Application of Monotonicity

Monotonicity is an important concept in calculus. It is useful in understanding the relationship between curves and their slopes. The monotonic function is either increasing or decreasing. These concepts of monotonicity have been broadly applied in branches of mathematics, physics, engineering, economics, and biology.

JEE Main 2025: Sample Papers | Mock Tests | PYQs | Study Plan 100 Days

JEE Main 2025: Maths Formulas  | Study Materials

JEE Main 2025: Syllabus | Preparation Guide | High Scoring Topics 

In this article, we will cover the concept of Monotonicity. This topic falls under the broader category of Calculus, which is a crucial chapter in Class 11 Mathematics. This is very important not only for board exams but also for competitive exams, which even include the Joint Entrance Examination Main and other entrance exams: SRM Joint Engineering Entrance, BITSAT, WBJEE, and BCECE. A total of five questions have been asked on this topic in JEE Main from 2013 to 2023, including one question in 2014, one in 2020, one question in 2021, one in 2022, and one in 2023.

Application of Monotonicity

A function is said to be monotonic if it is either increasing or decreasing in its entire domain. By a monotonic function $f$ in an interval I , we mean that $f$ is either increasing in the Given domain or decreasing in a given domain.

Increasing Function

A function $f(x)$ is increasing in $[a, b]$ if $f\left(x_2\right) \geq f\left(x_1\right)$ for all $x_2>x_1$, where $\mathrm{x}_1, \mathrm{x}_2 \in[\mathrm{a}, \mathrm{b}]$.

If a function is differentiable, then $\frac{d}{d x}(f(x)) \geq 0 \quad \forall x \in(a, b)$
A function is said to be increasing if it is increasing in its entire domain.
Example:
- $f(x)=x$ is increasing in $R$. (As $f^{\prime}(x)=1$, so $f^{\prime}(x) \geq 0$ for all values of $x$ in $R$, so it is an increase in $R$ ).
- $f(x)=\tan ^{-1} x$, is also an increasing function on $R$ as $f^{\prime}(x) \geq 0$ for all real values of $x$.
- $f(x)=[x]$ is also an increasing function on R. Its differentiation is not defined at all points, but from its graph, we can see that by giving the higher value of $x$ to this function, it returns the equal or higher value of $y$. For this function $x_2>$ $x_1$ implies $f\left(x_2\right) \geq f\left(x_1\right)$. Hence it is an increasing function.

Decreasing Function

A function $f(x)$ is decreasing in the interval $[a, b]$ if $f\left(x_2\right) \leq f\left(x_1\right)$ for all $x_2>x_1$, where $x_1, x_2 \in[a, b]$

If a function is differentiable, then $\frac{d}{d x}(f(x)) \leq 0 \quad \forall \quad x \in(a, b)$
Example
- $f(x)=-x$ is decreasing in $R\left(\right.$ As $f^{\prime}(x)=-1$, so $f^{\prime}(x)<0$ for all real values of $x$. We can also see that it is decreasing from its graph)
- $f(x)=e^{-x}$ is decreasing in $R$ (As $f^{\prime}(x)=-e^{-x}$, so $f^{\prime}(x)<0$ for all real values of $x$. We can also see that it is decreasing from its graph)
- $f(x)=\cot (x)$ is decreasing in $(0, \pi)$
- $f(x)=\cot ^{-1}(x)$ is decreasing in $R$

(a) Finding Range of the function

We can use the concept of monotonicity to find the range of the function

Illustration

Find the range of the function

$
f(x)=x \sin x-\frac{1}{2} \cos ^2 x \text { for } x \in[0, \pi / 2]
$

Given, $f(x)=x$ sin $x-\frac{1}{2} \cos ^2 x$
$\therefore \quad f^{\prime}(x)=\sin (x)+x \cos (x)+\frac{\sin (2 x)}{2}$
$\Rightarrow \quad f^{\prime}(x)=\sin x(1+\cos x)+x \cos x$
Clearly, $f^{\prime}(x)>0$ for $x \in[0, \pi / 2]$
$\therefore \quad f(x)$ is strictly increasing function
$\therefore \quad$ Range of $\mathrm{f}(\mathrm{x})$ is $[f(0), f(\pi / 2)]=[-1 / 2, \pi / 2]$

(b) Finding Number of Roots of a function

Illustration

Find the number of solutions of the equation

$
f(x)=x^3+2 x+\cos x+\tan x=0 \text { for } x \in(-\pi / 2, \pi / 2)
$

Given, $f(x)=x^3+2 x+\cos x+\tan x=0$

$
\begin{array}{ll}
\therefore & f^{\prime}(x)=3 x^2+(2-\sin x)+\sec ^2 x \\
\Rightarrow & f^{\prime}(x)>0
\end{array}
$

So, $\mathrm{f}(\mathrm{x})$ is an increasing function

$
\therefore \quad f(0)=1 \text { and } f(-1)<0
$

Hence, graph of $\mathrm{f}(\mathrm{x})$ cut the x -axis at single point in $(-\pi / 2, \pi / 2)$ So, there is only 1 root in the given domain

(c) Inequalities Using Monotonicity

We can prove inequalities using the concept of monotonicity. To prove $\mathrm{f}(\mathrm{x})$ $<g(x)$ in some interval, we consider another function $h(x)=g(x)-f(x)$ or $f(x)-g(x)$ and then we study the nature of $h(x)$, i.e., monotonicity of $h(x)$ in the given interval.

Illustration

For all x in $(0,1)$, which one of the following is correct
1. $e^x<1+x$
2. $\log (1+x)<x$
3. $\sin x>x$
4. $\log x>x$

Sol.

Check all the option one-by-one

1. Let $f(x)=e^x-1-x$

$
\Rightarrow f^{\prime}(x)=e^x-1>0, \forall x \in(0,1)
$

So, $f(x)$ is increasing, when $0<x<1$

$
\begin{aligned}
& \Rightarrow \quad f(x)>f(0) \\
& \Rightarrow \quad e^x-1-x>0 \\
& \Rightarrow \quad e^x>1+x
\end{aligned}
$

Hence, (a) is false.

2. $
\begin{aligned}
& \text { Let } g(x)=\log (1+x)-x \\
& \Rightarrow g^{\prime}(x)=\frac{1}{1+x}-1=\frac{-x}{1+x}<0, \forall x \in(0,1)
\end{aligned}
$

So, $g(x)$ is decreasing, when $0<x<1$

$
\Rightarrow \quad g(x)<g(0) \Rightarrow \log (1+x)-x<0 \Rightarrow \log (1+x)<x
$

Hence, $(b)$ is true.

3. Let $h(x)=\sin x-x \Rightarrow h^{\prime}(x)=\cos x-1<0, \forall x \in(0,1)$ So, $h(x)$ is decreasing, when $0<x<1 \Rightarrow h(x)<h(0)$

$
\begin{aligned}
& \Rightarrow \quad \sin x-x<0 \\
& \Rightarrow \quad \sin x<x
\end{aligned}
$

Hence, $(c)$ is false.

4. Let $g(x)=\log x-x \quad \Rightarrow g^{\prime}(x)=\frac{1}{x}-1$

$
\begin{aligned}
& \therefore g^{\prime}(x)>0, \forall x \in(0,1) \text { or } g(x)<g(1) \\
& \Rightarrow \quad \log x-x<-1
\end{aligned}
$

$\Rightarrow x>\log x+1$ which means $x>\log x$ Hence, (d) is false.
Thus, (b) is the correct answer.

Recommended Video Based on Applications of Monotonicity

Solved Examples

Example 1: If $a_a$ is the greatest term in the sequence $a_n=\frac{\mathrm{n}^3}{\mathrm{n}^4+147}, \mathrm{n}=1,2,3, \ldots$, then a is equal to

[JEE Main 2023]
1) 0.158
2) 0.231
3) 0.367
4) 0.436

Key Concepts

Solution

$
\begin{aligned}
& f(x)=\frac{x^3}{x^4+147} \\
& f^{\prime}(x)=\frac{\left(x^4+147\right) 3 x^2-x^3\left(4 x^3\right)}{\left(x^4+147\right)^2} \\
& =\frac{3 x^6+147 \times 3 x^2-4 x^6}{+v e}=x^2\left(44-x^4\right) \\
& f^{\prime}(x)=0 \text { at } x^6=147 \times 3 x^2 \\
& x^2=0, x^4=147 \times 3 \\
& x=0, x^2= \pm \sqrt{147 \times 3} \\
& x^2= \pm 21 \\
& x= \pm \sqrt{21}
\end{aligned}
$

fmax at $f(4)$ or $f(5)$

$
\begin{aligned}
& f(4)=\frac{64}{403} \simeq 0.158 \quad f(5)=\frac{125}{772} \simeq 0.161 \\
& \therefore a=5
\end{aligned}
$

Example 2: If $f(x)=\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x-1, x \in R$, then the equation $f(x)=0$ has: [JEE Main 2014]

1) no solution
2) one solution
3) two solutions
4) more than two solutions

Solution
derivative of $y=\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x-1$

$
\begin{aligned}
& =\left(\frac{3}{5}\right)^x \ln \left(\frac{3}{5}\right)+\left(\frac{4}{5}\right)^x \ln \left(\frac{4}{5}\right)-0 \\
& =\ln \left(\frac{3}{5}\right)\left(\frac{3}{5}\right)^x+\ln \left(\frac{4}{5}\right)\left(\frac{4}{5}\right)^x
\end{aligned}
$

derivative of $y>0$

Hence it has only one root or none

$
\begin{aligned}
& y=\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x-1=0 \\
& \left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x=1 \\
& (3)^x+(4)^x=5^x
\end{aligned}
$

From here $\mathrm{x}=2$
Hence, the answer is the option 2.

Example 3: Let $f(x)=x \cos ^{-1}(-\sin |x|), x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, then which of the following is true? [JEE Main 2020]
1) $f^{\prime}(0)=-\frac{\pi}{2}$
2) $f^{\prime}$ is decreasing in $\left(-\frac{\pi}{2}, 0\right)$ and increasing in $\left(0, \frac{\pi}{2}\right)$
3) $f$ is not differentiable at $x=0$
4) $f^{\prime}$ is increasing in $\left(-\frac{\pi}{2}, 0\right)$ and decreasing in $\left(0, \frac{\pi}{2}\right)$

Solution

$\begin{aligned}
& f^{\prime}(x)=x\left(\pi-\cos ^{-1}(\sin |x|)\right) \\
& =x\left(\pi-\left(\frac{\pi}{2}-\sin ^{-1}(\sin |x|)\right)\right) \\
& =x\left(\frac{\pi}{2}+|x|\right) \\
& f(x)= \begin{cases}x\left(\frac{\pi}{2}+x\right) & x \geq 0 \\
x\left(\frac{\pi}{2}-x\right) & x<0\end{cases} \\
& f^{\prime}(x)= \begin{cases}\frac{\pi}{2}+2 x & x \geq 0 \\
\frac{\pi}{2}-2 x & x<0\end{cases}
\end{aligned}$

$f^{\prime}(x)$ is increasing in $\left(0, \frac{\pi}{2}\right)$ and decreasing in $\left(\frac{-\pi}{2}, 0\right)$
Hence, the answer is the option 2.

Example 4: The number of real solutions of $\mathrm{x}^7+5 \mathrm{x}^3+3 \mathrm{x}+1=0$ is equal to $\qquad$ . [JEE Main 2022]
1) 0
2) 1
3) 3
4) 5

Solution

$\begin{aligned}
& \mathrm{f}(\mathrm{x})=\mathrm{x}^7+5 \mathrm{x}^3+3 \mathrm{x}+1 \\
& \mathrm{f}^{\prime}(\mathrm{x})=7 \mathrm{x}^6+15 \mathrm{x}^2+3>0 \forall \mathrm{x} \in \mathrm{R}
\end{aligned}$

$\Rightarrow f(x)$ is strictly increasing
$\Rightarrow \mathrm{f}(\mathrm{x})$ will have exactly 1 real root.
Hence, the answer is the option 2.

Example 5: A wire of length 36 m cut into two pieces, one of the pieces is bent to form a square, and the other is bent to form a circle. If the sum of the areas of the two figures is minimum, and the circumference of the circle is $k$ (meter), then $\left(\frac{4}{\pi}+1\right) k$ is equal to $\qquad$
[JEE
Main 2021]
1) 36
2) 25
3) 49
4) 6

Solution

Let x length be used for circle \& 36 -x for square

$
\begin{aligned}
& \text { Area } A=\pi\left(\frac{x}{2 \pi}\right)^2+\left(\frac{36-x}{4}\right)^2 \\
& A=\frac{x^2}{4 \pi}+\frac{(36-x)^2}{16} \\
& \frac{d A}{d x}=0 \\
& \Rightarrow \frac{2 x}{4 \pi}+\frac{2(36-x)(-1)}{16}=0 \\
& \Rightarrow x=\frac{36 \pi}{4+\pi}=\text { circum ference } k \\
& \therefore k\left(\frac{4+\pi}{\pi}\right)=36
\end{aligned}
$

Hence, the answer is the option 1.

Summary

Monotonic functions are important in calculus and analysis because they simplify the study of functions and make them useful for various mathematical and real-world applications. This kind of function is either increasing or decreasing.