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

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 $$4$$ .Maths

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