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1

MCQ (Single Correct Answer)

Let R be the real line. Consider the following subsets of the plane $$R \times R$$ :

$$S = \left\{ {(x,y):y = x + 1\,\,and\,\,0 < x < 2} \right\}$$

$$T = \left\{ {(x,y): x - y\,\,\,is\,\,an\,\,{\mathop{\rm int}} eger\,} \right\}$$,

$$S = \left\{ {(x,y):y = x + 1\,\,and\,\,0 < x < 2} \right\}$$

$$T = \left\{ {(x,y): x - y\,\,\,is\,\,an\,\,{\mathop{\rm int}} eger\,} \right\}$$,

Which one of the following is true ?

A

Neither S nor T is an equivalence relation on R

B

Both S and T are equivalence relation on R

C

S is an equivalence relation on R but T is not

D

T is an equivalence relation on R but S is not

Given $$S = \left\{ {\left( {x,y} \right):y = x + 1\,\,} \right.\,$$

and $$\,\,\,\left. {0 < x < 2} \right\}$$

As $$\,\,\,\,x \ne x + 1\,\,\,$$

for any $$\,\,\,x \in \left( {0,2} \right) \Rightarrow \left( {x,x} \right) \notin S$$

$$\therefore$$ $$S$$ is not reflexive.

Hence $$S$$ in not an equivalence relation.

Also $$\,\,\,T = \left\{ {x,\left. y \right)} \right.:x - y$$ is an integer $$\left. {} \right\}$$

as $$x - x = 0$$ is an integer $$\forall x \in R$$

$$\therefore$$ $$T$$ is reflexive.

If $$x-y$$ is an integer then $$y-x$$ is also an integer

$$\therefore$$ $$T$$ is symmetric

If $$x-y$$ is an integer and $$y - z$$ is an integer then

$$(x-y)+(y-z)=x-z$$ is also an integer.

$$\therefore$$ $$T$$ is transitive

Hence $$T$$ is an equivalence relation

and $$\,\,\,\left. {0 < x < 2} \right\}$$

As $$\,\,\,\,x \ne x + 1\,\,\,$$

for any $$\,\,\,x \in \left( {0,2} \right) \Rightarrow \left( {x,x} \right) \notin S$$

$$\therefore$$ $$S$$ is not reflexive.

Hence $$S$$ in not an equivalence relation.

Also $$\,\,\,T = \left\{ {x,\left. y \right)} \right.:x - y$$ is an integer $$\left. {} \right\}$$

as $$x - x = 0$$ is an integer $$\forall x \in R$$

$$\therefore$$ $$T$$ is reflexive.

If $$x-y$$ is an integer then $$y-x$$ is also an integer

$$\therefore$$ $$T$$ is symmetric

If $$x-y$$ is an integer and $$y - z$$ is an integer then

$$(x-y)+(y-z)=x-z$$ is also an integer.

$$\therefore$$ $$T$$ is transitive

Hence $$T$$ is an equivalence relation

2

MCQ (Single Correct Answer)

The conjugate of a complex number is $${1 \over {i - 1}}$$ then that complex number is

A

$${{ - 1} \over {i - 1}}$$

B

$${1 \over {i + 1}}\,$$

C

$${{ - 1} \over {i + 1}}$$

D

$${1 \over {i - 1}}$$

$$\left( {{1 \over {i - 1}}} \right) = {1 \over { - i - 1}} = {{ - 1} \over {i + 1}}$$

3

MCQ (Single Correct Answer)

If $$\,\left| {z + 4} \right|\,\, \le \,\,3\,$$, then the maximum value of $$\left| {z + 1} \right|$$ is

A

6

B

0

C

4

D

10

$$z$$ lies on or inside the circle with center $$(-4,0)$$ and radius $$3$$ units.

From the Argand diagram maximum value of $$\left| {z + 1} \right|$$ is $$6$$

From the Argand diagram maximum value of $$\left| {z + 1} \right|$$ is $$6$$

4

MCQ (Single Correct Answer)

The value of $$\sum\limits_{k = 1}^{10} {\left( {\sin {{2k\pi } \over {11}} + i\,\,\cos {{2k\pi } \over {11}}} \right)} $$ is

A

i

B

1

C

- 1

D

- i

$$\sum\limits_{k = 1}^{10} {\left( {\sin {{2k\pi } \over {11}} + i\cos {{2k\pi } \over {11}}} \right)} $$

$$ = i\sum\limits_{k = 1}^{10} {\left( {\cos {{2k\pi } \over {11}} - i\,\sin {{2k\pi } \over {11}}} \right)} $$

$$ = i\sum\limits_{k = 1}^{10} {{e^{ - {{2k\pi } \over {11}}}}} i = i\left\{ {\sum\limits_{k = 0}^{10} {{e^{ - {{2k\pi } \over {11}}}}} - 1} \right\}$$

$$ = i\left[ {1 + {e^{ - {{2\pi } \over {11}}i}} + e - {{4\pi } \over {11}}i + .....11\,\,terms} \right] - i$$

$$ = i\left[ {{{1 - {{\left( {{e^{ - {{2\pi } \over {11}}}}} \right)}^{11}}} \over {1 - {e^{ - {{2\pi } \over {11}}i}}}}} \right] - i$$

$$ = i\left[ {{{1 - {e^{ - 2\pi i}}} \over {1 - {e^{ - {{2\pi } \over {11}}i}}}}} \right] - i$$

$$ = i \times 0 - i$$

[as $$\,\,\,\,\,\,$$ $${e^{ - 2\pi i}} = 1$$ ]

$$ = - i$$

$$ = i\sum\limits_{k = 1}^{10} {\left( {\cos {{2k\pi } \over {11}} - i\,\sin {{2k\pi } \over {11}}} \right)} $$

$$ = i\sum\limits_{k = 1}^{10} {{e^{ - {{2k\pi } \over {11}}}}} i = i\left\{ {\sum\limits_{k = 0}^{10} {{e^{ - {{2k\pi } \over {11}}}}} - 1} \right\}$$

$$ = i\left[ {1 + {e^{ - {{2\pi } \over {11}}i}} + e - {{4\pi } \over {11}}i + .....11\,\,terms} \right] - i$$

$$ = i\left[ {{{1 - {{\left( {{e^{ - {{2\pi } \over {11}}}}} \right)}^{11}}} \over {1 - {e^{ - {{2\pi } \over {11}}i}}}}} \right] - i$$

$$ = i\left[ {{{1 - {e^{ - 2\pi i}}} \over {1 - {e^{ - {{2\pi } \over {11}}i}}}}} \right] - i$$

$$ = i \times 0 - i$$

[as $$\,\,\,\,\,\,$$ $${e^{ - 2\pi i}} = 1$$ ]

$$ = - i$$

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Complex Numbers

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