Question-206746

Question Number 206746 by sonukgindia last updated on 23/Apr/24

Answered by A5T last updated on 23/Apr/24

f (2) + 2 f (- 1) = 2 \dots (i)

f (- 1) + 2 f (\frac{1}{2}) = - 1 \dots (i i)

f (\frac{1}{2}) + 2 f (2) = \frac{1}{2} \dots (i i i)

2 \times (i i i) - (i i) : 4 f (2) - f (- 1) = 2 \dots (i v)

2 \times (i v) + (i) : 9 f (2) = 6 \Rightarrow f (2) = \frac{2}{3}

Answered by mr W last updated on 24/Apr/24

f (x) + 2 f (\frac{1}{1 - x}) = x \dots (i)

f (\frac{1}{1 - x}) + 2 f (\frac{x - 1}{x}) = \frac{1}{1 - x} \dots (i i)

f (\frac{x - 1}{x}) + 2 f (x) = \frac{x - 1}{x} \dots (i i i)

(i i i) i n t o (i i) :

f (\frac{1}{1 - x}) + 2 [\frac{x - 1}{x} - 2 f (x)] = \frac{1}{1 - x}

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

t h i s i n t o (i) :

f (x) + 2 [4 f (x) + \frac{1}{1 - x} - \frac{2 (x - 1)}{x}] = x

\Rightarrow f (x) = \frac{1}{9} [x - \frac{2}{1 - x} + \frac{4 (x - 1)}{x}]

f (2) = \frac{1}{9} [2 - \frac{2}{1 - 2} + \frac{4 (2 - 1)}{2}] = \frac{2}{3} ✓

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