The domain of the function f(x) = log (sqrt { | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For the function $f(x) = \log (\sqrt {x - 4} + \sqrt {6 - x} )$ to be defined,the expression inside the logarithm must be positive,and the express

Vedclass · Accepted Answer

$[4, 6]$

Vedclass · Answer

(C) For the function $f(x) = \log (\sqrt {x - 4} + \sqrt {6 - x} )$ to be defined,the expression inside the logarithm must be positive,and the expressions inside the square roots must be non-negative.$1$. For $\sqrt {x - 4}$ to be defined,$x - 4 \ge 0$,which implies $x \ge 4$.$2$. For $\sqrt {6 - x}$ to be defined,$6 - x \ge 0$,which implies $x \le 6$.$3$. The sum $\sqrt {x - 4} + \sqrt {6 - x}$ must be strictly greater than $0$. Since both square root terms are non-negative,their sum is always $\ge 0$. The sum is $0$ only if $x-4=0$ and $6-x=0$ simultaneously,which is impossible ($x=4$ and $x=6$). Thus,the sum is always positive for $x \in [4, 6]$.Combining these conditions,we get $4 \le x \le 6$.Therefore,the domain is $[4, 6]$.

The domain of the function $f(x) = \log (\sqrt {x - 4} + \sqrt {6 - x} )$ is

Similar Questions

If $n(A) = 3$ and $n(B) = 6$ and $A \subseteq B$,then the number of elements in $A \cap B$ is equal to

If $A = \{x : x = \frac{n-1}{n+1}, n \in W\}$ and $\{n \leq 10\}$,point out the correct statement from the following:

In a town with a population of $5000$,$3200$ people are egg-eaters,$2500$ are meat-eaters,and $1500$ eat both egg and meat. How many are pure vegetarians?

Let $f:R \to R$ be defined by $f(x) = 2x + |x|$,then $f(2x) + f(-x) - f(x) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module