Question

All the values of m for which both roots of the equation $x^2-2mx+m^2-1=0$ are greater than -2 but 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)

$-1$ < $m$ < $3$

$x^2-2mx+m^2-1=0$

$(x-m{)}^2-1=0$ $(a^2-2ab+b^2=(a-b{)}^2)$

$(x-m-1)(x-m+1)=0$ $(a^2-b^2=(a+b\left)\right(a-b\left)\right)$

Thus ,

$x-m-1=0$ OR $x-m+1=0$

$x=m+1$ OR $x=m-1$

But,

$m+1>-2$ and $m-1<4$

$m>-3$ or $m<5$

$m+1<4$ or $m-1>-2$

$m<3$ or $m>-1$

Thus Thus

$(-2,-1,0,1,2)$ $(0,1,2,3,4)$

Taking common

Thus, $-1<m<3$