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Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục trên $\mathbb{R}$ và có đồ thị như hình vẽ dưới đây:
Có bao nhiêu giá trị nguyên của tham số $m$ không vượt quá $2022$ để bất phương trình
$\dfrac{m}{f\left( x \right)}-\sqrt{mf\left( x \right)}-1\ge \dfrac{3}{4}{{f}^{2}}\left( x \right)$ đúng với mọi $x\in \left[ -2; 3 \right]$ ?
A. $1875$
B. $1872$
C. $1874$
D. $1873$
$\dfrac{m}{f\left( x \right)}-\sqrt{mf\left( x \right)}-1\ge \dfrac{3}{4}{{f}^{2}}\left( x \right)$ đúng với mọi $x\in \left[ -2; 3 \right]$ ?
A. $1875$
B. $1872$
C. $1874$
D. $1873$
Điều kiện: $mf\left( x \right)\ge 0$. Do $x\in \left[ -2; 3 \right]$ thì $f\left( x \right)\ge 0$ nên: $m\ge 0$.
Ta có: $\dfrac{m}{f\left( x \right)}-\sqrt{mf\left( x \right)}-1\ge \dfrac{3}{4}{{f}^{2}}\left( x \right)\Leftrightarrow \dfrac{m}{f\left( x \right)}-\sqrt{mf\left( x \right)}+\dfrac{{{f}^{2}}\left( x \right)}{4}\ge {{f}^{2}}\left( x \right)+1$
$\Leftrightarrow {{\left[ \sqrt{\dfrac{m}{f\left( x \right)}}-\dfrac{f\left( x \right)}{2} \right]}^{2}}\ge {{f}^{2}}\left( x \right)+1$
$\Leftrightarrow \left[ \begin{aligned}
& \sqrt{\dfrac{m}{f\left( x \right)}}-\dfrac{f\left( x \right)}{2}\ge \sqrt{{{f}^{2}}\left( x \right)+1} \\
& \sqrt{\dfrac{m}{f\left( x \right)}}-\dfrac{f\left( x \right)}{2}\le -\sqrt{{{f}^{2}}\left( x \right)+1} \\
\end{aligned} \right.$
Nên: $\sqrt{m}\ge \sqrt{\left[ {{f}^{2}}\left( x \right)+1 \right]f\left( x \right)}+\dfrac{1}{2}f\left( x \right)\sqrt{f\left( x \right)}$ $\vee $ $\sqrt{m}\le -\sqrt{\left[ {{f}^{2}}\left( x \right)+1 \right]f\left( x \right)}+\dfrac{1}{2}f\left( x \right)\sqrt{f\left( x \right)}$
$\Rightarrow \left[ \begin{aligned}
& \sqrt{m}\ge \underset{\left[ -2;3 \right]}{\mathop{\max }} \left\{ \sqrt{\left[ {{f}^{2}}\left( x \right)+1 \right]f\left( x \right)}+\dfrac{1}{2}f\left( x \right)\sqrt{f\left( x \right)} \right\} \\
& \sqrt{m}\le \underset{\left[ -2;3 \right]}{\mathop{\min }} \left\{ -\sqrt{\left[ {{f}^{2}}\left( x \right)+1 \right]f\left( x \right)}+\dfrac{1}{2}f\left( x \right)\sqrt{f\left( x \right)} \right\} \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& \sqrt{m}\ge 4+2\sqrt{17} \\
& \sqrt{m}\le 4-2\sqrt{17} \\
\end{aligned} \right.$
Nên: $m\ge {{\left( 4+2\sqrt{17} \right)}^{2}}\approx 149,96$. Kết hợp với $m\in \mathbb{Z}$ thì có $1873$ giá trị $m$ thỏa mãn.
Ta có: $\dfrac{m}{f\left( x \right)}-\sqrt{mf\left( x \right)}-1\ge \dfrac{3}{4}{{f}^{2}}\left( x \right)\Leftrightarrow \dfrac{m}{f\left( x \right)}-\sqrt{mf\left( x \right)}+\dfrac{{{f}^{2}}\left( x \right)}{4}\ge {{f}^{2}}\left( x \right)+1$
$\Leftrightarrow {{\left[ \sqrt{\dfrac{m}{f\left( x \right)}}-\dfrac{f\left( x \right)}{2} \right]}^{2}}\ge {{f}^{2}}\left( x \right)+1$
$\Leftrightarrow \left[ \begin{aligned}
& \sqrt{\dfrac{m}{f\left( x \right)}}-\dfrac{f\left( x \right)}{2}\ge \sqrt{{{f}^{2}}\left( x \right)+1} \\
& \sqrt{\dfrac{m}{f\left( x \right)}}-\dfrac{f\left( x \right)}{2}\le -\sqrt{{{f}^{2}}\left( x \right)+1} \\
\end{aligned} \right.$
Nên: $\sqrt{m}\ge \sqrt{\left[ {{f}^{2}}\left( x \right)+1 \right]f\left( x \right)}+\dfrac{1}{2}f\left( x \right)\sqrt{f\left( x \right)}$ $\vee $ $\sqrt{m}\le -\sqrt{\left[ {{f}^{2}}\left( x \right)+1 \right]f\left( x \right)}+\dfrac{1}{2}f\left( x \right)\sqrt{f\left( x \right)}$
$\Rightarrow \left[ \begin{aligned}
& \sqrt{m}\ge \underset{\left[ -2;3 \right]}{\mathop{\max }} \left\{ \sqrt{\left[ {{f}^{2}}\left( x \right)+1 \right]f\left( x \right)}+\dfrac{1}{2}f\left( x \right)\sqrt{f\left( x \right)} \right\} \\
& \sqrt{m}\le \underset{\left[ -2;3 \right]}{\mathop{\min }} \left\{ -\sqrt{\left[ {{f}^{2}}\left( x \right)+1 \right]f\left( x \right)}+\dfrac{1}{2}f\left( x \right)\sqrt{f\left( x \right)} \right\} \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& \sqrt{m}\ge 4+2\sqrt{17} \\
& \sqrt{m}\le 4-2\sqrt{17} \\
\end{aligned} \right.$
Nên: $m\ge {{\left( 4+2\sqrt{17} \right)}^{2}}\approx 149,96$. Kết hợp với $m\in \mathbb{Z}$ thì có $1873$ giá trị $m$ thỏa mãn.
Đáp án D.