If the roots of $x^{2} - 2 m x + m^{2} - 1 = 0$ lie between $- 2$ and $4$ , then

- 3 \leq m \leq 3

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3 \leq m \leq 5

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- 1 \leq m \leq 5

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- 1 \leq m \leq 3

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Solution

The correct option is D $- 1 \leq m \leq 3$
$x^{3} - 2 m x + m^{2} - 1 = 0 \Rightarrow x = \frac{2 m \pm \sqrt{4 m^{2} - 4 (m^{2} - 1)}}{2} = \frac{2 m + 2}{2} \Rightarrow x = m \pm 1$
Now $- 2 < x < 4 \Rightarrow - 2 < m \pm 1 < 4$
$\Rightarrow - 3 < m < 3$ and $- 1 < m < 5 \Rightarrow - 1 < m < 3$

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