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/
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
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f(x)=a●b= 2cos(x/2)* √2*sin(x/2+π/4)+ tan(x/2+π/4)* tan(x/2-π/4)
=2√2 sin(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
=2√2 sin(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
1年前
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