Number Theory Test 26

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Number Theory Test 26

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Question 1
1 / -0

If $${z}_{1},{z}_{2}$$ are two complex numbers and $$c>0$$ such that $${ \left| { z }_{ 1 }+{ z }_{ 2 } \right| }^{ 2 }\le \left( 1+c \right) { \left| { z }_{ 1 } \right| }^{ 2 }+k{ \left| { z }_{ 2 } \right| }^{ 2 },$$ then $$k=$$

A
$$1-c$$

B
$$c-1$$

C
$$1+{c}^{-1}$$

D
$$1-{c}^{-1}$$

Solution

Since Re $$\displaystyle \left( { z }_{ 1 }{ \overline { z } }_{ 2 } \right) \le \left| { z }_{ 1 }{ \overline { z } }_{ 2 } \right| $$

$$\displaystyle \therefore { \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 2 } \right| }^{ 2 }+Re\left( { z }_{ 1 }{ \overline { z } }_{ 2 } \right) \le { \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 2 } \right| }^{ 2 }+2\left| { z }_{ 1 }{ \overline { z } }_{ 2 } \right| $$

$$\displaystyle \Rightarrow { \left| { z }_{ 1 }+{ z }_{ 2 } \right| }^{ 2 }\le { \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 2 } \right| }^{ 2 }+2\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| $$ ...(1)

Also, Since $$A.M.\ge G.M.$$

$$\displaystyle \therefore \frac { { \left( \sqrt { c } \left| { z }_{ 1 } \right| \right) }^{ 2 }+{ \left( \frac { 1 }{ \sqrt { c } } \left| { z }_{ 2 } \right| \right) }^{ 2 } }{ 2 } \ge { \left\{ c.{ \left| { z }_{ 1 } \right| }^{ 2 }.\frac { 1 }{ c } { \left| { z }_{ 2 } \right| }^{ 2 } \right\} }^{ \frac { 1 }{ 2 } }\left( \because c>0 \right) $$

$$\displaystyle \Rightarrow c{ \left| { z }_{ 1 } \right| }^{ 2 }+\frac { 1 }{ c } { \left| { z }_{ 1 } \right| }^{ 2 }\ge 2\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| $$

$$\displaystyle \therefore { \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 1 } \right| }^{ 2 }+2\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| \le { \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 2 } \right| }^{ 2 }+c{ \left| { z }_{ 1 } \right| }^{ 2 }+\frac { 1 }{ c } { \left| { z }_{ 2 } \right| }^{ 2 }$$

$$\displaystyle \Rightarrow { \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 2 } \right| }^{ 2 }+2\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| \le \left( 1+{ c }^{ -1 } \right) \left( { \left| { z }_{ 2 } \right| }^{ 2 } \right) $$ ...(2)

From (1) and (2), we get

$$\displaystyle { \left| { z }_{ 1 }+{ z }_{ 2 } \right| }^{ 2 }\le \left( 1+c \right) { \left| { z }_{ 1 } \right| }^{ 2 }+\left( 1+{ c }^{ -1 } \right) { \left| { z }_{ 2 } \right| }^{ 2 }$$

$$\displaystyle \therefore k=1+{ c }^{ -1 }$$
Question 2
1 / -0

Find the product and write the answer in standard form.
$$\left( 2-4i \right) \left( 3+7i \right) $$

A
$$34+2i$$

B
$$-5-3i$$

C
$$5+3i$$

D
$$5-3i$$

Solution

Given $$z=(2-4i)(3+7i)$$

$$z=[2(3+7i)-4i(3+7i)]$$

$$=[6+14i-12i-28i^2]$$

$$=[34+2i]$$............[since, $$i^2=-1$$]
Question 3
1 / -0

The product of $$(3-2i)$$ and $$\left(\dfrac { 5 }{ 2 } -4i\right)$$, if $$i=\sqrt { -1 } $$ , is:

A
$$-\dfrac { 1 }{ 2 } -17i$$

B
$$14+\dfrac{9}{2}i$$

C
$$2-8i-14{i}^{2}$$

D
$$i\left(8+\dfrac{9}{2}\right)$$

Solution

$$(3 - 2i) \times \left(\dfrac{5}{2} - 4i\right)$$

$$= 3\left(\dfrac{5}{2} - 4i\right) - 2i\left(\dfrac{5}{2} - 4i\right)$$

$$= \dfrac{15}{2} - 12i - 5i - 8$$

$$= \dfrac{-1}{2} - 17i$$
Question 4
1 / -0

The resultant complex number when $$(4+6i)$$ is divided by $$(10-5i)$$ is

A
$$\dfrac {2}{25} + \dfrac {16}{25}i$$

B
$$\dfrac {2}{25} - \dfrac {16}{25}i$$

C
$$\dfrac {2}{5} + \dfrac {6}{5}i$$

D
$$\dfrac {2}{5} - \dfrac {6}{5}i$$

Solution

The value of $$\cfrac{4 + 6i}{10 - 5i}$$
$$= \cfrac{4 + 6i}{10 - 5i} \times \cfrac{10 + 5i}{10 + 5i}$$
$$= \cfrac{(4 + 6i)(10 + 5i)}{(10 - 5i)(10 + 5i)}$$
$$= \cfrac{40 + 20i + 60i - 30}{100 + 25}$$
$$= \cfrac{10 + 80i}{125}$$
$$= \cfrac{2}{25} + \cfrac{16}{25}i$$
Question 5
1 / -0

Find the product. Write the answer in standard form.
$$i\left( 6-2i \right) \left( 7-5i \right) $$

A
$$52+16i$$

B
$$10{i}^{3}+44{i}^{2}+42i$$

C
$$-44-32i$$

D
$$44+32i$$

Solution

Given $$z=i(6-2i)(7-5i)$$

$$z=i[6(7-5i)-2i(7-5i)]$$

$$=i[42-30i-14i+10i^2]$$

$$=i[32-44i]$$............[since, $$i^2=-1$$]

$$=-44i^2+32i$$

$$z=44+32i$$
Question 6
1 / -0

Let $${ X }_{ n }=\left\{ z=x+iy:{ \left| z \right| }^{ 2 } \le \dfrac { 1 }{ n } \right\} $$ for all integers $$n\ge 1$$. Then, $$\displaystyle\bigcap _{ n=1 }^{ \infty }{ { X }_{ n } } $$ is

A
A singleton set

B
Not a finite set

C
An empty set

D
A finite set with more than one element

Solution

Given, $${ X }_{ n }=\left\{ z=x+iy:{ \left| z \right| }^{ 2 }\le \dfrac { 1 }{ n } \right\} $$
$$=\left\{ { x }^{ 2 }+{ y }^{ 2 }\le \dfrac { 1 }{ n } \right\} $$
$$\therefore { X }_{ 1 }=\left\{ { x }^{ 2 }+{ y }^{ 2 }\le 1 \right\} $$
$${ X }_{ 2 }=\left\{ { x }^{ 2 }+{ y }^{ 2 }\le \dfrac { 1 }{ 2 } \right\} $$
$${ X }_{ 3 }=\left\{ { x }^{ 2 }+{ y }^{ 2 }\le \dfrac { 1 }{ 3 } \right\} $$
.............................................
.............................................
.............................................
$${ X }_{ \infty }=\left\{ { x }^{ 2 }+{ y }^{ 2 }\le 0 \right\} $$
$$\therefore \bigcap _{ n=1 }^{ \infty }{ { X }_{ n } } ={ X }_{ 1 }\cap { X }_{ 2 }\cap { X }_{ 3 }\cap \dots \cap { X }_{ \infty }$$
$$=\left\{ { x }^{ 2 }+{ y }^{ 2 }=0 \right\} $$
Hence, $$\bigcap _{ n=1 }^{ \infty }{ { X }_{ n } } $$ is a singleton set.
Question 7
1 / -0

The modulus of $$\dfrac { 1-i }{ 3+i } +\dfrac { 4i }{ 5 } $$ is

A
$$\sqrt { 5 } $$ unit

B
$$\dfrac { \sqrt { 11 } }{ 5 } $$ unit

C
$$\dfrac { \sqrt { 5 } }{ 5 } $$ unit

D
$$\dfrac { \sqrt { 12 } }{ 5 } $$ unit

Solution

Let $$z=\dfrac { 1-i }{ 3+i } +\dfrac { 4i }{ 5 } $$
$$=\dfrac { 5-5i+12i-4 }{ 5\left( 3+i \right) } =\dfrac { 1+7i }{ 5\left( 3+i \right) } $$
$$=\dfrac { \left( 1+7i \right) \left( 3-i \right) }{ 5\left( 9+1 \right) } $$
$$=\dfrac { 10+20i }{ 50 } =\dfrac { 1+2i }{ 5 } $$
$$\therefore \left| z \right| =\sqrt { { \left( \dfrac { 1 }{ 5 } \right) }^{ 2 }+{ \left( \dfrac { 2 }{ 5 } \right) }^{ 2 } } =\dfrac { 1 }{ 5 } \sqrt { 1+4 } =\dfrac { \sqrt { 5 } }{ 5 } $$
Question 8
1 / -0

If $$A = (3 - 4i)$$ and $$B = (9 + ki)$$, where $$k$$ is a constant.
If $$AB - 15 = 60$$, then the value of $$k$$ is

A
$$6$$

B
$$24$$

C
$$12$$

D
$$3$$

Solution

Given, $$A=3-4i, B=9+ki, Ab-15=0$$

$$\therefore AB= (3-4i)(9+ki)$$

$$\therefore 27 + 3ki – 36i – 4ki^2-15=60$$

$$\therefore -48+3ki-36i+4k=0$$

Separate the real and imaginery part equal to zero, then we get the value of $$k$$,

$$\therefore -48 + 4k = 0, 3k – 36 = 0$$

$$\therefore -48 = -4k, 3k = 36$$

$$\therefore k = 12, k = 12$$

So, the value of $$ k$$ is $$12$$.
Question 9
1 / -0

The simplest form of $$\sqrt {-18} \times \sqrt {-50}$$ is

A
$$-30$$

B
$$-30i$$

C
$$30$$

D
$$30i$$

Solution

We know that $$i^2=-1$$
So, $$\sqrt{18i^2}\times\sqrt{50i^2}$$
$$=$$ $$i\sqrt{18}\times i\sqrt{50}$$
$$=$$ $$i^2\sqrt{18\times50}$$
$$=$$ $$-\sqrt{900}$$
$$= - 30$$
Question 10
1 / -0

Simplify $$(2+8i)(1-4i)-(3-2i)(6+4i)$$
(Note$$:i=\sqrt{-1}$$)

A
$$8$$

B
$$26$$

C
$$34$$

D
$$50$$

Solution

The value of $$\left ( 2+8i \right )\left ( 1-4i \right )-\left ( 3-2i \right )\left ( 6+4i \right )$$
$$=$$ $$\left ( 2+8i-8i-32i^{2} \right )-\left ( 18-12i+12i-8i^{2} \right )$$
$$=$$ $$2-32i^{2}-18+8i^{2}$$
Put the given value of $$i=\sqrt{-1}$$, we get
$$=$$ $$2-32(\sqrt{-1})^{2}-18+8(\sqrt{-1})^{2}$$
$$=$$ $$2-32(-1)-18+8(-1)$$
$$=$$ $$2+32-18-8$$
$$=$$ $$34-26=8$$

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Re-Attempt Weekly Quiz Competition

Number Theory Test 26

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Number Theory Test 26

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SHARING IS CARING

Question 1

$$1-c$$

$$c-1$$

$$1+{c}^{-1}$$

$$1-{c}^{-1}$$

Solution

Question 2

$$34+2i$$

$$-5-3i$$

$$5+3i$$

$$5-3i$$

Solution

Question 3

$$-\dfrac { 1 }{ 2 } -17i$$

$$14+\dfrac{9}{2}i$$

$$2-8i-14{i}^{2}$$

$$i\left(8+\dfrac{9}{2}\right)$$

Solution

Question 4

$$\dfrac {2}{25} + \dfrac {16}{25}i$$

$$\dfrac {2}{25} - \dfrac {16}{25}i$$

$$\dfrac {2}{5} + \dfrac {6}{5}i$$

$$\dfrac {2}{5} - \dfrac {6}{5}i$$

Solution

Question 5

$$52+16i$$

$$10{i}^{3}+44{i}^{2}+42i$$

$$-44-32i$$

$$44+32i$$

Solution

Question 6

A singleton set

Not a finite set

An empty set

A finite set with more than one element

Solution

Question 7

$$\sqrt { 5 } $$ unit

$$\dfrac { \sqrt { 11 } }{ 5 } $$ unit

$$\dfrac { \sqrt { 5 } }{ 5 } $$ unit

$$\dfrac { \sqrt { 12 } }{ 5 } $$ unit

Solution

Question 8

$$6$$

$$24$$

$$12$$

$$3$$

Solution

Question 9

$$-30$$

$$-30i$$

$$30$$

$$30i$$

Solution

Question 10

$$8$$

$$26$$

$$34$$

$$50$$

Solution

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