Question

please help me to solveit​

Answer 1

Answer:

This is the answer of your question.

Answer 2

$\large\underline{\sf{Given \:Question - }}$

If a function satisfies the relation

$\rm :\longmapsto\:2f(x) + f\bigg[\dfrac{1}{x} \bigg] = - \: x \: \: \: x \ne \: 0$

then f(2) equals

$\purple{\large\underline{\sf{Solution-}}}$

Given function is

$\rm :\longmapsto\:2f(x) + f\bigg[\dfrac{1}{x} \bigg] = - \: x - - - (1)$

Now, change

$\red{ \boxed{ \sf{ \: \:x \: \to \: \frac{1}{x} \: \: we \: get \: \: }}}$

$\rm :\longmapsto\:2f\bigg[\dfrac{1}{x} \bigg] + f(x) = - \: \dfrac{1}{x} - - - - (2)$

On multiply equation (1) by 2, we get

$\rm :\longmapsto\:4f(x) +2 f\bigg[\dfrac{1}{x} \bigg] = - \:2 x - - - (3)$

On Subtracting, equation (2) from (3), we get

$\rm :\longmapsto\:3f(x) =\dfrac{1}{x} - \:2 x$

$\rm :\longmapsto\:3f(x) =\dfrac{1 - 2 {x}^{2} }{x}$

$\rm :\longmapsto\:f(x) =\dfrac{1 - 2 {x}^{2} }{3x}$

Hence, On substituting x = 2, we get

$\rm :\longmapsto\:f(2) =\dfrac{1 - 2 {(2)}^{2} }{3(2)}$

$\rm :\longmapsto\:f(2) =\dfrac{1 - 8 }{6}$

$\bf :\longmapsto\:f(2) =\dfrac{ - 7 }{6}$