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Question

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


A
1<m<3
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B
1<m<4
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C
2<m<0
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D
m>3
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Solution

The correct option is B $$-1 < m < 3$$
For both roots to exist. $$D>0$$

$$b^{2}-4ac>0  \Rightarrow 4m^{2}-4(m^{2}-1)>0$$

$$=4>0$$

$$f(-2)>0$$  and $$f(4)>0$$

$$\Rightarrow 4+4m+m^{2}-1>0    \ \ \ and \ \    16-8m+m^{2}-1>0$$

$$\Rightarrow m^{2}+4m+3>0     \ \ \ \ \ \  and\ \ \ \ \ \         m^{2}-8m+15>0$$ 

$$\Rightarrow (m+3)(m+1)>0    \ \ \ \ \ \ \   (m-5)(m-3)>0$$

$$\Rightarrow m\in (-\infty ,-3)\bigcup (-1,\infty ) \ \  and\     m\in (-\infty ,3)\bigcup (5,\infty )$$

taking common of above intervals 

$$m\in (-1,3)$$ 

$$\therefore$$ $$m$$ should belong to $$(-1, 3)$$ for roots to be between $$-2$$ and $$4$$ .

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