RD Sharma Solutions Class 12 Mathematics Chapter 16 VSA

Kuldeep MauryaUpdated on 21 Jan 2022, 01:36 PM IST

The Class 12 RD Sharma chapter exercise VSA solution is that one book that needs to be used by every student who is preparing for the class 12th board examinations. Innumerable students struggle when it comes to the subject of maths. Chapters like Increasing and Decreasing function especially are pretty complex and require a lot of effort to solve. In this case, extra resources are required for these chapters so that students can prepare at home through self-practice and self-study. The RD Sharma class 12th exercise VSA solution will act as a secondary material to aid students in their daily practice and build-up their problem-solving skill

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RD Sharma Class 12 Solutions Chapter 16 VSA Increasing and Decreasing Functions - Other Exercise
Increasing and Decreasing Functions Excercise: VSA
RD Sharma Chapter-wise Solutions

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

Increasing and Decreasing Functions Excercise: VSA

Increasing and Decreasing Function Exercise Very short Answer type Question 1

Answer:

$a>1$
Hint:
If $f(x)$ is increasing, then $f'(x)>0$
Given:

$f(x)=a^{x}$
Explanation:
It is given that,

$f(x)=a^{3}$

$\Rightarrow f'(x)=a^{x}\cdot \log a$

$\because f'(x)>0$ as $f(x)$ is increasing

$\Rightarrow a^{x}\cdot \log a>0$

$\Rightarrow a>1$ $[\because \log 1=0]$

Thus, value of $a$ has to be $a >1$

Increasing and Decreasing Function Exercise Very Short Answer type Question 2

Answer:
Value of a will be $0<a<1$
Hint:
If $f(x)$ is decreasing, then $f'(x)<0$
Given:
$f(x)=a^{x}$
Explanation:
It is given that,
$f(x)=a^{x}$
$\Rightarrow f'(x)=a^{x}\cdot \log a$
$\because f'(x)<0$ as $f(x)$ is decreasing
$\Rightarrow a^{x}\cdot \log a<0$
$\Rightarrow 0<a<1$ $[\because \log 0=1]$
Thus, the value of $a$ will be $0<a<1$

Increasing and Decreasing Function Excercise Very short Answer type Question 3

Answer:
The set of values of $a$ for the given function is $a>1$
Hint:
If $f(x)$ is increasing, then $f'(x)>0$
Given:
$f(x)=\log a^{x}$
Explanation:
It is given that,
$f(x)=\log a^{x}$
$\Rightarrow$ $a^{f(x)}=x$ … (i)
Differentiate equation (i) with respect to $x$
$\Rightarrow$ $a^{f(x)}\log a=f'(x)=1$
$\Rightarrow$ $f'(x)=\frac{1}{a^{f(x)}\log a}$
$\Rightarrow$ $\frac{1}{a^{x}\log a}>0$ as $f(x)$ is increasing
$\Rightarrow$ $a>1$ $[\because \log 1=0]$
So the function is increasing if $a>1$

Increasing and Decreasing Function Excercise Very Short Answer type Question 4

Answer:
The function is decreasing if $0<a<1$
Hint:
If $f(x)$ is decreasing, then $f'(x)<0$
Given:
$f(x)=\log a^{x}$
Explanation:
It is given that,
$f(x)=\log a^{x}$
$\Rightarrow$ $a^{f(x)}=x$ … (i)
Differentiate equation (i) with respect to $x$
$\Rightarrow$ $a^{f(x)}\log a=f'(x)=1$
$\Rightarrow$ $f'(x)=\frac{1}{a^{f(x)}\log a}$
$\Rightarrow$ $f'(x)=\frac{1}{a^{x}\log a}$
$\Rightarrow$ $\frac{1}{a^{x}\log a}<0$ as $f(x)$ is decreasing
$\Rightarrow$ $a<1$ $[\because \log 0=1]$
Thus, the function is decreasing if $0<a<1$

Increasing and Decreasing Function Excercise Very Short Answer type Question 5

Answer:
The function is increasing if $a\epsilon (0,\infty )$
Hint:
If $f(x)$ is increasing, then $f'(x)>0$
Given:
$f(x)=a(x+sinx)+a$
Explanation:
It is given that,
$f(x)=a(x+sinx)+a$
$\Rightarrow$ $f'(x)=a(1+cosx)$
Now, $a(1+cosx)>0$ as $f(x)$ is increasing
$\Rightarrow$ $a>0$
Thus, $f(x)$ is increasing if $a\epsilon (0,\infty )$

Increasing and Decreasing Function Excercise Very Short Answer type Question 6

Answer:
The function is increasing if $a\epsilon (-\infty ,-1)$
Hint:
If $f(x)$ is increasing, then $f'(x)>0$
Given:
$f(x)=\sin x-ax+4$
Explanation:
It is given that,
$f(x)=\sin x-ax+4$
$\Rightarrow$ $f'(x)=\cos x-a$
Now, $\cos x-a>0$ as $f(x)$ is increasing
$\Rightarrow$ $\cos x>a$
For $a\epsilon (-\infty ,-1)$, $\cos x>0$
Thus, $f(x)$ is increasing for $a\epsilon (-\infty ,-1)$

Increasing and Decreasing Function Excercise Very Short Answer type Question 7

Answer:
The function is decreasing if $b\epsilon (-\infty ,0)$
Hint:
If $f(x)$ is decreasing, then $f'(x)<0$
Given:
$f(x)=b(x+\cos x)+4$
Explanation:
It is given that,
$f(x)=b(x+\cos x)+4$
$\Rightarrow$ $f'(x)=b(1-\sin x)$
Now, $b(1-\sin x)>0$ as $f(x)$ is decreasing
$\Rightarrow$ $1-\sin x>0$ for all values of $x$
Thus, $f(x)$ is decreasing for $b\epsilon (-\infty ,0)$

Increasing and Decreasing Function Excercise Very Short Answer type Question 8

Answer:
The function is increasing if $a\epsilon (0,\infty )$
Hint:
If $f(x)$ is increasing, then $f'(x)>0$
Given:
$f(x)=x+\cos x+ax+b$
Explanation:
It is given that,
$f(x)=x+\cos x+ax+b$
$\Rightarrow$ $f'(x)=1-\sin x+a$
$\Rightarrow$ $f'(x)=(1+a)-\sin x$
Now, $(1+a)-\sin x>0$ as $f(x)$ is increasing
$\Rightarrow$ $(1+a)>\sin x$
$\Rightarrow$ $a>\sin x-1>0$
Thus, $f(x)$ is increasing for $a\epsilon (0,\infty )$

Increasing and Decreasing Function Excercise Very Short Answer type Question 9

Answer:
The function is increasing if $k\epsilon (1,\infty )$
Hint:
If $f(x)$ is increasing, then $f'(x)>0$
Given:
$f(x)=kx-\sin x$
Explanation:
It is given that,
$f(x)=kx-\sin x$
$\Rightarrow$ $f'(x)=k-\cos x$
Now, $k-\cos x>0$ as $f(x)$ is increasing
$\Rightarrow$ $k>\cos x$
$\Rightarrow$ $k\epsilon (1,\infty )$
Thus, $f(x)$ is increasing if $k\epsilon (1,\infty )$

Increasing and Decreasing Function Exercise Very Short Answer type Question 10

Answer:
The given function is decreasing.
Hint:
If $f(x)$ is increasing, then $f'(x)>0$
If $f(x)$ is decreasing, then $f'(x)<0$
Given:
$f(x)=\tan ^{-1}g(x)$
Where $g(x)$ is decreasing on $R$.
Explanation:
It is given that,
$f(x)=\tan ^{-1}g(x)$ … (i)
Differentiate equation (i) with respect to $x$
$\Rightarrow$ $f'(x)=\frac{1}{1+[g(x)]^{2}}g'(x)$
Since $g(x)$ is decreasing, $g'(x)<0$
Also $1+[g(x)]^{2}>0$
Thus, $f'(x)<0$
Thus, the function is decreasing.

Increasing and Decreasing Function Excercise Very Short Answer type Question 11

Answer:
The values of $a$ for which the function is decreasing if $a\epsilon (-\infty ,0)$
Hint:
If $f(x)$ is decreasing, then $f'(x)<0$
Given:
$f(x)=ax+b$
Explanation:
It is given that,
$f(x)=ax+b$
$\Rightarrow$ $f'(x)=a$
Now, $f'(x)<0$ as $f(x)$ is decreasing
$\Rightarrow$ $a<0$
$\Rightarrow$ $a\epsilon (-\infty ,0)$
Thus, $f(x)$ is decreasing for $a\epsilon (-\infty ,0)$

Increasing and Decreasing Function Excercise Very Short Answer type Question 12

Answer:
The interval in which the function increasing is $\left [ 0,\frac{\pi }{4} \right ]$
Hint:
If $f(x)$ is increasing, then $f'(x)>0$
Given:
$f(x)=\sin x+\cos x$
Explanation:
It is given that,
$f(x)=\sin x+\cos x$
$\Rightarrow$ $f'(x)=\cos x-\sin x$
Now, $\cos x-\sin x>0$ as $f(x)$ is increasing
$\Rightarrow$ $\cos x>\sin x$
$\Rightarrow$ $\tan x<1$
$\therefore$ $x\epsilon \left [ 0,\frac{\pi }{4} \right ]$
The interval in which the function increasing is $\left [ 0,\frac{\pi }{4} \right ]$

Increasing and Decreasing Function Excercise Very Short Answer type Question 13

Answer:
The function $f(x)$ is increasing in its domain.
Hint:
$f(x)$ is increasing, if $f'(x)>0$
$f(x)$ is decreasing, if $f'(x)<0$
Given:
$f(x)=\tan x-x$
Explanation:
It is given that,
$f(x)=\tan x-x$
Differentiate equation (i) with respect to $x$
$\Rightarrow$ $f'(x)=\sec ^{2}x-1$
$\because$ $\sec ^{2}x-1>0$ for all $x$
$\therefore$ $f(x)$ is an increasing function.
Thus, the given function is increasing in its domain.

Increasing and Decreasing Function Excercise Very Short Answer type Question 14

Answer:
The set of values of $a$ for which $f(x)$ is strictly increasing is $a\epsilon (-\infty ,-1]\cup [1,\infty )$
Hint:
If $f(x)$ is increasing, then $f'(x)>0$
Given:
$f(x)=\cos x+a^{2}x+b$
Explanation:
It is given that,
$f(x)=\cos x+a^{2}x+b$
$\Rightarrow$ $f'(x)=-\sin x+a^{2}$
$\therefore$ $f'(x)>0$ as $f(x)$ is increasing
$\Rightarrow$ $-\sin x+a^{2}>0$
$\Rightarrow$ $a^{2}>\sin x$
$\Rightarrow$ $a\epsilon (-\infty ,-1]\cup [1,\infty )$
The set of values of $a$ for which $f(x)$is strictly increasing is $a\epsilon (-\infty ,-1]\cup [1,\infty )$

Therefore, the RD Sharma class 12th exercise VSA for the chapter Increasing and Decreasing order is absolutely a must-have for all students in class 12. The questions and answers contained in the RD Sharma class 12 solution chapter 16 exercise VSA are predominantly short answers. Students can use these answers to enhance their analytical skills and increase their speed. They are super effective in teaching students the entire syllabus in concise time and build their confidence to sit for the exam. The solution has only 14 answers in the RD Sharma class 12th exercise VSA which follows the exercise questions of chapter 16. The solutions of the book cover concepts like :-

Proving the monotonicity
Finding the intervals
Increasing and decreasing functions

The RD Sharma class 12 solution of Increasing and decreasing function exercise VSA is a coveted secondary material that is known for its various benefits. Some of them are given below:-

The solutions cover all questions provided in the particular exercise section and RF Sharma Solutions have a dedicated book for all chapters.
The RD Sharma class 12 chapter 16 ex VSA will be beneficial for board exams as it contains some helpful tips to solve problems easily.
The RD Sharma class 12th exercise VSA answers are provided by skilled staff of RD Sharma Solutions who are experts in maths. The books is trustworthy and reliable.
RD Sharma class 12 solutions chapter 16 exercise VSA is also a guide for teachers to give homeworks . Many teachers refer to RD Sharma class 12th exercise VSA to set of question papers for school tests as well.
The RD Sharma class 12th exercise VSA can be found for free of cost at the website of Career360.

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