Number Theory Test 34

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Number Theory Test 34

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Question 1
1 / -0

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

A
Servlets

B
Hyperlink

C
Hypertext

D
Active X controls

Solution

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

An ActiveX control is a component program object that can be re-used by many application programs within a computer or among computers in a network. The technology for creating ActiveX controls is part of Microsoft's overall ActiveX set of technologies, chief of which is the Component Object Model (COM).

ActiveX controls can be downloaded as small programs or animations for Web pages, but they can also be used for any commonly-needed task by an application program in the latest Windows and Macintosh environments. In general, ActiveX controls replace the earlier OCX (Object Linking and Embedding custom controls). An ActiveX control is roughly equivalent in concept and implementation to the Java applet.

An ActiveX control can be created in any programming language that recognizes Microsoft's Component Object Model. The distributed support for COM is called the Distributed Component Object Model (DCOM). In implementation, an ActiveX control is a dynamic link library (DLL) module. An ActiveX control runs in what is known as a container, an application program that uses the Component Object Model program interfaces.
Question 2
1 / -0

Which of the following is a prime number?

A
$$161$$

B
$$221$$

C
$$373$$

D
$$437$$

Solution

A prime number is a whole number greater than $$1$$ whose only factors are $$1$$ and itself.

$$\Rightarrow$$ $$\sqrt{437} > 22$$

$$\Rightarrow$$  All prime numbers less than $$22$$ are $$: 2, 3, 5, 7, 11, 13, 17, 19.$$

$$\Rightarrow$$  $$161$$ is divisible by $$7$$, and $$221$$ is divisible by $$13$$.

$$\Rightarrow$$  $$373$$ is not divisible by any of the above prime numbers.

$$\therefore$$ $$373$$ is prime number
Question 3
1 / -0

If z satisfies $$\left| {z - 1} \right| < \left| {z + 3} \right|$$ then $$w = 2z + 3 - i$$ , ( where $$w = 2z + 3 - i$$ ) satisfies:

A
$$\left| {w - 5 - i} \right| < \left| {w + 3i} \right|$$

B
$$\left| {w - 5} \right| < \left| {w + 3} \right|$$

C
$$\left( {iw} \right) > 1$$

D
$$\left| {\arg \left( {w - 1} \right)} \right| < {\pi \over 2}$$

Solution
Question 4
1 / -0

If $$z_1 \, z_2$$ be two distinct complex numbers and let z = (1 - t) $$z_1$$ + t$$z_2$$ for some real number t with 0 < t < 1. If arg $$(\omega)$$ denotes the principal argument of a non-zero complex number $$(\omega)$$, then

A
$$|z \, - \, z_1| \, + \, |z \, - \, z_2| \, = \, |z_2\, - \, z_1|$$

B
$$arg(z \, - \, z_1) \, = \, arg (z \, - \, z_2)$$

C
$$\begin{vmatrix}

z \, - \, z_1 & \bar z \, - \, \bar{z_1}\\

z_2 \, - \, z_1 & \bar{z_2} \, - \, \bar{z_1}

\end{vmatrix}$$

D
$$arg(z \, - \, z_1) \, = \, arg (z_2 \, - \, z_1)$$

Solution

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,
$$|z-z_{1}|+|z-z_{2}|=|z_{1}-z_{2}|$$
Question 5
1 / -0

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

A
$$i$$

B
$$2i$$

C
$$1 - i$$

D
$$1 - 2i$$

Solution

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.
Question 6
1 / -0

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

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

B
$$1$$

C
$$\sqrt { 2 }$$

D
$$2$$

Solution

$$ \left| z+\dfrac { 1 }{ z } \right| ^{ 2 }={ r }^{ 2 }+\dfrac { 1 }{ { r }^{ 2 } } +2r.\dfrac { 1 }{ r } \cos { \left( \theta +\theta \right) } =4$$
or $${ r }^{ 2 }+\dfrac { 1 }{ { r }^{ 2 } } =4-2\cos { 2\theta }$$
$$\therefore\quad \left( r-\dfrac { 1 }{ r } \right) ^{ 2 }=2\left( 1-\cos { 2\theta } \right) =4\sin { ^{ 2 }\theta }$$
$$\left| { r }_{ 1 }-{ r }_{ 2 } \right| =\left| r-\dfrac { 1 }{ r } \right| =2\sin { \theta } =2\sin { \dfrac { \pi }{ 4 } } =\sqrt { 2 }$$
Question 7
1 / -0

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

A
$$-1/5$$

B
$$1/5$$

C
$$1/10$$

D
$$-1/10$$

Solution

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 $$-\dfrac{1}{5}$$.
Question 8
1 / -0

If $$\dfrac{2z_1}{3z_2}$$ is a purely imaginary number,then $$\left|\dfrac{z_1-z_2}{z_1+z_2}\right|=$$

A
$$3/2$$

B
$$1$$

C
$$2/3$$

D
$$4/9$$

Solution

$$\dfrac{2z_1}{3z_2}=k'i$$
$$\dfrac{z_1}{z_2}=\dfrac{3k'i}{2}$$
$$\dfrac{z_1}{z_2}=ki$$ where $$\dfrac{3k'}{2}=k$$
$$z_1=kiz_2$$
Thus $$\left|\dfrac{z_1-z_2}{z_1+z_2}\right|=\left|\dfrac{kiz_2-z_2}{kiz_2+z_2}\right|$$

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

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

$$=\dfrac{\sqrt{k^2+1^2}}{\sqrt{k^2+1^2}}=1$$
Question 9
1 / -0

Find the real number $$x$$ if $$(x-2i)(1+i)$$ is purely imaginary.

A
$$2$$

B
$$-2$$

C
$$4$$

D
$$-4$$

Solution

$$(x-2i)(1+i)$$
$$ = x+ix -2i-2i^2$$
$$= (x+2) -i(x-2)$$ is purely imaginary if (x+2)=0
$$\therefore x=-2$$
Question 10
1 / -0

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

A
$$ - 1 - i$$

B
$$ - 1 + i$$

C
$$ 1 - i$$

D
$$1 + i$$

Solution

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, $$a={ i }^{ 0 }=1$$
$$r=i$$
$$n={ 102 }$$
$${ i }^{ 102 }t=\cfrac { 1\left( { i }^{ 102 }-1 \right) }{ i-1 } =\cfrac { { i }^{ 2 }-1 }{ i-1 } =\cfrac { -2\left( i+1 \right) }{ \left( i-1 \right) \left( i+1 \right) } $$
$${ i }^{ 102 }t=i+1\Rightarrow { i }^{ 2 }t=i+1$$
$$t=-i-1$$

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Re-Attempt Weekly Quiz Competition

Number Theory Test 34

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Number Theory Test 34

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SHARING IS CARING

Question 1

Servlets

Hyperlink

Hypertext

Active X controls

Solution

Question 2

$$161$$

$$221$$

$$373$$

$$437$$

Solution

Question 3

$$\left| {w - 5 - i} \right| < \left| {w + 3i} \right|$$

$$\left| {w - 5} \right| < \left| {w + 3} \right|$$

$$\left( {iw} \right) > 1$$

$$\left| {\arg \left( {w - 1} \right)} \right| < {\pi \over 2}$$

Solution

Question 4

$$|z \, - \, z_1| \, + \, |z \, - \, z_2| \, = \, |z_2\, - \, z_1|$$

$$arg(z \, - \, z_1) \, = \, arg (z \, - \, z_2)$$

$$\begin{vmatrix}z \, - \, z_1 & \bar z \, - \, \bar{z_1}\\ z_2 \, - \, z_1 & \bar{z_2} \, - \, \bar{z_1} \end{vmatrix}$$

$$arg(z \, - \, z_1) \, = \, arg (z_2 \, - \, z_1)$$

Solution

Question 5

$$i$$

$$2i$$

$$1 - i$$

$$1 - 2i$$

Solution

Question 6

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

$$1$$

$$\sqrt { 2 }$$

$$2$$

Solution

Question 7

$$-1/5$$

$$1/5$$

$$1/10$$

$$-1/10$$

Solution

Question 8

$$3/2$$

$$1$$

$$2/3$$

$$4/9$$

Solution

Question 9

$$2$$

$$-2$$

$$4$$

$$-4$$

Solution

Question 10

$$ - 1 - i$$

$$ - 1 + i$$

$$ 1 - i$$

$$1 + i$$

Solution

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$$\begin{vmatrix}

z \, - \, z_1 & \bar z \, - \, \bar{z_1}\\

z_2 \, - \, z_1 & \bar{z_2} \, - \, \bar{z_1}

\end{vmatrix}$$