扬州拾遗(一):竞赛中的函数

一、函数问题基本方法

1.数型结合法

例1:求方程\(\left|x-1\right|=\frac{1}{x}\)的正根的个数。
解:作图像得,正根个数为\(1\)。

例2:求函数\(f\left(x\right)=\sqrt{x^{4}-3x^{2}-6x+13}-\sqrt{x^{4}-x^{2}+1}\)的最大值。
解:\(f\left(x\right)=\sqrt{\left(x^{2}-2\right)^{2}+\left(x-3\right)^{2}}-\sqrt{\left(x^{2}-1\right)^{2}+x^{2}}\)即表示点\(P\left(x,x^{2}\right)\)到点\(A\left(3,2\right),B\left(0,1\right)\)的距离之差,点\(P\)在抛物线\(y=x^{2}\)上。易得点\(P\)在直线\(l_{AB}:x-3y+3=0\)上,因此得到\(P\left(\frac{1-\sqrt{37}}{6},\frac{19-\sqrt{37}}{18}\right)\),故\(f\left(x\right)_{max}=\sqrt{\left[\left(\frac{1-\sqrt{37}}{6}-3\right)^{2}+\left(\frac{19-\sqrt{37}}{18}-2\right)^{2}\right]-\left[\left(\frac{1-\sqrt{37}}{6}-0\right)^{2}+\left(\frac{19-\sqrt{37}}{18}-1\right)^{2}\right]}=\frac{2\sqrt{245+34\sqrt{37}}}{9}\)

2.函数性质的应用

例3:设\(x,y\in\mathbf{R}\),且满足\(\left\{\begin{matrix}
\left ( x-1 \right )^{3}+1997\left ( x-1 \right )=-1\\
\left ( y-1 \right )^{3}+1997\left ( y-1 \right )=1
\end{matrix}\right.\),求\(x+y\)。
解:考虑函数\(f\left(s\right)=s^{3}+1997s\),该函数既是奇函数又是增函数。又\(f\left(x-1\right)=-f\left(y-1\right)\),得\(\left(x-1\right)+\left(y-1\right)=0\),因此\(x+y=2\)。

例4:奇函数\(f\left(x\right)\)在定义域\(\left(-1,1\right)\)内是减函数,又\(f\left(1-a\right)+f\left(1-a^{2}\right)<0\),求\(a\)的取值范围。 解:首先由函数的定义域得\(\left\{\begin{matrix}
-1<1-a<1\\ -1<1-a^{2}<1 \end{matrix}\right.\),解得\(0< a< \sqrt{2}\)。下面进行分类讨论: \(1^{\circ}\left | 1-a \right |< 1-a^{2}\Rightarrow 0< a< 1\) \(2^{\circ}\left | 1-a^{2} \right |< 1-a\Rightarrow -2< a< 0\)(不符题意) 综上所述,\(a\in\left ( 0,1 \right )\)。 例5:设\(f\left(x\right)\)是定义在\(\left(-\infty,+\infty\right)\)上以\(2\)为周期的函数,对\(k\in\mathbf{Z}\),用\(I_{k}\)表示区间\(\left(2k-1,2k+1\right]\),已知当\(x\in I_{0}\)时,\(f\left(x\right)=x^{2}\),求\(f\left(x\right)\)在\(I_{k}\)上的解析式。
解:根据图像的平移,易得\(f\left(x\right)=\left(x-2k\right)^{2}\)。

例6:解方程\(\left(3x-1\right)\left(\sqrt{9x^{2}-6x+5}+1\right)+\left ( 2x-3 \right )\left ( \sqrt{4x^{2}-12x+13}+1 \right )=0\)。
解:首先进行因式分解:\(\left(3x-1\right)\left(\sqrt{\left ( 3x-1 \right )^{2}+4}+1\right)+\left ( 2x-3 \right )\left ( \sqrt{\left ( 2x-3 \right )^{2}+4}+1 \right )=0\)
考察函数\(f\left ( s \right )=s\left ( \sqrt{s^{2}+4}+1 \right )\),它既是奇函数又是增函数。因此上述方程转化为\(\left ( 3x-1 \right )+\left ( 2x-3 \right )=0\),解得\(x=\frac{4}{5}\)。

3.配方法

例7:求函数\(y=x+\sqrt{2x+1}\)的值域。
解:首先考虑定义域得\(D_{y}=\left [ -\frac{1}{2},+\infty \right )\)。又\(y=\left ( \sqrt{x+\frac{1}{2}}+\sqrt{\frac{1}{2}} \right )^{2}-1\geq -\frac{1}{2}\)。因此该函数的值域为\(R_{y}=\left [ -\frac{1}{2},+\infty \right )\)。

4.换元法

例8:求函数\(y=\left ( \sqrt{1+x}+\sqrt{1-x}+2 \right )\left ( \sqrt{1-x^{2}}+1 \right ),x\in\left [ 0,1 \right ]\)的值域。
解:令\(s=\sqrt{1+x}+\sqrt{1-x}\),则\(s^{2}=2\left ( \sqrt{1-x^{2}}+1\right )\),且\(s\in\left [ -\sqrt{2},\sqrt{2} \right ]\)。代入得\(y=\frac{1}{2}\cdot \left ( s+2 \right )\cdot s^{2}=\frac{1}{2}s^{3}+s^{2}\),易得该函数的值域为\(R_{y}=\left [ 2+\sqrt{2},8 \right ]\)。

5.关于反函数

例9:若函数\(y=f\left ( x \right )\)定义域、值域均为\(\mathbf{R}\),且存在反函数。若\(y=f\left ( x \right )\)在\(\left(-\infty,+\infty\right)\)上递增,求证:\(y=f^{-1}\left ( x \right )\)在\(\left(-\infty,+\infty\right)\)上递增。
解:不妨设\(x_{1}>x_{2}\),则\(f\left ( x_{1} \right )>f\left ( x_{2} \right )\),因此\(f^{-1}\left ( f\left ( x_{1} \right ) \right )-f^{-1}\left ( f\left ( x_{2} \right ) \right )=x_{1}-x_{2}>0\),所以\(y=f^{-1}\left ( x \right )\)在\(\left(-\infty,+\infty\right)\)上递增。证毕。

例10:设函数\(f\left ( x \right )=\sqrt[4]{\frac{4x+1}{3x+2}}\),解方程\(f\left ( x \right )=f^{-1}\left ( x \right )\)。
解:即\(\sqrt[4]{\frac{4x+1}{3x+2}}=x\),整理得\(\left ( x-1 \right )\left ( 3x^{4}+5x^{3}+5x^{2}+5x+1 \right )=0\),显然\(3x^{4}+5x^{3}+5x^{2}+5x+1>0\),故\(x=1\)。

二、典型例题

题1:已知不等式\(\sin^{2}x+a\cos x+a^{2}\geq 1+\cos x\)对一切\(x\in\mathbf{R}\)恒成立,求\(a\)的取值范围。
解:\(\cos^{2}x+\left ( 1-a \right )\cos x – a^{2}\leq 0\),其中\(\cos x\in\left [ -1,1 \right ]\)。令\(f\left ( x \right )=\cos^{2}x+\left ( 1-a \right )\cos x – a^{2}\),则\(\left\{\begin{matrix}
f\left ( -1 \right )\leq 0\\
f\left ( 1 \right )\leq 0
\end{matrix}\right.\Rightarrow \left\{\begin{matrix}
a^{2}-a\geq 0\leq 0\\
a^{2}-a-2\geq 0\leq 0
\end{matrix}\right.\Rightarrow a\in\left ( -\infty,-2 \right ]\cup\left [ 1,+\infty \right )\)。

题2:已知函数\(f\left ( x \right )\)的定义域是\(\left [ -\frac{1}{2},\frac{1}{2} \right ]\),求函数\(g\left ( x \right )=f\left ( ax \right )+f\left ( \frac{x}{a} \right )\)的定义域\(\left ( a> 0\right )\)。
解:根据函数函数\(f\left ( x \right )\)的定义域很容易得到以下不等式\(\left\{\begin{matrix}
-\frac{1}{2}\leq ax\leq \frac{1}{2}\\
-\frac{1}{2}\leq \frac{x}{a}\leq \frac{1}{2}
\end{matrix}\right.\)。即\(\left\{\begin{matrix}
-\frac{1}{2a}\leq x\leq \frac{1}{2a}\\
-\frac{a}{2}\leq x\leq \frac{a}{2}
\end{matrix}\right.\)分类讨论得:
\(1^{\circ}0< a\leq 1,D_{g}=\left [ -\frac{a}{2},\frac{a}{2} \right ]\) \(2^{\circ}a> 1,D_{g}=\left [ -\frac{1}{2a},\frac{1}{2a} \right ]\)

题3:已知\(f\left(x\right)\)是定义在\(\mathbf{R}\)上的函数,\(f\left(1\right)=1\),且对任意\(x\in\mathbf{R}\)都有\(f\left(x+5\right)\geq f\left(x\right) + 5,f\left(x+1\right)\leq f\left(x\right)+1\),且\(g\left ( x \right )=f\left ( x \right )+1-x\)求\(g\left(2012\right)\)。
解:\(f\left ( x+5 \right )\leq f\left ( x+4 \right )+1\leq \cdots \leq f\left ( x \right )+5\),又\(f\left ( x \right )\geq f\left ( x \right )+5\),故\(f\left ( x \right )= f\left ( x \right )+5\)。运用数学归纳法,并结合\(f\left ( 1 \right )=1\),不难得到\(f\left ( x \right )=x\)。因此\(g\left ( 2012 \right )=f\left ( 2012 \right )+1-2012=1\)。

题4:设函数\(f:\mathbf{N^{*}}\rightarrow \mathbf{N^{*}}\),且严格递增。当\(m.n\)互质时,\(f\left ( m\cdot n \right )=f\left ( m \right )\cdot f\left ( n \right )\)。若\(f\left ( 19 \right )=19\),求\(f\left ( f\left ( 19 \right ) \right )\cdot f\left ( 98 \right )\)的值。
解:由题意得,当\(x\leq 19\)时,有\(f\left ( x \right )=x\)。又\(f\left ( 17\cdot 19 \right )=f\left ( 17 \right )\cdot f\left ( 19 \right )=323\),则当\(x\leq 323\)时,有\(f\left ( x \right )=x\)。所以\(f\left ( f\left ( 19 \right ) \right )\cdot f\left ( 98 \right )=f\left ( 19 \right )\cdot f\left ( 98 \right )=19*98=1862\)。

题5:设函数\(f:\mathbf{N^{*}}\rightarrow \mathbf{N^{*}}\),且严格递增。\(f\left ( f\left ( n \right ) \right )=3n\)。求\(f\left ( 1 \right )+f\left ( 9 \right )+f\left ( 36 \right )\)。
解:首先可以得到\(f\left ( f\left ( 1 \right ) \right )=3\)。不妨假设\(f\left ( 1 \right )=1\),代入得\(f\left ( f\left ( 1 \right ) \right )=f\left ( 1 \right )=1\neq 3\),矛盾。因此易证\(f\left ( 1 \right )=2\)。于是可以列出:\(f\left ( 2 \right )=f\left ( f\left ( 1 \right ) \right )=3,f\left ( 3 \right )=f\left ( f\left ( 2 \right ) \right )=6,f\left ( 6 \right )=f\left ( f\left ( 3 \right ) \right )=9,f\left ( 9 \right )=f\left ( f\left ( 6 \right ) \right )=18\),于是得到\(f\left ( 9 \right )=18\)。由\(f\left ( 3 \right )=6,f\left ( 6 \right )=9\)再结合题中所给条件,得\(f\left ( 4 \right )=7\)。因此\(f\left ( 7 \right )=f\left ( f\left ( 4 \right ) \right )=12,f\left ( 12 \right )=f\left ( f\left ( 7 \right ) \right )=21,f\left ( 21 \right )=f\left ( f\left ( 12 \right ) \right )=36,f\left ( 36 \right )=f\left ( f\left ( 21 \right ) \right )=63\),于是得到\(f\left ( 36 \right )=63\)。将以上的结果相加得\(f\left ( 1 \right )+f\left ( 9 \right )+f\left ( 36 \right )=2+18+63=83\)。

题6:已知函数\(f\left ( x \right )=\log_{x}{\left ( x+1 \right )},x\in\left ( 1,+\infty \right )\),试比较\(f\left ( x \right ),f\left ( x+1 \right )\)的大小。
解:\(\frac{f\left ( x+1 \right )}{f\left ( x \right )}=\frac{\log{\left (x+2 \right )}}{\log{\left (x+1 \right )}}\cdot \frac{\log{\left (x+1 \right )}}{\log{x}}=\frac{\log{\left ( x+2 \right )}}{\log{x}}> 1\),所以\(f\left ( x+1 \right )> f\left ( x \right )\)。

题7:已知\(3^{a}+13^{b}=17^{a},5^{a}+7^{b}=11^{b}\),试判断实数\(a,b\)的大小关系,并证明之。
解:当\(a=1\)时,\(b> 1\),得\(b> a\)。假设\(b\leq a\),易得\(\left\{\begin{matrix}
17^{a}\leq 3^{a}+13^{a}\\
5^{b}+7^{b}\leq 11^{b}
\end{matrix}\right.\Rightarrow \left\{\begin{matrix}
1\leq \left ( \frac{3}{17} \right )^{a}+\left ( \frac{13}{17} \right )^{a}\\
1\geq \left ( \frac{5}{11} \right )^{b}+\left ( \frac{7}{11} \right )^{b}
\end{matrix}\right.\),不妨令\(f\left ( x \right )=\left ( \frac{3}{11} \right )^{x}+\left ( \frac{13}{17} \right )^{x},g\left ( x \right )=\left ( \frac{5}{11} \right )^{x}+\left ( \frac{7}{11} \right )^{x}\),易得\(f\left ( 1 \right )=\frac{16}{17}< g\left ( 1 \right )=\frac{12}{11}\)。不成立。综上所述\(b> a\)。

题8:设\(f\left ( x \right )=x^{n}+ax^{2}+bx+c\),\(n\)为自然数。已知\(f\left ( -1 \right )=0,f\left ( 1 \right )=-6,f\left ( 2 \right )=-9,f\left ( 3 \right )=-4,f\left ( 6 \right )=119\),求\(f\left ( x \right )\)。
解:将所给条件代入得\(\left\{\begin{matrix}
\left ( -1 \right )^{n}+a-b+c=0\\
1+a+b+c=-6\\
2^{n}+4a+2b+c=-9\\
3^{n}+9a+3b+c=-4\\
6^{n}+36a+6b+c=119
\end{matrix}\right.\)对\(n\)进行分类讨论:
\(1^{\circ}n\)为奇数,上述方程组可化简得\(\left\{\begin{matrix}
-1+a-b+c=0\\
1+a+b+c=-6\\
2^{n}+4a+2b+c=-9\\
3^{n}+9a+3b+c=-4\\
6^{n}+36a+6b+c=119
\end{matrix}\right.\),解得\(\left\{\begin{matrix}
n=3\\
a=-2\\
b=-4\\
c=-1
\end{matrix}\right.\)。
\(2^{\circ}n\)为偶数,上述方程组可化简得\(\left\{\begin{matrix}
1+a-b+c=0\\
1+a+b+c=-6\\
2^{n}+4a+2b+c=-9\\
3^{n}+9a+3b+c=-4\\
6^{n}+36a+6b+c=119
\end{matrix}\right.\),不成立。
综上所述\(f\left ( x \right )=x^{3}-2x^{2}-4x-1\)。

题9:已知\(a,b,c\)为非零实数,\(f\left ( x \right )=\frac{ax+b}{cx+d},x\in\mathbf{R}\),且\(f\left ( 19 \right )=19,f(97)=97\)。若当\(x\neq -\frac{d}{c}\)时,对于任意实数\(x\),均有\(f\left ( f\left ( x \right ) \right )=x\),试求出\(f\left ( x \right )\)值域以外的唯一数。
解:易知\(x=19,x=97\)即为不动点,又为稳定点,又\(f\left(x\right)\)是经过反比例函数平移得到,很容易得到\(f\left ( x \right )=\frac{58x-1843}{x-58}\),故所求数为\(58\)。

题10:设有函数\(f\left ( x \right )=\sin{\left ( x+a_{1} \right )}+\frac{1}{1\times 2}\sin{\left ( x+a_{2} \right )}+\cdots +\frac{1}{\left ( n-1 \right )\times n}\sin{\left ( x+a_{n} \right )}\),其中\(a_{1},a_{2},\cdots ,a_{n}\)为常数,证明:(1)至少有一个实数,使\(f\left ( x_{0} \right )\neq 0\);(2)如果\(f\left ( x_{1} \right )=f\left ( x_{2} \right )=0\),则\(x_{1}-x_{2}=m\pi\)(\(m\)是一个整数)。
解:(1)令\(x_{0}=\frac{\pi}{2}-a_{1}\),则\(f\left ( x_{0} \right )\geq 1-\left ( \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{\left ( n-1 \right )\times n} \right )=1-\frac{n-1}{n}=\frac{1}{n}> 0\)。
(2)\[f\left ( x \right )=\sin{x}\cos{a_{1}}+\cos{x}\sin{a_{1}}+\sum_{i=2}^{n}{\frac{1}{\left ( i-1 \right )\times i}\left ( \sin{x}\cos{a_{i}}+\cos{x}\sin{a_{i}} \right )}\\=\left ( \cos{a_{1}}+\sum_{i=2}^{n}{\frac{1}{\left ( i-1 \right )\times i}\cdot \cos{a_{i}}} \right )\cdot \sin{x}+\left ( \sin{a_{1}}+\sum_{i=2}^{n}{\frac{1}{\left ( i-1 \right )\times i}\cdot \sin{a_{i}}} \right )\cdot \cos{x}=A\sin{\left (x+\varphi \right ) }\\A=\sqrt{\left ( \cos{a_{1}}+\sum_{i=2}^{n}{\frac{1}{\left ( i-1 \right )\times i}\cdot \cos{a_{i}}} \right )^{2}+\left ( \sin{a_{1}}+\sum_{i=2}^{n}{\frac{1}{\left ( i-1 \right )\times i}\cdot \sin{a_{i}}} \right )^{2}},\tan{\varphi }=\frac{\left ( \cos{a_{1}}+\sum_{i=2}^{n}{\frac{1}{\left ( i-1 \right )\times i}\cdot \cos{a_{i}}} \right )}{\left ( \sin{a_{1}}+\sum_{i=2}^{n}{\frac{1}{\left ( i-1 \right )\times i}\cdot \sin{a_{i}}} \right )}\]。由\(A\sin{x+\varphi }=0\)得\(x+\varphi=k\pi\),于是\(x_{1}-x_{2}=\left(k_{1}-k_{2}\right)\pi=m\pi\)。证毕。

题11:设正实数\(x,y\)满足\(xy=1\),求函数\(f\left ( x,y \right )=\frac{x+y}{\left [ x \right ]\cdot \left [ y \right ]+\left [ x \right ]+\left [ y \right ]+1}\)的值域(这里\(\left [ z \right ]\)表示不超过\(z\)的最大整数)。
解:\(1^{\circ}x=y=1,f\left ( 1,1 \right )=\frac{1}{2}\)
\(2^{\circ}\)不妨设\(x>1\),则\(\left [ y \right ]=0\),记\(x=m+\lambda\),其中\(m=\left [ x \right ],\lambda =\left \{ x \right \}\)。则\(f\left ( x,y \right )=\frac{x+\frac{1}{x}}{\left [ x \right ]+1}=\frac{x+\frac{1}{x}}{m+1}=\frac{m+\lambda +\frac{1}{m+\lambda }}{m+1}\)。令\(a_{m}=\frac{m+\frac{1}{m}}{m+1}\),则\(a_{m}\leq f\left ( x,y \right )< \frac{m+1+\frac{1}{m+1}}{m+1}\)。又\(a_{m+1}-a_{m}=\frac{m-2}{\left ( m-2 \right )m\left ( m+1 \right )}\Rightarrow a_{1}> a_{2}< a_{3}< \cdots < a_{m}\)。因此\(\frac{5}{6}\leq f\left ( x,y \right )< 1 + \frac{1}{\left ( m+1 \right )^{2}}\leq \frac{5}{4}\)。故值域为\(R_{f}=\left [ \frac{5}{6},\frac{5}{4} \right )\)。 三、二次函数

例1:已知\(f\left ( x \right )=ax^{2}+bx\),满足\(1\leq f\left ( -1 \right )\leq 2\)且\(2\leq f\left ( 1 \right )\leq 4\),求\(f\left ( -2 \right )\)的取值范围。
解:
法一:由题设得,\(\left\{\begin{matrix}
1\leq a-b\leq 2 & \left ( \textrm{i} \right )\\
2\leq a+b\leq 4 & \left ( \textrm{ii} \right )
\end{matrix}\right.\),由\(3\times \left ( \textrm{i} \right )+\left ( \textrm{ii} \right )\)得\(5\leq f\left ( -2 \right )\leq 10\)。
法二:\(\left\{\begin{matrix}
f\left ( -1 \right )=a-b\\
f\left ( 1 \right )=a+b
\end{matrix}\right.\Rightarrow \left\{\begin{matrix}
a=\frac{1}{2}\cdot \left [ f\left ( 1 \right ) + f\left ( -1 \right ) \right ]\\
b=\frac{1}{2}\cdot \left [ f\left ( 1 \right ) – f\left ( -1 \right ) \right ]
\end{matrix}\right.\),因此\(f\left ( -2 \right )=f\left ( 1 \right )+3f\left ( -1 \right )\in\left [ 5,10 \right ]\)。

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