Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

If $$\alpha $$ and $$\beta $$ are the roots of the equation $${x^2} - x + 1 = 0,$$ then $${\alpha ^{2009}} + {\beta ^{2009}} = $$

A

$$\, - 1$$

B

$$\, 1$$

C

$$\, 2$$

D

$$\, - 2$$

$${x^2} - x + 1 = 0$$

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

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

$$\alpha = {1 \over 2} + i{{\sqrt 3 } \over 2} = - {\omega ^2}$$

$$\beta = {1 \over 2} - {{i\sqrt 3 } \over 2} = - \omega $$

$${\alpha ^{2009}} + {\beta ^{2009}} = {\left( { - {\omega ^2}} \right)^{2009}} + {\left( { - \omega } \right)^{2009}}$$

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

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

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

$$\alpha = {1 \over 2} + i{{\sqrt 3 } \over 2} = - {\omega ^2}$$

$$\beta = {1 \over 2} - {{i\sqrt 3 } \over 2} = - \omega $$

$${\alpha ^{2009}} + {\beta ^{2009}} = {\left( { - {\omega ^2}} \right)^{2009}} + {\left( { - \omega } \right)^{2009}}$$

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

2

MCQ (Single Correct Answer)

If the roots of the equation $$b{x^2} + cx + a = 0$$ imaginary, then for all real values of $$x$$, the expression $$3{b^2}{x^2} + 6bcx + 2{c^2}$$ is :

A

less than $$4ab$$

B

greater than $$-4ab$$

C

less than $$-4ab$$

D

greater than $$4ab$$

Given that roots of the equation

$$b{x^2} + cx + a = 0$$ are imaginary

$$\therefore$$ $${c^2} - 4ab < 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

Let $$y = 3{b^2}{x^2} + 6bc\,x + 2{c^2}$$

$$ \Rightarrow 3{b^2}{x^2} + 6bc\,x + 2{c^2} - y = 0$$

As $$x$$ is real, $$D \ge 0$$

$$ \Rightarrow 36{b^2}{c^2} - 12{b^2}\left( {2{c^2} - y} \right) \ge 0$$

$$ \Rightarrow 12{b^2}\left( {3{c^2} - 2{c^2} + y} \right) \ge 0$$

$$ \Rightarrow {c^2} + y \ge 0$$

$$ \Rightarrow y \ge - {c^2}$$

But from eqn. $$(i),$$ $${c^2} < 4ab$$

or $$ - {c^2} > - 4ab$$

$$\therefore$$ we get $$y \ge - {c^2} > - 4ab$$

$$y > - 4ab$$

$$b{x^2} + cx + a = 0$$ are imaginary

$$\therefore$$ $${c^2} - 4ab < 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

Let $$y = 3{b^2}{x^2} + 6bc\,x + 2{c^2}$$

$$ \Rightarrow 3{b^2}{x^2} + 6bc\,x + 2{c^2} - y = 0$$

As $$x$$ is real, $$D \ge 0$$

$$ \Rightarrow 36{b^2}{c^2} - 12{b^2}\left( {2{c^2} - y} \right) \ge 0$$

$$ \Rightarrow 12{b^2}\left( {3{c^2} - 2{c^2} + y} \right) \ge 0$$

$$ \Rightarrow {c^2} + y \ge 0$$

$$ \Rightarrow y \ge - {c^2}$$

But from eqn. $$(i),$$ $${c^2} < 4ab$$

or $$ - {c^2} > - 4ab$$

$$\therefore$$ we get $$y \ge - {c^2} > - 4ab$$

$$y > - 4ab$$

3

MCQ (Single Correct Answer)

If $$\,\left| {z - {4 \over z}} \right| = 2,$$ then the maximum value of $$\,\left| z \right|$$ is equal to :

A

$$\sqrt 5 + 1$$

B

2

C

$$2 + \sqrt 2 $$

D

$$\sqrt 3 + 1$$

Given that $$\left| {z - {4 \over z}} \right| = 2$$

Now $$\left| z \right| = \left| {z - {4 \over z} + {4 \over { - z}}} \right| \le \left| {z - {4 \over z}} \right| + {4 \over {\left| z \right|}}$$

$$ \Rightarrow \left| z \right| \le 2 + {4 \over {\left| z \right|}}$$

$$ \Rightarrow {\left| z \right|^2} - 2\left| z \right| - 4 \le 0$$

$$ \Rightarrow \left( {\left| z \right| - {{2 + \sqrt {20} } \over 2}} \right)\left( {\left| z \right| - {{2 - \sqrt {20} } \over 2}} \right) \le 0$$

$$\left( {\left| z \right| - \left( {1 + \sqrt 5 } \right)} \right)\left( {\left| z \right| - \left( {1 - \sqrt 5 } \right)} \right) \le 0$$

$$ \Rightarrow \left( { - \sqrt 5 + 1} \right) \le \left| z \right| \le \left( {\sqrt 5 + 1} \right)$$

$$ \Rightarrow {\left| z \right|_{\max }} = \sqrt 5 + 1$$

Now $$\left| z \right| = \left| {z - {4 \over z} + {4 \over { - z}}} \right| \le \left| {z - {4 \over z}} \right| + {4 \over {\left| z \right|}}$$

$$ \Rightarrow \left| z \right| \le 2 + {4 \over {\left| z \right|}}$$

$$ \Rightarrow {\left| z \right|^2} - 2\left| z \right| - 4 \le 0$$

$$ \Rightarrow \left( {\left| z \right| - {{2 + \sqrt {20} } \over 2}} \right)\left( {\left| z \right| - {{2 - \sqrt {20} } \over 2}} \right) \le 0$$

$$\left( {\left| z \right| - \left( {1 + \sqrt 5 } \right)} \right)\left( {\left| z \right| - \left( {1 - \sqrt 5 } \right)} \right) \le 0$$

$$ \Rightarrow \left( { - \sqrt 5 + 1} \right) \le \left| z \right| \le \left( {\sqrt 5 + 1} \right)$$

$$ \Rightarrow {\left| z \right|_{\max }} = \sqrt 5 + 1$$

4

MCQ (Single Correct Answer)

The quadratic equations $${x^2} - 6x + a = 0$$ and $${x^2} - cx + 6 = 0$$ have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then the common root is

A

1

B

4

C

3

D

2

Let the roots of equation $${x^2} - 6x + a = 0$$ be $$\alpha $$

and $$4$$ $$\beta $$ and that of the equation

$${x^2} - cx + 6 = 0$$ be $$\alpha $$ and $$3\beta .$$ Then

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha + 4\beta = 6;\,\,\,\,\,\,\,4\alpha \beta = a$$

and $$\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha + 3\beta = c;\,\,\,\,\,\,\,3\alpha \beta = 6$$

$$ \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a = 8$$

$$\therefore$$ The equation becomes

$${x^2} - 6x + 8 = 0$$

$$ \Rightarrow \left( {x - 2} \right)\left( {x - 4} \right) = 0$$

$$ \Rightarrow $$ roots are $$2$$ and $$4$$

$$ \Rightarrow \alpha = 2,\beta = 1$$

$$\therefore$$ Common root is $$2.$$

and $$4$$ $$\beta $$ and that of the equation

$${x^2} - cx + 6 = 0$$ be $$\alpha $$ and $$3\beta .$$ Then

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha + 4\beta = 6;\,\,\,\,\,\,\,4\alpha \beta = a$$

and $$\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha + 3\beta = c;\,\,\,\,\,\,\,3\alpha \beta = 6$$

$$ \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a = 8$$

$$\therefore$$ The equation becomes

$${x^2} - 6x + 8 = 0$$

$$ \Rightarrow \left( {x - 2} \right)\left( {x - 4} \right) = 0$$

$$ \Rightarrow $$ roots are $$2$$ and $$4$$

$$ \Rightarrow \alpha = 2,\beta = 1$$

$$\therefore$$ Common root is $$2.$$

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