5. (5 分) 设函数 f(x)=\left\{\begin{array}{ll}1+\log _{2}(2-x), & x

Question

5. (5 分) 设函数 f(x)=\left\{\begin{array}{ll}1+\log _{2}(2-x), & x<1 \ 2^{x-1}, & x \geqslant 1\end{array}ight., 则 f(-2)+f\left(\log _{2} 12ight)=( )

Accepted Answer

Answer: C. 9 解：函数 f(x)=\left\{\begin{array}{l}1+\log _{2}(2-x), x<1 \ 2^{x-1}, x \geqslant 1\end{array}ight., 即有 f(-2)=1+\log _{2}(2+2)=1+2=3, f\left(\log _{2} 12ight)=2^{\log _{2} 12-1}=2 \log _{2} 12 	imes \frac{1}{2}=12 	imes \frac{1}{2}=6 则有 f(-2)+f\left(\log _{2} 12ight)=3+6=9. 故选: C.

5. (5 分) 设函数 $f(x)=\left\{\begin{array}{ll}1+\log _{2}(2-x), & x<1 \\ 2^{x-1}, & x \geqslant 1\end{array}\right.$, 则 $f(-2)+f\left(\log _{2} 12\right)=($ )

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