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Question Number 189053 by mathlove last updated on 11/Mar/23

f i n d f (x)

1 : f (\frac{x + 1}{x - 1}) = x + 3; x \neq 1

2 : f (\frac{2 x + 1}{x - 1}) = x^{2} + 2 x; x \neq 1

3 : f (x + 1) + f (x - y) = 2 f (x) c o s y \forall x, y

f (0) = f (\frac{π}{2}) = 1

Answered by cortano12 last updated on 11/Mar/23

(1) f (\frac{x + 1}{x - 1}) = x + 3, x \neq 1

let u = \frac{x + 1}{x - 1} \Rightarrow x = \frac{u + 1}{u - 1}

f (u) = \frac{u + 1}{u - 1} + 3 = \frac{4 u - 2}{u - 1}

∴ f (x) = \frac{4 x - 2}{x - 1}, x \neq 1

Answered by aleks041103 last updated on 11/Mar/23

2 .

\frac{2 x + 1}{x - 1} = y

2 x + 1 = y x - y

(2 - y) x = - y - 1 \Rightarrow x = \frac{y + 1}{y - 2}

f (y) = x^{2} + 2 x + 1 - 1 = {(x + 1)}^{2} - 1 =

= \frac{{(y + 1)}^{2}}{{(y - 2)}^{2}} - 1 = \frac{y^{2} + 2 y + 1 - y^{2} + 4 y - 4}{y^{2} - 4 y + 4} =

= \frac{6 y - 3}{{(y - 2)}^{2}} = \frac{3 (2 y - 1)}{{(y - 2)}^{2}}

\Rightarrow f (x) = \frac{3 (2 x - 1)}{{(x - 2)}^{2}}

Answered by aleks041103 last updated on 11/Mar/23

3 .

x = - 1 \Rightarrow f (0) + f (- y - 1) = 2 f (- 1) c o s (y)

1 + f (z) = 2 f (- 1) c o s (- z - 1)

f (z) = 2 f (- 1) c o s (z + 1) - 1

\Rightarrow f (- 1) = 2 f (- 1) - 1 \Rightarrow f (- 1) = 1

\Rightarrow f (z) = 2 c o s (z + 1) - 1

b u t f (π / 2) = 2 c o s (\frac{π}{2} + 1) - 1 \neq 1

\Rightarrow t h e r e i s n o s o l u t i o n