Question

If $2f(x + 1) + f\left( {\dfrac{1}{{x + 1}}} \right) = 2x$, then $f(2)$ is equal to

Answer 1

Hint: First we will put $x = 1$ in the given functional equation. After simplifying we will find the value of $f\left( {\dfrac{1}{2}} \right)$ in terms of $f(2)$. Then will put $x = - \dfrac{1}{2}$ then after simplification will put the value of $f\left( {\dfrac{1}{2}} \right)$ to get the value of $f(2)$

Complete step by step solution: Given, functional equation is $2f(x + 1) + f\left( {\dfrac{1}{{x + 1}}} \right) = 2x$
Put $x = 1$ in the functional equation
\[2f(1 + 1) + f\left( {\dfrac{1}{{1 + 1}}} \right) = 2(1)\]
We will get after solving above expression
\[2f(2) + f\left( {\dfrac{1}{2}} \right) = 2\]
Shifting $2f(2)$ to the other side to get the value of $f\left( {\dfrac{1}{2}} \right)$
$f\left( {\dfrac{1}{2}} \right) = 2 - 2f(2)$
Put $x = - \dfrac{1}{2}$
\[2f\left( {\left( {\dfrac{{ - 1}}{2}} \right) + 1} \right) + f\left( {\dfrac{1}{{\dfrac{{ - 1}}{2} + 1}}} \right) = 2\left( { - \dfrac{1}{2}} \right)\]
We will get after solving above expression
\[2f\left( {\dfrac{1}{2}} \right) + f\left( {\dfrac{1}{{\dfrac{1}{2}}}} \right) = - 1\]
After solving
\[2f\left( {\dfrac{1}{2}} \right) + f\left( 2 \right) = - 1\]
Putting the value of $f\left( {\dfrac{1}{2}} \right)$ in the above equation
$2\left( {2 - 2f(2)} \right) + f(2) = - 1$
After simplification
$4 - 4f(2) + f(2) = - 1$
Shifting constant on other side
$ - 3f(2) = - 1 - 4$
On solving, we will get
$ - 3f(2) = - 5$
Dividing both sides by $ - 3$
$f(2) = \dfrac{5}{3}$
Hence, the value of $f(2)$ is $\dfrac{5}{3}$.

Note: Students should choose the values according to our needs. If they choose the wrong value of x they will not get the correct answer. And should do all the calculations carefully without any error to get the required answer. After getting the two equations in two variables we can solve by different methods.