A) \[-3\le m\le 4\]
B) \[-3\le m\le 5\]
C) \[-1\le m\le 5\]
D) \[-1\le m\le 3\]
Correct Answer: C
Solution :
\[{{x}^{2}}-2mx+{{m}^{2}}-1=0\] or \[x=\frac{2m\pm \sqrt{4{{m}^{2}}-4({{m}^{2}}-1)}}{2}\] \[=\frac{2m\pm 2}{2}\] \[=m\pm 1\] Since, the roots lie between \[-2\] and\[4\], therefore \[-2\le m\pm 1\le 4\] or \[-3\le m\le 3\] and \[-1\le m\le 5\]
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