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Graduate Aptitude Test in Engineering

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Engineering Mathematics

General Aptitude

1

Let $$\alpha $$ and $$\beta $$ be two roots of the equation x^{2} + 2x + 2 = 0 , then $$\alpha ^{15}$$ + $$\beta ^{15}$$ is equal to :

A

-256

B

512

C

-512

D

256

Given equation,

x^{2} + 2x + 2 = 0

$$ \therefore $$ x = $${{ - 2 \pm \sqrt {4 - 4.1.2} } \over {2.1}}$$

x = $$-$$ 1 $$ \pm $$ i

$$ \therefore $$ $$\alpha $$ = $$-$$ 1 + i

and $$\beta $$ = $$-$$ 1 $$-$$ i

__Note __ :

x + iy = r (cos$$\theta $$ + isin$$\theta $$)

$$ \therefore $$ (x + iy)^{n} = r^{n} (cosn$$\theta $$ + isinn$$\theta $$)

$$ \therefore $$ $$-$$ 1 + i = $$\sqrt 2 $$ [cos$${{3\pi } \over 4}$$ + isin$${{3\pi } \over 4}$$ ]

$$ \Rightarrow $$ ($$-$$ 1 + i)^{15} = $${\left( {\sqrt 2 } \right)^{15}}$$ [cos$$\left( {{{15.3\pi } \over 4}} \right) + i\sin \left( {{{15.3\pi } \over 4}} \right)$$]

And $$-$$1 $$-$$ i = $$\sqrt 2 $$ $$\left[ {\cos \left( { - {{3\pi } \over 4}} \right) + i\sin \left( { - {{3\pi } \over 4}} \right)} \right]$$

= $$\sqrt 2 \left[ {\cos {{3\pi } \over 4} - \sin {{3\pi } \over 4}} \right]$$

$$ \therefore $$ ($$-$$1 $$-$$ i)^{15} = $${\left( {\sqrt 2 } \right)^{15}}\left[ {\cos \left( {{{15.3\pi } \over 4}} \right) - i\sin \left( {{{15.3\pi } \over 4}} \right)} \right]$$

Now

$$\alpha $$^{15} + $$\beta $$^{15}

= ($$-$$1 + i)^{15} + ($$-$$ 1 $$-$$ i)^{15}

= $${\left( {\sqrt 2 } \right)^{15}}$$ $$\left[ {2\cos \left( {{{15.3\pi } \over 4}} \right)} \right]$$

= $${\left( {\sqrt 2 } \right)^{15}}\left[ {2\cos \left( {11\pi + {\pi \over 4}} \right)} \right]$$

= $${\left( {\sqrt 2 } \right)^{15}}\left[ {2\left( { - \cos {\pi \over 4}} \right)} \right]$$

= $${\left( {\sqrt 2 } \right)^{15}} \times 2 \times - {1 \over {\sqrt 2 }}$$

= $$ - {\left( {\sqrt 2 } \right)^{14}}.2$$

= $$-$$ 2^{7} $$ \times $$ 2

= $$-$$ 2^{8}

= $$-$$ 256

x

$$ \therefore $$ x = $${{ - 2 \pm \sqrt {4 - 4.1.2} } \over {2.1}}$$

x = $$-$$ 1 $$ \pm $$ i

$$ \therefore $$ $$\alpha $$ = $$-$$ 1 + i

and $$\beta $$ = $$-$$ 1 $$-$$ i

x + iy = r (cos$$\theta $$ + isin$$\theta $$)

$$ \therefore $$ (x + iy)

$$ \therefore $$ $$-$$ 1 + i = $$\sqrt 2 $$ [cos$${{3\pi } \over 4}$$ + isin$${{3\pi } \over 4}$$ ]

$$ \Rightarrow $$ ($$-$$ 1 + i)

And $$-$$1 $$-$$ i = $$\sqrt 2 $$ $$\left[ {\cos \left( { - {{3\pi } \over 4}} \right) + i\sin \left( { - {{3\pi } \over 4}} \right)} \right]$$

= $$\sqrt 2 \left[ {\cos {{3\pi } \over 4} - \sin {{3\pi } \over 4}} \right]$$

$$ \therefore $$ ($$-$$1 $$-$$ i)

Now

$$\alpha $$

= ($$-$$1 + i)

= $${\left( {\sqrt 2 } \right)^{15}}$$ $$\left[ {2\cos \left( {{{15.3\pi } \over 4}} \right)} \right]$$

= $${\left( {\sqrt 2 } \right)^{15}}\left[ {2\cos \left( {11\pi + {\pi \over 4}} \right)} \right]$$

= $${\left( {\sqrt 2 } \right)^{15}}\left[ {2\left( { - \cos {\pi \over 4}} \right)} \right]$$

= $${\left( {\sqrt 2 } \right)^{15}} \times 2 \times - {1 \over {\sqrt 2 }}$$

= $$ - {\left( {\sqrt 2 } \right)^{14}}.2$$

= $$-$$ 2

= $$-$$ 2

= $$-$$ 256

2

Let

A = $$\left\{ {\theta \in \left( { - {\pi \over 2},\pi } \right):{{3 + 2i\sin \theta } \over {1 - 2i\sin \theta }}is\,purely\,imaginary} \right\}$$

. Then the sum of the elements in A is :

A = $$\left\{ {\theta \in \left( { - {\pi \over 2},\pi } \right):{{3 + 2i\sin \theta } \over {1 - 2i\sin \theta }}is\,purely\,imaginary} \right\}$$

. Then the sum of the elements in A is :

A

$${5\pi \over 6}$$

B

$$\pi $$

C

$${3\pi \over 4}$$

D

$${{2\pi } \over 3}$$

Given complex number,

$${{3 + 2i\sin \theta } \over {1 - 2i\sin \theta }}$$

$$ = {{\left( {3 + 2i\sin \theta } \right)\left( {1 + 2i\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$$

$$ = {{3 + 6i\sin \theta + 2i\sin \theta - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }}$$

$$ = {{\left( {3 - 4{{\sin }^2}\theta } \right) + i\left( {8\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$$

As complex number is purely imaginary, So real part of this complex number is zero.

$$ \therefore $$ $${{3 - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }}$$ = 0

$$ \Rightarrow $$ $$3 - 4{\sin ^2}\theta = 0$$

$$ \Rightarrow $$ $$\sin \theta = \pm {{\sqrt 3 } \over 2}$$

as $$\theta $$ $$ \in $$ $$\left( { - {\pi \over 2},\pi } \right)$$

$$ \therefore $$ $$\theta $$ $$=$$ $$-$$ $${\pi \over 3},{\pi \over 3},{{2\pi } \over 3}$$

$$ \therefore $$ Sum of those values of A is

$$ = - {\pi \over 3} + {\pi \over 3} + {{2\pi } \over 3}$$

$$ = {{2\pi } \over 3}$$

$${{3 + 2i\sin \theta } \over {1 - 2i\sin \theta }}$$

$$ = {{\left( {3 + 2i\sin \theta } \right)\left( {1 + 2i\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$$

$$ = {{3 + 6i\sin \theta + 2i\sin \theta - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }}$$

$$ = {{\left( {3 - 4{{\sin }^2}\theta } \right) + i\left( {8\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$$

As complex number is purely imaginary, So real part of this complex number is zero.

$$ \therefore $$ $${{3 - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }}$$ = 0

$$ \Rightarrow $$ $$3 - 4{\sin ^2}\theta = 0$$

$$ \Rightarrow $$ $$\sin \theta = \pm {{\sqrt 3 } \over 2}$$

as $$\theta $$ $$ \in $$ $$\left( { - {\pi \over 2},\pi } \right)$$

$$ \therefore $$ $$\theta $$ $$=$$ $$-$$ $${\pi \over 3},{\pi \over 3},{{2\pi } \over 3}$$

$$ \therefore $$ Sum of those values of A is

$$ = - {\pi \over 3} + {\pi \over 3} + {{2\pi } \over 3}$$

$$ = {{2\pi } \over 3}$$

3

Let z_{0} be a root of the quadratic equation, x^{2} + x + 1 = 0, If z = 3 + 6iz$$_0^{81}$$ $$-$$ 3iz$$_0^{93}$$, then arg z is equal to :

A

$${\pi \over 4}$$

B

$${\pi \over 6}$$

C

$${\pi \over 3}$$

D

0

1 + x + x^{2} = 0

x = $${{ - 1 \pm \sqrt {1 - 4} } \over 2} = {{ - 1 \pm i\sqrt 3 } \over 2}$$

z_{0} = w, w^{2}

Now

z = 3 + 6iz$$_0^{81}$$ $$-$$ 3iz$$_0^{93}$$

z = 3 + 6iw^{81} $$-$$ 3iw^{93} (w^{93} = w^{81} = 1)

$$ \Rightarrow $$ z = 3 + 3i

then arg(z) = tan^{$$-$$1}$$\left( {{3 \over 3}} \right)$$ = tan^{$$-$$1} (1) = $${\pi \over 4}$$

x = $${{ - 1 \pm \sqrt {1 - 4} } \over 2} = {{ - 1 \pm i\sqrt 3 } \over 2}$$

z

Now

z = 3 + 6iz$$_0^{81}$$ $$-$$ 3iz$$_0^{93}$$

z = 3 + 6iw

$$ \Rightarrow $$ z = 3 + 3i

then arg(z) = tan

4

Let z_{1} and z_{2} be any two non-zero complex numbers such that $$3\left| {{z_1}} \right| = 4\left| {{z_2}} \right|.$$ If $$z = {{3{z_1}} \over {2{z_2}}} + {{2{z_2}} \over {3{z_1}}}$$ then -

A

$${\rm I}m\left( z \right) = 0$$

B

$$\left| z \right| = \sqrt {{17 \over 2}} $$

C

$$\left| z \right| =$$ $${1 \over 2}\sqrt {9 + 16{{\cos }^2}\theta } $$

D

Re(z) $$=$$ 0

Given, $$3\left| {{z_1}} \right| = 4\left| {{z_2}} \right|$$

$$ \Rightarrow $$ $${{\left| {{z_1}} \right|} \over {\left| {{z_2}} \right|}} = {4 \over 3}$$

$$ \Rightarrow $$ $${{\left| {3{z_1}} \right|} \over {\left| {2{z_2}} \right|}} = {4 \over 3} \times {3 \over 2} = 2$$

As we know, for any compled number

$${{3{z_1}} \over {2{z_2}}} = {{\left| {3{z_1}} \right|} \over {\left| {2{z_2}} \right|}}$$(cos$$\theta $$ + i sin$$\theta $$)

= 2(cos$$\theta $$ + i sin$$\theta $$)

$$ \therefore $$ $${{2{z_2}} \over {3{z_1}}}$$ = $${1 \over {2\left( {\cos \theta + i\sin \theta } \right)}}$$

= $${1 \over {2\left( {\cos \theta + i\sin \theta } \right)}} \times {{\left( {\cos \theta - i\sin \theta } \right)} \over {\left( {\cos \theta - i\sin \theta } \right)}}$$

= $${\left( {{1 \over 2}\cos \theta - {i \over 2}\sin \theta } \right)}$$

Now, given

$$z = {{3{z_1}} \over {2{z_2}}} + {{2{z_2}} \over {3{z_1}}}$$

= 2(cos$$\theta $$ + i sin$$\theta $$) + $${\left( {{1 \over 2}\cos \theta - {i \over 2}\sin \theta } \right)}$$

= $${{5 \over 2}\cos \theta + {3 \over 2}i\sin \theta }$$

So, |z| = $$\sqrt {{{25} \over 4}{{\cos }^2}\theta + {9 \over 4}{{\sin }^2}\theta } $$

= $${1 \over 2}\sqrt {9 + 16{{\cos }^2}\theta } $$

z is neither purely real nor purely imaginary and |z| depends on $$\theta $$.

$$ \Rightarrow $$ $${{\left| {{z_1}} \right|} \over {\left| {{z_2}} \right|}} = {4 \over 3}$$

$$ \Rightarrow $$ $${{\left| {3{z_1}} \right|} \over {\left| {2{z_2}} \right|}} = {4 \over 3} \times {3 \over 2} = 2$$

As we know, for any compled number

$${{3{z_1}} \over {2{z_2}}} = {{\left| {3{z_1}} \right|} \over {\left| {2{z_2}} \right|}}$$(cos$$\theta $$ + i sin$$\theta $$)

= 2(cos$$\theta $$ + i sin$$\theta $$)

$$ \therefore $$ $${{2{z_2}} \over {3{z_1}}}$$ = $${1 \over {2\left( {\cos \theta + i\sin \theta } \right)}}$$

= $${1 \over {2\left( {\cos \theta + i\sin \theta } \right)}} \times {{\left( {\cos \theta - i\sin \theta } \right)} \over {\left( {\cos \theta - i\sin \theta } \right)}}$$

= $${\left( {{1 \over 2}\cos \theta - {i \over 2}\sin \theta } \right)}$$

Now, given

$$z = {{3{z_1}} \over {2{z_2}}} + {{2{z_2}} \over {3{z_1}}}$$

= 2(cos$$\theta $$ + i sin$$\theta $$) + $${\left( {{1 \over 2}\cos \theta - {i \over 2}\sin \theta } \right)}$$

= $${{5 \over 2}\cos \theta + {3 \over 2}i\sin \theta }$$

So, |z| = $$\sqrt {{{25} \over 4}{{\cos }^2}\theta + {9 \over 4}{{\sin }^2}\theta } $$

= $${1 \over 2}\sqrt {9 + 16{{\cos }^2}\theta } $$

z is neither purely real nor purely imaginary and |z| depends on $$\theta $$.

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