Question

If for nonzero x, \[af(x) + bf\left( {\dfrac{1}{x}} \right) = \dfrac{1}{x} - 5\] where \[a \ne b\] , then \[f(2) = \]
A. \[\dfrac{{3(2b + 3a)}}{{2\left( {{a^2} - {b^2}} \right)}}\]
B. \[\dfrac{{3(2b - 3a)}}{{2\left( {{a^2} - {b^2}} \right)}}\]
C. \[\dfrac{{3(3a - 2b)}}{{2\left( {{a^2} - {b^2}} \right)}}\]
D. \[\dfrac{6}{{a + b}}\]

Answer 1

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 \[f\left( {\dfrac{1}{2}} \right)\] .