Topic
Find all values of \(m\) so that the function \(y = \frac{1}{{\sqrt {{x^2} – 4x + 6m – 1} }}\) has a defined set of \ (\mathbb{R}\).
Solution method – See details
\(\frac{1}{{\sqrt {f(x)} }}\) defines when \(f(x) > 0\)
Trim \(f\left( x \right) = a{x^2} + bx + c > 0\;\forall x \in \mathbb{R} \Leftrightarrow \left\{ \begin{array}{l }a > 0\\\Delta < 0\end{array} \right.\)
Detailed explanation
The function \(y = \frac{1}{{\sqrt {{x^2} – 4x + 6m – 1} }}\) determines when \({x^2} – 4x + 6m – 1 > 0 \)
Therefore, the function has a defined set of \(\mathbb{R}\)\( \Leftrightarrow {x^2} – 4x + 6m – 1 > 0\)\(\forall x \in \mathbb{R} \)
\( \Leftrightarrow \left\{ \begin{array}{l}a > 0\\\Delta < 0\end{array} \right.\)
Where \(a = 1 > 0,\Delta = {\left( { – 4} \right)^2} – 4.1.\left( {6m – 1} \right) = – 24m + 20\)
Hence \(
\Leftrightarrow – 24m + 20 < 0 \Leftrightarrow m > \frac{5}{6}\)So \(m > \frac{5}{6}\)
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