2个回答
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f(x) = msinx + √2 cosx , (m>0)
= √(m² + 2)sin(x+θ)
∵ √(m² + 2) = 2 (m>0)
∴ m = √2
f(x) = √2 sinx + √2 cosx
= 2(√2/2 sinx +√2/2 cosx)
= 2(sinx cos45° +sin45° cosx)
= 2 sin(x + 45°)
= 2 sin(x + π/4)
0<x<π (题目没给出括号是( ),还是[ ]的,或其他情况,我当它是( )的);
π/4 < x + π/4 < 5π/4
x + π/4 = 5π/4时,(取不到的,假设能取到)
f(x)取最小值:2 sin(5π/4) = -√2
∴f(x) > -√2
x + π/4 = π/2时
f(x)取最大值:2 sin(π/2) = 2
∴f(x) ≤ 2
f(x)值域:
(-√2 , 2]
= √(m² + 2)sin(x+θ)
∵ √(m² + 2) = 2 (m>0)
∴ m = √2
f(x) = √2 sinx + √2 cosx
= 2(√2/2 sinx +√2/2 cosx)
= 2(sinx cos45° +sin45° cosx)
= 2 sin(x + 45°)
= 2 sin(x + π/4)
0<x<π (题目没给出括号是( ),还是[ ]的,或其他情况,我当它是( )的);
π/4 < x + π/4 < 5π/4
x + π/4 = 5π/4时,(取不到的,假设能取到)
f(x)取最小值:2 sin(5π/4) = -√2
∴f(x) > -√2
x + π/4 = π/2时
f(x)取最大值:2 sin(π/2) = 2
∴f(x) ≤ 2
f(x)值域:
(-√2 , 2]
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