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$(0,\infty )$
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:

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 $0<a<1$
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:

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 $x_{1}<x_{2}$
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:

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]
$\begin{aligned} &\Rightarrow \quad \frac{1}{1+x_{1}^{2}}<\frac{1}{1+x_{2}^{2}} \\ &\Rightarrow \quad f\left(x_{1}\right)<f\left(x_{2}\right) \end{aligned}$
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.

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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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