Question

All the values of $m$ for which both the roots of $x^2-2mx+m^2-1=0$ are greater than $-2$ but less than $4$, lies in the interval

• A. $(-1,3)$
• B. $(3,\infty )$
• C. $(-2,0)$
• D. $(-2,4)$

Answer 1

The correct option is A $(-1,3)$

Let $f\left(x\right)=x^2-2mx+m^2-1$, whose roots are $\alpha ,\beta$
Given: $-2<\alpha \le \beta <4$


Conditions:
$\left(i\right)\Delta \ge 0\Rightarrow 4m^2-4(m^2-1)\ge 0\Rightarrow 4>0\therefore m\in R\dots \left(1\right)$

$\left(ii\right)f(-2)>0\Rightarrow m^2+4m+3>0\Rightarrow (m+1)(m+3)>0\Rightarrow m<-3\text{or}m>-1\dots \left(2\right)$

$\left(iii\right)f\left(4\right)>0\Rightarrow m^2-8m+15>0\Rightarrow (m-3)(m-5)>0\Rightarrow m<3\text{or}m>5\dots \left(3\right)$

$\left(iv\right)-2<-\frac{b}{2a}<4\Rightarrow -2<m<4\dots \left(4\right)$

From $\left(1\right),\left(2\right),\left(3\right)$ and $\left(4\right),$ we get
$m\in (-1,3)$