Question

The number of integral values of $m$, for which the roots of $x^2-2mx+m^2-1=0$ will lie between $-2$ and $4$ is

• A. $2$
• B. $0$
• C. $3$
• D. $1$

Answer 1

The correct option is C $3$

Let $y=x^2-2mx+m^2-1$
For roots to lie between -2 and 4,
We must have,
(i) $D\ge 0\Rightarrow 4m^2-4(m^2-1)=4>\forall m\in R$
(ii)$-2<-\frac{b}{2a}<4\Rightarrow -2<m<4$
(iii)$y(-2)>0\Rightarrow m^2+4m+3>0\Rightarrow (m+1)(m+3)>0\Rightarrow m<-3,m>-1$
(iv)$y\left(4\right)>0\Rightarrow m^2-8m+15>0\Rightarrow (m-5)(m-3)>0\Rightarrow m>5,m<3$
Considering all the cases, permissible integral values of $m$ are $0,1,2$.