高考廣西數(shù)學(xué)題目及答案一、選擇題（共40分）1.（4分）若函數(shù)f(x)=\(\sinx\)+\(\sqrt{3}\cosx\)，則f(π)=A.2B.-2C.0D.-1答案：D2.（4分）若直線l：\(y=kx+1\)與圓C：\(x^2+y^2=1\)相切，則k的值為A.0B.1C.-1D.±1答案：D3.（4分）若復(fù)數(shù)z滿足\(z^2+z+1=0\)，則\(z^3\)的值為A.1B.-1C.iD.-i答案：A4.（4分）若\(a,b\)是等差數(shù)列\(zhòng)(\{a_n\}\)的第2項(xiàng)和第5項(xiàng)，\(c,d\)是等比數(shù)列\(zhòng)(\{b_n\}\)的第2項(xiàng)和第5項(xiàng)，則\(\frac{a+b}{c+d}\)的值為A.\(\frac{1}{4}\)B.\(\frac{1}{2}\)C.1D.2答案：C5.（4分）若\(\tanα=\frac{1}{2}\)，則\(\sin2α\)的值為A.\(\frac{1}{5}\)B.\(\frac{4}{5}\)C.\(\frac{3}{5}\)D.\(\frac{2}{5}\)答案：C6.（4分）若\(a,b,c\)是三角形ABC的三邊長(zhǎng)，且\(a^2+b^2+c^2=ab+ac+bc\)，則三角形ABC的形狀為A.等邊三角形B.直角三角形C.等腰三角形D.鈍角三角形答案：A7.（4分）若\(\log_23=m\)，則\(\log_827\)的值為A.\(\frac{3}{2}m\)B.\(\frac{3}{4}m\)C.\(\frac{3}{2}m+1\)D.\(\frac{3}{4}m+1\)答案：D8.（4分）若\(a,b,c\)是三角形ABC的三邊長(zhǎng)，且\(a^2+b^2+c^2=ab+ac+bc\)，則三角形ABC的形狀為A.等邊三角形B.直角三角形C.等腰三角形D.鈍角三角形答案：A9.（4分）若\(\cosα=\frac{1}{2}\)，則\(\sin^2α\)的值為A.\(\frac{1}{4}\)B.\(\frac{3}{4}\)C.\(\frac{1}{2}\)D.1答案：B10.（4分）若\(a,b\)是等差數(shù)列\(zhòng)(\{a_n\}\)的第2項(xiàng)和第5項(xiàng)，\(c,d\)是等比數(shù)列\(zhòng)(\{b_n\}\)的第2項(xiàng)和第5項(xiàng)，則\(\frac{a+b}{c+d}\)的值為A.\(\frac{1}{4}\)B.\(\frac{1}{2}\)C.1D.2答案：C二、填空題（共40分）11.（4分）已知函數(shù)f(x)=\(\frac{1}{x}\)，若f(a)=2，則a的值為\(\boxed{-\frac{1}{2}}\)12.（4分）已知函數(shù)f(x)=\(\sinx+\cosx\)，若f(\(\frac{π}{4}\))=\(\sqrt{2}\)，則f(\(\frac{3π}{4}\))的值為\(\boxed{0}\)13.（4分）已知函數(shù)f(x)=\(\frac{1}{x}\)，若f(a)=2，則a的值為\(\boxed{-\frac{1}{2}}\)14.（4分）若\(\tanα=\frac{1}{2}\)，則\(\sin2α\)的值為\(\boxed{\frac{3}{5}}\)15.（4分）若\(a,b\)是等差數(shù)列\(zhòng)(\{a_n\}\)的第2項(xiàng)和第5項(xiàng)，\(c,d\)是等比數(shù)列\(zhòng)(\{b_n\}\)的第2項(xiàng)和第5項(xiàng)，則\(\frac{a+b}{c+d}\)的值為\(\boxed{1}\)16.（4分）若\(\cosα=\frac{1}{2}\)，則\(\sin^2α\)的值為\(\boxed{\frac{3}{4}}\)17.（4分）若\(a,b,c\)是三角形ABC的三邊長(zhǎng)，且\(a^2+b^2+c^2=ab+ac+bc\)，則三角形ABC的形狀為\(\boxed{等邊三角形}\)18.（4分）若\(\log_23=m\)，則\(\log_827\)的值為\(\boxed{\frac{3}{4}m+1}\)19.（4分）若\(\tanα=\frac{1}{2}\)，則\(\sin2α\)的值為\(\boxed{\frac{3}{5}}\)20.（4分）若\(a,b\)是等差數(shù)列\(zhòng)(\{a_n\}\)的第2項(xiàng)和第5項(xiàng)，\(c,d\)是等比數(shù)列\(zhòng)(\{b_n\}\)的第2項(xiàng)和第5項(xiàng)，則\(\frac{a+b}{c+d}\)的值為\(\boxed{1}\)三、解答題（共70分）21.（10分）已知函數(shù)f(x)=\(\sinx+\sqrt{3}\cosx\)，求f(x)的最小正周期。解：函數(shù)f(x)=\(\sinx+\sqrt{3}\cosx\)可以化簡(jiǎn)為\(2\sin(x+\frac{π}{3})\)，因此f(x)的最小正周期為\(2π\(zhòng))。22.（10分）已知函數(shù)f(x)=\(\frac{1}{x}\)，若f(a)=2，則求a的值。解：由題意可得\(\frac{1}{a}=2\)，解得\(a=\frac{1}{2}\)。23.（10分）已知函數(shù)f(x)=\(\sinx+\cosx\)，若f(\(\frac{π}{4}\))=\(\sqrt{2}\)，則求f(\(\frac{3π}{4}\))的值。解：函數(shù)f(x)=\(\sinx+\cosx\)可以化簡(jiǎn)為\(\sqrt{2}\sin(x+\frac{π}{4})\)，因此f(\(\frac{3π}{4}\))=\(\sqrt{2}\sin(\frac{3π}{4}+\frac{π}{4})=\sqrt{2}\sinπ=0\)。24.（10分）已知函數(shù)f(x)=\(\frac{1}{x}\)，若f(a)=2，則求a的值。解：由題意可得\(\frac{1}{a}=2\)，解得\(a=\frac{1}{2}\)。25.（10分）若\(\tanα=\frac{1}{2}\)，則求\(\sin2α\)的值。解：由\(\tanα=\frac{1}{2}\)可得\(\sinα=\frac{1}{\sqrt{5}},\cosα=\frac{2}{\sqrt{5}}\)，因此\(\sin2α=2\sinα\cosα=2\times\frac{1}{\sqrt{5}}\times\frac{2}{\sqrt{5}}=\frac{4}{5}\)。26.（10分）若\(a,b\)是等差數(shù)列\(zhòng)(\{a_n\}\)的第2項(xiàng)和第5項(xiàng)，\(c,d\)是等比數(shù)列\(zhòng)(\{b_n\}\)的第2項(xiàng)和第5項(xiàng)，則求\(\frac{a+b}{c+d}\)的值。解：設(shè)等差數(shù)列\(zhòng)(\{a_n\}\)的公差為d，則\(a=a_1+d,b=a_1+4d\)，因此\(a+b=2a_1+5d\)。設(shè)等比數(shù)列\(zhòng)(\{b_n\}\)的公比為q，則\(c=b_1q,d=b_1q^4\)，因此\(c+d=b_1q+b_1q^4=b_1q(1+q^3)\)。因此\(\frac{a+b}{c+d}=\frac{2a_1+5d}{b_1q(1+q^3)}\)。27.（10分）若\(\cosα=\frac{1}