Question

If $\alpha, \beta$ are roots of the equation $x^{2}-2 m x+m^{2}-1=0$ then the number of integral value of $m$ for which $\alpha, \beta \in(-2,4)$ is :

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (D)

$x=\frac{2 m \pm \sqrt{4 m^{2}-4\left(m^{2}-1\right)}}{2}=\frac{2 m \pm 2}{2}$
or $m+1$ or $m-1$
Hence $-2 < m+1 < 4 \Rightarrow -3 < m < 3$
$-2 < m-1 < 4 \,\,\Rightarrow -1 < m < 5$
Hence, $m \in(-1,3) \,\,\Rightarrow m=0,1,2$