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Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

If $${\left( {{{1 + i} \over {1 - i}}} \right)^x} = 1$$ then

A

x = 2n + 1, where n is any positive integer

B

x = 4n , where n is any positive integer

C

x = 2n, where n is any positive integer

D

x = 4n + 1, where n is any positive integer.

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

$$ \Rightarrow $$ $${\left[ {{{\left( {1 + i} \right)\left( {1 + i} \right)} \over {\left( {1 - i} \right)\left( {1 + i} \right)}}} \right]^x} = 1$$

$$ \Rightarrow $$ $${\left[ {{{{{\left( {1 + i} \right)}^2}} \over {1 - {i^2}}}} \right]^x} = 1$$

$$ \Rightarrow $$ $${\left[ {{{1 + 2i + {i^2}} \over {1 + 1}}} \right]^x} = 1$$

$$ \Rightarrow $$ $${\left[ {{{1 + 2i - 1} \over 2}} \right]^x} = 1$$

$$ \Rightarrow {\left( i \right)^x} = 1$$

We know $$i = \sqrt { - 1} $$

$$\therefore$$ $${i^2} = - 1$$

$$ \Rightarrow $$ $${i^3} = - 1 \times i = - i$$

$$ \Rightarrow $$ $${i^4} = - i \times i = - {i^2} = - \left( { - 1} \right) = 1$$

So when power of $$i$$ is 4 or multiple of 4 then it's value is = 1

$$\therefore$$ $${\left( i \right)^x} = 1$$ $$ = {\left( i \right)^{4n}}$$ where n is a positive integer.

$$ \Rightarrow $$ $${\left[ {{{\left( {1 + i} \right)\left( {1 + i} \right)} \over {\left( {1 - i} \right)\left( {1 + i} \right)}}} \right]^x} = 1$$

$$ \Rightarrow $$ $${\left[ {{{{{\left( {1 + i} \right)}^2}} \over {1 - {i^2}}}} \right]^x} = 1$$

$$ \Rightarrow $$ $${\left[ {{{1 + 2i + {i^2}} \over {1 + 1}}} \right]^x} = 1$$

$$ \Rightarrow $$ $${\left[ {{{1 + 2i - 1} \over 2}} \right]^x} = 1$$

$$ \Rightarrow {\left( i \right)^x} = 1$$

We know $$i = \sqrt { - 1} $$

$$\therefore$$ $${i^2} = - 1$$

$$ \Rightarrow $$ $${i^3} = - 1 \times i = - i$$

$$ \Rightarrow $$ $${i^4} = - i \times i = - {i^2} = - \left( { - 1} \right) = 1$$

So when power of $$i$$ is 4 or multiple of 4 then it's value is = 1

$$\therefore$$ $${\left( i \right)^x} = 1$$ $$ = {\left( i \right)^{4n}}$$ where n is a positive integer.

2

MCQ (Single Correct Answer)

Let $${Z_1}$$ and $${Z_2}$$ be two roots of the equation $${Z^2} + aZ + b = 0$$, Z being complex. Further , assume that the origin, $${Z_1}$$ and $${Z_2}$$ form an equilateral triangle. Then

A

$${a^2} = 4b$$

B

$${a^2} = b$$

C

$${a^2} = 2b$$

D

$${a^2} = 3b$$

Given quadratic equation,

$${Z^2} + aZ + b = 0$$

and two roots are $${Z_1}$$ and $${Z_2}$$.

$$\therefore$$ $${Z_1}$$ + $${Z_2}$$ = $$-a$$ and $${Z_1}$$$${Z_2}$$ = $$b$$

Question says,

There are three complex numbers:

1. Origin (0)

2. $${Z_1}$$

3. $${Z_2}$$

and they form an equilateral triangle. So They are the vertices of the triangle.

[**Important Point :** If $${Z_1}$$, $${Z_2}$$ and $${Z_3}$$ are the vertices of an equilateral triangle then -

$$Z_1^2$$ + $$Z_2^2$$ + $$Z_3^2$$ = $${Z_1}{Z_2}$$ + $${Z_2}{Z_3}$$ + $${Z_3}{Z_1}$$ ]

In this question,

$${Z_1}$$ = 0, $${Z_2}$$ = $${Z_1}$$ and $${Z_3}$$ = $${Z_2}$$

By putting those values in the equation we get,

$${0^2}$$ + $$Z_1^2$$ + $$Z_2^2$$ = $$0$$ + $${Z_1}{Z_2}$$ + 0

$$ \Rightarrow $$ $$Z_1^2$$ + $$Z_2^2$$ = $${Z_1}{Z_2}$$

$$ \Rightarrow $$ $$Z_1^2$$ + $$Z_2^2$$ = $$b$$ [ as $${Z_1}$$$${Z_2}$$ = $$b$$ ]

$$ \Rightarrow $$ $${\left( {{Z_1} + {Z_2}} \right)^2}$$ - $$2{Z_1}{Z_2}$$ = $$b$$

$$ \Rightarrow $$ $${\left( {{Z_1} + {Z_2}} \right)^2}$$ - $$2b$$ = $$b$$

$$ \Rightarrow $$ $${\left( {{Z_1} + {Z_2}} \right)^2}$$ = $$3b$$

$$ \Rightarrow $$ $${\left( { - a} \right)^2}$$ = $$3b$$

$$ \Rightarrow $$ $${a^2}$$ = $$3b$$

So Option (D) is correct.

[**Note :** This question is asked to check if you know the following formula -

"If $${Z_1}$$, $${Z_2}$$ and $${Z_3}$$ are the vertices of an equilateral triangle then -

$$Z_1^2$$ + $$Z_2^2$$ + $$Z_3^2$$ = $${Z_1}{Z_2}$$ + $${Z_2}{Z_3}$$ + $${Z_3}{Z_1}$$" ]

$${Z^2} + aZ + b = 0$$

and two roots are $${Z_1}$$ and $${Z_2}$$.

$$\therefore$$ $${Z_1}$$ + $${Z_2}$$ = $$-a$$ and $${Z_1}$$$${Z_2}$$ = $$b$$

Question says,

There are three complex numbers:

1. Origin (0)

2. $${Z_1}$$

3. $${Z_2}$$

and they form an equilateral triangle. So They are the vertices of the triangle.

[

$$Z_1^2$$ + $$Z_2^2$$ + $$Z_3^2$$ = $${Z_1}{Z_2}$$ + $${Z_2}{Z_3}$$ + $${Z_3}{Z_1}$$ ]

In this question,

$${Z_1}$$ = 0, $${Z_2}$$ = $${Z_1}$$ and $${Z_3}$$ = $${Z_2}$$

By putting those values in the equation we get,

$${0^2}$$ + $$Z_1^2$$ + $$Z_2^2$$ = $$0$$ + $${Z_1}{Z_2}$$ + 0

$$ \Rightarrow $$ $$Z_1^2$$ + $$Z_2^2$$ = $${Z_1}{Z_2}$$

$$ \Rightarrow $$ $$Z_1^2$$ + $$Z_2^2$$ = $$b$$ [ as $${Z_1}$$$${Z_2}$$ = $$b$$ ]

$$ \Rightarrow $$ $${\left( {{Z_1} + {Z_2}} \right)^2}$$ - $$2{Z_1}{Z_2}$$ = $$b$$

$$ \Rightarrow $$ $${\left( {{Z_1} + {Z_2}} \right)^2}$$ - $$2b$$ = $$b$$

$$ \Rightarrow $$ $${\left( {{Z_1} + {Z_2}} \right)^2}$$ = $$3b$$

$$ \Rightarrow $$ $${\left( { - a} \right)^2}$$ = $$3b$$

$$ \Rightarrow $$ $${a^2}$$ = $$3b$$

So Option (D) is correct.

[

"If $${Z_1}$$, $${Z_2}$$ and $${Z_3}$$ are the vertices of an equilateral triangle then -

$$Z_1^2$$ + $$Z_2^2$$ + $$Z_3^2$$ = $${Z_1}{Z_2}$$ + $${Z_2}{Z_3}$$ + $${Z_3}{Z_1}$$" ]

3

MCQ (Single Correct Answer)

If $$z$$ and $$\omega $$ are two non-zero complex numbers such that $$\left| {z\omega } \right| = 1$$ and $$Arg(z) - Arg(\omega ) = {\pi \over 2},$$ then $$\,\overline {z\,} \omega $$ is equal to

A

$$- i$$

B

1

C

- 1

D

$$i$$

Given that,

$$\left| {z\omega } \right| = 1$$

$$ \Rightarrow $$ $$\left| z \right|\left| \omega \right|$$ = 1

$$ \Rightarrow $$ $$\left| z \right|$$ = $${1 \over {\left| \omega \right|}}$$

and $$Arg(z) - Arg(\omega ) = {\pi \over 2}$$

$$ \Rightarrow $$ $$Arg\left( {{z \over \omega }} \right)$$ $$= {\pi \over 2}$$

When argument of a complex number is $${\pi \over 2}$$, it means it is making an angle of $${\pi \over 2}$$ with the real axis in the counterclockwise, so it is along the imaginary axis and positive side of imaginary axis.

So, $${{z \over \omega }}$$ is a purely imaginary number that means there is no real part in this complex number.

So we can assume,

$${{z \over \omega }}$$ = $$ki$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| {ki} \right|$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| k \right|\left| i \right|$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$k$$ [ as $$\left| i \right|$$ = 1 ]

$$ \Rightarrow $$ $$\left| z \right|$$$$ \times $$$${1 \over {\left| \omega \right|}}$$ = $$k$$

$$ \Rightarrow $$ $$\left| z \right|$$$$ \times $$ $$\left| z \right|$$ = $$k$$ [ as $${1 \over {\left| \omega \right|}}$$ = $$\left| z \right|$$ ]

$$ \Rightarrow $$ $${\left| z \right|^2}$$ = $$k$$

$$ \Rightarrow $$ $$\left| z \right|$$ = $$\sqrt k $$

$$\therefore$$ $$\left| \omega \right|$$ = $${1 \over {\sqrt k }}$$

As $${{z \over \omega }}$$ is imaginary so we can write,

$${{z \over \omega }}$$ = $$ - {{\overline z } \over {\overline \omega }}$$

[ When $$z$$ is imaginary then $$z$$ = $$-\overline z $$ ]

$$ \Rightarrow $$ $$\overline z \omega $$ = $$ - z\overline \omega $$

$$ \Rightarrow $$ $$\overline z \omega $$ = $$-{{z \over \omega }}$$.$$\overline \omega $$.$$\omega $$

$$ \Rightarrow $$ $$\overline z \omega $$ = $$-{{z \over \omega }}$$.$${\left| \omega \right|^2}$$

$$ \Rightarrow $$ $$\overline z \omega $$ = $$-ki$$.$${\left( {{1 \over {\sqrt k }}} \right)^2}$$

$$ \Rightarrow $$ $$\overline z \omega $$ = $$-ki$$.$${1 \over k}$$

$$ \Rightarrow $$ $$\overline z \omega $$ = $$-i$$

**Method 2 :**

Given that,

$$\left| {z\omega } \right| = 1$$

$$ \Rightarrow $$ $$\left| z \right|\left| \omega \right|$$ = 1

$$ \Rightarrow $$ $$\left| z \right|$$ = $${1 \over {\left| \omega \right|}}$$

and $$Arg(z) - Arg(\omega ) = {\pi \over 2}$$

$$ \Rightarrow $$ $$Arg\left( {{z \over \omega }} \right)$$ $$= {\pi \over 2}$$

When argument of a complex number is $${\pi \over 2}$$, it means it is making an angle of $${\pi \over 2}$$ with the real axis in the counterclockwise, so it is along the imaginary axis and positive side of imaginary axis.

So, $${{z \over \omega }}$$ is a purely imaginary number that means there is no real part in this complex number.

So we can assume,

$${{z \over \omega }}$$ = $$ki$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| {ki} \right|$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| k \right|\left| i \right|$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$k$$ [ as $$\left| i \right|$$ = 1 ]

$$ \Rightarrow $$ $$\left| z \right|$$$$ \times $$$${1 \over {\left| \omega \right|}}$$ = $$k$$

$$ \Rightarrow $$ $$\left| z \right|$$$$ \times $$ $$\left| z \right|$$ = $$k$$ [ as $${1 \over {\left| \omega \right|}}$$ = $$\left| z \right|$$ ]

$$ \Rightarrow $$ $${\left| z \right|^2}$$ = $$k$$

$$ \Rightarrow $$ $$\left| z \right|$$ = $$\sqrt k $$

$$\therefore$$ $$\left| \omega \right|$$ = $${1 \over {\sqrt k }}$$

(1) Magnitude of $$\overline z \omega $$

= $$\left| {\overline z } \right|\left| \omega \right|$$

= $$\left| z \right|\left| \omega \right|$$ [ as $$\left| z \right|$$ = $$\left| {\overline z } \right|$$ ]

= $$\sqrt k $$.$${{1 \over {\sqrt k }}}$$

= 1

$$\therefore$$ The distance from the origin of $${\overline z \omega }$$ is 1.

(2) Argument of $${\overline z \omega }$$ = $$Arg\left( {\overline z \omega } \right)$$

= $$Arg\left( {\overline z } \right) + Arg\left( \omega \right)$$

= $$-Arg\left( z \right) + Arg\left( \omega \right)$$

= $$ - \left( {Arg\left( z \right) - Arg\left( \omega \right)} \right)$$

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

$$\therefore$$ $${\overline z \omega }$$ is at (0, -1) on the negative side of imaginary axis and making an angle of $${\pi \over 2}$$ clockwise.

$$\therefore$$ $${\overline z \omega }$$ = 0 + (-1)$$ \times $$$$i$$ = $$-i$$

$$\left| {z\omega } \right| = 1$$

$$ \Rightarrow $$ $$\left| z \right|\left| \omega \right|$$ = 1

$$ \Rightarrow $$ $$\left| z \right|$$ = $${1 \over {\left| \omega \right|}}$$

and $$Arg(z) - Arg(\omega ) = {\pi \over 2}$$

$$ \Rightarrow $$ $$Arg\left( {{z \over \omega }} \right)$$ $$= {\pi \over 2}$$

When argument of a complex number is $${\pi \over 2}$$, it means it is making an angle of $${\pi \over 2}$$ with the real axis in the counterclockwise, so it is along the imaginary axis and positive side of imaginary axis.

So, $${{z \over \omega }}$$ is a purely imaginary number that means there is no real part in this complex number.

So we can assume,

$${{z \over \omega }}$$ = $$ki$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| {ki} \right|$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| k \right|\left| i \right|$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$k$$ [ as $$\left| i \right|$$ = 1 ]

$$ \Rightarrow $$ $$\left| z \right|$$$$ \times $$$${1 \over {\left| \omega \right|}}$$ = $$k$$

$$ \Rightarrow $$ $$\left| z \right|$$$$ \times $$ $$\left| z \right|$$ = $$k$$ [ as $${1 \over {\left| \omega \right|}}$$ = $$\left| z \right|$$ ]

$$ \Rightarrow $$ $${\left| z \right|^2}$$ = $$k$$

$$ \Rightarrow $$ $$\left| z \right|$$ = $$\sqrt k $$

$$\therefore$$ $$\left| \omega \right|$$ = $${1 \over {\sqrt k }}$$

As $${{z \over \omega }}$$ is imaginary so we can write,

$${{z \over \omega }}$$ = $$ - {{\overline z } \over {\overline \omega }}$$

[ When $$z$$ is imaginary then $$z$$ = $$-\overline z $$ ]

$$ \Rightarrow $$ $$\overline z \omega $$ = $$ - z\overline \omega $$

$$ \Rightarrow $$ $$\overline z \omega $$ = $$-{{z \over \omega }}$$.$$\overline \omega $$.$$\omega $$

$$ \Rightarrow $$ $$\overline z \omega $$ = $$-{{z \over \omega }}$$.$${\left| \omega \right|^2}$$

$$ \Rightarrow $$ $$\overline z \omega $$ = $$-ki$$.$${\left( {{1 \over {\sqrt k }}} \right)^2}$$

$$ \Rightarrow $$ $$\overline z \omega $$ = $$-ki$$.$${1 \over k}$$

$$ \Rightarrow $$ $$\overline z \omega $$ = $$-i$$

Given that,

$$\left| {z\omega } \right| = 1$$

$$ \Rightarrow $$ $$\left| z \right|\left| \omega \right|$$ = 1

$$ \Rightarrow $$ $$\left| z \right|$$ = $${1 \over {\left| \omega \right|}}$$

and $$Arg(z) - Arg(\omega ) = {\pi \over 2}$$

$$ \Rightarrow $$ $$Arg\left( {{z \over \omega }} \right)$$ $$= {\pi \over 2}$$

When argument of a complex number is $${\pi \over 2}$$, it means it is making an angle of $${\pi \over 2}$$ with the real axis in the counterclockwise, so it is along the imaginary axis and positive side of imaginary axis.

So, $${{z \over \omega }}$$ is a purely imaginary number that means there is no real part in this complex number.

So we can assume,

$${{z \over \omega }}$$ = $$ki$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| {ki} \right|$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| k \right|\left| i \right|$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$k$$ [ as $$\left| i \right|$$ = 1 ]

$$ \Rightarrow $$ $$\left| z \right|$$$$ \times $$$${1 \over {\left| \omega \right|}}$$ = $$k$$

$$ \Rightarrow $$ $$\left| z \right|$$$$ \times $$ $$\left| z \right|$$ = $$k$$ [ as $${1 \over {\left| \omega \right|}}$$ = $$\left| z \right|$$ ]

$$ \Rightarrow $$ $${\left| z \right|^2}$$ = $$k$$

$$ \Rightarrow $$ $$\left| z \right|$$ = $$\sqrt k $$

$$\therefore$$ $$\left| \omega \right|$$ = $${1 \over {\sqrt k }}$$

(1) Magnitude of $$\overline z \omega $$

= $$\left| {\overline z } \right|\left| \omega \right|$$

= $$\left| z \right|\left| \omega \right|$$ [ as $$\left| z \right|$$ = $$\left| {\overline z } \right|$$ ]

= $$\sqrt k $$.$${{1 \over {\sqrt k }}}$$

= 1

$$\therefore$$ The distance from the origin of $${\overline z \omega }$$ is 1.

(2) Argument of $${\overline z \omega }$$ = $$Arg\left( {\overline z \omega } \right)$$

= $$Arg\left( {\overline z } \right) + Arg\left( \omega \right)$$

= $$-Arg\left( z \right) + Arg\left( \omega \right)$$

= $$ - \left( {Arg\left( z \right) - Arg\left( \omega \right)} \right)$$

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

$$\therefore$$ $${\overline z \omega }$$ is at (0, -1) on the negative side of imaginary axis and making an angle of $${\pi \over 2}$$ clockwise.

$$\therefore$$ $${\overline z \omega }$$ = 0 + (-1)$$ \times $$$$i$$ = $$-i$$

4

MCQ (Single Correct Answer)

The locus of the centre of a circle which touches the circle $$\left| {z - {z_1}} \right| = a$$ and$$\left| {z - {z_2}} \right| = b\,$$ externally

($$z,\,{z_1}\,\& \,{z_2}\,$$ are complex numbers) will be

($$z,\,{z_1}\,\& \,{z_2}\,$$ are complex numbers) will be

A

an ellipse

B

a hyperbola

C

a circle

D

none of these

Let the circle be $$\left| {z - {z_3}} \right| = r.$$

Then according to given conditions

$$\left| {{z_3} - {z_1}} \right| = r + a$$ (Shown in the image)

and $$\left| {{z_3} - {z_2}} \right| = r + b.$$ (Shown in the image)

Eliminating $$r,$$ we get

$$\left| {{z_3} - {z_1}} \right| - \left| {{z_3} - {z_2}} \right| = a - b.$$

$$\therefore$$ Locus of center $${z_3}$$ is

$$\left| {z - {z_1}} \right| - \left| {z - {z_2}} \right| = a - b$$ = constant.

Definition of hyperbola says, when difference of distance between two points is constant from a particular point then that particular point will lie on a hyperbola.

Here distance of z_{1} from z_{3} is = $$r + a$$ and distance of z_{2} from z_{3} is = $$r + b$$

Now their difference = ($$r + a$$) - ($$r + b$$) = $$a - b$$ = a constant

$$\therefore$$ Locus of z_{3} is a hyperbola.

Then according to given conditions

$$\left| {{z_3} - {z_1}} \right| = r + a$$ (Shown in the image)

and $$\left| {{z_3} - {z_2}} \right| = r + b.$$ (Shown in the image)

Eliminating $$r,$$ we get

$$\left| {{z_3} - {z_1}} \right| - \left| {{z_3} - {z_2}} \right| = a - b.$$

$$\therefore$$ Locus of center $${z_3}$$ is

$$\left| {z - {z_1}} \right| - \left| {z - {z_2}} \right| = a - b$$ = constant.

Definition of hyperbola says, when difference of distance between two points is constant from a particular point then that particular point will lie on a hyperbola.

Here distance of z

Now their difference = ($$r + a$$) - ($$r + b$$) = $$a - b$$ = a constant

$$\therefore$$ Locus of z

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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

Differentiation

Application of Derivatives

Indefinite Integrals

Definite Integrals and Applications of Integrals

Differential Equations