解答
1.
令\(f(x)=a^{2|\;x|\;}+3x^2+|\;x|\;=1\)
\(f(-x)=f(x)\)
\(f(x)\)為偶函數
\(\alpha,\beta\)為相反數
\(\cases{\displaystyle \alpha+\beta=0\ldots(1)\cr \alpha-3\beta=\frac{4}{3}}\ldots(2)\)
\(\displaystyle \alpha=\frac{1}{3},\beta=-\frac{1}{3}\)
令\(\displaystyle x=\frac{1}{3}\)代入原式
\(\displaystyle\alpha^{\frac{2}{3}}+3\cdot \frac{1}{9}+\frac{1}{3}=1\)
\(\displaystyle a^{\frac{2}{3}}=\frac{1}{3}\)
\(\displaystyle a=(\frac{1}{3})^{\frac{3}{2}}=\sqrt{\frac{1}{27}}=\frac{\sqrt{1}}{{3\sqrt{3}}}=\frac{\sqrt{3}}{9}\)
2.
\(\cases{y=log_2(kx^2)+\frac{3}{4}x\cr y=2^{|\;x|\;}+\frac{3}{4}x}\)
\(log_2(kx^2)+\frac{3}{4}x=2^{|\;x|\;}+\frac{3}{4}x\)
\(log_2(kx^2)=2^{|\;x|\;}\)
\(log_2(kx^2)-2^{|\;x|\;}=0\)
令\(f(x)=log_2(kx^2)-2^{|\;x|\;}=0\)
\(f(-x)=f(x)\)
\(f(x)\)為偶函數
兩根互為相反數
令兩根為\(\alpha,-\alpha(\alpha>0)\)代入\(y=2^{|\;x|\;}+\frac{3}{4}x\)
\(A(\alpha,2^{\alpha}+\frac{3}{4}\alpha),B(-\alpha,2^{\alpha}-\frac{3}{4}\alpha)\)
\(\overline{AB}=\sqrt{(2\alpha)^2+(\frac{3}{2}\alpha)^2}=10\)
\(\displaystyle \frac{5}{2}\alpha=10\)
\(\alpha=4\)
\(A(4,19),B(-4,13)\)
將\(A(4,19)\)代入\(y=log_2(kx^2)+\frac{3}{4}x\)
\(log_2 16k+3=19\)
\(4+log_2 k=16\)
\(log_2k=12\)
\(k=2^{12}=4096\)