InterviewSolution
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InterviewSolution
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If the expression`([s in(x/2)+cos(x/2)-i t a n(x)])/([1+2is in(x/2)])`is real, then the set of all possible values of `x`is......... |
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Answer» Correct Answer - x = `2 n pi + 2 alpha , alpha = tan^(-1) k ` , where k `in (1 ,2 )` or x = 2 `n pi` `((sin x/2+cos x/2)- I tan x )/(1+2 i sin x/2)in R` `=((sin"" x/2+ cos""x/2-i tanx)(1-2 i sin ""x/2))/(1+4 sin^2""x/2)` Since , it is real so imaginary part will be zero `therefore -2 sin ""x/2(sin ""x/2+ cos ""x/2)-tan x=0` `rArr -2 sin ""x/2(sin ""x/2+ cos ""x/2)cos x+2 sin ""x/2cos ""x/2=0` `rArr sin ""x/2[(sin""x/2cos ""x/2)(cos^2""x/2- sin ^2""x/2)+cos""x/2]=0` `therefore sin ""x/2=0` `rArr x= 2 npi` or `(sin ""x/2+cos ""x/2)(cos^2""x/2- sin ^2""x/2)+cos""x/2=0` on dividing by `cos^3""x/2` we get `(tan ""x/2+1)(1-tan^2""x/2)+(1+tan^2""x/2)=0` `rArr tan^3""x/2-tan""x/2-2=0` Let `tan""x/2=t` and `f(t)=t^3-t-2` then `f(1)=-2 lt 0` and `f(2)=4 gt 0 ` Thus f(t) changes sign from negative to positve in the interval (1,2) `therefore` Let t= k be the root for which f(k)=0 and `k in (1,2)` `therefore` t=k or `tan ""x/2=k = tan alpha` `rArr x//2 = npi+alpha` `rArr {:{(x=2npi+2alpha ","alpha=tan^(-1)k",where k" in "(1,2)"),(" "or x=2npi):}` |
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