Question

If the values of $m$ for which exactly one root of the equation $x^2-2mx+m^2-1=0$ lies in the interval $(-2,4)$ belong to set $A$, then the number of integral points common to set $A$ and $R$ are

Answer 1

let f(x)= $x^2-2mx+m^2-1$ as exactly one root of f(x)=0 lies in the interval, we can take D>0 and f(-2) f(4)<0
Consider D>0: ${(-2m)}^2-4.1(m^2-1)>0$
$\Rightarrow 4>0\therefore mϵR$ ....................(1)
consider f(-2) f(4) <0:
$(4+4m+m^2-1)(16-8m+m^2-1)<0$
$\Rightarrow (m^2+4m+3)(m^2-8m+15)<0$
$\Rightarrow (m+1)(m+3)(m-3)(m-5)<0$
$\Rightarrow (m+3)(m+1)(m-3)(m-5)<$
$\therefore mϵ(-3,-1)\cup (3,5)$ ..........(2)
Hence, the values of m satisfying (1) and (2) at the same time are $\therefore mϵ(-3,-1)\cup (3,5)$.