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### AIEEE 2007

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

## Explanation

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

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

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

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

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

$$\Rightarrow a \in \left( { - 3,3} \right)$$
2

### AIEEE 2006

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
$${17 \over 7}$$

## Explanation

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

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

$$D \ge 0$$ 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

### AIEEE 2006

MCQ (Single Correct Answer)
All the values of $$m$$ for which both roots of the equation $${x^2} - 2mx + {m^2} - 1 = 0$$ are greater than $$- 2$$ but less then 4, lie in the interval
A
$$- 2 < m < 0$$
B
$$m > 3$$
C
$$- 1 < m < 3$$
D
$$1 < m < 4$$

## Explanation

Equation $${x^2} - 2mx + {m^2} - 1 = 0$$

$${\left( {x - m} \right)^2} - 1 = 0$$

or $$\left( {x - m + 1} \right)\left( {x - m - 1} \right) = 0$$

$$x = m - 1,m + 1$$

$$m - 1 > - 2$$ and $$m + 1 < 4$$

$$\Rightarrow m > - 1$$ and $$m<3$$

or $$\,\,\, - 1 < m < 3$$
4

### AIEEE 2006

MCQ (Single Correct Answer)
If the roots of the quadratic equation $${x^2} + px + q = 0$$ are $$\tan {30^ \circ }$$ and $$\tan {15^ \circ }$$, respectively, then the value of $$2 + q - p$$ is
A
2
B
3
C
0
D
1

## Explanation

$${x^2} + px + q = 0$$

Sum of roots $$= \tan {30^ \circ } + \tan {15^ \circ } = - p$$

Products of roots $$= \tan {30^ \circ }.\tan {15^ \circ } = q$$

$$\tan {45^ \circ } = {{\tan {{30}^ \circ } + \tan {{15}^ \circ }} \over {1 - \tan {{30}^ \circ }.\tan {{15}^ \circ }}}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {{ - p} \over {1 - q}} = 1$$

$$\Rightarrow - p = 1 - q \Rightarrow q - p = 1$$

$$\therefore$$ $$2 + q - p = 3$$

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