Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

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 $$2.$$

2

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 $$-1.$$

3

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

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

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

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

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

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

4

MCQ (Single Correct Answer)

If $$x$$ is real, the maximum value of $${{3{x^2} + 9x + 17} \over {3{x^2} + 9x + 7}}$$ is

A

$${1 \over 4}$$

B

$$41$$

C

$$1$$

D

$${17 \over 7}$$

$$y = {{3{x^2} + 9x + 17} \over {3{x^2} + 9x + 7}}$$

$$3{x^2}\left( {y - 1} \right) + 9x\left( {y - 1} \right) + 7y - 17 = 0$$

$$D \ge 0$$ as $$x$$ is real

$$81{\left( {y - 1} \right)^2} - 4 \times 3\left( {y - 1} \right)\left( {7y - 17} \right) \ge 0$$

$$ \Rightarrow \left( {y - 1} \right)\left( {y - 41} \right) \le 0$$

$$ \Rightarrow 1 \le y \le 41$$

$$\therefore$$ Max value of $$y$$ is $$41$$

$$3{x^2}\left( {y - 1} \right) + 9x\left( {y - 1} \right) + 7y - 17 = 0$$

$$D \ge 0$$ as $$x$$ is real

$$81{\left( {y - 1} \right)^2} - 4 \times 3\left( {y - 1} \right)\left( {7y - 17} \right) \ge 0$$

$$ \Rightarrow \left( {y - 1} \right)\left( {y - 41} \right) \le 0$$

$$ \Rightarrow 1 \le y \le 41$$

$$\therefore$$ Max value of $$y$$ is $$41$$

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