MCQ Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 7

The number which have only two factors

$(1$ and the number itself) are called as ________.

0%
Composite
0%
Even
0%
Prime
0%
None of these

Computers use the .... number system to store data and perform calculations.

0%
binary
0%
octal
0%
decimal
0%
hexadecimal

$z_1, z_2$ are two complex numbers such that

$arg(z_1+z_2)=0$ and

$Im(z_1z_2)=0$ , then.

0%
$z_1=-z_2$
0%
$z_1=z_2$
0%
$z_1=\vec{z_2}$
0%
none of these

Explanation

Let

$z_1=a+ib$

and

$z_2=c+id$

Now

$z_1+z_2=(a+c)+i(b+d)$

Given

$arg(z_1+z_2)=0$

$\Rightarrow tan^{-1}[\dfrac{b+d}{a+c}]=0$

$\Rightarrow b+d=0$ .........

$(1)$

Now,

$z_1z_2=(ac-bd)+i(ad+bc)$

Given

$Im(z_1z_2)=0$

$\Rightarrow ad+bc=0$ .........

$(2)$

Putting

$b=-d$ from

$(1)$ in

$(2)$ we get

$a=c$

Hence

$z_1=a+ib=c-id=$ conjugate of

$z_2=z_2^{\rightarrow}$

Hence option C is correct.

$z + \sqrt {2}|z + 1| + i = 0$ and

$z = x + iy$ , then

0%
$x = -2$
0%
$x = 2$
0%
$y = -2$
0%
$y = 1$

Explanation

On putting

$z=x+iy$ , We get

=>

$x+iy+\sqrt{2(x+1)^2+2y^2} +i=0$

=>

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

=> On comparing real part and imaginary part, We get

=>

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

$y+1=0$

So,

$y=-1$

and ,

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

=>

${2(x+1)^2+2} =x^2$

=>

$x=-2$

${z}_{1}=-3+5i;{z}_{2}=-5-3i$ and

$z$ is a complex number lying on the line segment joining

${z}_{1}$ and

${z}_{2}$ , then

$arg(z)$ can be:

0%
$-\cfrac { 3\pi }{ 4 }$
0%
$-\cfrac { \pi }{ 4 }$
0%
$\cfrac { \pi }{ 6 }$
0%
$\cfrac { 5\pi }{ 6 }$

The complex numbers

$z=x+iy$ which satisfy the equation

$\left| \cfrac { z-5i }{ z+5i } \right| =1$ lie on:

0%
the x-axis
0%
straight line $y=5$
0%
a circle through the origin
0%
none of these

Explanation

$|z-5i|$

$=|z+5i|$

$|x+iy-5i|$

$=|x+iy+5i|$

$\sqrt{x^2+(y-5)^2}$

$=\sqrt{x^2+(y+5)^2}$

$\implies x^2+y^2+25-10y$

$=x^2+y^2+25+10y$

$\implies y=0$

Hence

$x$ axis.

Which of the following is not a binary number?

0%
001
0%
101
0%
202
0%
110

The memory of a computer is commonly expressed in terms of Kilobytes or Megabytes. A byte is made up of

0%
eight decimal digits
0%
eight binary digits
0%
two binary digits
0%
two decimal digits

What is the byre capacity of a drum which is

$5$ inch high,

$10$ inch diameter, and which has

$60$ tracks per inch and bit density of

$800$ bits per inch?

0%
$942000$ bytes
0%
$9712478$ bytes
0%
$192300$ bytes
0%
$14384$ bytes
0%
None of the above

What digits are representative of all binary numbers?

0%
$0$
0%
$1$
0%
Both (a) and (b)
0%
$3$
0%
None of the above

$\left| {z - \dfrac{4}{z}} \right| = 2$ , then the maximum value of

$\left| z \right|$ is

0%
$\sqrt 5$
0%
$\sqrt 5 + 1$
0%
$\sqrt 5 - 1$
0%
$1 - \sqrt 5$

$z_1, z_2, z_3$ are three points lying on the circle |z| =2, then the minimum value of

$|z_1 + z_2|^2 + | z_2 + z_3|^2 + | z_3 + z_1|^2$ is equal to

0%
$6$
0%
$12$
0%
$15$
0%
$24$

An output device that converts data from a binary format in main storage to coded hole patterns punched into a paper tape is?

0%
Paper tape punch
0%
Punched paper tape
0%
Magnetic disk
0%
Magnetic tape
0%
None of the above

The

$2$ 's complement number of

$110010$ is?

0%
$001101$
0%
$110011$
0%
$010011$
0%
All of the above
0%
None of the above

Instructions and memory addresses are represented by.

0%
Character codes
0%
Binary codes
0%
Binary word
0%
Parity bit
0%
None of the above

What is the highest address possible if

$16$ bits are used for each address?

0%
$65536$
0%
$12868$
0%
$16556$
0%
$643897$
0%
None of the above

Multiplication of

$111_2$ by

$101_2$ is?

0%
$110011_2$
0%
$100011_2$
0%
$111100_2$
0%
$000101_2$
0%
None of the above

The

$0$ and

$1$ in the binary numbering system are called binary digits or ____________.

0%
Bytes
0%
Kilobytes
0%
Decimal bytes
0%
Bits
0%
Nibbles

The real and imaginary part of the complex number

$1 + \sqrt {i}$ where

$i = \sqrt {-1}$ are

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

Explanation

$\sqrt { i } ={ cis\left( \dfrac { \pi }{ 2 } \right) }^{ 0.5 }$

From De moivres theorem,

$\sqrt { i } ={ cis\left( \dfrac { \pi }{ 2 } \right) }^{ 0.5 }=\pm cis\left( 0.5\frac { \pi }{ 2 } \right) =\dfrac { 1 }{ \sqrt { 2 } } +i\dfrac { 1 }{ \sqrt { 2 } }$ ,

$\therefore$ the real and imaginary parts are

$1+\dfrac { 1 }{ \sqrt { 2 } } ,\dfrac { 1 }{ \sqrt { 2 } }$ and

$1-\dfrac { 1 }{ \sqrt { 2 } } ,-\dfrac { 1 }{ \sqrt { 2 } }$

Find a complex number z satisfying the equation

$z+\sqrt{2}|z+1|+i=0$

0%
$2-i$
0%
$-2-i$
0%
$\sqrt{2}-i$
0%
None of these

Explanation

Let

$z=x+iy$ . Then,

$z+\sqrt{2}|z+1|+i=0$

$x+iy+\sqrt{2}|(x+1)+iy|+i=0$

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

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

$x+\sqrt{2(x+1)^2+2}=0$ and

$y=-1$

$\sqrt{2(x+1)^2+2}=-x$ and

$y=-1$

$2(x+1)^2+2=x^2$ and

$y=-1$

$x^2+4x+4=0$ and

$y=-1$

$(x+2)^2=0$ and

$y=-1$

$x=-2$ and

$y=-1$

Hence,

$z=x+iy=-2-i$

${ a }^{ 2 }+{ b }^{ 2 }=1$ , then

$\dfrac {\left( 1+b+ia \right) }{\left( 1+b-ia \right)}$ is

0%
$1$
0%
$2$
0%
$b+ia$
0%
$a+ib$

Explanation

$a^2+b^2=1$

$\implies 1-a^2=b^2$

$\dfrac{1+b+ia}{1+b-ia}$

$=\dfrac{1+b+ia}{1+b-ia}\times \dfrac{1+b+ia}{1+b+ia}$

$=\dfrac{(1+b)^2-a^2+2ia(1+b)}{(1+b)^2+a^2}$

$=\dfrac{b^2+2b+1-a^2+2ia(1+b)}{1+2b+b^2+a^2}$

$=\dfrac{2b^2+2b+2ia(1+b)}{2+2b}$

$=\dfrac{(2b+2ia)(1+b)}{2(1+b)}$

$=b+ia$

Answer-(C)

$\left| {z - 1} \right| = 2$ , then the value of

$z\overline z - z - \overline z$ is equal to:

0%
$-3$
0%
$4$
0%
$3$
0%
$-4$

Explanation

Given

$|z-1|=2$

we know that

$|z|^{2}=z.\bar{z}$

so taking square root on both side.

$|z-1|^{2}=4$

$(z-1)(\bar{z-1})=4$

$(z-1)(\bar{z}-1)=4$

$z.\bar{z}-\bar{z}-z+1=4$

$z.\bar{z}-z-\bar{z}=3$

so answer is 3.

The total number of even prime numbers is?

0%
$0$
0%
$1$
0%
$2$
0%
Unlimited

Explanation

The number

$2$ is the only prime number, which is even. So, the number of even primes is equal to

$1$ .

The real part of

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

0%
$\displaystyle\frac{1}{3+5\cos\theta}$
0%
$\displaystyle\frac{1}{5-3\cos\theta}$
0%
$\displaystyle\frac{1}{3-5\cos\theta}$
0%
$\displaystyle\frac{1}{5+3\cos\theta}$

Explanation

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

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

$=\dfrac{1}{(2\sin^2\dfrac{\theta}{2} +4i \sin\dfrac{\theta}{2}.\cos \dfrac{\theta}{2})}$

$=\dfrac{1}{2\sin\dfrac{\theta}{2}}\dfrac{1}{(\sin\dfrac{\theta}{2} +2i \cos\dfrac{\theta}{2})}$

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

$=\dfrac{1}{\sin\dfrac{\theta}{2}}\dfrac{\left(\sin\dfrac{\theta}{2}-2i\cos \dfrac{\theta}{2}\right)}{(2\sin^2\dfrac{\theta}{2} +8 \cos^2\dfrac{\theta}{2})}$

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

$=\dfrac{\left(1-2i\cot \dfrac{\theta}{2}\right)}{5+3\cos \theta}$

So the real part is

$\dfrac{1}{5+3\cos \theta}$ ..

$\dfrac {lz_{2}}{mz_{1}}$ is purely imaginary number, then

$\left |\dfrac {\lambda z_{1} + \mu z_{2}}{\lambda z_{1} - \mu z_{2}}\right |$ is equal to

0%
$\dfrac {l}{m}$
0%
$\dfrac {\lambda}{\mu}$
0%
$\dfrac {-\lambda}{\mu}$
0%
$1$

Explanation

$\cfrac { { lz }_{ 2 } }{ m{ z }_{ 1 } } =ki\\ \cfrac { { z }_{ 2 } }{ { z }_{ 1 } } =\cfrac { mki }{ l } \\ \implies |\cfrac { { \lambda z }_{ 1 }+\mu { z }_{ 2 } }{ { \lambda z }_{ 1 }-\mu { z }_{ 2 } } |\\ \implies |\cfrac { \lambda +\mu { z }_{ 2 }/{ z }_{ 1 } }{ \lambda -\mu { z }_{ 2 }/{ z }_{ 1 } } |\\ \implies |\cfrac { \lambda +\mu \cfrac { mki }{ l } }{ \lambda -\mu \cfrac { mki }{ l } } |\\ \implies|\cfrac { \lambda l+\mu mki }{ \lambda l-\mu mki } |$

$\implies|\cfrac{re^{i\theta}}{re^{-i\theta}}|$

$\implies |e^{2i\theta}|$

$\implies 1$

The complex number

$\dfrac{1+2i}{1-i}$ lies in which quadrant of the compiles plan

0%
First
0%
Second
0%
Third
0%
Fourth

Explanation

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

Multiply the coyugate of

$(1-i)$ is numerator

$\times$ denominator

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

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

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

$=\dfrac{-3+3i}{2}$

$=\dfrac{-3}{2}+\dfrac{3}{2}i$

$\therefore$ It is of the form

$-a+3i (-a, b)$

$\therefore \left(\dfrac{-3}{2}, \dfrac{3}{2}\right)$

$\therefore 2^{nd}$ Quandrant

In binary synchronous, communication ............code is used by the receiver to check the validity of the message recovered.

0%
$OK$
0%
$ACK$
0%
$BCC$
0%
$SOH$

Binary number are used because:

0%
Decimal system can not be represented on motherboard
0%
Binary data needs just 2 wires for the transmission, one for 1 & other for 0
0%
Binary data is easier to represent using 'on'(1) & 'off' (0) states of switches.
0%
All of these

Which of the following is NOT a binary system?

0%
EBCDIC
0%
ASCII
0%
HEX
0%
None of these

..............is a binary synchronous data transmission.

0%
Each bit is sent over a separate wire
0%
Sequence numbers are sent in numbered frames
0%
Each character is prefixed with DLE
0%
Character are used for control purpose

Which are not property of Binary relations

0%
reflexive relation, symmetric relation, anti symmetrical relation
0%
reflexive, transitive, equivalence relations
0%
transitive, partial ordering relation, symmetric
0%
reflexive, partial, chain relation

............are the Pentium binary program that can be embedded in a web page.

0%
Servlets
0%
Hyperlink
0%
Hypertext
0%
Active X controls

If Z satisfied the equation

$\left ( \dfrac{Z- 2}{Z + 2} \right ) \, \left ( \dfrac {{\bar Z} - 2}{{\bar Z} + 2} \right )$ then minimum, value of

$|Z|$ is equal to

0%
0
0%
2
0%
4
0%
6

Explanation

$(Z - 2) [{\bar Z}- {\bar2}](Z -+2) [{\bar Z} -2]$

$|Z - 2|^ 2 \, = | Z \, + \, 2| \Rightarrow |Z - 2| = |Z + \, 2|$
Line on imaginary axis

$|Z_1|min = 0$
1062254_993217_ans_79c992a59bc44b9bbd05a1ac02211936.png

$\mid{z_1}\mid=2$ ,

$\mid{z_2}\mid=3$ ,

$\mid{z_3}\mid=4$ and

$\mid{z_1+z_2+z_3}\mid=2$ , then the value of

$\mid{4z_2z_3+9z_3z_1+16z_1z_2}\mid$ .

0%
24
0%
48
0%
96
0%
120

Explanation

$z_1\bar{z_1}=|z_1|^2=1$

$,z_2\bar{z_2}=|z_2|^2=9$

$,z_3\bar{z_3}=|z_3|^2=6$

$|4z_2z_3+9z_3z_1+16z_1z_2|$

$\implies |\bar{z_1}z_1z_2z_3$

$+\bar{z_2}z_1z_2z_3$

$+\bar{z_3}z_1z_2z_3|$

$\implies |z_1||z_2||z_3||\bar{z_1}+\bar{z_2}+\bar{z_3}|$

$\implies 24\times$

$|\bar{z_1}+\bar{z_2}+\bar{z_3}|$

$\implies 48$

Evaluate:

${ \left( \dfrac { cos\dfrac { \pi }{ 8 } -isin\dfrac { \pi }{ 8 } }{ cos\dfrac { \pi }{ 8 } +isin\dfrac { \pi }{ 8 } } \right) }^{ 4 }$

0%
$1$
0%
$-1$
0%
$2$
0%
$\dfrac { 1 }{ 2 }$

Explanation

We know that

$e^{i\theta}=cos\theta+isin\theta$ and

$e^{-1\theta}=cos\theta-isin\theta$

${ \left( \dfrac { cos\dfrac { \pi }{ 8 } -isin\dfrac { \pi }{ 8 } }{ cos\dfrac { \pi }{ 8 } +isin\dfrac { \pi }{ 8 } } \right) }^{ 4 }$

$=\left(\dfrac{e^{-i(\pi/8)}}{e^{i(\pi/8)}}\right)^4=(e^{-i(\pi/4)})^4=e^{-i\pi}=cos\pi-isin\pi=-1$

$| \frac{z_1 - 2z_2}{2 - z_1\bar{z}_2} | = 1$ and

$|z_2| \neq 1$ then the value of

$|z_1|$ is

0%
4
0%
2
0%
1
0%
$\frac{1}{2}$

Explanation

$\left| \dfrac{z_{1}-2z_{2}}{2-z_{1}\bar {z_{2}}} \right|= 1, |z_{2}| \neq 1$

$\Rightarrow z_{1} \bar {z_{2}} \neq 2, |z_{1}- 2z_{2}|= |2-z_{1}\bar{z_{2}}|$

$\Rightarrow |z_{1}-2z_{2}|^{2}= |2- z_{1} \bar{z_{2}}|^{2}$

$\Rightarrow |z_{1}|^{2}+4 |z_{2}|^{2}- (2 \bar{z_{1}z_{2}}+2z_{1} \bar{z_{2}})$

$= 4+ |z_{1}|^{2}+ |\bar{z_{2}}|^{2}- (2 \bar{z_{1}z_{2}}+2 z_{1} \bar{z_{2}})$

$\Rightarrow |z_{1}|^{2} (1- |z_{2}|^{2})+4 (|z_{2}|^{2}-1)=0$

$[\because |\bar{z_{2}}|= |z_{2}|]$

$\Rightarrow (|z_{1}|^{2}-4)(1-|z_{2}|^{2})=0$

$|z_{2}| \neq 1 \Rightarrow |z_{2}|=2$

The complex number

$x+iy$ whose modulus is unity,

$y\neq 0$ , can be represented as

$x+iy=\dfrac { a+i }{ a-i }$ , where

$a$ is real number.

0%
True
0%
False

Explanation

$\cfrac{a+i}{a-i}$

$=\cfrac{(a+i)^2}{a^2+1}=\cfrac{a^2-1+2ai}{a^2+1}$

$\cfrac{a^2-1}{a^2+1}+\cfrac{2ai}{a^2+1}=z$ (Let)

$|z|=\sqrt{\cfrac{(a^2-1)^2}{(a^2+1)^2}+(\cfrac{2a}{a^2+1})^2}$

$\implies \cfrac{\sqrt{(a^2+1)^2}}{a^2+1}=1$

Hence true.

Simplify

$\left ( \dfrac{2i}{1 \, + \, i} \right )^2$

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

Explanation

We have,

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

$\Rightarrow \dfrac{4{{i}^{2}}}{1+{{i}^{2}}+2i}$

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

$\Rightarrow \dfrac{-4}{2i}$

$\Rightarrow \dfrac{-2}{i}$

$\Rightarrow \dfrac{-2i}{{{i}^{2}}}$

$\Rightarrow \dfrac{-2i}{-1}$

$\Rightarrow 2i$

Hence, this is the answer.

$z_1 \, z_2$ be two distinct complex numbers and let z = (1 - t)

$z_1$ + t

$z_2$ for some real number t with 0 < t <If arg

$(\omega)$ denotes the principal argument of a non-zero complex number

$(\omega)$ , then

0%
$|z \, - \, z_1| \, + \, |z \, - \, z_2| \, = \, |z_2\, - \, z_1|$
0%
$arg(z \, - \, z_1) \, = \, arg (z \, - \, z_2)$
0%
$\begin{vmatrix} z \, - \, z_1 & \bar z \, - \, \bar{z_1}\\ z_2 \, - \, z_1 & \bar{z_2} \, - \, \bar{z_1} \end{vmatrix}$
0%
$arg(z \, - \, z_1) \, = \, arg (z_2 \, - \, z_1)$

Explanation

Given equation:

$z=(1-t)z_{1}+tz_{2}$

So we can write,

$|z-z_{1}|=t|z_{2}-z_{1}|=t|z_{1}-z_{2}|$ -------(1)

$z-z_{2}=(1-t)(z_{1}-z_{2})$

$|z-z_{2}|=(1-t)|(z_{1}-z_{2})|=(1-t)|z_{2}-z_{1}|$ ------(2)

From (1) and (2) we can say,

$|z-z_{1}|+|z-z_{2}|=|z_{1}-z_{2}|$

If z is a complex number such that

$\left|\dfrac{z-3i}{z+3i}\right|=1$ then z lies on?

0%
The real axis
0%
The line Im(z) $=3$
0%
A circle
0%
None of these

Explanation

Given

$\left| \cfrac { z-3i }{ z+3i } \right| =1$

$\Rightarrow \left| \cfrac { z-3i }{ z+3i } \right| =1$

$\Rightarrow z$ is equidistant from

$(0,3)$ &

$(0,-3)$

It lies on real axis.

Let $z$ be a complex number such that $\left| z+\dfrac { 1 }{ z } \right| =2$ .

If $\left| z \right| ={ r }_{ 1 }$ and $\left| \dfrac { 1 }{ z } \right| =$ ${r}_{2}$ for $\arg z=\dfrac { \pi }{ 4 }$ then

$\left| { r }_{ 1 }-{ r }_{ 2 } \right| =$

0%
$\dfrac { 1 }{ \sqrt { 2 } }$
0%
$1$
0%
$\sqrt { 2 }$
0%
$2$

Explanation

$\left| z+\dfrac { 1 }{ z } \right| ^{ 2 }={ r }^{ 2 }+\dfrac { 1 }{ { r }^{ 2 } } +2r.\dfrac { 1 }{ r } \cos { \left( \theta +\theta \right) } =4$
or

${ r }^{ 2 }+\dfrac { 1 }{ { r }^{ 2 } } =4-2\cos { 2\theta }$

$\therefore\quad \left( r-\dfrac { 1 }{ r } \right) ^{ 2 }=2\left( 1-\cos { 2\theta } \right) =4\sin { ^{ 2 }\theta }$

$\left| { r }_{ 1 }-{ r }_{ 2 } \right| =\left| r-\dfrac { 1 }{ r } \right| =2\sin { \theta } =2\sin { \dfrac { \pi }{ 4 } } =\sqrt { 2 }$

${z}_{1}=1+2i,\ {z}_{2}=2+3i,\ {z}_{3}=3+4i$ , then

${z}_{1},\ {z}_{2}$ and

${z}_{3}$ are collinear.

0%
True
0%
False

Explanation

Line formula from

$z_3$ and

$z_1$

$(y-y_1)=\cfrac{y_3-y_1}{x_3-x_1}$

$\times(x-x_1)$

$\implies (y-2)=\cfrac{(4-2)(x-1)}{3-1}$

$\implies y-2=x-1\\ \implies y=x+1$

$z_2=(2,3)$

$y=x+1$ satisfies

$z_2$ .

Hence

$z_1,z_2$ and

$z_3$ are collinear.

$\dfrac{2z_1}{3z_2}$ is a purely imaginary number,then

$\left|\dfrac{z_1-z_2}{z_1+z_2}\right|=$

0%
$3/2$
0%
$1$
0%
$2/3$
0%
$4/9$

Explanation

$\dfrac{2z_1}{3z_2}=k'i$

$\dfrac{z_1}{z_2}=\dfrac{3k'i}{2}$

$\dfrac{z_1}{z_2}=ki$ where

$\dfrac{3k'}{2}=k$

$z_1=kiz_2$

Thus

$\left|\dfrac{z_1-z_2}{z_1+z_2}\right|=\left|\dfrac{kiz_2-z_2}{kiz_2+z_2}\right|$

$=\left|\dfrac{z_2(ki-1)}{z_2(ki+1)}\right|$

$=\dfrac{|ki-1|}{|ki+1|}$

$=\dfrac{\sqrt{k^2+1^2}}{\sqrt{k^2+1^2}}=1$

If z satisfies

$\left| {z - 1} \right| < \left| {z + 3} \right|$ then

$w = 2z + 3 - i$ , ( where

$w = 2z + 3 - i$ ) satisfies:

0%
$\left| {w - 5 - i} \right| < \left| {w + 3i} \right|$
0%
$\left| {w - 5} \right| < \left| {w + 3} \right|$
0%
$\left( {iw} \right) > 1$
0%
$\left| {\arg \left( {w - 1} \right)} \right| < {\pi \over 2}$

Find the real number

$x$ if

$(x-2i)(1+i)$ is purely imaginary.

0%
$2$
0%
$-2$
0%
$4$
0%
$-4$

Explanation

$(x-2i)(1+i)$

$= x+ix -2i-2i^2$

$= (x+2) -i(x-2)$ is purely imaginary if (x+2)=0

$\therefore x=-2$

Real part of

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

0%
$-1/5$
0%
$1/5$
0%
$1/10$
0%
$-1/10$

Explanation

Given

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

$=\dfrac{2i}{3 - i}$ [Since

$(i+1)^2=2i$ ]

$=\dfrac{2i(3+i)}{3^2 - i^2}$ [After rationalizing the denominator]

$=\dfrac{2i(3+i)}{10}$

$=\dfrac{(3i-1)}{5}$ .

Now real part of the given complex number is

$-\dfrac{1}{5}$ .

$i \, \log \left(\dfrac{x - i}{x + i}\right)$ is equal to

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$2i\log (x-i)-i\log (x^2+1)$
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$2i\log (x-i)+i\log (x^2+1)$
0%
$2i\log (x+i)-3i\log (x^2+1)$
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$2i\log (x-i)-i\log (x^2+i)$

Explanation

$i\log \left ( \cfrac{x-i}{x+i} \right )$

$=i\log \left ( \cfrac{x-i}{x+i}\times \cfrac{x-i}{x-i}\right )$

$=i\log \left ( \cfrac{(x-i)^{2}}{x^{2}-i^{2}} \right )$

$=i\log \left ( \cfrac{(x-i)^{2}}{x^{2}+1} \right )$

$=i\log (x-i)^{2}-i\log \left ( x^{2}+1 \right )$

$=2i\log (x-i)-i\log \left ( x^{2}+1 \right )$

On the complex plane locus of a point

$z$ satisfy inequality

$2\le \left| z-1 \right| <3$ denotes

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region between, the concentric circles of radii $3$ and $1$ centered at $(1,0)$
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region between the concentric circles of radii $3$ and $2$ centered at $(1,0)$ excluding the inner and outer boundaries.
0%
region between the concentric circles of radii $3$ and $2$ centered at $(1,0)$ including the inner and outer boundaries.
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region between, the concentric circles of radii $3$ and $2$ centered at $(1,0)$ including he inner boundary and excluding the outer boundary.

The value of

$\dfrac{1}{i} + \dfrac{1}{{{i^2}}} + \dfrac{1}{{{i^3}}} + ... + \dfrac{1}{{i^{102}}}$ is equal to

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

Explanation

Let

$t=\cfrac { 1 }{ i } +\cfrac { 1 }{ { i }^{ 2 } } +.....+\cfrac { 1 }{ { i }^{ 102 } }$

${ i }^{ 102 }t={ i }^{ 101 }+{ i }^{ 100 }+....+{ i }^{ 0 }$

Using sum of GP,

$a={ i }^{ 0 }=1$

$r=i$

$n={ 102 }$

${ i }^{ 102 }t=\cfrac { 1\left( { i }^{ 102 }-1 \right) }{ i-1 } =\cfrac { { i }^{ 2 }-1 }{ i-1 } =\cfrac { -2\left( i+1 \right) }{ \left( i-1 \right) \left( i+1 \right) }$

${ i }^{ 102 }t=i+1\Rightarrow { i }^{ 2 }t=i+1$

$t=-i-1$

$i^2= -1$ , then

$1+ i^2+ i^4 +i^6+i^8 +.............to ( 2n +1)$ terms is equal to

0%
$0$
0%
$1$
0%
$3i$
0%
$4i$

Explanation

$1+ i^2 + i^4 + ......(2n+1)terms$

First term

$a = i^2$

Common ratio

$r = i^2$

$1 + \cfrac{a(r^n-1)}{r-1}$

$= 1 - \cfrac{i^2(i^{2(2n)} - 1)}{i-1}$

$= 1 - \cfrac{i^2(i^4-1)}{i-1}$ Since

$i^4 = 1$

$=1$

CBSE Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 7 - 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 7 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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