If $$\tan^{-1}(\alpha+ i\beta) = x+iy,$$ then $$x$$ is equal to

$$\frac{1}{2}\tan^{-1}\left ( \dfrac{2\alpha}{1-\alpha^2-\beta^2}\right)$$

$$\frac{1}{2}\tan^{-1}\left ( \dfrac{2\alpha}{1+\alpha^2+\beta^2}\right)$$

$$\tan^{-1}\left ( \dfrac{2\alpha}{1-\alpha^2-\beta^2}\right)$$

If $$(x+iy)(2-3i)=4+i$$ then (x, y) =

$$\left ( 1,\dfrac{1}{13} \right )$$

MCQ Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 1

$(\dfrac{3-z_{1}}{2-z_{1}})(\dfrac{2-z_{2}}{3-z_{2}})=k$ , then point

$A(z_{1}, z_{2}), C(3, 0)$ and

$D(2, 0)$ (taken in clockwise sense ) will

0%
lie on a circle only for $k>0$
0%
lie on a circle only for $k=0$
0%
lie on a circle $\forall k\in R$
0%
be vertices of a square $\forall k\in (0, 1)$

$\tan^{-1}(\alpha+ i\beta) = x+iy,$ then

$x$ is equal to

0%
$\frac{1}{2}\tan^{-1}\left ( \dfrac{2\alpha}{1-\alpha^2-\beta^2}\right)$
0%
$\frac{1}{2}\tan^{-1}\left ( \dfrac{2\alpha}{1+\alpha^2+\beta^2}\right)$
0%
$\tan^{-1}\left ( \dfrac{2\alpha}{1-\alpha^2-\beta^2}\right)$
0%
None of the above

The complex numbers

${ z }_{ 1 },{ z }_{ 2 },{ z }_{ 2 }$ satisfying

$\dfrac { { z }_{ 1 }+{ z }_{ 3 } }{ { z }_{ 2 }-{ z }_{ 3 } } =\dfrac { 1-i\sqrt { 3 } }{ 2 }$

0%
If area zero
0%
right angled isosceles
0%
equilateral
0%
obtuse angled

$z$ is a complex number such that

$| z | = 1 , z \neq 1 ,$ then the real part of

$\frac { z - 1 } { z + 1 }$ is

0%
$\frac { 1 } { | z + 1 | ^ { 2 } }$
0%
$\frac {- 1 } { | z + 1 | ^ { 2 } }$
0%
$\frac { \sqrt 2 } { | z + 1 | ^ { 2 } }$
0%
0

Evaluate:

$\dfrac {3+2 i\sin \theta}{1-2i \sin \theta},\theta \epsilon \left(-\dfrac {\pi}{2},\pi\right)$ is

0%
$2\pi/3$
0%
$\pi/3$
0%
$4\pi/3$
0%
$\pi$

Find the modules and amplitude for each of the following complex numbers

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

Let

$\alpha,\ \beta$ be real and

$\mathrm{z}$ be a complex number. If

$\mathrm{z}^{2}+\alpha \mathrm{z}+\beta=0$ has two distinct roots on the line

$Re(z) =1$ , then it is necessary that:

0%
$\beta\in(0,1)$
0%
$\beta\in(-1,0)$
0%
$|\beta|=1$
0%
$\beta\in(1, \infty)$

Which of the following is/are not twin prime(s)?

0%
$(2, 3)$
0%
$(41,43)$
0%
$(17, 19)$
0%
$(59, 61)$

Explanation

$3 - 2 = 1$
They don't differ by

$2$ .

Rest other options are prime numbers that differ by

$2.$

So, option

$A$ is correct.

Sum of first three prime numbers which end at 3 is:

0%
20
0%
16
0%
39
0%
40

Explanation

First three prime numbers which end with

$3$ are

$3,13,23$ .

$Sum=3+13+23=39$

Which of the following is the factor of all prime numbers?

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

Explanation

Option

$A$ is correct because

$1$ is the factor of all prime numbers.

What is the sum of the smallest prime and the largest prime less than

$10$ ?

0%
$7$
0%
$9$
0%
$10$
0%
$11$
0%
$12$

Explanation

Smallest prime number less than

$10$ is

$2$ and largest prime number less than

$10$ is

$7$ and their sum is

$2+7=9$

For

$i=\sqrt{-1}$ , what is the sum

$\left(7+3i\right) + \left(-8+9i\right)$ ?

0%
$-1+12i$
0%
$-1-6i$
0%
$15+12i$
0%
$15-6i$

Explanation

Given:

$(7+3i)+(-8+9i)$

$= (7+(-8))+(3i+9i)$

$=(7-8)+12i$

$=-1+12i$

Option A is correct.

Perform the indicated operations:

$(8-2i)-(-2-6i)$

0%
$6+4i$
0%
$10+4i$
0%
$10+8i$
0%
$10-8i$

Explanation

The value of

$(8-2i)-(-2-6i)$

$= 8+2+i(6-2)$

$= 10+4i$

$|\mathrm{z}^{2}-1|=|\mathrm{z}|^{2}+1$ , then

$\mathrm{z}$ lies on

0%
the real axis
0%
the imaginary axis
0%
a circle
0%
an ellipse

Explanation

$|\mathrm{z}^{2}-1|=|\mathrm{z}|^{2}+1$

$|\mathrm{z}^{2}-1|^2=(|\mathrm{z}|^{2}+1)^2\Rightarrow (z^2-1)(\bar{z}^2-1)=|z|^4+2|z|^2+1$

$\Rightarrow |z|^4-\bar{z}^2-z^2+1=|z|^4+2|z|^2+1$

$\Rightarrow \bar{z}^2+z^2+2|z|^2=0\Rightarrow \bar{z}^2+z^2+2z\bar{z}=0$

$\Rightarrow (z+\bar{z})^2=0\Rightarrow z+\bar{z}=0$

$\Rightarrow z$ is purely imaginary. Hence

$z$ lies on imaginary axis.

$z = x + iy$ and

$\omega = \dfrac{(1 -iz)}{(z-i)}$ , then

$\left|\omega\right| = 1$ implies that in the complex plane

0%
z lies on the imaginary axis
0%
z lies on the real axis
0%
z lies on the unit circle
0%
none of these

Explanation

Given

$w=\dfrac { 1-iz }{ z-i }$ and

$\left| w \right| =1$

$\Rightarrow \left| \dfrac { 1-iz }{ z-i } \right| =1$ ...(1)

Substitute

$z=x+iy$ in equation (1)

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

$\Rightarrow \left| 1+y-ix \right| =\left| x+i\left( y-1 \right) \right| \\ \Rightarrow { \left( 1+y \right) }^{ 2 }+{ x }^{ 2 }={ x }^{ 2 }+{ \left( y-1 \right) }^{ 2 }\\ \Rightarrow y=0$

Therefore z lies on the real axis.

Ans: B

$(x+iy)(2-3i)=4+i$ then (x, y) =

0%
$\left ( 1,\dfrac{1}{13} \right )$
0%
$\left ( -\dfrac{5}{13},\dfrac{14}{13} \right )$
0%
$\left ( \dfrac{5}{13},\dfrac{14}{13} \right )$
0%
$\left ( -\dfrac{5}{13},-\dfrac{14}{13} \right )$

Explanation

$(x+iy)(2-3i)=4+i$

$2x-(3x)i+(2y)i-3yi^{2}=4+i$

$\underbrace{2x+3y}_{Real }+\underbrace{(2y-3x)}_{Imaginary }i=4+i$

Comparing the real & imaginary parts,

$2x+3y=4$ --------------------------(1)

$2y-3x=1$ ----------------------------(2)
Solving eq(1) & eq(2),

$4x+6y=8$

$-9x+6y=3$

$13x=5\Rightarrow x=\dfrac{5}{13}$

$y=\dfrac{14}{13}$

$\therefore (x,y)=\left (\dfrac{5}{13},\dfrac{14}{13} \right )$

$z =3+5i$ , then

$z^3+z+198=$

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

Explanation

$z=3+5i$

$z^{3}=(3+5i)^{3}$

$=3^{3}+3.3^{2}(5i)+3.3(5i)^{2}+(5i)^{3}$

$=27-125i+135i-225$

$=-225+27+(135-125)i$

$=-198+10i$

$\therefore z^{3}+z+198$

$=-198+10i+3+5i+198$

$=3+15i$

The modulus of

$\sqrt{2}i-\sqrt{-2}i$ is:

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

Explanation

Let

$z=\sqrt{2}i-\sqrt{-2}i$

$\Rightarrow z=\sqrt{2}i-\sqrt{2}i^2$

$\Rightarrow z=\sqrt{2}+\sqrt{2}i$
Now,

$|z|=\sqrt{2+2}=2$

$z=2-3i$ then

$z^2-4z+13=$

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

Explanation

$z=2-3i$

$z^{2}=2^{2}-3^{2}-12i$

$=-5-12i$

$\therefore z^{2}-4z+13$

$=(-5-12i)-4(2-3i)+13$

$=-5-12i-8+12i +13$

$=-13+13$

$=0$

The complex number

$\displaystyle \frac{1+2i}{1-i}$ lies in the quadrant :

0%
I
0%
II
0%
III
0%
IV

Explanation

Let

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

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

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

$=\dfrac{1+3i-2}{1+1}$ ............

$[\because i^2 = -1]$

$\Rightarrow z =\dfrac{-1+3i}{2}$

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

$=x+iy$
Clearly

$x<0$ and

$y>0$
Hence

$z$ lies in

$\text{II}$ quadrant

$\sqrt{-3}\sqrt{-75}=$

0%
$15$
0%
$15i$
0%
$-15$
0%
$-15i$

Explanation

$\sqrt{-3}\times\sqrt{-75}=\sqrt{3\times(-1)}\sqrt{75\times(-1)}$

$=\sqrt{3}i\times\sqrt{75}i$

$=\sqrt{225}i^{2}$

$= - 15$

The sum of two complex numbers

$a + ib$ and

$c +id$ is a real number if

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

Explanation

It is given that

$z_{1}=a+ib$ and

$z_{2}=c+id$

Then

$z_{1}+z_{2}=(a+c)+i(b+d)$

Now

$(z_{1}+z_{2})$ is purely real.

Then the imaginary part has to be

$0$ .

Hence

$b+d=0$ .

The locus of complex number z such that z is purely real and real part is equal to - 2 is

0%
Negative y-axis
0%
Negative x-axis
0%
The point (-2, 0)
0%
The point (2, 0)

Explanation

$z=x+iy$

$(x,y)$

$z$ is purely real and the real part equals

$-2$

$\therefore y=0$ &

$x=-2$

$z=-2$
Hence, this would be represented by the point (-2,0) on the Argand Plane.

$\dfrac{1}{i-1}+\dfrac{1}{i+1}$ is

0%
positive rational number
0%
purely imaginary
0%
positive Integer
0%
negative integer

Explanation

Let

$Z= \dfrac{1}{i-1}+\dfrac{1}{i+1}$

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

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

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

$\therefore Z=-i$

The sum of two complex numbers

$a + ib$ and

$c+ id$ is purely imaginary if

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

Explanation

It is given that

$z_{1}=a+ib$ and

$z_{2}=c+id$

$z_{1}+z_{2}=(a+c)+i(b+d)$

$z_{1}+z_{2}$ is purely imaginary. (Given)
Then the real part has to be

$0$ .
Hence

$a+c=0$ .

Which of the following is not a prime number ?

0%
5
0%
7
0%
8
0%
11

Which of the following is not a composite number?

0%
$4$
0%
$6$
0%
$7$
0%
$8$

The number of prime numbers between 0 and 20 is

0%
7
0%
8
0%
6
0%
9

Explanation

The prime numbers between

$0$ and

$20$ are

$2, 3, 5, 7, 11, 13, 17$ and

$19$ .
Hence, the number of prime numbers between

$0$ and

$20$ is

$8$ .

The units digit of every prime number (other than

$2$ and

$5$ ) must be necessarily

0%
$1, 3$ or $5$
0%
$1, 3, 7$ or $9$
0%
$7$ or $9$
0%
$1$ or $7$

Explanation

All even numbers (other than

$2$ ) are composite, so prime numbers cannot end with any of the digits

$0,2,4,6,8$ .

Also, a number with unit digit

$5$ will be divisible by

$5.$

Therefore, the unit's digit of every prime number(other than

$5$ ) must be necessarily

$1, 3, 7$ or

$9$ .

Which of the following is not a composite number?

0%
$352+6$
0%
$357+7$
0%
$352+7$
0%
$353+1$

Explanation

$A:$

$352+6 = 358$ which is even, and have factors other than

$1$ and itself, So it is a Composite number.

$B:$

$357+7 = 364$ which is even, and have factors other than

$1$ and itself, So it is a Composite number.

$C:$

$352+7 = 359$ which is odd, and does not have factors other than

$1$ and itself, So it is a Prime number.

$D:$

$353+1 = 354$ which is even, and have factors other than

$1$ and itself, So it is a Composite number.

Hence, option

$C.$

The total number of prime numbers between

$120$ and

$140$ is

0%
$7$
0%
$6$
0%
$5$
0%
$4$

Explanation

The prime numbers must be from the odd numbers between

$120$
and

$140$ .
The prime numbers are

$127, 131, 137, 139$ .
Therefore, there are 4 prime numbers between

$120$ and

$140$ .

If the square of

$(a + ib)$ is real, then

$ab=$

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

Explanation

$(a+ib)^{2}=a^{2}-b^{2}+2iab$ is given to be real

$\Rightarrow ab=0$

Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique apart from the order in which the prime factors occur.

0%
True
0%
False
0%
Neither
0%
Either

Explanation

This is true as any composite number can be expressed in terms of its prime factors. The prime factors for 2 numbers can be same but their combination is unique.
Example,

$12=2^2*3$ and

$18=2*3^2$
Both have prime factors as 2 and 3 but the combination that 12 has is different from that of 18.

The number of composite numbers between

$101$ and

$120$
(excluding both) are

0%
$11$
0%
$12$
0%
$13$
0%
$14$

Explanation

There are

$9$ even numbers which are all composite numbers.
Amongst the odd numbers

$105, 111, 115, 117, 119$ are composite numbers.
So, there are

$14$ composite numbers in the given range.

What is the remainder obtained when a prime number greater than

$6$ is divided by

$6$ ?

0%
$1 \ or\ 3$
0%
$1 \ or\ 5$
0%
$3\ or\ 5$
0%
$4 \ or\ 5$

Explanation

We can take an example. Prime numbers greater than $6$ are $7, 11, 13,$ and so on. When $7$ is divided by $6,$ remainder is $1$ . When $11$ is divided by $6$ , remainder is $5$ . Similarly, when $13$ is divided by $6$ , remainder is $1$ . The remainder obtained is either $1$ or $5$ .

So the correct answer will be option $B$

A number other than one which is either divisible by

$1$ or itself is called a

0%
composite number
0%
prime number
0%
comprime number
0%
none of these

Explanation

Given that a number is divisible by itself and $1$ .
$\therefore$ If x is the number then its factors are x and $1$ . So this fits in the condition for a number to be prime.
$\therefore$ It is a prime number.

Write two pairs of twin primes between

$20$ and

$50.$

0%
$16$ and $26, 19$ and $29$
0%
$29$ and $31, 41$ and $43$
0%
$11$ and $21, 61$ and $71$
0%
$15$ and $17, 37$ and $39$

Explanation

A twin prime is a set of prime numbers that must be either

$2$ less or

$2$ more than another prime number. Example

$5$ and

$7, 11$ and

$13.$

$29$ and

$31, 41$ and

$43$ are two pairs of twin primes between

$20$ and

$50.$

Find the value of

$x$ of the equation

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

0%
$1$
0%
$2$
0%
$0$
0%
none of these

Explanation

Then

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

$\Rightarrow |{(1-i)}|^{x} =|2|^{x}$

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

$\Rightarrow { \left( \sqrt { 2 } \right) }^{ x }={ 2 }^{ x }$

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

$\Rightarrow 2x=x\Rightarrow x=0$

Hence, option C is correct.

Which of the following is a prime number?

0%
$889$
0%
$997$
0%
$899$
0%
$1147$

Explanation

A prime can be written as

$6k+1$ or

$6k-1$ provided the number is greater than

$6$ .

$997=166*6+1$
Therefore,

$997$ is a prime number.

Write all prime numbers between

$20$ and

$50.$

0%
$21, 29, 33, 37, 41, 43$ and $47$
0%
$29, 33, 37, 39, 41, 43$ and $47$
0%
$23, 29, 31, 37, 41, 43$ and $47$
0%
None of these

Explanation

All prime numbers between

$20$ and

$50$ are

$23, 29, 31, 37, 41, 43$ and

$47.$

Solve

$\displaystyle \left ( 1-i \right )x+\left ( 1+i \right )y= 1-3i,$

0%
$\displaystyle x= -1, y= 2.$
0%
$\displaystyle x= 2, y= -1.$
0%
$\displaystyle x= 2, y= 1.$
0%
$\displaystyle x= 1, y= 2.$

Explanation

$\displaystyle \left ( 1-i \right )x+\left ( 1+i \right )y= 1-3i.$
Equating real and imaginary parts, we get

$\displaystyle x+y= 1$ and

$\displaystyle -x+y= -3.$
Adding both equations we get

$y=-1$

Substituting this value of

$y$ in any of the 2 equations, we get

$x=2$

$\therefore \displaystyle x= 2, y= -1.$

Ans: B

Evaluate :

$\sqrt{-25} + 3 \sqrt{-4} +2 \sqrt{-9}$

0%
$-17i$
0%
$5i$
0%
$17i$
0%
$6i$

Explanation

$\sqrt{-25} + 3 \sqrt{-4} +2 \sqrt{-9}$

$=5\sqrt{-1}+6\sqrt{-1}+6\sqrt{-1}$ we know,

$\sqrt{-1}=i$

$= 5i + 6i + 6i = 17i$

How many prime numbers are there between

$0$ and

$30$ ?

0%
$9$
0%
$10$
0%
$8$
0%
$11$

Explanation

Prime numbers:- The numbers which are divisible by

$1$ and themselves, or the number which has only two factors,

$1$ and the number itself

e.g.

$7$

Note:-

$1$ is not a prime number, the smallest prime number is

$2$

Prime numbers between

$1$ and

$30$ are

$2,~3,~5,~7,~11,~13,~17,~19,~23,~29.$

Total ten numbers are there.

An example for twin primes is

0%
$5, 11$
0%
$3, 5$
0%
$11, 17$
0%
$3, 7$

Explanation

Twin primes are those prime which has only one even number between them.
Out of given options clearly

$3$ and

$5$ are twin primes as only even number

$4$ lie between them.
Option

$B$ is correct.

The number which is neither prime nor composite is

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

Explanation

A prime number is a number which has only two factors,

$1$ and the number itself. So, a prime number has 2 factors.

A composite number is a number that has factors other than

$1$ and the number itself. A composite number has more than 2 factors.

The number which is neither prime nor composite is

$1$ as it is divisible only by

$1$ . It has only one factor.

Hence, Option B is correct answer.

The two consecutive prime numbers with difference

$2$ are called

0%
co-primes
0%
twin primes
0%
composite
0%
none of these

Explanation

Twin prime numbers are two prime numbers whose difference is

$2.$
Example :

$3$ and

$5,$

$17$ and

$19 ,41$ and

$43$ etc.

Number of prime numbers from 1 to 50 are

0%
$18$
0%
$12$
0%
$15$
0%
$20$

Explanation

There are 15 prime numbers from 1 to 50. They are

$2, 3, 5, 7, 11, 13, 17,19, 23, 29, 31,37,41,43,47.$

The numbers which have more than two factors are called

0%
even
0%
prime
0%
composite
0%
none of these

All prime numbers are

0%
even numbers
0%
odd numbers
0%
composite numbers
0%
odd numbers except 2

Explanation

Prime numbers are

$2,3,5,7,11,13,....$

Only

$2$ is an even number and others are odd numbers.

So, all prime numbers are odd numbers except for

$2$ .

Hence, correct answer is option D.

The prime number that comes just after 47 is ......

0%
57
0%
51
0%
53
0%
none of these

Explanation

After

$47$ ,

$48,49,50,51,52$ are composite numbers; but,

$53$ is a prime number, as it has no factor other than 1 and itself.

So correct answer will be option C

CBSE Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 1 - 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 1 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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