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If `phi(x)=lim_(x->oo)(x^(2n)(f(x)+g(x)))/(1+x^(2n))` then which of the following is correctA. `phi (x)=g(x) "for all"x in R`B. `phi (x)=f(x)"for all" x in R `C. `phix={(g(x)"for" -1ltxlt1,,),(f(x)"for"|x|ge1,,):}`D. `phix={(g(x)"for" |x|lt1,,),(f(x)"for"|x|gt1,,),((f(x)+g(x))/(2)"for"|x|=1,,):}` |
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Answer» Correct Answer - D We have, `lim_(xtooo)x^(2n)={(0if|x|lt1,,),(ooif|x|gt1,,),(1if|x|=1,,):}` Thus, we have the following cases: ,Brgt CASE I When `-1lt x lt 1` In this case, we have `lim_(x^2n=0`. ` therefore phi(x)=lim_(nto oo) (x^2nf(x)+g(x))/(1+x^2x)=g(x)` CASE II When `|x|gt 1` In this case, we have `lim_(nto oo) (1)/(x^2n)=0` `therefore phi(x)=lim_(xtooo) (x^2nf(x)+g(x))/(1+x^2n)` `rArr phi(x)=lim_(xtooo) (f(x)+(g(x))/(x^(2n)))/(1+(1)/(x^(2n)))=(f(x)+0)/(1+0) =f(x)` CASE III When `|x|=1` In this case, we have `x^(2n)=1`. `therefore phi(x)=(f(x)+g(x))/(2)` . |
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