【已知向量a=(2cos(x/2),tan(x/2+π/4)),b=(根号2sin(x/2+π/4),tan(x/2-π/4)),令f(x)=a×b,是否存在实数x∈[0,π]使f(x)+f’(x)=0,其中f’(x)是f(x)的导函数?若存在,则求出x的值,若不存在,则证明之】
已知向量a=(2cos(x/2),tan(x/2+π/4)),b=(根号2sin(x/2+π/4),tan(x/2-π/4)),令f(x)=a×b,是否存在实数x∈[0,π]使f(x)+f’(x)=0,其中f’(x)是f(x)的导函数?若存在,则求出x的值,若不存在,则证明之
f(x)=a●b=2cos(x/2)*√2*sin(x/2+π/4)+tan(x/2+π/4)*tan(x/2-π/4)
=2√2sin(x/2+π/4)*cos(x/2)+tan(x/2+π/4)*tan(x/2-π/4)
=√2(sin(x+π/4)+sin(π/4))+[cos(π/2)-cos(x+π/2)]/[cos(x+π/2)+cos(π/2)]
=√2*(sin(x)*cos(π/4)+cos(x)*sin(π/4)+1/2*√2)+[-cos(x+π/2)]/[cos(x+π/2)]
=√2*(sin(x)*1/2*√2+cos(x)*1/2*√2+1/2*√2)-1
=sin(x)+cos(x)
f’(x)=cos(x)-sin(x)
则f(x)+f’(x)=2*cos(x)=0,x=π/2
即存在实数x∈[0,π]使f(x)+f’(x)=0,且x=π/2
