Question

If $af\left( x+1 \right)+bf\left( \dfrac{1}{x+1} \right)=x,x\ne 1,a\ne b$ then f(2) is equal to
a) $\dfrac{2a+b}{\left( {{a}^{2}}-{{b}^{2}} \right)}$
b) $\dfrac{a}{{{a}^{2}}-{{b}^{2}}}$
c) $\dfrac{a+2b}{{{a}^{2}}-{{b}^{2}}}$
d) ab

Answer 1

Hint: Observe the given expression and put x = x – 1 to get the expression in form of f(x), $f\left( \dfrac{1}{x} \right)$ only. Now, form two equations using this equation, by getting one another equation by replacing x by $\dfrac{1}{x}$. Now, find f(x) and put x = 2 to get the value of f(2).

Complete step-by-step answer:
We have
$af\left( x+1 \right)+bf\left( \dfrac{1}{x+1} \right)=x,x\ne 1,a\ne b$ ……………… (i)
As we need to find the value of f(2) from the given expression in equation (i). So, we need to find the value of f(x) from the given relation by applying some operations with it and hence by putting x = 2 in f(x), we will get the value of f(2).
Now, we know that the equation (i) can be written in form of f(x), if we replace ‘x’ by ‘x-1’ in the whole equation in the following ways:
Replacing ‘x’ by ‘x – 1’ in equation (i), we get
$af\left( x-1+1 \right)+bf\left( \dfrac{1}{x-1+1} \right)=x,$
$af\left( x \right)+bf\left( \dfrac{1}{x} \right)=x-1$ …………….. (ii)
Now, we can observe that f(x) and $f\left( \dfrac{1}{x} \right)$ are just reverse functions w.r.t the variable ‘x’. So, we can get another equation just similar to equation (ii), if we replace ‘x’ in equation (ii) by $'\dfrac{1}{x}'$. So, we get
$\begin{align}
  & \Rightarrow af\left( \dfrac{1}{x} \right)+bf\left( \dfrac{1}{\left( \dfrac{1}{x} \right)} \right)=\dfrac{1}{x}-1 \\
 & \Rightarrow af\left( \dfrac{1}{x} \right)+bf\left( x \right)=\dfrac{1}{x}-1 \\
 & \\
\end{align}$
$\Rightarrow bf\left( x \right)+af\left( \dfrac{1}{x} \right)=\dfrac{1}{x}-1$ ………………. (iii)
Now, we have two equations ,f(x) and $f\left( \dfrac{1}{x} \right)$.
So, we can find value of $f\left( \dfrac{1}{x} \right)$ with the help elimination approach i.e. by eliminating the term $f\left( \dfrac{1}{x} \right)$ from both the equations.
Now, multiply the whole equation (ii) by ‘a’ and equation (iii) by ‘b’; hence, we get equation (ii) as
 $\Rightarrow a\left( af\left( x \right)+bf\left( \dfrac{1}{x} \right) \right)=\left( x-1 \right)a$
$\Rightarrow {{a}^{2}}f\left( x \right)+abf\left( \dfrac{1}{x} \right)=ax-a$…………….. (iv)
Similarly, equation (ii) can be given after multiplication of ‘b’ to the whole equation
$\Rightarrow b\left( bf\left( x \right)+af\left( \dfrac{1}{x} \right) \right)=b\left( \dfrac{1}{x}-1 \right),$

$
\Rightarrow {{b}^{2}}f\left( x \right)+abf\left( \dfrac{1}{x} \right)=\dfrac{b}{x}-b$ …………….. (v)
Now, subtract equation (iv) and (v) to get value f(x) and eliminate $f\left( \dfrac{1}{x} \right)$ as
$\begin{align}
  & \left[ {{a}^{2}}f\left( x \right)+abf\left( \dfrac{1}{x} \right) \right]-\left[ {{b}^{2}}f\left( x \right)+abf\left( \dfrac{1}{x} \right) \right]=\left( ax-a \right)-\left( \dfrac{b}{x}-b \right), \\
 & {{a}^{2}}f\left( x \right)+abf\left( \dfrac{1}{x} \right)-{{b}^{2}}f\left( x \right)-abf\left( \dfrac{1}{x} \right)=\left( ax-a \right)-\left( \dfrac{b}{x}-b \right), \\
 & {{a}^{2}}f\left( x \right)-{{b}^{2}}f\left( x \right)=ax-a-\dfrac{b}{x}+b, \\
 & f\left( x \right)\left( {{a}^{2}}-{{b}^{2}} \right)=ax-a-\dfrac{b}{x}+b, \\
\end{align}$
$f\left( x \right)=\left( ax-a-\dfrac{b}{x}+b \right)\dfrac{1}{{{a}^{2}}-{{b}^{2}}}$………………… (vi)
Now, we can put x = 2 to the above expression to get the value of f(2), as now we got the equation of f(x). So, put x = 2
$\begin{align}
  & f\left( 2 \right)=\left( 2a-a-\dfrac{b}{2}+b \right)\dfrac{1}{{{a}^{2}}-{{b}^{2}}}, \\
 & f\left( 2 \right)=\left( \dfrac{4a-2a-b+2b}{2} \right)\dfrac{1}{{{a}^{2}}-{{b}^{2}}}, \\
 & f\left( 2 \right)=\dfrac{2a+b}{2\left( {{a}^{2}}-{{b}^{2}} \right)} \\
\end{align}$
Hence, value of f(2) is $\dfrac{2a+b}{2\left( {{a}^{2}}-{{b}^{2}} \right)}$

Note: Put x = 1 to the given expression get expression as $af\left( 2 \right)+bf\left( \dfrac{1}{x} \right)=1$ and put $x=\dfrac{-1}{2}$ to the given expression and get expression as $af\left( \dfrac{1}{2} \right)+bf\left( 2 \right)=1$. Now, one can solve them and get the value of f(2) directly. Hence, observing the question of these kinds will make our solution shorter.
Replacing x by another value will apply with the term ‘x’ only. Don’t replace any other constant term by another value. As ‘x’ is the variable of the given expression, so, we can replace it by any other variable in the whole expression. But don’t change any other term.
One may solve the equations in f(x), $f\left( \dfrac{1}{x} \right)$ by substituting method as well or by another method.