Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

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

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

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

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

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

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

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

2

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 $$2.$$

3

MCQ (Single Correct Answer)

**STATEMENT - 2 :** For every natural number $$n \ge 2,$$,
$$$\sqrt {n\left( {n + 1} \right)} < n + 1.$$$

A

Statement - 1 is false, Statement - 2 is true

B

Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for statement - 1

C

Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1

D

Statement - 1 is true, Statement - 2 is false

Statements $$2$$ is $$\sqrt {n\left( {n + 1} \right)} < n + 1,n \ge 2$$

$$ \Rightarrow \sqrt n < \sqrt {n + 1} ,n \ge 2$$ which is true

$$ \Rightarrow \sqrt 2 < \sqrt 3 < \sqrt 4 < \sqrt 5 < - - - - - - \sqrt n $$

Now $$\sqrt 2 < \sqrt n \Rightarrow {1 \over {\sqrt 2 }} > {1 \over {\sqrt n }}$$

$$\sqrt 3 < \sqrt n \Rightarrow {1 \over {\sqrt 3 }} > {1 \over {\sqrt n }};$$

$$\sqrt n \le \sqrt n \Rightarrow {1 \over {\sqrt n }} \ge {1 \over {\sqrt n }}$$

Also $${1 \over {\sqrt 1 }} > {1 \over {\sqrt n }}$$

$$\therefore$$ Adding all, we get

$${1 \over {\sqrt 1 }} + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }} + ....... + {1 \over n} > {n \over {\sqrt n }} = \sqrt n $$

Hence both the statements are correct and statement $$2$$ is a correct explanation of statement $$-1.$$

$$ \Rightarrow \sqrt n < \sqrt {n + 1} ,n \ge 2$$ which is true

$$ \Rightarrow \sqrt 2 < \sqrt 3 < \sqrt 4 < \sqrt 5 < - - - - - - \sqrt n $$

Now $$\sqrt 2 < \sqrt n \Rightarrow {1 \over {\sqrt 2 }} > {1 \over {\sqrt n }}$$

$$\sqrt 3 < \sqrt n \Rightarrow {1 \over {\sqrt 3 }} > {1 \over {\sqrt n }};$$

$$\sqrt n \le \sqrt n \Rightarrow {1 \over {\sqrt n }} \ge {1 \over {\sqrt n }}$$

Also $${1 \over {\sqrt 1 }} > {1 \over {\sqrt n }}$$

$$\therefore$$ Adding all, we get

$${1 \over {\sqrt 1 }} + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }} + ....... + {1 \over n} > {n \over {\sqrt n }} = \sqrt n $$

Hence both the statements are correct and statement $$2$$ is a correct explanation of statement $$-1.$$

4

MCQ (Single Correct Answer)

If the difference between the roots of the equation $${x^2} + ax + 1 = 0$$ is less than $$\sqrt 5 ,$$ then the set of possible values of $$a$$ is

A

$$\left( {3,\infty } \right)$$

B

$$\left( { - \infty , - 3} \right)$$

C

$$\left( { - 3,3} \right)$$

D

$$\left( { - 3,\infty } \right)$$

Let $$\alpha $$ and $$\beta $$ are roots of the equation $${x^2} + ax + 1 = 0$$

So, $$\alpha + \beta = - a$$ and $$\alpha \beta = 1$$

given $$\left| {\alpha - \beta } \right| < \sqrt 5 $$

$$ \Rightarrow \sqrt {{{\left( {\alpha - \beta } \right)}^2} - 4\alpha \beta } < \sqrt 5 $$

(as $${\left( {\alpha - \beta } \right)^2} = {\left( {\alpha + \beta } \right)^2} - 4\alpha \beta $$ )

$$ \Rightarrow \sqrt {{a^2} - 4} < \sqrt 5 $$

$$ \Rightarrow {a^2} - 4 < 5$$

$$ \Rightarrow {a^2} - 9 < 0 \Rightarrow {a^2} < 9$$

$$ \Rightarrow - 3 < a < 3$$

$$ \Rightarrow a \in \left( { - 3,3} \right)$$

So, $$\alpha + \beta = - a$$ and $$\alpha \beta = 1$$

given $$\left| {\alpha - \beta } \right| < \sqrt 5 $$

$$ \Rightarrow \sqrt {{{\left( {\alpha - \beta } \right)}^2} - 4\alpha \beta } < \sqrt 5 $$

(as $${\left( {\alpha - \beta } \right)^2} = {\left( {\alpha + \beta } \right)^2} - 4\alpha \beta $$ )

$$ \Rightarrow \sqrt {{a^2} - 4} < \sqrt 5 $$

$$ \Rightarrow {a^2} - 4 < 5$$

$$ \Rightarrow {a^2} - 9 < 0 \Rightarrow {a^2} < 9$$

$$ \Rightarrow - 3 < a < 3$$

$$ \Rightarrow a \in \left( { - 3,3} \right)$$

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Complex Numbers

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