Cho hàm số f(x)=2^x-2^-x Số giá trị nguyên

* Hướng dẫn giải

Đáp án C

Có $f\left(-x\right)=2^{-x}-2^x=-\left(2^x-2^{-x}\right)=-f\left(x\right)$

$f^{\text{'}}\left(x\right)=2^x\ln 2+2^{-x}\ln 2>0,\forall x\Rightarrow f\left(x\right)$ là hàm đồng biến trên $\mathrm{ℝ}$

Do đó

$f\left(\left|x^3-2x^2+3x-m\right|\right)+f\left(2x-2x^2-5\right)<0,\forall x\in \left(0;1\right)$

$\begin{array}{l}\iff f\left(\left|x^3-2x^2+3x-m\right|\right)<-f\left(2x-2x^2-5\right)=f\left(2x^2-2x+5\right),\forall x\in \left(0;1\right)\\ \iff \left|x^3-2x^2+3x-m\right|<2x^2-2x+5,\forall x\in \left(0;1\right)\\ \iff -\left(2x^2-2x+5\right)<x^3-2x^2+3x-m<2x^2-2x+5,\forall x\in \left(0;1\right)\\ \iff \left\{\begin{array}{l}m>x^3-4x^2+5x-5,\forall x\in \left(0;1\right)\\ m<x^3+x+5,\forall x\in \left(0;1\right)\end{array}\right.\end{array}$

• Xét $g\left(x\right)=x^3-4x^2+5x-5,\forall x\in \left(0;1\right)$
• $g^{\text{'}}\left(x\right)=3x^2-8x+5;g^{\text{'}}\left(x\right)=0\iff \left[\begin{array}{l}x=1\\ x=\frac{5}{3}\end{array}\right.$

• Xét $h\left(x\right)=x^3+x+5,\forall x\in \left(0;1\right)$
• $h^{\text{'}}\left(x\right)=3x^2+1>0,\forall x\in \left(0;1\right)$

Vậy $-3\le m\le 5.$