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} + 3 x^{2} + 9 x + 6 \sin x, then

find the number of real roots of the

equation

\frac{1}{x - f (1)} + \frac{2}{x - f (2)} + \frac{3}{x - f (3)} = 0 .

Answered by ajfour last updated on 27/Aug/17

l e t f (1) = a, f (2) = b, f (3) = c

g i v e n e q u a t i o n i m p l i e s :

x^{2} - (b + c) x + b c + 2 x^{2} - 2 (c + a) x + 2 c a

+ 3 x^{2} - 3 (a + b) x + 3 a b = 0

6 x^{2} - (5 a + 4 b + 3 c) x + (3 a b + b c + 2 c a) = 0

D = {(5 a + 4 b + 3 c)}^{2} - 24 (3 a b + b c + 2 c a)

= 25 a^{2} + 16 b^{2} + 9 c^{2} - 32 a b - 18 c a

= 16 {(a - b)}^{2} + 9 {(c - a)}^{2} > 0

\Rightarrow D > 0, H e n c e t w o r e a l r o o t s .

Commented by Tinkutara last updated on 27/Aug/17

Thank you very much Sir!

Commented by ajfour last updated on 27/Aug/17

y e s .

Commented by $@ty@m last updated on 27/Aug/17

∵ f (1) \neq f (2) \neq f (3)

∴ a \neq b \neq c

\Rightarrow D \neq 0

\Rightarrow t w o d i s t i n c t r e a l r o o t s .

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