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

• Increasing and Decreasing Functions exercise 16.1

Increasing and Decreasing Functions exercise 16.1 , question 1

Answer:
$f(x)$ is increasing in$(0,\mathrm{\infty })$
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 $(0,\mathrm{\infty })$
Solution:
Let us consider $x_1,x_2\in (0,\mathrm{\infty })$ and $x_1<x_2$
Then,
$x_1<x_2$
Taking loge on both sides,
$\Rightarrow {\log }_e{x}_1<{\log }_e{x}_2$
$\Rightarrow f\left(x_1\right)<f\left(x_2\right)$ for all $x_1,x_2\in (0,\mathrm{\infty })$
So, $f(x)$ is increasing in $(0,\mathrm{\infty })$

Increasing and Decreasing Functions exercise 16.1 , question 2

Answer:
$f(x)$ is decreasing on$(0,\mathrm{\infty })$if $0<a<1$
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 $(0,\mathrm{\infty })$, if $a>1$ and decreasing on $(0,\mathrm{\infty })$ , if $0<a<1$
Solution:
Here we have two cases:
Case I: when $a>1$
Let us consider $x_1,x_2ϵ(0,\mathrm{\infty })$
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$(0,\mathrm{\infty })$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,
$\begin{array}{rlr}& \Rightarrow & \log x_1<\log x_2\\ & \Rightarrow & \frac{\log x_1}{\log a}>\frac{\log x_2}{\log a}\\ & \Rightarrow & f\left(x_1\right)>f\left(x_2\right)\end{array}$

Thus, $f(x)$ is decreasing on $(0,\mathrm{\infty })$ 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 $x_1<x_2ϵ(a,b)$.
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 $x_1,x_2ϵR$Such that $x_1<x_2$
Then,
$\begin{array}{rl}& x_1<x_2\\ & \Rightarrow a{x}_1<a{x}_2[\because a>0]\end{array}$
$\Rightarrow a{x}_1+b<a{x}_2+b$ [Given that $b$ is constant]
$\Rightarrow f\left(x_1\right)<f\left(x_2\right)$
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 $x_1,x_2ϵ(a,b)$.
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 $x_1,x_2ϵR$ Such that $x_1<x_2$
Then,
$\begin{array}{rl}& x_1<x_2\\ \Rightarrow & a{x}_1>a{x}_2\end{array}[\because a<0]$
$\Rightarrow a{x}_1+b>a{x}_2+b$ [Given that b is constant]
$\Rightarrow f\left(x_1\right)>f\left(x_2\right)$
Thus, $x_1<x_2\Rightarrow f\left(x_1\right)>f\left(x_2\right)$ for all $x_1,x_2ϵR$
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 $(0,\mathrm{\infty })$
Solution:
Let $x_1,x_2\in (0,\mathrm{\infty })$ 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 $x_1,x_2\in (0,\mathrm{\infty })$.
So, $f(x)$is a decreasing function.

Increasing and Decreasing Functions exercise 16 .1 , question 6

Answer:

$f(x)$ is a decreasing function on $(0,\mathrm{\infty }]$.
$f(x)$ is increasing on $[-\mathrm{\infty },0)$
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 $(0,\mathrm{\infty }]$ and increases in the interval $[-\mathrm{\infty },0)$.
Solution:
Here we have two cases:
Case I: if $x\in [0,\mathrm{\infty })$
Let $x_1,x_2\in [0,\mathrm{\infty })$ 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]
$\begin{array}{rl}& \Rightarrow \frac{1}{1+x_1^2}>\frac{1}{1+x_2^2}\\ & \Rightarrow f\left(x_1\right)>f\left(x_2\right)\end{array}$
Thus, $x_1<x_2\Rightarrow f\left(x_1\right)>f\left(x_2\right)$ for all $x_1,x_2\in [0,\mathrm{\infty })$.
So, $f(x)$is a decreasing function.
Case II: if $x\in (-\mathrm{\infty },0]$
Let $x_1,x_2\in (-\mathrm{\infty },0]$ 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]
$\begin{array}{rl}& \Rightarrow \frac{1}{1+x_1^2}<\frac{1}{1+x_2^2}\\ & \Rightarrow f\left(x_1\right)<f\left(x_2\right)\end{array}$
Thus, $x_1<x_2\Rightarrow f\left(x_1\right)<f\left(x_2\right)$ for all $x_1,x_2\in (-\mathrm{\infty },0]$.
So, $f(x)$is increasing function on$(-\mathrm{\infty },0]$.

Increasing and Decreasing Functions exercise 16.1 , question 7

Answer:

$f(x)$ 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 $x\in [0,\mathrm{\infty })$
Let $x_1,x_2\in [0,\mathrm{\infty })$ 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]
$\begin{array}{rl}& \Rightarrow \frac{1}{1+x_1^2}<\frac{1}{1+x_2^2}\\ & \Rightarrow f\left(x_1\right)<f\left(x_2\right)\end{array}$
Thus, $x_1>x_2\Rightarrow f\left(x_1\right)<f\left(x_2\right)$ for all $x_1,x_2\in [0,\mathrm{\infty })$.
So, $f(x)$is a decreasing function on $[0,\mathrm{\infty })$.
Case II: if $x\in (-\mathrm{\infty },0]$
Let $x_1,x_2\in (-\mathrm{\infty },0]$ 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]
$\begin{array}{rl}& \Rightarrow \frac{1}{1+x_1^2}>\frac{1}{1+x_2^2}\\ & \Rightarrow f\left(x_1\right)>f\left(x_2\right)\end{array}$
Thus, $x_1>x_2\Rightarrow f\left(x_1\right)>f\left(x_2\right)$ for all $x_1,x_2\in (-\mathrm{\infty },0]$.
So, $f(x)$is increasing function on$(-\mathrm{\infty },0]$.
In both cases we get, $f(x)$is a decreasing function on$[0,\mathrm{\infty })$and increasing function on$(-\mathrm{\infty },0]$.
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 $(0,\mathrm{\infty })$
Hint:
Use $f(x)=|x|=\{\begin{array}{l}x,\text{ if }x>0\\ -x,\text{ if }x<0\end{array}$
Given:
$f(x)=|x|$
To prove:
We have to prove that function $f(x)=|x|$ is strictly increasing in $(0,\mathrm{\infty })$.
Solution:
Let $x_1,x_2\in (0,\mathrm{\infty })$ 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 $x_1,x_2\in (0,\mathrm{\infty })$.
So, $f(x)$is strictly increasing in $(0,\mathrm{\infty })$

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

Answer:

$f(x)$is strictly increasing in $(-\mathrm{\infty },0)$
Hint:
Use $f(x)=|x|=\{\begin{array}{l}x,\text{ if }x>0\\ -x,\text{ if }x<0\end{array}$
Given:
$f(x)=|x|$
To prove:
We have to prove that function $f(x)=|x|$ is strictly increasing in $(-\mathrm{\infty },0)$.
Solution:
Let $x_1,x_2\in (-\mathrm{\infty },0)$ and $x_1>x_2$
Since$x_1,x_2<0$, therefore the function of $x_1,x_2$ is
$\begin{array}{rl}& -x_1<-x_2\\ & f\left(x_1\right)<f\left(x_2\right)\end{array}$
Thus, $x_1>x_2\Rightarrow f\left(x_1\right)<f\left(x_2\right)$ for all $x_1,x_2\in (-\mathrm{\infty },0)$.
So, $f(x)$is strictly increasing in $(-\mathrm{\infty },0)$

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,
$f(x)=7x-3$
To prove:
We have to prove that function$f(x)=7x-3$ is strictly increasing function on R.
Solution:
Here given that,
$f(x)=7x-3$
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,
$\begin{array}{ll}\Rightarrow & 7x_1-3<7x_2-3\\ \Rightarrow & f\left(x_1\right)<f\left(x_2\right)\end{array}$
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.

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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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RD Sharma Chapter-wise Solutions

• Chapter 1 - Relations
• Chapter 2 - Functions
• Chapter 3 - Inverse Trigonometric Functions
• Chapter 4 - Algebra of Matrices
• Chapter 5 - Determinants
• Chapter 6 - Adjoint and Inverse of a Matrix
• Chapter 7 - Solution of Simultaneous Linear Equations
• Chapter 8 - Continuity
• Chapter 9 - Differentiability
• Chapter 10 - Differentiation
• Chapter 11 - Higher Order Derivatives
• Chapter 12 - Derivative as a Rate Measurer
• Chapter 13 - Differentials, Errors and Approximations
• Chapter 14 - Mean Value Theorems
• Chapter 15 - Tangents and Normals
• Chapter 16 - Increasing and Decreasing Functions
• Chapter 17 - Maxima and Minima
• Chapter 18 - Indefinite Integrals
• Chapter 19 - Definite Integrals
• Chapter 20 - Areas of Bounded Regions
• Chapter 21 - Differential Equations
• Chapter 22 - Algebra of Vectors
• Chapter 23 - Scalar Or Dot Product
• Chapter 24 - Vector or Cross Product
• Chapter 25 - Scalar Triple Product
• Chapter 26 - Direction Cosines and Direction Ratios
• Chapter 27 - Straight Line in Space
• Chapter 28 - The Plane
• Chapter 29 - Linear programming
• Chapter 30- Probability
• Chapter 31 - Mean and Variance of a Random Variable