求自变量的取值范围时需要注意的角度
前言
以下情形,其实质是不等式性质的灵活应用;在求解函数自变量的取值范围时,我们能想到用不等式的性质,让我们感觉比较难的是,列不等式组时如何确保所有的角度我们都能考虑到,不至于遗漏造成错误。
扇形面积计算
分析:设扇形的弧长为\(l\),半径为\(R\),则\(C=2R+l\),即\(l=C-2R\);
由\(\left\{\begin{array}{l}{l>0}\\{R>0}\end{array}\right.\),即\(\left\{\begin{array}{l}{C-2R>0}\\{R>0}\end{array}\right.\),
解得,\(0<R<\cfrac{C}{2}\);
由\(S=\cfrac{1}{2}\cdot l\cdot R=\cfrac{1}{2}\cdot (C-2R)\cdot R\)
\(=-R^2+\cfrac{C}{2}R=-(R-\cfrac{C}{4})^2+\cfrac{C^2}{16}\)
故当\(R=\cfrac{C}{4}\)时,\(S_{max}=\cfrac{C^2}{16}\);
此时,\(l=C-2R=\cfrac{C}{2}\),即此时\(\alpha=\cfrac{l}{R}=2 (rad)\)。
均值不等式求面积
分析:设矩形的长为\(x\),则宽\(y=\cfrac{C}{2}-x\),
由\(\left\{\begin{array}{l}{x>0}\\{y>0}\end{array}\right.\),即\(\left\{\begin{array}{l}{x>0}\\{\cfrac{C}{2}-x>0}\end{array}\right.\),
解得,\(0<x<\cfrac{C}{2}\);
又由于\(S=xy=x\cdot (\cfrac{C}{2}-x)\leq [\cfrac{x+(\cfrac{C}{2}-x)}{2}]^2=\cfrac{C^2}{16}\),
当且仅当\(x=\cfrac{C}{2}-x\),即\(x=y=\cfrac{C}{4}\in (0,\cfrac{C}{2})\)时取得等号。
二次型函数值域求解
分析:由于\(sinx+siny=\cfrac{1}{3}\),
所以\(sin=\cfrac{1}{3}-siny\),
由于\(\left\{\begin{array}{l}{-1\leq siny\leq 1}\\{-1\leq sinx\leq 1}\end{array}\right.\)
即就是,\(\left\{\begin{array}{l}{-1\leq siny\leq 1}\\{-1\leq \cfrac{1}{3}-siny\leq 1}\end{array}\right.\)
解得\(-\cfrac{2}{3}\leq siny\leq 1\);
又由于\(M=\cfrac{1}{3}-siny-cos^2y=(siny-\cfrac{1}{2})^2-\cfrac{11}{12}\),
则当\(siny=-\cfrac{2}{3}\),\(sinx=1\)时,\(M_{max}=\cfrac{4}{9}\);
当\(siny=\cfrac{1}{2}\),\(sinx=-\cfrac{1}{6}\)时,\(M_{min}=-\cfrac{11}{12}\);
故最大值与最小值的差为\(\cfrac{4}{9}-(-\cfrac{11}{12})=\cfrac{49}{36}\)。
角的范围求解
分析:本题先将\(\cfrac{c}{b}=\cfrac{sinC}{sinB}=2cosB\),
接下来的难点是求\(B\)的范围,注意列不等式的角度,锐角三角形的三个角都是锐角,要同时限制
由\(\begin{cases} &0<A<\cfrac{\pi}{2} \\ &0<B<\cfrac{\pi}{2} \\ &0<C<\cfrac{\pi}{2}\end{cases}\)得到,\(\begin{cases} &0<\pi-3B<\cfrac{\pi}{2} \\ &0<B<\cfrac{\pi}{2} \\ &0<2B<\cfrac{\pi}{2}\end{cases}\)
解得\(B\in (\cfrac{\pi}{6},\cfrac{\pi}{4})\),故 \(2cosB \in (\sqrt{2},\sqrt{3})\) 。
分析:由题目\((a-b)(sinA-sinB)=(c-b)sinC\),角化边得到,\((a-b)(a+b)=(c-b)c\),整理得到\(b^2+c^2-a^2=bc\),
由余弦定理可知,\(cosA=\cfrac{b^2+c^2-a^2}{2bc}=\cfrac{bc}{2bc}=\cfrac{1}{2}\),又\(A\in (0,\pi)\),故\(A=\cfrac{\pi}{3}\);
又已知\(a=\sqrt{3}\),则有\(2R=\cfrac{a}{sinA}=2\),则\(b=2RsinB=2sinB\),\(c=2RsinC=2sinC\),
则\(b^2+c^2=(2sinB)^2+(2sinC)^2\)
\(=4sin^2B+4sin^2(\cfrac{2\pi}{3}-B)\)
\(=2\cdot 2sin^2B+2\cdot 2sin^2(\cfrac{2\pi}{3}-B)\)
\(=2(1-cos2B)+2[1-cos(\cfrac{4\pi}{3}-2B)]\)
\(=2-2cos2B+2+2cos(2B-\cfrac{\pi}{3})\)
\(=4+\sqrt{3}sin2B-cos2B=4+2sin(2B-\cfrac{\pi}{6})\),
又由于三角形为锐角三角形,则三个角都是锐角,
故满足\(\left\{\begin{array}{l}{0<B<\cfrac{\pi}{2}}\\{0<C<\cfrac{\pi}{2}}\end{array}\right.\),即\(\left\{\begin{array}{l}{0<B<\cfrac{\pi}{2}}\\{0<\cfrac{2\pi}{3}-B<\cfrac{\pi}{2}}\end{array}\right.\),
解得,\(\cfrac{\pi}{6}<B<\cfrac{\pi}{2}\),故\(\cfrac{\pi}{6}<2B-\cfrac{\pi}{6}<\cfrac{5\pi}{6}\),则\(\cfrac{1}{2}<sin(2B-\cfrac{\pi}{6})\leq 1\)
故\(b^2+c^2=4+2sin(2B-\cfrac{\pi}{6})\in (5,6]\),故选\(A\)。
随机事件的概率
分析:由于任一事件的概率的取值范围是\(0\leq P(A)\leq 1\),且\(A\) 、\(B\) 发生的概率均不等于 \(0\),
故此题目中 \(0<P(A)<1\), \(0<P(B)<1\),且由互斥事件的概率满足 \(P(A)+P(B)\leq 1\) ,
故其应该满足条件如下:
\(\left\{\begin{array}{l}{0<2-a<1}\\{0<4a-5<1}\\{(2-a)+(4a-5)\leq 1}\end{array}\right.\),化简得\(\left\{\begin{array}{l}{1<a<2}\\{\cfrac{5}{4}<a<\cfrac{3}{2}}\\{a\leq\cfrac{4}{3}}\end{array}\right.\),
解得\(\cfrac{5}{4}<a\leq\cfrac{4}{3}\),故选\(D\)。
三角形边之比
分析:又已知及三角形三边关系得到,\(\left\{\begin{array}{l}{a<b+c\leqslant 3a}\\{a+b>c}\\{a+c>b}\end{array}\right.\),
得到,\(\left\{\begin{array}{l}{1<\cfrac{b}{a}+\cfrac{c}{a}\leqslant 3}\\{1+\cfrac{b}{a}>\cfrac{c}{a}②}\\{1+\cfrac{c}{a}>\cfrac{b}{a}③}\end{array}\right.\),将②③合写为一个双连不等式,
得到,\(\left\{\begin{array}{l}{1<\cfrac{b}{a}+\cfrac{c}{a}\leqslant 3}\\{-1<\cfrac{c}{a}-\cfrac{b}{a}<1}\end{array}\right.\),
两式相加,得到\(0<2\times \cfrac{c}{a}<4\),得到\(\cfrac{c}{a}\in (0,2)\),故选\(B\).