Question

If $af(x + 1) + bf\left( {\dfrac{1}{{x + 1}}} \right) = x$ $x \ne - 1,a \ne b$ then $f\left( 2 \right)$ is equal to
A. $\dfrac{{2a + b}}{{2\left( {{a^2} - {b^2}} \right)}}$
B. $\dfrac{a}{{{a^2} - {b^2}}}$
C. $\dfrac{{a + 2b}}{{{a^2} - {b^2}}}$
D. None of these

Answer 1

Hint: First, we shall analyze the given information. The given equation is $af(x + 1) + bf\left( {\dfrac{1}{{x + 1}}} \right) = x$
We are asked to calculate the value of $f\left( 2 \right)$
Now, we need to replace $x + 1$ by $\dfrac{1}{{x + 1}}$ in the given equation so that we can obtain two equations. By simplifying those two equations, we can obtain the final equation.
We can get the value of $f\left( 2 \right)$ when we substitute $x = 1$ in the final equation which is the required answer too.

Complete step by step answer:
 The given equation is $af(x + 1) + bf\left( {\dfrac{1}{{x + 1}}} \right) = x$ $x \ne - 1,a \ne b$…………..$\left( 1 \right)$
We are asked to calculate the value of $f\left( 2 \right)$
Now, we need to replace $x + 1$ by $\dfrac{1}{{x + 1}}$ in the given equation.
$ \Rightarrow x = \dfrac{1}{{x + 1}} - 1$
Now, our new equation is as follows.
\[af\left( {\dfrac{1}{{x + 1}}} \right) + bf\left( {x + 1} \right) = \dfrac{1}{{x + 1}} - 1\]
\[ \Rightarrow af\left( {\dfrac{1}{{x + 1}}} \right) + bf\left( {x + 1} \right) = \dfrac{{1 - x - 1}}{{x + 1}}\]
\[ \Rightarrow af\left( {\dfrac{1}{{x + 1}}} \right) + bf\left( {x + 1} \right) = \dfrac{{ - x}}{{x + 1}}\] ……………$\left( 2 \right)$
Here, we got two equations.
Now, we need to multiply the first equation by a, and the second equation by b.
When we multiply the first equation by a, we get
${a^2}f(x + 1) + abf\left( {\dfrac{1}{{x + 1}}} \right) = ax$……………$\left( 3 \right)$
When we multiply the second equation by b, we get
\[ \Rightarrow abf\left( {\dfrac{1}{{x + 1}}} \right) + {b^2}f\left( {x + 1} \right) = \dfrac{{ - bx}}{{x + 1}}\]…………………$\left( 4 \right)$
Now, we have to subtract the third and the fourth equations, we get
${a^2}f(x + 1) + abf\left( {\dfrac{1}{{x + 1}}} \right) - abf\left( {\dfrac{1}{{x + 1}}} \right) - {b^2}f\left( {x + 1} \right) = ax - \dfrac{{ - bx}}{{x + 1}}$
$ \Rightarrow {a^2}f(x + 1) + abf\left( {\dfrac{1}{{x + 1}}} \right) - abf\left( {\dfrac{1}{{x + 1}}} \right) - {b^2}f\left( {x + 1} \right) = ax + \dfrac{{bx}}{{x + 1}}$
$ \Rightarrow {a^2}f(x + 1) - {b^2}f\left( {x + 1} \right) + abf\left( {\dfrac{1}{{x + 1}}} \right) - abf\left( {\dfrac{1}{{x + 1}}} \right) = ax + \dfrac{{bx}}{{x + 1}}$
$ \Rightarrow {a^2}f(x + 1) - {b^2}f\left( {x + 1} \right) = ax + \dfrac{{bx}}{{x + 1}}$
Now, we shall pick the common terms.
\[ \Rightarrow \left( {{a^2} - {b^2}} \right)f(x + 1) = ax + \dfrac{{bx}}{{x + 1}}\]
On taking LCM on the right-hand side, we get
\[ \Rightarrow \left( {{a^2} - {b^2}} \right)f(x + 1) = \dfrac{{ax\left( {x + 1} \right) + bx}}{{x + 1}}\]
\[ \Rightarrow \left( {{a^2} - {b^2}} \right)f(x + 1) = \dfrac{{a{x^2} + ax + bx}}{{x + 1}}\]
\[ \Rightarrow f(x + 1) = \dfrac{1}{{\left( {{a^2} - {b^2}} \right)}} \times \dfrac{{a{x^2} + ax + bx}}{{x + 1}}\] , $x \ne - 1,a \ne b$………..$\left( 5 \right)$
We are asked to calculate the value of$f\left( 2 \right)$
We can get the value of$f\left( 2 \right)$when we substitute $x = 1$in $\left( 5 \right)$
\[f(1 + 1) = \dfrac{1}{{\left( {{a^2} - {b^2}} \right)}} \times \dfrac{{a{1^2} + a + b}}{{1 + 1}}\]
\[ \Rightarrow f(2) = \dfrac{1}{{\left( {{a^2} - {b^2}} \right)}} \times \dfrac{{a + a + b}}{2}\]
\[ \Rightarrow f(2) = \dfrac{1}{{\left( {{a^2} - {b^2}} \right)}} \times \dfrac{{2a + b}}{2}\]
\[ \Rightarrow f(2) = \dfrac{{2a + b}}{{2\left( {{a^2} - {b^2}} \right)}}\]

So, the correct answer is “Option A”.

Note: The given equation is $af(x + 1) + bf\left( {\dfrac{1}{{x + 1}}} \right) = x$
We are asked to calculate the value of $f\left( 2 \right)$
It can be solved in another method. That is we shall substitute $x = 1$ in the given equation.
Again, we need to substitute $x = - \dfrac{1}{2}$ in the given equation.
Now, solving the obtained equations as we followed above, we can obtain the value of $f\left( 2 \right)$