• # question_answer 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     AIEEE  Solved  Paper-2006 A) $m>3$                                B) $-1<m<3$         C)        $1<m<4$           D) $-2<m<0$

Method I Given equation is${{x}^{2}}-2mx+{{m}^{2}}-1$. Since, both roots are real and greater than -2 and less than 4. i.e.,        $-2<x<4$ $\Rightarrow$ $-2<\frac{-b\pm \sqrt{D}}{2a}<4$$\Rightarrow$ $-2<m\pm 1<4$ [$\because$$D=4{{m}^{2}}-4{{m}^{2}}+4$] $\Rightarrow$ $-2<m\pm 1<4$ Either    $-3<m<3$                        ...(i) and        $-1<m<5$                       ...(ii) The common set of values of 'm' satisfying the condition given is $-1<\text{ }m<\text{ }3$ Method II 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$$4{{m}^{2}}-4{{m}^{2}}+4\ge 0\Rightarrow m\in R$                  ...(i) and $-2<\frac{-b}{2a}<4\Rightarrow -2<\left( \frac{2m}{2.1} \right)<4$ $\Rightarrow$$-2<m<4$                                                     ...(ii) Also, $f(4)>0\Rightarrow 16-8m+{{m}^{2}}-1>0$ $\Rightarrow$ ${{m}^{2}}-8m+15>0$ $\Rightarrow$ $(m-3)(m-5)>0$ $\Rightarrow$ $-\infty <m<3$ and            $5<m<\infty$                                                             ...(iii) Also,$f(-2)>0\Rightarrow 4+4m+{{m}^{2}}-1>0$ $\Rightarrow$${{m}^{2}}+4m+3>0$ $\Rightarrow$$(m+3)(m+1)>0$ $\Rightarrow$ $-\infty <m<-3$and$-1<m<\infty$    ...(iv) From Eqs. (i), (ii), (iii) and (iv), we get m lies between -1 and 3