Đề bài: Cho hàm số : $y = \frac{{x^2\cos \alpha – 2x + \cos\alpha }}{{x^2 – 2x\cos\alpha + 1}},\alpha \in (0,\pi )$Tìm miền giá trị của hàm số $y$
Lời giải
Giải
Ta có:
$y^2 – 1 = (\frac{{{x^2}c{\rm{os}}\alpha – 2{\rm{x}} + c{\rm{os}}\alpha }}{{{x^2}
– 2{\rm{x}}c{\rm{os}}\alpha + 1}})^2 – 1$
$ = \frac{{{{\left( {{x^2}c{\rm{os}}\alpha – 2{\rm{x}} +
c{\rm{os}}\alpha } \right)}^2} – {{\left( {{x^2} – 2{\rm{x}}c{\rm{os}}\alpha + 1}
\right)}^2}}}{{{{\left( {{x^2} – 2{\rm{x}}c{\rm{os}}\alpha + 1} \right)}^2}}}$
$ = \frac{{ – \left( {{{\sin }^2}\alpha } \right){x^4} – {{\sin
}^2}\alpha + 2{{\rm{x}}^2}\left( {1 – c{\rm{o}}{{\rm{s}}^2}\alpha } \right)}}{{{{\left( {{x^2}
– 2{\rm{x}}c{\rm{os}}\alpha + 1} \right)}^2}}}$
$ = \frac{{ – \left( {{{\sin }^2}\alpha } \right){x^4} – {{\sin
}^2}\alpha+ 2{{\rm{x}}^2}{{\sin }^2}\alpha }}{{{{\left( {{x^2} – 2{\rm{x}}c{\rm{os}}\alpha +
1} \right)}^2}}}$
$ = \frac{{{{\sin }^2}\alpha \left( {{x^4} – 2{{\rm{x}}^2} + 1}
\right)}}{{{{\left( {{x^2} – 2{\rm{x}}c{\rm{os}}\alpha + 1} \right)}^2}}} = \frac{{ – {{\sin
}^2}\alpha {{\left( {{x^2} – 1} \right)}^2}}}{{{{\left( {{x^2} – 2{\rm{x}}c{\rm{os}}\alpha + 1}
\right)}^2}}} \le 0$
$\Leftrightarrow y^2 \leq 1 \Leftrightarrow |y| \leq 1 \Leftrightarrow -1 \leq y \leq 1$
$\Rightarrow $ Miền giá trị : $T = [-1, 1].$
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