Question

If af(x) + bf(1/x)=x-1 x is not equal to zero and a is not equal to b then f2=

Answer 1

Step-by-step explanation:

¥unnecessary question

Answer 2

Answer:

$Given\: x \cancel =0\:and\:a \cancel =b\:such\:that$

$\sf \blue {✎\:af \: (x) + bf( \dfrac{1}{x} ) = \dfrac{1}{x} - 5....(1)}$

$\sf \red {✭Substitute \: \dfrac{1}{x} in \: place \: of \: x \: ,we \: get}$

$\sf \red \implies \orange {af( \dfrac{1}{x}) + bf( \dfrac{1}{( \dfrac{1}{x}) } ) = \dfrac{1}{( \dfrac{1}{x}) }}$

$\sf \purple \implies \pink { af \dfrac{1}{x} + bf (x) = x - 5 ....(2) }$

$\sf \green{➯\:On \: adding \: equations \: (1) \: and \: (2) \: we \: get}$

$\sf \orange \implies \red {af(x) + bf( \dfrac{1}{x} ) + af( \dfrac{1}{x} ) + bf(x) = \dfrac{1}{x} - 5 + x - 5}$

$\sf \pink \implies \purple {af(x) + bf( x ) + af( \dfrac{1}{x} ) + bf( \dfrac{1}{x} ) = x + \dfrac{1}{x} - 10}$

$\sf \green \implies \blue {(a + b) \: f(x) + (a + b) \: f( \dfrac{1}{x} ) = x + \dfrac{1}{x} - 10}$

$\sf \red \implies \orange {(a + b) \: f (x) + f (\dfrac{1}{x} ) = x + \dfrac{1}{x} - 10}$

$\bf \therefore f(x) + f (\dfrac{1}{x} ) = \dfrac{1}{a + b} (x + \frac{1}{x} - 10)...(3)$

$\sf \purple \implies \pink {af(x) + bf( \dfrac{1}{x} ) - af( \dfrac{1}{x}) + bf(x) = \dfrac{1}{x} - 5 - (x - 5)}$

$\sf \blue \implies \green {af(x) + bf( \dfrac{1}{x} ) - af( \dfrac{1}{x} ) - bf( x ) = \dfrac{1}{x} - 5 - x + 5}$

$\sf \orange \implies \red {af(x) - bf(x) - af( \dfrac{1}{x} ) + bf( \dfrac{1}{x} ) = \dfrac{1}{x} - x}$

$\sf \pink \implies \purple {(a + b) \: f(x) - (a - b) \: f (\dfrac{1}{x} ) = \dfrac{1}{x} - x}$