Question

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 :

• A. $m > 3$
• B. $-1 < m < 3$
• C. $1 < m < 4$
• D. $-2 < m < 0$

Answer 1

Answer: (B)

$-1 < m < 3$

Answer 2

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.