Application of Monotonicity - Practice Questions & MCQ

Application of Monotonicity (Part 1)

(a) Finding Range of the function

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

Illustration

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

$
\begin{array}{ll}
\therefore & f^{\prime}(x) \\
\Rightarrow & f^{\prime}(x) \\
\Rightarrow & =\sin (x)+x \cos (x)+\frac{\sin (2 x)}{2} \\
\end{array}
$
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 the Number of Roots of a function

Illustration
Find the number of solutions to 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, the graph of $f(x)$ cuts the $x$-axis at a single point in $(-\pi / 2, \pi / 2)$ So, there is only 1 root in the given domain.

Application of Monotonicity (Part 2)

(c) Inequalities Using Monotonicity

We can prove inequalities using the concept of monotonicity. To prove $f(x)<g(x)$ 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 options 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.

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)
$
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.