
If $2f(x + 1) + f\left( {\dfrac{1}{{x + 1}}} \right) = 2x$, then $f(2)$ is equal to
Answer
163.2k+ views
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.
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.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

Instantaneous Velocity - Formula based Examples for JEE

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series
