Answer

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**Hint:**In the given equation \[af(x) + bf\left( {\dfrac{1}{x}} \right) = \dfrac{1}{x} - 5\] , first put x=2 and then put \[x = \dfrac{1}{2}\] you will get two equations containing \[f(2)\] and \[f\left( {\dfrac{1}{2}} \right)\] and then eliminate \[f\left( {\dfrac{1}{2}} \right)\] to get the value of \[f(2)\] .

**Complete step by step answer:**

As discussed in the hint let us do the same.

We will put x=2 in \[af(x) + bf\left( {\dfrac{1}{x}} \right) = \dfrac{1}{x} - 5\] and will get

\[af(2) + bf\left( {\dfrac{1}{2}} \right) = \dfrac{1}{2} - 5........................(i)\]

Also if we put \[f\left( {\dfrac{1}{2}} \right)\] we will surely get

\[af\left( {\dfrac{1}{2}} \right) + bf\left( 2 \right) = 2 - 5........................(ii)\]

Now in the equations we have both \[f(2)\] and \[f\left( {\dfrac{1}{2}} \right)\]

So let us eliminate \[f\left( {\dfrac{1}{2}} \right)\]

For that we have to first find the value of \[f\left( {\dfrac{1}{2}} \right)\] in terms of \[f(2)\]

Let us get the value of \[f\left( {\dfrac{1}{2}} \right)\] from equation (ii)

\[\begin{array}{l}

\therefore af\left( {\dfrac{1}{2}} \right) + bf\left( 2 \right) = 2 - 5\\

\Rightarrow af\left( {\dfrac{1}{2}} \right) + bf\left( 2 \right) = - 3\\

\Rightarrow af\left( {\dfrac{1}{2}} \right) = - 3 - bf\left( 2 \right)\\

\Rightarrow f\left( {\dfrac{1}{2}} \right) = \dfrac{{ - 3 - bf\left( 2 \right)}}{a}

\end{array}\]

Now as we have the value of \[f\left( {\dfrac{1}{2}} \right)\] let us put the same value in equation (i) we will get it as

\[\begin{array}{l}

\therefore af(2) + bf\left( {\dfrac{1}{2}} \right) = \dfrac{1}{2} - 5\\

\Rightarrow af(2) + b\left( {\dfrac{{ - 3 - bf\left( 2 \right)}}{a}} \right) = \dfrac{{1 - 10}}{2}\\

\Rightarrow af(2) - \dfrac{{3b}}{a} - \dfrac{{{b^2}f(2)}}{a} = - \dfrac{9}{2}\\

\Rightarrow af(2) - \dfrac{{{b^2}f(2)}}{a} = - \dfrac{9}{2} + \dfrac{{3b}}{a}\\

\Rightarrow f(2)\left( {\dfrac{{{a^2} - {b^2}}}{a}} \right) = \dfrac{{ - 9a + 6b}}{{2a}}\\

\Rightarrow f(2)\left( {{a^2} - {b^2}} \right) = \dfrac{{ - 9a + 6b}}{2}\\

\Rightarrow f(2) = \dfrac{{ - 9a + 6b}}{{2\left( {{a^2} - {b^2}} \right)}}\\

\Rightarrow f(2) = \dfrac{{3(2b - 3a)}}{{2\left( {{a^2} - {b^2}} \right)}}

\end{array}\]

**So, the correct answer is “Option B”.**

**Note:**we can also do this question by elimination method rather than substitution method just by multiplying the whole equation (ii) by \[\dfrac{b}{a}\] which will make the coefficient of \[f\left( {\dfrac{1}{2}} \right)\] equal to that of we have in equation (i) from where we can eliminate \[f\left( {\dfrac{1}{2}} \right)\] .

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