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Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục và xác định trên $\mathbb{R}$ và có đồ thị như hình vẽ. Có bao nhiêu giá trị nguyên của m để bất phương trình ${{3.12}^{f\left( x \right)}}+\left[ {{f}^{2}}\left( x \right)-1 \right]{{.16}^{f\left( x \right)}}\ge \left( {{m}^{2}}+3m \right){{.3}^{2f\left( x \right)}}$ có nghiệm với mọi x?
A. 5.
B. 7.
C. Vô số.
D. 6.
A. 5.
B. 7.
C. Vô số.
D. 6.
Ta có: $\begin{aligned}
& \ \ \ \ \ {{3.12}^{f\left( x \right)}}+\left[ {{f}^{2}}\left( x \right)-1 \right]{{.16}^{f\left( x \right)}}\ge \left( {{m}^{2}}+3m \right){{.3}^{2f\left( x \right)}},\forall x\in \mathbb{R} \\
& \Leftrightarrow \left[ {{f}^{2}}\left( x \right)-1 \right]{{\left( \dfrac{16}{9} \right)}^{f\left( x \right)}}+3.{{\left( \dfrac{4}{3} \right)}^{f\left( x \right)}}\ge {{m}^{2}}+3m,\forall x\in \mathbb{R}\ \ \ \left( 1 \right) \\
\end{aligned}$
Mà $f\left( x \right)\ge 1,\forall x\in \mathbb{R}$ nên $\left\{ \begin{aligned}
& \left[ {{f}^{2}}\left( x \right)-1 \right]{{\left( \dfrac{16}{9} \right)}^{f\left( x \right)}}\ge 0 \\
& 3.{{\left( \dfrac{4}{3} \right)}^{f\left( x \right)}}\ge 4 \\
\end{aligned} \right.,\forall x\in \mathbb{R}$.
Đặt $h\left( x \right)=\left[ {{f}^{2}}\left( x \right)-1 \right]{{\left( \dfrac{16}{9} \right)}^{f\left( x \right)}}+3.{{\left( \dfrac{4}{3} \right)}^{f\left( x \right)}}$.
Mà $h\left( x \right)\ge 4,\forall x\in \mathbb{R}$.
Suy ra $\underset{\mathbb{R}}{\mathop{\min }} h\left( x \right)=4\Leftrightarrow x=2$.
Khi đó ${{m}^{2}}+3m\le h\left( x \right),\forall x\in \mathbb{R}\Leftrightarrow {{m}^{2}}+3m\le \underset{\mathbb{R}}{\mathop{\min }} h\left( x \right)\Leftrightarrow {{m}^{2}}+3m\le 4\Leftrightarrow -4\le m\le 1$.
Vì $m\in \mathbb{Z}$ nên $m\in \left\{ -4;-3;-2;-1;0;1 \right\}$.
& \ \ \ \ \ {{3.12}^{f\left( x \right)}}+\left[ {{f}^{2}}\left( x \right)-1 \right]{{.16}^{f\left( x \right)}}\ge \left( {{m}^{2}}+3m \right){{.3}^{2f\left( x \right)}},\forall x\in \mathbb{R} \\
& \Leftrightarrow \left[ {{f}^{2}}\left( x \right)-1 \right]{{\left( \dfrac{16}{9} \right)}^{f\left( x \right)}}+3.{{\left( \dfrac{4}{3} \right)}^{f\left( x \right)}}\ge {{m}^{2}}+3m,\forall x\in \mathbb{R}\ \ \ \left( 1 \right) \\
\end{aligned}$
Mà $f\left( x \right)\ge 1,\forall x\in \mathbb{R}$ nên $\left\{ \begin{aligned}
& \left[ {{f}^{2}}\left( x \right)-1 \right]{{\left( \dfrac{16}{9} \right)}^{f\left( x \right)}}\ge 0 \\
& 3.{{\left( \dfrac{4}{3} \right)}^{f\left( x \right)}}\ge 4 \\
\end{aligned} \right.,\forall x\in \mathbb{R}$.
Đặt $h\left( x \right)=\left[ {{f}^{2}}\left( x \right)-1 \right]{{\left( \dfrac{16}{9} \right)}^{f\left( x \right)}}+3.{{\left( \dfrac{4}{3} \right)}^{f\left( x \right)}}$.
Mà $h\left( x \right)\ge 4,\forall x\in \mathbb{R}$.
Suy ra $\underset{\mathbb{R}}{\mathop{\min }} h\left( x \right)=4\Leftrightarrow x=2$.
Khi đó ${{m}^{2}}+3m\le h\left( x \right),\forall x\in \mathbb{R}\Leftrightarrow {{m}^{2}}+3m\le \underset{\mathbb{R}}{\mathop{\min }} h\left( x \right)\Leftrightarrow {{m}^{2}}+3m\le 4\Leftrightarrow -4\le m\le 1$.
Vì $m\in \mathbb{Z}$ nên $m\in \left\{ -4;-3;-2;-1;0;1 \right\}$.
Đáp án D.