Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

If $${z_1}$$ and $${z_2}$$ are two non-zero complex numbers such that $$\,\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$$, then arg $${z_1}$$ - arg $${z_2}$$ is equal to

A

$${\pi \over 2}\,$$

B

$$ - \pi $$

C

0

D

$${{ - \pi } \over 2}$$

Given that, $$\,\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$$

$$\,\left| {{z_1} + {z_2}} \right|$$ is the vector sum of $${z_1}$$ and $${z_2}$$. So $$\,\left| {{z_1} + {z_2}} \right|$$ should be $$<$$ $$\left| {{z_1}} \right| + \left| {{z_2}} \right|$$ but here they are equal so $${z_1}$$ and $${z_2}$$ are collinear.

S if $${z_1}$$ makes an angle $$\theta $$ with x axis then $${z_2}$$ will also make $$\theta $$ angle.

$$\therefore$$ arg $${z_1}$$ - arg $${z_2}$$ = $$\theta $$ - $$\theta $$ = 0

$$\,\left| {{z_1} + {z_2}} \right|$$ is the vector sum of $${z_1}$$ and $${z_2}$$. So $$\,\left| {{z_1} + {z_2}} \right|$$ should be $$<$$ $$\left| {{z_1}} \right| + \left| {{z_2}} \right|$$ but here they are equal so $${z_1}$$ and $${z_2}$$ are collinear.

S if $${z_1}$$ makes an angle $$\theta $$ with x axis then $${z_2}$$ will also make $$\theta $$ angle.

$$\therefore$$ arg $${z_1}$$ - arg $${z_2}$$ = $$\theta $$ - $$\theta $$ = 0

2

MCQ (Single Correct Answer)

If the cube roots of unity are 1, $$\omega \,,\,{\omega ^2}$$ then the roots of the equation $${(x - 1)^3}$$ + 8 = 0, are

A

$$ - 1, - 1 + 2\,\,\omega , - 1 - 2\,\,{\omega ^2}$$

B

$$ - 1, - 1, - 1$$

C

$$ - 1,1 - 2\omega ,1 - 2{\omega ^2}$$

D

$$ - 1,1 + 2\omega ,1 + 2{\omega ^2}$$

$${\left( {x - 1} \right)^3} + 8 = 0$$

$$ \Rightarrow \left( {x - 1} \right) = \left( { - 2} \right){\left( 1 \right)^{1/3}}$$

$$ \Rightarrow x - 1 = - 2\,\,\,$$ or $$\,\,\, - 2\omega \,\,\,\,$$ or $$\,\,\,\, - 2{\omega ^2}$$

or $$\,\,\,x = - 1\,\,\,$$ or $$\,\,\,1 - 2\omega \,\,\,$$ or $$\,\,\,1 - 2{\omega ^2}.$$

$$ \Rightarrow \left( {x - 1} \right) = \left( { - 2} \right){\left( 1 \right)^{1/3}}$$

$$ \Rightarrow x - 1 = - 2\,\,\,$$ or $$\,\,\, - 2\omega \,\,\,\,$$ or $$\,\,\,\, - 2{\omega ^2}$$

or $$\,\,\,x = - 1\,\,\,$$ or $$\,\,\,1 - 2\omega \,\,\,$$ or $$\,\,\,1 - 2{\omega ^2}.$$

3

MCQ (Single Correct Answer)

If $$\,\omega = {z \over {z - {1 \over 3}i}}\,$$ and $$\left| \omega \right| = 1$$, then $$z$$ lies on

A

an ellipse

B

a circle

C

a straight line

D

a parabola

Given $$\,\omega = {z \over {z - {1 \over 3}i}}\,$$ and $$\left| \omega \right| = 1$$

$$\therefore$$ $${{\left| z \right|} \over {\left| {z - {1 \over {\sqrt 3 }}i} \right|}} = \left| \omega \right|$$

$$ \Rightarrow $$ $${{\left| z \right|} \over {\left| {z - {1 \over {\sqrt 3 }}i} \right|}} = 1$$

$$ \Rightarrow $$ $$\left| z \right| = \left| {z - {1 \over {\sqrt 3 }}i} \right|$$ ..........equation (1)

$$\left| z \right|$$ represent distance of $$z$$ from point (0, 0) and

$$\left| {z - {1 \over {\sqrt 3 }}i} \right|$$ represent distance of $$z$$ from point $$\left( {0,{1 \over {\sqrt 3 }}} \right)$$.

According to the equation (1) the distance of $$z$$ from point (0, 0) and $$\left( {0,{1 \over {\sqrt 3 }}} \right)$$ is equal. Only if z is on a straight line then it will be equal distance from the both the points.

$$\therefore$$ $${{\left| z \right|} \over {\left| {z - {1 \over {\sqrt 3 }}i} \right|}} = \left| \omega \right|$$

$$ \Rightarrow $$ $${{\left| z \right|} \over {\left| {z - {1 \over {\sqrt 3 }}i} \right|}} = 1$$

$$ \Rightarrow $$ $$\left| z \right| = \left| {z - {1 \over {\sqrt 3 }}i} \right|$$ ..........equation (1)

$$\left| z \right|$$ represent distance of $$z$$ from point (0, 0) and

$$\left| {z - {1 \over {\sqrt 3 }}i} \right|$$ represent distance of $$z$$ from point $$\left( {0,{1 \over {\sqrt 3 }}} \right)$$.

According to the equation (1) the distance of $$z$$ from point (0, 0) and $$\left( {0,{1 \over {\sqrt 3 }}} \right)$$ is equal. Only if z is on a straight line then it will be equal distance from the both the points.

4

MCQ (Single Correct Answer)

If $$\,\left| {{z^2} - 1} \right| = {\left| z \right|^2} + 1$$, then z lies on

A

an ellipse

B

the imaginary axis

C

a circle

D

the real axis

Given $$\,\left| {{z^2} - 1} \right| = {\left| z \right|^2} + 1$$,

By squaring both sides we get,

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

$$ \Rightarrow $$ $$\left( {{z^2} - 1} \right)$$$$\overline {\left( {{z^2} - 1} \right)} $$ = $${\left( {{{\left| z \right|}^2} + 1} \right)^2}$$ [ as $${{{\left| z \right|}^2}}$$ = $$z\overline z $$ ]

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

$$ \Rightarrow $$ $${\left( {z\overline z } \right)^2}$$ $$-$$ $${{z^2}}$$ $$-$$$${{{\left( {\overline z } \right)}^2}}$$ $$+$$ 1 = $${\left| z \right|^4}$$ $$+$$ 2$${{{\left| z \right|}^2}}$$ $$+$$ 1

$$ \Rightarrow $$ $${\left| z \right|^4}$$ $$-$$ $${{z^2}}$$ $$-$$$${{{\left( {\overline z } \right)}^2}}$$ $$+$$ 1 = $${\left| z \right|^4}$$ $$+$$ 2$${{{\left| z \right|}^2}}$$ $$+$$ 1

$$ \Rightarrow $$ $${{z^2}}$$ $$+$$$${{{\left( {\overline z } \right)}^2}}$$ $$+$$ 2$${z\overline z }$$ = 0

$$ \Rightarrow $$ $${\left( {z + \overline z } \right)^2}$$ = 0

$$ \Rightarrow $$ $${z + \overline z }$$ = 0

$$ \Rightarrow $$ $$z$$ = $$-$$ $${\overline z }$$

If $$z$$ = x + iy

then $${\overline z }$$ = x - iy

$$\therefore$$ x + iy = - (x - iy)

$$ \Rightarrow $$ x + iy = - x + iy

$$ \Rightarrow $$ x = 0

$$\therefore$$ z is purely imaginary.

So, it is lie on the imaginary axis.

By squaring both sides we get,

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

$$ \Rightarrow $$ $$\left( {{z^2} - 1} \right)$$$$\overline {\left( {{z^2} - 1} \right)} $$ = $${\left( {{{\left| z \right|}^2} + 1} \right)^2}$$ [ as $${{{\left| z \right|}^2}}$$ = $$z\overline z $$ ]

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

$$ \Rightarrow $$ $${\left( {z\overline z } \right)^2}$$ $$-$$ $${{z^2}}$$ $$-$$$${{{\left( {\overline z } \right)}^2}}$$ $$+$$ 1 = $${\left| z \right|^4}$$ $$+$$ 2$${{{\left| z \right|}^2}}$$ $$+$$ 1

$$ \Rightarrow $$ $${\left| z \right|^4}$$ $$-$$ $${{z^2}}$$ $$-$$$${{{\left( {\overline z } \right)}^2}}$$ $$+$$ 1 = $${\left| z \right|^4}$$ $$+$$ 2$${{{\left| z \right|}^2}}$$ $$+$$ 1

$$ \Rightarrow $$ $${{z^2}}$$ $$+$$$${{{\left( {\overline z } \right)}^2}}$$ $$+$$ 2$${z\overline z }$$ = 0

$$ \Rightarrow $$ $${\left( {z + \overline z } \right)^2}$$ = 0

$$ \Rightarrow $$ $${z + \overline z }$$ = 0

$$ \Rightarrow $$ $$z$$ = $$-$$ $${\overline z }$$

If $$z$$ = x + iy

then $${\overline z }$$ = x - iy

$$\therefore$$ x + iy = - (x - iy)

$$ \Rightarrow $$ x + iy = - x + iy

$$ \Rightarrow $$ x = 0

$$\therefore$$ z is purely imaginary.

So, it is lie on the imaginary axis.

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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