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Question: Let \(f\left( x \right) = {e^x} + \sin x\) be defined on the interval \(x \in \left[ { - 4,0} \right...

Let f(x)=ex+sinxf\left( x \right) = {e^x} + \sin x be defined on the interval x[4,0]x \in \left[ { - 4,0} \right], the odd extension of f(x) in the interval
[-4, 4]
A. ex+sinx,x(0,4) B. ex+sinx,x(0,4) C. exsinx,x(0,4) D. excosx,x(0,4)  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) \\\

Explanation

Solution

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(x)=ex+sinx, x[4,0]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\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.