Question

If $\alpha ,\beta$ are the roots of the equation $x^2+\alpha x+\beta =0$ such that $\alpha \ne \beta$ and $||x-\beta |-\alpha |<1,$ then

• A. inequality is satisfied by exactly three integral values of $x$
• B. inequality is satisfied by all values of $x\epsilon (-4,-2)$
• C. roots of the equation are opposite in sign
• D. All of the above

Answer 1

The correct option is D All of the above

$\Rightarrow \alpha +\beta =-\alpha$ and $\alpha \beta =\beta \Rightarrow \alpha =1$

$\Rightarrow \beta =-2$

$\therefore (\alpha ,\beta )=(1,-2)$

$\Rightarrow ||x+2|-1|<1$

$\Rightarrow -1<|x+2|-1<1$

Let us inequality on left side first

$\Rightarrow -1<|x+2|-1$

$\Rightarrow |x+2|>0\Rightarrow x\in (-\infty ,\infty )$

Right inequality ;

$\Rightarrow |x+2|-1<1$

$\Rightarrow |x+2|<2$

$\Rightarrow -2<x+2<2$

$\Rightarrow x\in (-4,0)$

Combining both we get $x\in (-4,0)$

$(-3,-2,-1)$ three integers satisfy the equation and all values between $(-4,0)$ satisfy the equation

$\alpha ,\beta$ are of different sign