

i.e

$2\sin\theta=0$

$\Rightarrow \sin \theta=0$

$\Rightarrow \theta=n\pi$

Let

$z_{r}(1 \le r \le 4)$ be complex numbers such that

$|z_{r}|=\sqrt {r+1} and |30\ z_{1}+20\ z_{2}+15 z_{3}+12\ z_{4}|=k|z_{1}z_{2}z_{3}+z_{2}z_{3}z_{4}+z_{3}z_{4}z_{1}+z_{4}z_{1}z_{2}|$ . Then value of

$k$ equals ?

0%
$|z_{1}z_{2}z_{3}|$
0%
$|z_{2}z_{3}z_{4}|$
0%
$|z_{3}z_{4}z_{1}|$
0%
$|z_{4}z_{1}z_{2}|$

Explanation

$|z_1| = \sqrt 2, \ |z_2| = \sqrt 3, \ |z_3| = \sqrt 4, \ |z_4| = \sqrt 5$

$| 30z_1 + 20 z_2 + 15 z_3 + 12 z _4 | = k | z_1z_2z_3 + z_2z_3z_4+z_3z_4z_1+z_4z_1z_2|$

$\overline{|30z_1 + 20 z _2 + 15 z_3 + 12 z_4|} =k | z_1z_2z_3+z_2z_3z_4+z_3z_4z_1+z_4z_1z_2|$

$| 30 \bar {z_1} + 20 \bar {z_2} + 15 \bar {z_3} + 12 \bar {z_4} | = k|z_1z_2z_3+z_2z_3z_4+z_3z_4z_1+z_4z_1z_2$

$z_1 \bar{z_1} = 2, \ z_2 \bar { z_2} = 3 , \ z_3 \bar { z_3} = 4, \ z_4\bar{z_4} = 5$

$60 \left | \dfrac{ 1}{z_1} + \dfrac{1}{z_2} + \dfrac{1}{z_3} + \dfrac{1}{z_4} \right | = k |z_1z_2z_3+z_2z_3z_4+z_3z_4z_1+z_4z_1z_2|$

$k = |z_4z_1z_2|$

For a complex number $z$ , the minimum value of $\left| z \right| + \left| {z - 1} \right|$ is

0%
1
0%
2
0%
3
0%
none of these

$\left| {{z_1}} \right| = = 1,\left| {{z_2}} \right| = 2,$ , then the value of

${\left| {{z_1} + {z_2}} \right|^2} + {\left| {{z_1} - {z_2}} \right|^2}$ is equal to

0%
$2$
0%
$3$
0%
$4$
0%
none of these

Explanation

$| z _1 + z _2 | ^2 + |z _1-z_2|^2 = (z _1+z_2 ) ( \bar{z_1} + \bar {z_2}) + (z_1-z_2)(\bar{z_1} - \bar {z_2})$

$\Rightarrow 2z_1 \bar{z_1} + 2 z_2 \bar {z_2} = 2 ( |z_1|^2 + |z_2|^2) = 2 ( 1 + 4 ) = 10$

$z_{1},\ z_{2}$ are two complex numbers such that

$arg\left( { z }_{ 1 }+{ z }_{ 2 } \right) =0$ and

$Im\left( { z }_{ 1 }{ z }_{ 2 } \right) =0$ , then

0%
$z_{1}=-z_{2}$
0%
$z_{1}=z_{2}$
0%
$z_{1}=\bar { { z }_{ 2 } }$
0%
$none\ of\ these$

Explanation

Let the complex numbers be,

$z_1=a+ib,z_2=x+iy$

$arg(z_1+z_2)=arg(a+x+i(b+y))=0$

$\tan^{-1}\left ( \dfrac{b+y}{a+x} \right )=0$

$\dfrac{b+y}{a+x}=0$

$b+y=0$ .....(1)

$Im(z_1z_2)=Im(ax-by+i(ay+bx))=0$

$(ay+bx)=0$

from (1),

$ay=xy\Rightarrow a=x$

$\Rightarrow b=-y,a=x$

$\therefore z_1=x-iy,z_2=x+iy$

Hence proved,

$z_1=\bar{z}_2$

$\sqrt{-2}\sqrt{-3}=$

0%
- $\sqrt{6}$
0%
$\sqrt{6}$
0%
$i\sqrt{6}$
0%
$-i\sqrt{6}$

Explanation

${\textbf{Step - 1: Writing the product in terms of i (iota)}}$

${\text{We know that, }}\sqrt {{\text{ - 1}}} {\text{ = i and }}\sqrt {{\text{ab}}} {\text{ = }}\sqrt {\text{a}} \sqrt {\text{b}}$

$\therefore {\text{ }}\sqrt {{\text{ - 2}}} {\text{ = }}\sqrt {{\text{ - 1}}} \sqrt {\text{2}} {\text{ = i}}\sqrt 2$ $\quad \quad \quad \text{.....eqn(i)}$

${\text{and }}\sqrt {{\text{ - 3}}} {\text{ = }}\sqrt {{\text{ - 1}}} \sqrt {\text{3}} {\text{ = i}}\sqrt {\text{3}}$ $\quad \quad \quad \text{.....eqn(ii)}$

${\textbf{Step - 2: Multiplying the terms}}$

$\sqrt {{\text{ - 2}}} \sqrt {{\text{ - 3}}} {\text{ = }}\left( {{\text{i}}\sqrt {\text{2}} } \right)\left( {{\text{i}}\sqrt {\text{3}} } \right)$ $\quad \quad \quad \textbf{[From eqn(i) and eqn(ii)]}$

$\Rightarrow {\text{ }}\sqrt {{\text{ - 2}}} \sqrt {{\text{ - 3}}} {\text{ = }}{{\text{i}}^{\text{2}}}\sqrt {{\text{3}} \times 2}$

$\Rightarrow {\text{ }}\sqrt {{\text{ - 2}}} \sqrt {{\text{ - 3}}} {\text{ = - }}\sqrt {\text{6}} {\quad \quad \quad \quad \quad \quad \textbf{ [}}\because {\text{ }}{{\textbf{i}}^{\textbf{2}}}{\textbf{ = - 1]}}$

$\mathbf{{\text{Thus, the value of the product }}\sqrt {{\text{ - 2}}} \sqrt {{\text{ - 3}}} {\text{ is -}}\sqrt {\text{6}} }$

The argument of the complex number

$\sin \dfrac {6\pi}{5}+i\left(1+\cos \dfrac {6\pi}{5}\right)$ is

0%
$\dfrac {6\pi}{5}$
0%
$\dfrac {5\pi}{6}$
0%
$\dfrac {9\pi}{10}$
0%
$\dfrac {2\pi}{5}$

Explanation

$\sin{\dfrac{6\pi}{5}}+i\left(1+\cos{\dfrac{6\pi}{5}}\right)$ is equal to

$\sin{\left(\dfrac{\pi}{5}\right)}+i\left(1-\cos{\left(\dfrac{\pi}{5}\right)}\right)$

Lies in the second quardent of complex plane and its argument is

$arg \left(n+iy\right)=\pi-{\tan}^{-1}{\left|\dfrac{y}{x}\right|}$

$=\pi-{\tan}^{-1}{\left|\dfrac{1-\cos{\left(\dfrac{\pi}{5}\right)}}{ \sin{\left(\dfrac{\pi}{5}\right)}}\right|}$

$=\pi-{\tan}^{-1}{\left|\dfrac{2{\sin}^{2}{\left(\dfrac{\pi}{10}\right)}}{ 2\sin{\left(\dfrac{\pi}{10}\right)\cos{\left(\dfrac{\pi}{10}\right)}}}\right|}$

$=\pi-{\tan}^{1}{\left|\dfrac{\sin{\left(\dfrac{\pi}{10}\right)}}{\cos{\left(\dfrac{\pi}{10}\right)}}\right|}$

$=\pi-{\tan}^{-1}{\left|\tan{\left(\dfrac{\pi}{10}\right)}\right|}$

$=\pi-{\tan}^{-1}{\left(\tan{\left(\dfrac{\pi}{10}\right)}\right)}$

$=\pi-\dfrac{\pi}{10}$

$=\dfrac{9\pi}{10}$

$\left| z \right| = 1$ , then

$\left| z - 1 \right|$ is

0%
< $\left| arg (z) \right|$
0%
> $\left| arg (z) \right|$
0%
= $\left| arg (z) \right|$
0%
None of these

Explanation

Given

$|z|=1\ \Rightarrow \ z=1\ e^{i\theta}$ where

$\theta =\arg (z)$

$|z-1|=|e^{i\theta}-1|=|(\cos \theta -1)+i\sin \theta|$

$=\sqrt {(i\cos \theta -1)^2+\sin^2 \theta}$

$=\sqrt {2(1-\cos \theta)}$

$=2\sin \dfrac {\theta}{2}$

For

$\theta \ge 0,\ \sin \theta \le \theta \ \Rightarrow \ |z-1| =2\sin \dfrac {\theta}{2}\le 2\dfrac {\theta }{2}$

Hence,

$|z-1| < |\arg (z)|$

$\therefore \$ Option

$A$ is correct

If z is a complex number such that

$|z|\ge 2$ , then the minimumm value of

$\left|z+\dfrac{1}{2}\right|$ :

0%
is equal to $\dfrac{5}{2}$
0%
lies in the interval $(1,2)$
0%
is strictly greater then $\dfrac{5}{2}$
0%
is strictly greater than $\dfrac{3}{2}$ but less than $\dfrac{5}{2}$

$|z|=1$ and

$|\omega -1| =1$ where

$z, \omega \in C$ , then the largest set of values of

$|2z - 1|^2 + | 2\omega -1|^2$ equals

0%
$[1, 9]$
0%
$[2, 6]$
0%
$[2, 12]$
0%
$[2, 18]$

Explanation

$|z|=1$ implies that Z in Q circle with radius with a center (0,0) in the complex plane.

$|w-1|=1$ implies that it in a circle with the radius unit, center

$(1,0)$ in the complex plane.

$|2z-1|^2+|2w-1|^2=4\left(\left|z-\dfrac{1}{2}\right|^2+\left|w-\dfrac{1}{2}\right|^2\right)$

It means it is the range of

$4$ times the sum of the square of the distance of points from

$(0.5,)$ from the respective circles of low of the z,w

The point

$(0.5,0)$ in symmetric with the both circles on the center of a radius on the X-axis so both of them get equal ranges so it simply becomes

$8$ times of annual circle range.

The min or max possible values of the distance from the point become the distance b/w the point and the vertices of the diameter on choice it lies that is

$\rightarrow \dfrac{1}{2},\dfrac{3}{2}$

$=8\times \left[\left(\dfrac{1}{2}\right)^2,\left(\dfrac{3}{2}\right)^2\right]=[2,18]$

2117502_1132651_ans_28e18deb27384e3283fccfdf8d12839f.png

$Z$ is a complex number such that

$|z| \ge 2$ ,
then the minimum value of

$\left|z + \dfrac{1}{2}\right|$

0%
Is equal to $\dfrac{5}{2}$
0%
Lies in the interval $(1, 2)$
0%
Is strictly grater than $\dfrac{5}{2}$
0%
Is strictly greater than $\dfrac{3}{2}$ but less than $\dfrac{5}{2}$

Explanation

R.E.F image

graph of

$|z|\geq 2$

everything outside the

circle and on the circle is

$|z|\geq 2$

Minimum distance bet

$^{n}$

$|z+1/2|$

$\Rightarrow$ distance

$AB$

By using distance formula

$\Rightarrow \dfrac 32$

$\therefore$ Option B is correct

1181346_1134501_ans_d6590265571e4db4af8ff1e9a0f0f07b.png

$z_1=3+7i$ then

$|z_1|$ is

0%
$\sqrt {28}$
0%
$\sqrt {58}$
0%
$\sqrt {68}$
0%
none of these

Explanation

Given

$z_1=3+7i$

$|z_1|=\sqrt {3^2+7^2}$

$|z_1|=\sqrt {9+49}$

$|z_1|=\sqrt {58}$

${z}_{1}$ and

${z}_{2}$ two complex numbers satisfying the equation

$\left| \dfrac { { z }_{ 1 }+{ iz }_{ 2 } }{ { z }_{ 1 }{ iz }_{ 2 } } \right| =1$ then

$\dfrac{{z}_{1}}{{z}_{2}}$ is a

0%
purely real
0%
of unit modulus
0%
purely imaginary
0%
none of these

Explanation

$|\dfrac{z_{1}+iz_{2}}{z_{1}iz_{2}}| = 1$

$z_{2}\left(\dfrac{1}{iz_{2}}+\dfrac{1}{z_{1}}\right) = 1\times z_{2}$

$\dfrac{1}{1}+\dfrac{z_{2}}{iz_{1}} = z_{2}$

$z_{1}+iz_{2} = iz_{1}+z_{2}$

Dividing the above

$eq^{n}$ by

$z_{2}$

$\dfrac{z_{1}}{z_{2}}+i = iz_{1}$

$\dfrac{z_{1}}{z_{2}} = i(z_{1}-1)$

= Purely imaginary

option (C) is correct

Mark the correct alternative of the following.
Which of the following is a prime number?

0%
$263$
0%
$361$
0%
$323$
0%
$324$

let

$|z+\bar{z}|+|z-\bar{z}|=2014$ . Then

$z$ lies on a

0%
Circle
0%
Straight line
0%
Square
0%
Rectangle

Explanation

$\left | z+\bar{z} \right |+\left | z-\bar{z} \right | = 2014$

$\left | z+\bar{z} \right |^2 = [ 2014 - \left | z-\bar{z} \right |]^2$

$\left | z \right |^2+\left | \bar{z} \right |^2 +2z\bar{z} = (2014)^2 - \left | z-\bar{z} \right |^2 -2(2014)(z-\bar{z})$

$z\bar{z}+\bar{z}\bar{z}+2z\bar{z} = (2014)^2 -\left | z \right |^2+\left | \bar{z} \right |^2 +2z\bar{z} -4028z+4028\bar{z}$

$2\left | \bar{z} \right |^2 + 2\left | z \right |^2+4028z-4028\bar{z} = (2014)^2$

which is similar to the

$eq^n$ of circle,

by letting

$z = x$ &

$y = \bar{z}$

$2x^2+2y^2+4028x-4028y = (2014)^2$

which is the equation of a circle with radius

$2014$ .

option (A) is correct.

Argument and modules of

$[\dfrac{1+i}{1-i}]^{2\pi i}$ are respectively.................

0%
$\dfrac{-\pi}{2}$ and $1$
0%
$\dfrac{\pi}{2}$ and $\sqrt{2}$
0%
$0$ and $\sqrt{2}$
0%
$\dfrac{\pi}{2}$ and $1$

Explanation

$(\dfrac{1+i}{1-i})^{2\pi i}$

let

$2 = (\dfrac{1+i}{1-i})^{2\pi i}$

Rationalizing the above, we get

$= \dfrac{(1+i)^2}{(1)^2-(i)^2} = \dfrac{1+(i)^2+2i}{1-(-1)} = \dfrac{1-1+2i}{1+1} = \dfrac{2i}{2} = i$

$z = 0+i = 1$

here x =0 ,y = 1

modules of

$2 = \sqrt{x^2 + y^2 } = \sqrt{0+1^2} = 1$

modules of z = 1

$z =i ,$ As per the question,

$(i)^{2\pi i }$

$z = r(cos\theta + sin\theta)$

$\Rightarrow (r(cos\theta + 1sin \theta))^{2\pi i}$

$= [r(cos(2\pi i))+\theta +i sin (2\pi i )\theta ]$

$(i)^{2\pi i} = (r)^{2\pi i}$

$\Rightarrow r = i$

$r^2 cos^2 \theta = 0 ; r^2 sin^2 \theta = 1$

$tan \theta = \dfrac{\pi}{2}$

$0+1 = r^2 cos^2\theta +r sin^2 \theta$

$1 = r^2(cos^2 \theta + sin^2 \theta)$

$1 = r^2 \Rightarrow r = 1$

argument of

$z = \dfrac{\pi}{2}$

Option (D) is correct.

1169585_1140014_ans_7099f96ac0f2472c82e25114b401bbaa.jpeg

Choose the composite numbers from the following numbers

$87, 67, 45, 34, 23, 27, 33$ .

0%
$45, 87, 34, 27, 33$
0%
$45, 87, 67, 33$
0%
$33, 27, 23, 34$
0%
All the above
0%
None of these

Explanation

$87, 45, 34, 33, 27$ are composite numbers

[A composite number define as except own and one which are divisible by any numbers such that (4,6,8,9,....)]

$\left| {\dfrac{{{z_1}}}{{{z_2}}}} \right| = 1$ and

$\arg \left( {{z_1}{z_2}} \right) = 0$ , then

0%
${z_1} = {z_2}$
0%
${\left| {{z_2}} \right|^2} = {z_1}{z_2}$
0%
${z_1}{z_2} = 1$
0%
${ z_{ 1 } }=-{ z_{ 2 } }$

Explanation

$\left|\dfrac{z_1}{z_2}\right|=1$ and

$arg(z_1z_2)=0$

$z_1=x_1+iy, z_2=x_2+iy_2$

then if

$\left|\dfrac{z_1}{z_2}\right|=1$

$\sqrt{x_1^2+y_1^2}=\sqrt{x_2^2+y_2^2}$

$x_1^2+y_1^2=x_2^2+y_2^2$

$arg(z_1z_2)=arg(x_1x_2-y_1y_2+i(x_1y_2+y_1x_2))=0$

$\dfrac{x_1y_2+y_1x_2}{x_1x_2-y_1y_2}=0$

$x_1y_2=-y_1x_2$

$\dfrac{x_1}{x_2}=\dfrac{-y_1}{y_2}$

$\dfrac{x_1}{y_2}=\dfrac{-x_2}{y_2}$

Both

$z_1z_2$ make

$180^o$ angle

$z_1+z_2=0$

if both have an equal modulus

$z_1+z_2=0$

$z_1=-z_2$

$D$ is correct

$x^{2}+y^{2}=1$ and

$x \neq -1$ then

$\dfrac {1+y+ix}{1+y-ix}$

0%
$1$
0%
$y+ix$
0%
$2$
0%
$x+ix$

$I_m$

$\left( {\sqrt {a + i\sqrt {{a^4} + {a^2} + 1} } } \right) =$

0%
$\dfrac{1}{2}\sqrt {{a^2} - a + 1}$
0%
$\sqrt {\dfrac{{{a^2} - a + a}}{2}}$
0%
$\dfrac{1}{2}\sqrt {{a^2} + a + 1}$
0%
$\sqrt {\dfrac{{{a^2} - a + 1}}{2}}$

Explanation

Let

$z=\sqrt{a+i\sqrt{a^4+a^2+1}}$

$x+iy=\sqrt{a+i\sqrt{a^4+a^2+1}}$

Find y which

$I_m(z)$

Squaring both sides

$x^2-y^2+2ixy=a+i\sqrt{a^4+a^2+1}$

$x^2-y^2=a$ ……..

$(1)$

$2xy=\sqrt{a^4+a^2+1}$

$2xy=\sqrt{(a^2+a+1)(a^2+1-a)}$

$xy=\sqrt{\left(\dfrac{a^2+a+1}{2}\right)\left(\dfrac{a^2-a+1}{2}\right)}$ ……

$(2)$

From

$(1)$ &

$(2)$ we get

$x=\sqrt{\dfrac{a^2+a+1}{2}}, y=\sqrt{\dfrac{a^2-a+1}{2}}$

$I_m(z)=\sqrt{\dfrac{a^2-a+1}{2}}$

D is correct.

If for complex number

$z_{1}and z_{2}arg(z_{1})-arg(z_{2})=0then \mid z_{ 1}-z_{2}\mid$ is equal to:

0%
$\mid z_{1}+z_{2}\mid$
0%
$\mid z_{1}\mid +\mid z_{2}\mid$
0%
$\parallel z_{1}\mid -\mid z_{2}\parallel$
0%
0

$z=(3+7i)(p+iq)$ where

$p,q\in I-\left\{ 0 \right\}$ , is purely imaginary then minimum value of

${ \left| z \right| }^{ 2 }$ is

0%
$0$
0%
$58$
0%
$\dfrac{3364}{3}$
0%
$3364$

$z(\neq - 1)$ is complex number such that

$\dfrac{z-1}{z+1}$ is purely imaginary, then

$|z|$ is equal to

0%
$1$
0%
$2$
0%
$3$
0%
$5$

Explanation

Given

$\dfrac{z-1}{z+1}$ is purely imaginary

Let

$z=x+iy$

Now

$\dfrac{z-1}{z+1}=\dfrac{x+iy-1}{x+iy+1}=\dfrac{(x-1)+iy}{(x+1)+iy}$

$=\dfrac{(x-1)+iy}{(x+1)+iy}\cdot \dfrac{(x+1)-iy}{(x+1)-iy}$

$=\dfrac{(x-1)(x+1)-i^2y^2+iy[x+1-(x-1)]}{(x+1)^2-i^2y^2}$

$=\dfrac{x^2-1+y^2+i2y}{(x+1)^2+y^2}$

$\Rightarrow \dfrac{z-1}{z+1}=\dfrac{x^2-1+y^2}{(x+1)^2+y^2}+i\dfrac{2y}{(x+1)^2y^2}$

But

$\dfrac{z-1}{z+1}$ is imaginary,

$\Rightarrow$ Re

$\left(\dfrac{z-1}{z+1}\right)=0$

$\therefore \dfrac{x^2-1+y^2}{(x+1)^2+y^2}=0$

$\Rightarrow x^2+y^2=1$

$\Rightarrow |z|=1$ .

$z=a+ib$ ,

$a,b,\in R$ ,

$b\ne 0$ and

$\left| z \right| =1$ , then

$z=\cfrac{c+i}{c-i}$ , where

$c$ is equal to

0%
$\cfrac{a}{b}$
0%
$\cfrac{a-1}{b}$
0%
$\cfrac{a+1}{b}$
0%
$\cfrac{a+1}{b+1}$

$z$ is a complex number, then

$z^{2}+\bar{z}^{2}=2$ represents-

0%
a circle
0%
a straight line
0%
a hyperbola
0%
an ellipse

Explanation

Let

$z=x+iy$

$\bar {z} =x-iy$

$\therefore z^{2}+{\bar {z}}^{2}={ \left( x+iy \right) }^{ 2 }+{ \left( x-iy \right) }^{ 2 }$

$=x^2+2xyi-y^2+x^2-2xyi-y^2=2$

$\therefore 2\left( { x }^{ 2 }-{ y }^{ 2 } \right) =2$

$\therefore \boxed {x^2-y^2=1}$

Equation of hyperbola.

The modulus of the complex number

$z=\frac { \left( 1-i\sqrt { 3 } \right) \left( \cos { \theta } +isin\theta \right) }{ 2\left( 1-i \right) \left( \cos { \theta } -isin\theta \right) }$ is-

0%
$\frac { 1 }{ 2\sqrt { 2 } }$
0%
$\frac { 1 }{ \sqrt { 3 } }$
0%
$\frac { 1 }{ \sqrt { 2 } }$
0%
$\frac { 1 }{ 2\sqrt { 3 } }$

$\left| {z + 2 - i} \right| = 5$ then the maximum value of

$\left| {3z + 9 - 7i} \right|$ is

0%
18
0%
19
0%
20
0%
8

$\left|z\right|=1$ and

$\varpi=\dfrac{z-1}{z+1}$ , where

$z\neq-1$ , then

$Re\left(\varpi\right)$ is

0%
$0$
0%
$-\dfrac{1}{|z+1|^{2}}$
0%
$\dfrac{1}{\sqrt{2}}{\left|z+1\right|^{2}}$
0%
$\dfrac{\sqrt{2}}{|z+1|^{2}}$

CBSE Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 8 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 8 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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