Question

Tìm m để phương trình sau có nghiệm:

\(x^2+\sqrt{4-x^2}=m\)

Answer 1

ĐKXĐ: \(x\in\left[-2;2\right]\)

Đặt \(f\left(x\right)=x^2+\sqrt{4-x^2}\)

\(f'\left(x\right)=2x-\dfrac{x}{\sqrt{4-x^2}}=0\Rightarrow\left[{}\begin{matrix}x=0\\\sqrt{4-x^2}=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow x=\left\{-\dfrac{\sqrt{15}}{2};0;\dfrac{\sqrt{15}}{2}\right\}\)

\(f\left(-2\right)=f\left(2\right)=4\) ; \(f\left(-\dfrac{\sqrt{15}}{2}\right)=f\left(\dfrac{\sqrt{15}}{2}\right)=\dfrac{17}{4}\) ; \(f\left(0\right)=2\)

\(\Rightarrow2\le f\left(x\right)\le\dfrac{17}{4}\)

\(\Rightarrow\) Pt có nghiệm khi \(2\le m\le\dfrac{17}{4}\)