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1

### AIEEE 2003

The value of '$$a$$' for which one root of the quadratic equation $$\left( {{a^2} - 5a + 3} \right){x^2} + \left( {3a - 1} \right)x + 2 = 0$$\$
is twice as large as the other is
A
$$- {1 \over 3}$$
B
$${2 \over 3}$$
C
$$- {2 \over 3}$$
D
$${1 \over 3}$$

## Explanation

Let the roots of given equation be $$\alpha$$ and $$2$$$$\alpha$$ then

$$\alpha + 2\alpha = 3\alpha = {{1 - 3a} \over {{a^2} - 5a + 3}}$$

and $$\alpha .2\alpha = 2{\alpha ^2} = {2 \over {{a^2} - 5a + 3}}$$

$$\Rightarrow \alpha = {{1 - 3a} \over {3\left( {{a^2} - 5a + 3} \right)}}$$

$$\therefore$$ $$2\left[ {{1 \over 9}{{{{\left( {1 - 3a} \right)}^2}} \over {{{\left( {{a^2} - 5a + 3} \right)}^2}}}} \right]$$

$$= {2 \over {{a^2} - 5a + 3}}$$

$${{{{\left( {1 - 3a} \right)}^2}} \over {\left( {{a^2} - 5a + 3} \right)}} = 9$$

or $$9{a^2} - 6a + 1$$

$$= 9{a^2} - 45a + 27$$

or $$39a = 26$$ or $$a = {2 \over 3}$$
2

### 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 $$H.P.$$
3

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

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

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

$$\Rightarrow ab + bc + ca < 1$$
4

### 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 $$q=-2.$$

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