Có bao nhiêu số nguyên dương a (a là tham số) để phương trình

* Hướng dẫn giải

Đáp án B

Điều kiện: $\left\{\begin{array}{l}2x-x^2>0\\ 2-x^2>0\end{array}\right.\iff 0<x<\sqrt{2}\Rightarrow D=\left(0;\sqrt{2}\right)$

Phương trình

$\iff \left(a^2+4a+5\right){\log }_3\left(2x-x^2\right)+\left(9a^2-6a+2\right){\log }_{11}\left(1-\frac{x^2}{2}\right)$

$={\log }_3\left(2x-x^2\right)+{\log }_{11}\left(1-\frac{x^2}{2}\right)$

$\iff f\left(x\right)={\left(a+2\right)}^2{\log }_3\left(2x-x^2\right)+{\left(3a-1\right)}^2{\log }_{11}\left(1-\frac{x^2}{2}\right)=0$

$\iff f\left(x\right)=\left(a^2+4a+4\right){\log }_3\left(2x-x^2\right)+\left(9a^2-6a+2\right){\log }_{11}\left(1-\frac{x^2}{2}\right)$

$={\log }_3\left(2x-x^2\right)+{\log }_{11}\left(1-\frac{x^2}{2}\right)=0\left(x\in \left(0;\sqrt{2}\right)\right)$

$\iff f\left(x\right)={\left(a+2\right)}^2{\log }_3\left(2x-x^2\right)+{\left(3a-1\right)}^2{\log }_{11}\left(1-\frac{x^2}{2}\right)=0$

Ta có:

$f\text{'}\left(x\right)={\left(a+2\right)}^2.\frac{2-2x}{\left(2x-x^2\right)\ln 3}+{\left(3a-1\right)}^2.\frac{1-x}{\left(1=\frac{x^2}{2}\right)\ln 11}=0$

$\iff x=1$

Ta có:

$\underset{x\to 0}{\lim }f\left(x\right)=-\infty ;f\left(1\right)=-{\left(3a-1\right)}^2{\log }_{11}2;\underset{x\to \sqrt{2}}{\lim }f\left(x\right)=-\infty \Rightarrow$phương trình đã cho có nghiệm duy nhất khi $-{\left(3a-1\right)}^2{\log }_{11}2=0\iff a=\frac{1}{3}\notin \mathbb{Z}$