Number Theory Test 28

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Number Theory Test 28

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Weekly Quiz Competition

Question 1
1 / -0

What is the approximate magnitude of $$8 + 4i$$?

A
4.15

B
8.94

C
12.00

D
18.64

E
32.00

Solution

Magnitude of $$8+4i$$ is
$$=\sqrt { { 8 }^{ 2 }+{ 4 }^{ 2 } } $$
$$=\sqrt { 64+16 } $$
$$=\sqrt { 80 }$$
$$ =8.94$$
Question 2
1 / -0

The real part of $${ \left( 1-\cos { \theta } +i\sin { \theta } \right) }^{ -1 }$$ is

A
$$\cfrac{1}{2}$$

B
$$\cfrac { 1 }{ 1+\cos { \theta } } $$

C
$$\tan { \cfrac { \theta }{ 2 } } $$

D
$$\cot { \cfrac { \theta }{ 2 } } $$

Solution

Given that: $${ \left( 1-\cos { \theta } +i\sin { \theta } \right) }^{ -1 }=\dfrac{1}{1-\cos\theta+i\sin\theta}$$

$$=\dfrac{1}{2\sin^2\frac{\theta}{2}+i\cdot 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}=\dfrac{1}{2\sin\frac{\theta}{2}}\cdot \dfrac{1}{\sin\frac{\theta}{2}+i\cos\frac{\theta}{2}}$$

$$=\dfrac{1}{2\sin\frac{\theta}{2}}\cdot \dfrac{1}{\sin\frac{\theta}{2}+i\cos\frac{\theta}{2}}\cdot \dfrac{\sin\frac{\theta}{2}-i\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}-i\cos\frac{\theta}{2}}$$, rationalize denominator

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

$$=\dfrac{1}{2}-\dfrac{i}{2}\cot\frac{\theta}{2}$$

The real part is $$\dfrac{1}{2}$$
Question 3
1 / -0

If $$z = \dfrac {(\sqrt {3} + i)^{3} (3i + 4)^{2}}{(8 + 6i)^{2}}$$, then $$|z|$$ is equal to

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

Uisng the identity $$|a+ib|=\sqrt{a^2+b^2}$$
Since $$\bigg|\dfrac{z_1z_2}{z_3} \bigg|=\dfrac{|z_1||z_2|}{|z_3|}$$
Therefore
$$|z| = \bigg|\dfrac {(\sqrt {3} + i)^{3} (3i + 4)^{2}}{(8 + 6i)^{2}}\bigg|$$
$$\quad =\dfrac{|(\sqrt 3+i)^3||(3i+4)^2|}{|(8+6i)^2|}$$
$$\quad =\dfrac{|\sqrt 3+i|^3|3i+4|^2}{|8+6i|^2}$$
$$\quad =\dfrac{2^3\cdot 5^2}{10^2}=2$$
Question 4
1 / -0

The value of $$(a+2i)(b-i)$$ is

A
$$a+b-i$$

B
$$ab+2$$

C
$$ab+(2b-a)i+2$$

D
$$ab-2$$

E
$$ab+(2b-a)i-2$$

Solution

The value of $$(a+2i)(b-i) $$
$$=a(b-i)+2i(b-i)$$
$$= ab-ai+2bi-2{i}^{2} $$
$$= ab+(2b-a)i+2$$
Question 5
1 / -0

The expression $$\dfrac{3-4i}{5+3i}$$ is equivalent to

A
$$\dfrac{27-29i}{34}$$

B
$$\dfrac{27-29i}{16}$$

C
$$\dfrac{3-29i}{34}$$

D
$$\dfrac{1}{8}$$

E
$$15-8i$$

Solution

Given, $$\dfrac { 3-4i }{ 5+3i } $$
Multiply and divide it with $$5-3i$$, we get
$$\dfrac { 3-4i }{ 5+3i } \times \dfrac { 5-3i }{ 5-3i } $$
$$=\dfrac {(3-4i)(5-3i)}{5^2-(3i)^2}$$
$$=\dfrac { 3-29i }{ 34 } $$
Question 6
1 / -0

The fraction $$\dfrac{1}{1+i}$$ is equivalent to

A
$$1-i$$

B
$$\dfrac{1+i}{2}$$

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

D
$$i$$

E
$$-i$$

Solution

Given, $$\dfrac{1}{1+i}$$
Multiply and divide it with $$1-i$$, we get
$$\dfrac{1}{1+i} \times \dfrac{1-i}{1-i} = \dfrac{1-i}{2}$$
Question 7
1 / -0

If $$(2 - i)\times(a - bi) = 2 + 9i$$, where i is the imaginary unit and a and b are real numbers, then a equals

A
$$3$$

B
$$2$$

C
$$1$$

D
$$0$$

E
$$-1$$

Solution

Given, $$(2-i)(a-bi) = 2a-2bi-ai-b $$
$$= (2a-b)+i(-a-2b)$$
$$ = 2+9i$$
By comparing, we get $$2a-b=2$$ and $$a+2b=-9$$
By solving, we get $$a=-1$$ and $$b=-4$$
Question 8
1 / -0

Perform the indicated operations:
$$(5+3i)(3-2i)$$

A
$$21-2i$$

B
$$19-3i$$

C
$$11-2i$$

D
$$21-i$$

Solution

The value of $$(5+3i)(3-2i) $$
$$=5(3-2i)+3i(3-2i)$$
$$= 15-10i+9i-6{i}^{2} $$
$$= 21-i$$
Question 9
1 / -0

If $$x + i y = \dfrac{3}{2 + cos \theta + i sin \theta}$$, then $$x^2 + y^2$$ is equal to

A
$$3x - 4$$

B
$$4x - 3$$

C
$$4x +3$$

D
None of these

Solution

$$x + i y = \dfrac{3}{2 + cos \theta + i sin \theta} = \dfrac{3 (2 + cos \theta - i sin \theta)}{(2 + cos \theta)^2 - i^2 sin^2 \theta}$$
$$= \dfrac{3(2+cos \theta-i sin \theta)}{(2+cos \theta)^2+sin \theta}$$

$$= \dfrac{6 + 3 cos \theta - 3 i sin \theta}{4 + cos^2 \theta + 4 cos \theta + sin^2 \theta}$$

$$= \dfrac{6 + 3 cos \theta - 3 i sin \theta}{5 + 4 cos \theta}$$

$$= \left( \dfrac{6 + 3 cos \theta}{5 + 4 cos \theta} \right ) + \left( \dfrac{-3 sin \theta}{5 + 4 cos \theta} \right )$$
On equating real and imaginary parts, we get
$$x = \dfrac{3 (2 + cos \theta)}{5 + 4 cos \theta}$$
and $$y = \dfrac{-3 sin \theta}{5 + 4 cos \theta}$$

$$\therefore x^2 + y^2 = \dfrac{9 [4 + cos^2 \theta + 4 cos \theta + sin^2 \theta]}{(5 + 4 cos \theta)^2}$$

$$= \dfrac{9}{5 + 4 cos \theta} = 4 \left( \dfrac{6 + 3 cos \theta}{5 + 4 cos^2 \theta} \right ) - 3$$
$$= 4x - 3$$
Hence, option B is correct.
Question 10
1 / -0

Let $$z = x + iy$$, where $$x$$ and $$y$$ are real. The points $$(x, y)$$ in the $$X-Y$$ plane for which $$\dfrac {z + i}{z - i}$$ is purely imaginary lie on

A
A straight line

B
An ellipse

C
A hyperbola

D
A circle

Solution

Let $$z = x + iy$$
$$\therefore \dfrac {z + i}{(z - i)}=\dfrac{x+iy+i}{x+iy-i}$$
$$= \dfrac {(x + i(y + 1))}{(x - i(1 - y))}\times \dfrac {(x + i(1 - y))}{(x + i(1 - y))}$$
$$=\dfrac {x^{2} + (y^{2} - 1)+2ix}{x^{2} + (1 - y)^{2}} $$
Since, $$\dfrac{z+i}{z-i}$$ should be purely imaginary
$$\therefore Re\left (\dfrac {z + i}{z - i}\right ) = 0$$
$$\implies \dfrac {x^{2} + (y^{2} - 1)}{x^{2} + (1 - y)^{2}} = 0$$
$$\implies x^{2}+y^{2}-1=0$$
$$\implies x^{2} + y^{2} = 1$$ which is the equation of a circle
Hence, $$(x,y)$$ lie on a circle.

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Re-Attempt Weekly Quiz Competition

Number Theory Test 28

Result

Number Theory Test 28

Score

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SHARING IS CARING

Question 1

4.15

8.94

12.00

18.64

32.00

Solution

Question 2

$$\cfrac{1}{2}$$

$$\cfrac { 1 }{ 1+\cos { \theta } } $$

$$\tan { \cfrac { \theta }{ 2 } } $$

$$\cot { \cfrac { \theta }{ 2 } } $$

Solution

Question 3

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 4

$$a+b-i$$

$$ab+2$$

$$ab+(2b-a)i+2$$

$$ab-2$$

$$ab+(2b-a)i-2$$

Solution

Question 5

$$\dfrac{27-29i}{34}$$

$$\dfrac{27-29i}{16}$$

$$\dfrac{3-29i}{34}$$

$$\dfrac{1}{8}$$

$$15-8i$$

Solution

Question 6

$$1-i$$

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

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

$$i$$

$$-i$$

Solution

Question 7

$$3$$

$$2$$

$$1$$

$$0$$

$$-1$$

Solution

Question 8

$$21-2i$$

$$19-3i$$

$$11-2i$$

$$21-i$$

Solution

Question 9

$$3x - 4$$

$$4x - 3$$

$$4x +3$$

None of these

Solution

Question 10

A straight line

An ellipse

A hyperbola

A circle

Solution

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