导数数以及三角函数题
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(11)

f(x)=acoswx-sinwx central symmetric about M( π/3,0), w>0

min f(x) = f(π/6)

1 possible value of a+w

Solution:

f(x)=acoswx-sinwx

f'(x) = -awsinwx- wcoswx

f'(π/6) = 0

-asin(πw/6)- cos(πw/6) =0 (1)

f(x)=acoswx-sinwx central symmetric about M(π/3,0)

f(0) = a

f(2π/3) = -f(0)

f(2π/3) = acos(2πw/3)-sin(2πw/3) =-a

a(1+ cos(2πw/3)) = sin(2πw/3)

a = tan(πw/3) (2)

sub (2) into (1)

sin(πw/6)sin(πw/3)+ cos(πw/6)cos(πw/3) =0

cos(πw/6) = 0

πw/6 = π/2, 3π/2,...

w = 3, 9, .

from (1)

-awsin(πw/6)- wcos(πw/6) =0

w=3

a=0

1 possible value a+w = 0+3 =3

(12)

y=f(x)=a^x ,( a>0, a≠1), symmetric about y=x

g(x) = f(x)[f(x)+f(2)-1]

y=g(x) is increasing [1/2,2]

Find: range of a

Solution:

y=f(x)=a^x ,( a>0, a≠1), symmetric about y=x

我不懂,不过我觉得题目有点问题.

y=f(x)=a^x ,( a>0, a≠1), 不可能对称于 y=x

因为

(0,1) 属于y=f(x)

(0,1)的相关对,y=x, 的称点是(1,0)

但是(1,0)不属于y=f(x)?

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