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Consider the function f(theta)=int_(0)^(1)(|sqrt(1-x^(2))-sintheta|)/(sqrt(1-x^(2)))dx, where 0le theta le (pi)/2, then |
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Answer» `f_("min")=sqrt(2)-1` Put `x=COS phi` `:.f(theta)=int_(0)^((pi)/2)|sinphi-sin theta|d phi` `=int_(0)^(theta)(sin theta-sin phi)d phi +int_(theta)^((pi)/2)(sin phi-sin theta)d phi` `=[phi sin theta+cos phi]_(0)^(theta)+[-cos phi-phi sin theta]_(theta)^(pi//2)` `=[theta sin theta+cos theta-0-1]-0-(pi)/2 sin theta +cos theta + theta sin theta` `=2(theta sin theta +cos theta)-(pi)/2 sin theta-1` `f'(theta)=2(theta cos theta+sin theta -sin theta)-(pi)/2 cos theta` `=2 (theta- (pi)/4) . cos theta` `f'(theta)=0` `:. theta =(pi)/4`, which is point of minima. `:. f((pi)/4)=2((pi)/4 1/(sqrt(2))+1/(sqrt(2)))-(pi)/2 1/(2sqrt(2))-1=sqrt(2)-1` `:. f(x)=1` and `f((pi)/2)=(pi)/2-1` So `f_("min")=sqrt(2)-1` and `f_("max")=1` |
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