## Solve Lesson 26 Page 52 SBT Math 10 – Kite>

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 $$m > \frac{5}{6}$$