RD Sharma Class 12 Exercise 16.1 Increasing and Decreasing Functions Solutions Maths

Q: Will I adequately download rd Sharma class 12th exercise 16.1?

To be sure, just a solitary tick is adequate to download RD Sharma Class 12th Exercise 16.1 Chapter 16 PDF from our website Career360.

Q: Which study material to follow to score high in boards?

There are many exam resources open watching out and the web, yet preparing with RD Sharma Book Solutions will help you unbound and score better checks in class 12 and ferocious exams.

Q: Where might I have the option to get class 12 rd Sharma chapter 16 exercise 16.1 solution for Class 12?

Candidates who want to download the Rd Sharma class 12 chapter 16 exercise 16.1 will find the free copy of the pdf at the Career360 website.

Q: Is handling rd Sharma class 12 solutions ex 16.1 enough for chapter 16 status?

For sure, handling the best books like Rd Sharma class 12 chapter 16 exercise 16.1 is the completed pack for your exam preparation.

Q: What sort of inquiries do RD Sharma Solutions oblige Class 12?

RD Sharma reference books offer concise response questions, short answer questions, different choice inquiries, and long answer questions, which in this way, help in utilizing time beneficially and cultivate fundamental derivation among the students.

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.

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

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

Answer:
$f(x)$ is increasing in $(0,\infty )$
Hint:
A function $f(x)$ is said to be a increasing function on $(a, b)$ , if $x_{1}<x_{2} \Rightarrow f\left(x_{1}\right)<f\left(x_{2}\right)$ 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,
$\Rightarrow \quad \log _{e} x_{1}<\log _{e} x_{2}$
$\Rightarrow \quad f\left(x_{1}\right)<f\left(x_{2}\right)$ 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

Answer:
$f(x)$ is decreasing on $(0,\infty )$ if $0<a<1$
Hint:

$f(x)$ is increasing on $(a,b)$ , if the values of $f(x)$ increase with the increase in the values of x.
$f(x)$ 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 $0<a<1$
Solution:
Here we have two cases:
Case I: when $a>1$
Let us consider $x_{1},x_{2}\epsilon (0,\infty )$
Such that $x_{1}<x_{2}$
$\begin{array}{ll} \Rightarrow & \log _{a} x_{1}<\log _{a} x_{2} \\ \Rightarrow & f\left(x_{1}\right)<f\left(x_{2}\right) \end{array}$
Thus, $f(x)$ is increasing in $(0,\infty )$ if $a>1$
Case II: when $0<a<1$
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 $x_{1}<x_{2}$
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 $0<a<1$

Increasing and Decreasing Functions exercise 16.1 , question 3

Answer:

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

Increasing and Decreasing Functions exercise 16.1 , question 4

Answer:

$f(x)=ax+b$ is a decreasing function on R.
Hint:
A function $f(x)$ is said to be a decreasing function on $(a,b)$ , if $x_{1}<x_{2}\Rightarrow f(x_{1})>f(x_{2})$ 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,
$\begin{aligned} & x_{1}<x_{2} \\ \Rightarrow \quad & a x_{1}>a x_{2} \end{aligned} \quad[\because a<0]$
$\Rightarrow \quad a x_{1}+b>a x_{2}+b$ [Given that b is constant]
$\Rightarrow \quad f\left(x_{1}\right)>f\left(x_{2}\right)$
Thus, $x_{1}<x_{2} \Rightarrow f\left(x_{1}\right)>f\left(x_{2}\right)$ 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

Answer:

$f(x)$ is a decreasing function.
Hint:
A function $f(x)$ is said to be a decreasing function on $(a,b)$ , if $x_{1}<x_{2} \Rightarrow f\left(x_{1}\right)>f\left(x_{2}\right)$ 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,
$\begin{array}{ll} & x_{1}<x_{2} \\ \Rightarrow & \frac{1}{x_{1}}>\frac{1}{x_{2}} \\ \Rightarrow & f\left(x_{1}\right)>f\left(x_{2}\right) \end{array}$
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 a decreasing function.

Increasing and Decreasing Functions exercise 16 .1 , question 6

Answer:

$f(x)$ is a decreasing function on $(0,\infty ]$ .
$f(x)$ is increasing on $[-\infty , 0)$
Hint:

$f(x)$ is increasing on $(a,b)$ , if the values of $f(x)$ increase with the increase in the values of x.
$f(x)$ 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 $x_{1}<x_{2}$
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]
$\begin{aligned} &\Rightarrow \quad \frac{1}{1+x_{1}^{2}}>\frac{1}{1+x_{2}^{2}} \\ &\Rightarrow \quad f\left(x_{1}\right)>f\left(x_{2}\right) \end{aligned}$
Thus, $x_{1}<x_{2} \Rightarrow f\left(x_{1}\right)>f\left(x_{2}\right)$ for all $x_{1},x_{2} \in [0,\infty)$ .
So, $f(x)$ is a decreasing function.
Case II: if $x \in (-\infty ,0]$
Let $x_{1},x_{2} \in (-\infty ,0]$ Such that $x_{1}<x_{2}$
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]
$\begin{aligned} &\Rightarrow \quad \frac{1}{1+x_{1}^{2}}<\frac{1}{1+x_{2}^{2}} \\ &\Rightarrow \quad f\left(x_{1}\right)<f\left(x_{2}\right) \end{aligned}$
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 $(-\infty ,0]$ .

Increasing and Decreasing Functions exercise 16.1 , question 7

Answer:

$f(x)$ is neither increasing nor decreasing on R.
Hint:

$f(x)$ is increasing on $(a,b)$ , if the values of $f(x)$ increase with the increase in the values of x.
$f(x)$ 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]
$\begin{aligned} &\Rightarrow \quad \frac{1}{1+x_{1}^{2}}<\frac{1}{1+x_{2}^{2}} \\ &\Rightarrow \quad f\left(x_{1}\right)<f\left(x_{2}\right) \end{aligned}$
Thus, $x_{1}>x_{2} \Rightarrow f\left(x_{1}\right)<f\left(x_{2}\right)$ for all $x_{1},x_{2} \in [0,\infty)$ .
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 $x_{1}<x_{2}$
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]
$\begin{aligned} &\Rightarrow \quad \frac{1}{1+x_{1}^{2}}>\frac{1}{1+x_{2}^{2}} \\ &\Rightarrow \quad f\left(x_{1}\right)>f\left(x_{2}\right) \end{aligned}$
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

Answer:

$f(x)$ 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

Answer:

$f(x)$ 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
$\begin{aligned} &-x_{1}<-x_{2} \\ &f\left(x_{1}\right)<f\left(x_{2}\right) \end{aligned}$
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 strictly increasing in $(-\infty,0)$

Increasing and Decreasing Functions exercise 16.1 , question 9

Answer:

$f(x)$ 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 $x_{1}<x_{2} \Rightarrow f\left(x_{1}\right)<f\left(x_{2}\right)$ 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 $x_{1}<x_{2}$
$x_{1}<x_{2}$
Multiplying 7 on both sides,
$\Rightarrow 7x_{1}<7x_{2}$
Subtracting 3 on both sides,
We get,
$\begin{array}{ll} \Rightarrow & 7 x_{1}-3<7 x_{2}-3 \\ \Rightarrow & f\left(x_{1}\right)<f\left(x_{2}\right) \end{array}$
Thus, $x_{1}<x_{2} \Rightarrow f\left(x_{1}\right)<f\left(x_{2}\right)$ for all $x_{1}, x_{2} \in R$
So, $f(x)$ is strictly increasing function on R.

RD Sharma Maths is one of the top-selling secondary materials books for Class 12. RD Sharma Class 12th Chapter 16 Exercise 16.1 meets an incredible essential of maths questions.

RD Sharma Class 12 Solutions Increasing and Decreasing Functions exercise 16.1 will help students to score brilliant grades and further join the flood of their tendency. In any case, the one subject that numerous students discover testing is Science. Students can't get optimal results without intensive practice.

The mathematical thoughts, issues, and solutions are referred to comprehensively in the book. Consequently, the students can have comprehensive data on maths and deal with mathematical inquiries with their correct answers.

RD Sharma Class 12th Exercise 16.1 Chapter 16 – Increasing and Decreasing Functions practicing these solutions can get their inquiries cleared immediately. RD Sharma Solutions are basic reference books to score high in Math board tests as in genuine hardships.

Rd Sharma class 12 chapter 16 exercise 16.1 solutions are reviewed every year to include the newest updates and materials. The updates help students to stay up to date with all their new syllabus and understand the formulae well. Class 12 RD Sharma chapter 16 exercise 16.1 solution has around nine inquiries like: -

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

Benefits :

Simple to understand the thought
Remarkable yet planned demonstration of the subjects
Point by point explanation of thoughts and formulae
Assists students with practicing with no issue

One book covers all chapters and is satisfactory for scoring extraordinary engravings.

Presently, students can download the RD Sharma class 12 solutions chapter 16 ex 16.1 in the PDF format from Career360.

Given that Science is a challenging subject for Class 12 students, the solutions book will teach them new techniques to understand Arithmetic better.

RD Sharma Chapter-wise Solutions

Frequently Asked Questions (FAQs)

1. Will I adequately download rd Sharma class 12th exercise 16.1?

To be sure, just a solitary tick is adequate to download RD Sharma Class 12th Exercise 16.1 Chapter 16 PDF from our website Career360.

2. Which study material to follow to score high in boards?

There are many exam resources open watching out and the web, yet preparing with RD Sharma Book Solutions will help you unbound and score better checks in class 12 and ferocious exams.

3. Where might I have the option to get class 12 rd Sharma chapter 16 exercise 16.1 solution for Class 12?

Candidates who want to download the Rd Sharma class 12 chapter 16 exercise 16.1 will find the free copy of the pdf at the Career360 website.

4. Is handling rd Sharma class 12 solutions ex 16.1 enough for chapter 16 status?

For sure, handling the best books like Rd Sharma class 12 chapter 16 exercise 16.1 is the completed pack for your exam preparation.

5. What sort of inquiries do RD Sharma Solutions oblige Class 12?

RD Sharma reference books offer concise response questions, short answer questions, different choice inquiries, and long answer questions, which in this way, help in utilizing time beneficially and cultivate fundamental derivation among the students.