Question

If both the distinct roots of the equation $x^2-ax-b=0(a,b\in R)$ are lying in between $-2$ and $2$, then

• A. $|a|<2-\frac{b}{2}$
• B. $|a|>2-\frac{b}{2}$
• C. $|a|<4$
• D. $|a|>\frac{b}{2}-2$

Answer 1

Answer: (A), (C), (D)

$|a|>\frac{b}{2}-2$


Required conditions are
$\left(1\right)f(-2)>0,\left(2\right)f\left(2\right)>0,\left(3\right)-2<-\frac{B}{2A}<2,\left(4\right)D>0$

$f(-2)>0$
$\Rightarrow 4+2a-b>0$
$\Rightarrow a>\frac{b}{2}-2$
$\Rightarrow -(2-\frac{b}{2})<a...\left(i\right)$

$f\left(2\right)>0$
$\Rightarrow 4-2a-b>0$
$\Rightarrow a<2-\frac{b}{2}...\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right),$
$-(2-\frac{b}{2})<a<2-\frac{b}{2}$
$\Rightarrow |a|<2-\frac{b}{2}$

Since, $|a|\ge 0$
$\Rightarrow 2-\frac{b}{2}>0$
$\Rightarrow \frac{b}{2}-2<0\le |a|$
$\Rightarrow |a|>\frac{b}{2}-2$

$-2<-\frac{B}{2A}<2$
$\Rightarrow -2<\frac{a}{2}<2$
$\Rightarrow -4<a<4$
$\Rightarrow a\in (-4,4)$
$\Rightarrow |a|<4$