RD Sharma Class 12 Exercise 16.1 Increasing and Decreasing Function Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 16.1 Increasing and Decreasing Function Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 21, 2022 01:56 PM IST

RD Sharma Class 12 Solutions Chapter 16 Exercise 16.1 Increasing and Decreasing Functions comprises issues on tracking down the homogeneous arrangement of natural conditions. RD Sharma Class 12th Exercise 16.1 issues are just tackled via CAREER360 specialists to accelerate the exam arrangement of students.

## RD Sharma Class 12 Solutions Chapter 16 Increasing and Decreasing Functions - Other Exercise

• Increasing and Decreasing Functions exercise 16.1

Increasing and Decreasing Functions exercise 16.1 , question 1

$f(x)$ is increasing in$(0,\infty )$
Hint:
A function $f(x)$ is said to be a increasing function on$(a, b)$, if $x_{1} for all $x_{1}, x_{2} \in(a, b)$.
Given:
$f(x)=\log _{e} x$
To prove:
Function $f(x)=\log _{e} x$ is increasing on $(0,\infty )$
Solution:
Let us consider $x_{1}, x_{2} \in(0, \infty)$ and $x_{1} < x_{2}$
Then,
$x_{1} < x_{2}$
Taking loge on both sides,
$\Rightarrow \quad \log _{e} x_{1}<\log _{e} x_{2}$
$\Rightarrow \quad f\left(x_{1}\right) for all $x_{1}, x_{2} \in(0, \infty)$
So, $f(x)$ is increasing in $(0,\infty )$

Increasing and Decreasing Functions exercise 16.1 , question 2

$f(x)$ is decreasing on$(0,\infty )$if $0
Hint:
1. $f(x)$is increasing on$(a,b)$, if the values of $f(x)$ increase with the increase in the values of x.
2. $f(x)$is decreasing on$(a,b)$, if the values of $f(x)$ decrease with the increase in the values of x.
Given:
Here given that,
$f(x)=\log _{a} x$
To prove:
Function $f(x)=\log _{a} x$ is increasing on $(0,\infty )$, if $a>1$ and decreasing on $(0,\infty )$ , if $0
Solution:
Here we have two cases:
Case I: when $a>1$
Let us consider $x_{1},x_{2}\epsilon (0,\infty )$
Such that $x_{1}
$\begin{array}{ll} \Rightarrow & \log _{a} x_{1}<\log _{a} x_{2} \\ \Rightarrow & f\left(x_{1}\right)
Thus, $f(x)$ is increasing in$(0,\infty )$if $a>1$
Case II: when $0
We have $f(x)=\log _{a} x$
We know that $\log _{a} x=\frac{\log x}{\log a}$
Therefore, $f(x)=\frac{\log x}{\log a}$
When $a<1$, then $a<0$
Let $x_{1}
Taking log on both sides,
\begin{aligned} &\Rightarrow \quad & \log x_{1}<\log x_{2} \\ &\Rightarrow & \frac{\log x_{1}}{\log a}>\frac{\log x_{2}}{\log a} \\ &\Rightarrow & f\left(x_{1}\right)>f\left(x_{2}\right) \end{aligned}

Thus, $f(x)$ is decreasing on $\left ( 0,\infty \right )$ if $0

Increasing and Decreasing Functions exercise 16.1 , question 3

$f(x)=ax+b$is an increasing function on R
Hint:
A function $f(x)$ is said to be a increasing function on $(a,b)$, if $x_{1} for all $x_{1}.
Given:
$f(x)=ax+b$ , where$a,b$are constant and $a>0$
To prove:
$f(x)=ax+b$ is an increasing function on R.
Solution:
Let $x_{1},x_{2}\epsilon R$Such that $x_{1}
Then,
\begin{aligned} &x_{1}0] \end{aligned}
$\Rightarrow \quad a x_{1}+b [Given that $b$ is constant]
$\Rightarrow \quad f\left(x_{1}\right)
Thus, $f(x)=ax+b$ is an increasing function on R.

Increasing and Decreasing Functions exercise 16.1 , question 4

$f(x)=ax+b$ is a decreasing function on R.
Hint:
A function $f(x)$is said to be a decreasing function on $(a,b)$, if $x_{1}f(x_{2})$ for all $x_{1},x_{2}\epsilon (a,b)$.
Given:
$f(x)=ax+b$, where $a,b$ are constant and $a<0$
To prove:
$f(x)=ax+b$ is a decreasing function on R.
Solution:
Let $x_{1},x_{2}\epsilon R$ Such that $x_{1}< x_{2}$
Then,
\begin{aligned} & x_{1}a x_{2} \end{aligned} \quad[\because a<0]
$\Rightarrow \quad a x_{1}+b>a x_{2}+b$ [Given that b is constant]
$\Rightarrow \quad f\left(x_{1}\right)>f\left(x_{2}\right)$
Thus, $x_{1}f\left(x_{2}\right)$ for all $x_{1},x_{2}\epsilon R$
So, $f(x)=ax+b$ is a decreasing function on R.

Increasing and Decreasing Functions exercise 16.1 , question 5

$f(x)$ is a decreasing function.
Hint:
A function $f(x)$ is said to be a decreasing function on $(a,b)$, if $x_{1}f\left(x_{2}\right)$ for all $x_{1}, x_{2} \in(a, b) \text { . }$
Given:
Here given that,
$f(x)=\frac{1}{x}$
To prove:
$f(x)=\frac{1}{x}$ is a decreasing function on $(0,\infty )$
Solution:
Let $x_{1}, x_{2} \in(0,\infty )$ Such that $x_{1}< x_{2}$
Then,
$\begin{array}{ll} & x_{1}\frac{1}{x_{2}} \\ \Rightarrow & f\left(x_{1}\right)>f\left(x_{2}\right) \end{array}$
Thus, $x_{1}f\left(x_{2}\right)$ for all $x_{1}, x_{2} \in(0,\infty )$.
So, $f(x)$is a decreasing function.

Increasing and Decreasing Functions exercise 16 .1 , question 6

$f(x)$ is a decreasing function on $(0,\infty ]$.
$f(x)$ is increasing on $[-\infty , 0)$
Hint:
1. $f(x)$ is increasing on$(a,b)$, if the values of$f(x)$ increase with the increase in the values of x.
2. $f(x)$ is decreasing on $(a,b)$, if the values of$f(x)$ decrease with the increase in the values of x.
Given:
$f(x)=\frac{1}{1+x^{2}}$
To prove:
$f(x)=\frac{1}{1+x^{2}}$ decreases in the interval $(0,\infty ]$ and increases in the interval $[-\infty , 0)$.
Solution:
Here we have two cases:
Case I: if $x \in [0,\infty)$
Let $x_{1},x_{2} \in [0,\infty)$ Such that $x_{1}
Then,
$x_{1}
$\Rightarrow x_{1}^{2} [Squaring on both sides]
$\Rightarrow 1+x_{1}^{2}<1+x_{2}^{2}$ [Adding 1 on both sides]
\begin{aligned} &\Rightarrow \quad \frac{1}{1+x_{1}^{2}}>\frac{1}{1+x_{2}^{2}} \\ &\Rightarrow \quad f\left(x_{1}\right)>f\left(x_{2}\right) \end{aligned}
Thus, $x_{1}f\left(x_{2}\right)$ for all $x_{1},x_{2} \in [0,\infty)$.
So, $f(x)$is a decreasing function.
Case II: if $x \in (-\infty ,0]$
Let $x_{1},x_{2} \in (-\infty ,0]$ Such that $x_{1}
Then
$x_{1}
$\Rightarrow x_{1}^{2}>x_{2}^{2}$ [Squaring on both sides]
$\Rightarrow 1+x_{1}^{2}>1+x_{2}^{2}$ [Adding 1 on both sides]
\begin{aligned} &\Rightarrow \quad \frac{1}{1+x_{1}^{2}}<\frac{1}{1+x_{2}^{2}} \\ &\Rightarrow \quad f\left(x_{1}\right)
Thus, $x_{1} for all $x_{1},x_{2} \in (-\infty ,0]$.
So, $f(x)$is increasing function on$(-\infty ,0]$.

Increasing and Decreasing Functions exercise 16.1 , question 7

$f(x)$ is neither increasing nor decreasing on R.
Hint:
1. $f(x)$ is increasing on$(a,b)$, if the values of$f(x)$ increase with the increase in the values of x.
2. $f(x)$ is decreasing on $(a,b)$, if the values of$f(x)$ decrease with the increase in the values of x.
Given:
$f(x)=\frac{1}{1+x^{2}}$
To prove:
$f(x)=\frac{1}{1+x^{2}}$ is neither increasing nor decreasing on R
Solution:
Here we have two cases:
Case I: if $x \in [0,\infty)$
Let $x_{1},x_{2} \in [0,\infty)$ Such that $x_{1}>x_{2}$
Then,
$x_{1}>x_{2}$
$\Rightarrow x_{1}^{2}>x_{2}^{2}$ [Squaring on both sides]
$\Rightarrow 1+x_{1}^{2}>1+x_{2}^{2}$ [Adding 1 on both sides]
\begin{aligned} &\Rightarrow \quad \frac{1}{1+x_{1}^{2}}<\frac{1}{1+x_{2}^{2}} \\ &\Rightarrow \quad f\left(x_{1}\right)
Thus, $x_{1}>x_{2} \Rightarrow f\left(x_{1}\right) for all $x_{1},x_{2} \in [0,\infty)$.
So, $f(x)$is a decreasing function on $[0,\infty )$.
Case II: if $x \in (-\infty ,0]$
Let $x_{1},x_{2} \in (-\infty ,0]$ Such that $x_{1}
Then
$x_{1}
$\Rightarrow x_{1}^{2}>x_{2}^{2}$ [Squaring on both sides]
$\Rightarrow 1+x_{1}^{2}>1+x_{2}^{2}$ [Adding 1 on both sides]
\begin{aligned} &\Rightarrow \quad \frac{1}{1+x_{1}^{2}}>\frac{1}{1+x_{2}^{2}} \\ &\Rightarrow \quad f\left(x_{1}\right)>f\left(x_{2}\right) \end{aligned}
Thus, $x_{1}>x_{2} \Rightarrow f\left(x_{1}\right)>f\left(x_{2}\right)$ for all $x_{1},x_{2} \in (-\infty ,0]$.
So, $f(x)$is increasing function on$(-\infty ,0]$.
In both cases we get, $f(x)$is a decreasing function on$[0,\infty )$and increasing function on$(-\infty ,0]$.
Thus, $f(x)$is neither increasing nor decreasing on R.

Increasing and Decreasing Functions exercise 16.1 , question 8 sub question

$f(x)$is strictly increasing in $(0,\infty)$
Hint:
Use $f(x)=|x|=\left\{\begin{array}{l} x, \text { if } x>0 \\ -x, \text { if } x<0 \end{array}\right.$
Given:
$f(x)=|x|$
To prove:
We have to prove that function $f(x)=|x|$ is strictly increasing in $(0,\infty)$.
Solution:
Let $x_{1},x_{2}\in (0,\infty )$ and $x_{1}>x_{2}$
Since$x_{1},x_{2}>0$, therefore the function of $x_{1},x_{2}$ is
$f\left(x_{1}\right)>f\left(x_{2}\right)$
Thus, $x_{1}>x_{2} \Rightarrow f\left(x_{1}\right)>f\left(x_{2}\right)$ for all $x_{1},x_{2}\in (0,\infty )$.
So, $f(x)$is strictly increasing in $(0,\infty)$

Increasing and Decreasing Functions exercise 16.1 , question 8 sub question b

$f(x)$is strictly increasing in $(-\infty,0)$
Hint:
Use $f(x)=|x|=\left\{\begin{array}{l} x, \text { if } x>0 \\ -x, \text { if } x<0 \end{array}\right.$
Given:
$f(x)=|x|$
To prove:
We have to prove that function $f(x)=|x|$ is strictly increasing in $(-\infty,0)$.
Solution:
Let $x_{1},x_{2}\in (-\infty,0 )$ and $x_{1}>x_{2}$
Since$x_{1},x_{2}<0$, therefore the function of $x_{1},x_{2}$ is
\begin{aligned} &-x_{1}<-x_{2} \\ &f\left(x_{1}\right)
Thus, $x_{1}>x_{2} \Rightarrow f\left(x_{1}\right) for all $x_{1},x_{2}\in (-\infty,0 )$.
So, $f(x)$is strictly increasing in $(-\infty,0)$

Increasing and Decreasing Functions exercise 16.1 , question 9

$f(x)$ is strictly increasing function on R.
Hint:
A function $f(x)$ is strictly increasing on$(a,b)$, if the value of$f(x)$ increase with the increase in the value of R.
If $x_{1}for all $x_{1}, x_{2} \in(a, b)$.
Given:
Here given that,
$f(x)=7 x-3$
To prove:
We have to prove that function$f(x)=7 x-3$ is strictly increasing function on R.
Solution:
Here given that,
$f(x)=7 x-3$
Let us consider$x_{1},x_{2}\in R$and $x_{1}
$x_{1}
Multiplying 7 on both sides,
$\Rightarrow 7x_{1}<7x_{2}$
Subtracting 3 on both sides,
We get,
$\begin{array}{ll} \Rightarrow & 7 x_{1}-3<7 x_{2}-3 \\ \Rightarrow & f\left(x_{1}\right)
Thus, $x_{1}for all $x_{1}, x_{2} \in R$
So, $f(x)$ is strictly increasing function on R.

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• Stringently increasing functions

• Stringently decreasing functions

• Discovering the stretches in which a function is increasing or decreasing

• Discovering the span wherein a function is increasing or decreasing

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