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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-

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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.
$$\mathrm{Let}\:{f}\left({x}\right)\:=\:{x}^{\mathrm{3}} \:+\:\mathrm{3}{x}^{\mathrm{2}} \:+\:\mathrm{9}{x}\:+\:\mathrm{6sin}{x},\:\mathrm{then} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation} \\ $$$$\frac{\mathrm{1}}{{x}\:−\:{f}\left(\mathrm{1}\right)}\:+\:\frac{\mathrm{2}}{{x}\:−\:{f}\left(\mathrm{2}\right)}\:+\:\frac{\mathrm{3}}{{x}\:−\:{f}\left(\mathrm{3}\right)}\:=\:\mathrm{0}. \\ $$
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.
$${let}\:{f}\left(\mathrm{1}\right)={a}\:,\:{f}\left(\mathrm{2}\right)={b}\:,\:{f}\left(\mathrm{3}\right)={c} \\ $$$${given}\:{equation}\:{implies}: \\ $$$${x}^{\mathrm{2}} −\left({b}+{c}\right){x}+{bc}+\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}\left({c}+{a}\right){x}+\mathrm{2}{ca} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:+\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}\left({a}+{b}\right){x}+\mathrm{3}{ab}=\mathrm{0} \\ $$$$\mathrm{6}{x}^{\mathrm{2}} −\left(\mathrm{5}{a}+\mathrm{4}{b}+\mathrm{3}{c}\right){x}+\left(\mathrm{3}{ab}+{bc}+\mathrm{2}{ca}\right)=\mathrm{0} \\ $$$${D}=\left(\mathrm{5}{a}+\mathrm{4}{b}+\mathrm{3}{c}\right)^{\mathrm{2}} −\mathrm{24}\left(\mathrm{3}{ab}+{bc}+\mathrm{2}{ca}\right) \\ $$$$\:\:=\mathrm{25}{a}^{\mathrm{2}} +\mathrm{16}{b}^{\mathrm{2}} +\mathrm{9}{c}^{\mathrm{2}} −\mathrm{32}{ab}−\mathrm{18}{ca} \\ $$$$\:\:=\mathrm{16}\left({a}−{b}\right)^{\mathrm{2}} +\mathrm{9}\left({c}−{a}\right)^{\mathrm{2}} \:>\:\mathrm{0} \\ $$$$\Rightarrow\:\boldsymbol{{D}}\:>\:\mathrm{0}\:,\:{Hence}\:{two}\:{real}\:{roots}.\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by Tinkutara last updated on 27/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
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.
$$\because{f}\left(\mathrm{1}\right)\neq{f}\left(\mathrm{2}\right)\neq{f}\left(\mathrm{3}\right) \\ $$$$\therefore{a}\neq{b}\neq{c} \\ $$$$\Rightarrow\boldsymbol{{D}}\neq\mathrm{0} \\ $$$$\Rightarrow{two}\:{distinct}\:{real}\:{roots}. \\ $$

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