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### 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 $$\,\,\, - 1 < m < 3$$
2

### 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 $$= \tan {30^ \circ }.\tan {15^ \circ } = q$$

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

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

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

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

### AIEEE 2005

MCQ (Single Correct Answer)
If the roots of the equation $${x^2} - bx + c = 0$$ be two consecutive integers, then $${b^2} - 4c$$ equals
A
$$-2$$
B
$$3$$
C
$$2$$
D
$$1$$

## Explanation

Let n and (n + 1) be the roots of x2 $$-$$ bx + c = 0.

Then, n + (n + 1) = b and n(n + 1) = c

$$\therefore$$ b2 $$-$$ 4c = (2n + 1)2 $$-$$ 4n(n + 1)

= 4n2 + 4n + 1 $$-$$ 4n2 $$-$$ 4n = 1

4

### AIEEE 2005

MCQ (Single Correct Answer)
The value of $$a$$ for which the sum of the squares of the roots of the equation
$${x^2} - \left( {a - 2} \right)x - a - 1 = 0$$ assume the least value is
A
$$1$$
B
$$0$$
C
$$3$$
D
$$2$$

## Explanation

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

Let $$\alpha$$ and $$\beta$$ are the roots of the equation.

$$\therefore$$ $$\alpha$$ + $$\beta$$ = $$a - 2$$

and $$\alpha$$$$\beta$$ = $$- a - 1$$

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

$$\Rightarrow$$ $${\alpha ^2} + {\beta ^2} = {\left( {a - 2} \right)^2} + 2\left( {a + 1} \right)$$

$$\Rightarrow$$ $${\alpha ^2} + {\beta ^2} = {a^2} - 2a + 6$$

$$\Rightarrow$$ $${\alpha ^2} + {\beta ^2} = {\left( {a - 1} \right)^2} + 5$$

$$\Rightarrow$$ The value of $${\alpha ^2} + {\beta ^2}$$ will be minimum, when $${a - 1}$$ = 0

$$\Rightarrow$$ $${a = 1}$$

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