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1

### WB JEE 2009

The quadratic equation whose roots are three times the roots of $$3a{x^2} + 3bx + c = 0$$ is

A
ax2 + 3bx + 3c = 0
B
ax2 + 3bx + c = 0
C
9ax2 + 9bx + c = 0
D
ax2 + bx + 3c = 0

## Explanation

$$3a{x^2} + 3bx + c = 0$$

Let roots of given equation are $$\alpha$$, $$\beta$$

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

$$\alpha \beta = {c \over {3a}}$$

$$\therefore$$ New roots are $$\gamma$$, $$\delta$$ $$\therefore$$ $$\gamma$$ = 3$$\alpha$$, $$\delta$$ = 3$$\beta$$

$$\gamma + \delta = 3(\alpha + \beta ) = - {{3b} \over a}$$

$$\gamma \delta = 3\alpha \,.\,3\beta = {{3c} \over a}$$

$$\therefore$$ Quadratic equation is $${x^2} - (\gamma + \delta )x + \gamma \delta = 0$$

$$\Rightarrow {x^2} + {{3b} \over a}x + {{3c} \over a} = 0 \Rightarrow a{x^2} + 3bx + 3c = 0$$

2

### WB JEE 2009

The sum of all real roots of the equation $$|x - 2{|^2} + |x - 2| - 2 = 0$$ is

A
7
B
4
C
1
D
5

## Explanation

$$|x - 2{|^2} + |x - 2| - 2 = 0$$

Let $$|x - 2| = t$$

$$\therefore$$ $${t^2} + t - 2 = 0 \Rightarrow (t + 2)(t - 1) = 0$$

$$\Rightarrow$$ t = $$-$$2 or 1

If t = $$-$$2 then $$|x - 2| = - 2$$, but modulus of any value is non negative, so it has no solution.

If t = 1, then $$|x - 2| = 1 \Rightarrow x - 2 = \pm 1 \Rightarrow x = 3$$ or 1

$$\therefore$$ Sum of roots = 3 + 1 = 4

3

### WB JEE 2009

If $$\alpha$$, $$\beta$$ be the roots of $${x^2} - a(x - 1) + b = 0$$, then the value of $${1 \over {{\alpha ^2} - a\alpha }} + {1 \over {{\beta ^2} - a\beta }} + {2 \over {a + b}}$$ is

A
$${4 \over {a + b}}$$
B
$${1 \over {a + b}}$$
C
0
D
$$-$$1

## Explanation

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

$$\Rightarrow {1 \over {{\alpha ^2} - a\alpha }} + {1 \over {{\beta ^2} - a\beta }} + {2 \over {a + b}} = 0$$

4

### WB JEE 2008

The equation $${x^2} - 3|x| + 2 = 0$$ has

A
No real root
B
One real root
C
Two real roots
D
Four real roots

## Explanation

$${x^2} - 3|x| + 2 = 0$$

$$\Rightarrow |x{|^2} - 3|x| - |x| + 2 = 0$$ ($$\because$$ $${x^2} = |x{|^2} = |x{|^2}$$)

$$\Rightarrow |x|(|x| - 2) - 1(|x| - 2) = 0 \Rightarrow (|x| - 1)(|x| - 2) = 0$$

$$\Rightarrow |x| - 1 = 0$$ or $$|x| - 2 = 0$$

If $$|x| - 1 = 0 \Rightarrow |x| = 1 \Rightarrow x = \pm 1$$

If $$|x| - 2 = 0 \Rightarrow |x| = 2 \Rightarrow x = \pm 2$$

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