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domain of the function f(x) = sqrt 1-sqrt x^2-4x+5 is.. what is the function

twisha , 7 Years ago
Grade 12th pass
anser 2 Answers
Deepak Kumar Shringi
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Last Activity: 7 Years ago
Samyak Jain
f(x) = \sqrt{}(1 – \sqrt{}(x2 – 4x + 5) ) 
For the domain of f(x), terrms inside sqr root should be positive. 
\therefore  x2 - 4x + 5 \geq 0       as well as    1 – \sqrt{}(x2 – 4x + 5) \geq 0
   x2 - 4x + 4 + 1 \geq 0
   (x – 2)2 + 1 \geq 0  or  (x – 2)2 \geq -1 
  i.e. x \epsilon R                                                 …............................(1)
\because 1 – \sqrt{}(x2 – 4x + 5) \geq 0   \therefore  \sqrt{}(x2 – 4x + 5) \leq 1 
    x2 – 4x + 5 \leq 1  or  x2 – 4x + 4 \leq 0
 i.e. (x – 2)2  \leq 0   \therefore  x = 2               ….................(2)
Taking intersection of (1) & (2), we get 
                       x = 2,   which is the domain of f(x).
Last Activity: 7 Years ago
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