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  • The domain of the function f defined by is equal to A. (– ∞, – 1) ∪ (1, 4] B. (– ∞, – 1] ∪ (1, 4] C. (– ∞, – 1) ∪ [1, 4] D. (– ∞, – 1) ∪ [1, 4)

The domain of the function f defined by f(x)=\sqrt{4-x}+\frac{1}{\sqrt{x^2-1}} is equal to
A. (- \infty, - 1) \cup (1, 4]

B. (- \infty, - 1] \cup (1, 4]
C. (- \infty, - 1) \cup [1, 4]
D. (- \infty, - 1) \cup [1, 4)

Answers (1)

Given data: f(x)=\sqrt{4-x}+\frac{1}{\sqrt{x^2-1}}

Now, we know that,

f(x) is defined when, 4-x\geq0\: \: and \: \: x^2-1>0

Thus, -x\geq-4 \: \: and \: \: (x-1)(x+1)>0

Thus, x\leq 4\: \: and\: \: x<-1/x>1

Thus, the domain of f(x) =(-\infty,-1)\cup(1,4]

Therefore, opt (a) is the correct answer.

 

Posted by

infoexpert21

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