Number Theory Test 25

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Number Theory Test 25

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Question 1
1 / -0

The locus represented by $$|z -1|=|z + i|$$ is

A
a circle of radius $$1$$ unit

B
an ellipse with foci at $$(1, 0)$$ and $$(0, 1)$$

C
a straight line through the origin

D
a circle on the line joining $$(1, 0)$$ and $$(0, 1)$$ as diameter

Solution

Given,
$$ |z−1|=|z+i|$$
To find the locus of the equation.
Solution,
Given that,
$$|z−1|=|z+i|$$
Say $$x=x+iy$$
$$ \left| x+iy-1 \right| =\left| x+iy+1 \right| $$
$$ \left| (x-1)+iy \right| =\left| x+i(y+1) \right| $$
$$\Rightarrow \sqrt{(x-1)^2+y^2}=\sqrt{x^2+(y+1)^2}$$
Squaring both sides,
$$x^2+1-2x+y^2=x^2+y^2+1+2y$$
$$\Rightarrow 1-2x=2y+1$$
$$\therefore \ x+y=0$$ [line passes through origin]
Hence, C is the right answer
Question 2
1 / -0

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

A
$$\dfrac{1}{2}$$

B
$$\dfrac{-1}{2}$$

C
$$1$$

D
$$-1$$

Solution

Given, $$\displaystyle |(1 + i)\frac{(2+i)}{(3 + i)}| $$=$$\displaystyle |(1 + i)|\frac{|(2+i)|}{|(3 + i)|}$$

Since, $$|Z_{1}Z_{2}|=|Z_{1}||Z_{2}|$$ and $$\displaystyle|\frac{Z_{1}}{Z_{2}}|=\frac{|Z_{1}|}{|Z_{2}|}$$

$$=\displaystyle \sqrt{2}\cdot \frac{\sqrt{5}}{\sqrt{10}}=1$$

$$\therefore \displaystyle |(1 + i)\frac{(2+i)}{(3 + i)}| = 1$$

Hence, option C.
Question 3
1 / -0

If $$\displaystyle z= (i)^{(i)^{(i)}}$$ where $$ i = \sqrt{-1}$$, then $$z$$ is equal to

A
$$-i$$

B
$$-1$$

C
$$1$$

D
$$i$$

Solution

$$Z=(i)^{(i)^{(i)}}$$
$$i=e^{i\pi/2}$$
$$i^i=e^{i.(i\pi/2)}=e^{-\pi/2}$$
$$i^{i^i}=e^{i(-\pi/2)}\\ \implies \cos(\pi/2)-i\sin(\pi/2)\\ \implies 0-i=-i$$
Question 4
1 / -0

If x and y are real then which one of the following is true

A
|x - y| = |x| - |y|

B
|x + y| $$\leq$$ |x| - |y|

C
|x - y| $$\geq$$ |x| - |y|

D
|x + y| = |x| + |y|

Solution

We know that $$|x+y|\leq |x|+|y|$$
$$\Rightarrow |x-y+y|\leq |x-y|+|y|$$
$$\Rightarrow |x|-|y|\leq |x-y|$$
$$\therefore |x-y|\geq |x|-|y|$$
Hence, option C.
Question 5
1 / -0

If $$|z -2 + i| = |z-3-i|$$, then the locus of z is

A
$$2x -4y -5 = 0$$

B
$$2x + 4y -5 = 0$$

C
$$x -2y + 5 = 0$$

D
none of these

Solution

Given: $$ |z-2+i|=|z-3-i|$$

To find the locus of: $$ z$$.

Solution:

Let, $$z=x+iy$$

$$|z-2+i|=|z-3-i|$$

$$|x+iy-2+i|=|x+iy-3-i|$$

$$ |(x-2)+i(y+1)|=|(x-3)+i(y-1)|$$

On squaring both sides, we get

$$ \Rightarrow { \left( \sqrt { { (x-2) }^{ 2 }+{ (y+1) }^{ 2 } } \right) }^{ 2 }={ \left( \sqrt { { (x-3) }^{ 2 }+{ (y-1) }^{ 2 } } \right) }^{ 2 } \quad ...\left\{ |x+iy|=\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \right\} $$

$$ \Rightarrow { x }^{ 2 }+4-4x+{ y }^{ 2 }+1+2y$$$$ = { x }^{ 2 }+9-6x+{ y }^{ 2 }+1-2y$$

$$ \Rightarrow 2x+4y-5=0$$
Question 6
1 / -0

Which of the following number is composite?

A
$$1$$

B
$$4$$

C
$$2$$

D
$$3$$

Solution

A composite number is a positive nteger that has at least one positive divisor other than one and number itself.
Also all the even numbers are composite.
Hence, answer will be B.
Question 7
1 / -0

Let $$\displaystyle \left| Z_{r}-r\right| \leq r,\forall r=1,2,3,.......n.$$ Then $$\displaystyle \left| \sum_{r=1}^{n} Z_{r} \right| $$ is less than

A
$$n$$

B
$$2n$$

C
$$n(n+1)$$

D
$$\displaystyle \frac{n(n+1)}{2}$$

Solution

$$|Z_r - r| \le r$$

$$\therefore Z_r \le 2r$$

or $$|Z_r| \le 2r$$

also $$\sum_{r=1}^n |Z_r| \le 2 \sum_{r=1}^n r$$

also $$|\sum_{r=1}^n Z_r| \le \sum_{r=1}^n |Z_r|$$ $$\{$$ extending triangle inequality $$\}$$

$$\therefore |\sum_{r=1}^n Z_r| \le 2\times \dfrac{n(n+1)}{2}$$

$$\therefore |\sum_{r=1}^n Z_r| \le {n(n+1)}$$

$$\therefore$$ option C is correct.
Question 8
1 / -0

If $$z_1, z_2$$ and $$z_3$$ are complex numbers such that $$|z_1| = |z_2| = |z_3| = \left | \displaystyle \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} \right | = 1,$$ then $$|z_1 + z_2 + z_3|$$ is

A
equal to 1

B
less than 1

C
greater than 3

D
equal to 3

Solution

$$|z_1| = |z_2| = |z_3| = 1$$ (given)
Now $$|z_1| = 1 \Rightarrow |z_r|^2 = 1$$
$$\Rightarrow z_1 \bar z_1 = 1$$
Similarly $$z_2 \bar z_2 = 1, z_3 \bar z_3 = 1$$
Now $$\displaystyle \left | \dfrac{1}{z_1} + \dfrac{1}{z_2} + \dfrac{1}{z_3} \right | = 1$$
$$\Rightarrow |\bar z_1 + \bar z_2 + \bar z_3| = 1$$
$$\Rightarrow |\overline{z_1 + z_2 + z_3}| = 1 \Rightarrow |z_1 + z_2 + z_3| = 1$$
Question 9
1 / -0

Fill in the blanks.
Every composite number can be expressed as a product of ____, and this factorization is unique except for the order in which the prime factors occur.

A
Co-primes

B
Twin primes

C
Primes

D
All of these

Solution
Question 10
1 / -0

The number of composite number between $$101$$ and $$120$$ are

A
$$11$$

B
$$12$$

C
$$13$$

D
$$None$$

Solution

A composite number is a positive integer that has atleast one positive divisor other than one or the number itself. In other words, a composite number is any integer greater than one that is not a prime number

Now between $$101$$ and $$120,$$
Numbers are
$$102,104,105,106,108,110,111,112,114,115,116,117,118,119$$

Total : $$14.$$

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Number Theory Test 25

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Number Theory Test 25

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Question 1

a circle of radius $$1$$ unit

an ellipse with foci at $$(1, 0)$$ and $$(0, 1)$$

a straight line through the origin

a circle on the line joining $$(1, 0)$$ and $$(0, 1)$$ as diameter

Solution

Question 2

$$\dfrac{1}{2}$$

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

$$1$$

$$-1$$

Solution

Question 3

$$-i$$

$$-1$$

$$1$$

$$i$$

Solution

Question 4

|x - y| = |x| - |y|

|x + y| $$\leq$$ |x| - |y|

|x - y| $$\geq$$ |x| - |y|

|x + y| = |x| + |y|

Solution

Question 5

$$2x -4y -5 = 0$$

$$2x + 4y -5 = 0$$

$$x -2y + 5 = 0$$

none of these

Solution

Question 6

$$1$$

$$4$$

$$2$$

$$3$$

Solution

Question 7

$$n$$

$$2n$$

$$n(n+1)$$

$$\displaystyle \frac{n(n+1)}{2}$$

Solution

Question 8

equal to 1

less than 1

greater than 3

equal to 3

Solution

Question 9

Co-primes

Twin primes

Primes

All of these

Solution

Question 10

$$11$$

$$12$$

$$13$$

$$None$$

Solution

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