NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

### AIEEE 2008

Let R be the real line. Consider the following subsets of the plane $$R \times R$$ :
$$S = \left\{ {(x,y):y = x + 1\,\,and\,\,0 < x < 2} \right\}$$
$$T = \left\{ {(x,y): x - y\,\,\,is\,\,an\,\,{\mathop{\rm int}} eger\,} \right\}$$,

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

## Explanation

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
2

### AIEEE 2008

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

## Explanation

$$\left( {{1 \over {i - 1}}} \right) = {1 \over { - i - 1}} = {{ - 1} \over {i + 1}}$$
3

### AIEEE 2007

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

## Explanation

$$z$$ 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 $$6$$
4

### AIEEE 2006

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

## Explanation

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

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

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

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

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

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

$$= i \times 0 - i$$

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

$$= - i$$

### Joint Entrance Examination

JEE Main JEE Advanced WB JEE

### Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

NEET

Class 12