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)

Since both roots of equation $x^2-2mx+m^2-1=0$ are greater than -2 but less than 4.
$\therefore D\ge 0$
$\Rightarrow 4m^2-4m^2+4\ge 0\Rightarrow mϵR...\left(i\right)$
$-2<\frac{-b}{2a}<4\Rightarrow -2<\left(\frac{2m}{2.1}\right)<4$
$\Rightarrow -2<m<\mathrm{4...}\left(ii\right)$
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.