Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

If $${z^2} + z + 1 = 0$$, where z is complex number, then value of $${\left( {z + {1 \over z}} \right)^2} + {\left( {{z^2} + {1 \over {{z^2}}}} \right)^2} + {\left( {{z^3} + {1 \over {{z^3}}}} \right)^2} + .......... + {\left( {{z^6} + {1 \over {{z^6}}}} \right)^2}$$ is

A

18

B

54

C

6

D

12

$${z^2} + z + 1 = 0 \Rightarrow z = \omega \,\,\,$$ or $$\,\,\,{\omega ^2}$$

So, $$z + {1 \over z} = \omega + {\omega ^2} = - 1$$

$${z^2} + {1 \over {{z^2}}} = {\omega ^2} + \omega = - 1,$$

$${z^3} + {1 \over {{z^3}}} = {\omega ^3} + {\omega ^3} = 2$$

$${z^4} + {1 \over {{z^4}}} = - 1,$$

$${z^5} + {1 \over {{z^5}}} = - 1$$

and $$\,\,\,\,{z^6} + {1 \over {{z^6}}} = 2$$

$$\therefore$$ The given sum $$ = 1 + 1 + 4 + 1 + 1 + 4 = 12$$

So, $$z + {1 \over z} = \omega + {\omega ^2} = - 1$$

$${z^2} + {1 \over {{z^2}}} = {\omega ^2} + \omega = - 1,$$

$${z^3} + {1 \over {{z^3}}} = {\omega ^3} + {\omega ^3} = 2$$

$${z^4} + {1 \over {{z^4}}} = - 1,$$

$${z^5} + {1 \over {{z^5}}} = - 1$$

and $$\,\,\,\,{z^6} + {1 \over {{z^6}}} = 2$$

$$\therefore$$ The given sum $$ = 1 + 1 + 4 + 1 + 1 + 4 = 12$$

2

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

3

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

4

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.

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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

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Properties of Triangle

Inverse Trigonometric Functions

Straight Lines and Pair of Straight Lines

Circle

Conic Sections

Functions

Limits, Continuity and Differentiability

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Application of Derivatives

Indefinite Integrals

Definite Integrals and Applications of Integrals

Differential Equations