Q. All the values of m for which both roots of the equation $x^{2}-2mx+m^{2}-1=0$ are greater than -2but less than 4 lie in the interval :
AIEEEAIEEE 2008
Solution:
Since both roots of equation $x^{2}-2mx+m^{2}-1=0$ are greater than -2 but less than 4.
$\therefore D \ge 0$
$\Rightarrow 4m^{2}-4m^{2}+4 \ge 0 \Rightarrow m\,\epsilon\,R ...\left(i\right)$
$-2 < \frac{-b}{2a} < 4 \Rightarrow -2 < \left(\frac{2m}{2.1}\right) < 4$
$\Rightarrow -2 < m < 4 ...\left(ii\right)$
Also $f \left(4\right) > 0 \Rightarrow 16-8m+m^{2}-1 > 0$
$\Rightarrow m^{2}-8m+15 > 0$
$\Rightarrow \left(m-3\right)\left(m-5\right) > 0$
$\Rightarrow -\infty < m < 3 $ and $5 < m < \infty ...\left(iii\right)$
Also $f \left(2\right) > 0 \Rightarrow 4+4m+m^{2}-1 > 0$
$\Rightarrow m^{2}-4m+3 > 0$
$\Rightarrow \left(m+3\right)\left(m+1\right) > 0$
$\Rightarrow -\infty < m < -3$ and $-1 < m < \infty ...\left(iv\right)$
$\therefore $ From $\left(i\right), \left(ii\right), \left(iii\right) and \left(iv\right)$
m lies between -1 and 3.
