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f(x) satisfies the relation f(x)-lamda int_(0)^(pi//2)sinxcostf(t)dt=sinx If f(x)=2 has the least one real root, then |
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Answer» `LAMDA EPSILON[1,4]` or `f(x)-lamda sinx int_(0)^(pi//2) cost f(t)dt=sinx` or `f(x)=Asinx=sinx` or `f(x)=(A+1)sinx` where `A=lamdaint_(0)^(pi//2) cost f(t)dt` or `A=lamda int_(0)^(pi//2) cos (A+1)sin dt` `=(lamda(A+1))/2int_(0)^(pi//2) sin 2tdt` `=(lamda(A+1))/2[(-cos 2t)/2]_(0)^(pi//2)` `=(lamda(A+1))/2` `:. A=(lamda)/(2-lamda)` `:.f(x)=((lamda)/(2-lamda)+1)sinx=(2/(2-lamda))sinx` `(2/(2-lamda))sinx=2` or `sinx=(2-lamda)` or `|2-lamda|le1` or `-1le lamda-2le 1` or `1 le lamda le 3` `int_(0)^(pi//2) f(x)dx=3` or `int_(0)^(pi//2) 2/(2-lamda) sinxdx=3` or `-[2/(2-lamda) cosx]_(0)^(pi//2) =3` or `2/(2-lamda)=3` |
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