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The least integral value of m for which every solution of inequality $1 \leq x\leq 2$ is solution of the inequality $x^2-mx+1 < 0$

$\begin{array}{1 1}(A)\;0&(B)\;1\\(C)\;2&(D)\;3\end{array} $

Can you answer this question?
 
 

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$m > 0$
$x^2-mx+1 < 0$
$\Rightarrow (x-\large\frac{m}{2})^2 < \large\frac{m^2}{4}$$-1$
Thus $m^2-4 > 0$
Or $m >2$
$\large\frac{m}{2}-\sqrt{\large\frac{m^2}{4}-1} < x < \large\frac{m}{2}+\sqrt{\large\frac{m^2}{4}-1}$
We should have
$\large\frac{m}{2}-\sqrt{\large\frac{m^2-4}{4}} < 1$ & $2 < \large\frac{m}{2}+\sqrt{\large\frac{m^2-4}{4}}$
$\Rightarrow m-2< \sqrt{m^2-4}$ & $4-m < \sqrt{m^2-4}$
$\Rightarrow m^2-4m+4 < m^2-4$ & $16-8m+m^2 < m^2-4$
$\Rightarrow m > 2$ and $m > \large\frac{5}{2}$
$\Rightarrow$ Least integral value of m is 3
Hence (D) is the correct answer.
answered Apr 16, 2014 by sreemathi.v
 

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