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Let f(x)={(sinx+cosx",",0 lt x lt (pi)/(...

Let `f(x)={(sinx+cosx",",0 lt x lt (pi)/(2)),(a",",x=pi//2),(tan^(2)x+"cosec"x",",pi//2 lt x lt pi):}`
Then its odd extension is

A

`{(-tan^(2)x-"cosec"x",",-pi lt x lt -(pi)/(2)),(-a",",x=-(pi)/(2)),(-sinx+cosx",",-(pi)/(2) lt x lt 0 ):}`

B

`{(-tan^(2)x+"cosec"x",",-pi lt x lt -(pi)/(2)),(-a",",x=-(pi)/(2)),(sinx-cosx",",-(pi)/(2) lt x lt 0 ):}`

C

`{(-tan^(2)x+"cosec"x",",-pi lt x lt -(pi)/(2)),(a",",x=-(pi)/(2)),(sinx-cosx",",-(pi)/(2) lt x lt 0 ):}`

D

`{(tan^(2)x+"cosec"x",",-pi lt x lt -(pi)/(2)),(-a",",x=-(pi)/(2)),(sinx+cosx",",-(pi)/(2) lt x lt 0 ):}`

Text Solution

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The correct Answer is:
To find the odd extension of the given function \( f(x) \), we will follow these steps: ### Step 1: Understand the function definition The function \( f(x) \) is defined as: - \( f(x) = \sin x + \cos x \) for \( 0 < x < \frac{\pi}{2} \) - \( f(x) = a \) for \( x = \frac{\pi}{2} \) - \( f(x) = \tan^2 x + \csc x \) for \( \frac{\pi}{2} < x < \pi \) ### Step 2: Define the odd extension The odd extension of a function \( f(x) \) is defined as: \[ f_{odd}(x) = \begin{cases} f(x) & \text{if } x \geq 0 \\ -f(-x) & \text{if } x < 0 \end{cases} \] ### Step 3: Find \( f(-x) \) We will evaluate \( f(-x) \) for the different intervals: 1. **For \( -x \) in \( (0, \frac{\pi}{2}) \)** (which means \( x \) is in \( (-\frac{\pi}{2}, 0) \)): \[ f(-x) = \sin(-x) + \cos(-x) = -\sin x + \cos x \] 2. **For \( -x = -\frac{\pi}{2} \)**: \[ f(-x) = a \] 3. **For \( -x \) in \( (\frac{\pi}{2}, \pi) \)** (which means \( x \) is in \( (-\pi, -\frac{\pi}{2}) \)): \[ f(-x) = \tan^2(-x) + \csc(-x) = \tan^2 x - \csc x \] ### Step 4: Substitute into the odd extension formula Now we substitute \( f(-x) \) into the odd extension formula: 1. **For \( x \) in \( (-\frac{\pi}{2}, 0) \)**: \[ f_{odd}(x) = -f(-x) = -(-\sin x + \cos x) = \sin x - \cos x \] 2. **For \( x = -\frac{\pi}{2} \)**: \[ f_{odd}(x) = -f(-x) = -a \] 3. **For \( x \) in \( (-\pi, -\frac{\pi}{2}) \)**: \[ f_{odd}(x) = -f(-x) = -(\tan^2 x - \csc x) = -\tan^2 x + \csc x \] ### Step 5: Combine the results The odd extension \( f_{odd}(x) \) can now be summarized as: \[ f_{odd}(x) = \begin{cases} \sin x - \cos x & \text{if } -\frac{\pi}{2} < x < 0 \\ -a & \text{if } x = -\frac{\pi}{2} \\ -\tan^2 x + \csc x & \text{if } -\pi < x < -\frac{\pi}{2} \end{cases} \]

To find the odd extension of the given function \( f(x) \), we will follow these steps: ### Step 1: Understand the function definition The function \( f(x) \) is defined as: - \( f(x) = \sin x + \cos x \) for \( 0 < x < \frac{\pi}{2} \) - \( f(x) = a \) for \( x = \frac{\pi}{2} \) - \( f(x) = \tan^2 x + \csc x \) for \( \frac{\pi}{2} < x < \pi \) ...
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