Question

if for non-zero x,
af(x) + bf(1/x)=1/x-5
where a not equal to b, then f (2) =?​

Answer 1

Given:

For non-zero x,

af(x) + bf(1/x)=1/(x-5)  , where a ≠ b

To Find:

The value of f at x  = 2 , that is f ( 2 ).

Solution :

Here we can solve the value of f(2) by converting this expression into a linear system of two equations.

   ( i ) Let x =2 .

• a f ( 2 ) + b f ( 1/2) = 1/(2-5) = -1/3 --(( 1 ))

    ( i i ) Let x = 1/2

• a f ( 1/2 ) + b f (2 ) = 1/(1/2 - 5 ) = -2/9 --((2))

Now similar to solving a linear equation of x and y ,

• Multiply (( 1 )) by a
• a²f(2) + abf(1/2) = -a/3 -- ((3))

• and multiply (( 2 )) by b .

• abf(1/2) + b²f(2) = -2b/9 --((4))

Subtracting (( 4 )) from (( 3 )) eliminates f(1/2) and gives f(2) .

• ((4 )) - (( 3 ))

• f ( 2 ) ( b² -  a² ) = -2b/9 + a/3 = ( 3a - 2b )/9
• f( 2 ) = (3a - 2b )/9(b² - a²)

Therefore value of f(2) is equal to  (3a - 2b )/9(b² - a²)

Answer 2

Answer:

$\sf \purple {✎\:Given \: x \cancel = 0 \: and \: a \cancel = b \: such \: that}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\bf \therefore f(x) - f( \dfrac{1}{x} ) = \dfrac{1}{a - b} ( \dfrac{1}{x} - x)...(4)$

$\sf \blue {✯\:On \: adding \: equations \: (3) \: and \: (4)}$

$\sf \green{we \: get}$

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

$\sf \purple \implies \pink {2f(x) = \dfrac{(a - b)(x + \dfrac{1}{x} - 10 ) + (a + b)( \dfrac{1}{x} - x) }{(a + b)(a - b)}}$

$\sf \green \implies \blue{2f(x) = \dfrac{1}{ {a}^{2} - {b}^{2} } (a - b)x + \dfrac{(a - b)}{x} - 10(a - b) + \dfrac{(a + b)}{x} - (a + b)x}$

$\sf \orange \implies \red {2f(x) = \dfrac{1}{ {a}^{2} - {b}^{2} } (a - b - a - b)x + \dfrac{(a - b + a + b)}{x} - 10(a - b)}$

$\sf \pink \implies \purple {2f(x) = \dfrac{1}{ {a}^{2} - {b}^{2} } - 2bx + \dfrac{2ax } {x} - 10(a - b)}$

$\sf \blue \implies \green {2f(x) = \dfrac{2}{ {a}^{2} - {b}^{2} } - bx + \dfrac{a}{x} - 5(a - b)}$

$\sf \red \implies \orange {f(x) = \dfrac{1}{ {a}^{2} - {b}^{2} } - bx + \dfrac{a}{x} - 5(a - b)}$

$\sf \purple \implies \pink {f(x) = \dfrac{1}{ {a}^{2} - {b}^{2} } - bx + \dfrac{a}{x} - \dfrac{5(a - b)}{ {a}^{2} - {b}^{2} }}$

$\sf \green \implies \blue {f(x) = \dfrac{1}{ {a}^{2} - {b}^{2} } - bx + \dfrac{a}{x} - \dfrac{5(a - b)}{( a+b )( a-b )}}$

$\bf \therefore \: f(x) = \dfrac{1}{ {a}^{2} - {b}^{2} } \dfrac{a}{x} - bx - \dfrac{5}{a + b}$

$\sf \red{✮\:Thus}$

$\sf \orange {☆\: f(x) = \dfrac{1}{ {a}^{2} - {b}^{2} } \dfrac{a}{x} - bx - \dfrac{5}{a + b} }$