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Let-f-x-x-3-3x-2-9x-6sinx-then-find-the-number-of-real-roots-of-the-equation-1-x-f-1-2-x-f-2-3-x-f-3-0-




Question Number 20367 by Tinkutara last updated on 26/Aug/17
Let f(x) = x^3  + 3x^2  + 9x + 6sinx, then  find the number of real roots of the  equation  (1/(x − f(1))) + (2/(x − f(2))) + (3/(x − f(3))) = 0.
Letf(x)=x3+3x2+9x+6sinx,thenfindthenumberofrealrootsoftheequation1xf(1)+2xf(2)+3xf(3)=0.
Answered by ajfour last updated on 27/Aug/17
let f(1)=a , f(2)=b , f(3)=c  given equation implies:  x^2 −(b+c)x+bc+2x^2 −2(c+a)x+2ca             +3x^2 −3(a+b)x+3ab=0  6x^2 −(5a+4b+3c)x+(3ab+bc+2ca)=0  D=(5a+4b+3c)^2 −24(3ab+bc+2ca)    =25a^2 +16b^2 +9c^2 −32ab−18ca    =16(a−b)^2 +9(c−a)^2  > 0  ⇒ D > 0 , Hence two real roots.
letf(1)=a,f(2)=b,f(3)=cgivenequationimplies:x2(b+c)x+bc+2x22(c+a)x+2ca+3x23(a+b)x+3ab=06x2(5a+4b+3c)x+(3ab+bc+2ca)=0D=(5a+4b+3c)224(3ab+bc+2ca)=25a2+16b2+9c232ab18ca=16(ab)2+9(ca)2>0D>0,Hencetworealroots.
Commented by Tinkutara last updated on 27/Aug/17
Thank you very much Sir!
ThankyouverymuchSir!
Commented by ajfour last updated on 27/Aug/17
yes.
yes.
Commented by $@ty@m last updated on 27/Aug/17
∵f(1)≠f(2)≠f(3)  ∴a≠b≠c  ⇒D≠0  ⇒two distinct real roots.
f(1)f(2)f(3)abcD0twodistinctrealroots.

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