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-If `a,b,c in R` then prove that the roots of the equation `1/(x-a)+1/(x-b)+1/(x-c)=0` are always real cannot have roots if `a=b=c`A. One roots lies in (a,b)B. One root lies in (b,c)C. One root must be non-realD. Three roots are real. |
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Answer» Correct Answer - A::B::D Simplifying `x(x-b)(x-c)+x(x-c)(x-a)+x(x-a)(x-b)-(x-a)(x-b)(x-c)=0` let `f(x)=x(x-b)(x-c)+x(x-c)(x-a)+x(x-a)(x-b)-(x-a)(x-b)(x-c)` `becausef(a)=a(a-b)(a-c)gt0` `f(b)=b(b-c)(b-a)gt0` `f(x)=c(c-a)(c-b)gt0` (assuming `altbltc)` If two roots are real, then the polynomial of degree three has the third root which must be real. |
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