Question

If \[af\left( x \right) + bf\left( {\dfrac{1}{x}} \right) = x - 1, x \ne 0\], and \[a \ne b\], then what is the value of \[f\left( 2 \right)\]?
A. \[\dfrac{{\left( {2a + b} \right)}}{{\left( {2\left( {{a^2} + {b^2}} \right)} \right)}}\]
B. \[\dfrac{{\left( {2a + b} \right)}}{{\left( {2\left( {{a^2} - {b^2}} \right)} \right)}}\]
C. \[\dfrac{{\left( {a + 2b} \right)}}{{\left( {2\left( {{a^2} + {b^2}} \right)} \right)}}\]
D. \[\dfrac{{\left( a \right)}}{{\left( {{a^2} - {b^2}} \right)}}\]

Answer 1

Hint In the given question, one equation is given. First, we will substitute \[x = 2\] and \[x = \dfrac{1}{2}\] in the given equation. Then, by simplifying the both equations, we will find the value of \[f\left( 2 \right)\].

Formula used
Elimination Method:
The Process of eliminating one of the variables in the system of linear equations using the addition or subtraction property of equality.

Complete step by step solution:
The given equation is \[af\left( x \right) + bf\left( {\dfrac{1}{x}} \right) = x - 1, x \ne 0\].
Let’s substitute \[x = 2\] in the above equation.
\[af\left( 2 \right) + bf\left( {\dfrac{1}{2}} \right) = 2 - 1\]
\[ \Rightarrow \]\[af\left( 2 \right) + bf\left( {\dfrac{1}{2}} \right) = 1\] \[.....equation\left( 1 \right)\]

Now substitute \[x = \dfrac{1}{2}\] in the above equation.
\[af\left( {\dfrac{1}{2}} \right) + bf\left( {\dfrac{1}{{\dfrac{1}{2}}}} \right) = \dfrac{1}{2} - 1\]
\[ \Rightarrow \]\[af\left( {\dfrac{1}{2}} \right) + bf\left( 2 \right) = - \dfrac{1}{2}\] \[.....equation\left( 2 \right)\]

Multiply equation \[\left( 1 \right)\] by \[a\] and equation \[\left( 2 \right)\] by \[b\].
\[{a^2}f\left( 2 \right) + abf\left( {\dfrac{1}{2}} \right) = a\] \[.....equation\left( 3 \right)\]
\[abf\left( {\dfrac{1}{2}} \right) + {b^2}f\left( 2 \right) = - \dfrac{b}{2}\] \[.....equation\left( 4 \right)\]

Now subtract equation \[\left( 4 \right)\] from equation \[\left( 3 \right)\].
\[{a^2}f\left( 2 \right) + abf\left( {\dfrac{1}{2}} \right) - abf\left( {\dfrac{1}{2}} \right) - {b^2}f\left( 2 \right) = a - \left( { - \dfrac{b}{2}} \right)\]
Simplify the above equation.
\[{a^2}f\left( 2 \right) - {b^2}f\left( 2 \right) = a + \dfrac{b}{2}\]
\[ \Rightarrow \]\[\left( {{a^2} - {b^2}} \right)f\left( 2 \right) = \dfrac{{2a + b}}{2}\]
\[ \Rightarrow \]\[f\left( 2 \right) = \dfrac{{2a + b}}{{2\left( {{a^2} - {b^2}} \right)}}\]
Hence the correct option is B.

Note: Some students solve the equation by putting \[x = 2\] in the equation \[af\left( x \right) + bf\left( {\dfrac{1}{x}} \right) = x - 1\] and solve it. They solve the equation \[af\left( 2 \right) + bf\left( {\dfrac{1}{2}} \right) = 2 - 1\] to get the value of \[f\left( 2 \right)\]. They get \[af\left( 2 \right) = \dfrac{{1 - bf\left( {\dfrac{1}{2}} \right)}}{2}\]. But this is not desire solution.