Question

All the values of $m$ for which both roots of the equation $x^2-2mx+m^2-1=0$ are $>-2$ but $<4$ lie in the interval

• A. $m>3$
• B. $–1<m<3$
• C. $1<m<4$
• D. $–2<m<0$

Answer 1

The correct option is B

$–1<m<3$

Explanation for the correct option:

Step 1. Since both roots of equation $x^2-2mx+m^2-1=0$are $>-2$ but $<4$.

$\therefore D\ge 0$

$\Rightarrow 4m^2-4m^2+4\ge 0;m\in R$ …..(i)

Now, $-2<\frac{-b}{2a}<4$

$\Rightarrow -2<\left(\frac{2m}{2\times 1}\right)<4$

$\Rightarrow -2<m<4$ .....(ii)

Step 2. Find the interval of $m$ for $f\left(4\right)>0$

$$\begin{gathered} \Rightarrow 16-8m+m^2-1>0 \\ \Rightarrow m^2-8m+15>0 \\ \Rightarrow (m-3)(m-5)>0 \end{gathered}$$

$\Rightarrow -\infty <m<3and5<m<\infty$ ....(iii)

Step 3. Find the interval of $m$ for$f(-2)>0$

$$\begin{gathered} \Rightarrow 4+4m+m^2-1>0 \\ \Rightarrow m^2+4m+3>0 \\ \Rightarrow (m+3)(m+1)>0 \end{gathered}$$
$\Rightarrow -\infty <m<-3and-1<m<\infty$ .....(iv)

$\therefore$ From (i),(ii),(iii) and (iv), we get

Thus, $m$ lies between $\mathbf{-}\mathbf{1}$ and $\mathbf{3}$

Hence, Option ‘B’ is Correct.