求解析式系的題目及答案

單項(xiàng)選擇題1.已知函數(shù)\(y=f(x)\)滿足\(f(1)=2\),且\(f(x+1)=3f(x)\),則\(f(x)\)的解析式為()A.\(f(x)=2\times3^{x}\)B.\(f(x)=2\times3^{x-1}\)C.\(f(x)=3\times2^{x}\)D.\(f(x)=3\times2^{x-1}\)答案：B解析:由\(f(x+1)=3f(x)\)可得\(\frac{f(x+1)}{f(x)}=3\),則\(\{f(x)\}\)是以\(f(1)=2\)為首項(xiàng),\(3\)為公比的等比數(shù)列,根據(jù)等比數(shù)列通項(xiàng)公式\(a_{n}=a_{1}q^{n-1}\),可得\(f(x)=f(1)\times3^{x-1}=2\times3^{x-1}\)。2.若二次函數(shù)\(y=ax^{2}+bx+c\)的圖象經(jīng)過(guò)點(diǎn)\((0,1)\),對(duì)稱軸為\(x=1\),且最小值為\(-1\),則其解析式為()A.\(y=2x^{2}-4x+1\)B.\(y=x^{2}-2x+1\)C.\(y=2x^{2}-2x+1\)D.\(y=2x^{2}-4x-1\)答案：A解析:因?yàn)閷?duì)稱軸為\(x=1\),最小值為\(-1\),設(shè)\(y=a(x-1)^{2}-1\),又過(guò)點(diǎn)\((0,1)\),代入可得\(a(0-1)^{2}-1=1\),解得\(a=2\),所以\(y=2(x-1)^{2}-1=2x^{2}-4x+1\)。多項(xiàng)選擇題1.若一次函數(shù)\(y=kx+b\)的圖象經(jīng)過(guò)\((1,3)\)和\((-1,1)\),則()A.\(k=1\)B.\(k=2\)C.\(b=1\)D.\(b=2\)答案:BD解析：把\((1,3)\)和\((-1,1)\)代入\(y=kx+b\)得\(\begin{cases}k+b=3\\-k+b=1\end{cases}\),兩式相加得\(2b=4\),\(b=2\),把\(b=2\)代入\(k+b=3\)得\(k=1\)。2.已知\(f(x)\)是二次函數(shù),且\(f(0)=1\),\(f(x+1)-f(x)=2x\),則\(f(x)\)的可能解析式為()A.\(f(x)=x^{2}-x+1\)B.\(f(x)=x^{2}+x+1\)C.\(f(x)=x^{2}-x-1\)D.\(f(x)=x^{2}+x-1\)答案：A解析:設(shè)\(f(x)=ax^{2}+bx+c\),由\(f(0)=1\)得\(c=1\),\(f(x+1)-f(x)=a(x+1)^{2}+b(x+1)+1-(ax^{2}+bx+1)=2ax+a+b=2x\),則\(\begin{cases}2a=2\\a+b=0\end{cases}\),解得\(a=1\),\(b=-1\),所以\(f(x)=x^{2}-x+1\)。判斷題1.若\(f(x+2)=f(x)\),則\(f(x)\)的周期為\(2\)。()答案：對(duì)解析:根據(jù)周期函數(shù)定義,若\(f(x+T)=f(x)\)(\(T\neq0\)),則\(T\)為周期,這里\(T=2\)滿足條件。2.二次函數(shù)\(y=ax^{2}+bx+c\)(\(a\neq0\))一定有最大值。()答案：錯(cuò)解析:當(dāng)\(a\gt0\)時(shí),二次函數(shù)有最小值;當(dāng)\(a\lt0\)時(shí),二次函數(shù)有最大值,所以不一定有最大值。簡(jiǎn)答題1.已知\(f(x)\)是一次函數(shù),且\(f[f(x)]=4x+3\),求\(f(x)\)的解析式。答案:設(shè)\(f(x)=kx+b\),則\(f[f(x)]=k(kx+b)+b=k^{2}x+kb+b\),所以\(\begin{cases}k^{2}=4\\kb+b=3\end{cases}\),解得\(\begin{cases}k=2,b=1\\k=-2,b=-3\end{cases}\),\(f(x)=2x+1\)或\(f(x)=-2x-3\)。2.已知二次函數(shù)\(y=f(x)\)的圖象過(guò)點(diǎn)\((0,3)\),對(duì)稱軸為\(x=2\),且最小值為\(-1\),求\(f(x)\)的解析式。答案：設(shè)\(f(x)=a(x-2)^{2}-1\),把\((0,3)\)代入得\(a(0-2)^{2}-1=3\),解得\(a=1\),所以\(f(x)=(x-2)^{2}-1=x^{2}-4x+3\)。討論題1.已知\(f(x)\)滿足\(f(x+3)=-f(x)\),判斷\(f(x)\)的周期性,并說(shuō)明理由。答案:因?yàn)閈(f(x+3)=-f(x)\),則\(f(x+6)=f((x+3)+3)=-f(x+3)=f(x)\),所以\(f(x)\)的周期是\(6\)。2.若二次函數(shù)\(y=ax^{2}+bx+c\)（\(a\neq0\)）與\(x\)軸有兩個(gè)交點(diǎn)