函數(shù)根式題目及答案

一、單項(xiàng)選擇題（每題2分，共10題）1.函數(shù)\(y=\sqrt{x-1}\)中，自變量\(x\)的取值范圍是（）A.\(x\geq1\)B.\(x\gt1\)C.\(x\leq1\)D.\(x\lt1\)2.若\(\sqrt{a-2}\)有意義，則\(a\)的取值范圍是（）A.\(a\gt2\)B.\(a\geq2\)C.\(a\lt2\)D.\(a\leq2\)3.函數(shù)\(y=\sqrt{3-x}\)的自變量\(x\)的取值范圍是（）A.\(x\gt3\)B.\(x\geq3\)C.\(x\lt3\)D.\(x\leq3\)4.函數(shù)\(y=\sqrt{2x+1}\)自變量\(x\)的取值范圍是（）A.\(x\geq-\frac{1}{2}\)B.\(x\gt-\frac{1}{2}\)C.\(x\leq-\frac{1}{2}\)D.\(x\lt-\frac{1}{2}\)5.若函數(shù)\(y=\frac{1}{\sqrt{x-3}}\)有意義，則\(x\)的取值范圍是（）A.\(x\gt3\)B.\(x\geq3\)C.\(x\lt3\)D.\(x\leq3\)6.函數(shù)\(y=\sqrt{x+2}+\frac{1}{x-1}\)中自變量\(x\)的取值范圍是（）A.\(x\geq-2\)B.\(x\geq-2\)且\(x\neq1\)C.\(x\gt-2\)D.\(x\gt-2\)且\(x\neq1\)7.對(duì)于函數(shù)\(y=\sqrt{5-x^2}\)，\(x\)的取值范圍是（）A.\(x\leq\sqrt{5}\)B.\(x\geq-\sqrt{5}\)C.\(-\sqrt{5}\leqx\leq\sqrt{5}\)D.\(x\leq-\sqrt{5}\)或\(x\geq\sqrt{5}\)8.函數(shù)\(y=\sqrt{x^2+1}\)自變量\(x\)的取值范圍是（）A.\(x\geq0\)B.\(x\leq0\)C.\(x\)為全體實(shí)數(shù)D.\(x\neq0\)9.若\(\sqrt{(2-x)^2}=2-x\)，則\(x\)的取值范圍是（）A.\(x\gt2\)B.\(x\geq2\)C.\(x\lt2\)D.\(x\leq2\)10.函數(shù)\(y=\sqrt{x-1}+\sqrt{2-x}\)中自變量\(x\)的取值范圍是（）A.\(1\leqx\leq2\)B.\(1\ltx\lt2\)C.\(x\geq1\)D.\(x\leq2\)二、多項(xiàng)選擇題（每題2分，共10題）1.以下函數(shù)中，自變量\(x\)的取值范圍是\(x\geq1\)的有（）A.\(y=\sqrt{x-1}\)B.\(y=\frac{1}{\sqrt{x-1}}\)C.\(y=\sqrt{(x-1)^2}\)D.\(y=\frac{1}{x-1}\)2.下列函數(shù)中，自變量取值范圍正確的是（）A.\(y=\sqrt{2x-1}\)，\(x\geq\frac{1}{2}\)B.\(y=\frac{1}{x+3}\)，\(x\neq-3\)C.\(y=\sqrt{x^2+1}\)，\(x\)為全體實(shí)數(shù)D.\(y=\frac{1}{\sqrt{4-x}}\)，\(x\lt4\)3.函數(shù)\(y=\sqrt{x+3}+\frac{1}{x-2}\)，自變量\(x\)需滿(mǎn)足（）A.\(x\geq-3\)B.\(x\neq2\)C.\(x\gt-3\)D.\(x\gt2\)4.若函數(shù)\(y=\sqrt{ax+b}\)有意義，則（）A.當(dāng)\(a\gt0\)時(shí)，\(x\geq-\frac{a}\)B.當(dāng)\(a\lt0\)時(shí)，\(x\leq-\frac{a}\)C.當(dāng)\(a=0\)且\(b\geq0\)時(shí)，\(x\)為全體實(shí)數(shù)D.當(dāng)\(a=0\)且\(b\lt0\)時(shí)，函數(shù)無(wú)意義5.下列關(guān)于函數(shù)\(y=\sqrt{9-x^2}\)說(shuō)法正確的是（）A.自變量\(x\)的取值范圍是\(-3\leqx\leq3\)B.函數(shù)圖象在\(x\)軸上方（包括與\(x\)軸交點(diǎn)）C.當(dāng)\(x=0\)時(shí)，\(y\)有最大值\(3\)D.它是一個(gè)二次函數(shù)6.對(duì)于函數(shù)\(y=\frac{\sqrt{x+1}}{x}\)，下列說(shuō)法正確的是（）A.自變量\(x\)的取值范圍是\(x\geq-1\)且\(x\neq0\)B.當(dāng)\(x=1\)時(shí)，\(y\)的值為\(\sqrt{2}\)C.函數(shù)圖象在第一、三象限D(zhuǎn).\(x\)越大，\(y\)的值越大7.以下函數(shù)中，\(x\)不能取\(1\)的有（）A.\(y=\sqrt{x-1}\)B.\(y=\frac{1}{x-1}\)C.\(y=\sqrt{1-x}\)D.\(y=(x-1)^2\)8.函數(shù)\(y=\sqrt{2x-3}+\sqrt{3-2x}\)，下列說(shuō)法正確的是（）A.自變量\(x\)的取值范圍是\(x=\frac{3}{2}\)B.函數(shù)值\(y=0\)C.該函數(shù)是一個(gè)常函數(shù)D.函數(shù)圖象是一個(gè)點(diǎn)9.若函數(shù)\(y=\sqrt{(x-a)^2}\)，當(dāng)\(x\lta\)時(shí)，\(y\)可化簡(jiǎn)為（）A.\(x-a\)B.\(a-x\)C.\(\vertx-a\vert\)D.\(-(x-a)\)10.函數(shù)\(y=\sqrt{x^2-4x+4}\)，可化簡(jiǎn)為（）A.\(y=\vertx-2\vert\)B.當(dāng)\(x\geq2\)時(shí)，\(y=x-2\)C.當(dāng)\(x\lt2\)時(shí)，\(y=2-x\)D.\(y=x-2\)三、判斷題（每題2分，共10題）1.函數(shù)\(y=\sqrt{x+5}\)中，自變量\(x\)的取值范圍是\(x\gt-5\)。（）2.若\(\sqrt{3-x}\)有意義，則\(x\)的取值范圍是\(x\leq3\)。（）3.函數(shù)\(y=\frac{1}{\sqrt{x-2}}\)中自變量\(x\)的取值范圍是\(x\geq2\)。（）4.函數(shù)\(y=\sqrt{x^2+4}\)自變量\(x\)的取值范圍是\(x\geq0\)。（）5.若\(\sqrt{(x-1)^2}=x-1\)，則\(x\geq1\)。（）6.函數(shù)\(y=\sqrt{2x+3}+\sqrt{3-2x}\)自變量\(x\)的取值范圍是\(-\frac{3}{2}\leqx\leq\frac{3}{2}\)。（）7.函數(shù)\(y=\frac{\sqrt{x-1}}{x-2}\)中，自變量\(x\)的取值范圍是\(x\gt1\)且\(x\neq2\)。（）8.函數(shù)\(y=\sqrt{9-x}\)，當(dāng)\(x=9\)時(shí)，函數(shù)值為\(0\)。（）9.函數(shù)\(y=\sqrt{(x+1)^2-4}\)自變量\(x\)的取值范圍是\(x\geq1\)或\(x\leq-3\)。（）10.函數(shù)\(y=\sqrt{5-2x}\)是一個(gè)一次函數(shù)。（）四、簡(jiǎn)答題（每題5分，共4題）1.求函數(shù)\(y=\sqrt{4-3x}\)自變量\(x\)的取值范圍。答：要使根式有意義，則\(4-3x\geq0\)，解不等式得\(3x\leq4\)，即\(x\leq\frac{4}{3}\)。2.函數(shù)\(y=\frac{\sqrt{x+2}}{x-1}\)中自變量\(x\)的取值范圍是多少？答：由\(x+2\geq0\)得\(x\geq-2\)，且\(x-1\neq0\)即\(x\neq1\)，所以\(x\)取值范圍是\(x\geq-2\)且\(x\neq1\)。3.若函數(shù)\(y=\sqrt{(x-3)^2}\)，當(dāng)\(x\lt3\)時(shí)，化簡(jiǎn)該函數(shù)。答：因?yàn)閈(x\lt3\)，則\(x-3\lt0\)，所以\(\sqrt{(x-3)^2}=\vertx-3\vert=3-x\)。4.已知函數(shù)\(y=\sqrt{2x-a}\)，當(dāng)\(x=3\)時(shí)函數(shù)有意義，求\(a\)的取值范圍。答：當(dāng)\(x=3\)時(shí)函數(shù)有意義，則\(2x-a=6-a\geq0\)，解得\(a\leq6\)。五、討論題（每題5分，共4題）1.討論函數(shù)\(y=\sqrt{x^2-1}\)自變量\(x\)的取值范圍對(duì)函數(shù)圖象的影響。答：由\(x^2-1\geq0\)得\(x\geq1\)或\(x\leq-1\)。取值范圍決定圖象在\(x\)軸上的分布，圖象在\(x\leq-1\)和\(x\geq1\)這兩個(gè)區(qū)間，且關(guān)于\(y\)軸對(duì)稱(chēng)。2.對(duì)于函數(shù)\(y=\frac{1}{\sqrt{x^2+1}}\)，討論其函數(shù)值的變化情況。答：因?yàn)閈(x^2\geq0\)，所以\(x^2+1\geq1\)，則\(\sqrt{x^2+1}\geq1\)，\(y=\frac{1}{\sqrt{x^2+1}}\)，當(dāng)\(x=0\)時(shí)，\(y\)有最大值\(1\)，\(\vertx\vert\)增大時(shí)，\(y\)值減小且趨近于\(0\)。3.討論函數(shù)\(y=\sqrt{3-x}+\sqrt{x-1}\)的最值情況。答：由\(3-x\geq0\)且\(x-1\geq0\)得\(1\leqx\leq3\)。\(y^2=(\sqrt{3-x}+\sqrt{x-1})^2=2+2\sqrt{(3-x)(x-1)}\)，當(dāng)\(x=2\)時(shí)，\((3-x)(x-1)\)有最大值\(1\)，此時(shí)\(y\)最大值為\(2\)；端點(diǎn)處\(x=1\)或\(x=3\)時(shí)，\(y\)最小值為\(\sqrt{2}\)。4.函數(shù)\(y=\sqrt{ax+b}\)（\(a\neq0\)），討論\(a\)、\(b\)的取值對(duì)自變量\(x\)取值范圍及函數(shù)圖象的影響。答：當(dāng)\(a\gt0\)，\(x\geq-\frac{a}\)，圖象在\(x=-\frac{a}\)右側(cè)；當(dāng)\(a\lt0\)，\(x\leq-\frac{a}\)，圖象在\(x=-\frac{a}\)左側(cè)。\(b\)影響\(x\)取值范圍的邊