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`f(x)=sinx+int_(-pi//2)^(pi//2)(sinx+tcosx)f(t)dt` The value of `int_(0)^(pi//2) f(x)dx` isA. `1`B. `-2`C. `-1`D. `2` |
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Answer» Correct Answer - C `f(x)=sinx+sin int_(-pi//2)^(pi//2) f(t)dt+cosx int_(-pi//2)^(pi//2) tf(t)dt` `=sinx (1+int_(-pi//2)^(pi//2) f(t)dt)+cosx int_(-pi//2)^(pi//2) tf(t)dt` `=Asinx+Bcosx` Thus, `A=1+int_(-pi//2)^(pi//2) f(t)dt` `=1+int_(-pi//2)^(pi//2) (Asint+Bcost)dt` `=1+2Bint_(0)^(pi//2) cost dt` `=A=1+2B` ............1 `B=int_(-pi//2)^(pi//2) tf(t)dt` `=int_(-pi//2)^(pi//2)t(Asint+Bcost)dt` `=2Aint_(0)^(pi//2) t sin tdt` `=2A[-tcost+sint]_(0)^(pi//2)` `:.B=2A` From equations 1 and 2 we get `A=-1//3, B=-2//3` `:.f(x)=-1/3(sinx+2cosx)` Thus, the range fo `f(x)` is `[-(sqrt(5))/3,(sqrt(5))/3]` `f(x)=-1/2(sinx+2cosx)` `=-(sqrt(5))/3 sin (x+tan^(-1)2)` `=-(sqrt(5))/3cos(x-"tan"^(-1)1/2)` `f(x)` in invertible if `-(pi)/2 le x=tan^(-1)2le(pi)/2` or `-(pi)/2-tan^(-1)2 le xle (pi)/2 -tan^(-1) 2` or `0lex-"tan"^(-1)1/2 le pi` or `"tan"^(-1)1/2 le x le pi+"tan"^(-1)1/2` or `pi le x-"tan"^(-1)1/2le 2pi` or `x epsilon [pi+cot^(-1)2, 2pi +cot^(-1) 2]` `int_(0)^(pi//2) f(x)dx=-1/3int_(0)^(pi//2) (sinx+2cosx)dx` `=-1/3[-cosx+2sinx]_(0)^(pi//2) =-1` |
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