Question

高一数学谢谢

第19题谢谢

Answer 1

(1)解:根号3sin2x ＋cos2x
=2[(根号3/2)sin2x +(1/2)cos 2x ]
=2(sin 2x cos 30°+cos 2x sin 30°)
上一行的30°可以替换成兀/6
=2sin (2x+兀/6)
A ＝2,,,,,W =2，，，φ=兀/6
(2)解:令2x +兀/6等于-兀/2+2k兀
2x等于负3兀/6+2k兀减兀/6
x =负3兀/12+k 兀减兀/12
x 等于负2兀/12+k 兀
x 等于-兀/6+k 兀
增加半个周期
x 加兀/2=-兀/6+k 兀加兀/2
=-兀/6＋k兀+3兀/6
=兀/3+k兀
f (x )增区间是[-兀/6+k兀，兀/3+k兀]
f (x )减区间是[兀/3+k兀，5兀/6＋k兀]
当x 属于[-兀/6，兀/3]时
f (x )最小值是x =-兀/6时是f (-兀/6)＝-f (兀/6)=-1
f (x )最大值是x =兀/3时是f (兀/3)＝1

Answer 2

f(x) = √3sin2x + cos2x
= 2(√3/2 sin2x + 1/2 cos2x)
= 2(cosπ/6 sin2x + sinπ/6 cos2x)
= 2sin(2x +π/6)
A = 2 , ω = 2 , ψ = π/6
x ∈ [-π/6 , π/3]
2x +π/6 ∈ [-π/6 , 5π/6]
f(x) = 2sin(2x +π/6) ∈ [-1/2 , 1]
x = -π/6 时，f(x) 取得最小值。此时： f(x) = -1/2
x = π/2 时，f(x)取得最大值。此时：f(x) = 1

Answer 3

利用公式转化