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### AIEEE 2003

If the sum of the roots of the quadratic equation $$a{x^2} + bx + c = 0$$ is equal to the sum of the squares of their reciprocals, then $${a \over c},\,{b \over a}$$ and $${c \over b}$$ are in
A
Arithmetic - Geometric Progression
B
Arithmetic Progression
C
Geometric Progression
D
Harmonic Progression

## Explanation

$$a{x^2} + bx + c = 0,$$ $$\alpha + \beta = {{ - b} \over a},\alpha \beta = {c \over a}$$

As for given condition, $$\alpha + \beta = {1 \over {{\alpha ^2}}} + {1 \over {{\beta ^2}}}$$

$$\alpha + \beta = {{{\alpha ^2} + {\beta ^2}} \over {{\alpha ^2}{\beta ^2}}} - {b \over a}$$

$$= {{{{{b^2}} \over {{a^2}}} - {{2c} \over a}} \over {{{{c^2}} \over {{a^2}}}}}$$

On simplification $$2{a^2}c = a{b^2} + b{c^2}$$

$$\Rightarrow {{2a} \over b} = {c \over a} + {b \over c}$$

$$\Rightarrow {c \over a},{a \over b},{b \over c}$$ are in $$A.P.$$

$$\therefore$$ $${a \over c},{b \over a},\,\,\& \,\,$$ are in $$H.P.$$
2

### AIEEE 2002

If $$a,\,b,\,c$$ are distinct $$+ ve$$ real numbers and $${a^2} + {b^2} + {c^2} = 1$$ then $$ab + bc + ca$$ is
A
less than 1
B
equal to 1
C
greater than 1
D
any real no.

## Explanation

As $$\,\,\,\,\,{\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} > 0$$

$$\Rightarrow 2\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) > 0$$

$$\Rightarrow 2 > 2\left( {ab + bc + ca} \right)$$

$$\Rightarrow ab + bc + ca < 1$$
3

### AIEEE 2002

If $$p$$ and $$q$$ are the roots of the equation $${x^2} + px + q = 0,$$ then
A
$$p = 1,\,\,q = - 2$$
B
$$p = 0,\,\,q = 1$$
C
$$p = - 2,\,\,q = 0$$
D
$$p = - 2,\,\,q = 1$$

## Explanation

$$p + q = - p$$ and $$pq = q \Rightarrow q\left( {p - 1} \right) = 0$$

$$\Rightarrow q = 0$$ or $$p=1.$$

If $$q = 0,$$ then $$p=0.$$ i.e.$$p=q$$

$$\therefore$$ $$p=1$$ and $$q=-2.$$
4

### AIEEE 2002

Difference between the corresponding roots of $${x^2} + ax + b = 0$$ and $${x^2} + bx + a = 0$$ is same and $$a \ne b,$$ then
A
$$a + b + 4 = 0$$
B
$$a + b - 4 = 0$$
C
$$a - b - 4 = 0$$
D
$$a - b + 4 = 0$$

## Explanation

Let $$\alpha ,\beta$$ and $$\gamma ,\delta$$ be the roots of the equations $${x^2} + ax + b = 0$$

and $${x^2} + bx + a = 0$$ respectively.

$$\therefore$$ $$\alpha + \beta = - a,\alpha \beta = b$$

and $$\gamma + \delta = - b,\gamma \delta = a.$$

Given $$\left| {\alpha - \beta } \right| = \left| {\gamma - \delta } \right|$$

$$\Rightarrow {\left( {\alpha - \beta } \right)^2} = {\left( {\gamma - \delta } \right)^2}$$

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

$$\Rightarrow {a^2} - 4b = {b^2} - 4a$$

$$\Rightarrow \left( {{a^2} - {b^2}} \right) + 4\left( {a - b} \right) = 0$$

$$\Rightarrow a + b + 4 = 0$$

( as $$a \ne b$$ )

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