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1

### AIEEE 2005

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

2

### AIEEE 2005

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

$$\Rightarrow$$ $${a = 1}$$
3

### AIEEE 2005

If both the roots of the quadratic equation $${x^2} - 2kx + {k^2} + k - 5 = 0$$ are less than 5, then $$k$$ lies in the interval
A
$$\left( {5,6} \right]$$
B
$$\left( {6,\,\infty } \right)$$
C
$$\left( { - \infty ,\,4} \right)$$
D
$$\left[ {4,\,5} \right]$$

## Explanation

both roots are less than $$5,$$

then $$(i)$$ Discriminant $$\ge 0$$

$$\left( {ii} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,p\left( 5 \right) > 0$$

$$\left( {iii} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{Sum\,\,of\,\,roots} \over 2} < 5$$

Hence $$\left( i \right)\,\,\,\,\,\,4{k^2} - 4\left( {{k^2} + k - 5} \right) \ge 0$$

$$4{k^2} - 4{k^2} - 4k + 20 \ge 0$$

$$4k \le 20 \Rightarrow k \le 5$$

$$\left( {ii} \right)\,\,\,\,\,f\left( 5 \right) > 0;25 - 10k + {k^2} + k - 5 > 0$$

or $${k^2} - 9k + 20 > 0$$

or $$k\left( {k - 4} \right) - 5\left( {k - 4} \right) > 0$$

or $$\left( {k - 5} \right)\left( {k - 4} \right) > 0$$

$$\Rightarrow k \in \left( { - \infty ,4} \right) \cup \left( { - \infty ,5} \right)$$

$$\left( {iii} \right)\,\,\,\,\,\,{{Sum\,\,of\,\,roots} \over 2}$$

$$= - {b \over {2a}} = {{2k} \over 2} < 5$$

The intersection of $$(i)$$, $$(ii)$$ & $$(iii)$$ gives

$$k \in \left( { - \infty ,4} \right).$$
4

### AIEEE 2005

In a triangle $$PQR,\;\;\angle R = {\pi \over 2}.\,\,If\,\,\tan \,\left( {{P \over 2}} \right)$$ and $$\tan \left( {{Q \over 2}} \right)$$ are the roots of $$a{x^2} + bx + c = 0,\,\,a \ne 0$$ then
A
$$a = b + c$$
B
$$c = a + b$$
C
$$b = c$$
D
$$b = a + c$$

## Explanation

$$\angle$$R = 90o $$\therefore$$ $$\angle$$P + $$\angle$$Q = 90o

$$\Rightarrow$$ $${P \over 2} + {Q \over 2} = {{90} \over 2} = 45$$o

$$\tan \left( {{P \over 2}} \right),\tan \left( {{Q \over 2}} \right)$$ are the roots of $$a{x^2} + bx + c = 0$$

$$\therefore$$ $$\tan \left( {{P \over 2}} \right) + \tan \left( {{Q \over 2}} \right) = - {b \over a},\,\,$$

and $$\tan \left( {{P \over 2}} \right).\tan \left( {{Q \over 2}} \right) = {c \over a}$$

$${{\tan \left( {{P \over 2}} \right) + \tan \left( {{Q \over 2}} \right)} \over {1 - \tan \left( {{P \over 2}} \right)\tan \left( {{Q \over 2}} \right)}}$$

$$= \tan \left( {{P \over 2} + {Q \over 2}} \right)$$= tan 45o = 1

$$\Rightarrow {{ - {b \over a}} \over {1 - {c \over a}}} = 1$$

$$\Rightarrow - {b \over a} = {a \over a} - {c \over a}$$

$$\Rightarrow - b = a - c$$ or $$c = a + b.$$

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