b. $\displaystyle a=b=0\nRightarrow \frac{b}{a} =-1$，故而错误

c. $\displaystyle 0^{2} =0$，故而错误

d. $\displaystyle a,b >1\Rightarrow ab >1$，为充分条件，正确

$$\begin{equation} \frac{y}{x} >\frac{y+z}{x+z} \quad \Leftrightarrow \quad y( x+z) >x( y+z) \quad \Leftrightarrow z( y-x) < 0 \end{equation}$$

矛盾，故而 $\displaystyle A$ 错误

$\displaystyle f( x) =x-\frac{1}{x}$ 为单调递增函数，故而 $\displaystyle B$ 正确

周期函数不可能严格单调，故而 $\displaystyle C,D$ 错误

同时，$\displaystyle f( x)$ 为幂函数 $\displaystyle \Rightarrow m^{2} -7m+13=1\Rightarrow ( m-3)( m-4) =0\Rightarrow m=3\ \text{或} \ 4$

显然 $\displaystyle m >2\ \nRightarrow m=3\ \text{或} \ 4$，$\displaystyle m=3\ \text{或} \ 4\ \Rightarrow m >2$，故而为必要不充分条件

$$\begin{gather} a=\log_{0.2} 5< \log_{0.2} 1=0\\ \notag\\ 0< b=0.2^{3} < 0.2^{0} =1\\ \notag\\ c=\left(\frac{1}{4}\right)^{-0.2} >\left(\frac{1}{4}\right)^{0} =1 \end{gather}$$

$\displaystyle \Rightarrow a< b< c$

$$\begin{equation} f\left( 3^{-0.1}\right) =f( 1-r) =f( 3+r) \end{equation}$$

这里 $\displaystyle r=1-3^{-0.1} \in ( 0,1)$

而 $\displaystyle \frac{f( x_{1}) -f( x_{2})}{x_{1} -x_{2}} >0,\ \forall x_{1} \neq x_{2} \in [ 2,+\infty ) \Leftrightarrow f( x)$ 在 $\displaystyle [ 2,+\infty )$ 上 $\displaystyle \nearrow$，因此：

$$\begin{gather} \log_{5} 64< 2< 3+r< 6^{\sqrt{2}}\\ \notag\\ \end{gather}$$

$\Longrightarrow a< b< c$

定义域 $\displaystyle \Rightarrow 8-4^{x+a} >0\Rightarrow 8-4^{2+a} >0\Rightarrow a< -\frac{1}{2}$

同时显然 $\displaystyle \min f( x) =a+1$，对数函数单调 $\displaystyle \Rightarrow$原条件等价于：

$$\begin{equation} \max g( x) -3< a+1\Rightarrow \log_{2}\left( 8-4^{a}\right) < a+4 \end{equation}$$

令 $\displaystyle t=8-4^{a}$，则 $\displaystyle a=\log_{4}( 8-t) \Rightarrow$：

$$\begin{gather} \log_{2} t< \log_{4}( 8-t) +4\Rightarrow \log_{4}\left(\frac{t^{2}}{8-t}\right) < 4\\ \notag\\ \Longrightarrow \frac{t^{2}}{8-t} < 256\Rightarrow t< \ -128+96\sqrt{2}\\ \notag\\ \Longrightarrow -\frac{1}{2}>a >1+\log_{2}\left( 3\sqrt{2} -4\right) \end{gather}$$

$$\begin{gather} x^{2} +bx+c=( x-3)( x-4) \Rightarrow \begin{cases} b=-7 & \\ c=12 & \end{cases}\\ \notag\\ \Rightarrow cx^{2} +bx+1=12x^{2} -7x+1=( 3x-1)( 4x-1) \end{gather}$$

$\displaystyle \Rightarrow cx^{2} +bx+1 >0$ 解集为 $\displaystyle \left( -\infty ,\frac{1}{4}\right)\bigcup \left(\frac{1}{3} ,+\infty \right)$

另一方面，在 $\displaystyle x^{2} +bx+c< 0$ 两边除以 $\displaystyle x^{2}$ 可得：

$$\begin{equation} c\left(\frac{1}{x}\right)^{2} +b\left(\frac{1}{x}\right) +1< 0 \end{equation}$$

故而不等式 $\displaystyle cx^{2} +bx+1< 0$ 的解集为 $\displaystyle \left(\frac{1}{4} ,\ \frac{1}{3}\right)$

$\displaystyle \Rightarrow cx^{2} +bx+1 >0$的解集为 $\displaystyle \left( -\infty ,\frac{1}{4}\right)\bigcup \left(\frac{1}{3} ,+\infty \right)$

当 $\displaystyle a^{2} \neq 4$ 时，不等式左侧为二次函数，解集为空 $\displaystyle \Rightarrow \ :$

$$\begin{gather} a^{2} -4< 0\ \text{且} \ \Delta < 0\\ \notag\\ \Rightarrow \begin{cases} a^{2} -4< 0 & \\ ( a+2)^{2} +4\left( a^{2} -4\right) < 0 & \end{cases} \Longrightarrow \begin{cases} ( a+2)( a-2) < 0 & \\ ( a+2)( 5a-6) < 0 & \end{cases} \end{gather}$$

故而 $\displaystyle a$ 的取值范围为 $\displaystyle \left[ -2,\frac{6}{5}\right)$

$$\begin{equation} \sin\left( \alpha +\frac{\pi }{6}\right) =\frac{\sqrt{6}}{3} \quad \Rightarrow \quad \tan\left( \alpha +\frac{\pi }{6}\right) =\sqrt{2} \end{equation}$$

$\displaystyle \Rightarrow \tan\left(\frac{\pi }{2}\right) =\tan\left( \alpha +\frac{\pi }{6} +\left(\frac{\pi }{3} -\alpha \right)\right) =\infty$

$$\begin{equation} \Rightarrow 1-\tan\left( \alpha +\frac{\pi }{6}\right)\tan\left(\frac{\pi }{3} -\alpha \right) =0 \end{equation}$$

$\displaystyle \Rightarrow \tan\left(\frac{\pi }{3} -\alpha \right) =\frac{\sqrt{2}}{2}$

$\displaystyle f([ 0,\pi ]) =\left[ 3,2\sqrt{3}\right] \Rightarrow f( 0) =1+a\in \left[ 3,2\sqrt{3}\right] \Rightarrow a\geq 2$，同时：

$$\begin{gather} \begin{aligned} f( x) & =2\sin\left( \omega x+\frac{\pi }{6}\right) +a\cos \omega x\\ & =\sqrt{3}\sin \omega x+( a+1)\cos \omega x \end{aligned}\\ \notag\\ \Longrightarrow \max f( x) =\sqrt{3+( a+1)^{2}} \leq 2\sqrt{3} \quad \Rightarrow a\leq 2 \end{gather}$$

$\displaystyle \Rightarrow a=2,\ f( x) =\sqrt{3}\sin \omega x+3\cos \omega x=2\sqrt{3}\sin\left( \omega x+\frac{\pi }{3}\right)$

随后 $\displaystyle \omega >0,\ f([ 0,\pi ]) =\left[ 3,2\sqrt{3}\right] \Rightarrow \ :$

$$\begin{gather} \left[\frac{\pi }{3} ,\frac{\pi }{2}\right]\left\{\omega x+\frac{\pi }{3}\Bigl| x\in [ 0,\pi ]\right\} \subseteq \left[\frac{\pi }{3} ,\frac{2\pi }{3}\right]\\ \notag\\ \Longrightarrow \omega \in \left[\frac{1}{6} ,\frac{1}{3}\right] \end{gather}$$

$$\begin{equation} \begin{aligned} f( x) & =\sin \omega x\cos \varphi +\cos \omega x\sin \varphi \\ & =\sin( \omega x+\varphi ) \end{aligned} \end{equation}$$

$\displaystyle \Rightarrow \max h( x) =\max(\sin( \omega x+\varphi ) -\omega \cos( \omega x+\varphi )) =\sqrt{1+\omega ^{2}} =\frac{2\sqrt{3}}{3}$

$\displaystyle \Rightarrow \omega =\frac{1}{\sqrt{3}}$，故而 A 错误

$\displaystyle f( 0) =\sin \varphi < 0\Rightarrow \varphi \in \left( -\frac{\pi }{2} ,0\right) ,\ g( 0) =\frac{1}{\sqrt{3}}\cos \varphi =\frac{1}{2\sqrt{3}} \Rightarrow \varphi =-\frac{\pi }{3} \Rightarrow$B 正确

由此可以得出：

$$\begin{equation} \begin{aligned} h( x) & =\sin( \omega x+\varphi ) -\omega \cos( \omega x+\varphi )\\ & =\sin\left(\frac{1}{\sqrt{3}} x-\frac{\pi }{3}\right) -\frac{1}{\sqrt{3}}\cos\left(\frac{1}{\sqrt{3}} x-\frac{\pi }{3}\right)\\ & =\frac{2}{\sqrt{3}}\sin\left(\frac{1}{\sqrt{3}} x-\frac{\pi }{2}\right) =-\frac{2}{\sqrt{3}}\cos\left(\frac{1}{\sqrt{3}} x\right) \end{aligned} \end{equation}$$

故而 $\displaystyle h( x) < 0,\ x\in \left( 0,\frac{\pi }{2}\right) \Rightarrow$C 正确，$\displaystyle h\left(\frac{\sqrt{3}}{2} \pi \right) =0\Rightarrow h( x)$ 关于点 $\displaystyle \left(\frac{\sqrt{3}}{2} \pi ,0\right)$ 对称，D 正确

$$\begin{equation} \begin{aligned} f( x) & =\frac{\sqrt{3}}{2}\sin \omega x+\cos^{2}\frac{\omega x}{2} -\frac{1}{2}\\ & =\frac{\sqrt{3}}{2}\sin \omega x+\frac{\cos \omega x+1}{2} -\frac{1}{2}\\ & =\sin\left( \omega x+\frac{\pi }{6}\right) \end{aligned} \end{equation}$$

显然 $\displaystyle A$ 错误，$\displaystyle \sin\left( 2x+\frac{\pi }{6}\right) \neq \sin\left( 2\left( x-\frac{\pi }{12}\right)\right) \Rightarrow B$ 错误

$\displaystyle x=\frac{\pi }{9}$ 为对称轴，$\displaystyle \omega \in ( 0,5)$ $\displaystyle \Rightarrow \ :$

$$\begin{gather} \frac{\pi \omega }{9} +\frac{\pi }{6} =\left(\frac{1}{2} +k\right) \pi ,\ \omega \in ( 0,5) \Rightarrow \omega =3+9k,\ \omega \in ( 0,5) \Rightarrow \omega =3\notag\\ \notag\\ \Longrightarrow \left\{\omega x+\frac{\pi }{6}\Bigl| x\in \left(\frac{4\pi }{9} ,\frac{7\pi }{9}\right)\right\} =\left(\frac{3\pi }{2} ,\frac{5\pi }{2}\right) \end{gather}$$

$\displaystyle \Rightarrow C$正确

区间 $\displaystyle ( 0,2\pi )$ 上没有零点 $\displaystyle \Rightarrow 2\pi \omega +\frac{\pi }{6} \leq \pi \Rightarrow \omega \in \left( 0,\frac{5}{12}\right] \Rightarrow D$ 正确

$\displaystyle f( x)$ 为奇函数，$\displaystyle f( x) =4^{x} +2x-1,\ x\in [ 0,1] \Rightarrow \ :$

$$\begin{gather} f( x) \ \text{在} \ [ -1,1] \ \text{上} \ \nearrow ，f([ -1,1]) =[ -5,5]\\ \notag\\ f( x) \ \text{在} \ [ 1,3] \ \text{上} \ \searrow ，f([ 1,3]) =[ -5,5] \end{gather}$$

画图可知，$\displaystyle f( x)$ 与 $\displaystyle x-1$ 有五个交点

$$\begin{gather} \begin{cases} 3\log_{3} x_{1} =-m & \\ 3\log_{3} x_{2} =m & \end{cases} \Longrightarrow \quad 3(\log_{3} x_{1} +\log_{3} x_{2}) =0\Rightarrow x_{1} x_{2} =1\\ \notag\\ x_{3} ,x_{4} \ \text{为} \ x^{2} -10x+24-m=0\ \text{的二根} \ \Rightarrow \begin{cases} x_{3} +x_{4} =10 & \\ x_{3} x_{4} =24-m & \end{cases} \end{gather}$$

$\displaystyle \Rightarrow \frac{( x_{3} -4)( x_{4} -4)}{x_{1} x_{2} x_{3}} =\frac{-m}{x_{3}}$

$\displaystyle x_{1} ,x_{2} \in ( 0,3] ,x_{3} >3\Rightarrow m\in ( 0,3) ,\ \frac{-m}{x_{3}} >-1$，同时当 $\displaystyle m\rightarrow 0$ 时，$\displaystyle x_{3}\rightarrow 5$，故而 B 正确

$$\begin{equation} P_{k} =\left( 2^{k} ,0\right) \quad \quad A_{k} =\left( 3\cdot 2^{k-1} ,2^{k}\right) \quad k=0,2,\dotsc \end{equation}$$

而图像为连接这些点的折线段，画图可知 A 正确

考虑方程 $\displaystyle f( x) =\frac{1}{2}$，显然此方程仅有三解，故而 B 错误

$\displaystyle A_{k} =\left( 3\cdot 2^{k-1} ,2^{k}\right) \Rightarrow$每个 $\displaystyle A_{k}$ 均被函数 $\displaystyle g( x) =\frac{3}{2x}$ 所经过，对图像进行分析可知 $\displaystyle f( x) \leq \frac{3}{2x}$，故而 $\displaystyle C$ 正确

区间 $\displaystyle \left[ 2^{n-1} ,2^{n}\right]$ 内的三角形，底为 $\displaystyle 2^{n-1}$，高为 $\displaystyle 2^{-( n-1)}$，易得 $\displaystyle S=\frac{1}{2}$，故而 D 正确

$\displaystyle \forall x,y\in I\ s.t.\ x >y,$ $\displaystyle \frac{x-y}{1-xy} >0\Rightarrow f( x) -f( y) =f\left(\frac{x-y}{1-xy}\right) >0\Rightarrow D$ 正确

若 $\displaystyle f( x)$ 为偶函数，则：

$$\begin{equation} f( x) -f( -x) =f\left(\frac{2x}{1-x^{2}}\right) =0 \end{equation}$$

将 $\displaystyle x=0.1$ 代入 $\displaystyle \Rightarrow f\left(\frac{0.2}{0.99}\right) =0$，矛盾，故而 $\displaystyle B$ 错误

若 $\displaystyle f( x) +f( y) =f\left(\frac{x+y}{1+xy}\right)$，则 $\displaystyle f( x) +f( -x) =0\Rightarrow \ :$

$$\begin{equation} f( x) -f( -x) =2f( x) =f\left(\frac{2x}{1-x^{2}}\right) \end{equation}$$

但另一方面：

$$\begin{equation} 2f( x) =f( x) +f( x) =f\left(\frac{2x}{1+x^{2}}\right) \end{equation}$$

矛盾，故而 $\displaystyle C$ 错误

$$\begin{equation} \log_{4} x^{2} =\frac{\log_{2} x^{2}}{\log_{2} 4} =\frac{1}{2} \cdot 2\log_{2} x^{2} =\log_{2} x \end{equation}$$

知道原条件可化为：

$$\begin{gather} \log_{2}( x+3y) =\log_{2} x+\log_{2}( 2y) \Rightarrow x+3y=2xy\\ \notag\\ \Longrightarrow \frac{3}{x} +\frac{1}{y} =2 \end{gather}$$

随后由基本不等式

$$\begin{gather} \begin{aligned} 2( 3x+y) & =\left(\frac{3}{x} +\frac{1}{y}\right)( 3x+y)\\ & =10+\frac{3y}{x} +\frac{3x}{y} \geq 16 \end{aligned}\\ \notag\\ \Longrightarrow 3x+y\geq 8 \end{gather}$$

且当 $\displaystyle x=y=2$ 时取得最小值

B. 实际上：

$$\begin{gather} x+4y=4xy\Rightarrow \frac{4}{x} +\frac{1}{y} =4\geq 2\sqrt{\frac{4}{xy}}\\ \notag\\ \Longrightarrow xy\geq 1 \end{gather}$$

故而 B 正确

C. 由 Cauchy 不等式：$\displaystyle \left( x^{2} +y^{2}\right)( 1+1) \geq ( x+y)^{2} =9\Rightarrow x^{2} +y^{2} \geq \frac{9}{2}$，C 正确

D. 由 Cauchy 不等式：$\displaystyle ( x+y+z)\left(\frac{1}{x} +\frac{1}{y} +\frac{1}{z}\right) \geq ( 1+1+1)^{2} =9$，故而 D 正确

$$\begin{equation} \sin B\cos A+\sin A\cos B=\sin^{2} C\quad \quad \Longrightarrow \sin( A+B) =\sin C=\sin^{2} C \end{equation}$$

$\displaystyle \Rightarrow \sin C=0\ \text{或} \ 1\Rightarrow \vartriangle$ABC 为直角三角形

C. $\displaystyle \alpha \bot \beta ,\ m\bot \alpha \Rightarrow m//\beta$；$\displaystyle n//\beta ,\ m//\beta \nRightarrow m\bot n$

D. 若已有 $\displaystyle m\bot \alpha$，取 $\displaystyle n$ 为平面 $\displaystyle \alpha ,\beta$ 的交线即满足以上条件，但这并不会对 $\displaystyle \alpha ,\beta$ 提出要求，遑论垂直

B. 即为常见的垂直判定法，显然正确

同时：

$$\begin{equation} \overrightarrow{AP} =\overrightarrow{AB_{1}} +\overrightarrow{B_{1} P} ,\ AB_{1} ,B_{1} P\bot CD\Rightarrow AP\bot CD \end{equation}$$