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**Hint:**AS we know that the above question consists of a functional equation. A functional equation is any equation in which the unknown represents the function. We know that this type of function assigns exactly one output to each specified type. It is common to name the functions $f(x)$ or $g(x)$. These functional equations have a common technique for solving the value of $f(x)$. We will first replace $x$ by $\dfrac{1}{x}$.

**Complete step by step solution:**

As per the given question we have $af(x) + bf\left( {\dfrac{1}{x}} \right) = \dfrac{1}{x} - 5$ and we have to find the value of $f(x)$.

We will first replace the $x$ by $\dfrac{1}{x}$ and we have: $af\left( {\dfrac{1}{x}} \right) + bf\left( x \right) = x - 5$. We will now multiply the first equation with $a$ and the second equation by $b$, then we have ${a^2}f(x) + abf\left( {\dfrac{1}{x}} \right) = \dfrac{a}{x} - 5a$ (we call it the third equation), now the another equation is $abf\left( {\dfrac{1}{x}} \right) + {b^2}f(x) = \dfrac{b}{x} - 5b$(it is the fourth equation). After this we will subtract the equation fourth from the third equation, we have: ${a^2}f(x) + abf\left( {\dfrac{1}{x}} \right) - abf\left( {\dfrac{1}{x}} \right) - {b^2}f(x) = \dfrac{a}{x} - 5a - (\dfrac{b}{x} - 5b)$, solving this we get ${a^2}f(x) - {b^2}f(x) =

\dfrac{a}{x} - 5a - \dfrac{b}{x} + 5b$. In this equation we will take the common factor out and we have, $\left( {{a^2} - {b^2}} \right)f(x) = (a - b)\dfrac{1}{x} - 5a + 5b$. We know the algebraic

formula that $({a^2} - {b^2}) = (a + b)(a - b)$. So we can substitute this and we get,$(a - b)(a + b)f(x) = (a - b)\dfrac{1}{x} - 5(a - b)$$ \Rightarrow (a + b)f(x) = \dfrac{{1 - 5x}}{x}$ .

Now by isolating the term we get $f(x) = \dfrac{{(1 - 5x)}}{{x(a + b)}}$.

**Hence the required answer is $f(x) = \dfrac{{(1 - 5x)}}{{x(a + b)}}$.**

**Note:**Before solving this type of question we should the function equation, their formulas and method to solve it. We should also have the knowledge of the algebraic identities as they are very useful in calculation of this kind of problem. We should note that in $\dfrac{1}{x}$, $x$ is replaced also which can be written as $\dfrac{1}{{\dfrac{1}{x}}} = \dfrac{x}{1}$.

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