備戰(zhàn)2025年高考二輪復習數(shù)學中低檔大題規(guī)范練6

規(guī)范練6(分值:43分)學生用書P2351.(13分)(2024·江蘇連云港模擬)已知函數(shù)f(x)=ex-12x2(1)求函數(shù)f(x)在x=1處的切線方程;(2)證明:?x∈[0,+∞),f(x)>sinx.(1)解由f(x)=ex-12x2-x,可得f'(x)=ex-x-1,f'(1)=e1-1-1=e-2又f(1)=e1-12×12-1=e-所以函數(shù)f(x)在x=1處的切線方程為y-e+32=(e-2)(x-1),即(e-2)x-y+12=(2)證明由f(x)=ex-12x2-x,可得f'(x)=ex-x-令h(x)=ex-x-1,可得h'(x)=ex-1.當x∈[0,+∞)時,h'(x)=ex-1≥0,所以h(x)=ex-x-1在區(qū)間[0,+∞)上單調(diào)遞增,所以h(x)≥h(0)=e0-0-1=0,即f'(x)=ex-x-1≥0,所以f(x)=ex-12x2-x在區(qū)間[0,+∞)上單調(diào)遞增所以f(x)≥f(0)=e0-12×02-0=當x=0時,f(0)=1>sin0=0,當x>0時,f(x)>1≥sinx.綜上所述,?x∈[0,+∞),f(x)>sinx.2.(15分)(2024·山東泰安二模)已知函數(shù)f(x)=12sin2x-π3,△ABC的內(nèi)角A,B,C所對的邊分別為a,b,c,且fA2=34(1)求A;(2)若sinC(1+cosB)=sinB(2-cosC),求cb的值解(1)∵fA2=34,∴12sinA-π3=3∴sinA-π3=32∵0<A<π,∴-π3<A-π∴A-π3=π3(2)sinC(1+cosB)=sinB(2-cosC),(方法一)sinC+sinCcosB+sinBcosC=2sinB,∴sinC+sin(B+C)=2sinB,∴sinC+sinA=2sinB,根據(jù)正弦定理得c+a=2b.由余弦定理得a2=b2+c2-2bccosA=b2+c2+bc,①將a=2b-c代入①式,得3b2-5bc=0,∴3b=5c,∴(方法二)c(1+a2+c2-b22∴c+a2+c2-∴c+a=2b.由余弦定理得a2=b2+c2-2bccosA=b2+c2+bc,①將a=2b-c代入①式,得3b2-5bc=0,∴3b=5c,∴3.(15分)(2024·山東泰安模擬)如圖1,在直角梯形ABCD中,AD∥BC,∠BAD=π2,AB=BC=12AD,E是AD的中點,O是AC與BE的交點.將△ABE沿BE折起到△A1BE的位置,如圖圖1圖2(1)證明:平面BCDE⊥平面A1OC;(2)若平面A1BE⊥平面BCDE,求平面A1BC與平面A1CD夾角的余弦值.(1)證明在圖1中連接CE,因為AD∥BC,AB=BC=12AD,E是AD的中點,∠BAD=π所以四邊形ABCE為正方形,四邊形BCDE為平行四邊形,所以BE⊥AC,CD∥BE.圖1圖2即在圖2中,BE⊥OA1,BE⊥OC,OA1∩OC=O,OA1,OC?平面A1OC,所以BE⊥平面A1OC.又因為BE?平面BCDE,所以平面BCDE⊥平面A1OC.(2)解因為平面A1BE⊥平面BCDE,平面A1BE∩平面BCDE=BE,A1O⊥BE,A1O?平面A1BE,所以A1O⊥平面BCDE.又因為OC?平面BCDE,所以A1O⊥OC.又由(1)知BE⊥OC,BE⊥OA1,所以直線BE,OC,OA1兩兩垂直.如圖,以O(shè)為坐標原點,分別以O(shè)B,OC,OA1所在直線為x軸、y軸、z軸建立如圖所示的空間直角坐標系,設(shè)A1B=1,所以A1B=A1E=BC=ED=1.因為BC∥ED,所以B22,0,0,E-22,0,0,A10,0,22,C0,22,0.得BC=-22,22,0,A1C=0,22,-22,CD=BE設(shè)平面A1BC的法向量n1=(x1,y1,z1),平面A1CD的法向量n2=