Question

Values of ${}^{\prime }{m}^{\prime }$ for which both the roots of equation $x^2-2mx+m^2-1=0$ are less than $4$ are

• A. $(-\infty ,-3)$
• B. $[-3,\infty )$
• C. $(3,\infty )$
• D. $(-\infty ,-3]$

Answer 1

The correct option is A

$(-\infty ,-3)$

Conditions for both the roots to be less than a real value $x_0$, can be visualized through graph

Here, $a=1(>0),b=-2m,c=m^2-1$

Observe that
$\left(i\right)f\left(x_0\right)>0$

$\left(ii\right)x_0>\frac{-b}{2a}$

Also for real roots to exist $\left(iii\right)b^2\ge 4ac$

$i\left)f\right(4)>0\Rightarrow 16-8m+m^2-1>0$

$\Rightarrow m^2-8m+15>0$

$\Rightarrow (m-3)(m-5)>0$

$m<3$ or $m>5$ . . . $\left(1\right)$

$ii)4>m\Rightarrow m<4$ . . . $\left(2\right)$

$iii)D\ge 0$

$\Rightarrow b^2-4ac\ge 0$

$\Rightarrow 4m^2-4(m^2-1)\ge 0$

$\Rightarrow 4\ge 0$ (Always true) . . . $\left(3\right)$

Intersection of $i),ii)$ & $iii)$ gives,

$m\in (-\infty ,3)$