数学题求帮一下忙,各位学霸看看
1、f(x) = - [ 1 - 2(sinx)^2 ] + 1 + (b/2) * 2sinxcosx
= 1 - cos(2x) + (b/2)sin(2x) f(π/6)
= 1 - cos(π/3) + (b/2)sin(π/3) = 2
b = 2( 1 + 1/2 )/(√3/2) = 2√3;
f(x) = 1 + 2[ (√3/2)sin(2x) - (1/2)cos(2x) ]
= 1 + 2[ cos(π/6)sin(2x) - sin(π/6) cos(2x)
= 1 + 2sin( 2x - π/6 )
周期 T = 2π/2 = π;
故图像对称中心坐标为 ( kπ/2 + π/12,1 ), 对称轴 x = kπ/2 + π/3,k∈Z。
2、a = -3 时,g(x) = 2sin( 2x - π/6 ) - 2,图像在 x 轴以下,最大点切于x 轴,故 a > -3;
g(π/2) = 2sin( π - π/6 ) + 1 + a
= 2sin(π/6) + 1 + a
= 2 + a ≤ 0,a ≥ -2;
a 的取值范围是 -3 < a ≤ -2 。
f(x)=2sin²x+bsinxcosx
f(π/6)=2(1/2)²+(b/2)*(1/2)√3
1/2+(b/4)√3=2,(b/4)√3=3/2
√3*b=6,b=2√3★
f(x)=1-cos(2x)+√3*sin(2x)=2sin(2x-π/6)
sin(2x-π/6)=0,2x-π/6=πk,k是整数。
2x=π(k+1/6),x=πk/2+π/12
对称中心(πk/2+π/12,0)★
2*0-π/6=-π/6,2sin(-π/6)=-1
2*π/2-π/6=5π/6,2sin(5π/6)=1
∴-2<a≤-1★
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