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        1. 設(shè)二次函數(shù)f(x)=x2+bx+c(b,c∈R),已知不論α.β為何實(shí)數(shù)恒有f(sinα)≥0和f(2+cosβ)≤0.(1)求證:b+c=-1,(2)求證c≥3,(3)若函數(shù)f(sinα)的最大值為8.求b.c的值. 查看更多

           

          題目列表(包括答案和解析)

          設(shè)二次函數(shù)f(x)=x2+bx+c(b,c∈R),已知不論αβ為何實(shí)數(shù)恒有f(sinα)≥0和f(2+cosβ)≤0。

          (1)求證: b+c=-1;

          (2)求證c≥3;

          (3)若函數(shù)f(sinα)的最大值為8,求b,c的值.

          查看答案和解析>>

          設(shè)二次函數(shù)f(x)=x2+bx+c(b,c∈R),已知不論α、β為何實(shí)數(shù)恒有f(sinα)≥0和f(2+cosβ)≤0

          (Ⅰ)求證:b+c=-1;

          (Ⅱ)求證:c≥3;

          (Ⅲ)若函數(shù)f(sinα)的最大值為8,求b,c的值.

          查看答案和解析>>

          設(shè)二次函數(shù)f(x)=x2+bx+c(b,c∈R),已知不論α、β為何實(shí)數(shù)恒有f(sinα)≥0和f(2+cosβ)≤0。
          (1)求證: b+c=-1;
          (2)求證c≥3;
          (3)若函數(shù)f(sinα)的最大值為8,求b,c的值.

          查看答案和解析>>

          設(shè)二次函數(shù)f(x)=x2+bx+c(b,c∈R),已知不論α,β為何實(shí)數(shù),恒有f(sinα)≥0,且f(2+cosβ)≤0.

          (1)求證:b+c=-1;

          (2)求c的取值范圍;

          (3)若函數(shù)f(sinα)的最大值為8,求b,c的值.

          查看答案和解析>>

          設(shè)二次函數(shù)f(x)=x2+bx+c(b,c∈R),已知不論α、β為何實(shí)數(shù),恒有f(sinα)≥0和f(2+cosβ)≤0.

          (1)求證:b+c=-1;

          (2)求證:c≥3;

          (3)若函數(shù)f(sinα)的最大值為8,求b、c的值.

          查看答案和解析>>

          難點(diǎn)磁場(chǎng)

          證明:若x>0,則α+β6ec8aac122bd4f6eα、β為銳角,∴0<6ec8aac122bd4f6eαβ6ec8aac122bd4f6e;0<6ec8aac122bd4f6eβ6ec8aac122bd4f6e,∴0<sin(6ec8aac122bd4f6eα)<sinβ.0<sin(6ec8aac122bd4f6eβ)<sinα,∴0<cosα<sinβ,0<cosβ<sinα,∴0<6ec8aac122bd4f6e<1,0<6ec8aac122bd4f6e<1,∴f(x)在(0,+∞)上單調(diào)遞減,∴f(x)<f(0)=2.若x<0,α+β6ec8aac122bd4f6e,∵α、β為銳角,0<β6ec8aac122bd4f6eα6ec8aac122bd4f6e,0<α6ec8aac122bd4f6eβ6ec8aac122bd4f6e,0<sinβ<sin(6ec8aac122bd4f6eα),∴sinβ<cosα,0<sinα<sin(6ec8aac122bd4f6eβ),∴sinα<cosβ,∴6ec8aac122bd4f6e>1, 6ec8aac122bd4f6e>1,

          f(x)在(-∞,0)上單調(diào)遞增,∴f(x)<f(0)=2,∴結(jié)論成立.

          殲滅難點(diǎn)訓(xùn)練

          一、1.解析:函數(shù)y=-xcosx是奇函數(shù),圖象不可能是A和C,又當(dāng)x∈(0, 6ec8aac122bd4f6e)時(shí),

          y<0.

          答案:D

          2.解析:f(x)=cos2x+sin(6ec8aac122bd4f6e+x)=2cos2x-1+cosx

          =2[(cosx+6ec8aac122bd4f6e]-1.

          答案:D

          二、3.解:在[-π,π]上,y=|c(diǎn)osx|的單調(diào)遞增區(qū)間是[-6ec8aac122bd4f6e,0]及[6ec8aac122bd4f6e,π].而f(x)依|c(diǎn)osx|取值的遞增而遞減,故[-6ec8aac122bd4f6e,0]及[6ec8aac122bd4f6e,π]為f(x)的遞減區(qū)間.

          4.解:由-6ec8aac122bd4f6eωx6ec8aac122bd4f6e,得f(x)的遞增區(qū)間為[-6ec8aac122bd4f6e,6ec8aac122bd4f6e],由題設(shè)得

          6ec8aac122bd4f6e

          三、5.解:(1)∵-1≤sinα≤1且f(sinα)≥0恒成立,∴f(1)≥0

          ∵1≤2+cosβ≤3,且f(2+cosβ)≤0恒成立.∴f(1)≤0.

          從而知f(1)=0∴b+c+1=0.

          (2)由f(2+cosβ)≤0,知f(3)≤0,∴9+3b+c≤0.又因?yàn)?i>b+c=-1,∴c≥3.

          (3)∵f(sinα)=sin2α+(-1-c)sinα+c=(sinα6ec8aac122bd4f6e)2+c-(6ec8aac122bd4f6e)2,

          當(dāng)sinα=-1時(shí),[f(sinα)]max=8,由6ec8aac122bd4f6e解得b=-4,c=3.

          6.解:如圖,設(shè)矩形木板的長(zhǎng)邊AB著地,并設(shè)OA=x,OB=y,則a2=x2+y2-2xycosα≥2xy-2xycosα=2xy(1-cosα).

          6ec8aac122bd4f6e

          ∵0<απ,∴1-cosα>0,∴xy6ec8aac122bd4f6e (當(dāng)且僅當(dāng)x=y時(shí)取“=”號(hào)),故此時(shí)谷倉的容積的最大值V1=(6ec8aac122bd4f6exysinα)b=6ec8aac122bd4f6e.同理,若木板短邊著地時(shí),谷倉的容積V的最大值V2=6ec8aac122bd4f6eab2cos6ec8aac122bd4f6e,

          ab,∴V1V2

          從而當(dāng)木板的長(zhǎng)邊著地,并且谷倉的底面是以a為底邊的等腰三角形時(shí),谷倉的容積最大,其最大值為6ec8aac122bd4f6ea2bcos6ec8aac122bd4f6e.

          7.解:如下圖,扇形AOB的內(nèi)接矩形是MNPQ,連OP,則OP=R,設(shè)∠AOP=θ,則

          QOP=45°-θ,NP=Rsinθ,在△PQO中,6ec8aac122bd4f6e

          6ec8aac122bd4f6e

          PQ=6ec8aac122bd4f6eRsin(45°-θ).S矩形MNPQ=QP?NP=6ec8aac122bd4f6eR2sinθsin(45°-θ)=6ec8aac122bd4f6eR2?[cos(2θ-45°)-6ec8aac122bd4f6e]≤6ec8aac122bd4f6eR2,當(dāng)且僅當(dāng)cos(2θ-45°)=1,即θ=22.5°時(shí),S矩形MNPQ的值最大且最大值為6ec8aac122bd4f6eR2.

          工人師傅是這樣選點(diǎn)的,記扇形為AOB,以扇形一半徑OA為一邊,在扇形上作角AOP且使∠AOP=22.5°,P為邊與扇形弧的交點(diǎn),自PPNOANPQOAOBQ,并作OMOAM,則矩形MNPQ為面積最大的矩形,面積最大值為6ec8aac122bd4f6eR2.

          8.解:∵在[-6ec8aac122bd4f6e]上,1+sinx>0和1-sinx>0恒成立,∴原函數(shù)可化為y=

          log2(1-sin2x)=log2cos2x,又cosx>0在[-6ec8aac122bd4f6e]上恒成立,∴原函數(shù)即是y=2log2cosx,在x∈[

          6ec8aac122bd4f6e]上,6ec8aac122bd4f6e≤cosx≤1.

          ∴l(xiāng)og26ec8aac122bd4f6e≤log2cosx≤log21,即-1≤y≤0,也就是在x∈[-6ec8aac122bd4f6e]上,ymax=0,

          ymin=-1.

          6ec8aac122bd4f6e

          綜合上述知,存在6ec8aac122bd4f6e符合題設(shè).

           

           

           


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