
Let $f\left( x \right) = {e^x} + \sin x$ be defined on the interval $x \in \left[ { - 4,0} \right]$, the odd extension of f(x) in the interval
[-4, 4]
$
A.{\text{ }}{e^{ - x}} + \sin x,x \in \left( {0,4} \right) \\
B.{\text{ }} - {e^{ - x}} + \sin x,x \in \left( {0,4} \right) \\
C.{\text{ }} - {e^{ - x}} - \sin x,x \in \left( {0,4} \right) \\
D.{\text{ }} - {e^{ - x}} - \cos x,x \in \left( {0,4} \right) \\
$
Answer
627k+ views
Hint: In this question we have been given a function f(x) which is defined in a certain interval and we have to find the odd extension of f(x) in the interval [-4, 4]. Odd extension means that the function breaks into a piecewise function which is defined over a specific interval, so simply find the breaking point of the given f(x) in the interval in which the odd extension is to be taken out.
Complete step-by-step answer:
Given function
$f\left( x \right) = {e^x} + \sin x,{\text{ }}x \in \left[ { - 4,0} \right]$
Now we have to find out the odd extension of f(x) in the interval [-4, 4]
According to odd extension property the function break into piecewise function which is defined as in the interval [-a, a]
The odd extension of f(x) is the function
\[{f_o}\left( x \right) = \left\{
f\left( x \right),{\text{ }}x \in \left[ { - a,0} \right] \\
- f\left( { - x} \right),{\text{ }}x \in \left[ {0,a} \right] \\
\right.\] So, use this property to calculate the odd extension of the given function in the interval [-4, 4]
$f\left( x \right) = {e^x} + \sin x,{\text{ }}x \in \left[ { - 4,0} \right]$
Now replace x with (-x) we have,
$f\left( { - x} \right) = {e^{ - x}} + \sin \left( { - x} \right)$
Now as we know $\sin \left( { - \theta } \right) = - \sin \theta $ so, use this property in the above equation we have,
$f\left( { - x} \right) = {e^{ - x}} - \sin x$
Now multiply by (-) in above equation we have,
$ - f\left( { - x} \right) = - \left( {{e^{ - x}} - \sin x} \right) = - {e^{ - x}} + \sin x$
$ - f\left( { - x} \right) = - {e^{ - x}} + \sin x,{\text{ }}x \in \left[ {0,4} \right]$
Hence the odd extension of the given function in the interval [-4, 4] is
\[{f_o}\left( x \right) = \left\{
{e^x} + \sin x,{\text{ }}x \in \left[ { - 4,0} \right] \\
- {e^{ - x}} + \sin x,{\text{ }}x \in \left[ {0,4} \right] \\
\right.\]
Hence option (b) is correct.
Note – Whenever we face such type of problems the key concept is to have the basic understanding of the odd extension defined over a period of interval, make sure that the interval given is only a subset of the domain of the given function otherwise there may arise a case even that the function is not defined. These concepts will help you get on the right track to get the answer.
Complete step-by-step answer:
Given function
$f\left( x \right) = {e^x} + \sin x,{\text{ }}x \in \left[ { - 4,0} \right]$
Now we have to find out the odd extension of f(x) in the interval [-4, 4]
According to odd extension property the function break into piecewise function which is defined as in the interval [-a, a]
The odd extension of f(x) is the function
\[{f_o}\left( x \right) = \left\{
f\left( x \right),{\text{ }}x \in \left[ { - a,0} \right] \\
- f\left( { - x} \right),{\text{ }}x \in \left[ {0,a} \right] \\
\right.\] So, use this property to calculate the odd extension of the given function in the interval [-4, 4]
$f\left( x \right) = {e^x} + \sin x,{\text{ }}x \in \left[ { - 4,0} \right]$
Now replace x with (-x) we have,
$f\left( { - x} \right) = {e^{ - x}} + \sin \left( { - x} \right)$
Now as we know $\sin \left( { - \theta } \right) = - \sin \theta $ so, use this property in the above equation we have,
$f\left( { - x} \right) = {e^{ - x}} - \sin x$
Now multiply by (-) in above equation we have,
$ - f\left( { - x} \right) = - \left( {{e^{ - x}} - \sin x} \right) = - {e^{ - x}} + \sin x$
$ - f\left( { - x} \right) = - {e^{ - x}} + \sin x,{\text{ }}x \in \left[ {0,4} \right]$
Hence the odd extension of the given function in the interval [-4, 4] is
\[{f_o}\left( x \right) = \left\{
{e^x} + \sin x,{\text{ }}x \in \left[ { - 4,0} \right] \\
- {e^{ - x}} + \sin x,{\text{ }}x \in \left[ {0,4} \right] \\
\right.\]
Hence option (b) is correct.
Note – Whenever we face such type of problems the key concept is to have the basic understanding of the odd extension defined over a period of interval, make sure that the interval given is only a subset of the domain of the given function otherwise there may arise a case even that the function is not defined. These concepts will help you get on the right track to get the answer.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Draw a diagram of nephron and explain its structur class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

Chemical formula of Bleaching powder is A Ca2OCl2 B class 11 chemistry CBSE

Name the part of the brain responsible for the precision class 11 biology CBSE

The growth of tendril in pea plants is due to AEffect class 11 biology CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

