Question

Find this answer with working pls

Answer 1

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

$\rm :\longmapsto\:f(x) = 6x - 5$

$\rm :\longmapsto\:g(x) = \dfrac{x + k}{2}$

and

$\rm :\longmapsto\:g(f(x)) = f(g(x))$

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\:value \: of \: k$

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

Given that ,

$\rm :\longmapsto\:f(x) = 6x - 5$

$\rm :\longmapsto\:g(x) = \dfrac{x + k}{2}$

Now,

Consider,

$\red{\bf :\longmapsto\:f(g(x))}$

$\: \rm \: = \: \:f\bigg(\dfrac{x + k}{2} \bigg)$

$\: \rm \: = \: \:6\bigg(\dfrac{x + k}{2} \bigg) - 5$

$\: \rm \: = \: \:3x + 3k - 5$

$\red{\bf :\longmapsto\:f(g(x)) = 3x + 3k - 5 - - - (1)}$

Consider,

$\green{\bf :\longmapsto\:g(f(x))}$

$\: \rm \: = \: \:g(6x - 5)$

$\: \rm \: = \: \:\dfrac{6x - 5 + k}{2}$

$\green{\bf :\longmapsto\:g(f(x)) = \dfrac{6x + k - 5}{2} - - - (2)}$

Now,

It is given that

$\blue{\bf :\longmapsto\:g(f(x)) = f(g(x))}$

So, On substituting the values from equation (1) and (2), we get

$\rm :\longmapsto\:\dfrac{6x + k - 5}{2} = 3x + 3k - 5$

$\rm :\longmapsto\:6x + k - 5 = 6x + 6k - 10$

$\rm :\longmapsto\:k - 5 = 6k - 10$

$\rm :\longmapsto\:k - 6k = - 10 + 5$

$\rm :\longmapsto\: - 5k = - 5$

$\bf\implies \:k = 1$

Additional information :-

$\rm :\longmapsto\:If \: f(g(x)) = x \: then \: g(x) = {f}^{ - 1}(x)$

$\rm :\longmapsto\:If \: g(f(x)) = x \: then \: f(x) = {g}^{ - 1}(x)$