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Please help me with the Domain of the above mentioned function

Please help me with the Domain of the above mentioned function

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Grade:11

3 Answers

Arun
25750 Points
6 years ago
Dear Aarushi
 
1-x² >0 
x belongs to (-1, 1)
1- sqrt(1 -x²) > 0
1 > sqrt(1 -x²)
x² > 0
x belngs to R..
1 - sqrt (1 - sqrt(1-x²) ) > 0
x belngs to R
Hence
Finally x belngs to (-1, 1)
Samyak Jain
333 Points
5 years ago
\sqrt{}1 – \sqrt{}1 – \sqrt{}(1 – x2)
Here, terms inside sqr root must be greater than or equal to 0.
\therefore 1 – x2 \geq 0  or x2 – 1 \leq 0  or (x+1)(x-1) \leq 0
x \epsilon (-1, 1)                           …................(1)
1 – \sqrt{}(1 – x2) \geq 0  or  \sqrt{}(1 – x2) \leq 1 or 1 – x2 \leq 1  or x2 \geq 0 , which is always true.
x \epsilon R                          ….......................(2)
1 – \sqrt{}1 – \sqrt{}(1 – x2) \geq 0   or  1 – \sqrt{}(1 – x2)  \leq 1   or   \sqrt{}(1 – x2) \geq 0  
1 – x2 \geq 0    As i’ve written above    x \epsilon (-1,1)              ...................(3)
Take intersection of (1), (2) & (3) to get 
x \epsilon (-1,1), which is the domain of the function
Samyak Jain
333 Points
5 years ago
Sorry, I made a slight mistake in above answer. the correct answer is below.
\sqrt{}1 – \sqrt{}1 – \sqrt{}(1 – x2)
Here, terms inside sqr root must be greater than or equal to 0.
\therefore 1 – x2 \geq 0  or x2 – 1 \leq 0  or (x+1)(x-1) \leq 0
x \epsilon [-1, 1]                           …................(1)
1 – \sqrt{}(1 – x2) \geq 0  or  \sqrt{}(1 – x2) \leq 1 or 1 – x2 \leq 1  or x2 \geq 0 , which is always true.
x \epsilon R                          ….......................(2)
1 – \sqrt{}1 – \sqrt{}(1 – x2) \geq 0   or  1 – \sqrt{}(1 – x2)  \leq 1   or   \sqrt{}(1 – x2) \geq 0  
1 – x2 \geq 0    As i’ve written above    x \epsilon [-1,1]              ...................(3)
Take intersection of (1), (2) & (3) to get 
x \epsilon [-1,1], which is the domain of the function

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