Given-3xf-1-x-f-x-2x-2-and-f-3-f-9-f-a-three-first-term-in-AP-respectively-Find-the-value-of-a-

Question Number 128942 by bramlexs22 last updated on 11/Jan/21

Given : 3 x f (\frac{1}{x}) + f (x) = 2 x + 2

a n d f (3), f (9), f (a) three first

term in AP respectively . Find

the value of a ?

Answered by liberty last updated on 11/Jan/21

(1) 3 xf (\frac{1}{x}) + f (x) = 2 x + 2

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

3 f (x) + xf (\frac{1}{x}) = 2 + 2 x

9 f (x) + 3 xf (\frac{1}{x}) = 6 x + 6

substract (2) & (1)

8 f (x) = 4 x + 4; f (x) = \frac{x + 1}{2}

then {\begin{cases} f (3) = 2; f (9) = 5 \\ f (a) = \frac{a + 1}{2} \end{cases}

since f (3), f (9), f (a) in AP

\Leftrightarrow 2 f (9) = f (3) + f (a)

\frac{a + 1}{2} = 10 - 2 = 8

a + 1 = 16 \Rightarrow a = 15

Answered by mathmax by abdo last updated on 11/Jan/21

3 xf (\frac{1}{x}) + f (x) = 2 x + 2 \Rightarrow \frac{3}{x} f (x) + f (\frac{1}{x}) = \frac{2}{x} + 2 \Rightarrow {\begin{cases} f (x) + 3 xf (\frac{1}{x}) = 2 x + 2 \\ \frac{3}{x} f (x) + f (\frac{1}{x}) = \frac{2}{x} + 2 \end{cases}

Δ_{s} = | \begin{matrix} 1 3 x \\ \frac{3}{x} 1 \end{matrix} | = 1 - 9 = - 8 \neq 0 \Rightarrow f (x) = \frac{| \begin{matrix} 2 x + 2 3 x \\ \frac{2}{x} + 2 1 \end{matrix} |}{- 8}

= - \frac{1}{8} (2 x + 2 - 3 x (\frac{2}{x} + 2)) = - \frac{1}{8} (2 x + 2 - 6 - 6 x) = \frac{4 x + 4}{8} = \frac{x + 1}{2}

f (3), f (9) and f (a) are in a . p \Rightarrow f (3) + f (a) = 2 f (9) \Rightarrow

2 + \frac{a + 1}{2} = 10 \Rightarrow 4 + a + 1 = 20 \Rightarrow a = 20 - 5 = 15