Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

Let $$\alpha \,,\beta $$ be real and z be a complex number. If $${z^2} + \alpha z + \beta = 0$$ has two distinct roots on the line Re z = 1, then it is necessary that :

A

$$\beta \, \in ( - 1,0)$$

B

$$\left| {\beta \,} \right| = 1$$

C

$$\beta \, \in (1,\infty )$$

D

$$\beta \, \in (0,1)$$

As real part of roots is $$1$$

Let roots are $$1 + pi,1 + q$$

$$\therefore$$ sum of roots $$ = 1 + pi + 1 + qi = - \alpha $$

which is real $$ \Rightarrow q = - p\,\,$$

or root are $$1+pi$$ and $$1-pi$$

product of roots $$ = 1 + {p^2} = \beta \in \left( {1,\infty } \right)$$

$$p \ne 0$$ as roots are distinct.

Let roots are $$1 + pi,1 + q$$

$$\therefore$$ sum of roots $$ = 1 + pi + 1 + qi = - \alpha $$

which is real $$ \Rightarrow q = - p\,\,$$

or root are $$1+pi$$ and $$1-pi$$

product of roots $$ = 1 + {p^2} = \beta \in \left( {1,\infty } \right)$$

$$p \ne 0$$ as roots are distinct.

2

MCQ (Single Correct Answer)

The number of complex numbers z such that $$\left| {z - 1} \right| = \left| {z + 1} \right| = \left| {z - i} \right|$$ equals

A

1

B

2

C

$$\infty $$

D

0

Let $$z=x+iy$$

$$\left| {z - 1} \right| = \left| {z + 1} \right|{\left( {x - 1} \right)^2} + {y^2}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {x + 1} \right)^2} + {y^2}$$

$$ \Rightarrow {\mathop{\rm Re}\nolimits} \,z = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow x = 0$$

$$\left| {z - 1} \right| = \left| {z - i} \right|{\left( {x - 1} \right)^2} + {y^2}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {x^2} + {\left( {y - 1} \right)^2}$$

$$ \Rightarrow x = y$$

$$\left| {z + 1} \right| = \left| {z - i} \right|{\left( {x + 1} \right)^2} + {y^2}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {x^2} + {\left( {y - 1} \right)^2}$$

Only $$(0,0)$$ will satisfy all conditions.

$$ \Rightarrow $$ Number of complex number $$z=1$$

$$\left| {z - 1} \right| = \left| {z + 1} \right|{\left( {x - 1} \right)^2} + {y^2}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {x + 1} \right)^2} + {y^2}$$

$$ \Rightarrow {\mathop{\rm Re}\nolimits} \,z = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow x = 0$$

$$\left| {z - 1} \right| = \left| {z - i} \right|{\left( {x - 1} \right)^2} + {y^2}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {x^2} + {\left( {y - 1} \right)^2}$$

$$ \Rightarrow x = y$$

$$\left| {z + 1} \right| = \left| {z - i} \right|{\left( {x + 1} \right)^2} + {y^2}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {x^2} + {\left( {y - 1} \right)^2}$$

Only $$(0,0)$$ will satisfy all conditions.

$$ \Rightarrow $$ Number of complex number $$z=1$$

3

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

4

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

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

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

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Complex Numbers

Quadratic Equation and Inequalities

Permutations and Combinations

Mathematical Induction and Binomial Theorem

Sequences and Series

Matrices and Determinants

Vector Algebra and 3D Geometry

Probability

Statistics

Mathematical Reasoning

Trigonometric Functions & Equations

Properties of Triangle

Inverse Trigonometric Functions

Straight Lines and Pair of Straight Lines

Circle

Conic Sections

Functions

Limits, Continuity and Differentiability

Differentiation

Application of Derivatives

Indefinite Integrals

Definite Integrals and Applications of Integrals

Differential Equations