
The function $f\left( x \right) = \dfrac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}$ is
A. an increasing function
B. a decreasing function
C. an even function
D. None of these
Answer
164.4k+ views
Hint: To check a function increasing or decreasing, first order derivative test is used. Find the differentiation of the given function and equate it with zero to find the critical points. Use the concept that a function is increasing in the interval on which the first order derivative of the function is positive. Otherwise, it is decreasing.
Formula Used:
Quotient Rule: If $f\left( x \right)$ and $g\left( x \right)$ be two functions of $x$ then the differentiation of the function $\dfrac{{f\left( x \right)}}{{g\left( x \right)}}$ is $\dfrac{d}{{dx}}\left\{ {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right\} = \dfrac{{g\left( x \right)\dfrac{d}{{dx}}\left\{ {f\left( x \right)} \right\} - f\left( x \right)\dfrac{d}{{dx}}\left\{ {g\left( x \right)} \right\}}}{{{{\left\{ {g\left( x \right)} \right\}}^2}}}$, provided $g\left( x \right) \ne 0$
$\dfrac{d}{{dx}}\left( {{e^{mx}}} \right) = m{e^{mx}}$
$\dfrac{d}{{dx}}\left( c \right) = 0$, where $c$ is a constant.
Complete step by step solution:
The given function is $f\left( x \right) = \dfrac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}$
Differentiating the function with respect to $x$, we get
$f'\left( x \right) = \dfrac{{\left( {{e^{2x}} + 1} \right)\dfrac{d}{{dx}}\left( {{e^{2x}} - 1} \right) - \left( {{e^{2x}} - 1} \right)\dfrac{d}{{dx}}\left( {{e^{2x}} + 1} \right)}}{{{{\left( {{e^{2x}} + 1} \right)}^2}}}$
Now, $\dfrac{d}{{dx}}\left( {{e^{2x}} - 1} \right) = \dfrac{d}{{dx}}\left( {{e^{2x}}} \right) - \dfrac{d}{{dx}}\left( 1 \right) = 2{e^{2x}} - 0 = 2{e^{2x}}$
and $\dfrac{d}{{dx}}\left( {{e^{2x}} + 1} \right) = \dfrac{d}{{dx}}\left( {{e^{2x}}} \right) + \dfrac{d}{{dx}}\left( 1 \right) = 2{e^{2x}} + 0 = 2{e^{2x}}$
So, $f'\left( x \right) = \dfrac{{\left( {{e^{2x}} + 1} \right)\left( {2{e^{2x}}} \right) - \left( {{e^{2x}} - 1} \right)\left( {2{e^{2x}}} \right)}}{{{{\left( {{e^{2x}} + 1} \right)}^2}}}$
Simplify the expression.
Take the term $\left( {2{e^{2x}}} \right)$ as common from the numerator.
$ \Rightarrow f'\left( x \right) = \dfrac{{\left( {2{e^{2x}}} \right)\left\{ {\left( {{e^{2x}} + 1} \right) - \left( {{e^{2x}} - 1} \right)} \right\}}}{{{{\left( {{e^{2x}} + 1} \right)}^2}}}$
$ \Rightarrow f'\left( x \right) = \dfrac{{\left( {2{e^{2x}}} \right)\left( {{e^{2x}} + 1 - {e^{2x}} + 1} \right)}}{{{{\left( {{e^{2x}} + 1} \right)}^2}}}$
Cancel out the term ${e^{2x}}$ from the numerator.
$ \Rightarrow f'\left( x \right) = \dfrac{{\left( {2{e^{2x}}} \right)\left( 2 \right)}}{{{{\left( {{e^{2x}} + 1} \right)}^2}}} = \dfrac{{4{e^{2x}}}}{{{{\left( {{e^{2x}} + 1} \right)}^2}}}$
Value of every perfect square expression is always positive and an exponential function is also always positive.
So, clearly, $f'\left( x \right) > 0$ for all real values of $x$.
Thus, the given function is an increasing function.
Option ‘A’ is correct
Note: Many students get confused about the condition for a function to be increasing and decreasing. They should remember that a function is increasing if the first order derivative of the function is positive for all real values of $x$ but if the first order derivative of a function is negative for all real values of $x$ then the function is decreasing. A function is even if the value of the function remains same after replacement of $x$ by $\left( { - x} \right)$ i.e. $f\left( { - x} \right) = f\left( x \right)$ and a function is odd if $f\left( { - x} \right) = - f\left( x \right)$.
Formula Used:
Quotient Rule: If $f\left( x \right)$ and $g\left( x \right)$ be two functions of $x$ then the differentiation of the function $\dfrac{{f\left( x \right)}}{{g\left( x \right)}}$ is $\dfrac{d}{{dx}}\left\{ {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right\} = \dfrac{{g\left( x \right)\dfrac{d}{{dx}}\left\{ {f\left( x \right)} \right\} - f\left( x \right)\dfrac{d}{{dx}}\left\{ {g\left( x \right)} \right\}}}{{{{\left\{ {g\left( x \right)} \right\}}^2}}}$, provided $g\left( x \right) \ne 0$
$\dfrac{d}{{dx}}\left( {{e^{mx}}} \right) = m{e^{mx}}$
$\dfrac{d}{{dx}}\left( c \right) = 0$, where $c$ is a constant.
Complete step by step solution:
The given function is $f\left( x \right) = \dfrac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}$
Differentiating the function with respect to $x$, we get
$f'\left( x \right) = \dfrac{{\left( {{e^{2x}} + 1} \right)\dfrac{d}{{dx}}\left( {{e^{2x}} - 1} \right) - \left( {{e^{2x}} - 1} \right)\dfrac{d}{{dx}}\left( {{e^{2x}} + 1} \right)}}{{{{\left( {{e^{2x}} + 1} \right)}^2}}}$
Now, $\dfrac{d}{{dx}}\left( {{e^{2x}} - 1} \right) = \dfrac{d}{{dx}}\left( {{e^{2x}}} \right) - \dfrac{d}{{dx}}\left( 1 \right) = 2{e^{2x}} - 0 = 2{e^{2x}}$
and $\dfrac{d}{{dx}}\left( {{e^{2x}} + 1} \right) = \dfrac{d}{{dx}}\left( {{e^{2x}}} \right) + \dfrac{d}{{dx}}\left( 1 \right) = 2{e^{2x}} + 0 = 2{e^{2x}}$
So, $f'\left( x \right) = \dfrac{{\left( {{e^{2x}} + 1} \right)\left( {2{e^{2x}}} \right) - \left( {{e^{2x}} - 1} \right)\left( {2{e^{2x}}} \right)}}{{{{\left( {{e^{2x}} + 1} \right)}^2}}}$
Simplify the expression.
Take the term $\left( {2{e^{2x}}} \right)$ as common from the numerator.
$ \Rightarrow f'\left( x \right) = \dfrac{{\left( {2{e^{2x}}} \right)\left\{ {\left( {{e^{2x}} + 1} \right) - \left( {{e^{2x}} - 1} \right)} \right\}}}{{{{\left( {{e^{2x}} + 1} \right)}^2}}}$
$ \Rightarrow f'\left( x \right) = \dfrac{{\left( {2{e^{2x}}} \right)\left( {{e^{2x}} + 1 - {e^{2x}} + 1} \right)}}{{{{\left( {{e^{2x}} + 1} \right)}^2}}}$
Cancel out the term ${e^{2x}}$ from the numerator.
$ \Rightarrow f'\left( x \right) = \dfrac{{\left( {2{e^{2x}}} \right)\left( 2 \right)}}{{{{\left( {{e^{2x}} + 1} \right)}^2}}} = \dfrac{{4{e^{2x}}}}{{{{\left( {{e^{2x}} + 1} \right)}^2}}}$
Value of every perfect square expression is always positive and an exponential function is also always positive.
So, clearly, $f'\left( x \right) > 0$ for all real values of $x$.
Thus, the given function is an increasing function.
Option ‘A’ is correct
Note: Many students get confused about the condition for a function to be increasing and decreasing. They should remember that a function is increasing if the first order derivative of the function is positive for all real values of $x$ but if the first order derivative of a function is negative for all real values of $x$ then the function is decreasing. A function is even if the value of the function remains same after replacement of $x$ by $\left( { - x} \right)$ i.e. $f\left( { - x} \right) = f\left( x \right)$ and a function is odd if $f\left( { - x} \right) = - f\left( x \right)$.
Recently Updated Pages
Environmental Chemistry Chapter for JEE Main Chemistry

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

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

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

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

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

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Advanced 2025 Notes
