Given-f-x-1-1-f-x-1-f-x-f-2-2-and-2-2018-x-f-2018-dx-2-a-3-b-5-c-7-d-11-e-101-f-then-a-b-c-d-e-f-

Question Number 129009 by bramlexs22 last updated on 12/Jan/21

Given f (x + 1) = \frac{1 + f (x)}{1 - f (x)}; f (2) = 2

and \int_{2}^{2018} x . f (2018) dx = 2^{a} . 3^{b} . 5^{c} . 7^{d} . 11^{e} . 101^{f}

then a + b + c + d + e + f =__

Answered by liberty last updated on 12/Jan/21

\to {\begin{cases} x = 1 \to f (2) = \frac{1 + f (1)}{1 - f (1)} = 2; f (1) = \frac{1}{3} \\ x = 2 \Rightarrow f (3) = \frac{1 + f (2)}{1 - f (2)} = \frac{3}{- 1} = - 3 \\ x = 3 \Rightarrow f (4) = \frac{1 + f (3)}{1 - f (3)} = \frac{- 2}{4} = - \frac{1}{2} \\ x = 4 \Rightarrow f (5) = \frac{1 + f (4)}{1 - f (4)} = \frac{1}{3} \\ x = 5 \Rightarrow f (6) = \frac{1 + f (5)}{1 - f (5)} = \frac{4 / 3}{2 / 3} = 2 \end{cases}

\Rightarrow \frac{1}{3}; 2; - 3; - \frac{1}{2}; \frac{1}{3}; 2; \dots

then f (2018) = f (4 \times 504 + 2) = f (2) = 2

\int_{2}^{2018} xf (2018) dx = \int_{2}^{2018} 2 x dx = 2018^{2} - 2^{2}

= 2020 \times 2016 = 4 \times 505 \times 4 \times 504

= 2^{4} \times 5 \times 101 \times 3 \times 168

= 2^{4} \times 5 \times 101 \times 7 \times 2^{3} \times 3^{2}

= 2^{7} \times 3^{2} \times 5 \times 7 \times 101

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