MCQ Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 3

Which one of the following is not a prime number?

0%
$31$
0%
$61$
0%
$71$
0%
$91$

Explanation

$91$ is divisible by

$7$ and

$13$ . So, it is not a prime number.

Rest all the given numbers are prime numbers since they don't have any factors other than

$1$ and themselves.

What is the modulus of

$\cfrac { \sqrt { 2 } +i }{ \sqrt { 2 } -i }$ where

$i=\sqrt { -1 }$

0%
$3$
0%
$\dfrac{1}{2}$
0%
$1$
0%
None of the above

Explanation

$\left| \dfrac { \sqrt { 2 } +i }{ \sqrt { 2 } -i } \right| =\dfrac { \sqrt { ({ \sqrt { 2 } })^{2}+{ 1 }^{ 2 } } }{ \sqrt { (\sqrt { 2 })^{ 2 }+(-1)^{ 2 } } } =1$
Hence, option C is correct.

$\cos { \left( \log { { i }^{ 4i } } \right) } =a+ib$ , then

0%
$a=1,b=-1$
0%
$a=-1,b=1$
0%
$a=1,b=0$
0%
$a=1,b=2$

Explanation

Given,

$a+ib=\cos { \left( \log { { i }^{ 4i } } \right) }$

$=\cos { \left[ 4i\log { i } \right] } =\cos { \left[ 4i\log { \left( { e }^{ i\cfrac { \pi }{ 2 } } \right) } \right] } =\cos { \left[ \left( 4i \right) \left( i\cfrac { \pi }{ 2 } \right) \right] } =\cos { \left( -2\pi \right) } =1$

$\therefore a=1;b=0$

Which of the following numbers is prime?

0%
$119$
0%
$187$
0%
$247$
0%
$179$

Explanation

$119$ is divisible by

$7$

$187$ is divissible by

$11$

$247$ is divisible by

$13$

$179$ is not divisible by any number.

Hence

$179$ is a prime number.

$( 7 \times 11 \times 13 + 13 )$ is a/an

0%
Composite number
0%
Prime number
0%
Irrational number
0%
Imaginary number.

Explanation

$(7\times 11\times 13+13)=13\{(7\times11)+1\}=13\times 78=1014$

Since,

$1014$ is divided by

$13,78$ which is other than

$1$ and the number itself, so it is a composite number.

The reciprocal of the smallest prime number is _______.

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

Explanation

Prime no. are those which has only two factors.

One and the number itself.

$Note:$

$1$ is neither prime nor composite.

So, the smallest prime number is

$2$ .

Hence, reciprocal of

$2$

$= \dfrac{1}{2}$

Which of the following is not a prime number?

0%
$23$
0%
$29$
0%
$43$
0%
$21$

Explanation

Factors of

$23 = 23 \times 1$

Factors of

$29 = 29 \times 1$

Factors of

$43 = 43 \times 1$

Factors of

$21 = 3 \times 7$

So, ioption D

$23$ is not a prime number

Instructions and memory address are represented by

0%
Character code
0%
Binary codes
0%
Binary word
0%
parity bit

Bit stands for

0%
Binary digits
0%
bit of system
0%
a part of byte
0%
All of above

A byte is made up of

0%
Eight bytes
0%
Eight binary digits
0%
Two binary digits
0%
Tow decimal points

The roots of the equation

${ \left( z+\alpha \beta \right) }^{ 3 }={ \alpha }^{ 3 }$ represent the vertices of a triangle, one of whose sides is of length

0%
$\sqrt { 3 } \left| \alpha \beta \right|$
0%
$\sqrt { 3 } \left| \alpha \right|$
0%
$\sqrt { 3 } \left| \beta \right|$
0%
$None\ of\ these$

Explanation

$z_1=\alpha-\alpha \beta,z_2=\alpha\omega-\alpha \beta,z_3=\alpha\omega^2-\alpha \beta$

Let vertices be denoted by

$A,B,C$

Length of

$AB=|z_1-z_2|=|\alpha||1-\omega|=|\alpha|\left|1-\dfrac{-1-i\sqrt{3}}{2}\right|=\sqrt{3}|\alpha|$

Length of

$BC=|z_2-z_1|=|\alpha|\left|1-\dfrac{-1+i\sqrt{3}}{2}\right|=\sqrt{3}|\alpha|$

Length of

$BC=|z_3-z_1|=|\alpha||\omega|\left|\dfrac{-1+i\sqrt{3}}{2}\right|\left|1-\dfrac{-1+i\sqrt{3}}{2}\right|=\sqrt{3}|\alpha|$

Hence option B is correct.

How many options doe a BINARY choice offer

0%
None
0%
One
0%
Two
0%
it depends on the amount of memory on the computer
0%
It depends on the speed of the computer's processor

Which number system is usually followed in a typical 32-bit computer?

0%
Binary
0%
Decimal
0%
Hexadecimal
0%
Octal

Put the following in the form of A + iB :

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

0%
$\dfrac{3}{4} \, + \, \dfrac{9}{4} \, i$
0%
$\dfrac{63}{25} \, - \, \dfrac{16}{25} \, i$
0%
$\dfrac{5}{4} \, + \, \dfrac{9}{4} \, i$
0%
$\dfrac{1}{4} \, + \, \dfrac{7}{4} \, i$

Explanation

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

$\implies \cfrac{6+9i-4i+6}{2-i+4i+2}$

$\implies \cfrac{12+5i}{4+3i}$

$\implies \cfrac{(12+5i)(4-3i)}{4^2+3^2}$

$\implies \cfrac{48-36i+20i+15}{25}$

$\implies \cfrac{63}{25}$

$-i \cfrac{16}{25}$

............... code is used to implement data transparency in the binary synchronous protocol.

0%
$DLE$
0%
$NAK$
0%
$EOH$
0%
$EOT$

The number

$111111111$ is a

0%
Prime number
0%
Composite number
0%
Divisible by $\frac{10^7-1}{9}$
0%
None of these

Explanation

A whole number that can be divided exactly by numbers other than 1 or itself is known as composite numbers.

111111111 is an odd composite number. It is composed of three distinct prime numbers multiplied together.

Prime factorisation of

$111111111 = {3}^{2} \times 37 \times 333667$

In the complex numbers, where

$i = \sqrt {-1}$ , the conjugate of any value

$a + bi$ is

$a -ib$ . What is the result when you multiply

$2 + 7i$ by its conjugate?

0%
$45$
0%
$-45$
0%
$45i$
0%
$53$
0%
$53i$

Explanation

Conjugate of

$2 + 7i$ is

$2-7i$

$(2 + 7i) \times (2-7i) = 2^2 - (7i)^2$

$= 4 -49(-1)$

$= 4 + 49$

$= 53$

Evaluate:

$\left| \left( 1+i \right) \dfrac { \left( 2+i \right) }{ \left( 3+i \right) } \right| =$ ?

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

Explanation

$\displaystyle i^2 = -1$

$\displaystyle \left|\frac{(1+i)(2+i)}{(3+i)}\right| = \left|\frac{2+i+2i+i^2}{3+i}\right|=\left|\frac{1+3i}{3+i}\right|$

$\displaystyle \left|\frac{1+3i}{(3+i)}\times\frac{(3-i)}{(3-i)} \right| = \left|\frac{3-i+9i-3i^2}{9-i^2} \right| =\left|\frac{6+8i}{10}\right|$

$\displaystyle \Rightarrow |0.6+0.8i| = \sqrt{(0.6)^2 + (0.8)^2} = 1$

$(1+i)^{-20}=a+ib$ , then the values of

$a$ and

$b$ are

0%
$a=2^{-10}, b=-2^{-10}$
0%
$a=-2^{-10}, b=0$
0%
$a=2^{-10}, b=0$
0%
$none\ of\ these$

Explanation

$(1+i)^{-20}\\((1+i)^2){-10}\\(1+i^2+2i)^{-10}\\=(2i)^{-10}\\=(\frac{2^{-10}}{i^10})\\=-2^{-10}\\\therefore \>by\>comparison\>a\>=\>-2^{-10},\>b=0$

The maximum value of

$\left| {{\rm{3z + 9 - 7i}}\left| {{\rm{if}}} \right|{\rm{z + 2 - i}}\left. {} \right|} \right.{\rm{ = 5}}$ is

0%
$20$
0%
$15$
0%
$25$
0%
$5$

Explanation

Given ,

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

$=>\left|3z+6-3i +3-4i\right|$

$=>\left|3(z+2-i)+(3-4i)\right|$

$\leq3\left|z+2-i\right| +\left|3-4i\right|$

$\leq3\times{5}+ 5$

$\leq20$

hence,maximum value is 20.

Find the modulus of the complex number

$-2+2\sqrt{3}i$ .

0%
2
0%
4
0%
6
0%
8

Explanation

The given complex number is

$-2+2\sqrt{3}i$ .

we know that if

$z=x+iy\Rightarrow |z|=\sqrt{x^2+y^2}$

Thus, modulus =

$\sqrt{(-2)^2+(2\sqrt{3})^2}=\sqrt{4+12}=\sqrt{16}=4$ .

Find the which of the complex number has greatest modulus.

0%
$7-5i$
0%
$\sqrt{3}+\sqrt{2}i$
0%
$-8+15i$
0%
$-3(1-i)$

Explanation

$|7-5i| = \sqrt {54 }$

$|\sqrt {3} +i \sqrt 2| = \sqrt 5$

$| -8+15i| = \sqrt {289}$

$|-3+3i| = 3\sqrt 2$

Thus option C has the greatest modulus.

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

0%
$|z|=1$
0%
$|z|>1$
0%
$|z|<1$
0%
$|z|<2$

Explanation

$\cfrac{z-1}{z+1}$

Let

$z=x+iy$

$\cfrac{x+iy-1}{x+1+iy}=\cfrac{x-1+iy}{x+1+iy}\times \cfrac{x+1-iy}{x+1-iy}\\k=\cfrac{x^2+x-ixy-x+iy-1+ixy+iy+y^2}{(x+1)^2+y^2}\\k=\cfrac{x^2+y^2+2iy-1}{(x+1)^2+y^2}$

If k is purely imaginary then real part

$=0$

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

For any complex number

$z$ the minimum value of

$|z|+|z-2013i|$ is...

0%
$2010$
0%
$2011$
0%
$2013$
0%
$2012$

Explanation

Q:- minimum value of

$\left | z \right |+\left | z-2013i \right |$

Solution For ZEG

$\Rightarrow 2013 \leqslant \left | z \right |+\left | z-2013i \right |$

$\Rightarrow$ minimum value of

$\left | z \right |+\left | z-2013i \right |$ is

2013

which is attained when

$z = i$

so, c is correct option.

$x^{2}+y^{2} \leqslant 1$ &

$y^2 \leqslant 1-x .$

1120336_1140024_ans_b1d27da3ec984cf389d2244ff874fc8e.jpg

$z=x+iy(x,y\epsilon R,x\neq -1/2),$ the number of values of z satisfying

$\left | z \right |^{n}=z^{2}\left | z \right |^{n-2}+1.(n\epsilon N,n> 1)$ is

0%
0
0%
1
0%
2
0%
3

Explanation

$\begin{array}{l} we\, \, write\, , \\ { \left| z \right| ^{ n } }={ \left| z \right| ^{ n-2 } }\left( { { z^{ 2 } }+z } \right) +1\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left\{ { we\, know,\, \, all\, \, are\, real\, no. } \right. \\ so,\, \, \, \, { z^{ 2 } }+z={ \overline { z } ^{ 2 } }+\overline { z } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { where\, ,\, \, { z^{ 2 } }+z\, \, \, \, is\, real\, no.\, } \right. \\ \Rightarrow { z^{ 2 } }\, -{ \overline { z } ^{ 2 } }+z-\overline { z } =0 \\ \Rightarrow (z-\overline { z } )\, \, (z+\overline { z } +1)=0\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left| \begin{array}{l} \, z=x+i\, y \\ \, \overline { z } =x-i\, y \end{array} \right. \\ \, \, \therefore \, \, \, \, 2\, i\, y\, (2x+1)=0 \\ \, \, \, y=0\, \, \, \, \, \, \, \, \, \, \, or\, \, \, \, \, \, x=-\frac { 1 }{ 2 } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (Here,\, \, \, z\, \, is\, a\, \, \, real\, \, \, number. \\ Let\, see, \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left| z \right| =x \\ \left| \begin{array}{l} \, \, { x^{ n } }={ x^{ n-2 } }.{ x^{ 2 } }+1 \\ \, \, \, { \left| x \right| ^{ n-2 } }.\, x=-1 \end{array} \right. \\ \, \, \, \therefore \, \, \, \, x=-1\, \\ so,\, number\, of\, z\, =1\, \, and\, \, the\, correct\, option\, is\, B.\, \, \, \, \, \end{array}$

Which of the following is true

0%
$(3 + \sqrt{-5})(3 - \sqrt{-5}) = 14$
0%
$(-2 + \sqrt{-3})(-3 + 2\sqrt{-3}) = -7\sqrt{3}i$
0%
$(2 + 3i)^2 = (-5 + 12i)$
0%
$(\sqrt{5} - 7i)^2 = -44 - 14\sqrt{5}i$

Explanation

(a)

$(3+\sqrt{-5})(3-\sqrt{-5})=14$

L.H.S

$= (3+\sqrt{-5})(3-\sqrt{-5})$

$=((3)^2-(\sqrt{-5})^2) ((a-b)(a+b)=a^2-b^2)$

$=9+5$

$=14 = R.H.S$

(b)

$(-3+\sqrt{-3})(-3+2\sqrt{-3})=-7\sqrt{3i}$

L.H.S

$=(-2+\sqrt{-3}) (-3+2\sqrt{-3})$

$=(-3+\sqrt{3i})(-3+2\sqrt{3i})$

$=4-4\sqrt{3i}-3\sqrt{3i}-6=-7\sqrt{3i}=R.H.S$

(c)

$(2+3i)^2=(-5+12i)$

L.H.S

$= (2+3i)^2$

$= 4+9i^2+12i$

$=4-9+12i= -5+12i=R.H.S$

(d)

$(\sqrt{5}-7i)^2=-44- 14\sqrt{5}i$

L.H.S

$= (\sqrt{5}-7i)^2$

$=5-49-14\sqrt{5}i$

$= - 44- 14\sqrt{5}i = R.H.S$

$\therefore (a), (b), (c), (d)$ all are correct

$z_{1}$ and

$z_{2}$ are two non-zero complex numbers such that

$|z_{1}|=|z_{2}|$ and

$argz_{1}+argz_{2}=\pi$ , then

$z_{2}$ equals

0%
$\bar{z_{1}}$
0%
$-\bar{z_{1}}$
0%
$z_{1}$
0%
$-z_{1}$

Explanation

$-z_1$

as,

$arg z_+argz_=\pi$

$z=1+i\sqrt { 3, } \left| arg\left( z \right) \right| +\left| arg\left( \overline { z } \right) \right|$

0%
$\dfrac{\pi}{3}$
0%
$\dfrac{2\pi}{3}$
0%
$0$
0%
$\dfrac{\pi}{2}$

Explanation

$z=i+i\sqrt{3}$

$\bar{z}=1-i\sqrt{3}$

$arg z=\dfrac{\pi}{3}$

$arg(\bar{z})=\dfrac{\pi}{3}$

$|arg z|+|arg (\bar{z})|=\dfrac{2\pi}{3}$ .

1198025_1158272_ans_52c212d42bed426f9291b2a8f77b43ec.jpg

$\alpha$ and

$\beta$ are real then

$\left| \dfrac { \alpha +i\beta }{ \beta +i\alpha } \right|=$

0%
Lies betwen 0 and 1
0%
= 1
0%
>1
0%
2

Explanation

We know that

$|\dfrac{z_{1}}{z_{2}}|$

$=\dfrac{|z_{1}|}{|z_{2}|}$
Hence

$|\dfrac{\alpha+i\beta}{\beta+i\alpha}|$

$=\dfrac{|\alpha+i\beta|}{|\beta+i\alpha|}$

$=\dfrac{\sqrt{\alpha^{2}+\beta^{2}}}{\sqrt{\beta^{2}+\alpha^{2}}}$

$=1$ .

If arg

$\left( {{Z_1} + {Z_2}} \right) = 0\;$ and

$|{z_1}|\; = \;|{z_2}|\; = 1.$ then.

0%
${z_1} + {z_2} = 0$
0%
${z_1}{\overline z _2} = 1$
0%
${z_1} = {\overline z _2}$
0%
none

Explanation

We have,

$\arg \left( {{Z_1} + {Z_2}} \right) = 0\,\,\,\,\,\,\,\,\,\,\,\left| {{Z_1}} \right| = 1\,\,\,\,\left| {{Z_2}} \right| = 1$

$\begin{matrix} { Z_{ 1 } }+{ Z_{ 2 } }=k\, \, \left( { k>0 } \right) \\ { Z_{ 1 } }=\cos \theta +i\sin \theta \\ { Z_{ 2 } }=\cos \theta +i\sin \theta \\ \end{matrix}$

$\cos \theta + \cos \theta + \left( {i\sin \theta + i\sin \theta } \right) = k$

$\Rightarrow \sin \varphi = - \sin \theta$

$\Rightarrow \cos \varphi = - \cos \theta \,\,\,\,\,\,\,\,\,\,\,\,as\left| {{Z_2}} \right| = 1$

${Z_1} = \cos \theta \, + i\sin \theta$

${Z_2} = \cos \theta \, - i\sin \theta$

${Z_1} = \overline {{Z_2}}$

Then,

Option

$C$ is correct answer.

The value of

${\left( {1 + i} \right)^5} \times {\left( {1 - i} \right)^5}$ is

0%
$-8$
0%
$8i$
0%
$8$
0%
$32$

Explanation

We have,

${{\left( 1+i \right)}^{5}}{{\left( 1-i \right)}^{5}}$

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

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

$={{\left[ {{\left( 1+i \right)}^{2}} \right]}^{2}}{{\left[ {{\left( 1-i \right)}^{2}} \right]}^{2}}\left( {{1}^{2}}-\left( -1 \right) \right)\,\,\,\,\therefore {{i}^{2}}=-1$

$={{\left[ \left( {{1}^{2}}+{{i}^{2}}+2i \right) \right]}^{2}}{{\left[ \left( {{1}^{2}}+{{i}^{2}}-2i \right) \right]}^{2}}\left( 2 \right)$

$=2{{\left[ \left( {{1}^{2}}+\left( -1 \right)+2i \right) \right]}^{2}}{{\left[ \left( {{1}^{2}}+\left( -1 \right)-2i \right) \right]}^{2}}\,\,\,\,\,\,\therefore {{i}^{2}}=-1$

$=2{{\left[ 2i \right]}^{2}}{{\left[ -2i \right]}^{2}}$

$=32$

Hence, this is the answer.

How many two-digit prime numbers are there having the digit

$3$ in their units place?

0%
$10$
0%
$8$
0%
$6$
0%
$5$

Explanation

$2$ digit prime nos. having

$3$ in their units place are-

$13,23,43,53,73$

$\therefore$ There are

$5$ such nos.

Argument and modulus of

$\left[\dfrac {1+i}{1-i}\right]^{2013}$ 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

Let

$P={ [\cfrac { 1+i }{ 1-i } ] }^{ 2013 }={ [\cfrac { \sqrt { 2 } { e }^{ \cfrac { i\pi }{ 4 } } }{ \sqrt { 2 } { e }^{ \cfrac { -i\pi }{ 4 } } } ] }^{ 2013 }=({ e }^{ \cfrac { i2013\pi }{ 2 } })$

$P=\cos { (2013\cfrac { \pi }{ 2 } ) } +i\sin { (2013\cfrac { \pi }{ 2 } ) } =i\sin { (-\cfrac { \pi }{ 2 } ) }$

So, argument

$=-\cfrac { \pi }{ 2 }$

and modulus

$=1$

$z_1=\sqrt { 3 } -i,z_2=1+i\sqrt { 3 } ,$ then amp

$(z_1+z_2)=$

0%
$\dfrac { \pi }{ 12 }$
0%
$\dfrac { \pi }{ 15 }$
0%
$\dfrac { \pi }{ 6 }$
0%
$\dfrac { \pi }{ 4 }$

Explanation

$z_1=\sqrt { 3 } -i,z_2=1+i\sqrt { 3 }$

$$ z_1+z_2=(\sqrt { 3 }+1)+i(\sqrt{3}-1) $$

$z_1+z_2=(\sqrt { 3 }+1)+i(\sqrt{3}-1)$

$\left| z_1+z_2 \right|= \sqrt {(\sqrt{3}+1)^2+(\sqrt{3}-1)^2}=2\sqrt{2}$

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

$z=x+iy$ then which of the following relationship is correct:

0%
$x^{2}+y^{2}=1$
0%
$x^{2}+y^{2}=4$
0%
$x^{2}+y^{2}=2$
0%
$x^{2}+y^{2}=3$

Find the least positive value of n, if

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

0%
1
0%
2
0%
3
0%
4

Explanation

We have,

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

Therefore,

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

$i^n=1$

$\implies$ n is a multiple of 4.

The smallest positive value of n is 4.

Inequality

$a + i b > c + i d$ can be explained only when :

0%
$b=0,c=0$
0%
$b=0,d=0$
0%
$a=0,c=0$
0%
$a=0,d=0$

Explanation

$\begin{array}{l} a+ib>c+id \\ is\, \, valid\, \, only\, \, when\, \, b=0\, \, \& \, \, d=0\, \, as\, \, inequality\, \, \, in \\ complex\, no.\, \, is\, \, not\, \, defined. \end{array}$

$z_1=3+4i\\z_2=4-5i$ Then find

$z_1+z_2$

0%
7-i
0%
7+i
0%
7+9i
0%
None of these

Explanation

$z_1=3+4i\\z_2=4-5i\\z_1+z_2=3+4i+4-5i\\7-i$

$z_1=3+4i,z_2=2-i$ find

$z_2-z_1$

0%
-1-5i
0%
2-5i
0%
1+5i
0%
1-5i

Explanation

Given

$z_1=3+4i\\z_2=2-i\\z_2-z_1=2-i-3-4i=-1-5i$

The complex numbers

$z_1=8+9i, z_2=4-6i$ then

$z_1-z_2$

0%
$4+15i$
0%
$4-3i$
0%
$12+3i$
0%
$12-15i$

Explanation

$z_1=8+9i\\z_2=4-6i\\z_1-z_2=8+9i-4+6i=4+15i$

$z$ is a complex number such that

$|z|=1$ , then

$\left|\dfrac 1{\bar z}\right|$ is

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

Explanation

Let

$z=x+iy\\|z|=\sqrt{x^2+y^2}=1\\x^2+y^2=1\\\dfrac 1z=\dfrac1{x+iy}\\\dfrac{x-iy}{\sqrt{x^2+y^2}}=x-iy\\\left|\dfrac 1z\right|=\sqrt{x^2+y^2}=1$

The sum of prime numbers, out of the numbers

$17, 8, 21, 13, 41, 2, 27, 31, 51$ is:

0%
$125$
0%
$102$
0%
$104$
0%
$155$

Explanation

Prime numbers out of

$17,8,21,13,41,2,27,31,51$ are

$17,13,41,2,31$ .

Sum of prime numbers

$= 17+13+41+2+31=104$ .

$z_1=4+i,z_2=4-i$ find

$z_1z_2$

0%
$17$
0%
$16$
0%
$17-i$
0%
$16i$

Explanation

$z_1=4+i\\z_2=4-i\\z_1z_2=(4+i)(4-i)\\16-i^2=16+1=17$

$z_1=9+8i\ \ \ |z|=$

0%
$\sqrt {145}$
0%
$\sqrt {163}$
0%
$\sqrt {117}$
0%
$\sqrt {137}$

Explanation

$z=9+8i\\|z|=\sqrt {a^2+b^2}\\\\|z|=\sqrt {9^2+8^2} =\sqrt {81+64}=\sqrt {145}$

Mark against the correct answer in each of the following .

$i^{-38}=$ ?

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

Explanation

$i^{-38}\\=\dfrac {1}{i^{-38}}\times \dfrac {i^2}{i^2}\\=\dfrac {-1}{i^{40}}\\=\dfrac {-1}{(i^4)^{10}}\\=\dfrac {-1}{(1)^{10}}\\=\dfrac {-1}{1}\\=-1$

Mark against the correct answer in each of the following .

$i^{-75}=$ ?

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

Explanation

$i^{-75}\\=\dfrac {1}{i^{75}}\\=\dfrac {1}{i^{75}\times i}\times i\\=\dfrac {1}{i^{76}}\times i\\=\dfrac {1}{(i^{4})^{19}}\times i\\=1\times i\\=i$

Mark against the correct answer in each of the following .

$i^{124}=$ ?

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

Explanation

$i^{124}\\=(i^4)^{31}\\=(1)^{31}\\=1$

$(x+iy)(2-3i)=4+i\left ( \dfrac{1}{2} \right )$ then

$x + y =$

0%
$\dfrac{3}{2}$
0%
$\dfrac{1}{2}$
0%
$0$
0%
$\dfrac{2}{3}$

Explanation

$(x+iy)(2-3i)=4+i\dfrac{1}{2}$

$2x+2iy-3ix+3y=4+\dfrac{i}{2}$

$2x+3y+i(2y-3x)=4+\dfrac{i}{2}$
Hence

$2x+3y=4$ ................................(i)
And

$2y-3x=\dfrac{1}{2}$ .............................(ii)

Solving both the equations give us

$y=1$

$x=\dfrac{1}{2}$

Hence

$x+y$

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

$=\dfrac{3}{2}$ .

In the complex numbers, where

$i^{2} = -1$ , what is the value of

$5 + 6i$ multiplied by

$3- 2i$ ?

0%
$27$
0%
$27i$
0%
$27 + 8i$
0%
$15 + 8i$
0%
$15 - 18i$

Explanation

We know,

$i^2=-1$

The required product is

$(5 + 6i)(3-2i)$

$= 15-10i+18i+12$

$= 27 + 8i$

$m_1$ ,

$m_2$ ,

$m_3$ and

$m_4$ respectively denote the moduli of the complex numbers

$1 + 4i, 3 + i, 1 – i \ and\ 2 – 3i$ then the correct order among the following is :

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$m_1$ < $m_2$ < $m_3$ < $m_4$
0%
$m_4$ < $m_3$ < $m_2$ < $m_1$
0%
$m_3$ < $m_2$ < $m_4$ < $m_1$
0%
$m_3$ < $m_1$ < $m_2$ < $m_4$

Explanation

$m_{1}=|1+4i|=\sqrt{1+16}=\sqrt{17}$ .

$m_{2}=|3+i|=\sqrt{9+1}=\sqrt{10}$

$m_{3}=|1-i|=\sqrt{1+1}=\sqrt{2}$

$m_{4}=|2-3i|=\sqrt{4+9}=\sqrt{13}$
Hence

$m_{1}>m_{4}>m_{2}>m_{3}$

CBSE Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 3 - 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 3 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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